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Five pieces of wood have an average length of 124cm and a [#permalink]
24 Oct 2005, 23:08
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Five pieces of wood have an average length of 124cm and a median length of 140cm. What is the maximum possible length, in cm, of the shortest piece of wood?
Re: Mean, Median and 5 pieces of wood [#permalink]
04 Aug 2009, 22:38
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The median of 5 pieces is 140. Therefore, there are 2 pieces >=140. Since, we want to maximize the smallest piece, we want to limit the largest piece(s) to the lowest value possible, because the larger the largest pieces the smaller the smallest pieces will have to be. But since the median is 140, it is the floor limit on the size of the 2 largest pieces...so the two largest pieces will have to be 140.
(A+B+140+140+140)/2 = 124 [where, A and B are the smaller pieces]
Since the question is asking for the maximum size of the smallest piece while preserving the average and median, A and B must be equal, so, (2A+140+140+140)/2 = 124
A = 100.
Another way to think about it is, how averages are distributed among numbers. For every inch more the average, their has to be an inch less than the average. So, we have 3 numbers which are 16 each more than the average...in total 48 over the average. The two smaller pieces will have to be compensate this. And to get the maximum lowest value the compensation should be distributed evenly...each member should be 24 less than the average...124-24 = 100. _________________
The three most significant times in your life are: 1. When you fall in love 2. The birth of your first child 3. When you prepare for your GMAT
. Five pieces of wood have an average length of 124 cm and a median length of 140cm. what is the maximum possible length of the shortest piece of wood?
Need help. Thanks
Five pieces of wood in order of increasing length: A,B,C,D,E A+B+C+D+E = 124*5 = 620 C is the median and equals to 140 Because Maximum (A+B) occurs when D and E at their minimum; and D and E cannot be lower than 140, so min. D and E = 140 So max. A+B = 620 - 3*140 = 200 A+B = 200 and B > A So, max A = 99 when B = 101 if B=A is acceptable, max. A = 100
Re: Mean, Median and 5 pieces of wood [#permalink]
28 Sep 2009, 18:05
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tejal777 wrote:
Phew..NOT clear at all.. I struggled and made it through robertrdzak explanation but not clear in the last part of each pc being 100 and 100..why cant it be 90 and 110??
The smallest two pieces COULD be 90, 110. They could also be 80, 120. However, the question stem asks for what the MAX length could be for the smallest piece of wood. In both these situations the smallest piece is 90 and 80 i.e not maximised.
So {100, 100, 140, 140, 140} and {90,100,140,140,140} and many other sets satisfy the conditions for mean and median. But in order to maximise the smallest piece 100 would be the only option. Hope that makes sense.
Re: Mean, Median and 5 pieces of wood [#permalink]
23 Sep 2010, 04:50
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Expert's post
robertrdzak wrote:
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
I see no ambiguity in this question.
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 --> \(x_1+x_1+140+140+140=620\) --> \(x_1=100\).
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 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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Five pieces of wood have an average (arithmetic mean) length of 124cm and a median length of 140 cm. What is the maximum possible length, in centimetres, of the smallest piece of wood?
think there is something wrong with this Q. If the median is 140cm and AVRG is 124 then the total lenght of the woods is 620cm.Since median is 140cm the values above the median should be > or = to the median.Their minimum value is 140x3=420cm.The max possible lenght of the shortest piece is 200-124=76cm
May be i am wrong
The average of 5 pieces is 124, this means the total length of all pieces=620
We now the median is 140, which gives us that 140 is the middle value of all the fives pieces... which means two pieces are below 140 and two pieces above 140....
To get the maximum length of the smallest piece of wood we must minimize the length of the two largest pieces....
Remember we have the total length of 620... and the middle value 140... all we need is to distribute 380 (i.e. 620-140) in such a way that two of the values would be below 140 and two at least 140... so 100+100+140+140+140=620, the maximum value of the smallest piece is 100, since if we chose 110, two other values would only yield 130 which cannot be the case...
Hope i wrote it in understandable way...
Last edited by SimaQ on 21 Feb 2006, 06:49, edited 1 time in total.
think there is something wrong with this Q. If the median is 140cm and AVRG is 124 then the total lenght of the woods is 620cm.Since median is 140cm the values above the median should be > or = to the median.Their minimum value is 140x3=420cm.The max possible lenght of the shortest piece is 200-124=76cm May be i am wrong
Everything is perfect with your reasoning... you got that 3 values must be at least 140*3=420cm so we need to distribute 200 among the rest pieces (i.e 2) which would equal 100...
The way I solved the question as follows.
Median is 140
Average is 124
Total length is 620
Now based on the available information 2 Lengths <= 140 <= remaining two lengths.
Sum of these 4 lengths has to be 620 - 140 = 480
To maximize the shortest length, I considered that two short lengths are equal is size = x and remaining two lengths equal in size = 140 so now the addition of two short lengths is
2X = 480 - (140 + 140)
2X = 200
X = 100
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