ssy wrote:For the first question:
(Average weight)(No. of Boxes) = Total weight of boxes.
Hence, the total weight of the 3 boxes is 7kg x 3 = 21kg.
As there are 3 boxes, we know that the middle box weighs 9kg and one box weighs less than 9kg while the other box is equal to or weighs more than 9kg.
The total weight of the 2 unknown boxes is 21kg-9kg = 12kg.
Bearing in mind the criteria that one box weighs less than 9kg while the other box is equal to or weights more than 9kg, the combination of possible weights for the remaining 2 boxes are as follows:
Small box - 1 kg, Big Box - 11 kg
Small box - 2 kg, Big Box - 10 kg
Small Box -3 kg, Big Box - 9 kg
Hence, the answer is C. The maximum weight of the small box is 3kg.
If the small box were any larger than 3 kg, for example, if the small box is 4kg, the remaining larger box would be 8kg. As there is an odd number of boxes, the median weight of the 3 boxes would be 8kg instead which is not possible given the statement in the question.
Good overall explanation, but we can apply a general principle to reduce our work.
On the GMAT, when we're asked to maximize something, we generally want to minimize everything else. Conversely, when we're asked to minimize something, we want to maximize everything else.
Here, we're asked to maximize the weight of the lightest box; accordingly, we want to minimize the weight of the other two boxes.
Avg = 7, so total weight of the 3 boxes is 3*7 = 21.
Median = 9, so our set must be {x, 9, y} (in ascending order).
We know that y must be at least equal to 9. So, to make x as big as possible, let's set y=9.
x + 9 + 9 = 21
x = 3
