CAT Averages Questions & Solutions

A sample of real CAT Averages past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 30 Averages questions in all — sign up free to practise them timed.

1. Q1.CAT 2025

   The average salary of 5 managers and 25 engineers in a company is 60000 rupees. If each of the managers received 20% salary increase while the salary of the engineers remained unchanged, the average salary of all 30 employees would have increased by 5%. The average salary, in rupees, of the engineers is

   • 45000

   • 50000

   • 54000

   • 40000

   Show solution

   Let the average salary of managers be $x$ and let the average salary of engineers be $y$ . The total salary of all the employees will be $(5+25)*60000 = 1800000$

   We have, $5x+25y = 1800000$ .... (1)

   If the average salary of all the employees increases by $5\%$ , the total salary of all the employees will also increase by $5\%$ , because the total number of employees remains the same. The new total salary will be, $1800000\times 1.05 = 1890000$

   Also, the average salary of all the managers has increased by $20\%$ and has become $1.2x$ , we have

   $5*1.2x + 25y = 1890000$ ,  or  6x + 25y  = 1890000 .... (2)

   Subtracting equation (1) from equation (2), we get,

   $x = 90000$

   Which gives $25y = 1800000 - 90000*5$   or  $y = \dfrac{1350000}{25} = 54000$

   Therefore, the correct answer is option C.

2. Q2.CAT 2024

   There are four numbers such that average of first two numbers is 1 more than the first number, average of first three numbers is 2 more than average of first two numbers, and average of first four numbers is 3 more than average of first three numbers. Then, the difference between the largest and the smallest numbers, is

   Answer: 15

   Show solution

   Let us assume the four numbers to be a, b, c and d in ascending order. 

   Average of first two numbers is 1 more than the first number
   $\frac{\left(a+b\right)}{2}=a+1$
   $b-a=2$
   $b=a+2$
   
   

   Average of first three numbers is 2 more than average of first two numbers
   $\frac{\left(a+b+c\right)}{3}=\frac{\left(a+b\right)}{2}+2$
   $2c=a+b+12$
   Substituting the value for b
   $2c=a+a+2+12$
   $2c=2a+14$
   $c=a+7$

   Average of first four numbers is 3 more than average of first three numbers.
   $\frac{\left(a+b+c+d\right)}{4}=\frac{\left(a+b+c\right)}{3}+3$
   $3d=a+b+c+36$
   Substituting the value of b and c
   $3d=a+a+2+a+7+36$
   $3d=3a+45$
   $d=a+15$

   d is the largest and a is the smallest and we know that d=a+15

   Hence the difference between the smallest and the largest values is 15. 

3. Q3.CAT 2023

   In a company, 20% of the employees work in the manufacturing department. If the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company, then the ratio of the average salary obtained by the manufacturing employees to the average salary obtained by the nonmanufacturing employees is

   • 6:5

   • 4:5

   • 5:4

   • 5:6

   Show solution

   Let the number of total employees in the company be 100x, and the total salary of all the employees be 100y.

   It is given that 20% of the employees work in the manufacturing department, and the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company.

   Hence, the total number of employees in the manufacturing department is 20x, and the total salary received by them is (100y/6)

   Average salary in the manufacturing department = (100y/6*20x) = 5y/6x

   Similarly, the total number of employees in the nonmanufacturing department is 80x, and the total salary received by them is (500y/6)

   Hence, the average salary in the nonmanufacturing department = (500y/6*80x) = 25y/24x

   Hence, the ratio is:- (5y/6x): (25y/24x) 

   => 120: 150 = 4:5

   The correct option is B

4. Q4.CAT 2022

   The average of three integers is 13. When a natural number n is included, the average of these four integers remains an odd integer. The minimum possible value of n is

   • 3

   • 4

   • 5

   • 1

   Show solution

   It is given that average of three numbers is 13.

   Sum = 3*13 = 39

   It is given,  $\ \frac{\ 39+n}{4}$ is a odd number.

   Minimum value  $\ \frac{\ 39+n}{4}$ can take such that n is a natural number is 11

   $\ \frac{\ 39+n}{4}=11$

   n = 5

   The answer is option C.

5. Q5.CAT 2022

   Five students, including Amit, appear for an examination in which possible marks are integers between 0 and 50, both inclusive. The average marks for all the students is 38 and exactly three students got more than 32. If no two students got the same marks and Amit got the least marks among the five students, then the difference between the highest and lowest possible marks of Amit is

   • 22

   • 21

   • 24

   • 20

   Show solution

   The average marks for all the students is 38.

   Sum = 5*38 = 190

   To find the minimum marks scored by Amit, we need to maximise the score of remaining students.

   Maximum scores sum of remaining students = 50 + 49 + 48 + 32 = 179

   Minimum possible score of Amit = 190 - 179 = 11

   It is given, Amit scored least. This implies maximum possible score of Amit is 31.

   Difference = 31 - 11 = 20

   The answer is option D.
   

6. Q6.CAT 2021

   Onion is sold for 5 consecutive months at the rate of Rs 10, 20, 25, 25, and 50 per kg, respectively. A family spends a fixed amount of money on onion for each of the first three months, and then spends half that amount on onion for each of the next two months. The average expense for onion, in rupees per kg, for the family over these 5 months is closest to

   • 26

   • 18

   • 16

   • 20

   Show solution

   Let us assume the family spends Rs. 100 each month for the first 3 months and then spends Rs. 50 in each of the next two months.

   Then amount of onions bought = 10, 5, 4, 2, 1, for months 1-5 respectively.

   Total amount bought = 22kg.

   Total amount spent = 100+100+100+50+50 = 400.

   Average expense =  $\frac{400}{22}=Rs.18.18\approx\ 18$

7. Q7.CAT 2020

   A batsman played n + 2 innings and got out on all occasions. His average score in these n + 2 innings was 29 runs and he scored 38 and 15 runs in the last two innings. The batsman scored less than 38 runs in each of the first n innings. In these n innings, his average score was 30 runs and lowest score was x runs. The smallest possible value of x is

   • 4

   • 3

   • 2

   • 1

   Show solution

   Given, \frac{\text{sum of scores in n matches+38+15}}{n+2}=29

   Given,  \frac{\text{sum of scores in n matches}}{n}=30

   => 30n + 53 = 29(n+2) => n=5

   Sum of the scores in 5 matches = 29*7 - 38-15 = 150

   Since the batsmen scored less than 38, in each of the first 5 innings. The value of x will be minimum when remaining four values are highest

   => 37+37+37+37 + x = 150

   => x = 2

8. Q8.CAT 2019

   Ramesh and Gautam are among 22 students who write an examination. Ramesh scores 82.5. The average score of the 21 students other than Gautam is 62. The average score of all the 22 students is one more than the average score of the 21 students other than Ramesh. The score of Gautam is

   • 53

   • 51

   • 48

   • 49

   Show solution

   Assume the average of 21 students other than Ramesh = a

   Sum of the scores of 21 students other than Ramesh = 21a

   Hence the average of 22 students = a+1

   Sum of the scores of all 22 students = 22(a+1)

   The score of Ramesh = Sum of scores of all 22 students - Sum of the scores of 21 students other than Ramesh = 22(a+1)-21a=a+22  = 82.5   (Given)

   => a = 60.5

   Hence, sum of the scores of all 22 students = 22(a+1) = 22*61.5 = 1353

   Now the sum of the scores of students other than Gautam = 21*62 = 1302

   Hence the score of Gautam = 1353-1302=51

9. Q9.CAT 2017

   An elevator has a weight limit of 630 kg. It is carrying a group of people of whom the heaviest weighs 57 kg and the lightest weighs 53 kg. What is the maximum possible number of people in the group?

   Answer: 11

   Show solution

   It is given that the maximum weight limit is 630. The lightest person's weight is 53 Kg and the heaviest person's weight is 57 Kg.

   In order to have maximum people in the lift, all the remaining people should be of the lightest weight possible, which is 53 Kg.

   Let there be n people.

   53 + n(53) + 57 = 630

   n is approximately equal to 9.8. Hence, 9 people are possible.

   Therefore, a total of 9 + 2 = 11 people can use the elevator.

10. Q10.CAT 2017

    The average height of 22 toddlers increases by 2 inches when two of them leave this group. If the average height of these two toddlers is one-third the average height of the original 22, then the average height, in inches, of the remaining 20 toddlers is

    • 30

    • 28

    • 32

    • 26

    Show solution

    Let the average height of 22 toddlers be 3x.
    Sum of the height of 22 toddlers = 66x
    Hence average height of the two toddlers who left the group = x
    Sum of the height of the remaining 20 toddlers = 66x - 2x = 64x
    Average height of the remaining 20 toddlers = 64x/20 = 3.2x
    Difference = 0.2x = 2 inches => x = 10 inches
    Hence average height of the remaining 20 toddlers = 3.2x = 32 inches
    

11. Q11.CAT 2002

    Amol was asked to calculate the arithmetic mean of 10 positive integers, each of which had 2 digits. By mistake, he interchanged the 2 digits, say a and b, in one of these 10 integers. As a result, his answer for the arithmetic mean was 1.8 more than what it should have been. Then |b - a| equals

    • 1

    • 2

    • 3

    • None of these

    Show solution

    Let the actual average be n. So, the new average is n + 1.8
    Actual total = 10n
    New total = 10n + 18

    Let the number which was miswritten = ab(a is the tenth's digit and b is the units digit) = 10a+b

    and reversed number ba = 10b+a

    
    So, 10b + a - (10a + b) = 18
    => 9(b-a) = 18
    => b-a = 2

12. Q12.CAT 2000

    Consider a sequence of seven consecutive integers. The average of the first five integers is n. The average of all the seven integers is:
    

    • n

    • n+1

    • kn, where k is a function of n

    • n+(2/7)

    Show solution

    The first five numbers could be n-2, n-1, n, n+1, n+2. The next two number would then be, n+3 and n+4, in which case, the average of all the 7 numbers would be $\frac{(5n+2n+7)}{7}$ = n+1

CAT Averages previous year questions with solutions — AthenaPrep