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OG 13 183

This topic has 4 expert replies and 5 member replies
oquiella Master | Next Rank: 500 Posts Default Avatar
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OG 13 183

Post Tue Dec 22, 2015 3:24 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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

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    Marty Murray Legendary Member
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    Post Tue Dec 22, 2015 8:34 pm
    Quote:
    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.

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    Last edited by Marty Murray on Thu Dec 24, 2015 5:33 am; edited 1 time in total

    Thanked by: duahsolo
    Post Wed Dec 23, 2015 5:06 am
    Quote:
    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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    Post Wed Dec 23, 2015 5:07 am
    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.

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    Top Member

    Post Wed Dec 23, 2015 3:36 pm
    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

    Post Wed Dec 23, 2015 3:57 pm
    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

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    Top Member

    Post Wed Dec 23, 2015 4:03 pm
    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?

    Post Thu Dec 24, 2015 5:34 am
    [quote="Mo2men"]
    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.

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    oquiella Master | Next Rank: 500 Posts Default Avatar
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    Post Thu Dec 24, 2015 6:06 am
    Marty Murray wrote:
    Quote:
    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?

    Marty Murray Legendary Member
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    Post Thu Dec 24, 2015 6:25 am
    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.

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