OG 13 183

Question

Seven pieces of rope have an average 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: D

Please explain reasoning

MartyMurray · Answer

Seven pieces of rope have an average 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

Average = Sum/Number

In this case 68 = Sum/7. So 7 x 68 = Sum = 476 = Total Length Of All The Ropes Together

We have 7 ropes with seven lengths, R1 R2 R3 R4 R5 R6 R7.

Since the median is 84, R4 = 84

Of the 6 remaining ropes, 3 must have lengths ≥ 84, and the other 3 must have lengths ≤ 84.

Since all the lengths must add up to 476, to maximize the length of the longest piece of rope, R7 = 4R1 + 14, maximize R1 and R7 and minimize all the others.

First fill in the ones ≥ 84. The minimum for those is 84.

R1 R2 R3 84 84 84 R7

Then minimize R2 and R3 by making them the same as R1.

R1 R1 R1 84 84 84 R7

R7 = 4R1 + 14 So we get the following.

R1 + R1 + R1 + 84 + 84 + 84 + (4R1 + 14) = 476

7R1 + 266 = 476

7R1 = 210

R1 = 30

R7 = (4 x 30) + 14 = 134

The correct answer is D.

Last edited by MartyMurray on Thu Dec 24, 2015 5:33 am, edited 1 time in total.

Marty Murray
Perfect Scoring Tutor With Over a Decade of Experience
MartyMurrayCoaching.com
Contact me at [email protected] for a free consultation.

GMATGuruNY · Answer

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 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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GMATGuruNY · Answer

Alternate approach:

The sum of the lengths = 7*68 = 476.

We can PLUG IN THE ANSWERS, which represent the maximum length of the longest piece.
The longest piece is 14 centimeters more than 4 TIMES the length of the shortest piece. Thus, when 14 is subtracted from the correct answer choice, the result will be a multiple of 4.
Subtracting 14 from each answer choice, we get:
68, 104, 106, 120, 138.
106 and 138 are not multiples of 4.
Eliminate C and E.

Since the question asks for the MAXIMUM possible length, we should start with the GREATEST of the remaining answer choices.
To MAXIMIZE the length of the longest piece, we must MINIMIZE the lengths of all of the other pieces.

Answer choice D: 134
Shortest piece = 120/4 = 30.
The 7 pieces are:
30, b, c, 84, e, f, 134.
The minimum possible value for b and c is 30, implying 3 pieces with a length of 30.
The minimum possible value for e and f is 84, implying 3 pieces with a length of 84.
Sum = 3(30) + 3(84) + 134 = 90+252+134 = 476.
Success!

The correct answer is D.

Mo2men · Answer

GMATGuruNY wrote:Alternate approach:

The sum of the lengths = 7*68 = 476.

We can PLUG IN THE ANSWERS, which represent the maximum length of the longest piece.
The longest piece is 14 centimeters more than 4 TIMES the length of the shortest piece. Thus, when 14 is subtracted from the correct answer choice, the result will be a multiple of 4.
Subtracting 14 from each answer choice, we get:
68, 104, 106, 120, 138.
106 and 138 are not multiples of 4.
Eliminate C and E.

Since the question asks for the MAXIMUM possible length, we should start with the GREATEST of the remaining answer choices.
To MAXIMIZE the length of the longest piece, we must MINIMIZE the lengths of all of the other pieces.

Answer choice D: 134
Shortest piece = 120/4 = 30.
The 7 pieces are:
30, b, c, 84, e, f, 134.
The minimum possible value for b and c is 30, implying 3 pieces with a length of 30.
The minimum possible value for e and f is 84, implying 3 pieces with a length of 84.
Sum = 3(30) + 3(84) + 134 = 90+252+134 = 476.
Success!

The correct answer is D.

I was afraid to use When I tried to use the same assumption that when 14 is subtracted from the correct answer choice, the result will be a multiple of 4. There is nothing in the prompt let us to drive this assumption. the rope length could be fraction We do not deal with human, cars,cats..etc.Is it applicable in every question like that?

Thanks

Brent@GMATPrepNow · Answer

amirhakimi wrote: 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

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

Mo2men · Answer

Thanks Brent for your reply. I solve this question with the same approach but still the question exists in the plug in solution.Can I assume that when subtracting 14 the result should be multiple f 4? if yes, why?

GMATGuruNY · Answer

Mo2men wrote:GMATGuruNY wrote:I was afraid to use When I tried to use the same assumption that when 14 is subtracted from the correct answer choice, the result will be a multiple of 4. There is nothing in the prompt let us to drive this assumption. the rope length could be fraction We do not deal with human, cars,cats..etc.Is it applicable in every question like that? Given that all of the numbers in the prompt -- 68, 84, 14 and 4 -- are integer values, it is almost guaranteed that all of the rope lengths will be integer values.

oquiella · Answer

Marty Murray wrote:Seven pieces of rope have an average 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 Average = Sum/Number In this case 68 = Sum/7. So 7 x 68 = Sum = 476 = Total Length Of All The Ropes Together We have 7 ropes with seven lengths, R1 R2 R3 R4 R5 R6 R7. Since the median is 84, R4 = 84 Of the 6 remaining ropes, 3 must have lengths ≥ 84, and the other 3 must have lengths ≤ 84. Since all the lengths must add up to 476, to maximize the length of the longest piece of rope, R7 = 4R1 + 14, maximize R1 and R7 and minimize all the others. First fill in the ones ≥ 84. The minimum for those is 84. R1 R2 R3 84 84 84 R7 Then minimize R2 and R3 by making them the same as R1. R1 R1 R1 84 84 84 R7 R7 = 4R1 + 14 So we get the following. R1 + R1 + R1 + 84 + 84 + 84 + (4R1 + 14) = 476 7R1 + 266 = 476 7R1 = 210 R1 = 30 R7 = (4 x 30) + 14 = 134 The correct answer is D. Why did you decide to make R1 all the same numbers?

MartyMurray · Answer

oquiella wrote:Why did you decide to make R1 all the same numbers?

The question asks for the maximum length of the longest piece.

We have to work with a total of 476. So the only way to maximize the length of the longest is to minimize the lengths of the others.

The shortest one is R1. While we know that R4 has to be 84, there is nothing to keep us from making R2 and R3 just as short as R1.

R1 is the shortest length. So the way to minimize R2 and R3 is to make them as short as R1.

Once we do that, R1 = R2 = R3. So R1 can be substituted for R2 and R3.

Meanwhile, R7 = 4R1 + 14. So 4R1 + 14 can be substituted for R7.

Once those substitutions have been made, there is just one variable with which to work and solving the problem is straightforward.

Marty Murray
Perfect Scoring Tutor With Over a Decade of Experience
MartyMurrayCoaching.com
Contact me at [email protected] for a free consultation.

Jeff@TargetTestPrep · Answer

oquiella wrote:Seven pieces of rope have an average 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 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

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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Jeff@TargetTestPrep · Answer

oquiella wrote:Seven pieces of rope have an average 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 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

Jeffrey Miller
Head of GMAT Instruction
[email protected]

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

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