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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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14 Jun 2014, 02:23
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?
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Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

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06 Jan 2015, 07:19
Bunuel 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 $$a+a+a+84+84+84+(4a+14)=7*68$$ --> $$a=30$$ --> $$g_{max}=4a+14=134$$.

Im unable to understand that how minimizing all other terms would maximize value of g, can you please explain?
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Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

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06 Jan 2015, 08:13
Im unable to understand that how minimizing all other terms would maximize value of g, can you please explain?[/quote]

Answer this: the sum of two positive integers is 10. What is the maximum possible value of the largest of the integers?[/quote]

I got your point, i don't know how i was thinking before, but i just got it . We must minimize all other values in order to get maximum value for g. Thanks
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Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

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18 Jul 2015, 00:15
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)

BrainLab, this is true only when you have an evenly spaced set such as 1,2,3,4,5,6,7 where mean = median = $$\frac{first No + last No}{2}$$

However, this is not the case with a unevenly spaced set such as 1, 2, 3, 10, 11, 12, 13
Notice that notice that in this case you would have different results
$$mean = \frac{Sum}{Total No's} = \frac{52}{7}$$
median = middle no = 10
$$\frac{first No + last No}{2} = \frac{1+13}{2}=7$$

you can even cross check with the soln that you have.
$$\frac{30+134}{2} = 82$$ and not 84.

Hope this brings clarity.
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Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

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28 Sep 2015, 04:13
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)

As Bunuel pointed out in an earlier post "The lengths of the pieces of the rope does not form an evenly spaced set to use (mean)=(first+last)/2."
So IMO, your solution does not hold good.

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

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20 Dec 2015, 11:44
mulhinmjavid wrote:
Bunuel 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 $$a+a+a+84+84+84+(4a+14)=7*68$$ --> $$a=30$$ --> $$g_{max}=4a+14=134$$.

Im unable to understand that how minimizing all other terms would maximize value of g, can you please explain?

Because the mean is the same, and the median is the size of the middle rope (from shortest to longest)
Therefore, if you minimize from one site of the median , you are going to maximize from the other.

Actually in this problem, as you have only integers, you can deduce that the size of the shortest has to be also an integer. So you can start working with the answers.

I like to start to work with the number C.
C) 120-14 = 106 (I cannot divide by 4, so its not the answer)
D) 134-14 = 120 (Bingo! I can divide by 4, lets check the last answer)
E) 152-14 = 138 (I cannot divide by 4)

Therefore the answer is letter D.
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Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

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19 May 2016, 04:41
option A : largest rope cannot be less than median ;so eliminate.
option C and option E : eliminated because of reasons below,
L= 14+4S
S=L-14/4
when you plugin L=118 or 152 you get a decimal value which will also make mean or median,which is clearly an integer as given, into decimal ; we are looking for an integer value of a measurement (cm)

remaining with option B and D :
both are ok but we need greatest value for rope ; so ans is 134 , i.e. option D , here we get smallest rope as 30 cms
check the ans by taking average of 3 ropes of 30cm(smallest one) , 3 ropes of 84cm(median) and 1 last rope of 134 cm(largest one); we get avg=68

incase if you are thinking why 3 rope of 30cms,1 rope of 84cm and 3 rope of 134cms cant be taken, think about the constraint -i.e. the avg which is 68 , if you take 3 ropes of 134cm , avg will be more than 68 cms.
thanks!
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Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

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19 Jun 2016, 23:33
the important thing to remember here is that it is a rope and not numbers otherwise...to obtain maximum rope length that is g.....we have to minimize all other lengths....a is smallest.....keep b and c equal to a keeping in mind inequality and e and f equal to median keeping in mind the equality...insert the values and use average formula to get the right answer
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Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

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01 Aug 2016, 10:41
GMATPrepNow 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

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.

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Hi Brent,

I just have one question :

Do we not have to keep the shortest term unique?? "Shortest".

I have got some questions wrong before because I kept the first few terms identical to maximise the largest and the explanation provided was that since the term "shortest" is mentioned, we cannot take same numbers.

I cannot trace the same question to illustrate further.

Can you tell me how do we identify?

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

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01 Aug 2016, 10:55
GMATPrepNow 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

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.

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Hi Brent,

I just have one question :

Do we not have to keep the shortest term unique?? "Shortest".

I have got some questions wrong before because I kept the first few terms identical to maximise the largest and the explanation provided was that since the term "shortest" is mentioned, we cannot take same numbers.

I cannot trace the same question to illustrate further.

Can you tell me how do we identify?

Thanks.

Even if we have 3 terms with the value of x and the others with the value greater than x, we will always call x the shortest.

e.g. Can you find out which is the shortest value among the below numbers?

1,1,1,2,3,4,5,6.

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

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01 Aug 2016, 12:49
Top Contributor

Hi Brent,

I just have one question :

Do we not have to keep the shortest term unique?? "Shortest".

I have got some questions wrong before because I kept the first few terms identical to maximise the largest and the explanation provided was that since the term "shortest" is mentioned, we cannot take same numbers.

I cannot trace the same question to illustrate further.

Can you tell me how do we identify?

Thanks.

No, we need not keep the shortest term unique.

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

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01 Aug 2016, 22:39
Hey everyone,

I can see the great explanations provided, but I just directly jumped to the POE method. Only in option 'D', i can solve the equation: 134 = 14+4x, x=30, whereas in all other options, 'x' doesn't yield an integer. Can such methods give correct solutions most of the time.?
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Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

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24 Oct 2016, 08:07
Bunuel 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 $$a+a+a+84+84+84+(4a+14)=7*68$$ --> $$a=30$$ --> $$g_{max}=4a+14=134$$.

What i do not understand. How are you getting to a = 30? I understand everything, however not this part Could please explain this to me?
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Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

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24 Oct 2016, 08:11
SamsterZ wrote:
Bunuel 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 $$a+a+a+84+84+84+(4a+14)=7*68$$ --> $$a=30$$ --> $$g_{max}=4a+14=134$$.

What i do not understand. How are you getting to a = 30? I understand everything, however not this part Could please explain this to me?

By solving $$a+a+a+84+84+84+(4a+14)=7*68$$.
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Re: Seven pieces of rope have an average (arithmetic mean) lengt  [#permalink]

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17 Dec 2016, 20:25
Excellent Question from the official Guide set.
Here is my solution to this one =>
Let the Rope pieces be ->
W1
W2
W3
W4
W5
W6
W7

In increasing order

Mean = 68

$$Mean = \frac{Sum}{#}$$

Hence Sum(7)=68*7 = 476 cms.

Now median = 84
#=7=> odd

Hence Median => Fourth term = W4

So W4=84

Now W7=14+4W1

We have to maximise the largest term i.e W7
For the we must minimise all other terms.

W2=W3=W1 each
W4=W5=W6=84each

Hence => 3W1+3*84+14+4W=476

=> 7W1=210
W1=30

Hence W7=14+4*30=134cms

Hence D

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

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09 Nov 2017, 03:57
Where could I find this concept of Maximum and Minimum? I never came across this in my studies so far.
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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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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.
Re: Seven pieces of rope have an average (arithmetic mean) lengt   [#permalink] 28 Dec 2017, 05:56

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