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

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Math Expert
Joined: 02 Sep 2009
Posts: 46217
Seven pieces of rope have an average (arithmetic mean) lengt [#permalink]

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09 Nov 2017, 04:04
santro789 wrote:
Where could I find this concept of Maximum and Minimum? I never came across this in my studies so far.

14. Min/Max Problems

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

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16 Dec 2017, 22:21
Bunuel Do you have similar problem's links to practice ?
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Seven pieces of rope have an average (arithmetic mean) lengt [#permalink]

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16 Dec 2017, 22:40

Did you look at 'similar topics' below end of your post?
Hope this helps!
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Posts: 46217
Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink]

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17 Dec 2017, 00:35
Bunuel Do you have similar problem's links to practice ?

Please check the post just above yours: https://gmatclub.com/forum/seven-pieces ... l#p1959061
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink]

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28 Dec 2017, 05:56
ronr34 wrote:
sunshinewhole wrote:
Bunuel wrote:
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 $$a+a+a+84+84+84+(4a+14)=7*68$$ --> $$a=30$$ --> $$g_{max}=4a+14=134$$.

to minimize y we took 84, we cud have valued all the rope lengths to our minimum=a.

It's a little confusing....
Since we have the equation $$g=4a+14$$, to maximize g, we also need to maximize a... isn't this the case?

Take Bunuel's analogy. x+y=10, to have max y e.g. 9 we have to minimise x to 1. Or seeing it another way y=10-x. So minimising x maximises y.
Doing the same thing in this example rearranging x+a+b+84+c+d+4x+14=7*68 gives us 4x+14=7*68-(x+a+b+84+c+d). So the more you minimise (x+a+b+84+c+d), the more you maximise 4x+14.
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Seven pieces of rope have an average (arithmetic mean) lengt [#permalink]

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04 Jan 2018, 08:46
JeffTargetTestPrep 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.

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

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04 Jan 2018, 09:28
1
dave13 wrote:
JeffTargetTestPrep 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.

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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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
Re: Seven pieces of rope have an average (arithmetic mean) lengt   [#permalink] 05 Jun 2018, 10:27

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