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Three boxes of supplies have an average (arithmetic mean) weight of 7 kilograms...

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Re: Three boxes of supplies have an average (arithmetic mean) weight of 7 kilograms...

by Pyunika » Tue Feb 04, 2020 12:59 am
Assuming x to be lightest , y median and z the heaviest.
So according to given information average is 7 so total sum for three box = 21
thus x+y+z = 21
Given y=9 (median weight)
thus x+z = 12
For maximum possible value for x is 3 such that z= 9
if x=4 then Z= 8(this changes median to 8)
So maximum possible value for lightest box is 3
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Re: Three boxes of supplies have an average (arithmetic mean) weight of 7 kilograms...

by Scott@TargetTestPrep » Sat Feb 08, 2020 7:19 pm
BTGmoderatorLU wrote: ↑
Fri Jan 31, 2020 3:22 pm
 Source: GMAT Prep

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

The OA is C
We are given that three boxes of supplies have an average (arithmetic mean) weight of 7 kilograms and a median weight of 9 kilograms. We must determine the maximum weight of the lightest box.

Since the average weight of the 3 boxes is 7, the sum of the weights of the 3 boxes is 3 x 7 = 21.

We can also define a few variables.

x = lightest box
y = second heaviest box
z = heaviest box

We can create the following equation:

x + y + z = 21

Since the median is 9, y must be 9. So we now have:

x + 9 + z = 21
x + z = 12

Remember, we need the value of x to be as large as possible, so we want to minimize the value of z. Since 9 is the median weight of the boxes, the smallest value that z could possibly be is also 9. Thus, the maximum value of x is 12 – 9 = 3.

Answer: C

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