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Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters.

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Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters.

by BTGmoderatorLU » Mon Apr 12, 2021 9:49 am

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Source: Official Guide

Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?

A. 82
B. 118
C. 120
D. 134
E. 152

The OA is D
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Re: Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeter

by Brent@GMATPrepNow » Tue Apr 13, 2021 7:17 am
BTGmoderatorLU wrote: ↑
Mon Apr 12, 2021 9:49 am
 Source: Official Guide

Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?

A. 82
B. 118
C. 120
D. 134
E. 152

The OA is D
So, we have 7 rope lengths.
If the median length is 84, then the lengths (arranged in ascending order) look like this: {_, _, _, 84, _, _, _}

The length of the longest piece of rope is 14 cm more than 4 times the length of the shortest piece of rope.
Let x = length of shortest piece.
This means that 4x+14 = length of longest piece.
So, we now have: {x, _, _, 84, _, _, 4x+14}

Our task is the maximize the length of the longest piece.
To do this, we need to minimize the other lengths.
So, we'll make the 2nd and 3rd lengths have length x as well (since x is the shortest possible length)
We get: {x, x, x, 84, _, _, 4x+14}

Since 84 is the middle-most length, the 2 remaining lengths must be greater than or equal to 84.
So, the shortest lengths there are 84.
So, we get: {x, x, x, 84, 84, 84, 4x+14}

Now what?

At this point, we can use the fact that the average length is 68 cm.
There's a nice rule (that applies to MANY statistics questions) that says:
the sum of n numbers = (n)(mean of the numbers)
So, if the mean of the 7 numbers is 68, then the sum of the 7 numbers = (7)(68) = 476

So, we now now that x+x+x+84+84+84+(4x+14) = 476
Simplify to get: 7x + 266 = 476
7x = 210
x=30

If x=30, then 4x+14 = 134
So, the longest piece will be 134 cm long.

Answer = D
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Re: Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeter

by Scott@TargetTestPrep » Mon Apr 26, 2021 6:47 am
BTGmoderatorLU wrote: ↑
Mon Apr 12, 2021 9:49 am
 Source: Official Guide

Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?

A. 82
B. 118
C. 120
D. 134
E. 152

The OA is D
Solution:

The total length of the 7 ropes is 68 x 7 = 476 cm. If we list the lengths of the ropes in ascending order, the 4th rope is 84 cm since the 4th rope has the median length. Since we want to find the maximum possible length of the longest rope, we assume the 7 ropes have the following lengths:

x, x, x, 84, 84, 84, y

where x is the length of the shortest rope and y is the length of longest rope.

We are given that y = 4x + 14 and we know that 3x + 3(84) + y = 476. Substituting y = 4x + 14 into 3x + 3(84) + y = 476, we have:

3x + 252 + 4x + 14 = 476

7x + 266 = 476

7x = 210

x = 30

Thus, the longest rope has length of 4(30) + 14 = 134 cm

Answer: D

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