Question

Three boxes of supplies have an average (arithmetic mean) weight of 7 kilograms and a median weight of 9 kilograms. What is the maximum possible weight, in kilograms, of the lightest box?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Answer 1

Answer:

(C) 3 kg is the maximum possible weight of the lightest box

Step-by-step explanation:

given data is:

1. mean = 7 kg
2. median = 9 kg
3. total number of boxes = 3

so since we know that the median is 9 and the total boxes are 3. We already know the weight of one of the three boxes.

so we can think of the sum of the three weights in ascending order as: A+9+B.

here A and B are the unknown weights of the remaining two boxes.

• We can first use all of the given data in the mean formula

mean = (sum of all weights)/(number of boxes)

put all the known values in the formula

$7 = \frac{A+9+B}{3}$

and rearrange the formula so one of A and B as the subject.

$A = 21 - (9+B)$

now you can use the given values in the Multiple choices from the question and find the maximum value. try each value to be equal to B, so you can find A.

but you need to be smart about it, and keep in mind that the only one of the value of A and B should be greater than the median i.e. 9, or else that will rearrange the way we have set up our boxes:: A + 9 + B (ascending order).

• start from (E), where B = 5

$A = 21 - (9+5)$

$A = 7$

here, no value is greater than the median: 3+9+7 (not ascending)

• now check (D), where B = 4.

$A = 21 - (9+4)$

$A = 8$

here no value is greater than the median: 4+9+8 (not ascending)

• now check (C), where B = 3

$A = 21 - (9+3)$

$A = 9$

here, one of the boxes is equal to the median: 3+9+9 (ascending)

you see a pattern here, the value of A is increasing, thus B is decreasing.

• check (B), where B = 2

$A = 21 - (9+2)$

$A = 10$

here, you can realize that B = 2 is not correct since this is not maximum possible weight of the lightest box: 2+9+10 (ascending, but the weight of the lightest box is not maximum possible)

Since, we want maximum possible weight for the lightest box, that is only possible when B = 3 (maximum possible)