Problem Solving

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


ans. C

thanks

A set of 15 different integers has a median of 25 and a range of 25. What is the greatest possible integer that could be in this set?

(A) 32
(B) 37
(C) 40
(D) 43
(E) 50

ans. D

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.

ssy thanks ..good explanation ...dint know it could be an equal value ..thanks

For the second question:

Since the median is 25, and the total numbers are odd, the middle number is 25. That means, there are 7 numbers on either side of number 25 in the set when they arranged in order. To maximize the highest number the numbers that are below 25 should be closer to 25 (so that in the range 25, we get the highest possible value.). So the seven numbrs below 25 will be 18 to 24. this leaves us the highest number = 18+range (i.e., 25) = 43.

Other way is to cross check with the options. Like, let's take the maximum value given in the options, 50. This means, all the numbers on the other side 25 must be 25s (range 25). This is not possible because the numbers are all different.


Thanks,
Anil

question 1 was really a good one...

The pace of the test...got the definition of MEDIAN wrong for me...

median is the middle number--> which does not imply that the other numbers are bigger or smaller than the median.

median is the middle number when all the numbers are arranged in ascending or decending order...

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