CAT 2022 Question Paper With Answers & Explanation

CAT 2022 Question Paper With Answers & Explanation

Section-1

Q. 1 The arithmetic mean of all the distinct numbers that can be obtained by rearranging the digits in 1421, including itself, is

1) 2222

2) 2442

3) 3333

4) 2592

Correct Answer
and
Explanation

Correct Answer

1

Explanation

The number of possible rearrangements of the
number 1421 = 4!/2! = 12
Now, consider only the units place.
1 occurs 6 times in the unit’s place.
2 occurs 3 times in the unit’s place.
4 occurs 3 times in the unit’s place.
The total sum of digits of the units place in all the 12 arrangements combined
= 1 × 6 + 2 × 3 + 4 × 3 = 24
Similarly, the sum of digits in ten’s, hundred’s or thousand’s place will also be 24.
So the sum of all the 12 rearrangements = 24 × 1111
Hence, the average of all these rearrangements = 24 × 1111/12 = 2222.

Q. 2 The average of all 3-digit terms in the arithmetic progression 38, 55, 72, ..., is

Correct Answer
and
Explanation

Correct Answer

548

Explanation

The common difference of the AP 38, 55, 72, … is 17.
The smallest 3-digit number in the AP is 106 and the largest 3-digit number in it is 990.
So 106 + 123 + … + 973 + 990
= 106 + (106 + 17) + (106 + 2 × 17) + … +
(106 + 52 × 17)
= 53 × 106 + 17(1 + 2 + 3 + … + 52)
= 53 × 106 + 17 × 52 × 53/2
= 53 × 106 + 17 × 26 × 53
Hence, average of all 3-digit numbers
= (53 × 106 + 17 × 26 × 53)/53 = 106 + 442 = 548.

Q. 3 Two ships are approaching a port along straight routes at constant speeds. Initially, the two ships and the port formed an equilateral triangle with sides of length 24 km. When the slower ship travelled 8 km, the triangle formed by the new positions of the two ships and the port became right-angled. When the faster ship reaches the port, the distance, in km, between the other ship and the port will be

1) 6

2) 4

3) 8

4) 12

Correct Answer
and
Explanation

Correct Answer

4

Explanation

Q. 4 The lengths of all four sides of a quadrilateral are integer valued. If three of its sides are of length 1 cm, 2 cm and 4 cm, then the total number of possible lengths of the fourth side is

1) 3

2) 6

3) 5

4) 4

Correct Answer
and
Explanation

Correct Answer

3

Explanation

less than the sum of the lengths of the other sides.
Let the fourth side be x.
Case 1: x is the largest side.
x < 1 + 2 + 4 or, x < 7
So x can take the values 4, 5, 6.
Case 2: 4 is the largest side and x is less than 4.
4 < x + 1 + 2 or, x > 1
So x can take the values of 2 or 3.
Hence, x can take 5 values i.e., 2, 3, 4, 5, 6.

Q. 5

1) 2

2) –1/2

3) 1/2

4) –2

Correct Answer
and
Explanation

Correct Answer

1

Explanation

Q. 6 A school has less than 5000 students and if the students are divided equally into teams of either 9 or 10 or 12 or 25 each, exactly 4 are always left out. However, if they are divided into teams of 11 each, no one is left out. The maximum number of teams of 12 each that can be formed out of the students in the school is

Correct Answer
and
Explanation

Correct Answer

150

Explanation

Let the number of students in the school be N.
N < 5000
N leaves a remainder of 4 when divided by 9, 10,
12, or 25.
N leaves a remainder of 4 when divided by LCM (9,
10, 12, 25) = 900.
N leaves a remainder of 4 when divided by 900.
N = 900x + 4
Since N < 5000, x can take range from 0 to 5.
But 900x + 4 is a multiple of 11 only when x = 2.
So N = 900 × 2 + 4 = 1804
Since 1804 = 12 × 150 + 4
Hence, when we divide these 1804 students into groups of 12, we get 150 groups.

Q. 7 A glass contains 500 cc of milk and a cup contains 500 cc of water. From the glass, 150 cc of milk is transferred to the cup and mixed thoroughly. Next, 150 cc of this mixture is transferred from the cup to the glass. Now, the amount of water in the glass and the amount of milk in the cup are in the ratio

1) 1 : 1

2) 3 : 10

3) 10 : 13

4) 10 : 3

Correct Answer
and
Explanation

Correct Answer

1

Explanation

Glass contains 500 cc of milk and cup contains 500 cc water.
After first transfer, glass contains 500 – 150 = 350 cc milk and the cup contains 150 cc milk and 500 cc water.
Now, the ratio of milk and water in the cup = 150 : 500 = 3 : 10
After second transfer, water in the glass = 150 × 3/13 cc and milk in the cup will be same i.e., 150 × 3/13 cc.
Hence, required ratio = 1 : 1.

Q. 8

1) 0

2) 4

3) 1

4) 3

Correct Answer
and
Explanation

Correct Answer

2

Explanation

Q. 9

1) – 70

2) – 60

3) – 50

4) 60

Correct Answer
and
Explanation

Correct Answer

3

Explanation

Q. 10 Suppose the medians BD and CE of a triangle ABC intersect at a point O. If area of triangle ABC is 108 sq. cm., then, the area of the triangle EOD, in sq. cm., is

Correct Answer
and
Explanation

Correct Answer

9

Explanation

Q. 11

Correct Answer
and
Explanation

Correct Answer

14

Explanation

Q. 12 A donation box can receive only cheques of Rs. 100, Rs. 250, and Rs. 500. On one good day, the donation box was found to contain exactly 100 cheques amounting to a total sum of Rs. 15250. Then, the maximum possible number of cheques of Rs. 500 that the donation box may have contained, is

Correct Answer
and
Explanation

Correct Answer

12

Explanation

Q. 13 Nitu has a an initial capital of Rs. 20,000. Out of this, she invests Rs. 8,000 at 5.5% in bank A, Rs. 5,000 at 5.6% in bank B and the remaining amount at x% in bank C, each rate being simple interest per annum. Her combined annual interest income from these investments is equal to 5% of the initial capital. If she had invested her entire initial capital in bank C alone, then her annual interest income, in rupees, would have been

1) 700

2) 800

3) 1000

4) 900

Correct Answer
and
Explanation

Correct Answer

2

Explanation

Q. 14 In a triangle ABC, AB = AC = 8 cm. A circle drawn with BC as diameter passes through A. Another circle drawn with center at A passes through B and C. Then the area, in sq. cm, of the overlapping region between the two circles is

1) 32(π – 1)

2) 16(π – 1)

3) 32π

4) 16π

Correct Answer
and
Explanation

Correct Answer

1

Explanation

Q. 15 In an examination, the average marks of students in sections A and B are 32 and 60, respectively. The number of students in section A is 10 less than that in section B. If the average marks of all the students across both the sections combined is an integer, then the difference between the maximum and minimum possible number of students in section A is

Correct Answer
and
Explanation

Correct Answer

63

Explanation

Q. 16

1) x ≤ r

2) x > r

3) x ≥ r

4) x ≠ r

Correct Answer
and
Explanation

Correct Answer

1

Explanation

For x = r,
f(r) = 2r – r = r
So f(f(r)) = f(r) = r
For x < r,
f(x) = r
f(f(x)) = f(r) = r
For x > r,
Let x = kr, where k is some number greater than 1.
f(x) = f(kr) = 2kr – r = (2k – 1)r
Since k > 0, 2k – 1 > k
So for any x > r, f(f(x)) ≠ f(f(r))
Hence, for f(f(x)) = f(x) to be satisfied, the necessary condition is, x ≤ r.

Q. 17 Two cars travel from different locations at constant speeds. To meet each other after starting at the same time, they take 1.5 hours if they travel towards each other, but 10.5 hours if they travel in the same direction. If the speed of the slower car is 60 km/hr, then the distance traveled, in km, by the slower car when it meets the other car while traveling towards each other, is

1) 90

2) 150

3) 100

4) 120

Correct Answer
and
Explanation

Correct Answer

1

Explanation

Speed of slower car = 60 km/h and they take 1.5 hours if they travel towards each other, to meet each other after starting at the same time.
In this case the distance covered by the slower car = 60 × 1.5 = 90 km.

Q. 18 Bob can finish a job in 40 days, if he works alone. Alex is twice as fast as Bob and thrice as fast as Cole in the same job. Suppose Alex and Bob work together on the first day, Bob and Cole work together on the second day, Cole and Alex work together on the third day, and then, they continue the work by repeating this three-day roster, with Alex and Bob working together on the fourth day, and so on. Then, the total number of days Alex would have worked when the job gets finished, is

Correct Answer
and
Explanation

Correct Answer

11

Explanation

Since Bob can finish a job in 40 days.
Then, Alex and Cole can finish this job in 20 and 60 days respectively.
Total work = LCM (20, 40, 60) = 120 units Alex, Bob and Cole can finish 6, 3 and 2 units of the job per day.
In 3 days, they can finish 2(6 + 3 + 2) = 22 units In 15 days, they can finish 5 × 22 = 110 units Out of remaining 10, 9 units of work can finish by Alex and Bob and 1 unit by Bob and Cole.
Hence, the total number of days Alex would have worked when the job gets finished, is 2 × 5 + 1 = 11 days.

Q. 19 A group of N people worked on a project. They finished 35% of the project by working 7 hours a day for 10 days. Thereafter, 10 people left the group and the remaining people finished the rest of the project in 14 days by working 10 hours a day. Then the value of N is

1) 36

2) 140

3) 23

4) 150

Correct Answer
and
Explanation

Correct Answer

2

Explanation

Q. 20 Suppose k is any integer such that the equation 2x² + kx + 5 = 0 has no real roots and the equation x² + (k – 5) x + 1 = 0 has two distinct real roots for x. Then, the number of possible values of k is

1) 13

2) 9

3) 8

4) 7

Correct Answer
and
Explanation

Correct Answer

2

Explanation

Q. 21 Moody takes 30 seconds to finish riding an escalator if he walks on it at his normal speed in the same direction. He takes 20 seconds to finish riding the escalator if he walks at twice his normal speed in the same direction. If Moody decides to stand still on the escalator, then the time, in seconds, needed to finish riding the escalator is

Correct Answer
and
Explanation

Correct Answer

60

Explanation

Q. 22 Consider six distinct natural numbers such that the average of the two smallest numbers is 14, and the average of the two largest numbers is 28. Then, the maximum possible value of the average of these six numbers is

1) 24

2) 23.5

3) 22.5

4) 23

Correct Answer
and
Explanation

Correct Answer

3

Explanation

The sum of the two smallest numbers = 14 × 2 = 28
The sum of the two largest numbers = 28 × 2 = 56 To maximize the average of all the 6 numbers, we must try to maximize the two numbers in between.
This is possible when the two largest numbers are 27 and 29.
The maximum average case is
a, b, 25, 26, 27, 29; where a + b = 28v Hence, the average of these 6 numbers
= (a + b + 25 + 26 + 27 + 29)/6
= (28 + 25 + 26 + 27 + 29)/6 = 22.5.

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