# CAT Average Questions PDF [Most Important]

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Averages is one of the important topics in the CAT Quant (Arithmetic) section. These questions are not very tough, and hence you should not miss out on the questions from Averages. Firstly, understand the concept of Averages, including all the Important Formulas. Solve & practice more questions from CAT Average so that you can easily solve these questions in the exam. You can check out these CAT Average questions from the CAT Previous year papers. In this post, we will look into some important Average Questions for CAT. These are a good source of practice for CAT preparation; If you want to practice these questions, you can download these Important Average Questions for CAT (with detailed answers) PDF along with the video solutions below, which is completely Free.

Question 1: Consider the set S = {2, 3, 4, …., 2n+1}, where n is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X – Y ?

a) 0

b) 1

c) (1/2)*n

d) (n+1)/2n

e) 2008

Solution:

The odd numbers in the set are 3, 5, 7, …2n+1

Sum of the odd numbers = 3+5+7+…+(2n+1) = $n^2 + 2n$

Average of odd numbers = $n^2 + 2n$/n = n+2

Sum of even numbers = 2 + 4 + 6 + … + 2n = 2(1+2+3+…+n) = 2*n*(n+1)/2 = n(n+1)

Average of even numbers = n(n+1)/n = n+1

So, difference between the averages of even and odd numbers = 1

Question 2: Three classes X, Y and Z take an algebra test.
The average score in class X is 83.
The average score in class Y is 76.
The average score in class Z is 85.
The average score of all students in classes X and Y together is 79.
The average score of all students in classes Y and Z together is 81.
What is the average for all the three classes?

a) 81

b) 81.5

c) 82

d) 84.5

Solution:

Let x , y and z be no. of students in class X, Y ,Z respectively.

From 1st condition we have

83*x+76*y = 79*x+79*y which give 4x = 3y.

Next we have 76*y + 85*z = 81(y+z) which give 4z = 5y .

Now overall average of all the classes can be given as $\frac{83x+76y+85z}{x+y+z}$

Substitute the relations in above equation we get,

$\frac{83x+76y+85z}{x+y+z}$  = (83*3/4 + 76 + 85*5/4)/(3/4 + 1 + 5/4) = 978/12 = 81.5

Question 3: Consider a sequence of seven consecutive integers. The average of the first five integers is n. The average of all the seven integers is:
[CAT 2000]

a) n

b) n+1

c) kn, where k is a function of n

d) n+(2/7)

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

Question 4: The average marks of a student in 10 papers are 80. If the highest and the lowest scores are not considered, the average is 81. If his highest score is 92, find the lowest.

a) 55

b) 60

c) 62

d) Cannot be determined

Solution:

Total marks = 80 x 10 = 800
Total marks except highest and lowest marks = 81 x 8 = 648
So Summation of highest marks and lowest marks will be = 800 – 648 = 152
When highest marks is 92, lowest marks will be = 152-92 = 60

Question 5: Total expenses of a boarding house are partly fixed and partly varying linearly with the number of boarders. The average expense per boarder is Rs. 700 when there are 25 boarders and Rs. 600 when there are 50 boarders. What is the average expense per boarder when there are 100 boarders?

a) 550

b) 580

c) 540

d) 560

Solution:

Let the fixed income be x and the number of boarders be y.

x + 25y = 17500

x + 50y = 30000

=> y = 500 and x = 5000

x + 100y = 5000 + 50000 = 55000

Average expense = $\frac{55000}{100}$ = Rs.550.

Question 6: A class consists of 20 boys and 30 girls. In the mid-semester examination, the average score of the girls was 5 higher than that of the boys. In the final exam, however, the average score of the girls dropped by 3 while the average score of the entire class increased by 2. The increase in the average score of the boys is

a) 9.5

b) 10

c) 4.5

d) 6

Solution:

Let, the average score of boys in the mid semester exam is A.
Therefore, the average score of girls in the mid semester exam be A+5.
Hence, the total marks scored by the class is $20\times (A) + 30\times (A+5) = 50\times A + 150$

The average score of the entire class is $\dfrac{(50\times A + 150)}{50} = A + 3$

wkt, class average increased by 2, class average in final term $= (A+3) + 2 = A + 5$

Given, that score of girls dropped by 3, i.e $(A+5)-3 = A+2$
Total score of girls in final term $= 30\times(A+2) = 30A + 60$

Total class score in final term $= (A + 5)\times50 = 50A + 250$

the total marks scored by the boys is $(50A + 250) – (30A – 60) = 20A + 190$
Hence, the average of the boys in the final exam is $\dfrac{(20G + 190)}{20} = A + 9.5$

Hence, the increase in the average marks of the boys is $(A+9.5) – A = 9.5$

Question 7: 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

a) 30

b) 28

c) 32

d) 26

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

Question 8: A CAT aspirant appears for a certain number of tests. His average score increases by 1 if the first 10 tests are not considered, and decreases by 1 if the last 10 tests are not considered. If his average scores for the first 10 and the last 10 tests are 20 and 30, respectively, then the total number of tests taken by him is

Solution:

Let the total number of tests be ‘n’ and the average by ‘A’
Total score = n*A
When 1st 10 tests are excluded, decrease in total value of scores = (nA – 20 * 10) = (nA – 200)
Also, (n – 10)(A + 1) = (nA – 200)
On solving, we get 10A – n = 190……….(i)
When last 10 tests are excluded, decrease in total value of scores = (nA – 30 * 10) = (nA – 300)
Also, (n – 10)(A – 1) = (nA – 300)
On solving, we get 10A + n = 310……….(ii)
From (i) and (ii), we get n = 60
Hence, 60 is the correct answer.

Question 9: The average of 30 integers is 5. Among these 30 integers, there are exactly 20 which do not exceed 5. What is the highest possible value of the average of these 20 integers?

a) 3.5

b) 5

c) 4.5

d) 4

Solution:

It is given that the average of the 30 integers = 5

Sum of the 30 integers = 30*5=150

There are exactly 20 integers whose value is less than 5.

To maximise the average of the 20 integers, we have to assign minimum value to each of the remaining 10 integers

So the sum of 10 integers = 10*6=60

The sum of the 20 integers = 150-60= 90

Average of 20 integers = $\ \frac{\ 90}{20}$ = 4.5

Question 10: 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

a) 53

b) 51

c) 48

d) 49

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

Question 11: 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

a) 4

b) 3

c) 2

d) 1

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

Question 12: The average age of three friends is 20 years and their ages are in the proportion 2:3:5, the age of the oldest friend is :

a) 12 years

b) 18 years

c) 30 years

d) 40 years

Solution:

Given average age of three friends is 20 years

Sum of their ages = $3\times\ 20$ = 60 years

Given, their ages are in ratio 2:3:5. Let their ages be 2x, 3x and 5x

2x + 3x + 5x = 60

10x = 60

x = 6

Their ages are 12 years, 18 years and 30 years

Age of oldest friend = 30 years

Question 13: The average temperature of Tuesday, Wednesday and Thursday was $45^\circ C$. The average temperature of Wednesday, Thursday and Friday was $46^\circ C$. If the temperature on Friday was $47^\circ C$, what was the temperature on Tuesday?

a) $44^\circ C$

b) $48^\circ C$

c) $40^\circ C$

d) $46^\circ C$

Solution:

Sum of temperatures on Wednesday, Thursday and Friday= 46*3 = 138

The temperature on friday= 47

So, the sum of temperatures on Wednesday and Thursday= 138-47 = 91

Sum of temperatures on Tuesday, Wednesday and Thursday= 45*3 = 135

The temperature on Tuesday= 135-91= 44

Question 14: In a football tournament, a player has played a certain number of matches and 10 more matches are to be played. If he scores a total of one goal over the next 10 matches, his overall average will be 0.15 goals per match. On the other hand, if he scores a total of two goals over the next 10 matches, his overall average will be 0.2 goals per match. The number of matches he has played is

Solution:

Let Total matches played be n and in initial n-10 matches his goals be x
so we get $\frac{\left(x+1\right)}{n}=0.15$
we get x+1 =0.15n                (1)
From condition (2) we get :
$\frac{\left(x+2\right)}{n}=0.2$
we get x+2 = 0.2n                  (2)
Subtracting (1) and (2)
we get 1 =0.05n
n =20
So initially he played n-10 =10 matches

Question 15: 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

a) 26

b) 18

c) 16

d) 20

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