# 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.7: In a class of 5 students, average weight of the 4 lightest students is 40 kgs, Average weight of the 4 heaviest students is 45 kgs. What is the difference between the the maximum and minimum possible average weight overall?
1. 2.8 kgs
2. 3.2 kgs
3. 3 kgs
4. 4 kgs

Choice C. 3 kgs

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## Detailed Solution

Lets say that the students are named a, b, c, d and e, in increasing order of weights. The average of a, b, c and d is 40 kg, whereas the average of b, c, d and e is 45 kg.
The sum of a, b, c and d is 160 kg, and the sum of b, c, d and e is 180 kg.

What is the total weight of all the students?
There are two ways of looking at this.
a) 160 + e
b) 180 + a
Or, e is 20 more than a.

The total weight is 160 + e. So, the highest value of e will correspond to the highest possible average. The highest possible value of e occurs when it is 20 higher than the highest possible value for a, which is 40 (all the first 4 scores are equal to 40).

So, the highest possible average is

160 + 60 / 5
= 44
This will be the case when the weights are 40 kgs, 40kgs, 40 kgs, 40 kgs and 60 kgs.

Conversely, the least possible value for the average occurs when a is the least. This happens when e is the least too (since a is 20 less than e).

The least possible value for e is 45 =
180 / 4
.
So, the least possible value for a would be 25.

The least possible average =
180 + 25 / 5
= 41

This will be the case when the weights are 25 kgs, 45kgs, 45 kgs, 45 kgs and 45 kgs. So, the difference between maximum possible and minimum possible average = 3 kgs.

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