# CAT Practice : Averages, Ratios, Mixtures

Averages

## Averages

Q.45: In the previous question, what can be the minimum and maximum value of A’s average if 5 teams are transferred from B to A?
1. 20.8 and 22.4
2. 20.6 and 22.6
3. 20.8 and 22.8
4. 20.8 and 21.8

Choice D. 20.8 and 21.8

## Detailed Solution

For minimum, teams with minimum score would be transferred from B to A
So, total score transferred = 24 x 5 = 120
A(Minimum Average) = $\frac{(20*20 + 120)}{25}$ = 20.8
For maximum, if we take teams with maximum score,
total score transferred = 32 x 5 = 160
However, if we assume that 5 teams scored the highest score, the average of remaining teams in B would come out to be,
$\frac{(625-160)}{20} = \frac{465}{20} = 23.25$ which is lower than the lowest score of B. So, 5 teams in B can’t have score of 32.
so, minimum total score which need to be on group B = 24 x 20 = 480
Score that can be transferred = 625 – 480 = 145
A(Maximum Average) = $\frac{(400 + 145)}{25} = \frac{545}{25} = 21.8$

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