Of the 5 numbers, the largest number is 4 greater than the : GMAT Data Sufficiency (DS)
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# Of the 5 numbers, the largest number is 4 greater than the

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Of the 5 numbers, the largest number is 4 greater than the [#permalink]

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10 Aug 2010, 02:05
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Of the 5 numbers, the largest number is 4 greater than the median. Is the mean greater than the median?
(1) The largest number plus the median is 34.
(2) The median minus the smallest number is 10

[Reveal] Spoiler:
B

The method i used was too tedious and turned out to be wrong at the end Can someone help me with a easy method / explanation please ?
[Reveal] Spoiler: OA
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10 Aug 2010, 02:54
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Expert's post
amijags wrote:
Of the 5 numbers, the largest number is 4 greater than the median. Is the mean greater than the median?
(1) The largest number plus the median is 34.
(2) The median minus the smallest number is 10

[Reveal] Spoiler:
B

The method i used was too tedious and turned out to be wrong at the end Can someone help me with a easy method / explanation please ?

Hi, and welcome to Gmat Club. Very good question, so +1 for it. Below is a solution:

Le these 5 numbers in ascending order be $$a$$, $$b$$, $$c$$, $$d$$, $$e$$: $$a\leq{b}\leq{c}\leq{d}\leq{e}$$.

Median would be the middle number - $$c$$ and the $$mean=\frac{a+b+c+d+e}{5}$$.

Given: $$e=c+4$$. Question: is $$\frac{a+b+c+d+e}{5}>c$$ --> is $$\frac{a+b+c+d+(c+4)}{5}>c$$ --> is $$a+b+d+4>3c$$

(1) The largest number plus the median is 34 --> $$e+c=34$$ --> $$c=15$$ and $$e=19$$ --> question becomes: is $$a+b+d+4>45$$. Now, if $$a=b=15$$ (max values possible for $$a$$ and $$b$$) and $$d=19$$ (max value possible for $$d$$) then answer would be YES but as min values of $$a$$ and $$b$$ are not limited at all then the answer could be NO as well. Not sufficient.

(2) The median minus the smallest number is 10 --> $$a=c-10$$ --> question becomes: is $$c-10+b+d+4>3c$$ --> $$b+d>2c+6$$. Now, max value of $$b$$ is $$c$$ and max value of $$d$$ is $$e=c+4$$, so max value of LHS (left hand side) would be $$LHS=b+d=c+c+4=2c+4$$, which is always less than right hand side: $$RHS=2c+6$$. Hence the answer to the question is NO. Sufficient.

Hope it's clear.
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10 Aug 2010, 03:24
Thank you very much

I get deflected at times am going to mark 2 sides henceforth for DS questions
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Re: Of the 5 numbers, the largest number is 4 greater than the [#permalink]

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21 Feb 2014, 04:46
Nice question +1

Just to add a conceptual approach to this one

Statement 1 is insufficient since we can only find the numerical values of the median and the largest number (M=15, L=19) But we still don’t know about the other three numbers. For instance, smallest numbers could be 11 and other two numbers equally spaced giving an evenly spaced set with Median= Mean. We have extra flexibility to place the other two numbers since we are told that the numbers are not necessarily integers.

Statement 2 is a bit more tricky. It says that the smallest number is 10 less than the median. So let's say that the median is 10 and the largest and smallest numbers are 14 and 0 respectively. Now, we will have 2 numbers to the left of 10 and two to the right. Even if we have 10 and 10 to the left and 10 and 10 to the right, the mean will ALWAYS be larger than the median. Therefore B is the correct answer

Hope this helps
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Re: Of the 5 numbers, the largest number is 4 greater than the [#permalink]

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18 May 2014, 09:35
Because St. 1 tells us : Median is 15.
So the five numbers can be 19, (one number between 19-15), 15, (2 numbers less than 15)

St. 2 gives us a range between the largest and the smallest number.
Now visualize the numbers on the numberline, the largest number is 4 greater than the median, and median in 10 greater than the smallest number. Now the Mean will be in the exact middle of this range. Therefore, it has to be less than the Median. IMO B
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Re: Of the 5 numbers, the largest number is 4 greater than the [#permalink]

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22 Apr 2016, 17:21
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