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# http://www.manhattangmat.com

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Senior Manager
Joined: 05 May 2003
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23 Feb 2004, 08:27
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

http://www.manhattangmat.com
Senior Manager
Joined: 05 May 2003
Posts: 424
Location: Aus
Followers: 2

Kudos [?]: 10 [0], given: 0

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23 Feb 2004, 09:19
How did you arrive at C ?

Other than {2,3,4,5,6} is there any other set which has range and average equal. I am struck there. What is easy way to crack this ?
Senior Manager
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23 Feb 2004, 09:24
Geethu wrote:
How did you arrive at C ?

Other than {2,3,4,5,6} is there any other set which has range and average equal. I am struck there. What is easy way to crack this ?

I think the statement B was kind of time consuming to crack. You have to try different combination. One thing we can think of is that we are already given that the range and the average are equal, With statement 2, we are also given that the numbers are different. So the numbers must be evenly spaced so that the middle number could be the average and the numbers are equally dispersed on both sides of the middle number. This was all just my hunch. Not sure if that is true.

Anyhow, from statement 2 , we have another combination 4, 6, 8, 10, 12

Even after this you will have to apply a little logic to arrive at the answer C. I would leave it for others and you to try.
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23 Feb 2004, 09:33
Now I agree with you. The answer should be C.
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23 Feb 2004, 10:01
I got C.

From statement 1)
range = 3 or 7
This is insufficient to answer the question.
with (1,2,3) you have range = 2 and average = 2
with (3,5,8,9,10) you have range = 7 and average = 7

Statement 2 is insufficient. You could have more than one combination of numbers that gives you a range and average that are equal to each other.

statements 1 & 2 together, range =7 and sum of integers in S will equal 35.

I can't think of any other way of solving this.
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