Problem Solving

Three boxes have an average weight of 7kg and a median weight of 9kg. What is the maximum weight of the smallest box?

A. 1
B. 2
C. 3
D. 4
E. 5

The OA is C.

Let's assume 3 boxes are X, Y, Z
x + y + z = 21
Median = 9 (middle number)

Be back solving
The smallest maximum weight
X Y(median) Z
1 9 11
2 9 10
3 9 9
4 9 8 (it can be the answer because Z value is below the Y (median value))
5 9 7 (it can be the answer because Z value is below the Y (median value))

So, the smallest maximum weight is 9. Option C.

Has anyone another strategic approach to solve this PS question? Regards!

AAPL wrote:Three boxes have an average weight of 7kg and a median weight of 9kg. What is the maximum weight of the smallest box?

A. 1
B. 2
C. 3
D. 4
E. 5

The OA is C.

Let's assume 3 boxes are X, Y, Z
x + y + z = 21
Median = 9 (middle number)

Be back solving
The smallest maximum weight
X Y(median) Z
1 9 11
2 9 10
3 9 9
4 9 8 (it can be the answer because Z value is below the Y (median value))
5 9 7 (it can be the answer because Z value is below the Y (median value))

So, the smallest maximum weight is 9. Option C.

Has anyone another strategic approach to solve this PS question? Regards!

As you say, the total weights of the boxes is 21, with the middle box being 9, meaning the heaviest and lightest sum to 12.

To maximize the lightest box, need to minimize the heaviest box. Since the median is 9, the minimum weight of the heaviest box is 9 also, leaving 3 kg left over for the lightest box

Given that the median is 9 that leaves us with the other two boxes with a combined weight of 12 kg. [Average * number = total --- 7 * 3 = 21. And total - median = sum of heaviest and smallest box = 12 kg].

Now 9 should be the median and the smallest box has to have the maximum weight. To leave the median as 9, the heaviest box should be at least 9kg. Which give us 3kg for the smallest box.

Hence, the weight of the box should be 3, 9, 9. Hence, the correct answer is C.