# Averages Questions for CAT

0
7516

Averages Questions for CAT:

From averages, ratio and proportion around 3 questions will appear in CAT exam. And sometimes the DI and LR questions can also be asked based on the topic of the averages. It is a basic concept to learn for CAT. So we have provided some important averages questions and answers for CAT. Download PDF of CAT averages questions with solutions and practice them.

Question 1: The average weight of a class of 10 students is increased by 2 kg when one student of 30kg left and another student joined. After a few months, this new student left and another student joined whose weight was 10 less than the student who left now. What is the difference between the final and initial averages?

a) 11
b) 1
c) 111
d) 121

Question 2: There are n weights in a bag measuring 1kg, 2kg and so on till n kg. These weights are divided into 3 parts. The first part contains the weights 1kg, 4kg, 7kg and so on. The second part contains the weights 2kg, 5kg, 8kg and so on. The third part contains the remaining weights. The average weights any two of the three parts is equal to the weight present in those parts but the average weight of the remaining one part is not equal to the weight present in that part. Which of the following can be a possible value of n?

a) 90
b) 93
c) 96
d) 88

Question 3: There are 7 members in a family whose average age is 25 years. Ram who is 12 years old is the second youngest in the family. Find the average age of the family in years just before Ram was born ?

a) 15.167
b) 18.2
c) 13
d) cannot be determined

Question 4: The average weight of a class is 54 kg. A student, whose weight is 145 kg, joined the class and the average weight of the class now becomes a prime number less than 72. Find the total number of students in the class now.

a) 7
b) 13
c) 15
d) Cannot be determined

Question 5: The average runs scored by Virat Kohli in four matches is 48. In the fifth match, Kohli scores some runs such that his average now becomes 60. In the 6th innings he scores 12 runs more than his fifth innings and now the average of his last five innings becomes 78. How many runs did he score in his first innings? (He does not remain not out in any of the innings)

a) 30
b) 50
c) 70
d) 90

Solutions for Averages questions for CAT:

Solutions:

Change in total weight of 10 students = difference in weight of the student who joined and the student who left
=> weigth of first student who left = 30 + (10×2) = 50
weight of the student who joined last = 50 – 10 = 40
Thus change in average weight = (40 – 30)/10 = 1
Alternate Solution:
The average weight increases by 2 when a student is replaced by another student. Since, there are 10 students, total change(increase as +2) in weight is 2 x10 = 20kg
Since, the weight of the new student is 30kg, the weight of the replaced student would be 30 – 20 = 10kg. Also after the 2nd replacement a new student having weight 10kg less than the 30kg student joins the class. Thus, the weight of the new student = 30 -10 = 20kg.
The difference in total weight between the first case and after two replacement is thus 20 – 10 = 10kg. Thus, difference is average weight = 10/10 = 1kg

We know that if in an AP the number of terms in a series is odd, the average of the terms of the series is equal to the middle term of the series. However if the number of terms in the series is even, the average of all the terms of the series is not equal to one of the terms of the series. Hence the three part contain terms 2x+1, 2x+1, 2x or 2x-1, 2x-1, 2x
Hence the total number of parts = 2x+1+2x+1+2x or 2x-1+2x-1+2x = 6x+2 or 6x-2
Among the options, the only number of the form 6x+2 or 6x-2 is 88. Hence 88 can be the required value of n.

In order to find the average age of the family before Ram was born, we need to know the age of the youngest member of the family. Thus, D is the right choice.

Let the original number of students in the class be N.
Total weight of the class = 54N
New total weight of the class = 54N + 145
New average weight of the class = (54N + 145)/(N+1) = (54N + 54)/(N+1) + 91/(N+1) = 54 + 91/(N+1).
Since the new average is an integer, (N+1) should be a factor of 91.
If N+1 = 7, the new average becomes 54 + 91/7 = 54 + 13 = 67
and if N+1 = 13, then the new average becomes 54 + 91/13 = 54 + 7 = 61
Both 67 and 61 are prime numbers less than 72. So, we cannot uniquely determine the number of students in the class.