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# Five logs of wood have an average length of 100 cm and a median 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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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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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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how do we know that the two smaller pieces have the same length? If the two smaller pieces have different values, then it would be possible to even increase the maximum length of the second smallest piece of wood?
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Re: Five logs of wood have an average length of 100 cm and a median length [#permalink]
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.

$$Σ_5 = 500cm$$

$$a, b , 116 , d , e = 500$$

Now, if $$d = e = 116$$ $$a + b + 116 + 116 + 116 = 500$$

Considering a = b we have 2a = 152

So, a = 76 , Answer must be (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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Bunuel VeritasKarishma, kindly approve the following method :

5 logs , say _ _ 116 _ _
116+ 4x = 500 (since average is 100) . So
4x = 384 or x = 96.

Now rewrite : 96 96 116 96 96. (here 116 is not the median yet)
Rewriting again to make 116 median,
76 76 116 116 116
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Re: Five logs of wood have an average length of 100 cm and a median length [#permalink]
Total logs of wood: 5

Median: 3rd log of wood = 116

Mean: 100 and therefore, Sum = Mean * Total = 100 * 5 = 500

We want the shortest log of maximum length and hence the 2 bigger logs should be 116 each [minimum possible].

=> Wood log 1 + Wood log 2 + 116 + 116 + 116 = 500

=> Wood log 1 + Wood log 2 = 500 - 348 = 152

Now shortest will have maximum value when both are equal and hence 2(Wood log short) = 152

=> Shortest wood log: $$\frac{152 }{ 2}$$ = 76

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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

Kudos for a correct solution.

ctouimi

We are given the mean. We need to maximise the shortest piece. It means the shortest piece should be as close to the mean as possible. Then the longest pieces should be as close to the mean as possible too.

116 is the median. So the largest two pieces must be at least 116.

_ _ 116, 116, 116

To make the mean 100, the extra of 3*16 = 48 must be adjusted in the two smallest pieces. So each piece must be 48/2 = 24 less than 100, the mean. So the two smallest pieces must be 76 each.

Note that if the two smallest pieces are not equal, the smallest piece would become smaller while the second smallest piece would become a bit larger e.g. 70 and 82 (the mean needs to stay at 100)
But if the smallest piece needs to take its maximum value then the two smallest pieces will need to be equal.

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Re: Five logs of wood have an average length of 100 cm and a median length [#permalink]
Because median is 116 out of 5 logs­, x3=116
x1+x2+116+x4+x5=500 (100*5(numberoflogs))
x1+x2+x4+x5=384
Because the maximum length needed for x1, we need to minimize x4 and x5 which lets say 116, so
x1+x2+116+116=384
x1+x2=152
Now x2 need to be minimized also to get the maximum amount for x1, so
152/2=76
x1=76

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