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Five logs of wood have an average length of 100 cm and a median length

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Five logs of wood have an average length of 100 cm and a median length [#permalink]

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Five logs of wood have an average length of 100 cm and a median length of 116 cm. What is the maximum possible length, in cm, of the shortest piece of wood?

(A) 50
(B) 76
(C) 84
(D) 96
(E) 100

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[Reveal] Spoiler: OA

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Re: Five logs of wood have an average length of 100 cm and a median length [#permalink]

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Bunuel wrote:
Five logs of wood have an average length of 100 cm and a median length of 116 cm. What is the maximum possible length, in cm, of the shortest piece of wood?

(A) 50
(B) 76
(C) 84
(D) 96
(E) 100.[/i]


Ans: B
Solution: we know avg length is 100 means total length is 500.
mediun is 116 means among the other four, 2 are less than 116 or equal ro 2 are greater or equal. but as we know avg is 100 means all of them can not be of same length.
now requirement is shortest piece with maximum length. means shortest piece can be as close as 116 but as other two bigger piec0e can be bigger or equal to 116 and we want to make shortest piece the largest we need to make those two bigger pieces = 116
now we know three length 116*3 = 348
500-348=152
so 152/2= 76
this can be the maximum shortest piece length
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Re: Five logs of wood have an average length of 100 cm and a median length [#permalink]

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New post 23 Aug 2015, 11:43
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Bunuel wrote:
Five logs of wood have an average length of 100 cm and a median length of 116 cm. What is the maximum possible length, in cm, of the shortest piece of wood?

(A) 50
(B) 76
(C) 84
(D) 96
(E) 100

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

First thing that comes to mind – median is the 3rd term out of 5 so the lengths arranged in increasing order must look like this:

___ ___ 116 ___ ___

The mean is given and we need to maximize the smallest number. Basically, the smallest number should be as close to the mean as possible. This means the greatest number should be as close to the mean as possible too (if the shortfall deviation is small, the excess deviation should by equally small).

If this doesn’t make sense, think of a set with mean 20:

19, 20, 21 (smallest number is very close to mean; greatest number is very close to the mean too)
1, 20, 39 (smallest number is far away from the mean, greatest number is far away too)

Using the same logic, let’s make the greater numbers as small as possible (so the smallest number can be as large as possible). The two greatest numbers should both be at least 116 (since 116 is the median). Now the lengths arranged look like this:

___ ___ 116 116 116

Since the mean is 100 and each of the 3 large numbers are already 16 more than 100 i.e. total 16*3 = 48 more than the mean (excess deviation is 48), the deviations of the two small numbers should be a total of 48 less than the mean. To make the smallest number as great as possible, each of the small numbers should be 48/2 = 24 less than the mean i.e. they both should be 76.

Answer (B).
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Re: Five logs of wood have an average length of 100 cm and a median length [#permalink]

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New post 24 Oct 2016, 05:44
Bunuel wrote:
Five logs of wood have an average length of 100 cm and a median length of 116 cm. What is the maximum possible length, in cm, of the shortest piece of wood?

(A) 50
(B) 76
(C) 84
(D) 96
(E) 100

Kudos for a correct solution.


in order to maximize the shortest piece of wood, we need to minimize the longest ones.
suppose we have 3 logs with a length of 116. in this case, the median is 116 too.
116 * 3 = 348.
since average of 5 is 100, the sum of the five must be 500.
500-348 = 152.
to have the maximum value for first one, we need to have first two of the same length...152/2=76.
maximum possible value for the shortest log is 76. B
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Re: Five logs of wood have an average length of 100 cm and a median length [#permalink]

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Here is my solution to this one=>


Let the Five pieces of wood be->

\(W(1\))
\(W(2)\)
\(W(3)\)
\(W(4)\)
\(W(5)\)




Mean(5)=100
Hence Sum(5)=500
Median => 3rd term => W(3)=116
To maximise W(1) => We would need to minimise the rest.

Notice median =116
Hence W(4) and W(5) cannot be smaller than 116.
Let they be each 116
Now W(2) cannot be less than W(1)
Let it be W(1)


Hence \(W(1)+W(1)+116+116+116=500\)

\(2*W(1)=152\)

Hence W(1)=76


Hence B
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Re: Five logs of wood have an average length of 100 cm and a median length [#permalink]

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Re: Five logs of wood have an average length of 100 cm and a median length   [#permalink] 14 Jan 2018, 11:16
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