D1541852097dmath-10_self_assessment_paper-1.pdf

MATHEMATICS

Time : 3 Hours Maximum Marks : 80

CBSE Solutions

1

Self Assessment Paper Section ‘A’

5. Distance covered in 1 revolution

= circumference of wheel

1. The given number ends in 5. Hence it is a multiple of 5. Therefore it is a composite number. [CBSE Marking Scheme, 2016] 1

OR 3 2 HCF of 3 × 5 and 3 × 52



= 32 × 5



= 9×5

(–5, 4)



(a, 2)



(–1, 0)

⇒



⇒

a =



⇒

a = – 3

1

Consonants are 21.

The probability of chosen a consonant =

1

21 . 26

2

1 2

Section ‘B’ 7. Given,



2





2

LCM (306, 1,314) = ?





Let

a = 306



= cos2 67 – cos2 67



b = 1,314



= 0



We know that







⇒



⇒

= cos 67 – {sin(90 – 67)}2 = cos 67 – {sin(90 – 67)}2 [\ sin(90 – q) = cos q]

2

\ cos 67 – sin 23 = 0

1

OR 2

× 1.26

[CBSE Marking Scheme, 2015] 1

4. cos 67 – sin 23 = cos 67 – {sin 23}



7

6. In English language there are 26 alphabets.

3. No, Angle included should be same.

2

22

[CBSE Marking Scheme, 2012]



−6 2

2

= 500 ×

= 1980 m.

1



Distance covered in 500 revolutions

= 500 × π × 1.26

R

−5 + ( −1) a= 2

2

= π × 1.26 m.

= 45 P Q 2. P is mid-point of QR

= πd

HCF (306, 1,314) = 18

a × b = LCM (a, b) × HCF (a, b) 306 × 1,314 = LCM (a, b) × 18

2



= tan 10° – cot 80°



= tan2 (90° – 80°) – cot2 80°





[\ tan (90° – q) = cot q] 2



= cot 80° – cot2 80°



= 0

1

1

LCM (a, b) =

306 × 1, 314 18

= 22, 338 \ LCM (306, 1314) = 22, 338

1

OSWAAL CBSE Sample Question Papers, MATHEMATICS, Class-X

2

8. AM is perpendicular bisector of BC.

2

A

3 3

B



1

h C

M

3 3 2

2 3 3 ( 3 3 )2 = h +    2 







⇒

27 = h 2 +

27 4



⇒

h2 = 27 −

27 4



⇒

h2 =

\



1 3× + 3+ 2 −1 3 = 1

h =

11. Area of the remaining cardboard = Area of rectangular cardboard – 2 × Area of circle 1 14 cm

7 cm

81 4 9 = 4.5 cm 2

d =

=

( 0 - 10 )

7

×

2

49 4

= 98 – 77

+ ( 12 - 12 )

2

100 + 0 = 10 units. 2

= 21 cm2

1

Commonly made error The required area is miscalculated.

[CBSE Marking Scheme, 2011]



44

 7 ×  7  2

22

1 = 98 -

2

7 cm

7 cm

= 14 ×7 - 2×

∴ Distance between (10, 12) and (0, 12)

1

= 5

2

9. The point on the y-axis is (0, 12)



1



= 1 + 3 + 2 – 1

In DAMB,





 1  3× + ( 3 )2 + 2 − 1  3  = (1)2

OR

Here AB =

( a + a )2 +( a + a )2 = 2 2 a



2 2 2 2 2 2 a + 3a − 2 3a + a + 2 3a + 3a



The student should carefully take the radius of

½

the circles as

( − a + 3 a )2 +( − a − 3 a )2 = 2 2 a ½

BC =



ANSWERING TIP...

Let A(a, a), B(– a, – a) and C ( - 3 a , 3 a )

= 8a2 = 2 2 a 2

( a + 3 a ) + ( a − 3a )

½ 2

= 2 2a

½

7 14 and not . 2 2 OR



Length of the cuboid so formed be l cm



∴ l = 5 + 5 = 10 cm, b = 5 cm ; h = 5 cm.



AC =

a 2 + 3a 2 + 2 3a 2 + a 2 + 3a 2 − 2 3a 2

= 2(10 × 5 + 5 × 5 + 5 × 10)



= 8a2 = 2 2 a Since AB = BC = AC, therefore ABC is an ½ equilateral triangle. [CBSE Marking Scheme, 2015]

= 2(50 + 25 + 50)

10.

2

2

3 tan 30° + tan 60° + cosec 30° − tan 45° cot 2 45°

Surface area = 2(l × b + b × h + h × l) 1

= 2 × 125 = 250 cm2.

1

[CBSE Marking Scheme, 2015] 12. Sample space (S) = {2, 3, 4 ...... 98, 99} \ n(S) = 98

Solutions

3

( i) An integer is divisible by 8 E = {8, 16, 24, 32 ...... 96} 96 − 8   n = 8 + 1  



\



Probability of chosen a number divisible by 8 n( E) P(E) = n(S)



n(E) = 12

12 6 = = 98 49 \ Probability of chosen a number divisible by 8 6 = 2 49



(ii) Probability of chosen a number is not divisible by 8 = P( E)







= 1 −







Probability of chosen a number is not divisible 43 by 8 = 2 49

P( E) = 1 – P(E)

P( E) =

6 49 − 6 = 49 49

43 49



1500 x + 15000 − 1500 x 1 = x( x + 100 ) 2 15000 1 = 2 x + 100 x 2



x2 + 100x = 30000 2 x + 100x – 30000 = 0 2 x + 600x – 500x – 3000 = 0 x(x + 600) – 500(x + 600) = 0 (x + 600)(x – 500) = 0 If x + 600 = 0 Þ x = – 600 speed can not be negative. If x – 500 = 0 Þ x = 500 \ Speed of the plane = 500 km/hr 1 15. We have, 4x – y = 4 ⇒ y = 4x – 4 1 x

0

1

2

y

–4

0

4



We have,



⇒

3x + 2y = 14 y =

x y

Section ‘C’

14 - 3x 2

1

0

2

4

7

4

1

Y

13. To find LCM (9, 12, 15) 9 = 3 × 3 12 = 2 × 2 × 3 15 = 3 × 5 ∴ LCM (9, 12, 15) = 3 × 3 × 2 × 2 × 5 = 180 minutes The bells will toll together after 180 minutes. 14. Let the speed of plane be x km/hr Time taken to cover 1500 km Distance 1500 = hr (t1) = speed x  Time taken to cover 1500 km When speed increased 100 km/hr 1500 hr (t2) = x + 100



1

24 20



1500 1500 1 − = x x + 100 2

4x – y = 4

16

1 1

12 8 (0, 7) 4

x'

1



(1, 0)

0



1 2 (0, – 4)

Y'

(2, 4) (4, 1) 3

4

x 5 6 3x + 2y = 14

1

As (2, 4) is common, therefore solution is (2, 4).

Commonly made error Points are incorrectly plotted on the graph and

Given :

t1 – t2 = 30 minutes 1 = hr 2 

28

that results in incorrect solution or no solution.

1

16. Since equation has equal roots, ∴ D = 0 i.e., b2 – 4ac = 0 2 {2(2k – 3)} – 4(k – 2)(5k – 6) = 0 1

OSWAAL CBSE Sample Question Papers, MATHEMATICS, Class-X

4

⇒ 4(4k2 – 12k + 9) – 4(k – 2)(5k – 6) = 0 2

⇒



k – 4k + 3 = 0

The student should carefully simplify the ratio of

2



⇒

k – 3k – k + 3 = 0



⇒

k(k – 3) –1(k – 3) = 0



⇒ (k – 3)(k – 1) = 0



\

k = 1, 3 1





[CBSE Marking Scheme, 2015]

17. Here given,

a3 = 9 ⇒ a + 2d = 9

the sum of both the APs and equate

⇒ (a + 7d) – (a + 4d) =6



⇒ 3d = 6



⇒



Substituting (ii) in (i), we get



⇒

a + 2(2) = 9



⇒

a = 5



So, AP is 5, 7, 9, 11, .....







A (7, 2)

E (x1, y1)

....(i)

F (x2, y2)

d = 2

C (1, 4)

B (9, 10)

....(ii) 1

1 1





18.

n–1 = m – 1. 2

a8 – a5 = 6





ANSWERING TIP...

⇒ 4k2 – 12k + 9 – 5k2 + 6k + 10k – 12 = 0 1



[CBSE Marking Scheme, 2015] OR Let a, A be the first terms and d, D be the common differences of two AP’s Then according to question, n [2 a + (n - 1)d ] Sn 7n + 1 2 = S 'n = n 4 n + 27 1 [2 A + (n - 1)D] 2 2 a + (n - 1)d 7n + 1 ⇒ = 2 A + (n - 1)D 4 n + 27  n - 1 a+ d 7n + 1  2  ⇒ = ½ 4 n + 27  n - 1 A+ D  2  n -1 Putting =m–1 2

∴ Co-ordinates of point E  9 + 7 10 + 2  , =   2 2 



of both the APs leads to equation n – 1 = m – 1, resulting in incorrect solution.

Length of EF =

(8 − 4 )

2

+ (6 - 3)

= ( 4 ) + (3) = 5 units 2



Length of BC =

(9 - 1)

2

2

2

...(i) 1

+ (10 - 4 )

2

= ( 8 ) + ( 6 ) = 10 units ...(ii) From equation (i) and (ii), we get 1 EF = BC. 2 Hence proved. 1 [CBSE Marking Scheme, 2015] 2

OR

2

B(5, –1)

A(2, –1)

n = 2m – 1 1 a + (m - 1)d 7(2 m - 1) + 1 = A + (m - 1)D 4(2 m - 1) + 27 am 14 m - 6 ⇒ Am = 8m + 23 ½

Incorrect simplification of the ratio of the sum

½

= (4, 3)

⇒

Commonly made error

½

= (8, 6) Co-ordinates of point F  7 + 1 2 + 4 , =   2 2 

O

D(2, 6)





C(5, 6)

AC =

(5 − 2 )2 + (6 + 1)2



=

32 + 7 2



=

9 + 49

Solutions

5

=

OR

1

58 

O





=

32 + 7 2



=

9 + 49



=

58



Since AC = BD =

10



m

(5 − 2 )2 + ( −1 − 6)2

0

BD =

1

58 cm, the diagonals of

rectangle ABCD are equal



 7 5 =  ,  2 2







 7 5 =  ,  2 2



Since the mid-point of diagonal AC



= mid-point of diagonal







∴ They bisect each other.

 2 + 5 6 + ( −1)  , Mid-point of BD =    2 2 

OL 1 = 100 2 ⇒ OL = 50 m OM = OL – ML = OL – FG = 50 – 20 = 30 m OM In ∆OMG, = sin 45° OG ⇒

 7 5 BD =  ,   2 2 

1 Hence Proved

19. Let OA be the electric pole and B be the point on the ground. Let BA = x AO = sin 45° AB

1

B

⇒



⇒

10 m

⇒

⇒ OG = 30 2 m. 1 Hence, distance of the bird from the girl is 30 2 m. [CBSE Marking Scheme, 2014]

ANSWERING TIP... The student should be careful in visualizing the

1

position of the bird, boy, girl and the building.

20.

45°



O

10 1 = x 2

C

8 A

x = 10 2



= 10 × 1.414 = 14.14 m

1

1 OM = 2 OG 1 30 = 2 OG

⇒

A

x

20 m

L F B 1 Let O be the position of the bird and B be the position of the boy. Let FG be the building and G be the position of the girl. OL In ∆OLB, = sin 30° BO

 2 + 5 −1 + 6  , Mid-point of AC =    2 2 





20 m

G

30°



In ∆ABO,

45°

M

1

[CBSE Marking Scheme, 2016]



6 2

B 2

AC = (8) + (6)2 = 100 ⇒ AC = 10 8 6 \ sin A = , cos A = 10 10 

1

OSWAAL CBSE Sample Question Papers, MATHEMATICS, Class-X

6

6 8 , cos C = 10 10  \ sin A cos C + cos A sin C 8 8 6 6 × + × = 10 10 10 10

and

A

sin C =

1

C

1

[CBSE Marking Scheme, 2016] 21. LHS = (cosec q – sin q) (sec q – cos q) (tan q + cot q)

 1   1   sin θ cos θ  + - sin θ  - cos θ  =   sin θ   cos θ   cos θ sin θ 



 1 - sin 2 θ   1 - cos2 θ   sin 2 θ + cos2 θ  =        sin θ   cos θ   sin θ.cos θ 



cos2 θ sin 2 θ  1  × × =  cos θ  sin θ.cos θ  sin θ



= 1



= RHS Hence proved.

3

r = 3, πrl = 47.1

∴

l =





h =





Volume of cone =

47.1 = 5 cm 3 × 3.14 5 2 - 3 2 = 4 cm





½

1

[CBSE Marking Scheme, 2016] OR

∠ADB = ∠ACB = 90° (Angle in a semi-circle) 1 ∴ DADB or DACB = × BD × AD 2 1 × 8 × 15 2







= 60 cm2



1

1 × 3.14 × 3 × 3 × 4 ½ 3

= 37.68 cm3.

=

∴ Area of shaded region 17 17   =  3.14 × × − 120  cm2 1 2 2   = (226.87 – 120) cm2 = 106.87 cm2. 1 [CBSE Marking Scheme, 2012]

Section ‘D’

[Q sin2 q + cos2 q = 1]



D

B

100 = = 1 100

22. Here

. O

Area of DACB = 60 cm2

∴ Area of shaded region = Area of circle – (Area of DADB + Area of DACB) 1



= Area of circle – (120 cm2),



17   2 2 cm ∵ AB = 15 + 8 = 17 cm ∴ radius =  2

23. Here, x = – 2 is the root of the equation 3x2 + 7x + p = 0 2 ⇒ 3(– 2) + 7(– 2) + p = 0 ⇒ p = 2 1 2 2 Root of the equation x + 4kx + k – k + 2 = 0 are equal. b2 – 4ac = 0 1 2 2 ⇒ 16k – 4(k – k + 2) = 0 1 ⇒ 3k2 + k – 2 = 0 ⇒ (3k – 2)(k + 1) = 0 1 2 ⇒ k = , − 1 1 3 [CBSE Marking Scheme, 2015] OR Let the total number of students be x. According to question, 3 x = 16 + x 1 8

3 x - 16 = ⇒ 8



⇒ 3x – 128 = 8 x



⇒ 3x – 8 x – 128 = 0



Let

x

1

x =y

3y2 – 8y – 128 = 0 ⇒ 3y2 – 24y + 16y – 128 = 0 ⇒ 3y(y – 8) + 16(y – 8) = 0 ⇒ (y – 8)(3y + 16) = 0 y = 8 or y = – 16/3 y = 8 ⇒ x = 64 as y ≠ – 16/3 Number of students = 64

1

1

Solutions

7

Commonly made error Incorrectly forms the equation involving square



⇒



⇒

root of x and incorrect simplification.

24. Let numerator be x. ∴ Denominator is x + 2.

25.

½

x Fraction = x+2

⇒



AD 2 16 = PE2 25 4 AD = 5 PE AD 4 = PE 5 

\

x x+2 34 + = ½ x+2 x 15 ⇒ 15(x2+ x2 + 4x + 4) = 34(x2 + 2x) ⇒ 30x2 + 60x + 60 = 34x2 + 68x 1 ⇒ 4x2 + 8x – 60 = 0 ⇒ x2 + 2x – 15 = 0 1 2 ⇒ x + 5x – 3x – 15 = 0 ⇒ x(x + 5) – 3(x + 5) = 0 ⇒ (x + 5)(x – 3) = 0 ⇒ x = 3 3 ∴ Fraction = 1 5 [CBSE Marking Scheme, 2012] ⇒

A

OR

A

P

C Q R D E B 1 Given : ∠A = ∠P ∠B = ∠C, ∠Q = ∠R Prove : Let ∠A = ∠P = x In ∆ABC, ∠A + ∠B + ∠C = 180 ⇒ x + ∠B + ∠B = 180 (Given, ∠B = ∠C) ⇒ 2∠B = 180 – x 180 - x ⇒ ∠B = ...(i) 1 2 Now, In ∆PQR ∠P + ∠Q + ∠R = 180 (Given, ∠Q = ∠R) ⇒ x + ∠Q + ∠Q = 180 ⇒ 2∠Q = 180 – x 180 - x ⇒ ∠Q = ...(ii) 2 In ∆ABC and ∆PQR, ∠A = ∠P (Given) ∠B = ∠Q (from eq. (i) and (ii)) ∆ABC ∼ ∆PQR (AA similarity) 2 ar ( ∆ABC ) AD ⇒ 1 = 2 ar ( ∆PQR ) PE

F

1

B E

G C

D



In trapezium ABCD, AB || DC and DC = 2AB. BE 4 Also, = EC 3 In trapezium ABCD, EF || AB || CD AF BE 4 ∴ = = FD EC 3 In ∆BGE and ∆BDC, ∠B = ∠B (Common) ∠BEG = ∠BCD (corresponding angles) ∴ ∆BGE ~ ∆BDC [AA similarity] 1 EG BE ⇒ = ...(i) CD BC BE 4 As, = EC 3 EC 3 ⇒ = BE 4 EC 3 ⇒ + 1 = + 1 1 BE 4 EC + BE 7 ⇒ = BE 4 BC 7 ⇒ = BE 4 BE 4 ⇒ = BC 7 EG 4 ⇒ from (i), = CD 7 4 ⇒ EG = CD ...(ii) 7 Similarly, ∆DGF ~ ∆DBA DF FG ⇒ = 1 DA AB

⇒

FG 3 = AB 7

OSWAAL CBSE Sample Question Papers, MATHEMATICS, Class-X

8

⇒

FG =

3 AB ...(iii) 7

27. A

 AF 4 BE  ∵ AD = 7 = BC    ⇒ EC = 3 = DE   BC 7 DA 

Adding eqns. (ii) and (iii), 4 3 EG + FG = CD + AB 7 7



⇒

EF =

=

4 3 ×(2AB) + AB 7 7 8 3 AB + AB 7 7

11 AB 7 ∴ 7EF = 11AB Hence proved. 1

=

26. Given : A circle of radius 3 cm with centre O and a point P at a distance of 7 cm from O. Construction : We have to construct the two tangents from P to the circle. Q

M

P

O

R

1 Steps of construction : 1. Draw a line segment PO = 7 cm. 2. From the point O, draw a circle of radius = 3 cm. 3. Draw a perpendicular bisector of PO. Let M be the mid-point of PO. 4. Taking M as centre and OM as radius a circle. 5. Let this circle intersect the given circle at the point Q and R. 6. Join PQ and PR. Thus PQ and PR are the required two tangents. 2 Length of the tangent 2

PQ − OQ









=

( 7 )2 − ( 3 )2



=

49 - 9 =

PQ = PR =

2

B

[CBSE Marking Scheme, 2015]

2x

C

Given : AD ⊥ BC 2DB = 3CD To prove : 5 AB2 = 5AC2 + BC2 Proof : 2DB = 3CD (Given) DB 3 ⇒ = 1 CD 2 ⇒ DB = 3x, CD = 2x and BC = 5x In ∆ADB, ∠BDA = 90° AB2 = AD2 + DB2 ⇒ AB2 = AD2 + (3x)2 ⇒ AB2 = AD2 + 9x2 ⇒ 5AB2 = 5AD2 + 45x2 ⇒ 5AD2 = 5 AB2 – 45x2 ...(i) 1 and AC2 = AD2 + CD2 ⇒ AC2 = AD2 + (2x)2 ⇒ AC2 = AD2 + 4x2 ⇒ 5AC2 = 5AD2 + 20x2 ⇒ 5AD2 = 5AC2 – 20x2 ...(ii) 1 On comparing eq. (i) and eq. (ii), 5AB2 – 45x2 = 5AC2 – 20x2 ⇒ 5AB2 = 5AC2 – 20x2 + 45x2 ⇒ 5AB2 = 5AC2 + 25x2 ⇒ 5AB2 = 5AC2 + (5x)2 \ 5AB2 = 5AC2 + BC2 [BC = 5x]  Hence proved. 1 28. E

A

F

2 cm



40 = 6.3 cm 1

D

3x



B

C

H

G

P ½ Let BPC is a hemisphere and ABC is a cone. Radius of hemisphere = Radius of cone 4 = = 2 cm 2 h = Height of cone = 2 cm 2 1 Volume of toy = πr 3 + πr 2 h 3 3 2 1 = × 3.14 × 2 3 + × 3.14 × 2 2 × 2 3 3

Solutions

9

= 25.12 cm3 ...(i) Let right circular cylinder EFGH circumscribe the given solid toy. 1½ Radius of cylinder = 2 cm, Height of cylinder = 4 cm Volume of right circular cylinder = πr2h = 3.14 × (2)2 × 4 cm3 ...(ii) = 50.24 cm3 1 ∴ Required volume = Volume of cylinder – Volume of toy = 50.24 – 25.12 = 25.12 cm3. 1 [CBSE Marking Scheme, 2012] 29. 

Weight (in kg)



Total number of possible products

120

More than or equal to 10

106

More than or equal to 20

89

More than or equal to 30

67

More than or equal to 40

41

More than or equal to 50

18

Products x.y which are less than 16 are {1 × 1, 1 × 4, 1 × 9, 2 × 1, 2 × 4, 3 × 1, 3 × 4, 4 × 1} 1

Daily Income (in `)

Number of Workers (f)

Cumulative Frequency (c.f.)

100 – 120

12

12

120 – 140

14

26

140 – 160

8

34

160 – 180

6

40

180 – 200

10

50

2

Less than Daily income in (`)

Number of Workers (f)

100

0

120

12

140

26

160

34

180

40

200

50

Scale : x-axis 1 cm = 10 units y-axis 1 cm = 10 units

100 90

(20, 89)

More than ogive (30, 67)

60

50 –

50



30 40 Lower limits

30. Given,

x = {1, 2, 3, 4}



10

20

50

n(x) = 4 y = {1, 4, 9, 16}

60

2

X

 ½

–

0

–

10

–

(50, 18)

–

30

⇒

•

45 – 40 – • 35 – 30 – • 1 cm= ` 20 on X axis 25 – 1 cm= 5 Workers on Y axis – 20 – 15 • 10 – 5 – • X 0 100 120 140 160 180 200

–

(40, 41)

40

20



Y

2

•

No. of Workers (f)

Cu m ul at i ve fr equen cy

80 70

1

OR

(0, 120) (10, 106)

8 1 = . 16 2

∴ Required probability =

Y

110

½

n (x.y) = 8

Plotting the points : 120

½

= 4 × 4 = 16.

Cumulative Frequency

More than or equal to 0

½

n(y) = 4

–





Daily income (`) →

2

nnn