Three boxes of supplies have an average (arithmetic mean)

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BTGmoderatorLU: Moderator; Posts: 2505; Joined: Sun Oct 15, 2017 1:50 pm; Followed by:6 members

Three boxes of supplies have an average (arithmetic mean)

by BTGmoderatorLU » Fri Aug 02, 2019 3:25 pm

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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 max possible weight, in kilograms, of the lightest box?

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

The OA is C

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Source: Beat The GMAT — Problem Solving | Fri Aug 02, 2019 3:25 pm

swerve: Legendary Member; Posts: 2499; Joined: Sun Oct 29, 2017 2:04 pm; Followed by:6 members

by swerve » Sat Aug 03, 2019 10:59 am

Let the boxes be
\(w_1\)
\(w_2\)
\(w_3\)

Mean \(= 7\)
Sum(3)\(=7\cdot 3=21\)

Hence \(w_1+w_2+w_3=21\)

Median\(=w_2=9\)
Hence \(w1+w3=12\)

Now to maximize \(w_1\) we must minimize \(w_3\)

Minimum value of \(w_3=\)median \(=9\)

Hence maximum \(w_1=12-9 = 3\)

Therefore, __C__

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by Scott@TargetTestPrep » Sun Aug 11, 2019 6:16 pm

BTGmoderatorLU wrote: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 max 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 of z is also 9. Thus, the maximum value of x is 12 - 9 = 3.

Answer: C

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