L1535194207d9789387504196.pdf

CHAPTER SECTION

B 1

INTEGERS

WORKSHEET-1 Solutions 1. We have –3 –2 A B C D E F G H I

J K L M N O

Now, completing the given number line is as follows : –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 A B C D E F G H I

J K L M N O

Hence, the position of required letters is shown in the following table :

Thus, the difference between two consecutive terms is – 4. Now, add this difference from first to last term i.e., 7 + (–4) = 7 – 4 = 3 ; 3 + (–4) = 3 – 4 = – 1; – 1 + (–4) = – 5 ; – 5 + (–4) = – 5 – 4 = – 9; –9 + (–4) = – 9 – 4 = – 13 and – 13 + (–4) = – 13 – 4 = – 17 Hence, the pattern is 7, 3, – 1, – 5, – 9, – 13, – 17. (b) We have, – 2, – 4, – 6, – 8, —, —, —· ∴ Difference = – 4 – (–2) = – 4 + 2 = – 2

Letter

Position

B

–6

Thus, the difference between two consecutive terms is – 2.

D

–4

Now, add this difference from first to last term

H

0

i.e., –2 + (–2) = – 2 – 2 = – 4;

J

2



M

5



–6 + (–2) = – 6 – 2 = – 8;

7



– 8 + (–2) = – 8 – 2 = – 10;

2. Arranging the given numbers 7, – 5, 4, 0 and – 4 in the ascending order, we have – 5, – 4, 0, 4 and 7. In order to represent these integers on the number line, we draw a line and mark a point on it almost in the middle (say 0) of it as shown. Now, we set off equal distances on the right hand side as well as on the left hand side to 0. Starting from 0 and proceeding 4 units on the right of it, we obtain 4 as marked by A and further proceeding 3 units from A to the right, we obtain 7 marked by B. Again, starting from 0 and count 4 units to the left of it, we obtain –4 and further count 1 unit from its left, we get –5.



– 10 + (–2) = – 10 – 2 = – 12 and



– 12 + (–2) = –12 – 2 = – 14

O

–5 –4

0

4

7

3. (a) We have, 7, 3, – 1, – 5, —, —, —. Difference between first two term = Second term – First term = 3 – 7 = – 4 Similarly, difference between another two terms = (– 1) – 3 = – 1 – 3 = – 4

– 4 + (–2) = – 4 – 2 = – 6;

Hence, the pattern is –2, –4, –6, –8, – 10, – 12, – 14.

(c) We have, 15, 10, 5, 0 —, —, —·

∴ Difference = 10 – 15 = – 5 Now, add this difference from first to last term i.e., 15 + (–5) = 15 – 5 = 10; 10 + (–5) = 10 – 5 = 5; 5 + (–5) = 5 – 5 = 0; 0 + (–5) = 0 – 5 = – 5; –5 + (–5) = – 10 and – 10 –5 = – 15 Hence, the pattern is 15, 10, 5, 0, – 5, – 10, – 15. (d) We have, –11, – 8, – 5, – 2, —, —, —· ∴ Difference = – 8 –(–11) = – 8 + 11 = 3 Now, add this difference from first to last term i.e., –11 + 3 = – 8; – 8 + 3 = – 5 –5 + 3 = – 2; – 2 + 3 = 1 1 + 3 = 4 and 4 + 3 = 7 Hence, the pattern is – 11, –8, –5, –2, 1, 4, 7.

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WORKSHEET-2 Solutions 1. Jack’s score in five successive rounds were given as 25, – 5, – 10, 15 and 10.

S OLUT I ONS



\ Jack’s total score = 25 + (– 5) + (– 10) + 15 + 10 = 25 – 5 – 10 + 15 + 10 = 50 – 15 = 35

P-1

2. At Srinagar, temperature on Monday = – 5°C Since, the temperature was dropped by 2°C on Tuesday, therefore, temperature on Tuesday = (– 5 – 2)°C = – 7°C Also, on Wednesday the temperature rose by 4°C. \ Temperature on Wednesday = (– 7 + 4)°C = – 3°C. 3. Here, the sea level is at 0 m and the plane is 5000 m above the sea level. ∴ Distance between plane and the sea level = 5000 m Also, the submarine is floating 1200 m below the sea level. ∴ Distance between the submarine and the sea level = 1200 m Hence, the vertical distance between the plane and the submarine = Distance between the plane and the sea level + Distance between the sea level and submarine = 5000 + 1200 = 6200 m 4. As withdrawing money from the bank is opposite to depositing money to the bank. Therefore, the deposit amount is represented by a positive (+ ve) integer. Now, Amount deposited = ` 2000 Amount withdrawn = – ` 1642 ∴ Remaining balance in Mohan‘s account = ` (2000 – 1642) = ` 358 5. The distance towards west will be represented by a negative integer. Rita’s movement is shown as under : West





C

– 30 + 20

B

East

– 20 – 10 A 10 20 Since, Rita moves 20 km towards east from point A, so she reaches B and then from B she moves 30 km towards west along the same road and reaches C. Now, distance moved towards West = – 30 km

and distance moved towards East = 20 km ∴ Rita’s final position from A = 20 + (–30) = 20 – 30 = – 10 km (i.e.West) Thus, her final position from A will be represented by the integer – 10. 6. (a) Temperature of Lahulspiti = – 8°C Temperature of Srinagar = – 2°C Temperature of Shimla = 5°C Temperature of Ooty = 14°C Temperature of Bangalore = 22°C (b)  T emperature of coldest places = – 8°C of Lahulspiti and temperature of hottest places = 22°C of Bangalore ∴ Temperature difference between hottest and coldest places = (22) – (– 8) = 22 + 8 = 30°C (c)  Temperature of Lahulspiti = – 8°C and temperature of Srinagar = – 2°C ∴ Temperature difference between Lahulspiti and Srinagar = – 2 – (– 8) = – 2 + 8 = 6°C (d) Temperature of Srinagar and Shimla together = Temperature of Srinagar + Temperature of Shimla = – 2°C + 5°C = 3°C  Temperature of Shimla = 5°C and temperature of Srinagar = – 2°C ∴ Therefore, the temperature of Srinagar and Shimla taken together is not less than temperature of Srinagar. But, it is less than the temperature of shimla.

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WORKSHEET-3 Solutions 1. (a) Given, a = 21 and b = 18 ∴

LHS = a – (–b) = 21 – (–18) = 21 + 18 = 39

and

RHS = a + b = 21 + 18 = 39

Hence, LHS = RHS (b) Given, a = 118 and b = 125 ∴ LHS = a – (–b) = 118 – (–125) = 118 + 125 = 243

P-2

and RHS = a + b = 118 + 125 = 243 Hence, LHS = RHS (c) Given, a = 75 and b = 84 ∴ LHS = a – (–b) = 75 – (–84) = 75 + 84 = 159 and RHS = a + b = 75 + 84 = 159 Hence, LHS = RHS (d) Given, a = 28 and b = 11 ∴ LHS = a – (–b) = 28 – (–11) = 28 + 11 = 39

M A T H E M A T I C S - VII

and RHS = a + b = 28 + 11 = 39 Hence, LHS = RHS 2. (a) We have, LHS = (–8) + (–4) = – 8 – 4 = – 12 and RHS = (–8) – (–4) = – 8 + 4 = – 4 Since, –12 < – 4 ∴ –8 + (– 4) < – 8 – (–4) (b) We have, LHS = –3 + 7 – 19 = – (19 + 3) + 7 = – 22 + 7 = – 15 and RHS = 15 – 8 + (–9) = 15 + (– 8 – 9) = 15 + (– 17) = – 2 Since, –15 < – 2 ∴ –3 + 7 – 19 < 15 – 8 + (– 9) (c) We have, LHS = 23 – 41 + 11 = 34 – 41 = – 7 and RHS = 23 – 41 – 11 = 23 – 52 = – 29 Since, = – 7 > – 29 ∴ 23 – 41 + 11 > 23 – 41 – 11 (d) We have, LHS = 39 + (–24) – 15 = 39 – 39 = 0 and RHS = 36 + (–52) – (–36) = 36 – 52 + 36 = 72 – 52 = 20 Since, 0 < 20 ∴ 39 + (–24) – 15 < 36 + (–52) – (–36) (e) We have, LHS = – 231 + 79 + 51 = – 231 + 130 = – 101 and RHS = – 399 + 159 + 81 = – 399 + 240 = – 159 Since, – 101 > – 159 ∴ –231 + 79 + 51 > – 399 + 159 + 81 3. Let the steps moved down be represented by positive integers and the steps moved up be represented by negative integers. (a) Initially, the monkey was at step = 1 After 1st jump, the monkey will be at step = 1 + 3 = 4 After 2nd jump,

the monkey will be at step = 4 + (– 2) = 2 After 3rd jump, the monkey will be at step = 2 + 3 = 5 After 4th jump, the monkey will be at step = 5 + (– 2) = 3 After 5th jump, the monkey will be at step = 3 + 3 = 6 After 6th jump, the monkey will be at step = 6 + (– 2) = 4 After 7th jump, the monkey will be at step = 4 + 3 = 7 After 8th jump, the monkey will be at step = 7 + (– 2) = 5 After 9th jump, the monkey will be at step = 5 + 3 = 8 After 10th jump, the monkey will be at step = 8 + (– 2) = 6 After 11th jump, the monkey will be at step = 6 + 3 = 9 Clearly, the monkey will be at water level (i.e., 9th step) after 11 jumps. (b) The monkey will reach back at the top step after 5 jumps. 4. On the completion of the table, we have Statement Observation (a) 17 + 23 = 40 Result is an integer (b) (–10) + 3 = – 7 Result is an integer (c) (– 75) + 18 = –57 Result is an integer (d) 19 + (– 25) = – 6 Result is an integer (e) 27 + (– 27) = 0 Result is an integer (f) (– 20) + 0 = – 20 Result is an integer (g) (– 35) + (– 10) = – 45 Result is an integer We observe that for any two integers a and b, a + b is always an integer. Yes, the sum of two integers is always an integer. No, we cannot find a pair of integers whose sum is not an integer.

Since, addition of integers gives integers, we can say that integers are closed under addition.

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WORKSHEET-4 Solutions 1. (a) Let a pair be – 10 and 2 \ sum = – 10 + 2 = – 8 which is a negative integer (b) Let a pair be – 4 and 4 \ Sum = – 4 + 4 = 0 (c) Let a pair be – 2 and – 1 \ sum = – 2 + (– 1) = – 2 – 1 = – 3, which is smaller than both the integers

S OLUT I ONS



(d) Let a pair be – 2 and 4

\ sum =– 2 + 4 = 2, which is smaller than only one of the integer

(e) Let a pair be 4 and 2

\ sum = 4 + 2 = 6, which is greater than both the integers. 2. (a) Let a pair be 2 and 4 \ Difference =2 – 4 = – 2, which is negative integer (b) Let a pair be 2 and 2

P-3

\ Difference =2 – 2 = 0



Statement

Observation

(c) Let a pair be – 2 and + 1



(a) 7 – 9 = – 2

Result is an integer



(b) 17 – ( – 21)



= 17 + 21 = 38

Result is an integer



(c) (– 8) – (– 14) = 6

Result is an integer



(d) (– 21) – (– 10)



= – 21 + 10 = – 11



(e) 32 – (– 17)



= 32 + 17 = 49







= – 18 + 18 = 0

Result is an integer



(g) (– 29) – 0 = – 29

Result is an integer



When we subtract two integers, the result obtained is also an integer.



No, there is no pair of integers whose difference is not an integer.



Since, subtraction of integers always gives an integers, we can say that integers are closed under subtraction.



\ Difference = – 2 – (+ 1) = – 2 – 1 = – 3, which is smaller than both the integers (d) Let a pair be – 8 and – 1 \ Difference =– 8 – (– 1) = – 8 + 1 = – 7, which is greater than one of the integer (e) Let a pair be 2 and – 1 \ Difference =2 – (– 1) = 2 + 1 = 3, which is greater than both the integers 3. (a) A pair of integers, whose sum is – 7, can be – 2 and – 5. ∴ Sum = – 2 + (– 5) = – 2 – 5 = – 7 (b) A pair of integers, whose difference is – 10, can be – 12 and – 2. ∴ Difference = – 12 – (– 2) = – 12 + 2 = – 10 (c) A pair of integers, whose sum is 0 can be – 3 and 3. ∴ Sum = – 3 + 3 = 0 4. (a) A pair of negative integers whose difference gives 8 can be – 12 and – 20 Q (– 12) – (– 20) = – 12 + 20 = 8. (b) A negative integer and a positive integer whose sum is – 5 can be – 13 and 8. Q (– 13) + 8 = – 13 + 8 = – 5 (c) A negative integer and a positive integer whose difference is – 3 can be – 1 and 2.

Q

5.

(– 1) – 2 = – 1 – 2 = – 3 Total score of team A = (– 40) + 10 + 0

and,

Result is an integer

(f) (– 18) – (– 18)

8. We can write 4 × (– 8) as

4 × (– 8) = (– 8) + (– 8) + (– 8) + (– 8)



It can be represented on the number line as under :







We have,

– 32

– 24

= – 40 + 10 + 0 = – 30

Result is an integer

– 16

–8

0

8 × (– 2) = (– 2) + (– 2) + (– 2) + (– 2) + (– 2) + (– 2) + (– 2) + (– 2)



It can be represented on the number line as under :



We have, 3 × (– 7) = (– 7) + (– 7) + (– 7)



It can be represented on the number line as under :







We have,







It can be represented on the number line as under :

total score of team B = 10 + 0 + (– 40)

= 10 + 0 – 40 = – 30 Since, the total scores of each team are equal. \ No team scored more than the other but each have equal score. Yes, integers can be added in any order and the result remains unaltered. For example, 10 + 0 + ( – 40) = – 30 = – 40 + 0 + 10 6. (a) (– 5) + (– 8) = (– 8) + (– 5) (b) – 53 + 0 = – 53 (c) 17 + (– 17) = 0 (d) [13 + (– 12)] + (– 7) = (13 + [(– 12) + (– 7)] (e) (– 4) + [15 + (– 3)] = [(– 4) + 15] + (– 3) 7. On completion of the table, we have

– 16 – 14 – 12 – 10 – 8 – 6 – 4 – 2

– 21

– 14

–7

0

0

10 × (– 1) = (– 1) + (– 1) + (– 1) + (– 1) + (– 1) + (– 1) + (– 1) +(– 1) + (– 1) + (– 1)

– 10 – 9

–8

–7

–6

–5

–4

–3

–2

–1

0

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WORKSHEET-5 Solutions

1. (a) 6 × (– 19) = – (6 × 19) = – 114

P-4

(b) 12 × (– 32) = – (12 × 32) = – 384 (c) 7 × (– 22) = – (7 × 22) = – 154.

M A T H E M A T I C S - VII

2. (a)

– 5 × 4 = – 20

and, 10 × 6 + 10 × (– 2) = 60 – 20



– 5 × 3 = – 15 = – 20 – (– 5)

= 40



– 5 × 2 = – 10 = – 15 – (– 5)

10 × [6 +(– 2)] = 10 × 6 +10 × (– 2)

\



– 5 × 1 = – 5 = – 10 – (– 5)





– 5 × 0 = 0 = – 5 – (– 5)

(b) We have,

Hence, it is true. (– 15) × [(– 7) + (– 1)] = (– 15) × (– 8) = 120



– 5 × – 1 = 5 = 0 – (– 5)





– 5 × – 2 = 10 = 5 – (– 5)

and, (– 15) × (– 7) + (– 15) × (– 1)



– 5 × – 3 = 15 = 10 – (– 5)

= 105 + 15



– 5 × – 4 = 20 = 15 – (– 5)

= 120



– 5 × – 5 = 25 = 20 – (– 5)

6. (a) (i) 15 × (– 16) = – (15 × 16) = – 240



– 5 × – 6 = 30 = 25 – (– 5)

(ii) 21 × (– 32) = – (21 × 32) = – 672

(b)

(– 6) × 3 = – 18





(– 6) × 2 = – 12 = – 18 – (– 6)



(– 6) × 1 = – 6 = – 12 – (– 6)

(b) (i)



(– 6) × 0 = 0 = – 6 – (– 6)

= – (25 × 21) = – 525,



(iii) (– 42) × 12 = – (42 × 12) = – 504 (iv) – 55 × 15 = – (55 × 15) = – 825 L.H.S. = 25 × (– 21)



(– 6) × –1 = 6 = 0 – (– 6)





(– 6) × – 2 = 12 = 6 – (– 6)



(– 6) × – 3 = 18 = 12 – (– 6)

= – (25 × 21) = – 525



(– 6) × – 4 = 24 = 18 – (– 6)



(– 6) × – 5 = 30 = 24 – (– 6)



(– 6) × – 6 = 36 = 30 – (– 6)



(– 6) × – 7 = 42 = 36 – (– 6)

3.

(– 31) × (– 100) =31 × 100

= 3100

(– 25) × (– 72) = 25 × 72

= 1800

(– 83) × (– 28) =83 × 28

= 2324 4. (a) We have,



10 × [6 – (– 2)] = 10 × [6 + 2]

= 10 × 8 = 80

10 × 6 – 10 × (– 2) = 60 + 10 × 2

= 60 + 20 = 80 \ 10 × [6 – (– 2)] = 10 × 6 – 10 × (– 2) (b) We have,

(– 15) × [(– 7) – (– 1)] = (– 15) × (– 7 + 1)

= (– 15) × (– 6)

R.H.S. = (– 25) × 21

\ L.H.S. = R.H.S. (ii) L.H.S. = (– 23) × 20 = – (23 × 20) = – 460

R.H.S. = 23 × (– 20)

= – (23 × 20) = – 460

\

L.H.S. = R.H.S.

7. The distributive property of multiplication is the property that states that multiplying a sum by a number is same as multiplying each added/ subtracting by the number and then adding/ subtracting the product. If a, b and c are real numbers. Then, a × (b + c) = a × b + a × c and

a × (b – c) = a × b – a × c

(a) We have, (– 49) × 18 = (– 49) × (20 – 2)

= (– 49) × 20 – (– 49) × (2) [by distributive property] = – 980 + 98 = – 882

(b) We have, (– 25) × (– 31) = (– 25) × [(– 30) + (– 1)]

= 90 and, – 15 × (– 7) – (– 15) × (– 1) = 105 – 15= 90 \ (– 15) × [(– 7) – (– 1)] = (– 15) × (– 7) – (– 15) × (– 1)

= (– 25) × (– 30) + (– 25) × (– 1)

5. (a) We have,

= 70 [(– 19)] + (– 1)]





10 × [6 + (– 2)] = 10 × 4 = 40

[by distributive property] = 750 + 25 = 775 (c) We have, 70 × (– 19) + (– 1) × 70 [by distributive property] = 70 × (– 20) = – 1400

qqq S OLUT I ONS

P-5

WORKSHEET-6 Solutions 1. (i) For any integer a, (– 1) × a = – a. (ii) We know that the product of any integer and (– 1) is the additive inverse of integer. To find the integer whose product with (– 1) is : (a) – 22 is its additive inverse, i.e., 22. (b) 37 is the additive inverse, i.e., – 37. (c) 0 is the additive inverse of 0. 2. (– 1) × 5 = – 5 (– 1) × 4 = – 4 = [– 5 – (– 1)] = – 5 + 1 (– 1) × 3 = – 3 = [– 4 – (– 1)] = – 4 + 1 (– 1) × 2 = – 2 = [– 3 – (– 1)] = – 3 + 1 (– 1) × 1 = – 1 = [– 2 – (– 1)] = – 2 + 1 (– 1) × 0 = 0 = [– 1 – (– 1)] = – 1 + 1 (– 1) × (– 1) = 1 = [0 – (– 1)] = 0 + 1 3. Initial room temperature = 40°C Temperature lowered every hour = (– 5)°C Temperature lowered in 10 hours = [(– 5) × 10]°C = – 50°C \ Room temperature after 10 hours = 40°C – 50°C = – 10°C 4. (a) We have, 18 × [7 + (– 3)] = 18 × 4 =72 and, [18 × 7] + [18 × (– 3)] = 126 – 54 = 72 \ 18 × [7 + (– 3)] = [18 × 7] + [18 × (– 3)] (b) We have, (– 21) × [(– 4) + (– 6)] = (– 21) × (– 4 – 6) = (– 21) (– 10) = 210 and, [(– 21) × (– 4)] + [(– 21) × (– 6)] = 84 + 126 = 210 \ (– 21) × [(– 4) + (– 6)] = [(– 21) × (– 4)] + [(– 21) × (– 6)] 5. (a) 3 × (– 1) = – 3 (b) (– 1) × 225 = – 225 (c) (– 21) × (– 30) = 630 (d) (– 316) × (– 1) = 316 (e) (– 15) × 0 × (– 18) = [(– 15) × 0] × (– 18) = 0 × (– 18) = 0 (f) (– 12) × (– 11) × (10) = [(– 12) × (– 11)] × (10) = (132) × (10) = 1320 (g) 9 × (– 3) × (– 6) = [9 × (– 3)] × (– 6) = (– 27) × (– 6) = 162 (h) (– 18) × (– 5) × (– 4) = [(– 18) × (– 5)] × (– 4) = 90 × (– 4) = – 360 (i) (–1) × (–2) × (–3) × 4 = [(– 1) × (– 2)] × [(– 3) × 4] = (2) × (– 12) = – 24 (j) (– 3) × (– 6) × (– 2) × (– 1) = [(– 3) × (– 6)] × [(– 2) × (– 1)] = (18) × (2) = 36

6. (a) We have, 26 × (–48) + (–48) × (–36) = [26 × (– 48) + (– 48) × (– 36)] = [26 + (– 36)] × (– 48) [by distributive property] = (– 10) × (– 48) = 480 (b) We have, 8 × 53 × (– 125) = 424 × (– 125) = – (424 × 125) = – [424 × (100 + 25)] = – (424 × 100 + 424 × 25) [by distributive property of multiplication over addition] = – (42400 + 10600) = – 53000 (c) We have, 15 × (– 25) × (– 4) × (– 10) = [(– 25) × (– 4)] × [(– 10) × 15] [By associative property for multiplication] = (25 × 4) × [– (10 × 15)] [ a × (–b) = – (a × b) and (– a) × (– b) = a × b] = 100 × (– 150) = – (100 × 150) = – 15000 (d) We have, (– 41) × 102 = – (41 × 102) [\(– a) × b = – (a × b)] = – [41 × (100 + 2)] = – (41 × 100 + 41 × 2) [by distributive property of multiplication over addition] = – [4100 + 82] = – [4182] = – 4182 (e) We have, 625 × (– 35) + (– 625) × 65 = 625 × (– 35) + 625 × (– 65) [ (– a) × b = a × (– b)] [by distributive property of multiplication over subtraction] = 625 × (– 100) = – (625 × 100) [ a × (– b) = – (a × b)] = – 62500 (f) We have, 7 × (50 – 2) = (7 × 50) – (7 × 2) [by distributive property of multiplication over subtraction] = (350 – 14) = 336 (g) We have, (– 17) × (– 29) = 17 × 29 [(– a) × (– b) = a × b] = 17 × (30 – 1) = (17 × 30) – (17 × 1) [by distributive property of multiplication over subtraction] = 510 – 17 = 493 (h) We have, (– 57) × (– 19) + 57 = 57 × 19 + 57 × 1 [(– a) × (– b) = a × b and a × 1 = a] = 57 × (19 + 1) [by distributive property of multiplication over addition] = 57 × 20 = 1140

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P-6

M A T H E M A T I C S - VII

WORKSHEET-7 Solutions 1. (a) (– 3) × (– 9) = 27 (b) 5 × (– 7) = (– 35) (c) 7 × (– 8) = (– 56) (d) (– 11) × (– 12) = 132 - 100 2. (a) We have, (– 100) ÷ 5 = – (100 ÷ 5) = = – 20 5





- 81 (b) We have, (– 81) ÷ 9 = – (81 ÷ 9) = =–9 9 (c) We have, (– 75) ÷ 5 = – (75 ÷ 5) =

- 75 = – 15 5

(d) We have, (– 32) ÷ 2 = – (32 ÷ 2) =

- 32 = – 16 2

3. (a) We have, 125 ÷ (– 25) = – (125 ÷ 25) = (b) We have, 80 ÷ (– 5) = – (80 ÷ 5) =

- 125 =–5 25

- 80 = – 16 5

- 64 (c) We have, 64 ÷ (– 16) = – (64 ÷ 16) = =–4 16 36 4. (a) We have, (– 36) ÷ (– 4) = 36 ÷ 4 = =9 4 (b) We have, (– 201) ÷ (– 3) = 201 ÷ 3 =

201 = 67 3

(c) We have, (– 325) ÷ (– 13) = 325 ÷ 13 =

325 = 25 13

5. (a) (– 30) ÷ 10 = – 3 (b) 50 ÷ (– 5) = – 10 (c) (– 36) ÷ (– 9) = 36 ÷ 9 = 4 (d) (–49) ÷ (49) = – 1 (e) 13 ÷ [(– 2) + 1] = 13 ÷ (– 2 + 1) = 13 ÷ (– 1) = – 13 (f) 0 ÷ (– 12) = 0 (g) (– 31) ÷ [(– 30) + (– 1)] = (– 31) ÷ (– 31) = 1 (h) [(– 36) ÷ 12] ÷ 3 = (– 3) ÷ 3 = – 1 (i) [(– 6) + 5] ÷ [(– 2) + 1] = (– 6 + 5) ÷ (– 2 + 1) = (– 1) ÷ (– 1) = 1 6. Profit on sale of 1 bag of white cement = ` 8 Loss on sale of 1 bag of grey cement = – ` 5 (a) Profit on sale of 3000 bags of white cement

= ` (3000 × 8)







= ` 24,000 Loss on sale of 5000 bags of grey cement





= ` (5000 × – 5)





= – ` 25,000



Different between the two





= ` 24,000 – ` 25,000





= – ` 1000



Hence, there is a loss of ` 1000.

(b) Loss on the sale of 6400 bags of grey cement = ` (6400 × 5)







= ` 32,000 In order to have neither profit nor loss, the profit on the sale of white cement should be ` 32,000. \ Number of white cement bags sold Total profit = Profit per bag 32000 = 4000 = 8



Hence, 4000 bags of white cement should be sold to have neither profit nor loss. 7. (a)  Marks awarded by 1 correct answer = + 5 ∴ Marks awarded by 4 correct answers = + 5 × 4 = + 20  Marks awarded by 1 wrong answer = – 2 ∴ Marks awarded by 6 wrong answers = – 2 × 6 = – 12 Total marks attained by Mohan = 20 + (– 12) = + 8 (b) Marks awarded by 5 correct answers = 5 × 5 = + 25 Marks awarded by 5 incorrect answers = (– 2) × 5 = – 10 ∴ Total marks awarded by Reshma = 25 + (– 10) = + 15 (c) Marks awarded by 2 correct answers = (2) × 5 = 10 Marks awarded by 5 incorrect answers = 5 × (– 2) = – 10 ∴ Total marks awarded by Heena = 10 + (– 10) = 0

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WORKSHEET-8 Solutions 1. Five pairs of integers (a, b) such that a ÷ b = – 3 are : (9, – 3), (12, – 4), (15, – 5), (18, – 6) and (– 21, 7)

S OLUT I ONS

Note : We may write many such pairs of integers. 2. Difference in temperature + 10°C and – 8°C



= [10 – (– 8)]°C

P-7







Decrease in temperature in one hour = 2°C



\ Number of hours taken to have temperature 8°C Total decrease 18 below zero = = =9 Decrease in one hour 2



So, at 9 P.M., the temperature will be 8°C below zero.



Temperature at mid-night = 10°C – (2 × 12)°C





3. (a)

= (10 + 8)°C = 18°C

a ÷ (b + c) = 12 ÷ [(– 4) + 2]



= 12 ÷ (– 4 + 2)





= 12 ÷ (– 2) = – 6



and,







\







and,







\

(a ÷ b) + (a ÷ c) = [12 ÷ (– 4) + [12 ÷ 2] = – 3 + 6 = 3 a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c) a ÷ (b + c) = (– 10) ÷ (1 + 1) = (– 10) ÷ 2 = – 5 (a ÷ b) + (a ÷ c) = [(– 10) ÷ 1] + [(– 10) ÷ 1] = (– 10) + (– 10) = – 20 a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)

4. Difference in heights at two position = 10 m – (– 350 m) = 360 m Rate of descent = 6 m/minute \ Time taken = (360 ÷ 6) minutes = 60 minutes i.e., one hour. 369 5. (a) We have, 369 ÷ 1 = = 369 [ a ÷ 1 = a] 1





(b) We have, (– 75) ÷ 75 =

- 75 = – 1 75

[ (– a) ÷ (– a) = 1]

- 87 = 87 -1

(d) We have, (– 87) ÷ (– 1) =

[Q (– a) ÷ (– 1) = a]

= 10°C – 24°C = – 14°C



(b)

[ (–a) ÷ a = – 11] - 206 (c) We have, (– 206) ÷ (– 206) = =1 - 206



(e) We have, (– 87) ÷ 1 =

- 87 = – 87 [Q (– a) ÷ 1 = – a] 1

(f) We have, (– 48) ÷ 48 =

- 48 = – 1 [Q (– a) ÷ a = – 1] 48

(g) We have, 20 ÷ (– 10) =

20 =–2 - 10 [ a ÷ (– a) = – a]

(h) We have, (– 12) ÷ 4 =

- 12 = – 3 [ (– a) ÷ a = – a] 4

6. (a) Marks given for 12 correct answers at the rate of + 3 marks for each answer = 3 × 12 = 36 Radhika’s score = 20 \ Marks deducted her for incorrect answers = 20 – 36 = – 16 Marks given for one incorrect answer = – 2 \ Number of incorrect answers = (– 16) ÷ (– 2) = 8 (b) Marks given for 7 correct answers at the rate of + 3 marks for each answer = 3 × 7 = 21 Mohini scores = – 5 \ Marks deducted her for incorrect answers = – 5 – 21 = – 26 Marks given for one incorrect answer = – 2 \ Number of incorrect answers = (– 26) ÷ (– 2) = 13

qqq

WORKSHEET-9

1. (b) 2. (c) 3. (a) 4. (d) 5. (a) 6. The integer itself. 7. Zero



8. We can have (– 2) + (– 5) = – 7

or – 2 and – 5 may be the required pair. 

9. (a) (– 30) ÷ 10 = – 3

(b) (– 36) ÷ (– 9) = 4

P-8





10. (a) (– 8) ÷ (– 2) = 4 (integer) (b) 3 ÷ (– 8) =

3 , which is not an integer. -8



11. We have {36 ÷ (– 9)} ÷ {(– 24) ÷ 6}  36   24  = -  ÷ -   9   6 

(- 4) 4 = (– 4) ÷ (–4) = - 4 = 4 = 1 ( ) M A T H E M A T I C S - VII

12. We have,

13. We have

27 – [5 + {28 – (29 – 7)}] = 27 – [5 + {28 – 22}]

48 – [18 – {16 – (5 – 4 − 1 )}]



= 48 – [18 – {16 – (5 – 3)}]

[Removing the innermost brackets]

= 27 – [5 + 6] [Removing Brackets] = 27 – 11 = 16

= 48 – [18 – {16 – 2}]

[Removing vinculum] [Removing parentheses]

= 48 – [18 – 14]

[Removing braces]

= 48 – 4 = 44

qqq

WORKSHEET-10 Solutions

1. (c)



2. (d)



3. (c)



4. (d)

5. (a) 6. Zero 7. Zero 8. (a) 13 ÷ [(– 2) + 1] = 13 ÷ [– 2 + 1] = 13 ÷ (– 1) = – 13 (b) 0 ÷ ( – 12) = 0 9. (a) Since, 1 + a = 3 or, or, (b) Since, or,

a = 3 – 1 a = 2 b – 2 = 1 b = 1 + 2

or, b = 3 10. (– 1) × (– 1) × (– 1) ...... 5 times If odd number of (–ve) integers are multiplied, then the sign of product will be negative. Since 5 is odd number. So, result = – 1. 11. (a) (– 20) × (– a) × 9 = 20 × a × 9 = 180a (b) (– 18) × (– 5) × (– 1) × 7 = – 18 × 5 × 1 × 7 = – 630 (c) (– 5) × (– 4) × (– 8) = – 5 × 4 × 8 = – 160 12. (a) 26 × (– 48) + (– 36) × (– 48) = [– (26 × 48)] + (36 × 48) [Qa × (– b) = – (a × b) (– a) × (– b) = (a × b)] = (– 1248) + 1728 = 480 OR 26 × (– 48) + (– 36) × (– 48) = [26 + (– 36)] × (– 48)

= (– 10) × (– 48), [by distributive property] = 480 (b) 15 × (– 25) + (– 4) × (– 25) = [15 + (– 4)] × (– 25) [by distributive property] = (11) × (– 25) = – 275 (c) 7 × (50 – 2) = 7 × 50 – 7 × 2 = 350 – 14 = 336 [distributively of multiplication over subtraction] 13. we have 1  222 -  {42 + (56 − 8 + 9}+ 108  3   1  = 222 -  {42 + (56 − 17}+ 108  3  

[Removing vinculum]

1  = 222 -  {42 + 39}+ 108  3  

[Removing parentheses]

 81  = 222 -  + 108  3  

[Removing braces]

= 222 – [27 + 108] = 222 – 135 = 87 14. We have 39 – [23 – {29 – (17 – 9 − 3 )}]

= 39 – [23 – {29 – (17 – 6)}]



[Removing vinculum]

= 39 – [23 – {29 – 11}]



[Removing parentheses]



= 39 – [23 – 18]



= 39 – 5



= 34

[Removing braces]

qqq S OLUT I ONS

P-9

WORKSHEET-11 Solutions

11. (a) Since 15 is odd, so the sign of product of 15 negative integers is negative. (b) Since product of integer is an integer and product of integer and zero is zero. \ Product = 0.



1. (b)



2. (c)



3. (c)



4. (c)

(c) Sign of product of 22 negative integers is positive, because 22 is even.



5. (a)



6. Zero 7. Multiplicative identity for integers is 1. as 1.a = a = a.1 8. (a) We have, 9 × (– 3) × (– 6) = {9 × (– 3)} × (– 6) = – (9 × 3) × (– 6) = – 27 × (– 6) = 27 × 6 = 162 (b) we have, (– 12) × (– 13) × (– 5) = {(– 12) × (– 13)} × (– 5) = (12 × 13) × (– 5) = 156 × (– 5) = – (156 × 5) = – 780 9. (a) Since the number of negative integers in the product is odd. Therefore, their product is negative. Thus, we have (– 1) × (– 2) × (– 3) × (– 4) × (– 5) = – (1 × 2 × 3 × 4 × 5) = – (2 × 3 × 4 × 5) = – (6 × 4 × 5) = – (24 × 5) = – 120 (b) Since the number of negative integers in the given product is even. Therefore, their product is positive. Thus, we have (– 3) × ( – 6) × (– 9) × (– 12) = (3 × 6 × 9 × 12) = (18 × 9 × 12) [Q 3 × 6 = 18] = (162 × 12) [Q 18 × 9 = 162] = 1944 10. LHS = (– 30) × [13 + (– 3)] = (– 30) × (13 – 3) = (– 30) × 10 = – 300 RHS = [(– 30) × 13] + [(– 30) × (– 3)] = (– 390) + 90 = – 300 or, LHS = RHS  Hence Proved

118 – [121 ÷ (11 × 11) – (– 4) – {3 – 9 − 2 }] = 118 – [121 ÷ (11 × 11) – (– 4) – {3 – 7}] = 118 – [121 ÷ 121 – (– 4) – {3 – 7}] = 118 – [1 – (– 4) – {– 4}] = 118 – [1 + 4 + 4] = 118 – 9 = 109

13.

Marks for one correct answer = + 5

Marks for one incorrect answer = – 2 Marks given for 4 correct answers = 5 × 4 = 20

Score of Jay = – 12

Marks received by him for incorrect answer = – 12 – 20 = – 32 Marks given for one incorrect answer = – 2 ∴ Number of incorrect answers = (– 32) ÷ (– 2) = (32) ÷ 2 = 16 Hence, total answers, Jay attempted which were incorrect is 16 answers. 14. (a) 25 km

West C

A

East B

35 km

Distance towards East denoted (+ve) sign and Distance towards West denoted (–ve) sign. Since first Ridhaan goes toward East 25 km So distance covered by him = + 25 km Again, he goes towards West. So, distance covered by him on same road = – 35 km. So, position of Ridhaan from A = + 25 – 35 = – 10 km (b) Value : Addition of Integers. (c) Value : A rolling stone can gathers no moss. 

P-10

12. We have

qqq

M A T H E M A T I C S - VII

CHAPTER SECTION

B 2

FRACTIONS AND DECIMALS

WORKSHEET-12 Solutions

(c) We have,

1. Given, length of a rectangular sheet of paper 1 l = 12 cm 2



12 × 2 + 1 25 = = cm 2 2

∴



Breadth of a rectangular sheet of paper 2 b = 10 cm 3 =

3 2 + 5 7

LCM of 5 and 7 = 5 × 7 = 35 3 2 3×7 + 2×5 + = 5 7 35

(d) We have,

10 × 3 + 2 32 = cm 3 3



9 4 − 11 15

 25 × 3 + 32 × 2  = 2  6  





[Q LCM of 2 and 3 = 6] 3)139( 46  75 + 64   139  12 = 2 = 2   6  6     19 18 139 1 = = 46 cm 1 3 3



3 2 3 2– = − 5 1 5



=



1 3

cm. 2. (a) We have,

(b) We have,

4+

=

2× 5 − 3×1 5 [Q LCM of 5 and 1 = 5] 10 − 3 7 = 5 5

7 4 7 = + 8 1 8



∴

135 − 44 165

=

91 165

7 2 3 + + 10 5 2 

2 10 , 5, 2 5 5, 5,1 1,1,1

7 2 3 7 + 4 + 15 + + = 10 5 2 10 =

26 13 = 10 5

[dividing numerator and denominator by 2] 2 1 2 1 (f) 2 +3 = 2+ +3+ 3 2 3 2















[Q LCM of 8 and 1 = 8]

(e) We have,

=

LCM of 10, 5 and 2 = 2 × 5 = 10



S OLUT I ONS

11 11, 15 15 1, 15 1, 1

9 4 9 × 15 − 4 × 11 − = 11 15 165



4 ×8 + 7 ×1 = 8 32 + 7 39 = = 8 8

21 + 10 31 = 35 35

LCM of 11 and 15 = 11 × 15 = 165 ∴



=



Now, perimeter of a rectangular sheet of paper  25 32  = 2(l + b) = 2  +  3   2

Hence, the perimeter of a rectangular sheet is 46

5 5, 7 7 1, 7 1, 1







2 3

= ( 2 + 3) + +

1 2

4 3 = 5+ + 

6 6 1 1× 3 3  2 2×2 4 Q 3 = 3 × 2 = 6 and 2 = 2 × 3 = 6 

7 6

= 5+ =5+ 6 6

6+1 6

1 6

=5+ +  1 6

1 6

= 5 1  6  

37 1 6 6 6

P-11



(g)

1 5 17 29 − 8 −3 = 2 8 2 8



1 8 × 2 + 1 16 + 1 17  = = Q8 2 = 2 2 2



and 3



=

5 3 × 8 + 5 29  = =  8 8 8

17 × 4 29 × 1 2×4 8×1

[Q L.C.M. of 2 and 8 = 8]





3. (a) We have,

=

68 29 68 - 29 = 8 8 8

= 39 = 4 7 8

8

2 2 8 , , 9 3 21

2 2 × 21 42 = = 3 3 × 21 63



and



Q 

(b) We have,



8 8 × 3 24 = = 21 21 × 3 63 14 < 24 < 42 [numerators of fractions]

5 7

1 3 7 , , 5 7 10

5, 7 , 5 1, 7 , 1 1, 1, 1

LCM of 5, 7 and 10 = 5 × 7 × 2 = 70

On converting given fractions into like fractions, we get 1 1 × 14 14 = = 5 5 × 14 70



3 3 × 10 30 = = 7 7 × 10 70



LCM of 9, 3, 21 = 3 × 3 × 7 = 63 3 9 , 3, 21 On converting given fractions into like 3 3, 1, 7 fractions, we get 7 1, 1, 7 2 2 × 7 14 = = 1, 1, 1 9 9 × 7 63

14 24 42 < < 63 63 63 2 8 2 Thus, < < 9 21 3 Hence, descending order of given fractions is 2 8 2 , , . 3 21 9  2 5, 7 , 10

∴

7 7 × 7 49 = = 10 10 × 7 70

and Q

14 < 30 < 49



[numerators of fractions] 14 30 49 < < 70 70 70

∴



1 3 7 < < 5 7 10 Thus, Hence, descending order of given fractions is 7 3 1 , , . 10 7 5

qqq

WORKSHEET-13 Solutions

1. Given, width of picture = 7

3 7×5+3 cm = 5 5

35 + 3 38 = = cm 5 5 3 7 × 10 + 3 and width of the frame = 7 cm = 10 10 73 70 + 3 = = cm 10 10 ∴ To fit the picture in frame, picture should be trimmed

P-12

= Width of the picture – Width of the frame

=

38 73 38 × 2 − 73 × 1 − = 5 10 10 

=

76 − 73 3 = cm 10 10

[Q LCM of 5 and 10 = 10]

3 Hence, cm picture should be trimmed. 10 2. Let an apple be 1 part. Ritu ate

3 part of an apple. 5

Part of the apple eaten by Somu 3 5

5 5

3 5

= 1− = − =

2 5







3 2 Since, > , therefore, Ritu had the larger share. 5 5

M A T H E M A T I C S - VII



3 2 1 Ritu had taken  −  = part of apple more than 5 5 5







her brother Somu. 7 hour. 3. Michael finished colouring the picture in 12

3 Vaibhav finished colouring the picture in hour, 4 9 i.e., hour. 12





=

9 7 (Q LCM of 4 and 12 = 12) 12 12





=

9 -7 12 





=

2 1 = 12 6



Hence, Vaibhav takes

4. (a) Given,















Q



2 2 3 5

1 hour more Michael. 6

BE = 2

3 2×4 + 3 = 4 4

8 + 3 11 = = 4 4 AE = 3



=

3 3×5 + 3 = 5 5

15 + 3 18 = 5 5

20 , 6 10 , 3 5, 3 5, 1 11

Now,



5 × 10 50 5 = = 2 2 × 10 20







18 18 × 4 72 = = 5 5 × 4 20





Perimeter of DABE = (AB + BE + EA)













11 18  +  cm 4 5 

 50 55 72  + +  cm  20 20 20 

=  =

177 cm 20

= 8

17 cm 20

(b) Given, length of rectangle BCDE, l = BE = CD 3 11 =2 cm = cm and breadth of the rectangle, b 4 4 = BC = DE =

7 cm. 6

Perimeter of rectangle BCDE = 2 (l + b)  3 7 = 2  2 +  cm  4 6



11 7  +  cm  4 6

 = 2

33 14  +  cm  12 12 

 = 2

= 2× =

47 cm 12

47 6 5 6

= 7 cm







Now, perimeter of ∆ ABC =



and perimeter of rectangle BCDE =





LCM of 20 and 6 = 2 × 2 × 3 × 5 = 60 On converting both fractions into like fractions, we get 177 × 3 177 531 = = cm 20 × 3 20 60



and



Since,



So,



Hence, the perimeter of the triangle is greater than that of the rectangle.

L.C.M. of 2, 4, 5 = 20

11 11 × 5 55 = = 4 4 × 5 20





9 7 > , therefore, Vaibhav worked longer. Since, 12 12 Difference between time taken by Vaibhav and Michael 3 7 = 4 12



5 2

=  +

177 cm 20 47 cm 6

47 × 10 47 470 = = cm 6 × 10 6 60

531 > 470 (numerators of fractions) 177 47 531 470 > > i.e., 20 6 60 60

qqq

S OLUT I ONS

P-13

WORKSHEET-14

Solutions 2 2×3 6 × 3 = = 7 7 7

1. (a)



5. (a) → (c)

(b)

9 9×6 54 5 × 6 = = =7 7 7 7 7

(c)

3×

(d)

13 13 × 6 78 1 × 6 = = =7 11 11 11 11



(b) → (a)

(c) → (b)

1 3×1 3 = = 8 8 8

2. We have, 2 ×

2 2 50 of 25 = × 25 = = 10 5 5 5

(c)

6. (a) → (d)

(b) → (b)

(c) → (a)

(d) → (c)



2 4 = 5 5

7. (a) 7 ×

3 7×3 21 = = 5 5 5

This is required lowest form.

The pictorial representation is shown as below :

Now, dividing 21 (numerator) by 5 (denominator) We get, quotient = 4 and remainder = 1

+

2 – 5 (i)

3. (a)





















(b)

P-14

= 5× =5× =

14 + 3 7 17 5 × 17 = 7 7

85 1 = 12 7 7

4 1× 9 + 4 1 ×6 = ×6 9 9



4. (a)

4 – 5 (iii)

3 2×7 + 3 = 5× 7 7



(b)

2 – 5 (ii)

5×2



=



= =

9+4 ×6 9 13 13 × 6 ×6 = 9 9

78 6 =8 = 9 9 1 1 of 10 = × 10 2 2

=

1 × 10 10 = =5 2 2

1 1 1 × 16 16 of 16 = × 16 = = =4 4 4 4 4

∴ Mixed fraction = Quotient

(b) 4 ×

5)21( 4 20 1

Remainder 1 =4 Divisor 5

1 4 ×1 4 = = 3 3 3

This is required lowest form. Now, dividing 4 (numerator) by 3 (denominator), We get, quotient = 1 and remainder = 1 ∴ Mixed fraction = Quotient

(c) 2 ×

3 )4 (1 3 1

Remainder 1 =1 Divisor 3

6 2×6 12 = = 7 7 7

This is the required lowest form. Now, dividing 12 (numerator) by 7 (denominator),

7 )12(1 7 5

We get, quotient = 1 and remainder = 5 ∴ Mixed fraction = Quotient

(d) 5 ×

Remainder 5 =1 Divisor 7

2 5×2 10 = = 9 9 9

This is the required lowest form. Now, dividing 10 (numerator) by 9 (denominator), We get, quotient = 1 and remainder = 1 ∴ Mixed fraction = Quotient

9 )10(1 9 1

Remainder 1 =1 Divisor 9

M A T H E M A T I C S - VII



(e)

We get, quotient = 6 and remainder = 2

2 2×4 8 ×4= = 3 3 3



This is the required lowest form. Now, dividing 8 (numerator) by 3 (denominator),

3)8( 2 6 2

We get, quotient = 2 and remainder = 2 ∴ Mixed fraction = Quotient (f)

Remainder 2 =2 Divisor 3

This is the required lowest form.

4 11 × 4 44 = = 7 7 7

4 20 × 4 80 16 = = = = 16 5 5 5 1

This is the required lowest form. 1 13 × 1 13 (i) 13 × = = 3 3 3

This is the required lowest form. Now, dividing 44 (numerator) by 7 (denominator),

(h) 20 ×

Remainder 2 =6 Divisor 7

This is the required lowest form. Now, dividing 13 (numerator) by 3)13( 4 12 3 (denominator), We get, quotient = 4 and remainder = 1 1 Remainder 1 ∴ Mixed fraction = Quotient =4 Divisor 3

5 5×6 30 15 ×6= = = = 15 2 2 2 1

(g) 11 ×



∴ Mixed fraction = Quotient

7 )44( 6 42 2

(j) 15 ×

3 15 × 3 45 9 = = = =9 5 5 5 1

This is the required lowest form.

qqq

WORKSHEET-15 Solutions

2 2 of 18 = × 18 = 2 × 6 = 12 3 3 2 2 of 27 = × 27 = 2 × 9 = 18 3 3

(b) (i)

1. The complete boxes are shown below





(a)

1 1×1 1 1 1 1×1 1 1 = × = = (b) × = 14 5×7 35 2 7 × 5 7 2 7

1×1 1 1×1 1 1 1 1 1 = = (c) × = (d) × = 7 × 2 14 7 × 5 35 7 5 7 2 1 4 1× 4 4 × = = 3 5 3 × 5 15

2. (a)



3. (a)



2 1 2×1 2 × = = 3 5 15 3×5

(b)

8 4 8 × 4 32 × = = 3 7 3 × 7 21



3 2 3×2 6 × = = . 4 3 4 × 3 12

(b)

4.

(ii)



(ii)

(d) (i)





5. (a) (i) (ii)

1 1 of 46 =  46  23 2 2

S OLUT I ONS

4 of 20 = 5 4 of 35 = 5



(ii)



(ii) (b) (c) 1 1 of 24 =  24  12 2 2

3 3 of 36 =  36  3  9  27 4 4



6. (a) (i)



(a)

3 3 of 16 = × 16 = 3 × 4 = 12 4 4

(c) (i)



4 × 20 = 4 × 4 = 16 5

4 × 35 = 4 × 7 = 28 5

1 3 1 3 1 2×4 + 3 of 2 = ×2 = × 2 4 2 4 2 4

=

1 8 + 3 1 11 × = × 2 4 2 4

=

3 1 × 11 11 = =1 2×4 8 8

2 1 2 1 1 4×9 + 2 of 4 = ×4 = × 2 9 2 9 2 9

=











=

1 36 + 2 × 2 9 1 38 19 1 × = =2 2 9 9 9

P-15

(b) (i)



5 5 5 5 5 3×6 + 5 of 3 = ×3 = × 8 6 8 6 8 6













5 18 + 5 5 23 × = × 8 6 8 6

=

115 19 =2 48 48

2 2 5 5 5 9×3+ 2 of 9 = × 9 = × 8 3 8 3 8 3

(ii)

=



5 27 + 2 = × 8 3



1 5 29 145 = × = = 6 8 3 24 24

7. (a) Given, total amount of water in a bottle = 5 L

Water consumed by Vidya 2 2 2×5 = of 5 L = ×5L= L=2L 5 5 5

[dividing numerator and denominator by 5]

So, Vidya drank 2 L of water.

(b) Water consumed by Pratap = Total amount of water – Water consumed by Vidya

= 5 – 2 = 3 L

Pratap =

8. (a)













(c)













(d)



=

=

3×5



Water consumed by Pratap Total amount of water 3 5

1 2×4 +1 7×2 = 7× 4 4

= 7× =

= 15











5×6

(b)

= 3× =

25 + 1 5

3 × 26 78 = 5 5

= 15

3 5

3 4

1 6×3+1 4×6 = 4 × 3 3

= 4×







= 4×





=









(f)



8+1 4

7 × 9 63 = 4 4





1 5×5 +1 = 3× 5 5

3 4



(e)

24 + 3 4

5 × 27 135 = 4 4

= 33

∴ Fraction of total quantity of water consumed by







= 5×

18 + 1 3 19 4 × 19 = 3 3

1 76 = 25 3 3

1 3×4 +1 3 ×6 = ×6 4 4

=

12 + 1 13 × 3 ×3 = 2 2

=

1 39 = 19 2 2

2 3×5 + 2 3 ×8 = ×8 5 5

=

15 + 2 17 × 8 ×8 = 5 5

=

1 136 = 27 5 5

qqq

3 6 + 4×3 = 5× 4 4

WORKSHEET-16 Solutions 1. (a) (i)

P-16



1 1 1 1 1 of = × = 4 4 4 4 16

(ii) (iii)

1 3 of = 4 5 1 4 of = 4 3

1 3 3 × = 4 5 20 1 4 1 × = 4 3 3

M A T H E M A T I C S - VII



1 2 1 2 2 of = × = 7 9 7 9 63 1 6 1 6 6 of = × = 7 5 7 5 35

(b) (i) (ii)



(iii)

2. (a)



1 3 1 3 3 of = × = 7 10 7 10 70  2×3+ 2  2 2 2 ×2 = ×   3 3 3 3  



=

6+2 2 ×   3  3 

=

7 2 8 2×8 16 × = = =1 3 3 9 9 3×3

2 7 2×7 2 (b) × = = 7 9 × 9 7 9

[dividing numerator and denominator by 7] 3 6 3×6 18 9 × = = = 8 4 8×4 32 16



(d)

[dividing numerator and denominator by 2] 9 3 9×3 27 × = = 5 5 5×5 25



(e)

(c)







(f)

1 15 1 × 15 15 5 × = = = 3 8 24 8 3×8 [dividing numerator and denominator by 3] 11 3 11 × 3 33 13 × = = =1 2 10 2 × 10 20 20

4 12 4 × 12 48 13 × = = =1 5 7 35 35 5×7 2 3 2 3 6 3. (a) We have of = × = 7 4 7 4 28 (g)

6÷2 3 = 28 ÷ 2 = 14

and,

[Dividing numerator and denominator by 2] 3 5 3 5 15 of = × = 5 8 5 8 40 15 ÷ 5 3 = = 40 ÷ 5 8







[dividing numerator and denominator by 5] We know that , if two fractions have the same numerator but different denominators, the fraction with smaller denominator is greater.



Since, So,



8 < 14. 3 3 > 8 14

3 5 2 3 ⇒  of  is greater than  of  . 7 4 5 8    

S OLUT I ONS

(b) We have, 1 6 1 6 of = × 2 7 2 7 6 6 ÷ 2 3 = = = 14 14 ÷ 2 7







2 3 2 3 2×3 of = × = 3 7 3 7 3×7 6 6 ÷ 3 2 = = = 21 14 ÷ 2 7 We know that, if two fractions have the same denominator but different numerators, the fraction with greater numerator is greater. 3 2 Since, 3 > 2. So, > 7 7 and,

6 3 1 2 ⇒  of  of greater than  of  . 2 7 3 7     4. Let four saplings be planted in a row at the points A, B, C and D, respectively such that distance between 3 two adjacent saplings is , 4 i.e.

AB = BC = CD = 3/4m

3 m 4

3/4m

3/4m

A B C D Distance between the first and the last sapling = AD = AB + BC + CD = 3 × AB [Q AB = BC = CD] 3 3×3 9 1 =3× = = =2 m 4 4 4 4

Hence, the distance between the first and last sapling 1 is 2 m. 4  5×4 +1 2 1 2 ×5 = ×   4 5 4 5    20 + 1  2 = ×   5  4 

5. (a) We have,



=

2 21 2 × 21 × = 5 4 5×4

=

42 21 = 20 10

[dividing numerator and denominator by 2]  6×5+ 2  2 7 7 (b) We have, 6 × =   × 5 5 9 9    30 + 2  7 =  ×  5  9

=

32 7 32 × 7 × = 5 9 5×9

=

224 44 =4 45 45

P-17



(c)

 5×3 +1 3 1 3 ×5 = ×   3 2 3 2  





(d)





(e)

3

17 4 17 × 4 × = 5 7 5×7

=

68 33 =1 35 35

=

 15 + 1  3 ×   2  3 

=

3 16 3 × 16 × = 2 3 2×3



=

48 =8 6

 10 + 3  =   ×3  5 

 2×7 + 3  5 3 5 ×2 = ×   7 6 7 6  





=

(f)

 2×5+ 3 3 × 3 =   ×3 5 5  

2



=

13 3 13 × 3 × = 5 1 5

=

39 4 =7 5 5

=

 14 + 3  5 ×   6  7 

=

5 17 5 × 17 × = 6 7 6×7



=

85 1 =2 42 42

 21 + 4  3 =   × 7 5  

 3×5 + 2  2 4 4 × =   × 5 5 7  7 

 15 + 2  4 =  ×  5  7

(g)

3

 3×7 + 4  4 3 3 × =   × 7 7 5  5 



=

25 3 25 × 3 × = 7 5 7×5

=

5×3 15 1 = =2 7 7 7

qqq

WORKSHEET-17 Solutions

(a)

6÷5

1 =6÷ 3

 53  1     3 

 15  1   = 6 ÷ 16 = 6 ÷  3 3   = 6 ×

3 16  16 3   reciprocal of is   3 16 

=

1 18 9 = = 1 16 8 8

[Dividing numerator and denominator by 2] (b) Do same as above.

4 13 =2 7 18 2. We know that to divide a whole number by any fraction, multiply that whole number by the reciprocal of that fraction.

P-18





6÷







2∏



5 35 1 = = 17 2 2 2

= 6× =

7 42 = 4 4

1 42 ∏ 2 21 = = 10 4 ∏2 2 2

8 8 = 2 × reciprocal of 9 9

7÷2



= 7×

4 4 = 6 × reciprocal of 7 7



(c)





(b)

2 2 = 7 × reciprocal of 5 5



1. (a) We have,





7÷



=2× =

9 18 = 8 8

1 18 ∏ 2 9 = = 2 8∏2 4 4

M A T H E M A T I C S - VII

3. Distance covered by a car using 1 litre of petrol = 16 km 3 Distance covered by car using 2 litres of petrol 4 3 11    =  16 × 2  km = 16 ×  km 4 4  

= (4 × 11) km = 44 km 4. (a) (i) Since, 10 ÷ 2 = 5 and 30 ÷ 3 = 10

\ (ii)

2 5 10 × = 5 10 30 5 5∏5 = 10 10 ∏ 5

1 [Q H.C.F. of 5 and 10 = 5] 2 (b) (i) Since, 24 ÷ 3 = and 75 ÷ 5 = 15

8 \ is the simplest form. 15 3 1 3 2 5. (a) ÷ = × 5 2 5 1



=

3×2 6 1 = =1 5×1 5 5 1 2  Qreciprocal of 2 is 1   





 5×6 +1 1 9 9 ÷ =   ÷ 6 6 2 2   31 9 = ÷ 6 2

1 3 1 5 1× 5 ÷ = × = 2 5 2 3 2×3

5



3 8 24 = × 75 5 15

(ii) Since, H.C.F. of 8 and 15 = 1

(b)

3 5  Qreciprocal of 5 is 3    (d)

=

\

3 5  5 Qreciprocal of 5 is 3  6    2×2 +1 1 3 3 (c) 2 ÷ =   ÷ 2 2 5 5   5 × 5 = 5 × 5 = 25 = 4 1 = 6 6 2×3 2×3 =

=

=

31 2 × 6 9 9 2  Qreciprocal of 2 is 9    62 31 4 = =1 54 27 27

[dividing numerator and denominator by 2] 6. Given, number of days = 6 days Lipika reads the book per day for 3 = 1 h 4 =

1× 4 + 3 7 = h 4 4

=

42 21 1 = = 10 h 4 2 2

∴ Required hours by her to read the book 7 6×7 =6× = 4 4 [Dividing numerator and denominator by 2]

qqq

WORKSHEET-18 Solutions 3 4 12 × 4 = 12 × = 4 3 3 48 = = 16 3

1. (a)

5 6  Qreciprocal of 6 is 5   

12 ÷



3 4  Qreciprocal of 4 is 3   



(b)

14 ÷

S OLUT I ONS

5 6 14 × 6 84 = 14 × = = 6 5 5 5



(c) 8 ÷

(d) 4 ÷

7 3 8×3 =8× = = 3 7 7  Qreciprocal of 

24 7 7 3 is  3 7

8 3 =4× 3 8 8 3  Qreciprocal of 3 is 8   

P-19

4×3 12 3 = = 8 2 8 [dividing numerator and denominator by 4]  2×3+1 1 (e) 3÷2 =3÷   3 3   =

 6 +1 7 =3÷   =3÷ 3 3   3 3×3 9 =3× = = 7 7 7 7 3  Q Reciprocal of 3 is 7     3×7 + 4  4 (f) 5÷3 =5÷   7 7    21+ 4  25 =5÷   =5÷ 7   7 7 =5× 25 25 7  Qreciprocal of 7 is 25   

2. (a)





(d) (e) (f)

5×7 35 7 = = 25 25 5

[dividing numerator and denominator by 5] 7 7 1 7 ∏2 = × = 3 3 2 6 4 2 4 3 2 ÷ = × = 9 3 9 2 3

(b) (c)

=



3 7 3 3 8 ÷ = × = 7 7 7 8 8



1 3 7 5 35 2 ÷ = × = 3 5 3 3 9



3

1 8 7 3 21 ÷ = × = 2 3 2 8 16 4



3. (a) Reciprocal of

3 31 31 1 31 ∏7 = ∏7 = × = 7 7 7 49 7

3 7 = , an improper fraction. 7 3

(b) Reciprocal of

5 8 = , an improper fraction. 8 5

(c) Reciprocal of

9 7 = , a proper fraction. 7 9

(d) Reciprocal of

6 5 = , a proper fraction. 5 6

(e) Reciprocal of

12 7 = , a proper fraction. 7 12

P-20

1 8 (f) Reciprocal of = = 8 , a whole number. 8 1 1 11 (g) Reciprocal of = = 11, a whole number. 11 1 4. (a)



2 1 2 1 4 ∏ = × = 5 2 5 1 5

(b)

4 2 4 3 12 2 ∏ = × = = 9 3 9 2 18 3

(c)

3 8 3 7 21 ∏ = × = 7 7 7 8 56









=

21 ∏ 7 3 = 56 ∏ 7 8

1 3 7 3 7 5 35 (d) 2 ∏ = ∏ = × = 3 5 3 5 3 3 9 1 8 7 8 7 3 21 (e) 3 ∏ = ∏ = × = 2 3 2 3 2 8 16 2 1 2 3 2 2 4 (f) ∏ 1 = ∏ = × = 5 2 5 2 5 3 15 1 2 16 5 16 3 48 (g) 3 ∏ 1 = ∏ = × = 5 3 5 3 5 5 25 1 1 11 6 11 5 (h) 2 ∏ 1 = ∏ = × 5 5 5 5 6 5 55 55 ∏ 5 11 = = = 30 30 ∏ 5 6

5. (a) 0.5 or 0.05 First, we compare the digits on the left of the decimal point, starting from the leftmost digit. Here, digits to the left of the decimal point i.e., 0 are same. So, we compare the digits on the right of the decimal point starting from the tenth place. Here, 5 > 0 So, 0.5 > 0.05 Hence, 0.5 is greater. (b) 0.7 or 0.5 First, we compare the digits on the left of the decimal point i.e., 0 is same. So, we compare the digits on the right of the decimal point starting from the tenth place. Here, 7 > 5 So, 0.7 > 0.5 Hence, 0.7 is greater. (c) 7 or 0.7 First, we compare the digits on the left of the decimal point starting from the leftmost digit. Here, 7 > 0 [Q 7 has no decimal point, so we can write 7 as 7.0]

M A T H E M A T I C S - VII

So, 7.0 > 0.7 Hence, 7 is greater. (d) 1.37 or 1.49 First, we compare the digits on the left of the decimal point starting from leftmost digit. Here, both digits to the left of the decimal point are same. i.e., 1. So, we compare the digits on the right of the decimal point starting from the tenth place. Here, 3 < 4 So, 1.37 < 1.49 Hence, 1.49 is greater. (e) 2.03 or 2.30 First, we compare the digits on the left of the decimal point starting from the leftmost digit. Here, both the digits of the left at the decimal point are same. i.e., 2. So, we compare the digits on the right of the decimal point starting from the tenth place.

Here, 0 < 3 So, 2.03 < 2.30 Hence, 2.30 is greater. (f) 0.8 or 0.88 First, we compare the digit on the left of the decimal point, starting from the leftmost digit. Here, both the digits to the left of the decimal point are same. i.e., 0. So, we compare the digits on the right of the decimal point starting from the tenth place. Here, 8 = 8 i.e., both digits on the tenth place are also same. Now, we compare the digits on hundredth place. We get, 0 < 8 So, 0.8 < 0.88 Hence, 0.88 is greater.

qqq

WORKSHEET-19 Solutions 1. (a) (b)



77 rupees 77 paise = ` 77 + ` 0.77 = ` 77.77 50 paise = `



(b) (c)

50 100

= ` 0.50 235 paise = `



2. (a)



= ` 7.07



(e)

7 = ` 0.07 100

7 rupees 7 paise = ` 7 + ` 0.07



(d)

7 paise = `



(c)

(c) 200.03 = 2 × 100 + 0 × 10 + 0 × 1 + 0 ×

235 100

= ` 2.35 200 2 200 g = kg= kg 1000 10   1  1 g = kg  1000   = 0.2 g 3470 kg = 3.470 kg 3470 g = 1000 4 kg 8 g = 4 kg + 0.008 kg = 4.008 g

3. (a) 20.03 = 2 × 10 + 0 × 1 + 0 × (b) 2.03 = 2 × 1 + 0 ×

1 1 + 3× 10 100

(d) 2.034 = 2× 1 + 0 ×

(c) 8. (a)

56.3 × 1000 = 56300 5 cm =

5 m = 0.05 m 100   1  1 cm = m  100  



5 cm =

5 km 100 × 1000   1  1 cm = km  100000  



S OLUT I ONS

1 1 1 + 3× + 4× 10 100 1000

4. (a) Place value of 2 in 2.56 is 2 ones. (b) Place value of 2 in 21.37 is 2 tens. (c) Place value of 2 in 10.25 is tenths. (d) Place value of 2 in 9.42 is 2 hundredths. (e) Place value of 2 in 63.352 is 2 thousandths. 5. On subtracting 28 km from 42.6 km, we get, 42.6 – 28 = 14.6 km So, 28 km is less than 42.6 km by 14.6 km. 6. (a) 2.7 × 4 = 10.8 (b) 1.8 × 1.2 = 2.16 (c) 2.3 × 435 = 10.005 7. (a) 0.3 × 10 = 3 (b) 1.2 × 100 = 120

1 1 + 3× 10 100

1 1 + 3× 10 100





=

5 km= 0.00005 km 100000

P-21

(b)

35 mm =

35 cm = 3.5 cm 10   1  1 mm = cm  10  













35 mm =

35 35 m= m 10 × 100 1000



1   Q1 g = 1000 kg    Hence, it is clear that 8.950 > 8. 550 ∴ Sarala bought more fruits than Shyama. 2 12 11. (a) 0.2 × 6 = ×6 = = 1.2 10 10 = 8 + 0.950 = 8.950 kg

8 × 4.6 = 8 ×

46 368 = = 36.8 10 10

  1  1 mm = m  1000  

(b)

= 0.035 m 35 35 km = 35 mm = km 10 × 100 × 1000 1000000

(c)

2.71 × 5 =

271 1355 ×5 = = 13.55 100 100

(d)

20.1 × 4 =

804 201 ×4 = = 80.4 10 10

(e)

0.05 × 7 =

35 5 ×7 = = 0.35 100 100

(f)

211.02 × 4 =

  1  1 mm = km  10 , 00 , 000   = 0.000035 km 9. Given, distance between A and B = 7.5 km Distance between B and C = 12.7 km ∴ Distance travelled by Dinesh = AB + BC = 7.5 + 12.7 = 20.2 km Distance between A and D = 9.3 km Distance between D and C = 11.8 km ∴ Distance travelled by Ayub = AD + DC = 9.3 + 11.8 = 21.1 km Hence, it is clear that 21.1 > 20.2. ∴ Ayub travelled more distances, they both travelled = (21.1 – 20.2) km = 0.9 km Hence, Ayub travelled 0.9 km more than Dinesh. 10. Total fruits bought by Shyama = 5 kg 300 g + 3 kg 250 g = 8 kg 550 g 550 = 8 kg + kg 1000

1   kg  = 8 + 0.550 = 8.550 kg Q1 g = 1000  

And total fruits bought by Sarala = 4 kg 800 g + 4 kg 150 g = 8 kg 950 g 950 = 8 kg + kg 1000

(g)

84408 21102 ×4 = = 844.08 100 100

2 × 0.86 = 2 ×

86 172 = = 1.72 100 100

12. Area of rectangle = Length × Breadth = 5.7 cm × 3 cm  57  2 171 cm 2 =  × 3  cm = 10 10   = 17.1 cm2 13. (a) 1.3 × 10 = 13 (b) 36.8 × 10 = 368 (c) 153.7 × 10 = 1537 (d) 168.07 × 10 = 1680.7 (e) 31.1 × 100 = 3110 (f) 156.1 × 100 = 15610 (g) 3.62 × 100 = 362 (h) 43.07 × 100 = 4307 (i) 0.5 × 10 = 5 (j) 0.08 × 10 = 0.8 (k) 0.9 × 100 = 90 (l) 0.03 × 1000 = 30

qqq

WORKSHEET-20 Solutions

2. (a) 35.7 » 3 =

1. (a) 235.4 » 10 = 23.54 (b) 235.4 » 100 = 2.354 (c) 235.4 » 1000 = 0.2354

P-22

=











=

357 357 1 ∏3 = × 10 10 3 357 × 1 1 357 = × 10 × 3 10 3 119 1 × 119 = = 11.9 10 10

M A T H E M A T I C S - VII

(b)

25.5 » 3 =

255 255 1 ∏3= × 10 10 3





=

255 × 1 1 255 = × 10 × 3 10 3





=

1 85 × 85 = = 8.5 10 10

3. (a) 43.15 » 5 =











(b)



=

4315 × 1 1 4315 = × 100 × 5 100 5

=

1 863 × 863 = = 8.63 100 100

82.44 » 6 =

















8244 8244 1 ÷6 = × 100 100 6

=

8244 × 1 1 8244 = × 100 × 6 100 6

=

1 × 1374 100

=

1374 = 13.74 100

4. (a) 15.5 » 5 =



4315 4315 1 ÷ 5= × 100 100 5

155 155 1 ÷5 = × 10 10 5

=

155 × 1 1 155 = × 10 × 5 10 5

=

31 1 × 31 = = 3.1 10 10

(b) Do same as part (a) 126.35 ÷ 7 = 18.05 5. (a)





7.75 775 25 ∏ = 7.75 » 0.25 = 0.25 100 100



(b)



=

775 100 775 × = = 31 100 25 25

42.8 428 2 = 42.8 » 0.02 = ∏ 0.02 10 100 =

428 100 × = 214 × 10 10 2











= 2140

(c)





5.6 56 14 ∏ = 5.6 » 1.4 = 1.4 10 10



S OLUT I ONS



=

56 10 56 × = =4 10 14 14

6. Distance covers by a two-wheeler in 1 L petrol = 55.3 km ∴ Distance covered in 10 L petrol = 55.3 × 10 km = 553 km 7. (a) 2.5 × 0.3 On multiplying 25 and 3, we get 25 × 3 = 75 Number of digits after decimal point in given decimal number = 1 + 1 = 2 So, count 2 digits from the rightmost digits and move towards left, then put the decimal point, we get 0.75. Hence, 2.5 × 0.3 = 0.75 (b) 0.1 × 51. 7 On multiplying 1 and 517, we get 1 × 517 = 517 Number of digits after decimal point in given decimal numbers = 1 + 1 = 2 So, count 2 digits from the right most digits and move towards left, then put the decimal point, we get 5.17. Hence, 0.1 × 51.7 = 5.17 (c) 0.2 × 316.8 On multiplying 2 and 3168, we get 2 × 3168 = 6336 Number of digits after decimal point in decimal numbers = 1 + 1 = 2 So, count 2 digits from the rightmost digits and move towards left, then put the decimal point, we get 63.36 Hence, 0.2 × 316.8 = 63.36 (d) 1.3 × 3.1 On multiplying 13 and 31, we get 13 × 31 = 403 Number of digits after decimal point in given decimal numbers = 1 + 1 = 2 So, count 2 digits from the rightmost digits and move towards left, then put the decimal point, we get 4.03. Hence, 1.3 × 3.1 = 4.03 (e) 0.5 × 0.05 On multiplying 5 and 5, we get 5 × 5 = 25 Number of digits after decimal point in given decimal point, we get 0.025. Hence, 0.5 × 0.05 = 0.025 (f) 11.2 × 0.15 On multiplying 112 and 15, we get 112 × 15 = 1680 Number of digits after decimal point in given decimal numbers = 1 + 2 = 3 So, count 3 digits from the rightmost digits and move towards left, then put the decimal point, we get 1.680. Hence, 11.2 × 0.15 = 1.680 (g) 1.07 × 0.02 On multiplying 107 and 2, we get 107 × 2 = 214

P-23

Number of digits after decimal point in given decimal numbers = 2 + 2 = 4 So, count 4 digits from the rightmost digits and move towards left, then put the decimal point, we get 0.0214. Hence, 1.07 × 0.02 = 0.0214 (h) 10.05 × 1.05 On multiplying 1005 and 105, we get 1005 × 105 = 105525 Number of digits after decimal point in given decimal numbers = 2 + 2 = 4 So, count 4 digits from the rightmost digit and move towards left, then put the decimal point, we get 10.5525. Hence, 10.05 × 1.05 = 10.5525 (i) 101.01 × 0.01 On multiplying 10101 and 1, we get

10101 × 1 = 10101 Number of digits after decimal point in given decimal numbers = 2 + 2 = 4 So, count 4 digits from the rightmost digit and move towards left, then put the decimal point, we get 110.011. Hence, 101.01 × 0.01 = 1.0101

(j) 100.01 × 1.1

On multiplying 10001 and 11, we get

10001 × 11 = 110011

Number of digits after decimal point in given decimal numbers = 2 + 1 = 3 So, count 3 digits from the rightmost digit and move towards left, then put the decimal point, we get 110.011. Hence, 100.01 × 1.1 = 110.011

qqq

WORKSHEET-21 Solutions

=

[dividing numerator and denominator by 2]

1. (a) 2.7 » 100 = 0.027

(b) 0.3 » 100 = 0.003 (c) 0.78 » 100 = 0.0078



(d) 432.6 » 100 = 4.326

(b)

0.35 ÷ 5 =

(e) 23.6 » 100 = 0.236 (f) 98.53 » 100 = 0.9853 2. (a) 7.9 » 1000 = 0.0079

=

(b) 26.3 » 1000 = 0.0263 (d) 128.9 » 1000 = 0.1289 (e) 0.5 » 1000 = 0.0005 3. Distance covered in 2.4 litres of petrol = 43.2 km Distance covered in 1 litre of petrol = (43.2 ÷ 2.4) km  432 24  ÷  km =  10 10  



4. (a)

 432 10    km =  10 24   = 18 km



(c)

2.48 ÷ 4 =

=

7 = 0.07 100

248 1 × 100 4 1  Qreciprocal of 4 is 4    248 62 = = 0.62 400 100

[dividing numerator and denominator by 4]

(d)

65.4 ÷ 6 =

654 1 × 10 6 1  Qreciprocal of 6 is 6   

4 1 × 0.4 ÷ 2 = 10 2 1  Qreciprocal of 2 is 2   

P-24

35 1 35 × = 100 5 500 1  Qreciprocal of 5 is 5   

[dividing numerator and denominator by 5]

(c) 38.53 » 1000 = 0.03853



4 2 = = 0.2 20 10



=

654 109 = = 10.9 60 10

[dividing numerator and denominator by 6]

M A T H E M A T I C S - VII



(e)

651.2 ÷ 4 =



=

6512 1 × 10 4 1  Qreciprocal of 4 is 4    6512 1628 = = 162.8 40 10

[dividing numerator and denominator by 4]

(f)

14.49 ÷ 7 =



=

1449 1 × 100 7 1  Qreciprocal of 7 is 7    1449 207 = = 2.07 700 100



(e)

272.23 ÷ 10 =





(g)

1  Qreciprocal of 4 is 4   



396 99 = = = 0.99 400 100 [dividing numerator and denominator by 4]

(h)

80 1 × 0.80 ÷ 5 = 100 5 1  Qreciprocal of 5 is 5    =

80 16 = = 0.16 500 100

[dividing numerator and denominator by 5]

5. (a) (b)

48 1 48 × 4.8 ÷ 10 = = = 0.48 10 10 100 52.5 ÷ 10 =

525 1 525 × = = 5.25 10 10 100

(c)

0.7 ÷ 10 =

7 1 7 × = = 0.07 10 10 100

(d)

33.1 ÷ 10 =

331 1 331 × = = 3.31 10 10 100

= 27.223

(f)

0.56 ÷ 10 =

56 1 56 × = = 0.056 100 10 1000

(g)

3.97 ÷ 10 =

397 1 397 × = = 0.397 100 10 1000

6. (a) (b)

7 » 3.5 = 7 ∏

35 10 70 =7× = =2 10 35 35

36 » 0.2 = 36 ∏

2 10 = 36 × 10 2

= 18 × 10 = 180

(c)

3.25 » 0.5 =

[dividing numerator and denominator by 7] 396 1 3.96 ÷ 4 = × 100 4

27223 1 27223 × = 100 10 1000



(d)

=

30.94 » 0.7 = =

325 5 325 10 ∏ = × 100 10 100 5 65 = 6.5 10 3094 7 ∏ 100 10 3094 10 442 × = 100 7 10

= 44.2

5 25 5 100 ÷ = × =2 10 100 10 25



(e)

0.5 » 0.25 =



(f)

7.75 » 0.25 =

775 25 775 ÷ = = 31 100 100 25



(g)

76.5 » 0.15 =

765 15 ∏ 10 100

=

765 100 ´ 10 15

= 51 × 10 = 510

(h)

37.8 » 1.4 =

378 14 378 10 ∏ = × = 27 10 10 10 14



(i)

2.73 » 1.3 =

273 13 ∏ 100 10



=

273 10 21 × = = 2.1 100 13 10

qqq

WORKSHEET-22 Solutions

1. (c)



2. (c)

S OLUT I ONS



3. (b)



4. (d)



5. (a)

P-25

3 4 3 × = 4 5 5



6.



7. (a) 2.5 ×0.3 = 0.75

(b) 0.1 × 51.7 = 5.17.

8. (a) 1 × 1 = 1

as reciprocal of 1 is 1 1 (b) 2 × =1 2 as reciprocal of 2 is 1/2 1 (c) × (– 4) = – 1 4 as reciprocal of – 1/4 is – 4

(d) Since – 2

1 5 =– 2 2

2 and its reciprocal = – 5 -5 -2 So, × = 1 2 5

9. In order to know who solved less part of the 2 4 exercise, we will compare and . 7 5

We have, LCM of denominators (i.e., 7 and 5), 7 × 5 = 35. Converting each fraction into an equivalent fraction having 35 as its denominator, we have 2 2 × 5 10 4 4 × 7 28 = = and = = 7 7 5 35 5 5 × 7 35 × Q 10 < 28 10 28 2 4 ∴ < ⇒ < 35 35 7 5 Hence, Mukul solved lesser part than Deeksha. 10. In order to know who worked longer, we will 7 3 compare fractions and . 12 4 We have, (LCM of 12 and 4) = 12

Converting each fraction into an equivalent fraction with 12 as denominator, we have 7 7 ×1 7 3 3×3 9 = = and = = 12 × 12 4 × 12 12 1 4 3 Q

7 < 9 7 9 7 3 ∴ < ⇒ < 12 12 12 4 Thus, Vaibhav finished colouring in longer time. 3 7 9 7 9− 7 2 1 Now, – = – = = = 4 12 12 12 12 6 12 Hence, Pramod finished colouring in



time than Suval. 11. We have, 2 4 2×4 8 = (a) × = 9 5 9 × 5 45



1 2 7 2 7 × 2 14 = (b) 2 × = × = 3 5 5 5 3 × 5 15





(c) 5

1 hour more 6

3 3 23 17 23 × 17 391 27 ×2 = × = = = 13 4 7 4 7 28 28 4×7

12. (a) No. of students like to study English 1 = × 40 5 = 8 (b) No. of students like to study Maths 2 = × 40 5 = 2 × 8 = 16 (c) No. of students who like to study Science = 40 – (8 + 16) = 40 – 24 = 16 In part (ii), we know 2 × 40 = 16 5 2 So, of the total number of students 5 like to study Science.

qqq

P-26

M A T H E M A T I C S - VII

CHAPTER SECTION

B 3

DATA HANDLING

WORKSHEET-23 Solutions 1. Let the heights (in cm) of 10 students in the class be

150, 152, 151, 148, 149, 149, 150, 151, 153, 147.



Arranging the heights in ascending order, we have



147, 148, 149, 149, 150, 150, 151, 151, 152, 153.



Range of height of students = 153 – 147 = 6 2. We know that, whole numbers are those, which start from zero (0). \ First five whole numbers are 0, 1, 2, 3, and 4. Sum of numbers \ Mean = Number of terms =

0+1+2+3+4 10 = =2 5 5

Hence, the mean of first five whole numbers is 2. 3.

Total runs = 58 + 76 + 40 + 35 + 46



5 1 5+4 + 9 9 3rd number = 12 3 = 12 = = 2 2 12 × 2 24

\

1 11 5 9 1 > > > > 2 24 12 24 3

5 11 10 + 11 + 24 12 24 24 4th number = = = 2 2 2

\



1 11 21 5 9 1 > > > > > 2 24 48 12 24 3

5 9 10 + 9 + 5th number = 12 24 = 24 2 2

+ 45 + 0 + 100 = 400



Number of observations = 8

\

400 Mean = = 50 8

\

4. Given number are



1st number =

1 1 and . 2 3 Sum of given numbers Number of digits

1 1 3+2 + 5 5 = 2 3 = 6 = = 2 2 6 × 2 12

21 21 = 24 × 2 48

=

=

10 + 9 19 = 24 × 2 48

1 11 21 5 19 9 1 > > > > > > 2 24 48 12 48 24 3

Hence, 5 numbers between

1 1 11 21 and are , , 2 3 24 48

5 19 9 , and . 12 48 24 5. Arranging the grades in a class assessment in a tabular form : Grades xi

Tally Marks

Frequency fi

1

I

1

1

2

II

2

4

3

I

1

3

4

III

3

12

5

llll

5

25

1 5 6+5 + 11 11 2 12 = = 12 = = 2 2 12 × 2 24

6

IIII

4

24

7

II

2

14

8

I

1

8

1 11 5 1 > > > 2 24 12 3

9

I

1

9

 \

[LCM of 2 and 3 = 6] 1 5 1 > > 2 12 3

2nd number =

\ Similarly,

Given number + 1st number 2

Sfi = 20

f ix i

Sfixi = 100

(a) Highest grade = 9 (b) Lowest grade = 1

S OLUT I ONS

P-27

(c) Range = 9 – 1= 8 Σfi xi 100 = (d) Arithmetic mean = =5 Σfi 20 6. (a) A’s average number of points scored per game

=

14 + 16 + 10 + 10 50 = 4 4

(b) C’s average points per game



=

8 + 11 + 0 + 13 32 = =8 4 4

Since, we are comparing the performance, so we divide by 4 to find the mean for C.

(c) B’s average points per game



=

0 + 8 + 6 + 4 18 = =4.5 4 4

(To find B’s average, we find the sum of all observations and divide this by the number of observations.)

(d) Since, 12.5 > 8 > 4.5

\ The best performer is A.

qqq

WORKSHEET-24 Solutions 1. Sum of the enrolment during six consecutive years = 1555 + 1670 + 1750 + 2013 + 2540 + 2820 = 12348 12348 Mean enrolment = = 2058 6 2. Given, marks obtained by a group of students in a science test = 85, 76, 90, 85, 39, 48, 56, 95, 81 and 75 Now, on rearranging the marks in ascending order, we get = 39, 48, 56, 75, 76, 81, 85, 85, 90, 95 (a) It is clear from the ascending order of marks that, highest marks = 95 and lowest marks = 39 (b) Range of the marks obtained = Highest marks – Lowest marks = 95 – 39 = 56 (c) We have, total marks = 39 + 48 + 56 + 75 + 76 + 81 + 85 + 85 + 90 + 95 = 730 Number of students = 10 Total marks obtained by group \ Mean marks = Number of students

730 = = 73 10 Hence, mean marks obtained by the group is 73. 3. (a) Arranging the rainfall during the week in the ascending order, we have 0.0, 0.0, 1.0, 2.1, 5.5, 12.2, 20.5 Range = 20.5 – 0.0 = 20.5 (b) Sum of the rainfall during the week = 0.0 + 0.0 + 1.0 + 2.1 + 5.5 + 12.2 + 20.5 = 41.3 Mean =

P-28

(c) For five days, the rainfall was less than the mean rainfall. 4. (a) We have, 2, 6, 5, 3, 0, 3, 2, 4, 5, 2, 4. On arranging the data in ascending order, we get 0, 2, 2, 2, 3, 3, 4, 4, 5, 5, 6 Now, we will arrange it in a table form

41.3 = 5.9 7

Numbers

Occurring time (Frequency)

0

One time

2

Three times

3

Two times

4

Two times

5

Two times

6

One time

From the above, it is clear that 2, occur three times. Hence, 2 is the mode of the given data. (b) We have, 2, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14. On arranging the data in ascending order, we get 2, 10, 12, 14, 14, 14, 14, 14, 14, 16, 16, 18 Now, we will arrange it in a table form Numbers

Occurring time (Frequency)

2

One time

10

One time

12

One time

14

Six times

16

Two times

18

One time

Q 14 occurs the maximum number of times i.e., 6 times Hence, mode of the data is 14.

M A T H E M A T I C S - VII



5. Let us put the data in a tabular form : Variate

Tally bars

Frequency

12

3

14

III IIII llll

15

llll llll

10

16

llll I

6

17

II

2

18

I

1

19

I

1

13

(b) from ascending order of heights, we observe that first term is minimum. \ Height of the shortest girl = 128 cm (c) From ascending order, we get Highest value = 151 Lowest value = 128 \ Range of the data = Highest value – Lowest value = 151 – 128 = 23 Sum of height of all girls (d) Mean height of the girls = Number of girls

4 5

Looking the at table, clearly, we can say 15 is the ‘mode’, since 15 has occurred the maximum number of times. 6. On arranging the data in ascending order, we get 128, 132, 135, 139, 141, 143, 146, 149, 150, 151 (a) We can observe from the ascending order of heights that last term is maximum. Thus, height of the tallest girl = 151 cm

128+132+135+139+141 +143+146+149+150+151 = 10 1414 = = 141.4 10 Hence, the mean height of the girls is 141.4 cm. (e) It is clear that, we have mean height = 141.4 cm 5 girls have more height than the mean height i.e., 143 cm, 146 cm, 149 cm, 150 cm and 151 cm.

qqq

WORKSHEET-25 Solutions 1. On arranging the data in ascending order, we get 12, 12, 13, 13, 14, 14, 14, 16, 19

Mode =14



Median = 14.

Looking at the table, clearly, we can say 163 cm is the ‘mode’, since 163 has occurred the maximum number of times. Mode gives that observation which occurs most frequently in the data.

2. The weights (in kg) of 15 students of a class are :

38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47



Ascending Order = 32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50

4. On arranging the data in ascending order, we get

32, 32, 34, 35, 35, 38, 42 Now, arrange the data in tabular form Number

Tally marks

Occurring time (Frequency)

32

II

2

34

I

1

35

II

2

38

I

1

(a) Mode = 38

Median = 40. (b) Yes, there is more than one mode, i.e. 43. 3. Let us put the data in a tabular form : Height

Tally bars

Number of children

160

IIII

4

161

I

1

162

IIII

4

163

llll IIII

9

164

III

3

165

III

3

168

I

1

S OLUT I ONS

42

1 I It is clear from the above table, that 32 and 35 occur two times \ Mode = 32 and 35 In the ascending order, the middle value of data is 35. \ Median = 35. Hence, current mode is 32 and 35 and median is 35. 5. On arranging the data in ascending order, we get 5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25 Now, arrange the data in tabular form

P-29



(c) The value of the middle observation is 15.

Numbers

Tally marks

Occurring time (Frequency)

5

I

1

No, mean, mode and median are not same.

9

I

1



10

I

1

12

I

1

(i) To decide upon the number of different sweets needed for 25 persons called for a feast.

15

I

1

16

I

1

19

I

1

20

IIII

4

23

I

1

I

1

24 25

\

Here, 20 occur more frequently i.e., 4 times. \ Mode = 20 Since, median is the middle observation and in ascending order middle value is 20. \ Median = 20 Yes, both mode and median are the same. 6. (a) We have 6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15 On arranging the data in ascending order, we get 6, 8, 10, 10, 15, 15, 15, 50, 80, 100, 120 Now, sum of the runs = 6 + 8 + 10 + 10 + 15 + 15 + 15 + 50 + 80 

+ 100 + 120 = 429

7. Let us consider the following examples :

(ii) A shopkeeper selling shoes has to decide how to replenish his stock. (iii) To judge the performance of a cricket player. (iv) If only one fruit can be bought for everyone going on a picnic, which fruit should be arranged.

2

II

(a) Let us consider the first statement. Suppose the number of sweets needed by each person be 3, 4, 3, 4, 3, 2, 3, 4, 3, 3, 5, 3, 3, 4, 3, 5, 5, 3, 4, 5, 3, 4, 6.

Clearly, the mode here is 3 sweets. If we use mode as the needed value for this data, then 3 × 25 = 75 sweets only, 3 for each person. Clearly, the sweets would be inadequate. So, here mean is the appropriate value. Clearly, in the third statements, average performance will be appropriate to judge the player. So, mean is appropriate.

(b) Let us consider the second statement. Suppose the number of pairs of shoes of different sizes sold by a shopkeeper is as under :

Number of players = 11 \ Mean =

Sum of the runs 429 = = 39 11 Number of players

(b) From Part (i), ascending order of runs is as follows :



6, 8, 10, 10, 15, 15, 15, 50, 80, 100, 120

Here, 15 occurs more frequently i.e., 3 times. \

Mode = 15

Median = 15

Size of shoes

1

2

3

4 5

6

7 8

9

No. of pairs sold

1

4

3

2 3

4

5 7

6

Since, the sale of size 8 is maximum. So, its mode is 8. Thus, it is appropriate to replenish his stock with pair of size 8. In the fourth statement, the best liked fruit should be selected. So, mode is appropriate value.

qqq

WORKSHEET-26 Solutions

(d) False,

1. (a) True, the mode is always one of the numbers in a data.



(b) False, mean is not one of the numbers in a data because

Mean =

Sum of observations Number of observations

(c) False, the median is not always one of the numbers in a data.

P-30

Mean = =

6 + 4 + 3 + 8 + 9 + 12 + 13 + 9 8 64 =8 8

2. Clearly, from the given bar chart : (a) The most popular pet is cat. (b) Eight children have a dog as a pet. 3. (a) A fraction of the number of watches that leaked to the number tested for each company are :

M A T H E M A T I C S - VII





10 1 = 40 4

700

15 3 = 40 8

500

For C,



Scale : 1 unit = 100 books

600

Number of books

For B,

5. Double bar graph of given data is as follows :

20 1 = 40 2

For A,

25 5 = 40 8 (b) Clearly, 10 < 15 < 20 < 25 10 15 20 25 or, < < < 40 40 40 40 and For D,

Thus, company with fraction

400 300 200 100 0 1995

1996 1997 1998 Year English books Hindi books

10 , i.e., company B 40

has better watches. 4. Clearly, from the given graph, we have (a) Number of books sold in the year 1989 : 170 (approx.) 1990 : 475 (approx.) 1992 : 225 (approx.) (b) In the year 1990, about 475 books were sold. In the year 1991, about 225 books were sold. (c) Fewer than 250 books were sold in the years 1989 and 1992. (d) From the bar graph, it is clear that number of books in the year 1989 is 180 (approx.). We can see in the bar graph that the number of book sold in the year 1989 is little less than 200, but greater that 175.

(a) The difference in the scale of the two language books in individual year is as follows : Year



Difference

1995

500 – 350 = 150

1996

525 – 400 = 125

1997

600 – 450 = 150

1998

650 – 620 = 30

Clearly, the difference in the scale of the two language books was least in the year 1998.

(b) Yes, because the bar graph of the sale of English books became longer, faster, compared to Hindi books. So, the demand for English books rose faster.

qqq

WORKSHEET-27

Solutions 1. (a) We start the scale from 0. The greatest value of data is 135, so end the scale at a value greater than 135 such as 140. Use equal division along axes as increment of 10. Here, we take 1 unit for 10 children. Now, the graph is shown as below :

Number of childrens

Y 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0

Scale : 1 unit = 10 children

(b) (i) From the graph,



Maximum number of children in fifth class i.e., 135.



And minimum number of children in tenth class i.e., 80.



(ii) From the graph,



Number of students in sixth class = 120



Number of students in eighth class = 100



\ The required ratio



Fifth

Sixth

S OLUT I ONS

Seventh Eighth Classes

Ninth

Tenth

X



=

Number of students in sixth class hth class Number of students in eigh

=

120 100

=

6 or 6 : 5 5

P-31

2. The graph is as under : Scale : 1 cm = 10 marks 1st Term 2nd Term

It is inferred that more people prefer cricket and less athletics.

100



(b) Most popular sport is cricket.

90



(c) Watching is more preferred than participating.

80

4. The graph is as under :

70

Scale : 1 cm = 10°C 50°

50

Maximum temperature

30 20 10 0

Scale : 1 cm = 100 persons

1200

Watching

Participating

1100

800 700

Mumbai

Jammu

Jaipur

Delhi

Cities

(a) The city Jammu has the largest difference in the minimum and maximum temperature on the given data.

(b) Jammu is the hottest city and Bangalore is the coldest city.



(c) Bangalore and Jaipur are two cities of such type.



(d) Mumbai has the least difference between its minimum and the maximum temperature.

1000 900

Chennai

0

(a) In Mathematics the child has improved his performance the most. (b) In Social Science the child has improved his performance the least. (c) Yes, the performance has gone down in Hindi. 3. (a) The graph is as under (on the next page) :

1300

10°

Science S.Science

Bangalore

Maths Subjects

20°

Amritsar

Hindi

30°

Ahmedabad

English

Persons

Minimum temperature

40°

40

Temperature

Performance

60

600 500 400 300 200

Athletics

Hockey

Swimming

Basketball

Cricket

100

Sports

qqq

WORKSHEET-28 Solutions 1. (d) 2. (c)

18

5

18

6

15

5. (a) Certain to happen.

3. Head : 51 times and Tail : 49 times. 4.

P-32

4

(b) Can happen but not certain.

Outcome

Number of occurrence

1

16

2

17

3

16



(c) Impossible.



(d) Can happen but not certain.



(e) Can happen but not certain.

6. Out of 6 marbles, one can be drawn in 6 ways. So, total number of events = 6.

M A T H E M A T I C S - VII

(a) The marble with number 2 can be obtained only in one way. 1 \ Required probability = 6 (b) The marble with number 5 can be obtained only in one way. \ Required probability =

1 6

7. On tossing of a coin, the possible outcomes are head (H) or tail (T). 1 2

8. The following table gives result of a die thrown 250 times. Number of the Die 1 2

llllllllllllllllllllllll llllllllllll II

47

4

llllllllllllllllllllllll llllllllllllllll II

52

5

llllllllllllllllllllllll llll III

38

6

llllllllllllllllllllllll llllllll

40

The bar graph for the above data is as drawn below :

No. of times

Tally Marks

llllllllllllllllllllllll III

33

llllllllllllllllllllllll llllllll

40

60 50

Number of times

\ Required probability =

3

40 30 20 10 1

2

3 4 Face of die

5

6

qqq

WORKSHEET-29 Solutions 1. (c) 2. (c) 3. (d)

25

98+99+100+0+1+2 300 = = 50 6 6 5. Frequency distribution of ages of 25 students :

29 30

IIII

27

III

28

4.

Age

Tally marks

14 15 16 17

IIII IIII III IIII IIII III

Total

Frequency 4 8 10 3 25

6. Frequency distribution of scores :

III I I

Total

5 3 3 1 1 40

7. We have,

11 =

x + (x + 2) + (x + 4 ) + (x + 6) + (x + 8) 5

or,

11 =

5x + 20 5

or,

55 = 5x + 20

or,

5x = 35 x = 7

Variate

Tally marks

Frequency

15 16

IIII IIII I

4 6

or,

18

IIII I

6

20

IIII I

6

24

IIII

5

x + ( x + 2 ) + ( x + 4 ) 3x + 6 = = = x + 2 = 7 + 2 = 9 3 3 [ x = 7]

S OLUT I ONS

∴ Mean of first three observations

P-33

8. (a) Mean =

37 + 38 + 35 + 40 + 45 + 52 + 40 7

287 = = 41 7



(b) Mean of given data.

(c) Value : Girl child has equal right to get proper education. There should be no discrimination between boys and girls.

9. In the tabular form, the above data can be represented as follows :



4

4 × 5 = 20

6

4

4 × 6 = 24

7

3

3 × 7 = 21

8

2

2 × 8 = 16

9

1

1×9=9

Σfi = 20

Σfi xi = 108



∴ Mean grade =

Σfi xi 108 = = 5.4 Σfi 20

Grade

Tally Bars

Frequency (fi)



Hence, the mean grade is 5.4

1

I

1



2

I

1

It is evident from the frequency table that the highest and lowest grades are 9 and 1 respectively.

3

I

1



∴ Range of the data = 9 – 1 = 8

4

III

3



5

IIII

4

6

IIII

4

7

III

3

8

II

2

9

I

1

10. (a)

In order to compute the arithmetic mean of grades i.e., average grade, we prepare the following table :



5

Computation of Arithmetic Mean

Weight of protein food (in gram)

Tally marks

Frequency

75

IIII

4

80

llll

5

85

IIII

4

90

llllII

7

110

llll

5

Total

n = 25

Grade (xi)

Frequency (fi)

x i × fi

1

1

1×1=1

2

1

1×2=2

3

1

1×3=3





4

3

3 × 4 = 12



A nation has to take one of its human resources.

(b) Tabulation of data.

(c) In India children need sufficient protein food for nutrition. OR

qqq

P-34

M A T H E M A T I C S - VII

CHAPTER SECTION

B 4

SIMPLE EQUATIONS

WORKSHEET-30 Solutions p

1. Let the values of y = 2, 4, 7, 9, 10



Put the values of y one by one in (10y – 20) For 2,

[10(2) – 20] = 20 – 20 = 0

For 4,

[10(4) – 20] = 40 – 20 = 20

For 7,

[10(7) – 20] = 70 – 20 = 50

For 9,

[10(9) – 20] = 90 – 20 = 70

For 10,

[10(10) – 20] = 100 – 20 = 80

Yes, we see a solution to 10y – 20 = 50, when put y = 7 in 10y – 20. 2. On putting the given value of x in L.H.S. If we get the value equal to R.H.S., then value of x satisfies the equation, otherwise not. S. Equation No.



Value

x+3=0 x=3

No (Q x + 3 = 3 + 3 = 6 ≠ 0)

(b)

x+ 3 = 0

No (Q x + 3 = 0 + 3 = 3 ≠ 0)

(c)

x+3=0 x=–3

Yes (Q x + 3 = – 3 + 3 = 0)

(d)

x–7=1

x=7

No (Q x – 7 = 7 – 7 = 0 ≠ 1)

(e)

x–7=1

x=8

Yes (Q x – 7 = 8 – 7 = 1)

(f)

15x = 25

x=0

No (Q 15x = 15 × (0) = 0)

(g) 15x = 25

x=5

No (Q 15x = 15 ×(5) = 75)

(h) 15x = 25

x=–5

No (Q 5x = 5 ×(– 5) = –25)

m=–6

No (Q

x=0

(i)

m =2 3

m = 2 = – 2) 3

(j)

m =2 3

m=0

m 0 No (Q = = 0) 3 3

(k)

m =2 3

m=6

Yes (Q

m 6 = = 2) 3 3

3. (a) Let us evaluate the L.H.S. and R.H.S. of the given equation for some values of p and continue to give new values till the L.H.S. becomes equal to the R.H.S.

The given equation is 5p + 2 = 17 i.e., we have L.H.S. = 5p + 2 and R.H.S. = 17

R.H.S.

Is L.H.S. = R.H.S. ?

1

5×1+2= 5+2=7

17

No

2

5 × 2 + 2 = 10 + 2 = 12

17

No

3

5 × 3 + 2 = 15 + 2 = 17

17

Yes

Clearly, L.H.S. = R.H.S. for p = 3. Hence. p = 3 is the solution of the given equation. (b) Let us evaluate the L.H.S. and R.H.S. of the given equation for some values of m and continue to give new values the L.H.S. becomes equal to the R.H.S. The given equation is 3m – 14 = 4, that is, 14 subtracted from 3 times m gives 4. So, we substitute 1 values which gives 3m > 14. We have, L.H.S. = 3 m – 14 and R.H.S. = 4.

Say, whether the equation is satisfied. (Yes/No)

(a)

L.H.S.

m

L.H.S.

R.H.S.

Is L.H.S. = R.H.S. ?

5

3 × 5 – 14 = 15 – 14 = 1

4

No

6

3 × 6 – 14 = 18 – 14 = 4

4

Yes

Clearly, L.H.S. = R.H.S. for m = 6. Hence, m = 6 is the solution of the given equation. 4. The equations for the given statements are : (a) x + 4 = 9 (b) y – 2 = 8 (c) 10a = 70 (d) b ÷ 5 = 6 3 (e) × t = 15 (f) 7m + 7 = 77 4 1 × x – 4 = 4, where x is the number 4



(g)



(h) 6y – 6 = 60

(i)

1 × z + 3 = 30 3

5. (a) When, n = 1 then n + 5 = 1 + 5 = 6 ≠ 19 So, n = 1 is not solution of the given equation. (b) When, n = – 2, then

7n + 5 = 7(– 2) + 5 = – 14 + 5

= – 9 ≠ 19 So, n = – 2 is not a solution of the given equation.

(c) When,

then

n = 2 7n + 5 = 7 × 2 + 5 = 14 + 5

= 19 So, n = 2 is the solution of the given equation.

S OLUT I ONS

P-35



(d) When,

then

p = 1 4p – 3 = 4(1) – 3 = 4 – 3

= – 19 ≠ 13 So, p = – 4 is not a solution of the given equation.

= 1 ≠ 13



So, p = 1 is not a solution of the given equation.

then 4p – 3 = 4(0) – 3 = 0 – 3 = 3 ≠ 13



(e) When,

then

p = – 4

(f) When, p = 0

So, p = 0 is not a solution of the given equation.

4p – 3 = 4(– 4) – 3 = – 16 – 3

qqq

WORKSHEET-31 Solutions 1. The statements for the given equations are : (a) The sum of numbers p and 4 is 15. (b) The difference of m and 7 is 3. (c) Two times m is 7. (d) The numbers m divided by 5 gives 3. (e) Three times of m divided by 5 is equal to 6. (f) Three times p plus 4 gives 25. (g) Four times p minus 2 gives 8. (h) p divided by 2 plus 2 gives 8. 2. (a) Let number of marbles with Parmit be m. Irfan has 7 marbles more than five times the marbles Parmit has, Irfan has 37 marbles. or, 5m + 7 So, we get equation 5m + 7 = 37 (b) Let Laxmi be of y years old His father is 49 years old His father is 4 years older then 3 times of Laxmi’s age or, 3y + 4 So, we get equation 3y + 4 = 49 (c) Let the lowest score be l Highest marks obtained is twice the lowest marks plus 7 or, 2l + 7 Highest marks are = 87 So, equation will be, 2l + 7 = 87 (d) Let the base angle be b. Then, the vertex angle = 2b. Since, sum of the angles of a triangle is 180° \ b + b + 2b = 180° or, 4b = 180° or, b = 45° 3. (a) x – 1 = 0 Steps : * First add 1 to LHS (x – 1) and also to RHS, to get x on LHS

P-36

* Hence, we get x = 1 or, x – 1 + 1 = 0 + 1 (Step 1) or, x = 1 (b) x + 1 = 0 Steps : * First subtract 1 from LHS (x + 1) and also from RHS, to get x on LHS * Hence, we get x = – 1 or, x + 1 = 0 or, x + 1 – 1 = 0 – 1 (Step 1) or, x = – 1 (c) x – 1 = 5 Steps : * First add 1 from LHS and also to RHS, to get x on LHS * Hence, we get x = 6 or, x – 1 = 5 or, x – 1 + 1 = 5 + 1 (Step 1) or, x = 6 (d) We have, x + 6 = 2 In order to solve this equation, we have to get x by itself on the LHS. To get x by itself on the LHS, we need to shift 6. This can be done by subtracting 6 from both sides of the given equation. x + 6 = 2 or, x + 6 – 6 = 2 – 6  [Subtracting 6 from both sides] or, x + 0 = – 4 [Q 6 – 6 = 0 and 2 – 6 = – 4] or, x = – 4 So, x = – 4 is the solution of the given equation. (e) We have, y – 4 = – 7 In order to solve this equation, we have to get x by itself on the LHS. To get x by itself on the LHS, we need to shift – 4. This can be done by adding 4 to both sides of the given equation. y – 4 = – 7 or, y – 4 + 4 = – 7 + 4  [Adding 4 to both sides] or, y + 0 = – 3 [Q – 4 + 4 = 0 and – 7 + 4 = – 3] So, y = – 3 is the solution of the given equation.

M A T H E M A T I C S - VII

(f) We have, y – 4 = 4 In order to solve this equation, we have to get x by itself the LHS. To get x be itself on the LHS, we need to shift – 4. This can be done by adding 4 to both sides of the given equation. y – 4 = 4 or, y – 4 + 4 = 4 + 4  [Adding 4 to both sides] or, y + 0 = 8 [Q – 4 + 4 = 0 and 4 + 4 = 8] or, y = 8 So, y = 8 is the solution of the given equation. (g) We have, y + 4 = 4 In order to solve this equation, we have to get x by itself on the LHS. To get x by itself on the LHS, we need to shift 4. This can be done by subtracting 4 from both sides of the given equation. y + 4 = 4 or, y + 4 – 4 = 4 – 4

 or, or,

[Subtracting 4 from both sides] y + 0 = 0 [Q 4 – 4 = 0] y = 0

So, y = 4 is the solution of the given equation.

(h) We have,

y + 4 = – 4

In order to solve this equation, we have to get x by itself on the LHS. To get x by itself on the LHS, we need to shift 4. This can be done by subtracting 4 from both sides of the given equation. or, 

y + 4 = – 4 y + 4 – 4 = – 4 – 4 [Subtracting 4 from both sides]

or,

y + 0 = – 8

[Q 4 – 4 = 0 and – 4 – 4 = – 8] or,

y = – 8

So, y = – 8 is the solution of the given equation.

qqq

WORKSHEET-32 Solutions



1. (a) We have, 3l = 42 In order to solve this equation, we have to get l by itself on the LHS. For this, 3 has to be removed from the LHS. This can be done by dividing both sides of the equation by 3. Thus, 3l = 42 3l 42 or, = [Dividing both sides by 3] 3 3  or,

l = 14 42 3 × 14  3l  Q 3 = l and 3 = 3 = 14   

 Thus, l = 14 is the solution of the given equation. b (b) We have, =6 2

In order to solve this equation, we have to get b by itself on the LHS. To get by itself on LHS, we have to remove 2 from LHS. This can be done by multiplying both sides of the equation by 2. Thus, we have b = 6 2 or,  or,

b ×2 =6×2 2 [Multiplying both sides by 2] b = 12  b  Q 2 × 2 = b and 6 × 2 = 12   

 So, b = 12 is the solution of the given equation.

S OLUT I ONS

(c) We have,

p = 4 7

In order to solve this equation, we have to get p by itself on the LHS. To get p by itself on LHS, we have to remove 7 from LHS. This can be done by multiplying both sides of the equation by 7. Thus, we have or,  or,

p = 4 7 p ×7 = 4 × 7 7 [Multiplying both sides by 7] p = 28  p  Q 7 × 7 = p and 4 × 7 = 28   

 So, p = 28 is the solution of the given equation. (d) We have, 4x = 25 In order to solve this equation, we have to get x by itself on the LHS. For this, 4 has to be removed from the LHS. This can be dividing both sides of the equation by 4. Thus, 4x = 25 4x 25 or, = 4 4   [Dividing both sides by 4] or,

x =

25 4 

 4x  Q 4 = x   

P-37

So, x =

25 is the solution of the given equation. 4



(g) We have,

a 7 = 5 15

(e) We have, 8y = 36 In order to solve this equation, we have to get y by itself on the LHS. For this, 8 has to be removed from the LHS. This can be done by dividing both sides of the equation by 8. Thus, 8y = 36 8y 36 or, = 8 8

In order to solve this equation, we have to get a by itself on the LHS. This can be done by multiplying both sides of the equation by 5.





[Dividing both sides by 8] 9 y = 2

or,

36 4 × 9 9   8y Q 8 = y and 8 = 4 × 2 = 2   

 So, y =

9 is the solution of the given equation. 2

z 5 = 3 4 In order to solve this equation, we have to get z by itself on the LHS. To get z by itself on LHS, we have to remove 3 from LHS. This can be done by multiplying both sides of the equation by 3. Thus, we have z 5 = 3 4 (f) We have,

or,  or,

 So, z =

z 5 ×3 = ×3 3 4 [Multiplying both sides by 3] 15 z = 4 5 15   z Q 3 × 3 = z and 4 × 3 = 4    15 is the solution of the given equation. 4

a 7 = 5 15



a 7 ×5 = ×5 5 15

or,

[Multiplying both sides by 5]

or,

a =

7 7 × 5 7  a Q 5 × 5 = a and 15 × 5 = 3 × 5 = 3   

 So, z =

7 3

7 is the solution of the given equation. 3

(h) We have,

20t = – 10

In order to solve this equation, we have to get t by itself on the LHS. For this, 20 has to be removed from the LHS. This can be done by dividing both sides of the equation by 20. Thus,

20t = – 10

or,

20t −10 = 20 20



[Dividing both sides by 20]

or,

t =

−1 2

−10 10 × −1 −1   20t Q 20 = t and 20 = 10 × 2 = 2   

 So, t =

−1 is the solution of the given equation. 2

qqq

WORKSHEET-33 Solutions



or,

1. (a) 3n – 2 = 46 Steps : * In order to separate the variable, first transpose (– 2) from LHS to RHS. * On transposing (– 2) becomes (+ 2). * Now, divide both LHS and RHS by 3, so that n can be separated. 3n –2 = 46 or, 3n = 46 + 2



or,

P-38

3n = 48 48 n = 3

∴ n = 16. (b) Steps : * In order to separate the variable, first transpose 7 from LHS to RHS. * On transposing 7 becomes (– 7). * Now, divide both LHS and RHS by 5, so that m can be separated.

M A T H E M A T I C S - VII

or, or,

or,

5m + 7 =17 5m = 17 – 7 5m = 10 10 m = 5

∴ m = 2 (c) Steps : * In order to separate the variable, first multiply both LHS and RHS by 3. * On multiplying by 3, we get 40 × 3 in RHS and 20p in LHS. * Now, divide both LHS and RHS by 20, so that p will be separated. 20 p = 40 3

or,



or,

or, ∴ (d) Steps :

20p = 40 × 3 40 × 3 p = 20 p = 2 × 3 p =6

* In order to separate the variable, first multiply both LHS and RHS by 10.

or, p = 9 So, p = 9 is the solution of the given equation. p (c) We have, = 5 4 or, 

[Multiplying both sides by 4]

or,

(d) We have, or, or,



(e) We have,

or,

* Now, divide both LHS and RHS by 3, so that p will be separated.

or,



or,



or,

or, ∴ 2. (a) We have, or,

3p = 6 × 10 p =

6 × 10 3

S OLUT I ONS

−p × – 3 = 5 × – 3 3 [Multiplying both sides by – 3] p = – 15 3p = 6 4 3p 4 4 × = 6 × 4 3 3 4   Multiplying both sides by 3    p = 2 × 4 = 8

So, p = 8 is the solution of the given equation.

(f) We have,

or,

3s = – 9 3s −9 = 3 3

[Dividing both sides by 3] or,

s = – 3

p = 2 × 10

So, s = – 3 is the solution of the given equation.

p = 20



(g) We have,

10p = 10 10 10 p = 10 10

 [Dividing both sides by 10] or, p = 1 So, p = 1 is the solution of the given equation. (b) We have, 10p + 10 = 100 or, 10p + 10 – 10 = 100 – 10  [Subtracting 10 from both sides] or, 10p = 90 10 p 90 = 10 10 or, 

−p = 5 3

So, p = – 15 is the solution of the given equation.



3p = 6 10

p = 20

So, p = 20 is the solution of the given equation.

* On multiplying by 10, we get 6 × 10 in RHS and 3p in LHS.



p ×4 =5×4 4

[Dividing both sides by 10]

or,

3a + 12 = 0 3a + 12 – 12 = 0 – 12 [Subtracting 12 from both sides]

or,

3a = – 12

or,

3a −12 = 3 3



[Dividing both sides by 3] or,

a = – 4

So, a = – 4 is the solution of the given equation.

(h) We have,

or,

3s = 0 3s 0 = 3 3

[Dividing both sides by 3]

P-39

or,

s = 0

So, s = 0 is the solution of the given equation.

(i) We have,

or,

2q = 6 6 2q = 2 2

(k) We have, or, or,

2a + 6 = 0 2a + 6 – 6 = 0 – 6 [Subtracting 6 from both sides] 2a = – 6 2a −6 = 2 2

[Dividing both sides by 2]

or,

or,

[Dividing both sides by 2] or, a = – 3 So, a = – 3 is the solution of the given equation. (l) We have, 2q + 6 = 12 or, 2q + 6 – 6 = 12 – 6 [Subtraction 6 from both sides] or, 2q = 6 2q 6 = 2 2 or,

q = 3

So, q = 3 is the solution of the given equation.

(j) We have,

or,

2q – 6 = 0 2q – 6 + 6 = 0 + 6

[Adding 6 to both sides] or,

2q = 6

or,

2a 6 = 2 2

[Dividing both sides by 2] or,

q = 3

So, q = 3 is the solution of the given equation.

[Dividing both sides by 2] or, q = 3 So, q = 3 is the solution of the given equation.

qqq

WORKSHEET-34 Solutions 1. For 1 st equation : Starting with x = 5 Multiplying both sides by 5, 5x = 25 Adding 5 to both sides, 5x + 5 = 30 Let us solve it, 5x + 5 – 5 = 30 – 5 [Subtraction 5 from both sides] or, 5x = 25 5x 25 or, = 5 5 or, x = 5, which is its solution. So, x = 5 is the solution of the equation 5x + 5 = 30. For 2nd equation : Starting with x = 5 Multiplying both sides by 3, 3x = 15 Subtracting 3 from both sides, 3x – 3 = 12 Let us solve it, 3x – 3 + 3 = 12 + 3 [Adding 3 to both sides] or, 3x = 15 3x 15 or, = 3 3 [Dividing both sides by 3] or, x = 5, which is its solution. Hence, x = 5 is the solution of equation 3x – 3 = 12.

P-40



2. (a)

2y +

5 37 = 2 2

or,

2y =

37 5 − 2 2

or,

2y =

32 2

or,

y =

16 2

\ (b) or, \

(c)

or, or, \

(d)

or,

y =8 5t + 28 = 10 5t = – 18 t = –

18 5

a + 3 = 2 5 a =–1 5 a = – 1 × 5 a = – 5 q + 7 = 5 4 q = 5 – 7 4

M A T H E M A T I C S - VII

q = – 2 × 4 q = – 8

\

5 2 2 x × = – 10 × 2 5 5 2   Multiplying both sides by 5   

or,

or, (f) We have,

6z + 10 = – 2

or,

6z = – 2 – 10



[Transposing 10 on RHS]

or,

6z = – 12 6z −12 = or, z = – 2 6 6

or, (i) We have,

3l 2 = 2 3 3l 2 2 2 × = × 2 3 3 3

or,

x = – 4 5 25 x = 2 4



5 2 25 2 x× = × 2 5 4 5

or,

2   Multiplying both sides by 5   



5 x = 2

or, (g) We have,



(h) We have,



5 x = – 10 2







q = – 2 4

or, or, (e) We have,

7m +

19 = 13 2

or, 14m + 19 = 26 [Multiplying by 2] or, 14m = 26 – 19 [Transposing 19 on RHS] 14 m 7 or, = 14 14 or,

[Dividing both sides by 14] 1 m= 2

2   Multiplying both sides by 3   

or,

l =

4 9

(j) We have, 2b – 5 = 3 3



2b = 3 + 5 3

or,

[Transposing – 5 on RHS]

or,

2b = 8 3

or,

2 3 3 b × = 8 × 3 2 2 3   Multiplying both sides by 2   

or,

b =4 × 3 = 12

qqq

WORKSHEET-35 Solutions 1. (a) First equation :

Starting with x = 2



Multiply both sides by 2, 2x = 4



Add 3 to both sides , 2x + 3 = 7

Third equation :

Start with x = 2



Dividing both sides by 5,



Subtract 2 from both sides, x 2 −2 = −2 5 5

Second equation :

Start with x = 2





Multiply both sides by – 3, – 3x = – 6

or,



Add 8 to both sides, 8 – 3x = 2

S OLUT I ONS

x 2 = 5 5



x −8 −2 = 5 5

P-41

(b) First equation : Starting with x = – 2 Multiply both sides by 4, 4x = – 8 Subtract 3 from both sides, 4x – 3 = – 11 Second equation : Start with x = – 2 Multiply both sides by – 5, – 5x = 10 Add 10 to both sides, 10 – 5x = 20 Third equation : Start with x = – 2 Divide both sides by 2, Add 3 to both sides,

2. (a)

or, or, or, \ (b) or, or, or, \ (c) or,

x =–1 2

x +3 =2 2

2(x + 4) = 12 x + 4 =

12 2

x + 4 = 6 x = 6 – 4 x = 2 3(n – 5) = 21 21 n – 5 = 3 n – 5 = 7 n = 7 + 5 n = 12 3(n – 5) = – 21 −21 n – 5 = 3

or, n – 5 = – 7 or, n = – 7 + 5 \ n = – 2 (d) 3 – 2(2 – y) = 7 or, 3 – 4 + 2y = 7 or, 2y = 8 y = 4 (e) – 4(2 – x) = 9 or, – 8 + 4x = 9 or, 4x = 17 17 or, x = 4 (f) or, or, or,

(g)

4(2 – x) = 9 8 – 4x = 9 – 1 = 4x -1 x = 4 4 + 5 (p – 1) = 34

or, 4 + 5p – 5 = 34 or, 5p = 35 or, p = 7 (h) 34 – 5 (p – 1) = 4 or, 34 – 5p + 5 = 4 or, 35 = 5p or, p = 7 3. (a) We have, 4 = 5(p – 2) or, 4 = 5p – 10  [Expanding the bracket on RHS] or, 

– 5p = – 10 – 4 [Transposing 4 on RHS and 5p on LHS]

or,

– 5p = – 14

or,

−5 p −14 = −5 −5



[Dividing both sides by – 5]

or,

(b) We have,

p =

14 5

– 4 = 5(p – 2)

or,

– 4 = 5p – 10



[Expanding the bracket on RHS]

or,  or, or,  or, (c) or, (d) or, or, (e) or, (f) or, or,

– 5p = – 10 + 4 [Transposing – 4 on RHS and 5p on LHS] – 5p = – 6 −5 p −6 = −5 −5 [Dividing both sides by – 5] 6 p = 5 – 16 = – 5 (2 – p) – 16 = – 10 + 5p – 6 = 5p -6 p = 5 10 = 4 + 3(t + 2) 10 = 4 + 3t + 6 0 = 3t t = 0 28 = 4 + 3t + 15 9 = 3t t = 3 0 = 16 + 4m – 24 8 = 4m m = 2

qqq

P-42

M A T H E M A T I C S - VII

WORKSHEET-36 Solutions 1. Let the number of mangoes in the each smaller box be x. Then, number of mangoes in 8 such boxes = 8x Therefore, number of mangoes in each larger box = 8x + 4 According to question, 8x + 4 = 100 or, 8x + 4 – 4 = 100 – 4 or, 8x = 96 8x 96 or, = 8 8 or, x = 12 Thus, the number of mangoes in the smaller box is 12. 2. (a) Let the number be x Then, according to the question, 6x – 5 = 7 or 6x = 7 + 5 = 12 6x 12 or = or x = 2 6 6 \ The number is 2. (b) Let the number be y Then, according to the question, y +5 =8 3 y or =8–5 3 y =3 3 or y = 3 × 3 = 9 \ The number is 9. 3. (a) Let x be the required number. Then, the required equation is 8x + 4 = 60 or, 8x + 4 – 4 = 60 – 4 or, 8x = 56 8x 56 or, = or, x = 7 8 8 (b) Let x be the required number, Then, the required x equation is − 4 = 3 5 x or, × 5 – 4 × 5 = 3 × 5 5

or, x – 20 = 15 or, x – 20 + 20 = 15 + 20 or, x = 35 (c) Let x be the required number. Then, the required 3x equation be + 3 = 21. 4 or, or, or, or,

4×

3x + 4 × 3 = 4 × 21 4

3x + 12 = 84 3x + 12 – 12 = 84 – 12 3x = 72 3x 72 = or, x = 24 3 3

or, (d) Let the required number be m. Then, the required equation is 2m – 11 = 15 or, 2m – 11 + 11 = 15 + 11 or, 2m = 26 or, m = 13 (e) Let Munna have x notebooks. Then, the required equation is 50 – 3x = 8 or, 50 – 3x – 50 = 8 – 50 or, – 3x = – 42 - 42 −3x = or, x = 14 -3 −3 or, (f) Let the number be x. Then, the required equation is x + 19 =8 5 x + 19 = 5 × 8 5× 5 or, or, x + 19 – 19 = 40 – 19 or, x = 21 (g) Let the number be x. Then, the required equation is 5n 11 −7 = 2 2 Multiplying both sides by 2, we get 5n – 14 = 11 or, 5n – 14 + 14 = 11 + 14 or,

5n = 25

or,

n = 5

or,

25 5n = 5 5

qqq

WORKSHEET-37 Solutions



1. (a) Let the lowest score be n.





2n = 80 n =

2n + 7 = 87



S OLUT I ONS

2n = 87 – 7



80 2

= 40

P-43

(b) ∆ ABC is an isosceles ∠ A = 40° (Given) AB = AC (Given) ∠ B = ∠ C (Given) ∠ A + ∠ B + ∠ C = 180° (Sum of three angles of a triangle is 180°) A

n



5x 30 = or x = 6 5 5

Hence, Parmit has 6 marbles.

40°

B

To solve this equation, transposing (+ 7) from LHS to RHS, we get 5x = 37 – 7 or 5x = 30 On dividing both sides by 5, we get



(b) Let Laxmi’s age be x years, then 3 times of Laxmi’s age be 3x.

∴ Laxmi’s father age = 3x + 4, but Laxmi’s father is a 49 years.

n C

40° + n + n = 180° 2n = 180° – 40° 2n = 140° or, n = 70° (c) Let runs scored by Rahul be x. Then, runs scored by Sachin = Twice of runs scored Runs scored by Rahul = 2x ∴ Sum of their runs = x + 2x = 3x Since, the sum of their runs be two short of a double century. Therefore, we get the equation 3x + 2 = 200 To solve this equation, transposing (+2) from LHS to RHS, we get 3x = 200 – 2 or, 3x = 198 3x 198 = [dividing both sides by 3] 3 3 or, or, x = 66 Hence, the runs scored by Rahul is 66 and by Sachin = 2 × 66 = 132 2. (a) Let Parmit has x marbles, then the five times of x be 5x. ∴ Number of marbles Irfan has = 5x + 7 But Irfan has 37 marbles, then we get the equation 5x + 7 = 37

Therefore, we get the equation

3x + 4 = 49

To solve this equation, transposing 4 from LHS to RHS, we get

3x = 49 – 4 [on transposing 4 becomes – 4]

or,

3x = 45

On dividing both sides by 3, we get = 15 Hence, Laxmi’s age is 15 years.

(c) Let the number of fruit trees planted be x, then three times of fruits trees be 3x.

∴ Number of non-fruit trees = 2 + Three times the number of fruit trees = 2 + 3x But number of non-fruit trees was 77. Therefore, we get the equation 2 + 3x = 77 To solve this equation, transposing 2 from LHS to RHS, we get

3x = 77 – 2 [on transposing + 2 becomes – 2]

or,

3x = 75

On dividing both sides by 3, we get

3x 75 or, x = 25 = 3 3

Hence, the number of fruit trees planted was 25.

qqq

WORKSHEET-38 Solutions



7. 5x + 3x = 2 −

1. (d) 2. (d) 3. (b) 4. (a) 5. (c) 6. (a)

P-44

1 3



8x =

6 −1 3



8x =

5 3



x =

5 5 = 24 3×8

M A T H E M A T I C S - VII



7. Since, (a)

x + 5 = 10 2



x = 10 – 5 2

or,

[On transposing 5 to RHS] x = 5 2

or,

x × 2 = 5 × 2 2 Multiplying both sides by 2 or, x = 10 (b) 3x – 5 = 7 or, 3x = 7 + 5 [On transposing 5 to RHS] or, 3x = 12 3x 12 = [Dividing by 3] 3 3 or, x = 4 8. Let the number be x. According to question, or, 6x + 10 = 40 or, 6x = 40 – 10 [On transposing 10 to RHS] or, 6x = 30 6x 30 = 6 6 Thus,

x = 5 2x − 1 x−2 = +1 3 3

9.

[On transposing or,

2x − 1 x − 2 = 1 − 3 3

x−2 to LHS] 3

or,

( 2 x − 1) − ( x − 2 ) = 1 3

or,

2x − 1 − x + 2 = 1 3

or,

x +1 = 1 3

or, x + 1 = 3 × 1 or, x + 1 = 3 or, x = 3 – 1 Thus, x = 2 10. Quantity of pure alcohol in 400 ml. of 15% 15 = 400 × 100 = 60 ml.

Now, we add x ml of pure alcohol to the sample.



So, total pure alcohol = (60 + x) ml.



But volume of new sample = (400 + x) ml.



∴ Percentage of pure alcohol in new sample

=

( 60 + x ) × 100 ( 400 + x )

which is equal to 32% or, or,

60 + x × 100 = 32 400 + x 60 + x 32 = 400 + x 100

or,

100(60 + x) = 32(400 + x)

or,

100x + 6000 = 32x + 12800

or,

100x – 32x = 12800 – 6000

or,

68x = 6800

or,

x =

6800 = 100 ml. 68

qqq

S OLUT I ONS

P-45

CHAPTER SECTION

B 5

LINES AND ANGLES

WORKSHEET-39 Solutions

1. We know that, the sum of two complementary angles is 90°.

(a) 90°– 45° = 45°

(b) 90°– 65° = 25°



(d) 90°– 50° = 30°

(c) 90°– 41° = 49°

2. We know that, the sum of two supplement any angels is 180°. ∴ (a) (180° –100°) = 80° (b) (180° – 90°) = 90° (c) (180° – 55°) = 125° (d) (180° – 125°) = 55°. 3. Let angles be x and (90° – x)



x – (90° – x) = 12°

or,

x – 90° + x = 12°

or,

2x – 90° = 12°

or,

2x = 90° + 12°

or,

2x = 102°

or,

x = 51°

∴ Angles are 51° and (90° – 51°) = 39°. 4. Let the smaller angle be x. Therefore, its supplementary angles = (x + 44)° Since the sum of two supplement angles = 180° ∴ x + (x + 44)° = 180° or, x+x+44°= 180° or, 2x+ 44° = 180° On transposing 44° from LHS to RHS, we get 2x = 180° – 44° or, 2x = 136° 2 x 136 On dividing both sides by 2, we get = = 68° 2 2 or, x = 68° Hence, the smaller angle is 68°. and its supplementary larger angle = 68° + 44° = 112°. 5. (a) Since is this pair, sum of two angles = 70° + 20° = 90°

So, this pair of angles is complementary. (b) Since in this pair, sum of two angles = 75° + 25° = 100° ≠ 90°. So, this pair of angles is not complementary. (c) Since in this pair, sum of two angles = 48° + 52° = 100° ≠ 90°. So, this pair of angles is not complementary. (d) Since in this pair, sum of two angles = 35° + 55° = 90°. So, this pair of angles is complementary. 6. (a) In this pair, measures of the given angles are 110° and 50°. ∴ Sum of the given angles = 110° + 50° = 160°, which is less than 180°. So, this pair of angles is not supplementary. (b) In this pair, measures of the given angles are 105° and 65°. ∴ Sum of the given angles = 105° + 65° = 170°, which is less than 180°. So, this pair of angles is not supplementary. (c) In this pair, measures of the given angles are 50° and 130°. ∴ Sum of the given pair = 50° + 130° = 180°. So, this pair of angles is supplementary. (d) In this pair, measures of the given angles are 45° and 45°. ∴ Sum of the given pair = 45° + 45° = 90°, which is less than 180°. So, this pair of angles is not supplementary. 7. (a) Yes (b) Yes (c) No, they have no common vertex. (d) Yes (e) Yes

qqq

WORKSHEET-40 Solutions 1. (a) Yes, they have common vertex. (b) No, they have no common vertex. 2. Since, the lines intersect at a point, therefore, ∠1 = ∠3  [Vertically opposite angles] or, ∠3 = 30° [Q ∠1 = 30°, given]

P-46

Clearly, 

∠1 + ∠2 = 180° [Q ∠1 and ∠2 are angle of a linear]

or,

30° + ∠2 = 180° or, ∠2 = 180° – 30°

or,

∠2 = 150°

Hence

∠2 = 150° and ∠3 = 30°.



3. (a) The complementary angle of 20° will be (90° – 20°) = 70°.

M A T H E M A T I C S - VII

70º

20º



In the (c) pair, sum of the angles = 90° + 80° = 170° ≠ 180°. So, this pair cannot form as linear pair. In the (d) pair, sum of the angles = 65° + 115° = 180°. So, this pair can form a linear pair.

(b) The complementary angle of 63° will be (90° – 63°) = 27°. 27º

63º

(c) The complementary angle of 57° will be (90° – 57°) = 33°.

57º

33º

4. Since, the sum of the measures of an angle and its supplement is 180°, therefore, (a) The supplement of an angle of measure 105° is the angle of (180° – 105°), i.e., 75° (b) The supplement of an angle of measure 87° is the angle of (180° – 87°), i.e., 93°. (c) The supplement of an angle of measure 154° is the angle of (180°– 154°), i.e., 26°. 5. In the (a) pair, sum of the angles = 140° + 40° = 180°. So, this pair can form a linear pair. In this (b) pair, sum of the angles = 60 ° + 60° = 120° ≠ 180°. So, this pair cannot form a linear pair.



6. (a) 65°, 115° On adding 65° and 115°, we get 65° + 115° = 180° Since sum is 180°, hence this pair of angles is supplementary. (b) 63°, 27° On adding 63° and 27°, we get 63° + 27° = 90° Since, sum of them is 90° hence this pair of angles is complementary. (c) 112°, 68° On adding 112° and 68°, we get 112° + 68° = 180° Since, sum of the above two angles is 180°, hence this pair of angles is supplementary. (d) 130°, 50° On adding 130° and 50°, we get 130° + 50° = 180° Since, sum of the above two angles is 180°, hence this pair of angles is supplementary. (e) 45°, 45° On adding 45° and 45°, we get 45° + 45° = 90° Since, sum of the two angles is 90° hence this pair of angles is complementary. (f) 80° + 10° = 90° Since sum of angles is 90° Hence angles are complementary.

qqq

WORKSHEET-41 Solutions 1. We know that, the sum of two complementary angles is 90°. So, if an angle is greater than 45° then its complementary angle would be less than 45°. 2. Let the measure of the angle be x°. Then, the measure of its complement is given to 90°– x°. Since, the angle is equal of its complement. x° = 90° – x° On transposing x° from RHS to LHS, we get x° + x° = 90° or, 2x° = 90° or, x° = 45° 3. Let the angles be x°. Therefore, its supplement be 180° – x°.

S OLUT I ONS

Since, the angle is equal to it supplement. ∴ x° = 180° – x° On transposing x° from RHS to LHS, we get x° + x° = 180° or, 2x° = 180° On dividing both by 2, we get 180 ∴ x = = 90° 2 4. We know that, the two angles are supplementary, if their sum is 180°.

∴

∠1 +∠2 = 180°

[given]

P-47

Now, if ∠1 is decreased, then ∠2 should be increased, so that both the angles still remain supplementary. i.e. ∠ 1 + ∠ 2 = 180° [Until ∠ 1 = ∠ 2 = 90°] 5. (a) Pair of vertical opposite angles are ∠1, ∠4; ∠5, ∠2 + ∠3. (b) Pair of linear angles are ∠1, ∠5; ∠4, ∠5. 6. (a) No, because sum of two acute angles is always less than 180°. Some two acute angles cannot be supplementary. (b) No, because sum of two obtuse angles is always greater than 180°. (c) Yes, because sum of two right angles is always equal to 180°. So, two right angles are supplementary.







7. (a) Yes, ∠ 1 and ∠ 2 are adjacent to each other. (b) No, ∠ AOC and ∠AOE are not adjacent to each other as their is no common arm between them. (c) No, ∠ COE and ∠ COD don’t form linear pairs as their sum is not equal to 180° and their non-common sides are not opposite rays. (d) Yes, ∠ BOD and ∠ DOA are supplementary as their sum is equal to 180°. (e) Yes, ∠ 1 and ∠ 4 are vertically opposite to each other as they are formed by the intersection of lines AB and CD. (f) ∠ 2 and ∠ 3 or ∠ 2 + ∠ 3 is the vertically opposite angle to angle ∠ 5. As they are formed by the intersection of lines AB and CD.

qqq

WORKSHEET-42 Solutions 1. No, they do not always intersect at a right angles. 2. (a) ∠x =55° [vertically opposite angles] ∠y + 55° = 180° [linear pair] ∴ ∠y = 180° – 55° = 125° ∠z = ∠y = 125° [vertically opposite angles] Hence, ∠x = 55° , ∠y = 125° and ∠z = 125° (b) ∠y + 40° = 180 [by linear pair] or, ∠y + 40° = 180°– 40° or, ∠y = 140° ∠z = 40° [vertically opposite angles]  ∠x + 25° + ∠z = 180° [by linear pair] ∴ ∠x +25° + 40° = 180° or, x = 180° –65° = 115° Hence, ∠x = 115°, ∠y = 140° and ∠z = 40° 3. Let ABC be an equilateral triangle. Since, all the angles of an equilateral triangle are equal. A

B

∴

C

∠A = ∠B = ∠C = x°

(say)

We know that, sum of all the angles of a triangle is 180°. ∴ ∠A + ∠B + ∠C = 180° or,

x° + x° + x° = 180°

or,

3x° = 180°

P-48

or, x° = 60° Hence, ∠A = ∠B = ∠C = 60° 4. Let ABCD be the rectangle. ABCD is a parallelogram with ∠A = 90° We have, ∠C = ∠A = 90° (opposite angles of a parallelogram are equal) Again, ∠A = ∠B = 180° ( ∠A and ∠B are adjacent angles of a parallelogram) A

B

D

C

or, 90° + ∠B = 180° or, ∠B = 180° – 90° = 90° ∴ ∠D = ∠B = 90° ( opposite angles of a parallelogram are equal) Hence, ∠A = ∠B = ∠C = ∠D = 90°. 5. (a) Obtuse vertically opposite angles are ∠AOD and ∠BOC. (b) Adjacent complementary angles are ∠BOA and ∠AOE. (c) Equal supplementary angles are ∠BOE and ∠EOD. (d) Unequal supplementary angles are ∠BOA and ∠AOD, ∠BOC and ∠COD, ∠EOA and ∠EOC. (e) Adjacent angles that do not form a linear pair are ∠AOB and ∠AOE, ∠AOE and ∠EOD; ∠EOD and ∠COD. 6. (a) If two angles are complementary, then the sum of their measures is 90°. (b) If two angles are supplementary, then the sum of their measures is 180°. (c) Two angles forming a linear pair are supplementary or adjacent angles.

M A T H E M A T I C S - VII



(d) If two angles are supplementary, they form a linear pair. (e) If two lines intersect at a point, then the vertically opposite angles are always equal to each other.

(f) If two lines intersect at a point, and if one pair of vertically opposite angles are acute angles, then the other pair of vertically opposite angles are obtuse angles.



qqq

WORKSHEET-43 ∴

Solutions 1. An infinite number of transversals can be drawn for two given lines.

y = 55° [alternate interior angles]



(c) Given, l1 and l2 any two non-parallel lines; t is a transversal.

∴

2. Three intersections are there.

∠1 ≠ ∠2

(d) Given, l || m and t is a transversal.



We know that, the sum of pair of interior angles on the same side of the transversal is 180°.



∴

∠z + 60° =180° ∠z =180°– 60° =120°

or,

(e) Given, l || m and t is a transversal. 3. (a) Since, alternate







Since, l || m and t angles are equal.

So, l || m (b) Since, ∠1 \ We see that here corresponding angles are equal. \ l || m

(c) = 180° – 130°

= 50° is a transversal

∴

p 60° a

x = 180° – 70° = 110°

(b) Alternate interior angles.



(c) Pair of interior angles on the same side of the transversal.



(d) Corresponding angles.



(e) Alternate interior angles.



(f) Linear pair

e

and

d

∠1 = 120° ∠d = ∠1 = 120° ∠1 + ∠c = 180°

( sum of the angles on same side of transversal is 180°) or,

x = 60°



b

(vertically opposite angles) Now,

5. (a) Given, l || m and t is a transversal. ∴

1

q

4. (a) Corresponding angles.



m

l

\ x + 70° = 180°

or,



[pair of corresponding angles]

(f) Given, p || q and t is a transversal. We know that, the sum of pair of interior angles on the same side of a transversal is 180°. ∴ ∠a + 60° = 180° or, ∠a = 180° – 60° = 120° and l || m and q is a transversal. ∴ ∠a = ∠1 (pair of corresponding angles)

(Sum of the interior angles on the same side of a transversal is 180° )

x = 120°

[alternate interior angles]

(b) Given, a || b and c is a transversal.

120° + ∠c = 180°

or,

∠c = 180° – 120° = 60°

or,

∠b = ∠c = 60°

(vertically opposite angles) Hence, ∠a = 120°, ∠b = 60°, ∠c = 60° and ∠d = 120°.

qqq S OLUT I ONS

P-49

WORKSHEET-44 angles. (c) ∠2, ∠5 and ∠3, ∠8 are two pairs of interior angles on the same side of the transversal.

Solutions

1. (a)



2. (a) DGC = 70° (AB || DC and BC is transversal

∴ DGC = ∠ ABC = 70°;

(corresponding angles)

(d) ∠1, ∠3; ∠2, ∠4; ∠5, ∠7 and ∠6, ∠8 are four pairs of vertically opposite angles. 6. Given, p || q

(b) DEF = 70°

Here,

∴ BC || EF and DE is transversal ∴ ∠ DEF = ∠ DGC = 70°

(corresponding angles)

3. (a) Pair of corresponding angles.

[Linear pair]

or,

∠e = 180° – 125° = 55°



∠f = ∠e = 55°

[Vertically opposite angles]

Since, p||q and t is a transversal. \ ∠a = ∠f [Alternate angles] = 55° [∠f = 55°] or, ∠d = 125°  [Corresponding angles] ∠c = ∠a = 55°  [Vertically opposite angles] and ∠b = ∠d = 125° [Vertically opposite angles]

(b) If alternate interior angles are equal, lines are parallel. (c) If co-interior angles are supplementary, lines are parallel. 4. (a) Since, l || m and t is a transversal. \ ∠x = (180° – 110°) [Corresponding angles, Linear pair] = 70° (b) x + 2x = 180° (Co-interior angle) 3x = 180° x = 60° (c) If l || m and a is a transversal. Then, ∠x = 100° [Corresponding angles] 5. (a) ∠1, ∠5; ∠2, ∠6; ∠3, ∠7 and ∠4, ∠8 are four pairs of corresponding angles. (b) ∠2, ∠8 and ∠3, ∠5 are two pairs of alternate interior

∠e + 125° = 180°

Hence, ∠a = 55°, ∠b = 125°, ∠c = 55°, ∠d = 125°, ∠e = 55° and ∠f = 55°. 7. (a) l is not parallel to m  44° + 126° = 170° ≠ 180°.

(b) l is not parallel to m. (c) l is parallel to m. (d) l is not parallel to m.

qqq

WORKSHEET-45 of linear pair.

Solutions 1. (c)

2. (d)



3. (b)



4. (b)



5. (c)



6. (b)

7. (b)

8. (b)



9. (a)



10. (b)



11. (d)



12. In the given figure, ∠AOB and ∠COB are the angles



So,



∠AOC + ∠BOC = 180°

(3x + 10°) + (2x – 30°) = 180°

or,

3x + 10° + 2x – 30° = 180°

or,

5x – 20° = 180°

or,

5x = 180° + 20°

or,

5x = 200° 200o 5

or,

x =

Thus,

x = 40°

Now,

∠AOB = 2x + 8°

= 2 (40°) + 8° = 80° + 8° = 88°.

qqq

P-50

M A T H E M A T I C S - VII

WORKSHEET-46 Solutions 1. (c) 2. (d) 3. (a) 4. (a) 5. (b) 6. (c) 7. Let ∠EFG = x Then ∠AFD = x It is given that CD intersects line AB at F. Therefore, ∠CFB = ∠AFD (vertically opposite angles) So, x = 50° But ∠EFA = ∠AFD, which gives ∠EFA = 50° Now ∠CFB + ∠EFA + ∠EFC = 180° [as AB is a straight line] or, 50° + 50° + ∠EFC = 180° or, ∠EFC = 180° – 100° ∠EFC = 80°

Thus,

8. Given, ∠1 = (2a + b)° and ∠6 = (3a – b)° ∠1 = ∠7

so,

so,

[alternate exterior angles]

we have ∠7 = (2a +b)°



∴



or, (3a – b)° + (2a + b)° = 180°

or,

∠6 + ∠7 = 180° 3a – b + 2a + b = 180°

or, 5a = 180°

on dividing both sides by 5, we get

[linear pair]



or,







or,

a =

180 = 36° 5

∠1 + ∠2 = 180°

[linear pair]

2a + b + ∠2 = 180°

or, 2 × 36° + b + ∠2 = 180°

or,

or,

∴

b + ∠2 = 180° – 72° b + ∠2 = 108° ∠2 = (108 – b)°

9. Given, PQ||RS||UT (a) Given, c = 57° and a = c/3  PQ||UT ∴ ∠UTP = ∠QPT [alternate interior angles] or, ∠c = ∠a + ∠b [ QPT = a + b] 57 or, 57° = + ∠b 3 57° – 19° = ∠b ∠b = 38° ∠b + ∠d = 180° ∠d = 180° – 38° = 142° 2 (b) Given, c = 75° and a = c, 5

or, or, ∴ or,

or,

c = 75° and a =

2 × 75° = 30° 5

∴ ∠c = ∠a + ∠b [alternate interior angles] or, 75° = 30° + ∠b or, 75° – 30° = ∠b or, ∠b = 45°

qqq

S OLUT I ONS

P-51

CHAPTER SECTION

B 6

THE TRIANGLE AND ITS PROPERTIES

WORKSHEET-47 1. The three sides AB, BC, CA and three angles ∠A, ∠B, ∠C of D ABC are together the six elements of the D ABC. 2. Let the other angle be x ∴ x + 25° = 70° or, x = 70° – 25° = 45° 3. Exterior angle = 60° + 80° = 140° 4. The diagram is incorrect as “exterior angle is equal to sum of interior opposite angles.” Hence exterior angle should be 50° + 50° = 100° and not 50° as shown.



(b) Classification of triangles by considering the measures of their angles is as under : Acute triangles : (i) and (iv) Right triangles : (ii) and (vi) Obtuse triangles : (iii) and (v) 8. (a) In ∆ ABC, BE is a median. The figure is as following :

Solutions

5. PM is the altitude from the vertex P to the opposite side QP of ∆ PQR.

B

A

C

E

(b) In ∆ PQR, PQ, PR are altitudes of the triangle in the mentioned figure on PQ and PR. P

PD is the median from the vertex P to the opposite side QR of ∆ PQR. QM ≠ MR as M is not the mid-point of QR .

6. (a) Side PR opposite to the vertex Q P

Q

R

(c) In ∆ XYZ, YL is an altitude in the exterior of the triangle in the mentioned figure. X

Q



R

(b) Angle ∠LNM opposite to the side LM L

Y

Z

L

M

N

(c) Vertex S opposite to side RT. R

9. Draw a line segment BC. By paper folding, locate the perpendicular bisector of BC . The folded crease meets BC at D, its mid-point. A

S

T

7. (a) Classification of triangles by considering the lengths of their sides is as under : Scalene triangles : (ii) Isosceles triangles : (i), (iii), (v) and (vi) Equilateral triangles : (iv)

P-52

B

D

C

Take any point A on this perpendicular bisector. Join AB and AC. The triangles thus obtained is an isosceles ∆ ABC in which AB = AC.

M A T H E M A T I C S - VII

Since, D is the mid-point of BC, so AD is its median. Also AD is the perpendicular bisector of BC. So, AD is the altitude of ∆ ABC.

Thus, it is verified that the median and altitude of an isosceles triangles are same.

qqq

WORKSHEET-48 Solutions



x° = 50° + 70°



x° = 120°

70º

xº

(b) Exterior angle = S um of opposite interior angles



30º 60º

50º



xº

1. (a)Exterior angle = Sum of opposite interior angles

x° = 65° + 45° = 110°

Therefore,

(a)

x + 50° = 115°

or,

65º

2. We know that in a triangle, an exterior angle is equal to the sum of the two interior opposite angles.

(b)

xº

x° = 30° + 40° = 70°



(d) (e)

40º

30º

(d) Exterior angle = Sum of opposite interior angles x° = 60° + 60°

= 120° 60º xº

60º

or,

x = 125° – 90°

or,

x = 120° – 60°

or,

x = 80° – 30°

or,

x = 75° – 35°

x = 60° x + 30° = 80°

(f)

x = 100° – 70°

x = 35° x + 60° = 120°

or,

or,

x = 30° x + 90° = 125°

or,

xº



(c)

or,

(c) Exterior angle = Sum of opposite interior angles



x = 115° – 50°

x = 65° x + 70° = 100°

or, 45º

or,

x = 50° x + 35° = 75°

or,

x = 40°



3. (a) The sum of interior angles of a triangle is 180°.



∴

∠A + ∠B + ∠C = 180°



or,

x + 50° + 60° = 180°



or,

x + 110° = 180°



or,

x = 180° – 110°



or,

x = 70° A

(e) Exterior angle = Sum of opposite interior angles x° = 50° + 50° = 100°

xº

50º

50º

60º

B

(b) The sum of interior angles of a triangle is 180°.

50º xº

(f) Exterior angle = Sum of opposite interior angles x° = 30° + 60° = 90°

S OLUT I ONS

C



∴

∠P + ∠Q + ∠R = 180°



or,

90° + 30° + x° = 180°



or,

120° + x° = 180°



or,

x = 180° – 120°

P-53



or,

x = 60°



P

or,

x = 65°

Q 30º

xº 50º

xº

xº

R (c) The sum of interior angles of a triangle is 180°.

∴

∠X + ∠Y + ∠Z = 180°



or,

30° + 110° + x = 180°



or,

140° + x = 180°

or,

x = 180° – 140°

or,

(e) The sum of all 3 interior angles of a triangle is 180°. ∴ x + x + x = 180° or, 3x = 180° 180° or, x = 3 or, x = 60°

x = 40°

xº

X 30º

Y 110º xº Z

(d) Since, the sum of all 3 interior angles of a triangle is 180°.

∴

50° + x + x = 180°



or,

50° + 2x = 180°



or,

2x = 180° – 50°



or,

2x = 130°



or,



x =

xº

xº

(f) The sum of all 3 interior angles of a triangle is 180°. ∴ 2x + x + 90° = 180° or, 3x + 90° = 180° or, 3x = 180° – 90° or, x = 90° 90° or, x = 3 or, x = 30° 2xº

130° 2

90º

xº

qqq

WORKSHEET-49 A

Solutions

1. Let ABC be a triangle such that ∠B = 30° and ∠C = 80°.

n

Then, we have to find the measure of the third angle A. Now,

∠B + ∠C = 30° + 80° = 110°

By the angle sum property of a triangle, we have or,

∠A +∠B + ∠C = 180° ∠A + 110° = 180°

[ ∠B + ∠C = 110°] or,

∠A = 180° – 110°

or,

∠A = 70°

2. The sum of all the angles of a triangle is 180° i.e., ∠ABC + ∠BCA + ∠CAB = 180° According to question,

P-54

80º B

n C

One angle is 80° and other two angles are equal (let both n), then 80° + n + n = 180° or, 80° + 2n = 180° or, 2n = 180° – 80° or, 2n = 100° 100° or, n = = 50° 2 or, n = 50° ∴ ∠BCA = ∠CAB = 50°

M A T H E M A T I C S - VII

3. ∠A + ∠B + ∠C = 180° 1 : 2 : 1 = 180° Therefore, 1 + 2 + 1 = 4 B 45º

45º

90º

A



C

∠ABC =



=

∠BCA =

1 × 180° 4 180° = 45° 4 2 × 180° 4

180° × 2 = 4 = 45° × 2 = 90° 1 180° ∴ ∠CAB = = 45° × 180° = 4 4 Classification : (i) Isosceles triangle (ii) Right Angle Triangle 4. (a) In a triangle, exterior angle and adjacent interior angle form a linear pair, therefore, y + 120° = 180° or, y = 180° – 120° = 60° Also, the sum of the angles in a triangle is 180. ∴ x + 50° + y = 180° or, x = 180° – 50° – y or, x = 130° – 60° = 70° [ y = 60°] Hence, x = 70° and y = 60° (b) Here, y = 80° [Vertically opposite angles] Also, the sum of the angles in a triangle is 180°.

∴ x + y + 50° = 180° or, x + 80° + 50° = 180° [ y = 80°] or, x + 130° = 180° or, x = 180° – 130° or, x = 50° Hence, x = 50° and y = 80°. (c) In a triangle, the sum of the angles in a triangle is 180°. ∴ y + 60° + 50° = 180° or, y + 110° = 180° or, y = 180° – 110° or, y = 70° Also, exterior angle = sum of the interior opposite angles ∴ x = 50° + 60° = 110° Hence, x = 110° and y = 70°. (d) Here, x = 60° [Vertically opposite angles] Also, the sum of the angles in a triangle is 180°. ∴ x + y + 30° = 180° or, 60° + y + 30° = 180° [* x = 60°] or, 90° + y = 180° or, y = 180° – 90° or, y = 90° Hence, x = 60° and y = 90°. (e) Here, y = 90° Vertically opposite angles] Also, the sum of the angles in a triangle is 180° ∴ x + x + y = 180° or, 2x + 90° = 180° or, 2x = 180° – 90° or, 2x = 90° or, x = 45° Hence, x = 45° and y = 90°. (f) Clearly y = x [Vertically opposite angles] Also, the sum of the angles in a triangles is 180° ∴ x + x + y = 180° or, x + x + x = 180° [ y = x] or, 3x = 180° or, x = 60° Hence, x = 60° and y = 60°

qqq

WORKSHEET-50 Solutions 1. (a) Since, 2 + 3 = 5 Since, the sum of lengths of two sides is equal to the length of third side so a triangle cannot be form with these sides. (b) We have, 3 + 6 > 7, 3 + 7 > 6 and 6 + 7 > 3 i.e., the sum of any two sides is greater than the third side. So, these side form a triangle. (c) We have, 3 + 2  < 6 So, the given side cannot form a triangle.

S OLUT I ONS



2. Using triangle inequality property in Ds OPQ, OQR and OPR, we have R

O P



(a) Yes, OP + OQ > PQ



(b) Yes, OQ + OR > QR



(c) Yes, OR + OP > RP

Q

P-55



(∆OPQ, ∆OQR and ∆OPR, the sum of the lengths of any two sides is always greater than the third side.)

3. Yes, given AM is a median of ∆ABC. So, M is the mid-point of BC and AM divides ∆ABC into two triangles, ∆ABC and ∆AMC. We know that, the sum of the lengths of any two sides of a triangle is greater than the length of third side. In ∆ABM,

AB + BM > AM

...(i)

In ∆ABC,

CA + MC > AM

...(ii)

On a adding Eqs. (i) and (ii), we get

AB + BM + CA + MC > 2AM

or,

AB + (BM + MC) + CA > 2AM

or,

AB + BC + CA > 2AM



[from the figure, BM + MC = BC]

4. Yes, given ABCD is a quadrilateral. Here, diagonal AC divides the quadrilateral ABCD into two triangles ABC and ∆ACD. We know that, the sum of lengths of any two sides of a triangle is greater than the length of third side. In ∆ABC, AB + BC > AC

...(i)

In ∆ACD, CD + DA > AC

...(ii)

On adding Eqs. (i) and (ii), we get

AB + BC + CD + DA > 2AC ...(iii)

Similarly, the diagonal BD divides the quadrilateral ABCD into two triangles ∆BCD and ∆ABD. In ∆BCD, BC + CD > BD

(iv)

In ∆ABD, AB + DA > BD

(v)

On adding eqs. (iv) and (v), we get

(BC + CD + AB + DA) > 2BD ...(vi)

Again, adding Eqs. (iii) and (vi), we get AB + BC + CD + DA + BC + CD + AB + DA > 2AC + 2BD or, 2(AB + BC + CD + DA) > 2(AC + BD) or, AB + BC + CD + DA > AC + BD 5. Yes, given ABCD be a quadrilateral and its diagonal AC and BD intersect each other at O. D

C O

A

or,

B

Also,

We know that, the sum of lengths of any two sides of a triangle is greater than the length of third side. In ∆AOB,

OA + OB > AB

Similarly, in ∆BOC, OB + OC > BC ...(ii) In ∆COD, OC + OD > CD ...(iii) and in ∆DOA, OD + OA > DA ...(iv) On adding Eqs. (i), (ii), (iii) and (iv), we get (OA + OB) + (OB + OC) + (OC + OD) + (OD + OA) > AB + BC + CD + DA or, 2(OA + OB + OC + OD) > AB + BC + CD + DA or, AB + BC + CD + DA < 2(OA + OB + OC + OD) or, AB + BC + CD + DA < 2[(OA + OC) + (OB + OD)] or, AB + BC + CD + DA < 2 (AC + BD) [AC = OA + OC and BD = OB + OD] 6. (a) Since, in a triangle an exterior angle and the interior adjacent angle form a linear pair, therefore, ∠C + 120° = 180° or, ∠C = 180° – 120° = 60° Since, ∆ ABC is isosceles with AB = AC ∴ ∠B = ∠C or, ∠B = 60° i.e., y = 60° By the angle sum property, we have ∠A + ∠B + ∠C = 180° or, x + y + 60° = 180° or, x + 60° + 60° = 180° or, x + 120° = 180° or, x = 180° – 120° = 60° Hence, x = 60° and y = 60° (b) Since, ∆ ABC is isosceles with AB = AC ∴ ∠B = ∠C or, ∠B = x Also, ∠B + ∠C = 90° or, x + x = 90° [∠B = x] or, 2x = 90° or, x = 45° But, ∠B + y = 180° or, x + y = 180° [ ∠B = x] or, y = 180° – 45° [ x + 45°] or, y = 135° Hence, x = 45° and y = 135° (c) Since, ∆ ABC is isosceles with AB = AC ∴ ∠B = ∠C i.e., ∠B = ∠C = x and. ∠A = 92° [Vertical opposite angles] By the angle sum property, we have ∠A + ∠B + ∠C = 180° or, 92° + x + x = 180° or, 2x = 180° – 92° = 88°

...(i)

or,

x = 44° ∠C + y = 180°

[Linear pair]

y = 180° – 44°

= 136° [ ∠C = x = 44°] Hence, x = 44° and y = 136°

qqq

P-56

M A T H E M A T I C S - VII

WORKSHEET-51 Hence, the distance of the foot of the ladder from the wall is 9 m.

Solutions 1. Given, ∆PQR, is right angled at P, in which PQ = 10 cm and PR = 24 cm

4. (a) Let a = 2.5 cm, b = 6.5 cm and c = 6 cm

Q

10 cm

R

24 cm

a2 + c2 = 42.25 and b2 = (6.5)2 = 42.25

or,

a2 + c2 = b2

Thus, the given sides form a right angled triangle and the right angle is opposite to side of length 6.5 cm.

QR2 = PQ2 + PR2 or, QR2 = (10)2 + (24)2



QR2 = 100 + 576 or, QR2 = 676 QR =

or,

∴

a2 + c2 = (2.5)2 + (6)2 or, a2 + c2 = 6.25 + 36

P

In ∆PQR, by using Pythagoras theorem, or,

∴

676 = 26

Hence, the length of QR is 26 cm.

(b) Let a = 2 cm, b = 2 cm and c = 5 cm ∴

a2 + b2 = (2)2 + (2)2 or, a2 + b2 = 4 + 4

or,

a2 + b2 = 8 and c2 = (5)2 = 25

or,

a2 + b2 ≠ c2

2. Given, ∆ABC is right angled at C, in which

Therefore, the given sides can not form a right angled triangle.



(c) Let a = 1.5 cm, b = 2 cm and c = 2.5

AB = 25 cm and AC = 7 cm A

a2 + b2 = (1.5)2 + (2)2 = 2.25 + 4 = 6.25 and



25

7 cm

cm

∴

B

c = (2.5)2 = 6.25

∴

2

a2 + b2 = c2



C

Therefore, the given sides form a right angled triangle and the right angle is opposite to the side of length 2.5 cm.

In ∆ABC, by using Pythagoras theorem,

AB2 = AC2 + BC2 or, BC2 = AB2 – AC2

or,

BC2 = (25)2 – (7)2 or, BC2 = 625 – 49

or,

BC2 = 576 or, BC =



576 = 24

5. Let ACB be the tree before it broken at the point C, and let its top A touch the ground at A’ after it broke. Then, DA’BC is a right triangle, right-angled at B such that A’B = 12 m, BC = 5 m.

Hence, the length of BC is 24 cm.

By Pythagoras theorem, we have

3. Let AC be the ladder and AB be the wall C is the foot



(A’C)2 = (A’B)2 + (BC)2

of the ladder and A be the top if ladder, which is at

or,

(A’C)2 = 122 + 52

the window.

= 144 + 25 = 169

A

A

15

cm

12 cm C

C

B

a

Here, height of window A from the ground i.e., AB = 12 cm In right angled ∆ABC, by using Pythagoras theorem,

2

2

2

2

2

AC = AB + BC or, BC = AC – AB



BC = (15) – (12) or, BC = 225 – 144

or,

BC2 = 81 or, BC =

or,

2

S OLUT I ONS

2

2

2

2

81 = 9

A'

12 m

B

or,

A’C =

169 = 13

or,

AC = A’C = 13 m



\

[AC = A’C] AB = AC + BC

= (13 + 5) m = 18 m

P-57

6. Let ABCD be the rhombus and AC and BD are its diagonals.

∴ By Pythagoras theorem, we have

Then, given AC = 30 cm and BD = 16 cm.



We know that, the diagonals of a rhombus bisect each other at right angle. 1 1 AC = × 30 cm = 15 cm 2 2

∴

OA =

and

1 1 OB = BD = × 16 = 8 cm 2 2

Also, ∠AOB is right angle. cm cm

15

8

cm

cm

15

8 O

In ∆AOB, by using Pythagoras theorem,

AB2 = AO2 + OB2 or, AB2 = (15)2 + (8)2

or,

AB2 = 225 + 64 or, AB2 = 289

or,

AB =

289 = 17

= 4 × AB = 4 × 17 = 68 cm Hence, the perimeter of the rhombus in 68 cm. 7. (a) DABC is right-angled at B.



or,

or,

AC2 = AB2 + BC2 2

x = 9 + 16 x =

x2 = 64 + 36 x =

or, x2 = 100

100 = 10

(c) DABC is right-angled at B.

∴ By Pythagoras theorem, we have

or,

or,

AC2 = AB2 + BC2

or, x2 = 82 + 152

2

or, x2 = 289

x = 64 + 225 x =

289 = 17

(e) Using Pythagoras theorem in right-angled Ds ALB and ALC, we have. BL2 = AB2 – AL2 or, BL2 = 372 – 122 2 or, BL = (37 + 12) (37 – 12)

or,

or, x2 = 32 + 42 2

or, x = 25

25 = 5

BL2 = 49 × 25 BL =

49 × 25 = 7 × 5 = 35

Similarly, CL = 35 ∴ or,

∴ By Pythagoras theorem, we have



or,

Perimeter of a rhombus = 4 × Side



or,

or, x2 = 82 + 62



B

A

or,

AC2 = AB2 + BC

(d) DABC is right-angled at B. ∴ By Pythagoras theorem, we have AC2 = AB2 + BC2 or, x2 = 242 + 72 2 or, x = 576 + 49 or, x2 = 625 or, x = 625 = 25

C

D



BC = BL + LC = 35 + 35 = 70 x = 70

(f) Using Pythagoras theorem in right-angled DALB, we have



or,

or,

BL2 = AB2 – AL2

or, x2 = 122 – 32

2

or, x2 = 135

x = 144 – 9 x =

(b) DABC is right-angled at B.

135  

qqq

WORKSHEET-52 Solutions 1. By angle sum property, we have ∠P + ∠Q + ∠R = 180° or, ∠P + ∠25° + 65° = 180° or, ∠P + 90° = 180° or, ∠P = 180° – 90° = 90° \ DPQR is a right triangle, right-angled at P. \ By Pythagoras theorem PR2 + PQ2 = QR2 Thus, (b) is true.

2. Let ABCD be a rectangle such that AB = 40 cm and AC = 41 cm.

In right angled DABC, right angled at B, by Pythagoras theorem, we have

P-58

D

C

41 A

m

40 m

B

2

BC = AC2 – AB2 = 412 – 402 = (41 + 40) (41 – 40) = 81 × 1 = 81 BC = 81 = 9 cm, ⇒ Now, perimeter of rectangle = 2(AB + BC) = 2 (40 + 9) cm = 2 × 49 cm = 98 cm

M A T H E M A T I C S - VII

3. (a) Since, ∆ ABC is isosceles with AB = AC. ∴ ∠B = ∠C or, x = 40° (b) Since, ∆ ABC is isosceles with AB = AC. ∴ ∠B = ∠C or 45° = ∠C Also, by the angle sum property, we have ∠A + ∠B + ∠C = 180° or, x + 45° + 45° = 180° or, x + 90° = 180° or, x = 180° – 90° or, x = 90° (c) Since, ∆ ABC is isosceles with AB = AC ∴ ∠B = ∠C or, x = 50° (d) Since, ∆ ABC is isosceles with AB = AC ∴ ∠B = ∠C or, x = ∠C By the angle sum property, we have ∠A + ∠B + ∠C = 180° or, 100° + x + x = 180° or, 2x = 180° – 100° or, 2x = 80°  80°  or, x =  = 40°  2  (e) Since, ∆ ABC is isosceles with AB = AC ∴ ∠B = ∠C or, x = ∠C By the angle sum property, we have or,

∠A + ∠B + ∠C = 180° 90° + x + x = 180°

or, or,

2x = 180° – 90°  90°  x =  = 45°  2 

(f) Since, ∆ ABC is isosceles with AB = AC ∴ ∠B = ∠C ⇒ x = ∠C By the angle sum property, we have ∠A + ∠B + ∠C = 180° or, 40° + x + x = 180° or, 2x = 180° – 40°  140°  or, x =  = 70°  2  (g) Since, ∆ ABC is isosceles with AB = AC ∴ ∠B = ∠C ⇒ x = ∠C Since, in a triangle an exterior angle and the interior adjacent angle form a linear pair, therefore, ∠C + 120° = 180° or, x = 180° – 120° = 60° (h) Since, ∆ ABC is isosceles with AB = AC ∴ ∠B = ∠C ⇒ ∠B = x Also, in a triangle an exterior angle and the interior adjacent angle form a linear pair. Therefore, 110° + ∠A = 180° or, ∠A = 180° – 110° = 70° By the angle sum property, we have ∠A + ∠B + ∠C = 180° or, 70° + x + x = 180° or, 2x = 180° – 70° or, 2x = 110° °

or,

 110  x =  = 55°  2 

(i) Since, ∆ ABC is isosceles with AB = AC ∴ ∠B = ∠C ⇒ ∠B = x Also, ∠B = 30° [Vertically opposite angles] ∴ x = 30° [∠B = x]

qqq

S OLUT I ONS

P-59

CHAPTER SECTION

B 7

CONGRUENCE OF TRIANGLES WORKSHEET-53

Solutions 1. The angle included between the sides DE and EF of ∆ DEF is ∠DEF. 2. the additional information needed to establish ∆ PQR ≅ ∆ FED is ∠P = ∠F. 3. Two coins or notes of the same denomination and two keys of the same lock are examples of the congruent shapes. 4. ∆ ABC ≅ ∆ FED means that ∆ ABC superposes on ∆ FED exactly such that the vertices of ∆ ABC fall on the vertices of ∆ FED in the following order A ↔ F, B ↔ E and C ↔ D. Then, we have the following six equalities : AB ↔ FE , BC ↔ ED , CA ↔ DF

(i.e., corresponding angles are congruent) 5. If ∆ DEF ≅ ∆ BCA, then D ↔ B, E ↔ C and F ↔ A Thus, the parts of ∆ BCA that correspond to : (a) ∠E is ∠C (b) EF is CA

(c) ∠F is ∠A (d) DF is BA

6. (a) Statement ∆ ABC ≅ ∆ ABD is not meaningfully written as AB = AB [Common] BC = BD [Incorrect] and, CA = DA [Incorrect] (b) Statement ∆ ABC ≅ ∆ BAD is written meaningfully as AB = AB [Common] BC = BD [Given] and, CA = DA [Given] 7. (a) they have the same length

(b) 70° (c) m∠B = m∠B 8. (a) Three pairs of equal parts in ∆ADB and ∆ADC are : AD = AD [Common] AB = AC [Given] and DB = DC

[ D is mid-point of BC ] (b) By SSS criterion of congruence, ∆ ADB ≅ ∆ ADC (c) Yes, ∠ B = ∠ C [ Corresponding parts of congruent triangles are equal] 9. (a) In ∆s ABC and PQR, we have AB = PQ (= 1.5 cm) BC = QR (= 2.5 cm) and, CA = RP (= 2.2 cm) So, by SSS criterion of congruence, ∆ ABC ≅ ∆ PQR (b) In ∆ DEF and NML, we have DE = NM ( = 3.2 cm) EF = ML ( = 3 cm) FD = LN ( = 3.5 cm) So, by SSS criterion congruence, ∆ DEF ≅ NML (c) ∆s ABC and PQR are not congruent. In ∆s ADB and ADC, we have AD = AD (Common) DB = DC (= 2.5 cm) and BA = CA (= 3.5 cm) By SSS criterion of congruence, ∆ ADB ≅ ∆ ADC

qqq

WORKSHEET-54 D

Solutions

1. The side MN is included between the angles M and N of ∆ MNP.



2. (a) Three pairs of equal parts in two triangles AOC and BOD are AO = OB, CO = OD and ∠ AOC = ∠BOD.



(b) (b) is True. 3. To establish ∆ DEF ≅ ∆ MNP, by using ASA congruence rule the additional information needed is about the equality of sides DF and MP.

P-60

E

M

F

N

P

4. (a) ∆s ABC and DEF are not congruent. (b) In ∆s ACB and RPQ, we have AC = RP = 2.5 cm

∠C = ∠P = 35°

M A T H E M A T I C S - VII

and,

CB = PQ = 3 cm

AB = EF = 3.5 cm and ∠B = ∠E = 60° ∴ By ASA criterion of congruence, ∆ ABC ≅ ∆ FED (b) Given triangles are not congruent.

∴ By SAS criterion of congruence, ∆ ACB ≅ ∆ RPQ



(c) In ∆ DEF and ∆ PQR, we have



DF = PQ ( = 3.5 cm)



∠ F = ∠ Q ( = 40°)





FE = QR ( = 3 cm)

So, by SAS criterion of congruence, ∆ DFE ≅ ∆ PQR



(d) In ∆s RSP and PQR, we have RS = PQ = 3.5 cm



∠ PRS = ∠ RPQ = 30°

∠Q = ∠N = 30° ∆ PQR ≅ ∆ MNL

(d) In ∆s ABC and BAD, we have ∠CAB = ∠DBA = 30°



AB = BA

[Common]

∠ABC = ∠BAD = 30° + 45° = 75°

and

∆ RSP ≅ ∆ PQR.

∴ By ASA criterion of congruence,

5. (a) Since, in ∆s ABC and FED, we have

∆ ABC ≅ ∆ BAD



∠A = ∠F = 40°



and



∴ By SAS criterion of congruence,

QR = NL = 6 cm



RP = PR [Common]



∠R = ∠L = 60°



∴ By ASA criterion of congruence,

and,

(c) In ∆s PQR and MNL, we have



qqq

WORKSHEET-55 By ASA criterion of congruence.

Solutions

The additional information needed is about equality of hypotenuses, i.e., AC = RQ. 2. (a) In ∆s ∆DEF and ∆PQR, we have ∠D = ∠Q (= 60°) DE = QR (= 5 cm) and, ∠F = ∠R (= 80°) ∴ By ASA criterion of congruence,

80° E



80°

60°

F

5 cm

Q

80° 60° 5 cm R

∆ DFE ≅ ∆ QRP (b) Since, side included between equal angles in the given triangles is not equal. ∴ Given ∆s are not congruent. E

P

m

6c 60° D

80°

6 cm

60° F

80°

Q

(c) In ∆s DEF and QPR, we have ∠E = ∠P (= 80°) EF = PR (= 5 cm) and,

S OLUT I ONS

∠F = ∠R (= 30°)

80°

30° F

P

5 cm

30° R

(b) By RHS criterion of congruence,



D

5 cm

∆ DEF ≅ ∆ QPR 3. (a) The three pairs of equal parts in ∆ CBD and ∆ BCE are CB = BC, ∠ CDB = ∠ BEC and BD = CE.

P

E

Q

D

1. To establish by using RHS congruence rule that ∆ ABC ≅ ∆ RPQ.

R

(c) Yes,

∆ CBD ≅ ∆ BCE ∠ DCB = ∠ EBC

 Corresponding parts of congruent triangles are equal. 4. (a) The three pairs of equal parts in triangles BAC and DAC are ∠ BAC = ∠ DAC, AC = AC and ∠ BCA = ∠ DCA. (b) Yes, by ASA, criterion of congruence, ∆ BAC ≅ ∆ DAC. (c) Yes, AB = AD [∴ Corresponding parts of congruent triangles are equal] (d) Yes, CD = CB [∴ Corresponding parts of congruent triangles are equal] 5. (a) ∆s PQR and DEF are not congruent.



(b) In ∆s CAB and DBA, we have



∠C = ∠D (= 90°)



AB = BA (= 3.5 cm)

and,

CA = DB (= 2 cm)

P-61

∴ By RHS criterion of congruence,

∆ CAB ≅ ∆ DBA.

(c) In ∆s ABC and ADC, we have

and,

AB = AD (= 3.6 cm)

∴ By RHS criterion of congruence,

∆ ABC ≅ ∆ ADC.

(d) In ∆s PSQ and PSR, we have



∠ PSQ = ∠ PSR (= 90°)

hypotenuse PQ = hypotenuse PR (= 3 cm) and,

PS = PS [Common]

∆ PSQ ≅ ∆ PSR.



∠ B = ∠ D (= 90°)

hypotenuse AC = hypotenuse AC [Common]



∴ By RHS criterion of congruence,



6. (a) The three pairs of equal parts in ∆ ADB and ∆ADC are AD = AD, ∠ ADB = ∠ ADC and AB = AC. (b) By RHS criterion of congruence. ∆ ADB ≅ ∆ ADC



∠ B = ∠ C

(c) Yes,

 Corresponding parts of congruent triangles are equal.

(d) Yes,

BD = CD

 Corresponding parts of congruent triangles are equal.

qqq

WORKSHEET-56 Solutions

A

P



1. Itemwise reasons are as under :



1. Given



3. Common



4. By SAS criterion of congruence.



2. Clearly from the given figure, we have ∆ RAT ≅ ∆ WON.





RA = WO (given)



∠ RAT = ∠ WON (given)



AT = ON (given)

Draw a ∆ ABC with ∠ A = 30°, ∠ B = 40°, ∠ C = 110° and BC = 2.5 cm (say).



3. Completing the congruence statement, we have ∆ BCA ≅ ∆ BTA, ∆ QRS ≅ ∆ TPQ.

Draw another ∆ PQR with ∠ P = 30°, ∠ Q = 40°, ∠ R = 110° and QR = 3.5 cm.



4. (a) In order to show that ∆ ART ≅ ∆ PEN using SSS criterion, we need to show that

(i) AR = PE (ii) RT = EN

Clearly, the triangles have their corresponding angles equal, but they are not congruent. Since, BC ≠ QR.

(iii) AT = PN

Hence, the student is not justified.



2. Given

(b) If m ∠ T = m ∠ N an to use SAS criterion, we need to show that

(i) RT = EN and (ii) PN = AT

(c) If AT = PN and to use ASA criterion, we need to show that

(i) ∠ RAT = ∠ EPN and (ii) ∠ ATR = ∠ PNE

30°

30°

5. In two triangles, if the three angles of one triangle are respectively equal to the three angles of the other, then the triangles are not necessarily congruent.

40° B



Q

3.5 cm

R

∆ ABC ≅ ∆ DEF ∆ PQR ≅ ∆ XYZ

(c) By ASA criterion of congruence,



C

110°

(b) By SAS criterion of congruence,



2.5 cm

40°

6. (a) By SSS criterion of congruence,



110°

∆ LMN ≅ ∆ GFH

(d) By RHS criterion of congruence,



∆ ABE ≅ ∆ CDB

qqq

P-62

M A T H E M A T I C S - VII

WORKSHEET-57 Solutions

2. ∆ Since (i) ∠ C = ∠ D ........ given



(ii) BC = DE ........ given



(iii) ∠ B = ∠ E ........ given

B

C

1  =  × 4 × 3  sq. units 2   = 6 sq. units 1 and area of ∆ DCA = × DC × DA 2

Hence, D ABC ≅ FED by ASA criterion of congruency

D

1. In order to prove that ∆ ABC ≅ ∆ PQR, the additional pair of corresponding parts are named as BC = QR.

Criterion used is ASA rule of congruence.

A

3. In some special cases (which depend on the lengths of the sides and the size of the angle involved), SAS is enough to show congruence. However, it is not always enough.

Consider the following triangles : D

A

B

E

C

F

1  =  × 3 × 4  sq. units 2   = 6 sq. units ∴ Area (∆ ABC) = Area (∆ CDA) In ∆s ABC and CDA, we have AB = CD ∠ B = ∠ D = 90° and, AC = AC Hypotenuse ∴ By RHS criterion of congruence, ∆ ABC ≅ ∆ CDA Thus, two triangles of equal areas are drawn such that these triangles congruent. D

A

Here side AB is congruent to side DE (S) side AC is congruent to side DF (S) angle C is congruent to angle F (A) But the triangles are not congruent, as we can see. What happens in this : If we draw a vertical line through point A in the first triangle, we can sort of “flip” side AB around this line to get the second triangle, If we were to lay one triangle on top of the other and draw the vertical line, this how it would look. A(D)

B

E

C(F)

Clearly, side DE is just side AB flipped around the line. So, we have not changed the length of the side, and the other side AC (or DF) is unchanged, as is angle C (or F). So, these two triangles that have the same SSA information, but they are not congruent.

4. (a)

1 Area of ∆ ABC = × BC × AB 2

S OLUT I ONS

B

L

C

1 (b) Here, area of ∆ ABC = × BC × AB 2  1  =  × 5 × 4  sq. units 2   = 10 sq. units 1 and area of ∆ DBC = × BC × DL 2  1  =  × 5 × 4  sq. units 2   = 10 sq. units ∴ Area (∆ ABC) = Area (∆ DBC) Thus, two triangles of equal areas are drawn such that these triangles are not congruent. In this case their perimeters are unequal.

qqq

P-63

WORKSHEET-58 \ Width of street, AB = AC + BC = 15 + 8 = 23 m 8.

Solutions 1. (b) 2. (c). 3. When they have equal sides. 4. SAS.

D 13 m 5m

5. In DABC and DBAD,

E

B

AC = BD



= hypotenuse



ÐA = ÐB = 90°

[given]

9m

[given]

A

AB = AB [common side] 6. In DABC and DDCB, AC = DB [given] ÐBAC = ÐCDB = 90° [given] BC = BC [common] So by R.H.S. congruency we have







DABC @ DDBC

or, by C.P.C.T.,

7. Let AB is the street and C be the foot of ladder. Let D and E be windows at heights of 8 m and 15 m respectively from the ground. E

or,

AB = EC = 9 m

and

BD = 13 m



DE = 14 – 9 = 5 m



BD2 = BE2 + DE2



132 = BE2 + 52

or,

BE2 = (13)2 – (5)2



D 17

15

8

A

C

In the above figure, AB and CD are two poles whose heights are 9 m and 14 m respectively.

Now in right DBDE, by Pythagoras

AB = DC

17

C

B

Then CD and CE are the positions of ladder. From

= 169 – 25 BE2 = 144

or,

BE = 144

or,

BE = 12 m.

Hence, distance between their feets = 12 m.

the right angle DDAC, by Pythagoras Theorem, CD2 = AC2 + AD2 or, AC2 = CD2 – AD2 = 172 – 82 = 289 – 64 = 225

9. Given, EG = FI

AC = 225 = 15 Again from right DEBC, by Pythagoras Theorem, CE2 = BC2 + BE2 BC2 = CE2 – BE2 = 172 – 152 = 289 – 225 = 64



or,

or,

BC = 64 = 8 m

14 m

9m

adding GF on both sides, we get ∴ or,

EG + GF = GF + FI EF = GI

...(i) [from figure]

In DDEF and DHIG, DE = IH [given] and

∠DEF = ∠HIG [given) EF = GI

[from (i)]

Here, we see that, if two sides and the angle included between them of a triangle are equal to two corresponding sides and the angle included them of other triangle, then the triangles are congruent by SAS congruence rule.

Hence,

DDEF ≅ DHIG

qqq

P-64

M A T H E M A T I C S - VII

CHAPTER SECTION

B 8

COMPARING QUANTITIES WORKSHEET-59

Solutions 1. In a computer lab, For every 6 students there are = 3 computers 3 So, for each student there is = computer 6 Hence, for 24 students needed 3 = × 24 computers 6

= 3 × 4 = 12 computers ∴ 12 computers are needed for 24 students. 2. (a) Population of Rajasthan = 570 lakh Area of Rajasthan = 3 lakh sq. km 570 \ Number of people per sq. km = 3 = 190 Also, Population of U.P. = 1660 lakh Area of U.P. = 2 lakh sq. km 1660 \ Number of people per sq. km = 2 = 830 (b) Since population of Rajasthan is less per sq. km as compared to that of U.P. ∴ State of Rajasthan is less crowded. 3. Height

Number of In Children Fraction

In Percentage

110 cm

22

22 100

120 cm

25

25 100

25%, 25 per cent

128 cm

32

32 100

32%, 32 per cent

130 cm

21

21 100

21%, 21 per cent

22%, 22 per cent

Total 100 4. Writing the data in the tabular form :

1.

Size

Number of Shoes

In Fraction

In Percentage

2

20

20 100

20%, 20 per cent

2.

3

30

30 100

30%, 30 per cent

3.

4

28

28 100

28%, 28 per cent

4.

5

14

14 100

14%, 14 per cent

5.

6

8

8 100

8%, 8 per cent

Total

100

5. (a) ` 5 to 50 paise To compare these numbers, firstly we will change them to a single or same unit i.e., ` 5 = 500 paise as ` 1 = 100 paise, So ` 5 = 5 × 100 paise. Now, their ratio = 500 paise : 50 paise = 500 : 50 = 10 : 1. (b) We have, 50 kg to 210 g. We know that, 1 kg = 1000 g ∴ 50 kg = 50 × 1000 = 50,000 g ∴ Required ratio = 50 kg : 210 g = 50,000 g : 210 g 50000 = 210 = 5000 : 21 (c) We have, 9 m to 27 cm We know that, 1 m = 100 cm ∴ 9 m = 9 × 100 cm = 900 cm ∴ Required ratio = 9 m : 27 cm = 900 cm : 27 cm = 900 : 27 900 100 = = 27 3 or 100 : 3 (d) 30 days to 36 hour To compare these numbers firstly we will change them to a single unit of time i.e., 30 days = ? hours Q 1 day = 24 hours ∴ 30 days = 30 × 24 hours = 720 hours Now, their ratio is 720 hours : 36 hours or 720 : 36 or 20 : 1

qqq S OLUT I ONS

P-65

WORKSHEET-60 Solutions 1.

8 × 25 = 2 children 100

2. 35% + 65% = 100% [100 – 35 = 65] 64% + 20% + 16% = 100% [ 64 + 20 = 84 and 100 – 84 = 16] 43% = 100% – 57% [ 100 – 43 = 57] 70% = 100% – 30% [ 70 + 30 = 100] 3. Percentage of students having cycle = 65% Percentage of students having no cycle = (100 – 65)% = 35% 4. Percentage of apples in the basket = 50% Percentage of oranges in the basket = 30%



Percentage of mangoes in the basket = (100 – 50 – 30)% = (100 – 80)% = 20% 5. Let the number be n 25 × n = 9 100



25n = 9 × 100 25n = 900 900 n = = 36. 25



6. Let the number be n 75 × n = 15 100





75n = 15 × 100 75n = 1500 1500 n = 75



n = 20 7. Putting the data in the tabular form : Gold

Silver

Total

Number

20

10

30

Fraction

20 30

100 10 100 × = 3 30 100 100

In Percentage

200 2 % = 66 % 3 3

100 1 % = 33 % 3 3

8. (a)

12 × 100 = 75% 16

(b) 3·5 = (c)

35 × 100 = 350% 10

49 × 100 = 98% 50

(d)

2 × 100 = 100% 2

(e) 0·05 = 9. (a)

5 × 100 = 5% 100

8 × 100 = 25% 32

(b)

16 × 100 = 64% 25

(c)

5 × 100 = 1% 500

(d)

90 120

× 100 = 75% 50% of 164 =

(b)

75% of 12 =

50 1 × 164 = × 164 = 82 100 2 75 3 × 12 = × 12 100 4

= 3 × 3 = 9

10 30

200 20 100 × = 3 30 100 100

10. (a)

Bangles

On hundred

(c)

1 12 1 2 × 64 = 25 × 1 × 64 12 % of 64 = 2 100 2 100

=

1 1 × × 64 = 8 2 4

qqq

WORKSHEET-61 Solutions 1. (a)

P-66

1 part is shaded. 4

1  Percentage of part shaded =  × 100  % = 25% 4 

M A T H E M A T I C S - VII



(b)

3 part is shaded. 5

3  Percentage of part shaded =  × 100  % 5  = (3 × 20)% = 60% 3 (c) part is shaded. 8 3  Percentage of part shaded =  × 100  % 8   3  75 =  × 25  % = % 2 2  = 37.5% 25 1   8 × 100  1 2. (a) =   = 2 8  100  100     12.5  =   = 12.5%  100  (b)

5   4 × 100   125  5 =   =   = 125% 4  100   100     

 3   15   40 × 100    3   (c) = =  2  40 100  100     7.5  =   = 7.5%  100 

(d)

 2  200   7 × 100    2 =   =  7  100 7 100        

 4  28  4 =  7  = 28 % 7 100     3. (a) (b) (c)

 0.65 × 100  65 0.65 =  = 65%  = 100 100    2.1 × 100  210 2.1 =  = 210%  = 100  100   0.02 × 100  2 0.02 =  = 2%  = 100  100 

 12.35 × 100  1235 (d) 12.35 =  = 1235%  = 100 100   4. (a)

15 15% of 250 = × 250 100

S OLUT I ONS



=

75 = 37·5 2

1 1% of 1 hour = × 1 hour 100 1 = hour = 0·01 hour 100

(b)

20 20% of ` 2500 = × 2500 100 20 × 25 = = ` 500 1

(c)

75 75% of 1 kg = × 1 kg 100 75 3×1 3 = ×1= = kg 4 4 100 (d)

= 0·75 kg. 5. (a) Let the whole quantity be a. Then, 5% of a = 600 5 1 or, × a = 600 or, × a = 600 100 20 or, a = 12000 Hence, the whole quantity is 12000. (b) Let the whole quantity be a. Then, 12% of a = ` 1080 12 or, × a = ` 1080 100 or,

100   a = `  1080 × 12  

or, a = ` (90 × 100) = ` 9000 Hence, the whole quantity is ` 9000. (c) Let the whole quantity be a. Then, 40% of a = 500 km 40 or, × a = 500 km 100 or,

100   a =  500 × km 40  

or, a = (25 × 50) = 1250 km Hence, the whole quantity is 1250 km. (d) Let the whole quantity be a. Then 70% of a = 14 minutes 70 or, × a = 14 minutes 100 or,

10   a =  14 ×  minutes 7  

or, a = 20 minutes Hence, the whole quantity is 20 minutes. (e) Let the whole quantity be a. Then, 8% of a = 40 litres 8 or, × a = 40 litres 100

P-67

or, or, or, 6. (a)

100   a =  40 × litres 8   a = (5 × 100) litres a = 500 litres 25 1 25% = 0.25 = = 100 4

(b)

150% = 1.50 =

150 3 = 100 2

(c)

20% = 0.20 =

20 1 = 100 5

(d)

5% = 0.05 =

5 1 = 100 20

qqq

WORKSHEET-62 Solutions 1. In a city, Total percent of females = 30% Total percent of males = 40% Total percent of children will be 100% – (30 + 40)% = 100% – 70% = 30% are children. 2. Percentage of voters who voted = 60% Percentage of voters who did not vote = (100 – 60)% = 40% Total number of voters = 15000 Number of voters who did not vote = 40% of 15000  40  × 15000  = 6000 =   100  3. Let Meeta’s salary be ` x. Then, 10% of ` x = ` 400 10 or, × x = 400 100 or,

1 × x = 400 10

or, x = (10 × 400) = 4000 ∴ Meeta’s salary is ` 4000. 4. Out of 100, 25% matches are won. Then, out of 20, the number of matches that the team won 25 1 = × 20 = × 20 = 5 100 4 5. Total number of sweets = 15 Number of sweets with Manu = 20% of 15  20  × 15  = 3 =   100  Number of sweets with Sonu = 80% of 15  80  × 15  = 12 =   100  6. Sum of the ratios = 2 + 3 + 4 = 9 Sum of the angles of a triangle = 180°

P-68

o o 2  3  ∴ The angles are  × 180  , i.e., 40°;  × 180  , 9  9  o 4  i.e., 60°; and  × 180  , i.e., 80°. 9 

7. Original price of petrol = ` 1 New price of petrol = ` 52 Change in price = ` (52 – 1) = ` 51 Therefore, the percentage of increase Change in price = × 100 Original Price =

51 × 100 = 5100% 1

8. Given, C.P. of an item = ` 50; Profit% = 12% ∴ Profit = 12% of C.P. of an item 12 600 = 12% of ` 50 = × 50 = =`6 100 100

∴ S.P. of an item = C.P. + Profit = ` 50 + ` 6 = ` 56 9. (a) Original price of shirt = ` 80 New price of shirt = ` 60 Change in price = ` (80 – 60) = ` 20 Therefore, the percentage of decrease Change in Price = × 100 Original Price =

20 × 100 = 25% 80

(b) Original marks = 20 New marks = 30 Change in marks = (30 – 20) = 10 Therefore, the percentage of increase Change in marks = × 100 Original marks =

10 × 100 = 50% 20

10. We have, C.P. of the chair = ` 375 S.P. of the chair = ` 400

M A T H E M A T I C S - VII

Since S.P. > C.P, so, gain is given by Gain = S.P. – C.P. = ` (400 – 375) = ` 25  Gain  × 100  % Now, Gain % =   C.P. 

25   20  =  × 100  % =   %  375   3  = 6

2 % 3

qqq

WORKSHEET-63 Solutions 1. We have, P = Principal = ` 10,000, R = Rate per cent per annum = 5% and, T = Time = 1 year. P×R×T ∴ Simple interest (S.I.) = 100  10000 × 5 × 1  = `   100   = ` 500 2. We have, P = Principal = ` 3500, R = rate per cent per annum = 7% and, T = Time = 2 years. P×R×T ∴ Simple interest (S.I.) = 100

6. We have, S.I. = ` 450, R = 5% p.a., T = 3 years. 100 × S.I. ∴ Principal = R×T or,

7. Given, S.P. of an article = ` 250; Profit % = 5% Let C.P. of an article be ` x. 5 5x Then, profit = 5% of C.P. of an article = ×x = 100 100 Now, S.P. of an article = C.P. + Profit 5x ∴ 250 = x + 100

 3500 × 7 × 2  = `   100  

or, or,

= ` 490. 3. We have, P = Principal = ` 6050, R = rate of interest per annum = 6.5% and, T = Time = 3 years. P×R×T ∴ Simple interest (S.I.) = 100

or,

 6050 × 6.5 × 3  = `   100   = ` 1179.75 Now, Amount = Principal + S.I. = ` (6050 + 1179.75) = ` 7229.75. 4. We, have, P = Principal = ` 7000, R = rate per cent per annum = 3.5% and, T = Time = 2 years. P×R×T ∴ Simple interest (S.I.) = 100  7000 × 3.5 × 2  = `   100   = ` 490 Now, Amount = Principal + S.I. = ` (7000 + 490) = ` 7490 5. We have, P = ` 2400, R = 5% p.a. and S.I. = ` 240. S.I. × 100 ∴ Time = P×R or,

 240 × 100  Time =   years = 2 years  2400 × 5 

S OLUT I ONS

 100 × 450  Principal = `   = ` 3000.  5×3 

250 100 x + 5x = 1 100 105x = 250 × 100 250 × 100 x = 105

5000 or, x = = ` 238.10 21 Hence, C.P. of an article is ` 238.10. 8. Given, S.P. of an item = ` 540; Loss % = 5% Let the C.P. of an item be ` x. Then, loss = 5% of C.P. of an item 5 5x = ×x=` 100 100 Now, S.P. = C.P. – Loss 5x   ∴ 540 =  x − 100   or,

 100 x − 5x  540 =   100  

95x or, 540 = 100 or, 95x = 540 × 100 540 × 100 10800 or, x = =` 95 19 8 or ` 568.42 19 Hence, C.P. of an item is ` 568.42. = ` 568

qqq

P-69

WORKSHEET-64 (c) We have,

Solutions 1. Original population of the city = 25,000 Decreased population of the city = 24,500 ∴

Decrease in population = 25,000 – 24,500

= 500  500  × 100  % ∴ Percentage of decrease =  25000   = 2%



C.P. = ` 2,500 and S.P. = ` 3,000.

Since S.P. > C.P., so, there is a profit given by

Profit = S.P. – C.P.

= ` (3000 – 2500) = ` 500 Now,

 Profit  × 100  % Profit % =   C.P. 

The percent increase be x.

 500  × 100  % =  2500   = 20%

Here, increase in price = ` 3,70,000 – ` 3,50,00

(d) We have,

= ` 20,000

Since C.P. > S.P., so, there is a loss given by

2. Cost price of car = ` 3,50,000 The price increased next week = ` 3,70,000

∴

20 , 000 x = × 100 3, 50 , 000



=

2 200 × 100 = 35 35

5 40 = = 5 %. 7 7

5 ∴ There is an increment of 5 % in price. 7

3. (a) We have,

C.P. = ` 250 and S.P. = ` 325.

Since S.P. > C.P., so, there is a profit given by

Profit = S.P. – C.P.

= ` (325 – 250) = ` 75 Now,

 Profit  × 100  % Profit % =   C.P. 



=

C.P. = ` 12,000 and



S.P. = ` 13,500.

Since S.P. > C.P., so, there is a profit given by Profit = S.P. – C.P.

= ` (13,500 – 12,000) = ` 1,500 Now,

 Profit  × 100  = % Profit % =   C.P. 

 1500  × 100  % =  12000   = 12.5%

P-70

Loss = C.P. – S.P.

= ` (250 – 150) = ` 100 Now,

 Loss  × 100  = % Loss % =  C.P.  

 100  × 100  % =  250   = 40% 4. (a) We have ratios as 3 : 1. 3 + 1 = 4 is the total of parts. This means parts are

3 1 and . 4 4

3  Then, in per cent these are  × 100  % = 75% and 4 

75 × 100 % = 30% 250

(b) We have,





C.P. = ` 250 and S.P. = ` 150

1   4 × 100  %, i.e., 25% respectively.  

(b) We have ratio as 2 : 3 : 5.

2 + 3 + 5 = 10 is the total of parts. This means parts are

2 3 5 , and . 10 10 10

 2  Then, in per cent these are  × 100  % = 20%,  10   3  5    10 × 100  % = 30% and  10 × 100  % = 50%     respectively.

(c) We have ratios as 1 : 4.

1 + 4 = 5 is the total of parts.

M A T H E M A T I C S - VII

This means parts are

1 4 and 5 5

1  Then, in per cent these are  × 100  % = 20% and 5   4   5 × 100  % = 80% respectively.  

(d) We have ratios as 1 : 2 : 5.

1 + 2 + 5 = 8 is the total of parts. This means parts are

1 2 5 , and . 8 8 8

1  Then, in per cent these are  × 100  % = 12.5%, 8  2 5    8 × 100  % = 25% and  8 × 100  % = 62.5%     respectively.

qqq

WORKSHEET-65 Solutions 1. We have, C.P. = ` 275 and loss = 15% (100 − Loss %) × C.P. Now, S.P. = 100  85 × 275  S.P. = `   = ` 233.75.  100 

or,

Thus, the selling price of book is ` 233.75. 2. We have, P = ` 56,000, S.I. = ` 280, T = 2 years S.I.× 100 Now, R = P ×T  280 × 100  Rate =   % = 0.25%  56000 × 2 

or,

Hence, rate of interest is 0.25% per annum 3. We have, S.I. = ` 45, R = 9% per annum and T = 1 year S.I.× 100 Now, P = R×T  45 × 100  P = `   = ` 500  9×1 

or,

Thus, the sum borrowed is ` 500. 4. Cost price of a T.V. = ` 10,000 Let selling price of it be x. Profit % = 20% ∴ I get for it will be, as we know, Profit Profit % = × 100 C.P.

20 =

P × 100 10 , 000

or, or, ∴ Hence,

20 =

P 100



P = 20 × 100 = ` 2,000 C.P. + P = S.P. S.P. = 10,000 + 2,000 = ` 12,000 I get for it = ` 12,000.

S OLUT I ONS

5. Juhi sells a machine for ` 13,500 She loses in bargain = 20% The price she bought it be x. As here, S.P. = ` 13,500 Loss % = 20% ∴ C.P. = ? Loss = C.P. – S.P. Loss = x – 13,500 Here, Loss % = 20% Loss Q Loss % = × 100 C.P. ∴

20% =

Loss × 100 x

20 x or, = Loss 100 Placing value of (2) in (1), we get 20 x = x – 13,500 100 or, or, or, or, or,

...(1)

...(2)

20x = (x – 13500)100 20x = 100x – 1350000 13,50,000 = (100 – 20)x 13,50,000 = 80x 13, 50 , 000 x = 80

\ x = ` 16,875. 6. (a) Given, ratio of calcium, carbon and oxygen in a chalk stick are 10 : 3 : 12. Sum of the parts = 10 + 3 + 12 = 25 ∴ Percentage of carbon in chalk stick  Part of carbon in chalk stick  × 100  % =  Tatal part of chalk    3  =  × 100  % = (3 × 4)% = 12% 25   Hence, the percentage of carbon in a chalk stick is 12%.

P-71

(b) Let the weight of the chalk stick be x g. Now, weight of carbon = 3 g (which is 12% of chalk) ∴ 12% of x = 3 12 or, × x = 3 100 3 × 100 or, x = = 25 g 12 Hence, the weight of the chalk stick is 25 g. 7. (a) we have, P = ` 1,200, R = 12% p.a. and T = 3 years. P × R ×T Now, S.I. = 100

or,

 1200 × 12 × 3  S.I. = `   = ` 432 100  

Now, Amount = P + S.I. = ` (1200 + 432) = ` 1,632 (b) We have, P = ` 7,500, R = 5% p.a. and T = 3 years P × R ×T S.I. = 100 or,

 7500 × 5 × 3  S.I. = `   = ` 1,125 100  

Now, Amount = P + S.I. = ` (7500 + 1125) = ` 8,625

qqq

WORKSHEET-66 14. Cost of 6 bowls is ` 90

Solutions

1. (c) 2. (b) 4. (c) 5. 60 6. Time, rate percent, Principal 8. multiply 9. left

3. (b) 7. ` 168 10. sum/addition

11. more. 12. Meena gives interest of ` 45 Time = 1 year Rate = 9% Let principal be x, then P × R ×T I = 100

45 =

x ×9×1 100

45 × 100 =x 9 or, or, 5 × 100 = x ∴ x = 500 Hence, sum she has borrowed is ` 500. 13. Principal = ` 56,000 Time = 2 years Interest = ` 280 p.a. Let rate of interest be x ∴ or, ∴

Hence, cost of 10 bowls = `

90 6 90 × 10 6

= ` 150. 15. or,

Total parts = 3 + 2 = 5 Son got =

3 part 5

3 or, So his percentage share = × 100 = 60% 5

His daughter got = 2 part 5

2 So, her percentage share = × 100 = 40%. 5 16. Let the total marks = x Then, marks for accent and presentation = 20% of x 20 x = ×x= 100 5 Remaining marks = x –

5x − x x 4x = = 5 5 5

R =

100I P ×T

Now, marks for subject matter = 60% of

x =

100 × 280 56 , 000 × 2

=

60 4 x 12 x × = 100 5 25

x =

1 4

Remaining marks =

5 × 4 x − 12 x × 1 4 x 12 x = − 25 5 25

=

20 x −12 x 8x = 25 25

x = 0·25%

Hence, rate of interest for 2 years = 0·25%.

P-72



Cost of 1 bowl = `

4x 5

M A T H E M A T I C S - VII

According to the question, there are 8 marks for rebuttal. 8x So, = 8 or, 8x = 25 × 8 25 or,

x =

25×8 = 25 8

17. (a) Mr. Narayan saves 20% of his salary ∴ His expense = 100 – 20 = 80% 80 or, His expense = 20,000 × 100 = ` 16,000 (b) Percentage. (c) Value : Saving is very important.

qqq

S OLUT I ONS

P-73

CHAPTER SECTION

B 9

RATIONAL NUMBERS WORKSHEET-67

Solutions

1. Yes, it is a rational number, but a negative rational number. 2. Ten rational numbers may be taken as

1 2 5 −7 −3 3 −4 5 11 110 , , , , , , , , and . 2 3 90 11 13 7 5 9 13 117 − − −

3. We know that 5 =

5 , where 5 and 1 are both 1

positive. So, 5 is a positive rational number.



4. Five more positive rational numbers may be taken 3 3 11 −15 −110 as , , , and . 2 4 17 −17 113 5. We know that – 8 =

−8 where 8 is negative and 1 1

is positive. \ – 8 is a negative rational number.

6. Five more negative rational numbers may be taken −3 3 −11 15 −110 as , , , and . 2 −4 17 −17 113



7. (a) We have,

25 −15 5 = = = 16 4

Here, the rational numbers equivalent to rational 5 5 5 × 4 20 number are as follows = = 4 4 4 × 4 16 [ 4 × 4 = 16, since given denominator of equivalent 5 5 × 5 25 rational number, so multiply by 4] = = 4 4 × 5 20 [ 5 × 5 = 25, since given numerator of equivalent rational number, so multiply by 5] Also,

5 × ( −3) −15 5 = = 4 4 × ( −3) −12

[ 5 × (–3) = –15, since given numerator of equivalent rational number, so multiply by (– 3)] So, the mission integers in the boxes are filled as



−15 20 25 5 = = = 16 −12 20 4

(b) We have,

9 −6 −3 = = = 14 7

Here, the rational number equivalent to rational −3 number are as follows 7

−3 −3 × 2 −6 = = 7 7 × 2 14

or

−3 × ( −3) 9 −3 = = 7 × ( − 3 ) − 21 7

Also,

−3 −3 × 2 −6 = = 7 7 × 2 14

So, the missing integers in the boxes are filled as



−6 9 −6 −3 = = = 14 21 14 − 7

−2 , numerator and denominator are of 3 −2 opposite signs. Therefore, is a negative rational 3 number. 5 (b) In , numerator and denominator are of same 7 signs. 5 Therefore, is a positive rational number. 7

8. (a) In

3 , numerator and denominator are of opposite −5 3 signs. Therefore, is a negative rational number. −5 (d) The number 0 is neither a positive nor a negative rational number. 6 (e) In , numerator and denominator are of same 11 signs. 6 Therefore, is a positive rational number. 11 −2 (f) In , numerator and denominator are of same −9 signs. −2 Therefore, is a positive rational number. −9

(c) In

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P-74

M A T H E M A T I C S - VII

WORKSHEET-68 On dividing numerator and denominator by their HCF, we get −12 −12 ÷ 6 −2 = = 18 18 ÷ 6 3

Solutions 1. We have,

−5 −5 × 8 −40 = = 7 7×8 56



−3 × 7 −3 −21 = = × 8 7 8 56

and,

Hence, the standard form of

We know that −40 −39 −38 −37 −22 −21 < <…< < < < 56 56 56 56 56 56 −5 −3 Hence, five rational numbers between and 7 8 −39 −38 −37 −36 −35 may be taken as and , , , . 56 56 56 56 56

2. (a) Four rational numbers equivalent to

−2 are 7

−2 × 2 −4 −2 × 3 −6 , = = , 14 7 × 2 7 × 3 21 −2 × 4 −2 × 5 −8 −10 = and = 7 × 4 28 7×5 35 5 (b) Four rational numbers equivalent to are −3 5×2 10 5 × 3 15 = , = , −3 × 2 −6 −3 × 3 −9 5×4 20 5×5 25 and = = . − 3 × 4 − 12 − 3 × 5 − 15 4 (c) Four rational number equivalent to are 9 12 4×2 8 4×3 , = , = 27 9 × 2 18 9 × 3 4×4 16 4×5 20 and = = . 9 × 4 36 9×5 45 −18 3. (a) We have, 45



∴ 18 = 2 × 3 × 3 and 45 = 3 × 3 × 5 ∴ HCF of 18 and 45 = 3 × 3 = 9 On dividing numerator and denominator by their HCF, we get −18 −18 ÷ 9 −2 = = 45 45 ÷ 9 5



Hence, the standard form of

(b) We have, –

−18 −2 is . 45 5

12 18

∴ 12 = 2 × 2 × 3 and 18 = 2 × 3 × 3 ∴ HCF of 12 and 18 = 2 × 3 = 6

S OLUT I ONS



4. (a) Given,

−12 −2 . is 18 3

−3 −6 −9 −12 , ... , , , 5 10 15 20

Here,

−3 is the rational number in standard form. 5

Now,

−3 −3 × 1 −6 −3 × 2 = , = , 5 5 × 1 10 5×2 −9 −3 × 3 −12 −3 × 4 = , = 15 5×3 20 5×4

or

−3 × 1 −3 −3 × 2 −6 = , = , 5 10 5×1 5×2 −3 × 3 −9 −3 × 4 −12 = , = 5×3 15 5 × 4 20

Thus, we observe a pattern in these numbers. The next four numbers are −3 × 5 −15 −3 × 6 −18 = , = , 5 × 5 25 5 × 6 30 −3 × 7 −21 −3 × 8 −24 = , = 5 × 7 35 5×8 40 Hence, the required four more rational numbers are −15 −18 −21 −24 , , and 25 30 35 40

(b) Given,

−1 −2 −3 , ... , , 4 8 12

Here,

−1 is the rational number in standard form. 4

Now,

−1 −1 × 1 −2 −1 × 2 −3 −1 × 3 = , = , = 4 8 4 ×1 4 × 2 12 4×3

or

−1 × 1 −1 −1 × 2 −2 −1 × 3 −3 = , = , = 4×4 4 4×2 8 4×3 12

Thus, we observe a pattern in these numbers. The next four numbers are −1 × 4 −4 −1 × 5 −5 −1 × 6 −6 = , = , = , 16 4 × 5 20 4 × 6 24 4×4 −1 × 7 −7 = 4×7 28 Hence, the required four more rational numbers are −4 −5 −6 −7 , , and 16 20 24 28

P-75

(c) Four more rational numbers in the pattern of −1 2 3 4 5 6 7 8 are and , , , , , . 6 − 12 − 18 − 24 − 30 − 36 − 42 − 48



(d) Four more rational numbers in the pattern

8 10 12 −2 2 4 6 14 , , are and , , , . − 12 − 15 − 18 3 − 3 − 6 − 9 − 21

5. (a) We have,



1 =

−1 −1 × 6 −6 = = 1 6 1×6

and,

0 =

0 0×6 0 = = 1 6 1×6

We know that – 6 < – 5 < – 4 < – 3 < – 2 < – 1 < 0 −6 −5 −4 −3 −2 −1 0 or < < < < < < 6 6 6 6 6 6 6 Hence, five rational numbers between – 1 and 0 are −5 −4 −3 −2 −1 and , , , . 6 6 6 6 6

(b) We have,

and,

– 2 = – 1 =

−2 −2 × 6 −12 = = 1 1×6 6 −1 −6 = 1 6

We know that – 12 < – 11 < – 10 < – 9 < – 8 < – 7 < – 6

Hence, five rational numbers between – 2 and – 1 are −11 −10 −9 −8 −7 and , , , . 6 6 6 6 6

(c) We have,

and,

−4 −4 × 9 −36 = = 5 45 5×9 −2 −2 × 15 −30 = = 3 3 × 15 45

We know that – 36 < – 35 < – 34 < – 33 < – 32 < – 31 < – 30 −36 −35 −34 −33 −32 −31 −30 or < < < < < < 45 45 45 45 45 45 45 −4 −2 and 5 3

Hence, five rational numbers between are −35 −34 −33 −32 −31 and , , , . 45 45 45 45 45

(d) We have,

−1 −1 × 3 −3 = = 2 2×3 6

2 2×2 4 = = 3 3×2 6 −3 −2 −1 0 1 2 3 4 Now, < < < < < < < . 6 6 6 6 6 6 4 6 and

−2 −1 0 1 2 and , , , 6 6 6 6 6 −1 2 numbers between and . 2 3

Thus,

−12 −11 −10 −9 −8 −7 −6 or < < < < < < 6 6 6 6 6 6 6

are five rational

qqq

WORKSHEET-69 Solutions

1. Clearly from the number line, we see that



P represents 2

1 2×3+1 6+1 7 = = = 3 3 3 3



Q represents 2

2 2×3+2 6+2 8 = = = 3 3 3 3



1×3+1 1 R represents −1 = 3 3

and S represents −1

3 means 3 parts out of 4 parts to the right of 4 0. Thus, point A on the number line represents the 3 rational number as shown below : 4

Now,

= −

−4 3+1 = 3 3

2 1×3+ 2 3+2 −5 =– = − = 3 3 3 3

2. (a) Firstly, draw a number line and 0, 1 at unit distance, divide the gap between 0 and 1 into 4 1 equal parts and show 1 part as . 4

P-76

A 0



1 4

2 4

3 4

1

−5 is less then 0 and greater than –1. So, it 8 will lie on the left of 0 on the number line at the 5 same distance as from 0 to the right. 8

(b) Here,

Firstly, draw the number line and mark 0, – 1 on it at unit distance divide the gap between 0 and – 1 −1 −5 into 8 equal parts and show 1 part, as . Now, 8 8 means 5 parts out of 8 parts to the left of 0.

M A T H E M A T I C S - VII

Thus, point B on the number line represents the −5 rational number as shown below. 8

−44 ( −44 ) ÷ 4 −11 = = 72 72 ÷ 4 18



Hence, the simplest form of –1 –7 –6 –5 –4 –3 –2 –1 0 8 8 8 8 8 8 8





−7 is greater than – 2 and less then – 1, so 4 it will lie between – 2 and – 1 to the left of 0 at the 7 same distance as from 0 to the right. 4

(c) Here,

Firstly, draw the number line and mark 0, – 1, – 2 on it at unit distance. Divide the gap between 0, –1, – 2 1 into 4 equal parts and show 1 part as . 4 Now,

−7 3  = −  1 +  means 1 unit distance and 3 4 4  

parts out of 4 parts (between – 1 and – 2) to the left of zero. –7 — 4

–1

1

Hence, the simplest form of

−8 −4 is . 10 5

4. (a) Expressing the given rational numbers in the standard from, we have −7 ÷ 7 −7 −1 = = 21 ÷ 7 21 3



3 3÷3 1 = = 9 9÷3 3

and,



−8 ( −8 ) ÷ 2 −4 = = 6 3 6÷2

Hence, the simplest form of



−8 −4 is . 6 3

−16 −16 ÷ 4 −4 = = 20 20 ÷ 4 5 20 ÷ ( −5) 20 −4 = ( −25) ÷ ( −5) = −25 5

and

Clearly, the standard form of the given two rational numbers is the same. Hence,





25 25 ÷ 5 5 = = 45 45 ÷ 5 9

Hence, the simplest form of

25 5 is . 45 9

(c) H.C.F. of 44 and 72 is 4.

−44 Dividing the numerator and denominator of 72 by 4, we get

S OLUT I ONS

−16 20 = . 20 −25

(c) Expressing the given rational numbers in the standard from, we have

(b) H.C.F. of 25 and 45 = 5.

25 Dividing the numerator and denominator of by 45 5, we get

−7 3 ≠ 21 9

(b) Expressing the given rational numbers in the standard form, we have



3. (a) H.C.F. of 8 and 6 = 2.

−8 Dividing the numerator and denominator of by 6 2, we get



−8 ( −8 ) ÷ 2 −4 = = 10 10 ÷ 2 5





−8 by 10

2, we get

Hence,

7 (d) Similarly can be represented as 8

0



Dividing the numerator and denominator of

0

7 – 8



(d) H.C.F. of 8 and 10 = 2.

Clearly, the standard form of the given two rational numbers is not the same.

–2





−44 −11 is . 72 18

and,

( −2 ) ÷ ( −1) −2 2 = ( −3) ÷ ( −1) = −3 3 2 2 = 3 3

Clearly, the standard form of the given two rational numbers is the same Hence,

−2 2 = −3 3

(d) Expressing the given rational numbers in the standard from, we have

−3 −3 = 5 5

P-77

−12 ( −12 ) ÷ 4 −3 = = 20 20 ÷ 4 5

and,

Clearly, the standard form of the given two rational numbers is the same. −3 −12 Hence, = 5 20 (e) Expressing the given rational numbers in the standard from, we have 8 8 ÷ ( −1) −8 = = −5 ( −5) ÷ ( −1) 5 −24 ( −24 ) ÷ 3 −8 = = 15 15 3 5 ÷ and, Clearly, the standard form of the given two rational numbers is the same. −24 −8 Hence, = 15 5 (f) Expressing the given rational numbers in the standard form, we have 1 1 = 3 3

−1 −1 = 9 9 Clearly, the standard form of the given two rational numbers is not the same. 1 −1 Hence, ≠ 3 9 and,



(g) Expressing the given rational numbers in the standard form, we have −5 ( −5) ÷ ( −1) 5 = = ( −9 ) ÷ ( −1) −9 9



5 5 ÷ ( −1) −5 = = −9 ( −9 ) ÷ ( −1) 9

and,

Clearly, the standard form of the given two rational numbers is not the same. −5 5 ≠ −9 −9

Hence,

qqq

WORKSHEET-70 Solutions



1. (a) Given,

2 5 , 3 2

 LCM of the denominators 3 and 2 = 6 2 2×2 4 ∴ = = 3 3×2 6

[an equivalent rational number] 5 5×3 15 = = 2 2×3 6

[an equivalent rational number] On comparing the numerators, we get 15 > 4 15 4 5 2 or > ∴ > 6 6 2 3 5 2 Hence, is greater than . 2 3

(b) Given,

−5 −4 , 6 3

 LCM of the denomination 6 and 3 = 6 −5 −5 × 1 −5 −4 −4 × 2 −8 ∴ = = and = = 6 6×1 6 3 3×2 6 On comparing the numerators, we get – 5 > – 8 −5 −8 −5 −4 ∴ or > > 6 6 6 3 −5 −4 Hence, is greater than . 6 3

P-78

(c) Given,

−3 2 , 4 −3

First, write the given rational numbers with positive 2 × ( −1) 2 −2 denominator, we have = = −3 × ( −1) −3 3 Thus, the given rational numbers with positive −3 −2 denominators are and . 4 3  LCM of the denominators 4 and 3 = 12 −8 −3 −3 × 3 −9 −2 −2 × 4 ∴ = = and = = 12 4 4×3 12 3 3×4 On comparing the numerators, we get – 9 < – 8 or

−9 −8 −3 −2 or < < 12 12 4 3

Hence,

(d) Given,

−2 −3 is greater than . 3 4 −1 1 , 4 4

We know that, every positive rational number is greater than every negative rational number. Hence,

1 −1 is greater than . 4 4

M A T H E M A T I C S - VII

On converting given rational numbers into −23 −19 improper fractions, we get , 7 5

14 14 × ( −1) −14 = = ( −16 ) × ( −1) −16 16 Now, L.C.M. of 8 and 16 is 16. −7 −7 × 2 −14 \ = = 8 8×2 16

 LCM of the denominators 7 and 5 = 35

and,



2 4 (e) Given, −3 , −3 7 5

−23 −23 × 5 −19 × 7 −115 = = and = = 7 7 × 5 5×7 35

−133 35

On comparing the numerators, we get – 115 > – 133 ∴ –115 > – 133 or

−23 −19 −115 −133 or > > 7 5 35 35 or −3

Hence, −3

2 4 > −3 7 5

2 4 is greater than −3 . 7 5

2. (a) Clearly,

−5 2 is a negative rational number and 7 3

is a positive rational number. We know that every negative rational number is less than every positive rational number. Therefore,

−5 2 < 7 3

(b) Clearly, denominator of the given rational numbers are positive. The denominators are 5 and 7. Their L.C.M. is 35. So, we first express each rational number with 35 as common denominator.

\

−4 × 7 −4 −28 = = × 5 7 5 35

and

−5 −5 × 5 −25 = = 7 7×5 35

Now, we compared the numerators of these rational numbers. We find that



\

– 28 < – 25 or

−28 −25 < 35 35

−4 −5 < 5 7

denominator of

−7 8

is positive. The

14 is negative. So, we express it −16

with positive denominator as follows :

S OLUT I ONS

−7 14 = 8 −16

\ (d) Clearly, the denominator of the given rational numbers are positive. The denominators are 5 and Their L.C.M. is 20. So, we first express each rational number with 20 as common denominator. −8 −8 × 4 −32 = = 5 20 5 × 4 −7 −7 × 5 −35 = = 4 4×5 20 Comparing the numerators, we find that −32 −35 – 32 > – 35 or > 20 20 and,

−8 −7 \ > 5 4 (e) First we write each one of the given rational numbers with positive denominator. 1 Clearly, denominator of is negative. So, −3 expressing it with positive denominator as follows : 1 1 × ( −1) −1 = = −3 ( −3) × ( −1) 3 −1 and the denominator of is positive 4 Now, L.C.M. of 3 and 4 is 12. −1 −1 × 4 −4 \ = = 3 3 × 4 12 −1 −1 × 3 −3 = = 4 4×3 12 Comparing the numerators, we find that −4 −3 – 4 < – 3 or < 12 12 and,

(c) First we write each one of the given rational numbers with positive denominator. Clearly, denominator of

−14 −14 = 16 16 Comparing the numerators, we find that – 14 = – 14 −14 −14 or = 16 16

\

1 −1 < −3 4

(f) First we write each one of the given rational numbers with positive denominator.

P-79

5 is negative. So, −11 expressing it with positive denominator as follows :

Clearly, denominator of

5 5 × ( −1) −5 = = −11 ( −11) × ( −1) 11



and the denominator of

−5 is positive. 11



– 5 = – 5 or

\

−5 5 = −11 11



−5 −5 = 11 11

(g) Since every negative rational number is less than 0,

−7 \ 0 > 6

Comparing the numerators, we find that

qqq

WORKSHEET-71 Solutions

1. The reciprocal of

− 6 11 is and, the reciprocal of 11 − 6

−8 5 is . 5 −8

2. (a) (b)

=

−13 6 −13 + 6 −7 + = = =–1 7 7 7 7

19  −7  +   = 19 + ( −7 ) = 19 − 7 = 12 5  5  5 5 5

3. (a)

−3 ( −3) × 7 −21 ×7 = = 5 5 5

−6 ( −6 ) × ( −2 ) 12 (b) × ( −2 ) = = 5 5 5 4. (a) We have, −3 1 ( −3) × 1 −3 = × = 4 7 4×7 28 (b) We have, 5. (a) We have, (b) We have,

 −1  1 ( −1) 11 = + additive inverse of   2 −  3 5 3 5 11 1 33 + 5 = + = 5 3 15 (b)

2 −5 2 × ( −5) −10 × = = 3 9 3×9 27 2 −7 1 × ( −7 ) −7 = = × 3 8 3×4 12 −6 5 ( −6 ) × 5 −30 = × = 7 7 7×7 49

 −3  3 −3 6. (a) The additive inverse of is –   = .  9 9 9



8. (a) To find

38 8 = 2 15 15

−3 2 + : 7 3

Clearly, denominators of the given numbers are positive. The L.C.M. of denominators 7 and 3 is 21. −3 2 and into forms in which Now, we express 7 3 both of them have the same denominator 21. −3 −3 × 3 −9 We have, = = 7 7×3 21 2 × 7 14 2 = = 3 3 × 7 21 −3 2 −9 14 −9 + 14 \ + = + = 7 3 21 21 21 5 = 21 and,



(b) To find

−5 ( −3) + : 6 11

Clearly, denominators of the given numbers are positive. The L.C.M. of denominators 6 and 11 is 66. Now, we express each rational number into forms in which both of them have the same denominators 66.



 −9  9 −9 (b) The additive inverse of is –   = .  11  11 11

We have

−5 −5 × 11 −55 = = 6 6 × 11 66



 5  −5 5 (c) The additive inverse of is −  7  = 7 .   7

and

−3 −3 × 6 −18 = = 11 11 × 6 66

7 2 7 ( −2 ) 7. (a) − = + 9 5 9 5 35 + ( −18 ) 17 = = 45 45

P-80

\

−5 ( −3) −55 ( −18 ) = + + 6 11 66 66



=

−55 + ( −18 ) 66

M A T H E M A T I C S - VII

=

9. (a)

(b)

−55 + ( −18 ) −73 = 66 66

5  −11  5 + ( −11) −6 −3 + = =  = 4  4  4 4 2 5 3 25 + 9 34 = + = 3 5 15 15

−9 22 −9 × 3 + 22 × 2 = + 10 15 30 −27 + 44 17 = = 30 30 (c)

(d)



−3 5 3 5 3 × 9 + 5 × 11 + = + = −11 9 11 9 99

=

27 + 55 82 = 99 99

−8 ( −2 ) ( −8 ) × 3 + ( −2 ) + = 19 57 57 −24 + ( −2 ) −26 = = 57 57 (e)

−2 −2 +0 = 3 3

(f)

1 3 −7 23 −7 × 5 + 3 × 23 = −2 + 4 = + 3 5 3 5 15 (g) −35 + 69 34 = = 15 15

qqq

WORKSHEET-72 1. (a) Clearly, the given rational numbers have a common and positive denominator. Arranging the numerators of given rational −3 −2 −1 numbers in ascending order, we get , , 5 5 5

or,

– 3 < – 2 < – 1 −3 −2 −1 < < 5 5 5

Hence, the given numbers when arranged in −3 −2 −1 ascending order are , , 5 5 5 (b) Clearly, the denominators of given rational numbers are positive. The denominators are 3, 9, 3. Their L.C.M. = 9. Writing the numbers so that they have a common denominator 9 as follows : −1 −1 × 3 −3 −2 −2 = = , = 3 9 9 9 3×3

−4 −4 × 3 −12 and = = 3 9 3×3 Arranging the numerators of these rational numbers in ascending order, we get – 12 < – 3 < – 2 −12 −3 −2 or, < < 9 9 9 −4 −1 −2 or, < < 3 3 9 Hence, the given numbers when arranged in −4 −1 −2 ascending order are , , . 3 3 9

S OLUT I ONS

(c) Clearly, the denominators of given rational numbers are positive. The denominators are 7, 2, 4. Their L.C.M. is 28. Writing the numbers so that they have a common denominators 28 as follows : −3 −3 × 4 −12 = = 7 × 7 4 28

Solutions

and,

−3 −3 × 14 −42 = = 2 2 × 14 28 − 3 × 7 −3 −21 = = 4×7 4 28

– 42 < – 21 < – 12 −42 −21 −12 or, < < 28 28 28 or,

−3 −3 −3 < < 2 4 7

Hence, the given numbers when arranged in −3 −3 −3 ascending order are , , . 2 4 7

2. (a)



7 × 3 − 17 × 2 7 17 = − 72 24 36

=

21 − 34 −13 = 72 72

5  −6  −  = 5 + 6 = 5+6×3 63  21  63 21 63 5 + 18 23 = = 63 63 (b)

−6  −7  −   = −6 + 7 = −6 × 15 + 7 × 13 13  15  13 15 195 −90 + 91 1 = = 195 195 (c)

P-81

−3 7 3 × 11  7 × 8 = − 8 11 88

(d)

=

1 −19 −19 − 6 × 9 −2 − 6 = −6= 9 9 9

(e)



89 1 =– 1 88 88

=

1 −19 − 54 −73 = = −8 9 9 9

9  −7  9 7 ×   = 9 × −7 = –  ×  2 4 2  4  2 4

3. (a)

=

−63 8

3 2 3×2 6 (e) = × = 11 5 11 × 5 55 3 × ( − 5) 3 −5 (f) × = =1 −5×3 −5 3



(b)

=

(d)

 27  =    10 



(e)



6 9   54  −6 9 × =–  ×  =–      55  5 11 5 11

(c)

=



−27 10

−54 55

−4 4 = −15 15

−1 3 −1 4 −1 ×4 −4 −1 ÷ = × = = = 8 4 8 3 8×3 24 6 −2 1 −2 7 −2× 7 −14 ÷ = × = = 13 7 13 × 1 13 13 1

−7 × 13 −7  −2  −7 13 ÷  = × = 12 ×( −2 ) 12  13  12 −2 −91 91 = = −24 24

(f)

3  −4  3 65 3 5 ÷  = × = × 13  65  13 −4 1 −4 3×5 15 15 = = =– −4 4 1 × ( −4 )

3 -2 3 2 −6 × =–  ×  = 7  5  7 5 35

(d)

−4 × 1 −4 −4 1 × ÷ – 3 = = 5 ×( −3) 5 5 −3

(c)

 3 9 3 3 −9 (b) × (– 9) = =  ×  ×  10 1  10 10 1

=

2 3 −4 × 3 −12 =–4× = = =–6 3 2 2 2 −3 −3 1 −3 × 1 −3 × = ÷ 2 = = 5 5 2 10 5×2

–4÷

4. (a)

(g)

qqq

WORKSHEET-73 Solutions

=–

18 35

1. negative

2. positive

3. 2/9

4. – 3/4

5. left

6. right

7. less

8. less

9. different

10. same

 11  5 ( - 11) × 5  -  × = 4 7 4×7 55 =– 28

11. – 2/3

12. – 1/5



13. – 1

14. 1/2



-5 x = 8 -32

15.

or, or,

16. (a)

P-82



(–5) × (–32) = 8x x =

5 × 32 = 20 8

6  3 6 × ( - 3) × -  = 7  5 7×5

(b)

17. (a)



 7 5 5 7 – -  = +   3 7 7 3

5×3 7×7 + 7 ×3 3×7 15 49 = + 21 21 15 + 49 = 21 =



=

64 21

M A T H E M A T I C S - VII



(b)

5 -3 5 3 -  = +  8 8 8  8 

4 7 18. (a) and 3 5 Similarly Clearly ∴ (b)

=

5 + ( - 3) 8

=

2 1 = 8 4

4 5 20 × = 3 5 15 7 3 21 × = 5 3 15 20 21 < 15 15 4 7 < 3 5

3 7 and 4 11

L.C.M. of 4 and 11 = 44 3 3 × 11 33 ∴ = = 4 4 × 11 44 7 7×4 28 and = = 11 11 × 4 44

33 > 44 3 > 4

Since,

28 44

7 ∴ . 11 19. Given pattern is 1 2 3 4 – ,– ,– ,– ..... 3 6 9 12 ( - 1) × 1 1 Here, – = 3 3×1 ( − 1) × 2 2 – = 3 3×2 ( − 1) × 3 1 – = 3 3×3 ( − 1) × 4 4 = 12 3×4 Hence, next four numbers are ( − 1) × 5 5 = – 3×5 15 and

–



( − 1) × 6 6 = – 18 3×6 ( − 1) × 7 7 = – 3×7 21



( − 1) × 8 8 = – . 3×8 24



qqq

WORKSHEET-74 Solutions 1. 0 2. 0 3. 5/2 4. – 1/1 5. b ÷ m 6. positive, negative 7. simplest 8. zero 9. 1 10. – 36 11. 12 12. – 1 13. 0 (zero) 14. 9/49 15. 1 16. Q Length of cloth = 27 m [given] Since, 12 shirts of equal size is to be prepared. ∴ Length of cloth required for each shirt

S OLUT I ONS

= 27 ÷ 12 m =

27 m 12

Q

27 = 3 × 3 × 3

and

12 = 2 × 2 × 3

3 3 3 0

27 9 3 1

2 12 2 6 3 3 1

∴ HCF of 27 and 12 = 3 On dividing numerator and denominator by their HCF, we get

27 ÷ 3 9 = = 2.25 m 12 ÷ 3 4

So, for each shirt 2.25 m cloth is required.

17. (a) Sequence

2 7 8 13 , , , 5 10 15 30

L.C.M. of 5, 10, 15, 30 = 30 Sequence be

P-83

2 × 6 7 × 3 8 × 2 13 , , , 5 × 6 10 × 3 15 × 2 30 12 21 16 13 or , , , 30 30 30 30 Its ascending order is

12 13 < < 30 30 2 13 < < 5 30



16 21 < 30 30

8 7 or < 15 10 (b) L.C.M. and to find ascending order. (c) In a class, the students should stand in ascending order of height.

qqq

WORKSHEET-75 6 5 4 3 2 1 –1<– <− <− <− <− <− <0 7 7 7 7 7 7

Solutions 1. (a) 2. (b) 3. (c) 4. (c) 5. (c) 6. < 7. > 8. < 9. < 10. = 11.

Hence six rational numbers between – 1 to 0 are 6 5 4 3 2 1 − ,− ,− ,− ,− ,− 7 7 7 7 7 7

3 4

(b) To find more rational numbers between two rational numbers.



(c) Value : We should give space to others.



14. (a)

7 × 3 − 17 × 4 7 17 = − 144 48 36

=

12. (a) Given,

−3 2 + 7 3

 LCM of 7 and 3 = 21 −3 −3 × 3 2×7 −9 2 14 So, = = and = = 21 7 7×3 3×7 3 21 −3 2 −9 14 −9 + 14 5 ∴ + = = = + 7 3 21 21 21 21 −5 −3 (b) Given, + 6 11  LCM of 6 and 11 = 66 −5 −5 × 11 −55 −3 −3 × 6 −18 ∴ = = and = = 6 6 × 11 66 11 6 × 11 66 Now,



−5 −3 −55 −18 −55 + ( −18 ) = = + + 6 11 66 66 66

−55 − 18 −73 = = 66 66 13. (a) We have 7 0 – 1 = – and 0 = 7 7 7 6 5 4 3 2 1 0 – <− <− <− < − <− <− < 7 7 7 7 7 7 7 7



(b)

21 − 68 47 = − 144 144

5  −6  −  = 5 + 6 63  21  63 21

=

5 + 6( 3) 63

=

5 + 18 23 = . 63 63



(c)

−6  −7  −   = −6 + 7 13  15  13 15

=

−6(15) + 7(13) 195

=

−90 + 91 1 = 195 195



(d)

−3 7 −3(11) − 7( 8 ) = − 8 11 88

=



1 −33 − 56 −89 = = −1 . 88 88 88

15. Let the total volume = 1; Body floats volume = Body submerged volume =

2 9

1 2 – 1 9

Q LCM of 1 and 9 = 9

P-84

M A T H E M A T I C S - VII

On multiplying numerator and denominator by their HCF, we get 1×9 9 = 1 × 9 9 9 2 7 ∴ Body submerged volume = – = 9 9 9

Body submerged volume : Body floats volume 2 7 : 9 9 On multiplying both sides by 9, we get ratio 7 2 × 9 : × 9 = 7 : 2 9 9

In rational number form =

7 2

qqq

S OLUT I ONS

P-85

CHAPTER SECTION

10 B PRACTICAL GEOMETRY WORKSHEET-76 Solutions

1. Steps for construction : (a) Draw a line AB and take a point C outside the line AB. (b) Draw a transversal line CD which intersect AB at any point D. L

X

C



3. Steps of Construction : (a) Draw a line l and taking a point Q on it and a point P outside it. (b) Join PQ and draw an arc taking Q as a centre which cuts the line l and PQ on N and S. R

m

T

P M

Y

L

A

N S Q D

P

(c) Draw any arc PQ taking D as a centre which meets AB and CD at P and Q respectively. (d) Draw the arc taking same radius and centre C which meets the line DC at N. (e) Taking P as centre and PQ as radius, draw an arc. (f) This same –—arc draw taking centre as C which cuts the arc LN at L. (g) Draw a line XY through CL. Line XY is parallel to AB. 2. Steps of Construction : (a) Draw a line l and take a point A on the line. (b) Taking A as a centre, draw an arc which cuts the line at point P and Q. (c) Taking the radius more than previous, draw two arcs taking P and Q as a centre, which cut to each other at B. (d) Draw the line AB which is ⊥ to l. (e) Taking arc of 4 cm and taking A as a centre, draw an arc which cuts the line AB at C. (f) Taking the same arc AQ. Draw an arc from point C which cuts the line AC at N. (g) Taking the arc QM. Draw an arc taking the point N as a centre which cuts the arc at L. (h) Draw a line m through CN. (i) Line m is the required line. C

L

m

N

Step I

: Firstly, we draw a rough sketch of triangle with given measures marked on it. X

Y

6

cm

5 cm

Z

Step II : Draw a line segment YZ of length 5 cm Y



5 cm

Z

i.e., YZ = 5 cm.

Step III : With Y as centre and radius 4.5 cm, draw an arc. 5 cm

Z

Step IV : With Z as centre and radius 6 cm, draw another arc to cut the previous arc at X.

M Q A

(c) Taking the same arc QN. Draw the arc taking P as a centre which meets PQ at L. (d) Taking the arc SN. Draw from L which cuts the previous arc at M. (e) Draw a line m through PM which is parallel to l. (f) Taking a point R on the line m and draw an arc of equal length QN from R which cuts the line m at T. (g) Taking T as centre draw an arc with a distance equal to us which cuts the previous arc at A. (h) Draw a line through AR which cuts the line l at B. (i) BQPR is a parallelogram. 4. Given, three sides of ∆XYZ are XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm. To construct a triangle, we use the following steps :

Y

B P

N

Q





P-86

l

B B

4.5 cm

A

l

Y

5 cm

Z

M A T H E M A T I C S - VII

Step V : Join XY and ZX.

Step II : Draw a line segment QR = 5.5 cm.

X

Q

4.5 cm

6

3.5 cm

cm

Q

Y

Z

5 cm

Hence, ∆XYZ is the required triangle. 5. Steps of Construction : (a) Draw a line segment BC = 5.5 cm. (b) With centre B and radius = 5.5 cm, draw an arc of the circle. (c) With centre C and radius = 5.5. cm, draw another arc intersecting the arc drawn in step 2 at A.

3.5 cm

Q

3.5 cm P

cm

m

(a) Draw a line segment BC = 6 cm.

(b) With centre B and radius = 2.5 cm, draw an arc of the circle. A

6.5

m

4c

C



(c) With centre C and radius = 6.5 cm, draw another arc intersecting the arc drawn in step 2 at A.



(d) Join AB and AC to obtain the required triangle.

m R

cm

6 cm

B

4c

R

7. Steps of Construction :

2.5 cm

5.5

cm



P

3.5 cm

m

4c

4c

5.5

3.5 cm

Thus, ∆PQR is the required isosceles triangle.

(d) Join AB and AC to obtain the required triangle. 6. Given, three sides of ∆PQR are PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. To construct a triangle with given sides, we use the following steps : Steps of Construction : Step I : Firstly, we draw a rough sketch with given measures marked on it.

Q

R

Step V : Join PQ and PR.

Q

C

5.5 cm

R

Step IV : Now, with R as centre and radius 4 cm, draw another arc to cut the previous arc at P.

A

B

R

Step III : With Q as centre and radius 4 cm, draw an arc.

On measuring, we find that ∠B= 90°.

qqq

WORKSHEET-77 Solutions



1. Since ∠E + ∠F = 110° + 80° = 190°, so this cannot be drawn as the sum of all the angles of a triangle is 180°.



(d) Join EF to obtain the required triangle DEF. X F

2. Steps of Construction :

(a) Draw a line segment DE = 5 cm.

(b) Draw ∠EDF = 90°



(c) With centre D and radius = 3 cm, draw an arc to intersect DX at F.

S OLUT I ONS

3 cm 90° D

5 cm

E

P-87



3. Steps of Construction :

Step II : Draw a line segment AB = 5.8 cm.



(a) Draw a line XY. Take a point A on the line.



(b) Taking radius 6 cm draw an arc which cuts the line at B. Z

A

B

5.8 cm

Step III : At point A, draw a ray AX making an angle of 60° with AB. X

C

6 cm 60º A X

A

6 cm

B

Y

(c) From B draw an angle of 110° i.e., ∠ABZ=110 °

B

5.8 cm

Step IV : At point B, draw a ray By making an angle of 30° with AB. X

(d) From point B draw an arc of radius 6 cm which cuts the line BZ at C.

Y

(e) Join AC to get the required triangle. 60º

4. Steps of Construction :

A



(a) Draw a line BX.



(b) Taking radius 7.5 cm draw an arc which cuts the line BX at C.

30º 5.8 cm

B

Step V : Let rays AX and BY intersect at C. X Y

(c) At point C draw an angle of 60°.

C

∠BCY = 60°.

60º Y

A A

30º 5.8 cm

B

Thus, ∆ABC is the required triangle. 6. PQ = 5 cm, ∠PQR = 105°, ∠QRP = 40°

5 cm

 Sum of angles of a triangle is 180° 60º B

X

C

7.5 cm



(d) At point C draw an arc of radius 5 cm which cuts the line CY at A.



(e) Join AB, ABC is required triangle.

5. Given, AB = 5.8 cm, ∠A = 60° and ∠B = 30°. To construct a triangle with one side and two angles, we use the following steps : Step I

: Firstly, we draw a rough sketch with measures marked on it.

\ ∠PQR + ∠QRP + ∠QPR = 180° ∴

105° + 40° + ∠QPR = 180°

∴

145° + ∠QPR = 180°



∠QPR = 180° – 145°



∠QPR = 35°

Steps of Construction : (a) Draw a line segment PQ of 5 cm. Y X R

C

60º A

P-88

105º

30º 5.8 cm

B

Q

35º 5 cm

P

M A T H E M A T I C S - VII



(b) From point P draw an angle of 35°, such that ∠QPX = 35°



(c) With R as centre and radius equal to hypotenuse 6 cm, draw an arc of intersect ray QX at P. X

(c) From point Q draw an angle of 105°, such that ∠PQY = 105°

P

(d) These lines PX and QY cut at R. PQR is required triangle.

6c

m

7. Steps of Construction :

(a) Draw a line segment QR = 8 cm.

90°



(b) Draw ∠XQR = 90°



(c) With R as centre and radius = 10 cm, draw an arc of intersect ray QX at P. X P

Q

R

4 cm

(d) Join RP to obtain the required triangle DPQR. 9. Steps of Construction : (a) Draw a line segment CB = 6 cm.( CB = AC = 6 cm) (b) Draw ∠BCX = 90° X

10

A

cm

90° 6 cm Q



8 cm

90°

(d) Join RP to obtain the required triangle DPQR. 8. Steps of Construction :

(a) Draw a line segment QR = 4 cm.

R

(b) Draw ∠XQR = 90°

C



6 cm

B

(c) With C as centre and radius = 6 cm, draw an arc to intersect CX at A. (d) Join BA to obtain the required triangle ABC.

qqq

S OLUT I ONS

P-89

CHAPTER SECTION

11 B PERIMETER AND AREA WORKSHEET-78 Solutions 1. We have, perimeter of a square park = 320 m  320  m = 80 m \ Side of the square park =   4  \ Area of the square park = (Side)2 = (80)2 m2 = 6400 m2 2. Let the breadth of the rectangular plot of land be b. It is given that length of the plot = 22 m and the area of the plot = 440 sq. m 22 × b = 440 440 or b = = 20 22 \ Breadth of the plot = 20 m Hence, perimeter of the plot = 2(length + breath) = 2(22 + 20) m = 84 m 3. Let the breadth of the rectangular sheet be b cm. It is given that its length is 35 cm and perimeter is 100 cm. \ 2(length + breadth) = 100 or 2(35 + b) = 100 or 35 + b = 50 or b = 50 – 35 = 15 \ Breadth of the sheet = 15 cm Hence, area of the sheet = length × breadth = (35 × 15) m2 = 525 m2 4. Given, length of a rectangular piece of land, l = 500 m and breadth of a rectangular piece of land, b = 300 m (a) Area of a rectangular piece of land = l × b = 500 × 300 = 150000 m2 Hence, the area of a rectangular piece of land is 150000 m2. (b) Cost of 1 m2 land = ` 10000 ∴ Cost of 150000 m2 land = ` 10000 × 150000 = ` 1500000000 Hence, the cost of 150000 m2 land at the rate of 10000 per m2 is ` 1500000000. 5. Area of the square park = (Side)­2 = (60)2 m2 = 3600 m2 Let the breadth of the rectangular park be b. The, its area = length × breadth = (90 × b) m2 \ Area of rectangle Park = Area of the square park \ 90 × b = 3600

3600 = 40 90 \ The breadth of the rectangular park is 40 m. 6. Example 1. Let the length of the rectangle = 5 cm and breadth of the rectangle = 3 cm ∴ Area of a rectangle = Length × Breadth = 5 × 3 = 15 cm2 and perimeter of a rectangle = 2(Length + Breadth) = 2(5 + 3) = 2 × 8 = 16 cm Again, let the length of the rectangle = 6 cm and breadth of the rectangle = 5 cm ∴ Area of the rectangle = Length × Breadth = 6 ×5 = 30 cm2 and perimeter of a rectangle = 2 (Length + Breath) = 2 (6 + 5) = 2 × 11 = 22 cm. Thus, for both rectangles, area increases as the perimeter increases. Example 2. Let the side of the square = 4 cm ∴ Perimeter of the square = 4 × Side = 4 × 4 = 16 cm and Area of the square = (Side × Side) = 4 × 4 = 16 sq cm Again, let the side of the square = 5 cm ∴ Perimeter of the square = 4 × Side = 4 × 5 = 20 cm and Area of the square = Side × Side = 5 × 5 = 25 sq cm Thus, for both squares, area increases as the perimeter increases. 7. Example 1. Let length of the rectangle = 4 cm and breadth of the rectangle = 3 cm ∴ Area of the rectangle = l × b = 4 × 3 = 12 cm2 and perimeter of the rectangle = 2(l + b) = 2 (4 + 3) = 2 × 7 = 14 cm2 again, let the length of a rectangle = 6 cm and breadth of a rectangle = 2 cm ∴ Area of a rectangle = l × b = 6 × 2 = 12 cm2 Example 2. Let Length = 12 cm and Breadth = 1 cm ∴ Area = 12 × 1 cm2 = 12 cm2 Perimeter = 2(12 + 1) = 2 × 13 = 26 cm or

b =

qqq

P-90

M A T H E M A T I C S - VII

WORKSHEET-79 Solutions 1. Let the length of the rectangle be l cm. But its perimeter is 130 cm and the breadth is 30 cm. \ 2(length + breadth) = perimeter or 2(l + 30) = 130 or l + 30 = 65 or l = 65 – 30 = 35 \ Length of the rectangle = 35 cm Area of the rectangle = length × breadth = (35 × 30) cm2 = 1050 cm2 2. Total area of the wall = length × breadth = (4.5 × 3.6) m2 = 16.2 m2

Area of the door = length × breadth

= (2 × 1) m2 = 2 m2 \ Area to be white washed = (16.2 – 2) m2 = 14.2 m2 Cost of white washing at the rate of ` 20 per m2 = ` (20 × 14.2) = ` 284 3. Since rectangles of the length 6 cm and breadth 4 cm composed of congruent polygons, therefore. Area of each polygon = Area of the rectangle = (l × b) = (6 × 4) cm2 = 24 cm2 4. Given, length of the rectangular shape = 40 cm and breadth of the rectangular shape = 22 cm ∴ Length of wire = Perimeter of the rectangular shape = 2 (l + b) = 2 (40 + 22) = 2 × 62 = 124 cm Since, this wire is rebent in the shape of square. ∴ Perimeter of the square = Length of wire or 4 × Side = 124 cm on dividing both sides by 4, we get,

124 Side = = 31 cm 4 Hence, the side of a square shape is 31 cm. ∴ area of a rectangular shape = l × b = 40 × 22 = 880 cm2 and area of square shape = (Side)2 = (31)2 = 961 cm2 Hence, area of square shape is more than the area of rectangular shape. 5. (a) Area = base × height = (8 × 3.5) cm2 = 28 cm2 (b) Area = base × height = (8 × 2.5) cm2 = 20 cm2 (c) Area of parallelogram ABCD = AB × length of from C on AB = (7.2 × 4.5) cm2 = 32.4 cm2 8. Let r be the radius of the circle. Then, circumference = 2pr and, circumference = length of the wire or 2pr = 44 22 or 2× × r = 44 7 r = 1 or r = 7 7 Area of the circle of radius 7 cm  22  2 =  × 7 × 7 cm  7  or

= 154 cm2 Let the side of the square be x. Then, perimeter of the square = length of the wire 4x = 44 or x = 11 \ Side of the square = 11 cm Area of the square = (11)2 cm2 = 121 cm2 Thus, the circle encloses more area than the square.

qqq

WORKSHEET-80 Solutions

1. (a) Area of the triangle =

1 × base × height 2

1  2 =  × 4 × 3 cm 2  = 6 cm2 1 (b) Area of the triangle = × base × height 2

S OLUT I ONS

1  2 =  × 5 × 3.2 cm 2  = 8 cm2 1 (c) Area of the triangle = × base × height 2 1  2 =  × 3 × 4 cm 2  = 6 cm2

P-91

A = b × h = (5 cm × 3) cm2 = 15 cm2 ∴ Area = 15 cm2

1 × base × height 2

(d) Area of the triangle =

1  2 =  × 3 × 2 cm 2  = 3 cm2

3 cm

2. We know that

Area of a parallelogram = Base × Height



Area of parallelogram \ Base = Height

and

Height =

5 cm

(c) Base of parallelogram = 2.5 cm Height of parallelogram = 3.5 cm ∴ Area of parallelogram, A = b × h = 2.5 cm × 3.5 cm = (2.5 × 3.5) cm2 ∴ Area = 8.75 cm2

Areaof parallelogram Base

Therefore, the missing values are calculated as shown : S. No.

Area of the parallelogram

Base

Height

a

20 cm

246 =12.3 cm 20

b

154.5 =10.3 cm 15

15 cm

154.5 cm2

c

48.72 = 5.8 cm 8.4

8.4 cm

48.72 cm2

d

15.6 cm

16.38 =1.05 cm 15.6

16.38 cm2

3. (a) Base of parallelogram

3.5 cm

246 cm2 2.5 cm

(d) Base of parallelogram = 5 cm Height of parallelogram = 4.8 cm ∴ Area of parallelogram, A = b × h = 5 cm × 4.8 cm = (5 × 4.8) cm2 = 24.0 cm2 ∴ Area = 24 cm2

= 7 cm 5 cm

Height of parallelogram = 4 cm ∴ Area of parallelogram,

A = b × h

= 7 cm × 4 cm = 28 cm2

4 cm

4.8 cm



(e) Base of parallelogram = 2 cm Height of parallelogram = 4.4 cm ∴ Area of parallelogram, A = b × h = 2 cm × 4.4 cm = (2 × 4.4) cm2 = 8.8 cm2 ∴ Area = 8.8 cm2

7 cm

2 cm

(b) Base of parallelogram = 5 cm Height of parallelogram = 3 cm ∴ Area of parallelogram,

4.4 cm

qqq

P-92

M A T H E M A T I C S - VII

WORKSHEET-81 Solutions

1. We know that



Area of a triangle = D =

i.e.,

1 × base × height 2 1 BH 2 2∆ H

or

B =

and,

2∆ H = B

Therefore, the missing values are calculated as shown : Base

Height

Area of triangle

15 cm

2×87 15 = 11.6 cm

87 cm2

31.4 mm

1256 mm2

2×170.5 22

170.5 cm2

2×1256 31.4

1  2 2 =  × 5 × 12 cm = 30 cm 2  Again, area of DABC =

1 × BC × AD 2

or

30 =

1 × BC × AD 2

or

AD =

2 × 30 60 = cm 13 13

= 4

5.

Area of D ABC =

8 cm 13

1 × BC × AD 2

1  2 2 =  × 9 × 6 cm = 27 cm 2 

= 80 mm 22 cm

4. Since D ABC is right triangle at A 1 \ Area of D ABC = × AB × AC 2

= 15.5 cm 2. (a) The area of the parallelogram PQRS = base × corresponding height = SR × QM = (12 × 7.6) cm2 = 91.2 cm2 (b) The area of the parallelogram PQRS = base × corresponding height or PS × QN = Area of ||gm PQRS or 8 × QN = 91.2 [Using part (a)] 91.2 or QN = 8 = 11.4 cm 3. We have, Area of ||gm ABCD = 1470 cm2, AB = 35 cm and AD = 49 cm Area of ||gm ABCD = AD × BM or 1470 = 49 × BM 1470 or BM = = 30 cm 49 Again, area of ||gm ABCD = AB × DL or 1470 = 35 × DL 1470 or DL = = 42 cm 35

or

Also, area of DABC =

27 =

1 × AB × corresponding 2 height

1 × 7.5 × height from C to AB 2

or height from C to AB i.e., CE =

2 × 27 = 7.2 cm 7.5

6. We know that the circumference C of a circle of radius r is given by C = 2pr. (a) Here, r = 14 cm \ C = Circumference = 2pr 22   × 14 cm =  2 ×   7 = 88 cm (b) Here, r = 28 mm \ C = Circumference = 2pr 22   × 28 mm = 176 mm =  2 ×   7 (c) Radius = 21 cm Circumference = 2pr 22 = 2 × × 21 7 = 2 × 22 × 3 = 132 cm

qqq S OLUT I ONS

P-93

WORKSHEET-82 Solutions

1. Length of the lace required

= Circumference of the circular table  1.5  = 2pr, where r =   = 0.75 m  2  = (2 × 3.14 × 0.75) m = 4.71 m Cost of the lace at the rate of ` 15 per metre = ` (15 × 4.71) = ` 70.65 2. Perimeter of the given figure

1  =  × 2 πr + 2r  cm, where r = 5 cm 2 



22 1  =  ×2× × 5 + 10 cm 2  7



 110   110 + 70  + 10 cm =  cm =   7    7 =

180 cm = 25.7 cm (approx.) 7

3. Area of the circular table-top = pr2, 1.6 where r = = 0.8 m 2

= (3.14 × 0.8 × 0.8) m2 = 2.0096 m2 Cost of polishing the circular table-top at the rate of ` 15 per sq. m = ` (15 × 2.0096) 14.08    211.2  =  15 ×  = `    7  7  = ` 30.144 = ` 30.14 (approx.) 4. We know that the area A of a circle of radius r is given by A = pr2 (a) Here, r = 14 mm \ A = Area = pr2  22  2 =  × 14 × 14 mm  7  = (22 × 2 × 14) mm2 = 616 mm2 49 (b) Here, diameter = 49 m or r = m 2 A = Area = pr2  22 49 49  2 × m =  ×  7 2 2 \

49  2  11 =  × 7 ×  m  1 2

 3773  2 2 =   m = 1886.5 m  2 

P-94

(c) Here, \

r = 5 cm A = Area = pr2  22  2 =  × 5 × 5 cm  7  =

550 cm 2 = 78.57 cm2 7

5. Let r be the radius of the circle. Then, Circumference = 154 m or 2pr = 154 [ C = 2pr] 22 2× × r = 154 7 or 7 or r = 154 × 44 49 = m = 24.5 m 2 Area of the circle = pr2  22 49 49  2 × m =  ×  7 2 2 49  2  11 =  × 7 ×  m  1 2 3773 2 m = 1886.5 m 2 2 6. Given, diameter of the circular garden = 21 m ∴ Circumference of the circular garden = πd 22 = × 21 = 66 m 7 ∵ Length of the rope needed to make one round of fence = Circumference of the circular garden = 66 m ∴ Length of the rope needed to make two rounds of fence = 2 × 66 = 132 m Now, cost of 1 m rope = ` 4 ∴ Cost of 132 m rope = ` (4 × 132) = ` 528 Hence, the cost of the rope is ` 528. 7. Given, radius of a circular sheet (outer circle), R = 4 cm and radius of the removed circular sheet (inner circle), (r) = 3 cm ∴ We know that, Area of a circular sheet = π × (Radius)2 ∴ Area of circular sheet of radius 4 cm = π × (4)2 = 3.14 × 16 = 50.24 cm2 and area of circular sheet of radius 3 cm = π × (3)2 = 3.14 × 9 = 28.26 cm2 ∴ Area of remaining sheet = Area of circular sheet of radius 4 cm – Area of circular sheet of radius 3 cm = 50.24 – 28.26 = 21.98 cm2 Hence, the area of the remaining sheet is 21.98 cm2

=

qqq

M A T H E M A T I C S - VII

WORKSHEET-83 Solutions

=

1. Area of the square aluminium sheet = (6)2 cm2 = 36 cm2 Area of the circle cut out from the sheet = (3.14 × 2 × 2) cm2 = 12.56 cm2 Area of the sheet left over = (36 – 12.56) cm2 = 23.44 cm2 2. Let the radius of the circle be r. Then, circumference = 31.4 cm or 2pr = 31.4 or 2 × 3.14 × r = 31.4 31.4 or r = =5 2 × 3.14 \ Radius = 5 cm Area of the circle = pr2 = (3.14 × 25) cm2 = 78.5 cm2 3. Let r be the radius of flower garden. Then, Area = 314 m2

or

pr2 = 314

or

3.14 × r2 = 314

or

 314  r2 =   3.14 

or

r2 = 100

or

r =

100

or

r = 10 cm

Since radius of the garden is less than the radius of the area covered by the sprinkler, therefore, the sprinkler can water the entire garden. 4. Here, outer radius, R = 19 m and

inner radius, r = (19 – 10) m = 9 m

Circumference of the inner circle = 2pr = (2 × 3.14 × 9) m = 56.52 m Circumference of the outer circle = 2pR = (2 × 3.14 × 19) m = 119.32 m 5. Distance covered by the wheel in one revolution 22   × 28 cm = 2pr =  2 ×   7 = 176 cm Number of revolutions taken to cover 352 m

S OLUT I ONS

352 × 100 = 200 176

6. We know that the minute hand describes one complete revolution in one hour. \ Distance covered by its tip = Circumference of the circle of radius 15 cm = (2 × 3.14 × 15) cm = 94.2 cm 7. Area of the remaining sheet = Area of the circular card sheet – 2 × Area of small circle – Area of the rectangle 22 22 = × 14 × 14 − 3 × 1 − 2 × × ( 3.5)2 7 7 = 616 – 3 – 77 = 536 cm2 8. Given, Radius of the flower bed = 33 m Radius of the flower bed including the path = (33 + 4) m = 37 m Area of the path = [p(37)2 – p(33)2] m2 = p(372 – 332) m2 = p(37 + 33) (37 – 33) m2 = (3.14 × 70 × 4) m2 = 879.20 m2 9. (a) 1 cm2 = 1 cm × 1 cm = 10 mm × 10 mm  (As 1 cm = 10 mm) 2 = 100 mm \ 50 cm2 = (50 × 100) mm2 = 5000 mm2 (b) We know that a square of side 100 m has an area of 1 hectare (ha). \ 1 hectare = 100 m × 100 m = 10000 m2 \ 2 hectares = 20000 m2 (c) 1 m2 = 1 m × 1 m = 100 cm × 100 cm  (As 1 m = 100 cm) 2 = 10000 cm \ 10 m2 = 10 × 10000 cm2 = 100000 cm2 (d) Since 10000 cm2 = 1 m2  1  × 1000 m 2 \ 1000 cm2 =   10000  =

1 2 2 m = 0.1 m 10

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P-95

WORKSHEET-84 Solutions 1. Let ABCD be the garden, and let PQRS be the external boundaries of the path. D 75 m

90 m

A P

85 cm

5m

R C

B

= 0.675 hectare 2. Let ABCD be the rectangular park, and let PQRS be the external boundaries of the path. 3m

S D

125 m

A P

B 131 m

1.5 m

D S

We have,

P-96

5 cm

5m

P A

C R

2m

Q 8m

5.5 m B

P

8m

Q

We have, AB = 5.5 m AD = 4 m PQ = (5.5 + 2.25 + 2.25) m = 10 m and, QR = (4 + 2.25 + 2.25) m = 8.5 m (a) Area of the verandah = Area of PQRS  – Area of ABCD = (10 × 8.5 × 5.5 × 4) m2 = (85 – 22) m2 = 63 m2

(b) Cost of cementing the floor of the verandah at the rate of ` 200 per m2 = ` (200 × 63) = ` 12600.

5. Let ABCD be a square garden, and let PQRS be the internal boundary of the path. 1m

D

C R

S 1m

Q

We have, AB = 125 m AD = 65 m PQ = (125 + 3 + 3) m = 131 m and, QR = (65 + 3 + 3) m = 71 m Now, Area of the path = Area of PQRS  – Area of ABCD = (131 × 71 – 125 × 65) m2 = (9301 – 8125) m2 = 1176 m2 3. Let ABCD be the cardboard, and let PQRS be the painting without margin.

1.5 m

4m

C 65 m

C

A

71 cm

3m

R

R

D

6.5 cm

100 m

Q

We have, AB = 90 m AD = 75 m PQ = (90 + 5 + 5) m = 100 m and, QR = (75 + 5 + 5) = 85 m Now, Area of the path = Area of PQRS  – Area of ABCD = (100 × 85 – 90 × 75) m2 = (8500 – 6750) m2 = 1750 m2 Area of the garden = Area of ABCD = (90 × 75) m2 = 6750 m2 6750 = hectare 10000

2.25 m

S

2.25 m

5m

S

and, PC = (5 – 1.5 – 1.5) cm = 2 cm Now, Area of margin = Area of ABCD  – Area of PQRS = (8 × 5 – 5 × 2) cm2 = (40 – 10) cm2 = 30 cm2 4. Let ABCD be the hall, and let PQRS be the external boundary of verandah.

B

AB = 8 cm BC = 5 cm PQ = (8 – 1.5 – 1.5) cm = 5 cm

28 m

Q

P A

B

30 m

We have,

AB = 30 m

and,

PQ = (30 – 1 – 1) m

= 28 m

(a) The area of the path = Area of square ABCD – Area of square PQRS 2

= (30 – 282) m2 = (30 + 28) (30 – 28) m2 = (58 × 2) m2 = 116 m2

(b) Area of the remaining portion of the garden

= Area of square PQRS = (28 × 28) m2 = 784 m2 Cost of planing the grass in PQRS at the rate of ` 40 per m2 = ` (40 × 784) = ` 31,360

M A T H E M A T I C S - VII

6. Let ABCD represents the rectangular park of length 700 m and breadth 300 m. Area of the cross roads is the area of shaded portion i.e., the area of the rectangle PQRS and the area of the rectangle EFGH. But here, area of square KLMN is common in both rectangles, so it will be subtracted. 10 m

= 300 × 10 + 700 × 10 – 10 × 10 = 3000 + 7000 – 100 = 10000 – 100 = 9900 m2 =

Q

P

D

of rectangle EFGH – Area of square KLMN = PS × PQ × EF × EH – KL × KN

C

E

L

K

10 m

H

N

M

S

R



F

10 m

300 m

G B

A

10 m 700 m

Also, area of the park excluding cross road

= Area of rectangular park – Area of cross road



= (700 × 300) m2 – 9900 m2



= 210000 m2 – 9900 m2



Now, and

PQ = 10 m PS = 300 m,





EH = 10 m and EF = 700 m





KL = 10 m and KN = 10 m



9900 hec = 0.99 hec. 10000 Q1hec = 10000 m 2    1 hec  ⇒ 1 m 2 = 10000  

= 200100 m2 = = 20.01 hec

200100 hec 10000 1   2 Q1 m = 10000 hec 



qqq

∴ Area of the road = Area of rectangle PQRS + Area

WORKSHEET-85 1. Length of the cord wrapped around the circular pipe = Circumference of the pipe = (2 × 3.14 × 4) cm = 25.12 cm Length of the cord wrapped around the square box = Perimeter of the square = 4 × side = (4 × 4) cm = 16 cm Clearly, 25.12 cm > 16 cm. Yes, (25.12 – 16) cm, i.e., 9.12 cm cord is left with Pragya. 2. (a) The area of the whole plot = (10 × 5) m2 = 50 m2 (b) The area of the flower bed = (3.14 × 2 × 2) m2 = 12.56 m2 (c) The area of the remaining portion of the plot = (50 – 12.56) m2 = 37.44 m2 (d) The circumference of the flower bed = (3.14 × 2 × 2) m = 12.56 m 3. Area of quadrilateral ABCD = Area (DABC) + Area (DACD)

S OLUT I ONS

1 1  2  × 22 × 3 + × 22 × 3 cm 2 2 = = (33 + 33) cm2 = 66 cm2 4. (a) Let ABCD and EFGH be the cross roads. We have,

AB = 90 m

and,

BC = 3 m

\ Area of the road ABCD = (90 × 3) m2 = 270 m2 Again,

EF = 60 m

and

FG = 3 m E H 3m

60 m

Solutions

D A

S

R

P

Q 3m

C B

F G 90 m

\ Area of the road EFGH = (60 × 3) m2 = 180 m2 Clearly, area PQRS is common to both the roads. We have, Area PQRS = (3 × 3) m2 = 9 m2

P-97

\ Total area used as roads = Area of road ABCD + Area of road EFGH – Area of PQRS = (270 + 180 – 9) m2 = 441 m2 (b) Cost of constructing the roads at the rate of ` 110 per cm2 is ` (110 × 441) = ` 48,510 5. (a) Area of the shaded region DCEFD = Area of road ABCD – Area of DEBC – Area of D EAF 1 1 = AB × BC – × EB × BC – × AE × AF 2 2 1 1   =  18 × 10 - × 8 × 10 − × 10 × 6  cm2 2 2  

= (180 – 40 – 30) cm2 = 110 cm2

(b) Area of the shaded portion TUQ

= Area of the square PQRS –Area (DTSU) – Area (DUSQ) – Area (DUSQ) 1 1 = (PQ)2 – × TS × SU – UR 2 2



× QR −

1 × PQ ×TP 2

1 1  =  20 × 20 − × 10 × 10 −  2 2



×10 × 20 −

= (400 – 50 – 100 – 100) cm2



= (400 – 250) cm2 = 150 cm2

1  × 20 × 10 cm 2  2

qqq

WORKSHEET-86 Solutions 1. 10,00,00,000 2. 2, 50,00,000 3. equal 4. Number & side 5. perimeter, Area 6. 18 cm2 7. 35 cm2 8. Let the radius of the circle be r. Then, 2πr = 33 33 33 7 21 i.e., r = = = × 2π 2 22 4 21 cm 4



Thus, radius is



693 22 21 21 So, area of the circle = πr = × × = 8 7 4 4 2

= 86.625 Thus, area of the circle is 86.625 cm2. 9. DE = EA + AD = (8 + 5) cm = 13 cm DE is the radius of the circle. Also, DB is the radius of the circle. Next, AC = DB [Since diagonals of a rectangle are equal in length] Therefore, AC = 13 cm. From ∆ADC, DC2 = AC2 – AD2 DC2 = 132 – 52 DC2 = 169 – 25 DC2 = 144

So,

DC = 12

Thus, length of DC is 12 cm. Hence, perimeter of the rectangle ABCD = 2(12 + 5) cm = 34 cm. 10.   Height = 3 cm









or,



or,



or,



or,

1 bh 2 1 36 = bh 2 1 36 = ×b×3 2 72 = b × 3

Area of triangle =

72 = b 3 b = 24 cm

∴ Base is 24 cm. 11. Area covered by swimming pool = 30 m × 20 m = 600 m2. Length of outer rectangle = (30 + 8 + 8) m = 46 m and its breadth = (20 + 5 + 5) m = 30 m So, the area of outer rectangle = 46 m × 30 m = 1380 m2 Area of cemented path = Area of outer rectangle – Area of swimming pool = (1380 – 600) m2 = 780 m2 Cost of cementing 1 m2 path = ` 200 So, total cost of cementing the path = 780 × ` 200 = ` 1,56,000

qqq

P-98

M A T H E M A T I C S - VII

CHAPTER SECTION

B 12

ALGEBRAIC EXPRESSION WORKSHEET-87

Solutions 1. Three expressions each having 4 terms may be taken as 2x + 2y + z – 4, 3x2 – 4y3 + z + 5, xy + yz + zx + 3. 2. Grouping the like terms together, we have 12x, – 25x, x; 12, – 25, 1 – 25y, 12y, y 3. In 7xy + 5, we first obtain xy, multiply it by 7 to get 7xy and add 5 to 7xy to get the expression 7xy + 5. In x2y, we obtain x2, and multiply it by y to get x2y. In 4x2 – 5x, we first obtain x2, and multiply it by 4 to get 4x2 and then obtain second term by multiplying x by the constraint 5. From 4x2, we subtract 5x to finally arrive at 4x2 – 5x. 4. The expression 5xy + 10 consists of two terms 5xy and 10. Then term 5xy is a product of 5, x and y i.e., 5, x and y are factors of the term 5xy. And, the term 10 has 2 factor i.e., 5 and 2. These terms and factors of the terms of an expression can be represented by a tree diagram as shown in the figure : 5xy + 10

5xy

5

x

10

y

5

8y + 3x2

3x2

8y

y

8

3

x

x

In 7mn – 4, the terms are 7mn and (– 4). To obtain 7mn, we multiply the variable m with another variable n to get mn and them multiply mn by 7. To obtain (– 4), we take the integer – 4. Its tree diagram is as under : 7 mn – 4

7 mn

7

m

–4

n

–4

In 2x2y, the only term is 2x2y. For term 2x2y, we first obtain x2, and multiply it by another variable y to get x2y and finally multiply x2y by 2 to get 2x2y. Its tree diagram is as on the next page : 2x2y

2

5. In 4x – 3y : The coefficient of x in 4x is 4. The coefficient of y in – 3y is – 3. In a + b + 5 : The coefficient of a is 1. The coefficient of b is 1. In 2y + 5 : The coefficient of y in 2y is 2. In 2xy : The coefficient of xy in 2xy is 2. The coefficient of y in 2xy is 2x. The coefficient of x in 2xy is 2y. 6. Monomials are a, xy and 7. Binomials are a + b, xy + 5 and 4mn + 7. Trinomials are ab + a + b, 5x2 – x + 2, 4pq – 3q + 5p and 4m – 7n + 10. 7. In 8y+ 3x2, the terms are 8y and 3x2. Then term 8y is obtained by multiplying y by the constant 8.

S OLUT I ONS

The term 3x2 is obtained by multiplying 3, x and x. Its tree diagram is as under :

y 2

x

x

8. The terms and their factors in the expressions are shown by tree diagrams as under : (a) (b) 1 + x + x2

x –3

x

–3

1

x

x

–3

1

x

(c)

x2

x

x

y – y3

– y3

y

y

–1

y

y

y

P-99

(d)

(c)

5xy 2+ 7x 2y

5y + 3y2

5y 3y

5 xy 2

5

x

(d)

7x 2y

y y

7 x

x

(e)

y

xy + 2x2y2

2b

–ab

–1



a

b

2

b

(f)

2

– 3a

b

–3

2

a

a

(b)

(a) (b)

Expression

Terms

Factors

– 4x + 5

– 4x

– 4, x

5

5

– 4x

– 4, x

5y

5, y

– 4x + 5y

3, y, y

xy

x, y

2x2y2

2, x, x, y, y

pq

p, q

q

q

1.2ab

1.2ab,

1.2, a, b,

– 2.4b

– 2.4b

– 2.4, b, 3.6, a

+ 3.6a

3.6a

pq + q

(e) –ab + 2b 2– 3a 2

5, y

2

(g)

3 1 x+ 4 4

(h)

0.1p2 +0.2q2

3 x 4

3 ,x 4

1 4

1 4

0.1p2

0.1, p, p

2

0.2, q, q

0.2q

qqq

WORKSHEET-88 Solutions 1. (a) In the given terms, like terms are grouped as under : – xy2, 2xy2; – 4yx2, 20x2y; 8x2, – 11x2, – 6x2; 7y, y; – 100x, 3x; and – 11yx, 2xy (b) In the given terms, like terms are grouped as under : 10pq, – 7qp, 78qp; 7p, 2405p; 8q, – 100q; – p2q2, 12q2p2; – 23, 41; – 5p2, 701p2; and 13p2q, qp2. 2. (a) (m – n) + (m + n) = m – n + m + n = m + m – n + n = (1 + 1)m + (– 1+ 1)n = 2m + 0n = 2m + 0 = 2m and, (m – n) – (m + n) = m – n – m – n = m – m – n – n = (1 – 1)m + (– 1 – 1)n = 0 m + (– 2)n = 0 – 2n = – 2n (b) (mn + 5 – 2) + (mn + 3)= (mn + 3) + (mn + 3) = 2mn + 6 (mn + 5 – 2) – (mn + 3) = (mn + 3) – (mn + 3) = 0

P-100

3. (a) Binomial (c) Trinomial (e) Trinomial (g) Binomial (i) Trinomial (k) Binomial 4. (a) like (c) unlike (e) unlike 5.

(b) Monomial (d) Monomial (f) Binomial (h) Monomial (j) Binomial (l) Trinomial (b) like (d) like (f) unlike

Expression (a)

5 – 3t2 2

3

Terms (which is not a constant)

Numerical coefficient

– 3t2

–3

(b)

1+t+t +t

t t2 t3

1 1 1

(c)

x + 2xy + 3y

x 2xy 3y

1 2 3

(d)

100m + 1000n

100m 1000n

100 1000

M A T H E M A T I C S - VII

(e)

– p2q2 + 7pq

– p2q2 7pq

–1 7

(f)

1.2a – 0.8b

1.2a – 0.8b

1.2 – 0.8

(g)

3.14r2

3.14r2

3.14

(h)

2 (l + b)

2l 2b

2 2

(i)

0.1y + 0.01y2

0.1y 0.01y2

0.1 0.01

6. (a)

Expression

Terms containing x

Coefficient

(i)

y2x + y

y2x

y2

(ii)

13y2 – 8yx

– 8yx

– 8y

(iii)

x+y+2

x

1

(iv)

5 + z + zx

zx

z

(v)

1 + x + xy

x xy

1 y

(vi)

12xy2 + 25

12xy2

12y2

(vii)

7x + xy2

7x & xy2

7 & y2

Expression

Terms containing y2

Coefficient of y2

8 – xy2

– xy2

–x

5y2

5

(b) (i) (ii)

2

5y + 7x

(iii)

2x2y – 15xy2 + 7y2

– 15xy2 7y2

– 15x 7

7. (a) 21b – 32 + 7b – 20b = 21b + 7b – 20b – 32 = (21 + 7 – 20) b – 32 = 8b – 32 (b) – z2 + 13z2 – 5z + 7z3 – 15z = 7z3 – z2 + 13z2 – 5z – 15z = 7z3 + (– 1 + 13) z2 + (– 5 – 15) z = 7z3 + 12z2 – 20z (c) p – (p – q) – q – (q – p) = p – p + q – q – q + p = (p – p + p) + (q – q – q) = p – q (d) 3a – 2b – ab – (a – b + ab) + 3ab + b – a = 3a – 2b – ab – a + b – ab + 3ab + b – a = 3a – a – a – 2b + b + b – ab – ab + 3ab =(3 – 1 – 1) a + (– 2 + 1 + 1) b + (– 1 – 1 + 3) ab = a + (0) b + ab = a + ab (e) 5x2y – 5x2 + 3yx2 – 3y2 + x2 – y2 + 8xy2 – 3y2 = 5x2y + 3yx2 – 5x2 + x2 – 3y2 – y2 – 3y2 + 8xy2 = (5 + 3) x2y + (– 5 + 1) x2 + (– 3 – 1 – 3) y2 + 8xy2 = 8x2y – 4x2 – 7y2 + 8xy2 (f) (3y2 + 5y – 4) – (8y – y2 – 4) = 3y2 + 5y – 4 – 8y + y2 + 4 = 3y2 + y2 + 5y – 8y – 4 + 4 = (3 + 1) y2 + (5 – 8) y + (– 4 + 4) = 4y2 – 3y

qqq

WORKSHEET-89 Solutions 1. Let P denote the required expression, then (3x2 – 4y2 + 5xy + 20) – P = – x2 – y2 + 6xy + 20 Hence, required expression P = (3x2 – 4y2 + 5xy + 20) – (– x2 – y2 + 6xy + 20) = 3x2 – 4y2 + 5xy + 20 + x2 + y2 – 6xy – 20 = 3x2 + x2 – 4y2 + y2 + 5xy – 6xy + 20 – 20 = (3 + 1) x2 + (– 4 + 1) y2 + (5 – 6) xy + (20 – 20) = 4x2 – 3y2 – xy 2. (a) Required expression is equal to the subtraction of x2 + xy + y2 from 2x2 + 3xy. Hence, required expression = (2x2 + 3xy) – (x2 + xy + y2) = 2x2 +3xy – x2 – xy – y2 = 2x2 – x2 + 3xy – xy – y2 = (2 – 1) x2 + (3 – 1) xy – y2 = x2 + 2xy – y2 (b) Let P denote the required expression, then (2a + 8b + 10) – P = – 3a + 7b + 16 Hence, required expression P

S OLUT I ONS

= (2a + 8b + 10) – (– 3a + 7b + 16) = 2a + 8b + 10 + 3a – 7b – 16 = 2a + 3a + 8b – 7b + 10 – 16 = (2 + 3) a + (8 – 7) b + (10 – 16) = 5a + b – 6 3. (a) The sum of 3x – y + 11 and – y – 11 is given by (3x – y + 11) + (– y – 11) = 3x – y + 11 – y – 11 = 3x – 2y Now, we have to subtract 3x – y – 11 from 3x – 2y. Required expression = (3x – 2y) – (3x – y – 11) = 3x – 2y – 3x + y + 11 = – y + 11 (b) The sum of 4 + 3x and 5 – 4x + 2x2 is given by (4 + 3x) + (5 – 4x + 2x2) = 4 + 3x + 5 – 4x + 2x2 = 9 – x + 2x2 The sum of 3x2 – 5x and – x2 + 2x + 5 is given by (3x2 – 5x) + ( – x2 + 2x + 5) = 3x2 – 5x – x2 + 2x + 5 = 2x2 – 3x + 5 Now, we have to subtract 2x2 – 3x + 5 from 9 – x + 2x2

P-101

∴ Required expression= (9 – x + 2x2) – (2x2 – 3x + 5) = 9 – x + 2x2 – 2x2 + 3x – 5 = 2x + 4. 4. (a) The required difference is given by y2 – (– 5x2) = y2 + 5x2 = y2 + 5x2 (b) The required difference is given by (– 12xy) – 6xy = (– 12 – 6) xy = – 18 xy (c) The required difference is given by (a + b) – (a – b) = a + b – a + b = a – a + b + b = (1 – 1) a + (1 + 1) b = 2b. (d) The required difference is given by b (5 – a) – a (b – 5) = 5b – ab – ab + 5a = 5a + 5b + (– 1 – 1) ab = 5a + 5b – 2ab (e) The required difference is given by (4m2 – 3mn + 8) – (– m2 + 5mn) = 4m2 – 3mn + 8 + m2 – 5mn = 4m2 + m2 – 3mn – 5mn + 8 = (4 + 1) m2 + (– 3 – 5) mn + 8 = 5m2 – 8mn + 8 (f) The required difference is given by (5x – 10) – (– x2 + 10x – 5) = 5x – 10 + x2 –10x + 5 = x2 + (5 – 10) x + (– 10 + 5) = x2 – 5x – 5 (g) The required difference is given by (3ab – 2a2 – 2b2) – (5a2 – 7ab + 5b2) = 3ab – 2a2 – 2b2 – 5a2 + 7ab – 5b2 = – 2a2 – 5a2 – 2b2 – 5b2 + 3ab + 7ab = (– 2 – 5) a2 + (– 2 – 5) b2 + (3 + 7) ab = – 7a2 – 7b2 + 10ab (h) The required difference is given by (5p2 + 3q2 – pq) – (4pq – 5q2 – 3p2) = 5p2 + 3q2 – pq – 4pq + 5q2 + 3p2 = 5p2 + 3p2 + 3p2 + 5q2 – pq – 4pq = (5 + 3) p2 + (3 + 5) q2 + (– 1 – 4) pq = 8p2 + 8q2 – 5pq 5. (a) Required sum = 3mn + (– 5mn) + 8mn + (– 4mn) = (3 – 5 + 8 – 4) mn = (11 – 9) mn = 2mn (b) Required sum = (t – 8tz) + (3tz – z) + (z – t) = t – 8tz + 3tz – z + z – t = t – t – 8tz + 3tz – z + z = (1 – 1) t + (– 8 + 3) tz + (– 1 + 1) z = (0) t + (– 5) tz + (0) z = 0 – 5tz + 0 = – 5tz

(c) Required sum = (– 7mn + 5) + (12mn + 2) + (9mn – 8) + (– 2mn – 3) = – 7mn + 5 + 12mn + 2 + 9mn – 8 – 2mn – 3 = – 7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3 = (– 7 + 12 + 9 – 2) mn + (5 + 2 – 8 – 3) = 12mn – 4 (d) Required sum = (a + b – 3) + (b – a + 3) + (a – b + 3) = a + b – 3 + b – a + 3 + a – b + 3 = (a – a + a) + (b + b – b) + (– 3 + 3 + 3) = (1 – 1 + 1) a + (1 + 1 – 1) b + 3 = a + b + 3 (e) Required sum = (14x + 10y – 12xy – 13) + (18 – 7x – 10y + 8xy) + 4xy = 14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy = 14x – 7x + 10y – 10y – 12xy + 8xy + 4xy – 13 + 18 = (14 – 7) x + (10 – 10) y + (– 12 + 8 + 4) xy + (– 13 + 18) = 7x + 5 (f) Required sum = (5m – 7n) + (3n – 4m + 2) + (2m – 3mn – 5) = 5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5 = 5m – 4m + 2m – 7n + 3n – 3mn + 2 – 5 = (5 – 4 + 2) m + (– 7 + 3) n – 3mn + (2 – 5) = 3m – 4n – 3mn – 3 (g) Required sum = 4x2y + (– 3xy2) + (– 5xy2) + 5x2y = 4x2y – 3xy2 – 5xy2 + 5x2y = 4x2y + 5x2y – 3xy2 – 5xy2 = (4 + 5) x2y + (– 3 – 5) xy2 = 9x2y – 8xy2 (h) Required sum = (3p2q2 – 4pq + 5) + (– 10p2q2) + (15 + 9pq + 7p2q2) = 3p2q2 – 4pq + 5 – 10p2q2 + 15 + 9pq + 7p2q2 = 3p2q2 – 10p2q2 + 7p2q2 – 4pq + 9pq + 5 + 15 = (3 – 10 + 7) p2q2 + (– 4 + 9) pq + (5 + 15) = (0) p2q2 + 5pq + 20 = 5pq + 20 (i) Required sum = (ab – 4a) + (4b – ab) + (4a – 4b) = ab – 4a + 4b – ab + 4a – 4b = ab – ab – 4a + 4a + 4b – 4b = (1 – 1) ab + (– 4 + 4) a + (4 – 4) b = (0) ab + (0) a + (0) b = 0 + 0 + 0 = 0 (j) Required sum = (x2 – y2 – 1) + (y2 – 1 – x2) + (1 – x2 – y2) = x2 – y2 – 1 + y2 – 1 – x2 + 1 – x2 – y2 = x2 – x2 – x2 – y2 + y2 – y2 – 1 – 1 + 1 = (1 – 1 – 1) x2 + (– 1 + 1 – 1) y2 + (– 1 – 1 + 1) = – x2 – y2 – 1

qqq

P-102

M A T H E M A T I C S - VII

WORKSHEET-90 Solutions 1. At x = 0, 2x2 + x – a = 5 (Given) or 2(0) + 0 – a = 5 or 0 + 0 – a = 5 or – a = 5 or a = – 5 2. When p = – 2, then, (a) 4p + 7 = 4 (– 2) + 7 = – 8 + 7 = – 1 (b) – 3p2 + 4p + 7 = – 3 (– 2)2 + 4 (– 2) + 7 = – 3 (4) – 8 + 7 = – 12 – 1 = – 13 (c) – 2p3 – 3p2 + 4p + 7 = – 2 (– 2)3 – 3 (– 2)2 + 4 (– 2) + 7 = – 2 (– 8) – 3 (4) – 8 + 7 = 16 – 12 – 8 + 7 = 23 – 20 = 3 3. When x = – 1, then, (a) 2x – 7 = 2 (– 1) – 7 = – 2 – 7 = – 9 (b) – x + 2 = – (– 1) + 2 = 1 + 2 = 3 (c) x2 + 2x + 1 = (– 1)2 + 2(– 1) + 1 = 1 – 2 + 1 = 2 – 2 = 0 (d) 2x2 – x – 2 = 2(– 1)2 – (– 1) – 2 = 2(1) + 1 – 2 = 2 + 1 – 2 = 1 4. When a = 2, b = – 2, then (a) a2 + b2 = (2)2 + (– 2)2 = 4 + 4 = 8 (b) a2 + ab + b2 = (2)2 + (2) (– 2) + (– 2)2 = 4 – 4 + 4 = 4 (c) a2 – b2 = (2)2 – (– 2)2 = 4 – 4 = 0 5. When a = 0, b = – 1, then, (a) 2a + 2b = 0 + 2(– 1) = – 2 (b) 2a2 + b2 + 1 = 0 + (– 1)2 + 1 = 1 + 1 = 2 (c) 2a2b + 2ab2 + ab = 0 + 0 + 0 = 0 (d) a2 + ab + 2 = 0 + 0 + 2 = 2 6. (a) When z = 10, then z3 – 3(z – 10) = (10)3 – 3(10 – 10) = 1000 – 3(0) = 1000 – 0 = 1000 (b) When p = – 10, then p2 – 2p – 100 = (– 10)2 – 2(– 10) – 100 = 100 + 20 – 100 = 20

7. When a = 5 and b = – 3, then 2(a2 + ab) + 3 – ab = 2a2 + 2ab + 3 – ab = 2a2 + ab + 3 At a = 5, b = – 3; 2a2 + ab + 3 = 2(5)2 + (5)(–3) + 3 = 2 × 25 – 15 + 3 = 50 – 15 + 3 = 38 8. When, m = 2, then, (a) m – 2 = 2 – 2 = 0 (b) 3m – 5 = 3 × 2 – 5 = 6 – 5 = 1 (c) 9 – 5m = 9 – 5 × 2 = 9 – 10 = – 1 (d) 3m2 – 2m – 7 = 3 (2)2 – 2 (2) – 7 = 3 (4) – 4 – 7 = 12 – 11 = 1 5m 5×2 (e) – 4 = –4=5–4=1 2 2 9. (a) x + 7 + 4(x – 5) = x + 7 + 4x – 20 = (x + 4x) + (7 – 20) = 5x – 13 At x = 2, 5x – 13 = 5(2) – 13 = 10 – 13 = – 3 (b) 3(x + 2) + 5x – 7 = 3x + 6 + 5x – 7 = (3x + 5x) + (6 – 7) = 8x – 1 At x = 2, 8x – 1 = 8(2) – 1 = 16 – 1 = 15 (c) 6x + 5(x – 2) = 6x + 5x – 10 = 11x – 10 At x = 2, 11x – 10 = 11 × 2 – 10 = 22 – 10 = 12. (d) 4(2x – 1) + 3x + 11 = 8x – 4 + 3x + 11 = (8x + 3x) + (– 4 + 11) = 11x + 7 At x = 2, 11x + 7 = 11 × 2 + 7 = 22 + 7 = 29 10. (a) 3x – 5 – x + 9 = 2x + 4 At x = 3, = 2(3) + 4 = 6 + 4 = 10 (b) 2 – 8x + 4x + 4 = 6 – 4x At x = 3, = 6 – 4(3) = 6 – 12 = – 6 (c) 3a + 5 – 8a + 1 = – 5a + 6 At a = – 1, = – 5(– 1) + 6 = 5 + 6 = 11 (d) 10 – 3b – 4 – 5b = 6 – 8b At b = – 2, = 6 – 8(– 2) = 6 + 16 = 22 (e) 2a – 2b – 4 – 5 + a = 3a – 2b – 9 At a = – 1, b = – 2, 3a – 2b – 9 = 3(– 1) – 2(– 2) – 9 = – 3 + 4 – 9 = – 12 + 4 = – 8

qqq S OLUT I ONS

P-103

WORKSHEET-91 Solutions

2. (a) We know that the number of segments required

1. The algebraic expression of the given cases are as follows : (a) Subtraction of z from y = y – z (b) We know that, sum of numbers x and y is (x + y). 1 Then, one-half of the sum = ( x + y ) 2 (c) The number z multiplied by itself = z × z = z2 (d) We know that, the product of numbers p and q is pq. 1 Then, one-fourth of the product = pq 4



(e) We know that, square of number x is x2 and y is y2. Then, their sum = x2 + y2 (f) We know that, the product of numbers m and n is mn and its three times is 3mn. Then, added 5 in this product, we get 3mn + 5 (g) We know that, the product of y and z is yz. Then, subtract yz from 10, we get 10 – yz (h) We know that, sum of numbers a and b is (a + b) and their product is ab. Then, subtract (a + b) from ab, we get ab – (a + b).

to form n digits of the kind

is (5n + 1).

Therefore, to form 5, 10, 100 digits of above kind the number of segments required are (5 × 5 + 1) = (25 + 1) = 26, (5 × 10 + 1) = 50 + 1 = 51 and (5 × 100 + 1) = 500 + 1 = 501, respectively.

(b) We know that the number of segments required to form n digits of the kind is (3n + 1). Therefore, to form 5, 10, 100 digits of above kind the number of segments required are (3 × 5 + 1) = 15 + 1 = 16, (3 × 10 + 1) = 30 + 1 = 31 and (3 × 100 + 1) = 300 = 1 = 301 respectively.



(c) We know that the number of segments required to form n digits of the kind

is (5n + 2). Therefore,

to form 5, 10, 100 digits of above kind the number of segments required are (5 × 5 + 2) = 25 + 2 = 27, (5 × 10 + 2) = 50 + 2 = 52 and (5 × 100 + 2) = 500 + 2 = 502 respectively.

3. The completed table of number patterns is as on the next page : S. No.

Expression

Terms 1st

2nd

3rd

4th

5th

...

10th

...

100th

...

(i)

2n – 1

1

3

5

7

9

—

19

—

199

—

(ii)

3n + 2

2

5

8

11

17

—

32

—

302

—

(iii)

4n + 1

5

9

13

17

21

—

41

—

401

—

(iv)

7n + 20

27

34

41

48

55

—

90

—

720

—

2

5

10

17

26

—

101

—

10,001

—

(v)

2

n +1

Because in (a) 100th term = 2 (100) – 1 = 200 – 1 = 199 (b) 5th term = 3 (5) + 2 = 15 + 2 = 17 10th term = 3 (10) + 2 = 30 + 2 = 32 and, 100th term = 3 (100) + 2 = 300 + 2 = 302 (Note : Here nth term is 3n + 2 and 1st term we took n = 0 so that 3 (0) + 2 = 2, for 2nd term we took n = 1 so that 3 (1) + 2 = 5 etc.) (c) 5th term = 4 (5) + 1 = 20 + 1 = 21 10th term = 4 (10) + 1 = 40 + 1 = 41 and, 100th term = 4 (100) + 1 = 400 + 1 = 401 (d) 5th term = 7 (5) + 20 = 35 + 20 = 55 10th term = 7 (10) + 20 = 70 + 20 = 90 and, 100th term = 7 (100) + 20 = 700 + 20 = 720 (e) 5th term = 52 + 1 = 25 + 1 = 26 10th term = 102 + 1 = 100 + 1 = 101

qqq

P-104

M A T H E M A T I C S - VII

WORKSHEET-92 Solutions



1. (c) 2. (a) 3. (c) 4. (d) 5. (b) 6. (b) 7. – a2 – 2b + 4. 8. – 3x3y2 + 2x2y3





– 3x3y2 – x2y3 – 5y4

Sum = – 3x3y2 – x2y3 – 5y4

Now,

x4 + x3y2 + x2y3 + y4



3 2



2 3

– 3x y – x y – 5y

(+) 4

(+) 3 2

2 1 2 4 2 y − y + 5 − y + y2 − 2 3 7 7 3

1 2 2 4 2 =  + y − + y + 5 − 2 3 3 7 7

3 6 = y2 − y + 3 3 7 6 = y2 − y + 3 7

– 3x2y3 – 5y4



=

4

(+) 2 3

Difference = x + 4x y +2x y + 6y4

9. (a) Value of a2 + b2 + 3ab at a = 1 and b = – 2

= (1)2 + (– 2)2 + 3(1)(– 2) = 1 + 4 – 6 = 5 – 6 = – 1 (b) Value of a3 + a2b + ab2 + b3 at a = 1 and b = – 2 = (1)3 + (1)2(– 2) + (1)(– 2)2 + (– 2)3 = 1 – 2 + 4 – 8 = 5 – 10 = – 5 2 2 1 2 4  2  10.  y − y + 5  −  y − y + 2  7 3 3  7 



6 = ( 2 )2 − ( 2 ) + 3 7 = 4−

12 +3 7

28 − 12 + 21 7 37 = . 7 =

[Q y = 2]

11. (a) Total amount deposited in account = (5x2 + 3x – 2) + (x2 – x + 5) = 5x2 + 3x – 2 + x2 – x + 5 ½ = 5x2 + x2 + 3x – x – 2 + 5 ½ = 6x2 + 2x + 3 ½ Balance amount = (6x2 + 2x + 3) – (2x2 – x – 3) = 6x2 + 2x + 3 – 2x2 + x + 3 ½ = 6x2 – 2x2 + 2x + x + 3 + 3 ½ = 4x2 + 3x + 6. ½ (b) Addition of algebraic expressions. ½ (c) Value : Economy is everywhere. ½

qqq

S OLUT I ONS

P-105

CHAPTER SECTION

13 B

EXPONENTS & POWERS WORKSHEET-93 5

2

128

2

64

In 4 , 4 is known as base and 4 as exponent. (c) 75 = 7 × 7 × 7 × 7 × 7 = 16807 5 In 7 , 7 is known as base and 5 as exponent. (d) 53 = 5 × 5 × 5= 125 3 In 5 , 5 is known as base and 3 as exponent. (e) 26 = 2 × 2 × 2 × 2 × 2 × 2 = 64 In 26, 2 is known as base and 6 as exponent. 5. (a) We have, 2 512 2 256 2 128 2 64 2 32 2 16 2 8 2 4 2 2 1 ∴ 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 29 (b) We have,

2

32

7

343

2

16

7

49

2

8

7

7

2

4

2

2

Solutions 1. (a) 26 = 2 × 2 × 2 × 2 × 2 × 2 = 64 (b) 93 = 9 × 9 × 9 = 729 (c) 112 = 11 × 11 = 121 (d) 54 = 5 × 5 × 5 × 5 = 625 2. (a) We have, 3

729

3

243

3

81

3

27

3

9

3

3

1 ∴ 729 = 3 × 3 × 3 × 3 × 3 × 3 = 36 (b) We have,

1 ∴ 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 27 (c) We have,

∴ (c) We have,

1 343 = 7 × 7 × 7 = 73 3

729

3

243

7

343

3

81

7

49

3

27

7

7

3

9

1

3

3

∴ 343 = 7 × 7 × 7 = 73 3. (a) 6 × 6 × 6 × 6 = 64 (b) t × t = t2 (c) b × b × b × b = b4 (d) 5 × 5 × 7 × 7 × 7 = 52 × 73 (e) 2 × 2 × a × a = 22 × a2 (f) a × a × a × c × c × c × c × d = a3 × c4 × d 4. Five examples are : (a) 34 = 3 × 3 × 3 × 3 = 81 In 34, 3 is known as base and 4 as exponent. (b) 45 = 4 × 4 × 4 × 4 × 4 = 1024

1 ∴ 729 = 3 × 3 × 3 × 3 × 3 × 3 = 36 (d) We have, 5

3125

5

625

5

125

5

25

5

5

1 ∴ 3125 = 5 × 5 × 5 × 5 × 5 = 55

qqq

P-106

M A T H E M A T I C S - VII

WORKSHEET-94 Solutions 1. (a) (– 4)3 = (– 4) × (– 4) × (– 4) = – 64 (b) (– 3) × (– 2)3 = (– 3) × (– 2) × (– 2) × (– 2) = (– 3) × (– 8) = 24

(c)

(– 3)2 × (– 5)2 = ( – 3 × – 5)2 = (15)2

= 15 × 15 = 225

(d)

(– 2)3 × (– 10)3 = (– 2 × – 10)3 = (20)3

= 20 × 20 × 20 = 8000 2. (a) (62)4 = 62 × 4 = 68 (b) (22)100 = 22 × 100 = 2200 (c) (750)2 = 750 × 2 = 7100 (d) (53)7 = 53 × 7 = 521 3. (a) We have,

2.7 × 1012 = 2.7 × 10 × 1011

= 27 × 1011, contains 13 digits and,

1.5 × 108 = 1.5 × 10 × 107

= 15 × 107, contains 9 digits Clearly, 2.7 × 1012 > 1.5 × 108

(b) We have, 4 × 1014, contains 15 digits and,

3 × 1017, contains 18 digits ∴

4 × 1014 < 3 × 1017

4. (i) 25 × 23 = 25 + 3 = 28 (ii)

p3 × p2 = p3 + 2 = p5

(iii)

43 × 42 = 43 + 2 = 45

(iv)

a3 × a2 × a7 = a3 + 2 + 7 = a12





(v)

53 × 57 × 512 = 53 + 7 + 12 = 522

(vi) (– 4)100 × (– 4)20 = (– 4)100 + 20 = (– 4)120 5. (a) We have 43 = 4 × 4 × 4 = 64 and, 34 = 3 × 3 × 3 × 3 = 81 Clearly, 81 > 64 So, 34 is greater. (b) We have, 53 = 5 × 5 × 5 = 125 and, 35 = 3 × 3 × 3 × 3 × 3 = 243 Clearly, 243 > 125 So, 35 is greater. (c) We have, 2 8 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256 and, 82 = 8 × 8 = 64 Clearly, 256 > 64 So, 28 is greater. (d) We have, 1002 = 100 × 100 = 10000 and 2100 = (210)10 = (2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2)10 = (1024)10 = [(1024)2]5 = (1024 × 1024)5 = (1048576)5

S OLUT I ONS

Clearly, 2100 is greater. (e) We have, 210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 and 102 = 10 × 10 = 100 Clearly, 1024 > 100 i.e., 210 is greater. 6. (a) 29 ÷ 23 = 29 – 3 = 26 (b) 108 ÷ 104 = 108 – 4 = 104 (c) 911 ÷ 97 = 911 – 7 = 94 15 (d) 20 ÷ 2013 = 2015 – 13 = 202 (e) 713 ÷ 710 = 713 – 10 = 73 7. (a) 43 × 23 = (4 × 4 × 4) × (2 × 2 × 2) = (4 × 2) × (4 × 2) × (4 × 2) = (4 × 2)3 = (8)3 (b) 25 × b5 = (2 × 2 × 2 × 2 × 2) × (b × b × b × b × b) = (2 × b) × (2 × b) × (2 × b) × (2 × b) × (2 × b) = (2 × b)5 = (2b)5 (c) a2 × t2 = (a × a) × (t × t) = (a × t) × (a × t) = (a × t)2 = (at)2 (d) 56 × (– 2)6 = (5 × 5 × 5 × 5 × 5 × 5) × (– 2 × – 2 × – 2 × – 2 × – 2 × – 2) = [5× (– 2)]6 = (– 10)6 (e) (– 2)4 × (– 3)4 = (– 2× – 2× – 2× – 2) × (– 3× – 3× – 3× – 3) 4 = [(– 2) × (– 3)4] = (6)4 8. (a) 45 ÷ 35 = =

(b)

25 ÷ b5 =

=

(c)

(– 2)3 ÷ b3 =

=

(d)

p4 ÷ q4 =

45 35

=

4 4 4 4 4 4 × × × × =   3 3 3 3 3 3 25 b

5

=

(e) 56 ÷ (– 2)6 =

5

2×2×2×2×2 b×b×b×b×b

2 2 2 2 2 2 × × × × =   b b b b b b

5

−2 × −2 × −2 b×b×b

 −2  −2 −2 −2 × × =   b b b  b  p4 q

4

p =   q

4×4×4×4×4 3×3×3×3×3

=

3

p×p×p×p q×q×q×q

4

56 ( −2 )6

P-107

=

5×5×5×5×5×5 −2 × −2 × −2 × −2 × −2 × −2

5 5 5 5 5 5 × × × × = × −2 −2 −2 −2 −2 −2 5 =    −2 

6

9. (a) To determine prime factorisation of 648, we use the division method as shown below : 2

648

2

324

2

162

3

81

3

27

3

9

3

3

1 ∴ 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3 = 23 × 34 (b) We use the division method as shown under :

3

45

3

15

5

5

1 ∴ 540 = 2 × 2 × 3 × 3 × 3 × 5 = 22 × 33 × 5. (d) We use the method of division as under : 2

3600

2

1800

2

900

2

450

3

225

3

75

5

25

5

5

1 ∴ 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 24 × 32 × 52 10. (a)

2 × 103 = 2 × 1000 = 2000

(b)

72 × 22 = (7 × 2)2 = (14)2

3

405

3

135

3

45



3

15

= 8 × 5 = 40

5

5

1 ∴ 405 = 3 × 3 × 3 × 3 × 5 = 34 × 5 (c) We use the division method as shown below : 2

540

2

270

3

135







(c) (d)

= 14 × 14 = 196

23 × 5 = 2 × 2 × 2 × 5 3 × 44 = 3 × 4 × 4 × 4 × 4

= 3 × 256 = 768

(e)

0 × 102 = 0



(f)

52 × 33 = 5 × 5 × 3 × 3 × 3

= 25 × 27 = 675 (g) 24 × 32 = 2 × 2 × 2 × 2 × 3 × 3 = 16 × 9 = 144 (h) 32 × 104 = 3 × 3 × 10 × 10 × 10 × 10 = 9 × 10000 = 90000

qqq

WORKSHEET-95 Solutions 1. (a) L.H.S. = 10 × 1011 = 101 + 11 = 1012 and, R.H.S. = 10011 = (102)11 = 102 × 11 = 1022 or

L.H.S. ≠ R.H.S.

So, statement is false. (b) and,

P-108

L.H.S. = 23 = 2 × 2 × 2 = 8 R.H.S. = 52 = 5 × 5 = 25

or L.H.S. ≠ R.H.S. So, statement is false. (c) L.H.S. = 23 × 32 = 8 × 9 = 72, R.H.S. = 6 × 6 × 6 × 6 × 6 = 7776 ∴ L.H.S. ≠ R.H.S. So, statement is false. (d) True. Because 30 = 1 and (1000)0 = 1 or L.H.S. = R.H.S. 210 × 7 3 210 × 7 3 ( 2 5 )2 × 7 3 = 9 2. (a) = 3 3 3 2 ×7 8 × 7 (2 ) × 7 = 210 – 9 × 73 – 1 = 21 × 72

M A T H E M A T I C S - VII

= 2 × 49 = 98 52 × 52 × t 8 25 × 5 2 × t 8 = (b) 3 4 3 4 10 × t ( 2 × 5) × t 54 × t 8 = 3 3 4 2 × 5 × t =

or

1

\ 108 × 192 = (2 × 2 × 3 × 3 × 3)  × (2 × 2 × 2 × 2 × 2 × 2 × 3) = 22 × 33 × 26 × 31 = 22 + 6 × 33 + 1 = 28 × 34 (b) We have, 2 270 3 135 3 45 3 15 5 5 1

270 = 2 × 3 × 3 × 3 × 5 = 2 × 33 × 5

(c) We have,



3 729 3 243 3 81 3 27 3 9 3 3 1



\

768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

= 28 × 3 It is the required prime factor product form. 4. (a) 32 × 34 × 38 = 32 + 4 + 8 = 314 (b) 615 ÷ 610 = 615 – 10 = 65 (c) a3 × a2 = a3 + 2 = a5 (d) 7x × 72 = 7x + 2

(52)3 ÷ 53 = 52 × 3 ÷ 53

(e)

= 56 ÷ 53 = 56 – 3 = 53

(f)

25 × 55 = (2 × 5)5 = (10)5



(g)

a4 × b4 = (ab)4



(h)



(i)

(34)3 = 34 × 3 = 312 20

15

(2 ÷ 2 ) × 23 = (220 – 15) × 23

= 25 × 23 = 25 + 3 = 28

(j)

5. (a)

8t ÷ 82 = 8t –2 23 × 34 × 4 23 × 34 × 2 2 = 3 × 32 3 × 25

[ 4 = 2 × 2 = 22 and 32 = 2 × 2 × 2 × 2 × 2 = 25] =

23 + 2 × 34 31 × 2 5

=

2 5 × 34 31 × 2 5

= 25 – 5 × 34 – 1 = 1 × 33 = 33

(b)

[(52)3 ×54)] ÷ 57 = (52 × 3 × 54) ÷ 57

= (56 × 54) ÷ 57 = 56 + 4 ÷ 57 = 510 ÷ 57 = 510 – 7 = 53

729 = 3 × 3 × 3 × 3 × 3 × 3 = 36

and,

6

64 = 2 × 2 × 2 × 2 × 2 × 2 = 2



2 768 2 384 2 192 2 96 2 48 2 24 2 12 2 6 3 3 1

= 20 × 33

2 64 2 32 2 16 2 8 2 4 2 2 1

\

S OLUT I ONS

(d) We have,

54 − 3 × t 8 − 4 23

5t 4 5 × t4 = = 8 23 5 5 5 3 × 10 × 25 3 × ( 2 × 5 )5 × 5 2 (c) = 7 5 5 ×6 57 × ( 2 × 3 )5 35 × 2 5 × 55 × 52 = 57 × 2 5 × 3 5 5–5 = 3 × 25 – 5 × 55 + 2 – 7 0 = 3 × 20 × 50 = 1 × 1 × 1 = 1 3. (a) We have, 2 192 2 108 2 96 2 54 2 48 3 27 2 24 3 9 2 12 3 3 2 6 1 3 3

\

729 × 64 = 36 × 26



(c)

254 ÷ 53 = (52)4 ÷ 53

= 52 × 4 ÷ 53 = 58 ÷ 53 = 58 – 3 = 55

P-109



(d)

3 × 7 2 × 118 3

21 × 11



=

3 × 7 2 × 118



3

3 × 7 × 11

= 3

1–1 0

×7

2–1

1

= 3 × 7 × 11

× 11

8–3

5

= 1 × 7 × 115 = 7 × 115

37

(e)

3 4 × 33

=

37 34 + 3

=

37 37

= 37 – 7 = 30 or 1 0

0

0



(f)

2 + 3 + 4 = 1 + 1 + 1 = 3



(g)

20 × 30 × 40 = 1 × 1 × 1 = 1



(h)

(30 + 20) × 50 = (1 + 1) × 1

= 2 × 1 = 2

(i)

28 × a5 4 3 × a3

=

28 × a5 2 3

(2 ) × a

3

=

28 × a5 2 6 × a3

= 28 – 6 × a5 – 3 = 22 × a2 = (2a)2

(j)

 a5   3  × a8 = (a5 – 3) × a8 a 

= a2 × a8 = a2 + 8 = a10

(k)

4 5 × a8b 3 4 5 × a 5b 2

= 45– 5 × a8 – 5 × b3 – 2

= 40 × a3 × b1 = 1 × a3 × b = a3b (l) (23 × 2)2 = (23 + 1)2 = (24)2 = 28

qqq

WORKSHEET-96 Solutions

1. (a) 172 = 1 × 100 + 7 × 10 + 2 × 1

= 1 × 102 + 7 × 101 + 2 × 100

(b) 5643 = 5 × 1000 + 6 × 100 + 4 × 10 + 3 × 1

= 5 × 103 + 6 × 102 + 4 × 101 + 3 × 100 (c) 56439 = 5 × 10000 + 6 × 1000 + 4 × 100 + 3 × 10 + 9 × 1 4 3 = 5 × 10 + 6 × 10 + 4 × 102 + 3 × 101 + 9 × 100 (d) 176428 = 1 × 100000 + 7 × 10000 + 6 × 1000 + 4 × 100 + 2 × 10 + 8 × 1 = 1 × 105 + 7 × 104 + 6 × 103 + 4 × 102 + 2 × 101 + 8 × 100 2. 279404 = 2 × 100000 + 7 × 10000 × 9 × 1000  + 4 × 100 + 0 × 10 + 4 × 1 = 2 × 105 + 7 × 104 + 9 × 103 + 4 × 102  + 0 × 101 + 4 × 100 3006194 = 3 × 1000000 + 0 × 100000 × 0 × 10000  + 6 × 1000 + 1 × 100 + 9 × 10 + 4 × 1 = 3 × 106 + 0 × 105 + 0 × 104 + 6 × 103  + 1 × 102 + 9 × 10 + 4 × 100 2806196 = 2 × 1000000 + 8 × 100000 + 0 × 10000  + 6 × 1000 + 1 × 100 + 9 × 10 + 6 × 1 = 2 × 106 + 8 × 105 + 0 × 104 + 6 × 103  + 1 × 102 + 9 × 10 + 6 × 10° 120719 = 1 × 100000 + 2 × 10000 + 0 × 1000  + 7 × 100 + 1 × 10 + 9 × 1 = 1 × 105 + 2 × 104 + 0 × 103 + 7 × 102  + 1 × 101 + 9 × 100

P-110

20068 = 2 × 10000 + 0 × 1000 + 0 × 100  + 6 × 10 + 8 × 1 = 2 × 104 + 0 × 103 + 0 × 102 + 6 × 101  + 8 × 100 3. (a) 50,000,000 = 5 × 10000000 = 5 × 107 (b) 70,000,000 = 7 × 10000000 = 7 × 107 (c) 3,186,500,000 = 3,1865 × 1,00,000 = 3.1865 × 109 (d) 390878 = 3.90878 × 100000 = 3.90878 × 105 (e) 39087.8 = 3.90878 × 10000 = 3.90878 × 104 (f) 3908.78 = 3.90878 × 1000 = 3.90878 × 103 4. (a) 8 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100 = 8 × 10000 + 6 × 1000 + 0 × 100  + 4 × 10 + 5 × 1 = 86045 (b) 4 × 105 + 5 × 103 + 3 × 102 + 2 × 100 = 4 × 100000 + 5 × 1000 + 3 × 100 + 2 = 405302 (c) 3 × 104 + 7 × 102 + 5 × 100 = 3 × 10000 + 7 × 100 + 5 × 1 = 30705 (d) 9 × 105 + 2 × 102 + 3 × 10 = 9 × 100000 + 2 × 100 + 30 = 900230 5. (a) The mean distance between Earth and Moon = 384000000 m = 3.84 × 100000000 m = 3.84 × 108 m

M A T H E M A T I C S - VII

(b) Speed of light in vacuum = 300000000 m/s = 3.0 × 100000000 m/s = 3.0 × 108 m/s (c) Diameter of the Earth = 12756000 m = 1.2756 × 10000000 m = 1.2756 × 107 m (d) Diameter of the Sun = 1400000000 m = 1.4 × 1000000000 m = 1.4 × 109 m (e) In a galaxy there are on an average = 100,000,000,000 stars = 1 × 100000000000 stars = 1 × 1011 stars (f) The universe is estimated to be 12,000,000,000 years old

= 1.2 × 10,000,000,000 = 1.2 × 1010 years (g) The distance of Sum from the centre of the Milky Way Galaxy is estimated to be = 300,000,000,000,000,000,000 = 3 × 100000000000000000000 = 3 × 1020 m (h) Number of molecules contained in a drop of water weighing 1.8 gm = 60,230,000,000,000,000,000,000 = 6.023 × 10000000000000000000000 = 6.023 × 1022 (i) The Earth has 1,353,000,000 cubic km of sea water i.e., 1.353 × 1000000000 = 1.353 × 109 cubic km. (j) The population of India was about in March 2001 = 1,027,000,000 = 1.027 × 1000000000 = 1.027 × 109

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WORKSHEET-97 Solutions 1. 44 2. 3 3. 11/15 4. 8 5. 12 6. 0 7. 32 8. 13/14 9. 11 10. 5 11. 6 12. 6 13. 3 14. 5.37 15. 8·888 16. 7 17. 8 18. False 19. True 20. True

21. True 22. True 23. True 24. False 25. True 26. True 27. False 28. False 29. False 30. True 31. True 32. True 33. False 34. False 35. True 36. True 37. True 38. False 39. True 40. False 41. True 42. False

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P-111

CHAPTER SECTION

14 B

SYMMETRY

WORKSHEET-98 Solutions 1. With respect to the given line(s) of symmetry, the other hole(s) are marked in the given figures as under : (a)

3. The axis of symmetry corresponding the given punched holes are shown by dotted lines in the figures a under : (a)

(b)





(c)



(c)

(d)

(d)



(b)



(e)



(f)

(e)





(g)

(h)

2. Corresponding to the given line of symmetry, the completed figure is as under. Their respective names are also given under the figure : (a)

(b)

(c)





(i)

Square



(d)

Circle

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Triangle

Rhombus

(e)

Pentagon

(f)

Octagon

(k)

(j)

(l)

4. The multiple lines of symmetry in respect of the given figures are shown by dotted lines as under :

M A T H E M A T I C S - VII



(a)



(c)

(b)

(d)



(e)



(g)

(f)

(h)

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WORKSHEET-99 Solutions 1. Another name for the line of symmetry of (a) An isosceles triangle is the median. (b) A circle is a diameter of the circle. 2. Let us mark the vertices of the square as A, B, C and D. Take the diagonal BD as a line of symmetry and shade a few more squares as shown to make the figure symmetric about this diagonal. G D

C



(a)

(b)

4. (a) The English alphabet letters having reflectional symmetry about a vertical mirror are : A, H, I, M, O, T, U, V, W, X, Y

E

F

B, C, D, E, H, I, O and X

A

B H

Yes, there are more than one way to make it symmetrical. It is also symmetrical about AC, EF and GH as shown in figure. Clearly, the figure is symmetric about the second diagonal AC. Hence, the figure is symmetric about both the diagonals. 3. Completed shape symmetric about the mirror line(s) are as under :

(b) The English alphabet having reflectional symmetry about a horizontal mirror are : (c) The English alphabet having reflectional symmetry about both horizontal and vertical mirrors are :

H, I, O and X 5. Three examples of shapes with no line of symmetry are : (a) A scalene triangle

(b) A parallelogram



(c) An irregular quadrilateral.

6. (a) In a full turn, there are precisely three position (on rotation through the angles 120°, 240° and 360°) when the triangle looks exactly the same.

Because of this, we can say that it has a rotational symmetry of order 3.

(b) There is only one position where the triangle looks exactly the same, when rotated about its centre.

(a)

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(b)

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7.

Number of lines of symmetry

Figure (a) Equilateral triangle (b) Isosceles triangle (c) Scalene triangle (d) Square

(e) Rectangle (f) Rhombus (g) Parallelogram (h) Quadrilateral (i) Regular hexagon (j) Circle

3 1 0 4

2 2 0 depends on the shape 6 Infinite

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WORKSHEET-100 (c) To find the order of rotation in respect of :

Solutions

P

1. We know that when a figure which on rotation through an angle, about a point, look the same as original, are said to have rotational symmetry.

P 90°

A 90°

90° P

90° P

Figure (i), (ii), (iii) and (iv) have rotational symmetry.

2. Figure (a), (b), (d), (e) and (f) have rotational symmetry of order more than 1. 3. (a) To find the order of rotation in respect of

P

90°



Clearly, this figure requires four rotations, each through an angle of 90°, about the marked point (×) to come back to its original position. Therefore, it has rotational symmetry of order 4. 4. Let us mark a point A on each figure and also indicated the angle through which the figure is to be rotated as under : A

A 90°

90°

A 180°

A

×

90°

A



(a) A

90°

×

(b) A

A





Let us mark the end of vertical rectangle as A. We find that figure required four rotations through 90° each about the point (×) to get back to the original position. Thus, this figure has a rotational symmetry of order 4.

120°

×



×

(c)

(d)

A

(b) To find the order of rotation in respect of : A

90°

A 90° 72°

120°

×

× A 120° A



(e)

(f)

A

A

120° 60°

A



Clearly, this figure requires three rotations, each through an angle of 120°, about the marked point (×) to come back to original position. Therefore, it has rotational symmetry of order 3.

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×



(g)

120°

(h)

M A T H E M A T I C S - VII

Now to find the rotational symmetry, we proceed as under : In figure (a) : It requires two rotations, each through an angle of 180°, about the marked point (×) to come back to its original position. So, its rotational symmetry is of order 2. In figure (b) : It requires four rotations, each through an angle of 90°, about the marked point (×) to come back to its original position. So, its rotational symmetry is of order 4. In figure (c) : It requires three rotations, each through an angle of 120° about the marked point (×) to come back to its original position. So, its rotational symmetry is of order 3. In figure (d) : The figure requires four rotations, each through an angle of 90°, about the marked point (×) to come back to its original position. So,

its rotational symmetry is of order 4. In figure (e) : The figure requires four rotations, each through an angle of 90°, about the marked point (×) to come back to its original position. So, its rotational symmetry is of order 4. In figure (f) : The regular pentagon requires five rotations, each through an angle of 72°, about the marked point to come back to its original position. So, it has rotational symmetry of order 5. In figure (g) : The figure requires six rotations, each through an angle of 60°, about the marked point to come back to its original position. So, it has rotational symmetry of order 6. In figure (h) : The figure requires three rotations each through an angle of 120°, about the marked point to come back to its original position. So, it has rotational symmetry of order 3.

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WORKSHEET-101 Solutions

R

Order of Rotation

Angle of Rotation

Square

Yes

4

90°

Rectangle

Yes

2

180°

Rhombus

Yes

2

180°

Equilateral Triangle

Yes

3

120°

Regular Hexagon

Yes

6

60°

Circle

Yes

Infinite

Any angle

Semicircle No No No 7. Rough sketches in each cases are as under : (a) Three lines of symmetry

0°

×

12

×

0°

Centre of Rotation

Shape

12

1. Two figures that have both line symmetry and rotational symmetry are an equilateral triangle and a circle. 2. The name of quadrilaterals having both line and rotational symmetry is square. 3. When a figure has two or more lines of symmetry, then the figure should have a rotational symmetry of order more than 1 e.g., square. 4. The other angles are 120°, 180°, 240°, 300°, 360°. 5. (i) Yes (ii) No. 6. The given table is filled as under :

×

×

120°

R



R

(b) One line of symmetry but no rotational symmetry of order more than 1. A

B

C D



(c) No line of symmetry but have a rotational symmetry of order 2 D

C

A



B

(d) One line of symmetry but no rotational symmetry. D

F

C

A

F

E A

B

D

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C

a rotational symmetry of order 4.

E

B

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WORKSHEET-102 Solutions



1. (b) 2. (b) 3. (d) 4. (a) 5.



C

9. A square has a rotational symmetry of order 4 about its centre, In this case : (a) The centre of rotation is the centre of the square. (b) The angle of rotation is 90°. (c) The direction of rotation is clock wise. (d) The order is 4. A

(a) 90º B

A D

90º

C

A (i) 90º

(b)

(ii) 180º A

A B 6. (a) Median of an isosceles triangle. (b) Diameter of a circle. 7. (a) A regular hexagon has six lines of symmetry. (b) A quadrilateral (parallelogram) in general has no line of symmetry. 8. No, generally trapezium have no line of symmetry, leaving isosceles trapezium. In isosceles trapezium, AD = BC. So, there is one line of symmetry. D

90º 90º A (iv) 360º

(iii) 270º

A

C

(v) A

B

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P-116

M A T H E M A T I C S - VII

CHAPTER SECTION

15 B

VISUALISING SOLID SHAPES

WORKSHEET-103 Solutions

2. (b), (c), (d) and (f) nets form cubes.

1. Second and third nets are correct nets to make a tetrahedron.

3. Their matching is as under : (a) ↔ (ii) (b) ↔ (iv) (c) ↔ (i) (d) ↔ (iii) (e) ↔ (vi) (f) ↔ (v) 4. Their matching is as under : (a) ↔ (ii) (b) ↔ (i) (c) ↔ (v) (d) ↔ (iii) (e) ↔ (iv) 5. Faces (F)

6

4

9

7

Edges (E)

12

6

16

15

Vertices (V)

8

4

9

10

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WORKSHEET-104 Solutions 1. Inserted suitable numbers in the blanks, are as follows : 1 3

1 2

4

5 6

5 4

2

3

6

2. No, because one pair of opposite faces will have 1 and 4 on them whose total is 5 (≠ 7), and another pair of opposite faces will have 3 and 6 on them whose total is 9 (≠ 7). 3. In the net, three faces are shown. The nets given below can be used to make cubes :

4. Matching is a under : (a) ↔ (ii) (b) ↔ (iii) (c) ↔ (iv) (d) ↔ (i) 5. There are (i) 2 × 4 × 3 = 24 cubes, (ii) 1 × 3 × 2 + 2 × 1 × 1 = 6 + 2 = 8 cubes and (iii) 1 × 3 × 2 + 3 × (1 × 1 × 1) = 6 + 3 = 9 cubes. 6. Total would be on the face opposite to : (a) 5 + 6 is 2 + 1, i.e., 3 (i) (b) 4 + 3 is 3 + 4, i.e., .7. 7.

2 cm = H

2 cm = B 2 cm

2 cm

L Let us place two cubes with 2 cm edge side by side as shown in the diagram. Its length only increases and becomes (2 + 2) cm. i.e., 4 cm. But breadth and height remain same as 2 cm.

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WORKSHEET-105 Solutions 1. (a) True (b) False. 2. The solid (s) that match the given shadow are listed below : (a) a ball, a plate etc.

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(b) a cube, a book etc. (c) a cone, a jocker’s cap etc. (d) A cylindrical pipe, a cuboid etc. 3. When a bulb is kept burning right above the solid named : (a) Ball : There is its shadow and look like a circle.

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(b) Cylindrical pipe : Its shadow is and look like a rectangle. (c) Book : There is its shadow and look like a rectangle. 4. The view of each solid as seen from the direction indicated by the arrow are as under :

(c)



5. (a) (1) → Top, (2) → Side and (3) → Front



(b) (1) → Top, (2) → Side and (3) → Front

(c) (1) → Side, (2) → Front and (3) → Top

(a)

(d) (1) → Side, (2) → Top and (3) → Front







(b)

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WORKSHEET-106 Solutions



14.



15. After matching the two-dimensional figures, their corresponding names are given below :

1. (c) 2. (b) 3. (c) 4. (d) 5. (c) 6. (c) 7. 6 8. 5 9. 12 10. Sphere 11. Cone. 12. (a) Triangular pyramid (b) six 13. For a pyramid, faces vertices are given by : F – E + V = 2 4 – E + 4 = 2 (F = 4 and V = 4) or 8 – E = 2 or E = 8 – 2 = 6

(a)

(i)

Rectangle

(b)

(ii)

Circle

(c)

(iii)

Triangle

(d)

(iv)

Square

(e)

(v)

Quadrilateral

Hence, there are 6 edges.

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WORKSHEET-107 1. (d) Sol. An isometric sketch of a cuboid formed by placing three cubes with 2 cm edge side by side is shown in the adjoining figure.

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M A T H E M A T I C S - VII

4. Oblique sketches are as under :

2. 2 cm = H

2 cm = B 2 cm

2 cm

L Let us place two cubes with 2 cm edge side by side as shown in the diagram. Its length only increases and becomes (2 + 2) cm. i.e., 4 cm. But breadth and height remain same as 2 cm. 3.

Name of solid

Name of the cross-sections in case of a Vertical cut Horizontal cut

(a) A brick

Rectangle

Rectangle

(b) A round apple

Circle

Circle

(c) A die

Square

Square

(d) A circular pipe

Circle

Rectangle

(e) An ice-cream cone

Triangle

Circle

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