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Five peices of wood have an average length of 124 inches and [#permalink]
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20 Nov 2011, 04:49
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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 Even though I got the correct answer (B) by doing maths, but it was a total guess work. This is how I did.
Let say shortest piece be x
x+x+3*140=5*124 2x+420=620 2x=200 x=100
I have no idea how I did it. So can someone be please kind enough to let me know the mathematical approach and the concept ?
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Re: Five peices of wood have an average length of 124 inches and [#permalink]
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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 peices of wood have an average length of 124 inches and [#permalink]
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20 Nov 2011, 21:32
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.
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Re: Five peices of wood have an average length of 124 inches and [#permalink]
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05 Feb 2012, 14:50
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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.
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Re: Five peices of wood have an average length of 124 inches and [#permalink]
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14 Jun 2013, 04:27



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Re: Five peices of wood have an average length of 124 inches and [#permalink]
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25 Aug 2013, 02:21
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 peices of wood have an average length of 124 inches and [#permalink]
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25 Aug 2013, 05:46
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 peices of wood have an average length of 124 inches and [#permalink]
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12 Jul 2014, 11:22
We can find the total length to be as 620 a+b+c+d+e = 620 Since the number of pipes is 5, the median is the middle number ie c. (c=140) a+b+d+e = 480 to find the max length of the smallest pipe. we should look to find the minimum length of other 3 pipes (we are already given length of one of the pipes ,c=140). d=e=140 ( we assume d and e to be 140 as it is given the median is 140 )
a+b = 200 the max length will be 100.



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Re: Five peices of wood have an average length of 124 inches and [#permalink]
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06 Aug 2015, 03:29
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. Hi Bunuel, Why can't the second piece also be 140?



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Five peices of wood have an average length of 124 inches and [#permalink]
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06 Aug 2015, 04:10
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GeeWalia wrote: Hi Bunuel,
Why can't the second piece also be 140? Lets say the 5 pieces are ABCDE A being the smallest and E being the largest. Per the question, A B 140 D E In order to maximise A, we need to minimise B,D,E with D and E having minimum value of 140, 140 (equal to the median). For B =140 , we will have A, 140, 140, 140, 140 = A+560 giving A = 60 but we need to MAXIMIZE A. Thus, the only we will be able to maximise B is to minimize B and as B \(\geq\) A , we can have the minimum value of B = A. The only scenario possible is: A, A , 140,140,140, > A = 100. For checking it, lets say A = 110 is the maximum value, then based on the above, 110,110,140,140,140 > 124*5 = 620 , thus this is not possible. Hope this helps.



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Re: Five peices of wood have an average length of 124 inches and [#permalink]
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06 Aug 2015, 06:25
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. The set could consist of pieces of wood that are L1, L2, 140, 140, 140. In this case, 140 would be the median (and mode). Similarly, the set could consist of pieces of wood that are lengths 100, 100, 140, 140, 140  which maximizes the lengtth of the shortest piece.
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Five peices of wood have an average length of 124 inches and [#permalink]
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15 Dec 2016, 13:17
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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