Five pieces of wood have an average (arithmetic mean) length of 124 centimeters and a median length of 140 centimeters. What is the maximum possible length, in centimeters, of the shortest piece of wood?

A. 90
B. 100
C. 110
D. 130
E. 140

Given: 5 peices of wood have an average length of 124 centimeters --> total length = 124*5=620. Also median = 140.

If a set has odd number of terms the median of a set is the middle number when arranged in ascending or descending order;
If a set has even number of terms the median of a set is the average of the two middle terms when arranged in ascending or descending order.

As we have odd # of pieces then 3rd largest piece \(x_3=median=140\).

So if we consider the pieces in ascending order of their lengths we would have \(x_1+x_2+140+x_4+x_5=620\).

Question: what is the MAX possible length of the shortest piece of wood? Or \(max(x_1)=?\)

General rule for such kind of problems:
to maximize one quantity, minimize the others;
to minimize one quantity, maximize the others.

So to maximize \(x_1\) we should minimize \(x_2\), \(x_4\) and \(x_5\). Min length of the second largest piece of wood, \(x_2\) could be equal to \(x_1\) and the min lengths of \(x_4\) and \(x_5\) could be equal to 140 --> \(x_1+x_1+140+140+140=620\) --> \(x_1=100\).

Answer: B.
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Sum of lengths is 124*5=620

Median is 140
Sum of the lengths of other four pieces= 620-140=480

The lengths of pieces-

L1, L2, 140, L4, L5

The sum of the four pieces is constant.

L4 and L5 have to be minimum for L1 to be maximum but median length must be 140.

The minimum possible values of L4 and L5 could be 140, hence the L1+L2 = 620 - 420 = 200.
The maximum possible value of L1 = 100= L2

For questions with a fixed sum (here 5*avg = 620), if you need to maximize one term, then you minimize all the others. Likewise, if asked to minimize one term, you maximize all the others.

I find it useful to create placeholders for the terms on paper, filling in with numbers/variables:

(In order, smallest to largest)
x / x / 140 / 140 / 140 = sum of 620

Five pieces of wood have an average (arithmetic mean) length of 124 centimeters and a median length of 140 centimeters. What is the maximum possible length, in centimeters, of the shortest piece of wood?

A. 90
B. 100
C. 110
D. 130
E. 140

Given: 5 peices of wood have an average length of 124 centimeters --> total length = 124*5=620. Also median = 140.

If a set has odd number of terms the median of a set is the middle number when arranged in ascending or descending order;
If a set has even number of terms the median of a set is the average of the two middle terms when arranged in ascending or descending order.

As we have odd # of pieces then 3rd largest piece \(x_3=median=140\).

So if we consider the pieces in ascending order of their lengths we would have \(x_1+x_2+140+x_4+x_5=620\).

Question: what is the MAX possible length of the shortest piece of wood? Or \(max(x_1)=?\)

General rule for such kind of problems:
to maximize one quantity, minimize the others;
to minimize one quantity, maximize the others.

So to maximize \(x_1\) we should minimize \(x_2\), \(x_4\) and \(x_5\). Min length of the second largest piece of wood, \(x_2\) could be equal to \(x_1\) and the min lengths of \(x_4\) and \(x_5\) could be equal to 140 --> \(x_1+x_1+140+140+140=620\) --> \(x_1=100\).

Answer: B.
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It is a nice question, a GMAT type question i.e. fun to work out, can be reasoned out fairly quickly but needs you to think a little. Follow my train of thought here (which finally takes just a few seconds when you start doing it on your own)

First thing that comes to mind - Median is the 3rd term out of 5 so the lengths arranged must look like:

___ _____ 140 _______ __________

The mean is given and I need to maximize the smallest number. Basically, it should be as close to the mean as possible. Which means the greatest number should be as close to the mean as possible too.
If this doesn't make sense, think of a set with mean 20:
19, 20, 21 (smallest number very close to mean, greatest very close to mean too)
10, 20, 30 (smallest number far away, greatest far away too)

Using the same logic, lets make the greatest number as small as possible. The two greatest numbers should both be at least 140 (since 140 is the median)

___ _____ 140 140 140

Since the mean is 124, the 3 greatest numbers are already 16 each more than 124 i.e. total 16*3 = 48 more than the mean. So the two smallest numbers should be a total of 48 less than mean, 124. To make the smallest number as great as possible, both the small numbers should be 24 each less than the mean i.e. they should be 100.
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Karishma
I don't know if the key word "shortest" means the second. It means the least. So the answer should be 99 practically because of the need to differentiate the first from the second- and be compatible with keyword. Your thoughts on this?

Posted from my mobile device

Hey gmat1220,

Smallest just means the smallest element. It doesn't necessarily mean that there should be a unique 'smallest number'.

Say {1, 2, 5, 9, 1, 3, 9}
Which is the smallest number here? 1 right? It doesn't matter even if it appears twice. If I arrange them in ascending order {1, 1, 2, 3 ....} .. the first and the second both are smallest (or shortest length).
So two pieces of wood could have the shortest length. It would be maximized only if their lengths are equal and both have a length of 100.
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Say we list the lengths of our pieces of wood in increasing order:

S, a, 140, b, L

We know that the sum of these lengths is 5*124 = 620. Now, we want to make S, the smallest length, as big as possible. To do that, we want the other unknown lengths to 'use up' as little of the sum of 620 as possible. That is, the smaller we make a, b and L, the larger we can make S. Since b and L must be at least as large as the median, the smallest possible values for b and L are 140. That gives us this set:

S, a, 140, 140, 140

The three largest values now add to 420, so the two smallest values must add to 620-420 = 200. Since making them equal will make a as small as possible (a cannot be less than S), the largest possible value of S is 200/2 = 100.
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Why couldn't x4 and x5 be bigger than 140 and thus making x1 and x2 even smaller?

We want to maximize \(x_1\), not to minimize.

Next, \(x_4\) and \(x_5\) cannot be less than the median, which is 140.
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Here is my solution to this Great Official Question =>

Let W1,W2,W3,W4,W5 be the 5 wooden pieces in increasing order of length.
Given Mean = 124

\(Mean = Sum/#\)

Hence Sum(5) = 124*5 = 620
Now Median = 140
As #=5
Hence median => 3rd term => W3
Hence W3=140

Now to maximise the smallest piece that is W1 we must minimise all the other pieces keeping in mind the following things =>
All values To the left of median must be either less than or equal to it.
All values to the right of median must be greater than or equal to it.
All values in set must be greater than or equal W1
All values in set must be less than or equal to W5

Hence W1+W1+140+140+140 = 620
W1=100
Hence Maximum length of smallest piece of food is 100 inches

Hence B
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Your question is not clear. What are you trying to achieve? The question asks: What is the maximum possible length, in centimeters, of the shortest piece of wood? The answer is 100 centimeters. You got 90 centimeters...[/quote]:

My bad, miss read the question. Apologies

Given: Five pieces of wood have an average (arithmetic mean) length of 124 centimeters and a median length of 140 centimeters.

Asked: What is the maximum possible length, in centimeters, of the shortest piece of wood?

Let the 5 pieces of wood have lengths = {s, s, 140, 140, 140}

2s + 140*3 = 124*5
2s = 620 - 420 = 200
s = 100

The maximum possible length, in centimeters, of the shortest piece of wood = 100 cms

IMO B
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need to maximize length of shortest wood
x+x+140+140+140 = 620
2x = 200
x=100
option b

Why the last two values should not be greater than 140? We have to maximize it so can't we go over 140. which even lessen the value of a1?

I think I addressed this here: https://gmatclub.com/forum/five-pieces- ... l#p1260389
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To get the maximum length of the shortest piece, we must let other values as little as possible.
That is, the values after the median should equal the median, and the value before the median
should be equal to each other.
Let the shortest one be x:
x+x+140*3=124*5
x=100