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Statistics OG PS # 183

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Statistics OG PS # 183

by amirhakimi » Sat Nov 02, 2013 6:21 am
Can anyone explain this problem?

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

Answer is D
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by Brent@GMATPrepNow » Sat Nov 02, 2013 7:01 am
amirhakimi wrote:Can anyone explain this problem?

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

Answer 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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Brent
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by GMATGuruNY » Sat Nov 02, 2013 7:08 am
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

OA: D

The book ( OG 13) posts a pretty confusing way to solve, would love to hear any other recommendations.

Thanks in advance for any guidance.
The sum of the lengths = 7*68 = 476.

Let the smallest piece = x.
Then the length of the longest piece = 4x+14.
Median piece = 84.
Let the remaining pieces be a, b, c, d.

Here are the 7 pieces, in ascending order:
x, a, b, 84, c, d, 4x+14.

To MAXIMIZE the value of 4x+14, we must MINIMIZE the values of a, b, c, and d.
The least possible value for a and b is x.
The least possible value for c and d is 84.
Here are the 7 pieces:
x, x, x, 84, 84, 84, 4x+14.

Since the sum of the lengths is 476, we get:
x + x + x + 84 + 84 + 84 + 4x+14 = 476
7x + 266 = 476
7x = 210
x = 30.

Thus:
Greatest possible value for the longest piece = 4x+14 = 4*30 + 14 = 134.

The correct answer is D.
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