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Five pieces of wood have an average (arithmetic mean) length of 124 [#permalink]
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04 Aug 2009, 23:38
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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
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Re: Five pieces of wood have an average (arithmetic mean) length of 124 [#permalink]
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30 Jun 2010, 13:42
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
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Five pieces of wood have an average (arithmetic mean) length of 124 [#permalink]
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23 Sep 2010, 05:50
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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Re: Five pieces of wood have an average (arithmetic mean) length of 124 [#permalink]
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09 Apr 2011, 21:01
AnkitK wrote: 5 pieces of wood have an average length of 124 cm and a median of 140 cm .what is the maximum possible length in cm of the shortest piece of wood? A.90 B.100 C.110 D.130 E.140 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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Re: Five pieces of wood have an average (arithmetic mean) length of 124 [#permalink]
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10 Apr 2011, 06:18
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?
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Re: Five pieces of wood have an average (arithmetic mean) length of 124 [#permalink]
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10 Apr 2011, 07:51
gmat1220 wrote: 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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Re: Five pieces of wood have an average (arithmetic mean) length of 124 [#permalink]
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12 Sep 2011, 20:54
socalboy429 wrote: 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 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 620420 = 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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Re: Five pieces of wood have an average (arithmetic mean) length of 124 [#permalink]
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20 Nov 2011, 07:21
Sum of lengths is 124*5=620 Median is 140 Sum of the lengths of other four pieces= 620140=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
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Re: Five pieces of wood have an average (arithmetic mean) length of 124 [#permalink]
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05 Feb 2012, 15:50
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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Re: Five pieces of wood have an average (arithmetic mean) length of 124 [#permalink]
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14 Jun 2013, 05:27



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Re: Five pieces of wood have an average (arithmetic mean) length of 124 [#permalink]
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25 Aug 2013, 03:21
Bunuel wrote: enigma123 wrote: Apologies blink005, if I am getting this wrong. But how did you get this?
The minimum possible values of L4 and L5 could be 140, hence the L1+L2 = 620  420 = 200. Below is step by step analysis of this question. Hope it helps. 5 peices of wood have an average length of 124 inches and a median of 140 inches. What is the MAX possible length 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 inches > 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. Why couldn't x4 and x5 be bigger than 140 and thus making x1 and x2 even smaller?



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Re: Five pieces of wood have an average (arithmetic mean) length of 124 [#permalink]
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25 Aug 2013, 06:46
Skag55 wrote: Bunuel wrote: enigma123 wrote: Apologies blink005, if I am getting this wrong. But how did you get this?
The minimum possible values of L4 and L5 could be 140, hence the L1+L2 = 620  420 = 200. Below is step by step analysis of this question. Hope it helps. 5 peices of wood have an average length of 124 inches and a median of 140 inches. What is the MAX possible length 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 inches > 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. 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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Re: Five pieces of wood have an average (arithmetic mean) length of 124 [#permalink]
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15 Dec 2016, 14:17
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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