Nat Cat Test Solutions

NAT CAT Test Solutions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

131313

4 3 5 3 5 3 2 1 2 1 2 4 4

A5

14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25.

1 2 1 2 1 3 1 4 3 3 2 3

A

1.

(4); Given S x S is a number having S in its unit place so the possible numbers are 1 x 1, 5 x 5, 6 x 6, 1 x 1 is not possible +2 +3 now P(1) Q R 5 OR P(1) Q R 6 X 5 X6 A A A 5 A A A 6 Here A can either be 2 or 7 hence here as P = 1 hence A would be either 7, 8 or 9. by trial and error S ≠ 5. by trial and error 1 2 9 6 X6 7 7 7 6 Hence, the remainder is 96.

2.

(3); Given |x + 2| + |x – 3| < 7 by trial and error taking x as 1, 2, 3, ………. When x ≥ 4 the inequation isn’t satisfied. Now taking x as -1, -2, -3, ……. When x -3 the inequation isn’t satisfied Hence answer is – 3 < x < 4

3.

(5); Here the middle digit can be any from 0 – 9 and the another two digits are the factorization of the digits into two parts. When middle digit is “0” it can be factorize as 1 x 0, 2 x 0 ………. 9 x 0 = 9 numbers. When middle digit is “1” only 1 possibility 111 or 1 x 1 = 1 numbers When middle digit is “2” it can be factorize as 2 x 1 or 1 x 2 = 2 numbers When middle digit is “3” it can be factorize as 3 x 1 or 1 x 3 = 2 numbers When middle digit is “4” it can be factorize as 4 x 1 or 1 x 4 or 2 x 2 = 3 numbers When middle digit is “5” it can be factorize as 5 x 1 or 1 x 5 = 2 numbers When middle digit is “6” it can be factorize as 6 x 1 or 2 x 3 or 1 x 6 or 3 x 2 = 4 numbers When middle digit is “7” it can be factorize as 7 x 1 or 1 x 7 = 2 numbers When middle digit is “8” it can be factorize as 8 x 1 or 1 x 8 or 4 x 2 or 2 x 4 = 4 numbers When middle digit is “9” it can be factorize as 9 x 1 or 1 x 9 or 3 x 3 = 3 numbers 32 numbers (3); Given a = bx = cy = dz

4.

b=

1 ax

1 y

1

, c=a , d=a z

loga (abcd) = 1 +

1 1 1 + + x y z

Now here x, y, z are less than 6. 1 1 1 So + + to be integer we have to go through trial and error: x y z And then the triplets are (1, 1, 1), (2, 2, 1) x 3, (3, 3, 3), (4, 4, 2) x 3. So there are 8 ordered triplets.

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A6

A

5.

(5);

A

D N R

P

2 10 Q S

B

C

Given 4AB = 3BC AB : BC = 3 : 4 Let AB = 3x & BC = 4x Then AC = 5x

6x 2 =x 6x So in right triangle RNS

hence inradius RP =

3x 2

2

=

x 2

2

SQ =

x 2

NR =

x x , NS = 2 10 & RS = x + 2 2

+ (2 10 ) 2

x= 2 5 so, area of rectangle = 12x2 = 240 square unit. The area of rectangle ABCD is 240 square unit. 6.

(3); The length of boundary is Given 2 (PO + OR) = 26 PO + OR = 13

1 (circumference of circle) + BR + PR + AP 4

OA + OB = 20 BR + AP = 7 & PR = OQ = radius of circle = 10 1 ∴ Length of boundary = π x 100 + 7 + 10 = 17 + 25π 4 7.

(2); Given in 1000 ml of liquid C, A and B are in the ratio 3 : 2 A = 600 ml. B = 400 ml Now 400 ml is vapored where A and B are in the ratio 4 : 1 A = 320 ml. B = 80 ml Hence in remaining 60% of liquid C, liquid A is 280 ml and liquid B is 320 ml If we fill the container with liquid B then A = 280 ml & B = 320 + 400 = 720 ml A : B = 7 : 18

8.

(1); Let Ram, Shyam and Ghanshyam take r, s, and g days respectively to complete the work rs + rg rs + sg + gr sg Then applying the condition, r = m or m = or m + 1 = s+g sg sg n+1=

rs + sg + gr rg

rs + sg + gr rs sg rg 1 1 1 rs So + + = + + =1 m + 1 n + 1 p + 1 rs + sg + gr rs + sg + gr rs + sg + gr p+1=

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A7

A

Directions for Qs. 9 – 11: Refer to the following information for the following solutions.

A

70

60 a

C

b

B

Suppose at first they meet at C according to question a covers CB in 49 sec and b covers CA in 36 sec Speed of a 36 6 So = = Speed of b 49 7 or speed of a & b are 6x m/s & 7x m/s respectively AC = 60 m & CB = 70 m

70 10 = m/s 49 7 60 5 & speed of b is = m/s 36 3

9.

(2); Speed of a is

10.

(1);

11.

(2); If b returns back to shore B. It covers 260 m in whole by the same time A will cover 260 x

‘b’ reaches at A 13 sec before. Then ‘a’ reaches at B. So he travels for 13 sec, before ‘a’ reverse the 5 325 direction it means both ‘a’ and ‘b’ then cover remaining distance viz. 130 − 13 x = meter with the speed 3 3 10 5 65 + = m/s 7 3 21 325 x 21 Hence the time taken is = 35 sec 3 x 65 In 35 sec ‘a’ covers 50 m from the shore B The distance between both the meeting points is 20 m

260 − 12.

6 m 7

260 x 6 260 = a’s distance from shore A = m 7 7

(4); Let S1 ≡ 3, 10, 17, 24, ……….. S2 ≡ 7, 12, 17, 22 ……… ∴ General term of S1 = 7n – 4 General term of S2 = 5m + 2 ∴ for common term 7n – 4 = 5m + 2 5m + 6 n= 7 ∴ From above the common terms are obtained at n = 3, 8, 13, 18, 23, 28, 33, 38, 43, 48 m = 3, 10, 17, 24, 31, 38, 45, 52, 59, 66 ∴ The number of 10th common term in the two series are 48 and 66 48 ∴ Ratio of 10th common term numbers = = 8 : 11 66

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A8

A

13.

(4); The given series is S = 1 x 2 + 5 x 22 + 9 x 23 + …………+ 61 x 216 2 x S = 1 x 22 + 5 x 23 + 9 x 24 + ….….+ 57 x 216 + 61 x 217 S – 2S = 1 x 2 + 4 x 22 + 4 x 23 + 4 x 24 + ………. ………….+ 4 x 216 – 61 x 217 – S = 2 + 4 x 22 [1 + 2 + 22 + ……… + 214] – 61 x 217 S = 61 x 217 – 2 – 24 [215 – 1] = 61 x 217 – 2 – 219 + 24 = 217 (61 – 4) + 16 – 2 = 217 x 57 + 14

14.

(1); A

4

3

5

S

4

8 P N

B 3

O

Q

6

C R X takes 5 hours to complete the rectangle ABCD = 2(8 + 6) = 28 km 28 hence X’s speed is km / h 5 4x3 1 1 From the given figure radius of the circle is SO × OP = ON × PS 5 2 2 D

Hence circumference of the circle is 2 x π x So speed of Y is 2 x

4x3 5

22 4 x 3 7 x x 7 5 11

48 km / h 5 X's speed 7 = hence Y's speed 12

=

15.

(2); Given speed of engine is 24 km / h reduction in speed ∝ wagons

= k wagons k = 2 given when wagons are 4 reduction in speed is 4 km/hr. the maximum number of wagons are 2 wagon < 24 wagon < 144 ∴ Number of max. wagons = 143 16.

(1); Let n families be there in year 1700 So total population is n2 According to the condition in 1720 no. of people are n2 – 50n + 670 = (n – 25)2 + 45 when n = 26, no. of people = 46 & when n = 25 it becomes the least i.e. 45

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A9

A

17.

(2); Let w be the apparent weight during purchase and w’ be apparent weight during sell and Rs. P/kg be the price of sugar when purchased by tradsman. x ∴ Case I: CP = Pw SP = P 1 + w' 100

P 1+ ∴

x w ′ − Pw 100 x 100 = x + 11 …….. (1) Pw

Case II: CP = Pw '

SP = P 1 +

x w 100

x w = Pw ' …….. (2) 100 on solving (1) & (2) x = 10% ∴ P 1+

Directions for Qs. 18 & 19: Refer to the following information for the following solutions. Let the set representations be TOI → A = 50 IE → B = 75 ET → C = 85 HT → D = 65 Also A ∪ B ∪ C ∪ D = 175 Now let, exactly 3 news paper readers = Exactly 2 newspaper readers = exactly 4 newspaper readers = x. ∴ At least 2 newspaper readers = 3x At least 3 newspaper readers = 2x ∴ n(A ∪ B ∪ C ∪ D) = n(A) + n(B) + n(C) + n(D) – {n(A ∩ B) + n(B ∩ C) + n(C ∩ A) + n(D ∩ A) + n(B ∩ D) + n(D ∩ C)} + {n(A ∩ B ∩ C) + n(B ∩ C ∩ D) + n(A ∩ B ∩ D) + n(A ∩ C ∩ D)} – n(A ∩ B ∩ C ∩ D) ∴ 175 = 50 + 75 + 85 + 65 – 3x + 2x – x x = 50 18.

(1)

19.

(3); Exactly one newspaper = 175 – 3 x 50 = 25

20.

(1); Statement I: n is a positive integer So 4n will always leave reminder 1 when divided by 3 hence 4n would not be divisible by 3. 1 Statement II: n is a real number when n is , 4 n + 1 = 3 2 Which is divisible by 3. 1 m When n is or other like number (4n + 1) would be divisible by 3. but when n is other than m ∈ N 4n + 1 2 2 would not be divisible by 3. hence by II it can not be answered.

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A 10

A

21.

(4); By Statement I the quadrilateral is a square of side 2 cm by statement II ∠ ABP is 30° If we use both the statements simultaneously we can get the answer. A By I: diagonal = 2 2 ∴ BO = 2 = AO ∠ BPC = 75°

P

∴ PO can be determined by tan 75° =

3 −1 D

1 AP X BO 2 Hence both together are sufficient

∴ Area of ∆ ABP =

(3);

2 O

3 +1

∴ AP can be found

22.

B

C

240 B

A

1 2 P's time for a dis tan ce 9 = 4= According to the question Q's time for the dis tan ce 2 8 P 's speed 8 = Q's speed 9 Going through option only (3) satisfied the ratio. But at the same time (5) may be the answer. So we will approve the answer by checking further condition which is in 3 hrs. P would cover 120 km and in 2 hrs would cover 90 km Now as they both cover the remaining 120 & 150 at the same time (in 3hrs.) hence it is approved. {Q’s speed would be 45 + 5 according to question} Hence

23.

(3);

A P D 75°

B

C Q According to the theorem if we join PQ it would pass through the center. It means ∠ QCP = 90° & ∠CQP = 75° hence ∠ CPQ = 15° ∠ QAC = 15° {angle by same segment}

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A 11

A

24.

(2);

A 6 cm r O

B

r D

8

10

8-r C ∆ ABC ∼ ∆ DOC AB BC ∴ = OD OC

6 10 = r 8−r 48 - 6r = 10r r=3 ∴ Volume overflowed = = 36 π Total volume =

4 π x 33 3

1 π x 62 x 8 3

= 96π ∴ Remaining volume = 96 π – 36 π = 60 π 36π 3 = ∴ Ratio = 60π 5 25.

(3);

5 4 3 2 1

5

4 3 2 1

11 2 3 4 5 2 3 4 5

The points on or interior are: Interior (3, 3) On x + y = 5 (0, 5) (2, 3) (3, 2), (1. 4) (4, 1) (5, 0) on x2 + y2 = 25 (3, 4) (4, 3) (0, 5) (5, 0) ∴ Total points = 9

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A 12

A