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605-655 (Medium)|   Min-Max Problems|   Statistics and Sets Problems|                              
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Bunuel
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Seven pieces of rope have an average (arithmetic mean) lengt 09 Nov 2017, 04:04
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santro789
Where could I find this concept of Maximum and Minimum? I never came across this in my studies so far.
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14. Min/Max Problems


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Seven pieces of rope have an average (arithmetic mean) lengt 16 Nov 2018, 00:49
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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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D
.
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Seven pieces of rope have an average (arithmetic mean) lengt 22 Jan 2021, 07:24
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Seven pieces of rope have an average (arithmetic mean) lengt 19 May 2021, 22:59
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Hi Bunuel

For some reason I thought each length must be different and this is what my equation looked like:

\(a+(a+1)+(a+2)+84+85+86+(g) = 476\)

g= seventh piece of rope/ longest piece.
\(g= 14+4(a)\)

\(a+(a+1)+(a+2)+84+85+86+(14+4a) = 476\) --- got "a' as 29.something. Found a and plugged the value in g=14+4a got an approx answer= 130.something

my question is: had it been the case, if we were given: each value/length/etc must be distinct. is this the right approach?

Thanks.
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Seven pieces of rope have an average (arithmetic mean) lengt 20 May 2021, 01:02
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ueh55406
Hi Bunuel

For some reason I thought each length must be different and this is what my equation looked like:

\(a+(a+1)+(a+2)+84+85+86+(g) = 476\)

g= seventh piece of rope/ longest piece.
\(g= 14+4(a)\)

\(a+(a+1)+(a+2)+84+85+86+(14+4a) = 476\) --- got "a' as 29.something. Found a and plugged the value in g=14+4a got an approx answer= 130.something

my question is: had it been the case, if we were given: each value/length/etc must be distinct. is this the right approach?

Thanks.
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Notice that we are not told that the lengths must be integers. So, if the length of the shortest piece is a, then the next piece won't be a + 1, it could be a + 0.00000000...0001. Now, if were were told that the lengths are distinct integers, then your solution would be correct but the question would be flawed because it does not yield an integer solution.
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Seven pieces of rope have an average (arithmetic mean) lengt 24 Feb 2022, 03:04
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Hi all! :blushing:

Question : 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?

Now that is a classic and elegant question to test your application skills of maximization and minimization.
This question shall also test your conceptual understanding of Mean and Median.
So all in all, if you are able to get this right under 2 min then you are perfectly on the road ahead!!
In case you have not, then lets try to have the necessary takeaways from the question and ensure that you would be able to use this learning in the next questions of Maximization & Minimization.

:idea: Core GMAT concept:
Minimize (n-1) elements to maximize 1 element out of n elements and vice versa.
Also know that the middle value in an arranged set is called the median.
Since Arithmetic Mean cannot give a precise data representation when there are extreme values in a set, the statistical tool of Median came into existence.
When n is 'odd' Median is the middle value
and
when n is 'even' median is he average of the two middle values in an arranged set(arranged in ascending or descending order)

When arranged in an ascending order, values to the left of median can be < or = the median and values to the right of median can be > or = the median.

Now coming back to the question, lets arrange these 7 pieces in ascending order of length.

:idea: GMAT Tip:
The lengths of the segments may not be integers and hence do not use the approach of moving through elimination.
So a < b < c < d < e < f < g where a is the shortest piece and g is the longest.
According to the Question Stem g = 14 + 4a

Eliminate option A as the answer choice as longest piece cannot be lesser than the median! So we have 4 options of which 3 are incorrect now.
Sum of the elements= Average * number of elements and here substituting the corresponding values we have Sum = 68 * 7
Since we have a constant sum, we have to maximize on g by minimizing on the remaining lengths.
The median of 7 elements(odd number) would be (7 + 1)/2 = 4th element which is 84.
Thus the minimum length we can assign to each of the 3 pieces to the left of the 4th piece are a, a, a each as a is the smallest in length according to our arrangement.
So we have a , a , a , 84 all arranged.
What about the next 3 pieces?
Since we intend to minimize, the next two pieces after 84 can be assigned with the same value of 84 each.
(Refer to the GMAT Core Concept explained above)

Now the arrangement is a , a , a , 84 ,84 , 84 , 14+4a (arrangement in ascending order)

The sum of these pieces is
a + a + a + 84 + 84 + 84 + 14 + 4a = 68 * 7
=> 7a + 3*84 + 14 = 68 * 7
Now divide the LHS & RHS by 7

:idea: GMAT Tip :
Try not to calculate immediately on GMAT but rather bring it to the last step and then check if you can minimize on calculations.

=> a + 3*12 + 2 = 68
=> a = 68 -38 = 30
And hence the maximum value or the length of the longest piece is 4a + 14
= 4 * 30 +14
= 134

(option d) :thumbsup:

:idea: GMAT is a combination of logic, strategies and decisive thinking and all these elements when applied in the right way, add up to make you a smart problem solver and all this through a finite box of concepts.
So try not to just solve problems but be flexible with the process and learn "thinking through"! This will help you sail through GMAT with ease and elegance.
Lets see you on the other side of green :)

Devmitra Sen
GMAT Quant SME
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Seven pieces of rope have an average (arithmetic mean) lengt 08 May 2022, 13:53
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avigutman
Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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avigutman

This is so helpful! You are a rockstar as usual.

My approach was close but slightly off...

Spot 1: x
Spot 2: x+1
Spot 3: x+2
Spot 4: 84
Spot 5: x+4
Spot 6: x+5
Spot 7: 14+4x

Then, I added spots 1 through 7 together divided by 7 and set the equation equal to 68. To confirm, this is incorrect because by doing this, I am assuming that each spot is 1 greater than the next, correct and nowhere in the problem does it tell us that?
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Seven pieces of rope have an average (arithmetic mean) lengt 08 May 2022, 18:13
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woohoo921
My approach was close but slightly off...

Spot 1: x
Spot 2: x+1
Spot 3: x+2
Spot 4: 84
Spot 5: x+4
Spot 6: x+5
Spot 7: 14+4x

Then, I added spots 1 through 7 together divided by 7 and set the equation equal to 68. To confirm, this is incorrect because by doing this, I am assuming that each spot is 1 greater than the next, correct and nowhere in the problem does it tell us that?
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That's right, woohoo921. For example, all we know about spots 2 and 3 is that they're between x and 84 (and spot 3 is at least as long as spot 2). We don't even know whether they're integers!
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Seven pieces of rope have an average (arithmetic mean) lengt 22 Dec 2023, 02:18
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BrainLab
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)
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@Bunnel is this correct?
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Seven pieces of rope have an average (arithmetic mean) lengt 22 Dec 2023, 05:39
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BrainLab
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)

@Bunnel is this correct?
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No. In a set with an odd number of terms, the median is the middle term when the terms are arranged in order. For a set with an even number of terms, the median is the average of the two middle terms when arranged in order. The median isn't necessarily the average of the first and last terms, though this can occur in certain cases, like with an evenly spaced set. However, this is not the situation in this question.
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Seven pieces of rope have an average (arithmetic mean) lengt 28 Nov 2025, 18:08
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Walkabout
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
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Solved it in 10 seconds.

The largest value of the longest rope will always be 4K + 14. In short, if we subtract 14 from the largest value, the result should be a multiple of 4. Only option D satisfies that condition.
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