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Seven pieces of rope have an average (arithmetic mean) lengt

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Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

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New post 04 Jan 2018, 09:28
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dave13 wrote:
JeffTargetTestPrep wrote:
Walkabout 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


We need to first recognize that we are working with a maximum problem. This means that of the seven pieces of rope, we must make 6 of those pieces as small as we possibly can, within the confines of the given information, and doing so will maximize the length of the 7th piece.

We are first given that seven pieces of rope have an average (arithmetic mean) length of 68 centimeters. From this we can determine the sum.

average = sum/quantity

sum = average x quantity

sum = 68 x 7 = 476

Next we are given that the median length of a piece of rope is 84 centimeters. Thus when we arrange the pieces of rope from least length to greatest, the middle length (the 4th piece) will have a length of 84 centimeters. We also must keep in mind that we can have pieces of rope of the same length. Let's first label our seven pieces of rope with variables or numbers, starting with the shortest piece and moving to the longest piece. We can let x equal the shortest piece of rope, and m equal the longest piece of rope.

piece 1: x

piece 2: x

piece 3: x

piece 4: 84

piece 5: 84

piece 6: 84

piece 7: m

Notice that the median (the 4th rope) is 84 cm long. Thus, pieces 5 and 6 are either equal to the median, or they are greater than the median. In keeping with our goal of minimizing the length of the first 6 pieces, we will assign 84 to pieces 5 and 6 to make them as short as possible. Similarly, we have assigned a length of x to pieces 1, 2, and 3.

We can plug these variables into our sum equation:

x + x + x + 84 + 84 + 84 + m = 476

3x + 252 + m = 476

3x + m = 224

We also given that the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope. So we can say:

m = 14 + 4x

We can now plug 14 + 4x in for m into the equation 3x + m = 224. So we have:

3x + 14 + 4x = 224

7x = 210

x = 30

Thus, the longest piece of rope is 4(30) + 14 = 134 centimeters.

Answer is D.


Hello :) one question, why did you denote the shortest as X as well 2, and 3 as x ? Shouldnt we need to give the shortest piece the unique name to diffirentitate from 2 and 3 ? if you denote the longest as m and m is the unique name
Thanks! :-)


x represents the shortest piece and to get the shortest rope, the 2nd and 3rd piece would be the shortest possible length, i.e. x. Similarly, the 5th and 6th would be the shortest possible length of 84 (the median). Note m can be the median, but since we are not sure, it is represented as a variable.
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Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

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New post 05 Jun 2018, 10:27
This is how i solved. I went with the options here.Do let me know your views.
let the pieces of rope be a, b , c,d,e,f,g in ascending order and g is the largest , g =14+4a
and we need to find the length of the largest piece of rope.
Now the options are given to be value of g. Now the value of g must be a multiple of 4
Ex : If we consider option A) 82=14+4a=g
82-14=68 which is a multiple if 4 and the value of a is 17. by continuing this method we find that there are only options A,B and D that satisfies the condition in which D gives the max value for a and hence the max value of g
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Re: Seven pieces of rope have an average length of 68  [#permalink]

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New post 23 Jul 2018, 10:55
By no means an expert (first reply/comment on this forum) so this may not be the quickest way to solve this but my approach was as follows;

Reading the full question you note a few key pieces of information, namely the mean 68, median 84 and that you're looking for the maximum possible value of the longest rope. You also know you're looking to link the length of the longest rope with that of the smallest.

There are 7 ropes so let's call them r1 through r7.

We know that the median is 84 so r4 (the middle rope) has to equal 84.
As we're looking for the max value of r7 we can also assign 84 to both r5 and r6. This is the least possible value that they could be.
Similarly, for r7 to be as large as possible r1, r2, r3 must be as small a value as possible so we can say r1=r2=r3.
We also know that r7 = (4*r1) + 14

So at this stage we know r1 + r2 + r3 + 3(84) + r7 = Number of ropes * the mean [7(68)]

As r1=r2=r3 and r7=(4*r1) + 14 we can simplify this to;

7*r1 = 7(68) - 3(84) - 14
7*r1 = 210
r1 = 30 therefore;

r7 = (4*30) + 14 = 134

Answer is C
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Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

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New post 20 Sep 2018, 03:15
Bunuel wrote:

No. Because we have fixed total length of the rope: 7*68 centimeters. If you increase a, you'd be decreasing g.


Thanks for this! Just one thing- if I increase a I will decrease G - it would be because i already have 84, 84, 84 and cannot decease them further to compensate?
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Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

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New post 07 Oct 2018, 10:47
BrainLab wrote:
For the solution we don't even need an arithmetic mean. The median has a property that (first number + last number)/2 = median (in case of odd number of values) --> we have smallest number=x and largest number = 4x + 14

(4x+14+x)/2=84 --> x ≈ 30 --> 4x+14=134 (D)


That is only true for consecutive numbers, i.e. 1,2, 3 ,4,5 -> (5+1)/2 = 3. For random numbers this does not work, i.e. 2,5, 9 ,15,23 -> (2+23)/2 IS NOT 9.
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Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

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New post 07 Oct 2018, 10:51
BrainLab wrote:
For the solution we don't even need an arithmetic mean. The median has a property that (first number + last number)/2 = median (in case of odd number of values) --> we have smallest number=x and largest number = 4x + 14

(4x+14+x)/2=84 --> x ≈ 30 --> 4x+14=134 (D)


That is only true for consecutive numbers, i.e. 1,2,3,4,5, where (5+1)/2=3.
For random numbers, i.e. 2,5,9,15,23 this does not hold ( (23+2)/2 is not 9).
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Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

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New post 16 Nov 2018, 00:49
We can start by referring to the lengths of ropes as:

\(a≤b≤c≤d≤e≤f≤g\)


We know that average length is \(68\) cm so:

Total length of rope \(= (7)(68) = 476 \ cm\)


We also know that the median length = 84cm, so segment d = 84 cm.

We also know that \(g = 4a + 14\).

We are asked for maximum possible length of \(g\); to do that, all of the other lengths must be as short as possible.

That would give us:

\(a = b = c\)

\(d = e = f\)


With a total length of 476 cm, we can create the following formula:

\(a+b+c+d+e+f+g = 476\\
a+a+a+84+84+84+4a+14 = 476\\
7a + 266 = 476\\
7a = 210\\
a = 30 cm\)


Thus, the longest possible rope is

\(g = (4)(30) + 14 = 134 \ cm\)


The final answer is .
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Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

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New post 03 Feb 2019, 19:06
Bunuel wrote:
Walkabout 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


Say the lengths of the pieces in ascending order are: a, b, c, d, e, f, g --> \(a\leq{b}\leq{c}\leq{d}\leq{e}\leq{f}\leq{g}\).

The average length = 68 centimeters --> the total length 7*68 centimeters.
The median = 84 centimeters --> d=84.
The length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece --> \(g=4a+14\).

We need to maximize g. Now, to maximize g, we need to minimize all other terms.

The minimum value of b and c is a and the minimum value of e and f is median=d.

Thus we have that \(a+a+a+84+84+84+(4a+14)=7*68\) --> \(a=30\) --> \(g_{max}=4a+14=134\).

Answer: D.



Could you please explain why we need to minimize all other terms to get a maximum value for the longest piece? I know it is probably very simple, but I can't seem to understand it.

Thank you
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Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

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New post 17 Mar 2019, 02:24
gmat1013 wrote:
I didn't know how to go about solving this sum. So, I took help of the answer choices. The important cue given in the question is that the the largest number = 4*Smallest Number+ 14. So if you subtract 14 from the answer choices, the number obtained should be divisible by 14.
A) 82 - Can be cancelled off as it is less than the median
B) 118 - 118-14=104
C)120 - 120-14 = 106
D)134 - 134-14 = 120
E)152 - 152 - 14 = 138
Using the divisibility test for 4, you can strike off the answer choices C and E.

Now you're left with B and D. Since the question asks for the largest possible number, it's best to try solving using choice D first. If 134 is the largest number then 4*30 + 14 gives us the smallest possible value of 30. The median is 84. To maximise the value of the largest number all other values lower than the median should be taken as 30 and higher than it as 84. So we have 30*3 + 84*3 + 134 which is equal to the sum of 68*7. Hence the answer is D. :-D

Whenever you're stuck at a PS question the next best approach is POE :)




Gmat1013 , I think using divisibility test for 4 in POE is not a good option as lengths of popes do not have to be integers
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