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Mean of Set A

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by Brent@GMATPrepNow » Wed Jan 14, 2009 4:59 pm
If x is a positive integer, and set A = {x, 3, 6x+1, 5x-6, 2-x, 2x, x^2), what is the average (arithmetic mean of set A?
(1) The mean is an integer less than 15
(2) If the number 10-16x is added to set A, the mean will be 1
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by cramya » Wed Jan 14, 2009 5:23 pm
I would go wiht C)

Stmt II

Add all the elements

x^2+13x+10-6x = 8 * mean(1)

x^2-3x+2 = 0

x can be 2 or 1
INSUFF


Stmt I

x^2 + 13x = 7 mean

x(x+13) = 7 * mean

x can be 1 or a multiple of 7 and the mean changes

INSUFF

Together

x can just be 1
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by Brent@GMATPrepNow » Wed Jan 14, 2009 6:11 pm
You're almost there, Cramya.
But the answer isn't C
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Re: Mean of Set A

by logitech » Wed Jan 14, 2009 6:22 pm
When we add them :

X^2+13X


(1) The mean is an integer less than 15

X^2+13X < 7x15

X^2+13X < 105 (x is positive)

So x < 7


(2) If the number 10-16x is added to set A, the mean will be 1

x^2-3x+2 = 0

X=2 or 1

Together we still don't know

Choose E
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by logitech » Wed Jan 14, 2009 6:24 pm
cramya wrote:
Stmt I

x^2 + 13x = 7 mean

x(x+13) = 7 * mean

x can be 1 or a multiple of 7 and the mean changes
Cramya, question says LESS than 15.
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by Brent@GMATPrepNow » Wed Jan 14, 2009 6:26 pm
I need to sign off, so were's my solution:
The mean of set A is (x + 3 + 6x+1 + 5x-6 + 2-x + 2x + x^2)/7 = (x^2+13x)/7 = x(x+13)/7
(1) If the mean is less than 15 then x(x+13)/7 < 15, which means that x(x+13) < 105
We can see that, if x(x+13) < 105 then x < 6 (i.e, x = 1, 2, 3, 4, 5 or 6)
Only one of the these values (x=1) yields a integer mean
So, x must equal 1 (sufficient)
(2) If we add 10-16x to set A we get: mean = (x + 3 + 6x+1 + 5x-6 + 2-x + 2x + x^2 + 10-16x)/8
Simplify to get: (x^2 - 3x + 10)/8 = 1 --> x^2 - 3x + 10 = 8 --> x^2 - 3x + 2 = 0 --> x=1 or 2 (insuff)

The answer is A
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by logitech » Wed Jan 14, 2009 6:28 pm
Brent Hanneson wrote:I need to sign off, so were's my solution:
The mean of set A is (x + 3 + 6x+1 + 5x-6 + 2-x + 2x + x^2)/7 = (x^2+13x)/7 = x(x+13)/7
(1) If the mean is less than 15 then x(x+13)/7 < 15, which means that x(x+13) < 105
We can see that, if x(x+13) < 105 then x < 6 (i.e, x = 1, 2, 3, 4, 5 or 6)
Only one of the these values (x=1) yields a integer mean
So, x must equal 1 (sufficient)
(2) If we add 10-16x to set A we get: mean = (x + 3 + 6x+1 + 5x-6 + 2-x + 2x + x^2 + 10-16x)/8
Simplify to get: (x^2 - 3x + 10)/8 = 1 --> x^2 - 3x + 10 = 8 --> x^2 - 3x + 2 = 0 --> x=1 or 2 (insuff)

The answer is A
Nasty :lol:

Thanks Brent! You ruined my night! LOL
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by dmateer25 » Wed Jan 14, 2009 6:35 pm
Your questions were brutal today!

Thanks for creating these!
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by Brent@GMATPrepNow » Wed Jan 14, 2009 7:07 pm
I'm always happy to ruin a perfectly good evening with an evil math question or two :lol:
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by cramya » Wed Jan 14, 2009 7:46 pm
Happily misssed the less than 15 condition :D

Good question Brent!
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by yalanand » Thu Jan 15, 2009 7:47 am
Wow ....good one
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