# CAT Practice : Averages, Ratios, Mixtures

Question to find the maximum and minimum possible values of the average of a few numbers given some other parts of the jigsaw.

## Mean and Range

Q.15: Consider 4 numbers a, b, c and d. Ram figures that the smallest average of some three of these four numbers is 30 and the largest average of some three of these 4 is 40. What is the range of values the average of all 4 numbers can take?

Range from 32.5 to 37.5

## Detailed Solution

We can assume a, b,c d are in ascending order (with the caveat that numbers can be equal to each other)
a + b + c = 90
b + c + d = 120

We need to find the maximum and minimum value of a + b + c + d.
a + b + c + d = 120 + a. So, this will be minimum when a is minimum. Given a + b + c = 90. a is minimum when b + c is maximum. If b + c is maximum, d should be minimum.
Given that b + c + d = 120, the minimum value d can take is 40 as d cannot be less than b or c.
The highest value b + c can take is 80, when b = c = d = 40. When b = c = d = 40, a = 10.
a + b + c + d = 130. Average = 32.5

Similarly, a + b + c + d = 90 + d. So, this will be maximum when d is maximum.
Given b + c + d = 120. d is maximum when b + c is minimum. If b + c is minimum, a should be maximum.
Given that a + b + c = 90, the maximum value a can take is 30 as a cannot be greater than b or c. The lowest value b + c can take is 60, when a = b = c = 30. When a = b = c = 30, d = 60.
a + b + c + d = 150. Average = 37.5
So, the average has to range from 32.5 to 37.5

Correct Answer: Range from 32.5 to 37.5

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