Mean?

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[email protected]: Master | Next Rank: 500 Posts; Posts: 117; Joined: Mon Oct 27, 2008 5:08 pm; Thanked: 1 times

by [email protected] » Tue Feb 10, 2009 6:23 am

Find the set of integers whose mean is less than the its median
[12, 15, 21, 33, 36, 63]

[12, 18, 21, 33, 36, 42]

[6, 9, 15, 21, 24, 33]

[3, 12, 15, 18, 36, 42]

[3, 9, 27, 33, 45, 54]

E
is there an easy fast way of doing this problem?

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Source: Beat The GMAT — Problem Solving | Tue Feb 10, 2009 6:23 am

hardik.jadeja: Legendary Member; Posts: 535; Joined: Fri Jun 08, 2007 2:12 amFollowed by: 5 members; Thanked: 87 times; GMAT Score:730

by hardik.jadeja » Thu Feb 12, 2009 1:52 pm

Yes, there is an easy way to solve this.

First find median of all 5 sets. That should be easy.

1) 27
2) 27
3) 18
4) 16.5
5) 30

Now, subtract first value of the set from the median (Let's call it x) and compare it with the value you get when you subtract median from the last value of the set(Let's call it y).

If x>y then median > mean. But if x< y then median < mean.

Basically what this means is that values on the right side of the median are very closely distributed to each other. Whereas on the left side, the values are not closely distributed to each other, which actually brings the value of mean down.

So lets make a table of x, y & median
1) x=27-12=15;..........y=63-27=36;..........So y>x
2) x=27-12=15;..........y=42-27=15;..........So y=x
3) x=18-06=12;..........y=33-18=15;..........So y>x
4) x=16.5-3=13.5;......y=42-16.5=25.5;.....So y>x
5) x=30-03=27;..........y=54-30=24;..........Bingo x>y

So the answer is E.

You don't have to do all the calculations here.. You can skip most of them as you can easily figure out x will not be greater than y for the first four options..

Plus, always start from the back when you are asked a question where you have to use the given answer choices to find the correct one.. I started with E and got the answer in no time..

HTH..

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