CAT Practice : Averages, Ratios, Mixtures

One needs to recognize that averages and mixtures are effectively two sides of the same coin. This question is designed to drive home that point.

Averages - Sets of numbers

    Q.14: Natural numbers 1 to 25 (both inclusive) are split into 5 groups of 5 numbers each. The medians of these 5 groups are A, B, C, D and E. If the average of these medians is m, what are the smallest and the largest values m can take?

 

  • Correct Answer
    9, 17

Detailed Solution

Let us try to construct a scenario for getting the smallest value of the average. We need to have the minimum value for each of the 5 medians.

Now, what is the lowest value the median of any of the 5 groups can take?

The median is the middle term among the 5 terms. So, it cannot be 1 or 2, but it can be 3. If the set were 1, 2, 3, 4, 5, the median would be 3. If we choose a sub-group such as this, the smallest median fro the next group will be 8. The next group could be 6, 7, 8, 9, 10. The idea is to create sub-groups in such a way so that the second group can have a smaller median than 8.

Now, even a sub-group that has the numbers 1, 2, 3, 24, 25 would have the median as 3. So, if we want to keep the medians as small as possible, we should choose sub-groups that have small numbers till the median, and then very large numbers. Because, these large numbers do not affect the median, we will still have small numbers to choose from for the next group. So, choose sub-groups such that the first 3 numbers are small, the final two are as large as possible.

1, 2, 3, 24, 25
4, 5, 6, 22, 23
7, 8, 9, 20, 21
10, 11, 12, 18, 19
13, 14, 15, 16, 17

would be an ideal set of groups. The medians would be 3, 6, 9, 12, 15, and the average of the medians would be 9.
Similar set of groups can be found to find the highest value of the average. The medians would be 23, 20, 17, 14, 11 and the average would be 17.

Correct Answer: Smallest value: 9, Highest Value: 17



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