sameerdrana wrote:

Three boxes of supplies have an average(arithmetic mean) weight of 7 kilograms and a median weight of 9kg. What is the max possible weight, in kg, of the lightest box?

A. 1

B. 2

C. 3

D. 4

E. 5

Anestists wrote:

What if the Boxes where Five (5) instead of three (3)? What would then be the appropriate approach regarding this problem?

Anestists ,the approach would be similar:

minimize everything other than the lightest box in order to maximize that very box.

I assume you intended a mean of 7, and a median of 9.

As before, we assume that the weights must be in integers.

1) Median of 9, five numbers: we have A, B, 9, D, E

2) Find total weight of all 5 boxes. A*n=S: (5 * 7) = 35

3) Minimize D and E. We have to subtract 9, D, and E from 35 first

Smaller numbers, subtracted from 35, will leave a larger amount left for boxes A and B

Let D and E both = 9 (they cannot be smaller than the median)

4) Now we have A, B, 9, 9, 9

5) Amount left for boxes A and B: 35 - 27 = 8 kg

6) What is the max possible weight, in kg, of the lightest box?

Minimize box B to maximize box A

A and B - Possibilities

1 and 7

2 and 6

3 and 5

4 and 4

I suspect most people would pick A = 3 and B = 5

The question asks for THE lightest box, implying there is only one.

In ordinal and ordinary terms, least implies lesser: A must weigh less than B to qualify as the "lightest."

ANSWER: "The maximum possible weight, in kg, of the lightest box, is 3."

Is that what you were wanting to know?

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