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BrentGMATPrepNow
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Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post  Updated on: 26 Aug 2020, 09:43
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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


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

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Originally posted by BrentGMATPrepNow on 26 Jul 2016, 06:31.
Last edited by BrentGMATPrepNow on 26 Aug 2020, 09:43, edited 1 time in total.
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post  01 Aug 2016, 09:41
GMATPrepNow 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


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 =
Show: ::
D


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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] New post  01 Aug 2016, 09:55
aditi2013 wrote:
GMATPrepNow 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


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 =
Show: ::
D


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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.

Answer to above question will clear your doubt.
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post  01 Aug 2016, 11:49
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aditi2013 wrote:

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] New post  09 Nov 2017, 02:57
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] New post  09 Nov 2017, 03:04
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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 length of 68 [#permalink] New post  23 Jul 2018, 09:55
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] New post  15 Nov 2018, 23:49
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
Show: ::
D
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post  20 Mar 2020, 18:46
Hi Bunuel, I have a doubt, as d<=e<=f<=g, and for g to be higher value, we can make e and f also to be higher right, as it is in ascending order and so obviously, g also has to be greater than f (where f itself we made it as higher value so, f<g)...pls help me where im going wrong..
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post  30 Apr 2020, 23:38
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.

Bunuel
From the highlighted part, we know, f could be greater or equal to e. So, can we consider e=84 and f=85?
Thanks__
So, can we
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post  30 Apr 2020, 23:44
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Asad wrote:
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.

Bunuel
From the highlighted part, we know, f could be greater or equal to e. So, can we consider e=84 and f=85?
Thanks__
So, can we


We need to maximize g. Now, try it yourself. Make f equal to 85, solve and see what value of g you get. Then check how does this value compare to \(g_{max}=134\) we got in the solution.
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post  05 Jul 2020, 18:01
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.



A length of rope doesn't need to be an integer. 106/4 = 26.5 which is a valid length. I don't believe this is a valid approach. Please correct me if I'm wrong.
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post  31 Jul 2020, 20:45
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aeglorre 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


They're telling us that:\(L = 4S + 14\), and they want us to maximize L, thus we plug in options to solve for S such that we maximize S (this maximizes L). So our answer is the option that gives us: \(S* = \frac{(L -14)}{4}\), where * stands for the maximum value of S given the options

The answer needs to be divisible by four after subtraction with 14. The only of the options that are capable of this are B and D (104/4 = 26 and 120/4 = 30). A is obviously not a viable option since the middle value is 84.

Of course, they're asking for "LONGEST possible value" so naturally you take the option that has the highest value of the two: So our answer is D.


Beautiful !! Quick, Simple & makes perfect sense !
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post  11 Aug 2020, 00:27
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



IN this kind of problems make spaces. Here we have to make seven spaces
In order to maximize the longest rope, We will make numbers after median equal to median and the first three numbers as same
Let the small rope be S and large rope be L

S S S 84 84 84 L
_ _ _ _ _ _ _

The total mean equals = 68 * 7 = 476 and 84*3 = 252
So L + 3*S = 224
L = 14 + 4*S ( from question stem)

Now you can either solve the equation or plug in answers and check. Start with option D since you want to maximize.
This option will work.

When you start with E and plug it in L = 14 + 4S, you will not get an integer value for S.
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Re: Seven pieces of rope have an average (arithmetic mean) lengt [#permalink] New post  22 Oct 2020, 21:47
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.



Why can we not chose a value less that the median, say some 40 or 50 for B and C rather than assigning the same value as A?
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