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Five peices of wood have an average length of 124 inches and [#permalink]

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20 Nov 2011, 05:49

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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

Re: Five peices of wood have an average length of 124 inches and [#permalink]

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05 Feb 2012, 15:50

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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 --> \(x_1+x_1+140+140+140=620\) --> \(x_1=100\).

Re: Five peices of wood have an average length of 124 inches and [#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?

Re: Five peices of wood have an average length of 124 inches and [#permalink]

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25 Aug 2013, 06:46

Expert's post

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. _________________

Re: Five peices of wood have an average length of 124 inches and [#permalink]

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12 Jul 2014, 12:22

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 )

Re: Five peices of wood have an average length of 124 inches and [#permalink]

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06 Aug 2015, 04:29

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 --> \(x_1+x_1+140+140+140=620\) --> \(x_1=100\).

Re: Five peices of wood have an average length of 124 inches and [#permalink]

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06 Aug 2015, 07:25

Expert's post

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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