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CAT Aptitude Questions PDF:
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Question 1:
The integers 1, 2, …, 40 are written on a blackboard. The following operation is then repeated 39 times: In each repetition, any two numbers, say a and b, currently on the blackboard are erased and a new number a + b – 1 is written. What will be the number left on the board at the end?
a) 820
b) 821
c) 781
d) 819
e) 780
Question 2:
The number of common terms in the two sequences 17, 21, 25,…, 417 and 16, 21, 26,…, 466 is
a) 78
b) 19
c) 20
d) 77
e) 22
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Question 3:
Rahim plans to drive from city A to station C, at the speed of 70 km per hour, to catch a train arriving there from B. He must reach C at least 15 minutes before the arrival of the train. The train leaves B, located 500 km south of A, at 8:00 am and travels at a speed of 50 km per hour. It is known that C is located between west and northwest of B, with BC at 60° to AB. Also, C is located between south and southwest of A with AC at 30° to AB. The latest time by which Rahim must leave A and still catch the train is closest to
a) 6: 15 am
b) 6: 30 am
c) 6:45 am
d) 7: 00 am
e) 7: 15 am
Question 4:
A truck traveling at 70 kilometers per hour uses 30% more diesel to travel a certain distance than it does when it travels at the speed of 50 kilometers per hour. If the truck can travel 19.5 kilometers on a liter of diesel at 50 kilometers per hour, how far can the truck travel on 10 liters of diesel at a speed of 70 kilometers per hour?
a) 130
b) 140
c) 150
d) 175
Question 5:
Let T be the set of integers {3,11,19,27,…451,459,467} and S be a subset of T such that the sum of no two elements of S is 470. The maximum possible number of elements in S is
a) 32
b) 28
c) 29
d) 30
Answers and Solutions for CAT Aptitude Questions PDF:
Solutions:
1) Answer (C)
Let the first operation be (1+40-1) = 40, the second operation be (2+39-1) = 40 and so on
So, after 20 operations, all the numbers are 40. After 10 more operations, all the numbers are 79
Proceeding this way, the last remaining number will be 781
2) Answer (C)
The terms of the first sequence are of the form 4p + 13
The terms of the second sequence are of the form 5q + 11
If a term is common to both the sequences, it is of the form 4p+13 and 5q+11
or 4p = 5q -2. LHS = 4p is always even, so, q is also even.
or 2p = 5r – 1 where q = 2r.
Notice that LHS is again even, hence r should be odd. Let r = 2m+1 for some m.
Hence, p = 5m + 2.
So, the number = 4p+13 = 20m + 21.
Hence, all numbers of the form 20m + 21 will be the common terms. i.e 21,41,61,…,401 = 20.
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3) Answer (B)
According to given conditions angle between AC and AB is 30 degrees and between AB and BC is 60 degrees. So the triangle formed is a 30-60-90 triangle.
So, total time taken by train is 5 hrs, hence the train reaches at 1 pm. Accordingly, Rahim has to reach C fifteen minutes before i.e. at 12:45 PM.
Time taken by Rahim to travel by car is around 6.2 hrs. So, the latest time by which Rahim must leave A and still be able to catch the train is 6:30 am.
4) Answer (C)
If the truck is being driven at 70 kmph, it takes 1.3 liters of diesel to travel 19.5 km.
Therefore, with 10 liters of diesel, the truck can travel 10/1.3 * 19.5 km = 150 km.
5) Answer (D)
No. of terms in series T , 3+(n-1)*8 = 467 i.e. n=59.
Now S will have atleast have of 59 terms i.e 29 .
Also the sum of 29th term and 30th term is less than 470.
Hence, maximum possible elements in S is 30.
We hope that this CAT Aptitude Questions PDF is very useful to you.
