Practice Test: Averages - 2 - CAT MCQ

Let A, B, C represent their respective weights. Then, we have:

A + B + C = (45 x 3) = 135 .... (i)

A + B = (40 x 2) = 80 .... (ii)

B + C = (43 x 2) = 86 ....(iii)

Adding (ii) and (iii), we get: A + 2B + C = 166 .... (iv)

Subtracting (i) from (iv), we get : B = 31.

 B's weight = 31 kg.

Sum of the present ages of husband, wife and child = (27 * 3 + 3 * 3) years = 90 years.
Sum of the present ages of wife and child = (20 * 2 + 5 * 2) years = 50 years.
Husband's present age = (90 - 50) years 
= 40 years

It is given that average of three numbers is 13.
Sum = 3*13 = 39
It is given, is a odd number.
Minimum valuecan take such that n is a natural number is 11 = 11
n = 5
The answer is option C.

a + 2b = 6
From the above equation, we can say that maximum value b can take is 3 and minimum value b can take is 0.
a + b + b = 6
a + b = 6 - b
a + b is maximum when b is minimum, i.e. b = 0
Maximum value ofa + b = 6- 0 = 6
a + b is minimum when b is maximum, i.e. b = 3
Minimum value ofa + b = 6- 3 = 3
Average = 
The answer is option D.

Given,

Given,

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

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 = 400/22 = Rs. 18.18 ≈ 18 

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

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 = 90/20 = 4.5