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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