# CAT 2022 QA - Slot 3

Get the complete CAT 2022 paper with answers and explanations

## 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

###### 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

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.

4

#### 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

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.

1

#### 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

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

###### 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.

2

3

9

14

12

2

1

63

#### Q. 16

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

###### 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

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.

2

2

60