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Joined: 16 May 2014
Posts: 592
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GRE 1: Q165 V161
GRE Math Challenge #15 - Set X={a, b, c}, where a < b < c. I
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30 Sep 2014, 04:33
30 Sep 2014, 04:33
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Question Stats:
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23% (01:55) wrong based on 21 sessions
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Set X={a, b, c}, where a < b < c. If the average (arithmetic mean) of a and b is 3x–13, and the average of b and c is 3x+11, what is the range of set X?
Show: :: Answer
Ans:48
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Re: GRE Math Challenge #15 - Set X={a, b, c}, where a < b < c. I
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Updated on: 25 Aug 2015, 08:34
Updated on: 25 Aug 2015, 08:34
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soumya1989 wrote:
Set X={a, b, c}, where a < b < c. If the average (arithmetic mean) of a and b is 3x–13, and the average of b and c is 3x+11, what is the range of set X?
Since a < b < c, the range = c - a
Let's deal with the given information.
The average (arithmetic mean) of a and b is 3x–13
This means that (a + b)/2 = 3x–13
Multiply both sides by 2 to get: a + b = 6x - 26
The average (arithmetic mean) of b and c is 3x+11
This means that (b + c)/2 = 3x+11
Multiply both sides by 2 to get: b + c = 6x + 22
So, how do we use the above information to find the value of c - a?
Here's how.
We have two equations:
a + b = 6x - 26
b + c = 6x + 22
Notice what happens when we subtract the top equation from the bottom equation. We get:
(b + c) - (a + b) = (6x + 22) - (6x - 26)
Simplify to get: c - a = 48
So, the range =
Show: ::
48
Cheers,
Brent
Originally posted by GreenlightTestPrep on 15 May 2015, 11:26.
Last edited by GreenlightTestPrep on 25 Aug 2015, 08:34, edited 1 time in total.
Last edited by GreenlightTestPrep on 25 Aug 2015, 08:34, edited 1 time in total.
Re: GRE Math Challenge #15 - Set X={a, b, c}, where a < b < c. I
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25 Aug 2015, 08:01
25 Aug 2015, 08:01
How is it 40,I Got an answer 48
