# CAT Model Paper 9 Questions and Answers with Explanation Part 3

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11. Given that

(i)

(ii)

(iii)

(iv)

(A) I and IV only

(B) II and III only

(C) I and III only

(D) I , II, III, IV

Answer: C

Solution:

Put n = 1, x = 1

Then LHS =

RHS

Should = 10

Plug in all possible answer values

Option I and III give the right answer as

13. How many 3 digit numbers contain 5 as one of their digits?

(A) 271

(B) 252

(C) 300

(D) 243

Answer: B

Solution:

Option: B

Total number of 3 digit numbers

Those not having 5s any of their digit

Those having 5 as one of their digits

Alternate Solution:

This can also be manually counted.

5 can come either in hundreds place or tens place or units place.

In hundreds place: 500 to 599: 100 numbers.

In tens place: x 5y where y can take value from 0 to 9 and x can take values from 1 to 9 (except 5) = 80 numbers.

In unit place: xy5 where y can take value from 0 to 9 (except 5) and x can take value from 1 to 9 (except 5) = 72 numbers.

Total = 100 + 80 + 72 numbers = 252 numbers.

14. The quadratic function is known to pass through the points (-1; 6) ; (7; 6) , and (1; -6) . Find the smallest value of the function.

(A) -6

(B) -10

(C) -20

(D) -26

Answer: B

Solution:

If the points (-1,6) , (7,6) , and (1, -6) lie on the graph of then

Solving this system, we get a = 1, b =-6, c =-1.

The graph of is a parabola that opens upward and the vertex is

15. Ahmed and sahil set out together on bicycle traveling at 15 and 12 kilometers per hour, respectively. After 40 minutes, Ahmed stops to fix a flat tire. If it takes Ahmed one hour to fix the tire and sahil continues to during this time, how many hours will it take Ahmed to catch up to sahil assuming he resumes his ride at 15 kilometers per hour? (consider Ahmed՚s deceleration/acceleration before/after the flat to be negligible)

(A) 4.5

(B)

(C) 3.5

(D) 4

Answer: B

Solution:

In of an hour

A travels 10 km in 40 min

B travels 8km in 40 min

After one hour , A would have still traveled only 10 km and B would have traveled 20 km

Representing this on a timeline:

Time in hours | 1 | 2 | 3 | 4 | 5 | 5: 20 | |

Distance | A | 10 | 10 | 25 | 40 | 55 | 60 |

B | 8 | 20 | 32 | 44 | 56 | 60 |

A gains this 10 km in hours.

16. In an automated plant assembly line, the rate of rejection of components was 10 % on july 1^{st} and 6 % on july 2^{nd} . The combined rate of rejection for the days was 9 % . The ration of production volumes on july 1^{st} and on july 2^{nd} is

(A) 2: 1

(B) 3: 2

(C) 3: 1

(D) None of these

Answer: C

Solution:

Let the production on july 1^{st} and on july 2^{nd} be X and Y units respectively.

Therefore, rejection on July 1^{st} = 0.1X

Rejection on july 2^{nd} = 0.06Y

Therefore, Total rejection

Hence, X: Y = 3: 1

Required ratio is 3: 1.

17. The sum to 2n terms of the series is?

(A)

(B)

(C)

(D) None of these

Answer: C

Solution:

At n = 1

2n = 2

Sum to 2n terms = 125 + 864 = 989

Substitute n = 1 in answer options.

Only option c gives 989.

18. If (a) and g (0) = 0, then g (n) , where n is a natural number is?

(A) ng (1)

(B)

(C) 0

(D)

Answer: A

Solution:

Solve by assumption

Put a = 1

Then the equation changes to

Look in the answer options on substituting

Eliminate answer options where you are not getting

Answer is Option (a) .

19. Find the greatest number of five digits. Which is exactly divisible by 7, 10,14, 15,21 and 28.

(A) 99840

(B) 99900

(C) 99960

(D) 99990

Answer:

Solution:

The number should be exactly divisible by 14 (7,2) , 15 (3,5) , 21 (3,7) , 28 (4,7) .

Hence, it is enough to check the divisibility for 3,4, 5 and 7.

Only (C) is divisible by all.

20. If then the solution set of ‘a’ for the inequality is

(A)

(B)

(C)

(D)

Answer: B

Solution:

Substitute a value for “a” which is there in 2 answer options and not there in the other two take a = 2

At a = 2 LHS < RHS.

Hence, Option (c) & (d) can be eliminated

Answer can be option (a) or (b)

Now substitute a value which is there in option (b) and not in option (a) put a = 0

.

Hence, option (a) can also be eliminated.