CAT #1 700-800 Level Question Recipriocal Averages

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EMAN: Master | Next Rank: 500 Posts; Posts: 162; Joined: Sun Aug 09, 2009 4:17 pm; Location: Minnesota; Thanked: 1 times

CAT #1 700-800 Level Question Recipriocal Averages

by EMAN » Sun Oct 11, 2009 1:30 pm

The number x is the average (arithmetic mean) of the positive numbers a and b, and the reciprocal of the number y is the average (arithmetic mean) of the reciprocals of a and b. In terms of a and b, x - y = ?

(A)A^2 + b^2 / 2(A+B)
(B)(A-B)^2 / 2(A+B)
(C)A+B / 2
(D)A-B / 2
(E)0

I answered this question incorrectly on the test. I understand how to put this in equation format to a certain point, but after that my algebra skills are insufficient. The explanation involves a lot of tedious algebra. Someone please let me know if there is an easy way to do this without a lot of complicated factoring and other manipulative BS.

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Source: Beat The GMAT — Problem Solving | Sun Oct 11, 2009 1:30 pm

JeffB: Master | Next Rank: 500 Posts; Posts: 153; Joined: Tue Jun 02, 2009 2:48 pm; Location: Ohio; Thanked: 16 times

Re: CAT #1 700-800 Level Question Recipriocal Averages

by JeffB » Sun Oct 11, 2009 1:51 pm

EMAN wrote:The number x is the average (arithmetic mean) of the positive numbers a and b, and the reciprocal of the number y is the average (arithmetic mean) of the reciprocals of a and b. In terms of a and b, x - y = ?

(A)A^2 + b^2 / 2(A+B)
(B)(A-B)^2 / 2(A+B)
(C)A+B / 2
(D)A-B / 2
(E)0

I answered this question incorrectly on the test. I understand how to put this in equation format to a certain point, but after that my algebra skills are insufficient. The explanation involves a lot of tedious algebra. Someone please let me know if there is an easy way to do this without a lot of complicated factoring and other manipulative BS.

Pick numbers:

A = 2
B = 4

A+B/2 = X
2+4/2 = 3

(1/a + 1/b) = 1/y
(1/2 + 1/4 )/2= 3/8

y = 8/3

x-y
3- 8/3 = 1/3

Can eliminate E,D,C

Try A:

A^2 + b^2 / 2(A+B)
4 + 16 / 2(2+4)
20/12 = 5/3

Try B:

(A-B)^2 / 2(A+B)
(2 -4)^2/ 2(2+4)
4/12= 1/3

Edited for Nihil...Thanks...

Last edited by JeffB on Sun Oct 11, 2009 5:31 pm, edited 1 time in total.

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EMAN: Master | Next Rank: 500 Posts; Posts: 162; Joined: Sun Aug 09, 2009 4:17 pm; Location: Minnesota; Thanked: 1 times

Great Idea

by EMAN » Sun Oct 11, 2009 2:06 pm

Awesome. Thanks!.

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nithi_mystics: Senior | Next Rank: 100 Posts; Posts: 55; Joined: Mon Sep 07, 2009 9:11 pm; Location: California; Thanked: 4 times

Re: CAT #1 700-800 Level Question Recipriocal Averages

by nithi_mystics » Sun Oct 11, 2009 4:45 pm

JeffB wrote:
EMAN wrote:The number x is the average (arithmetic mean) of the positive numbers a and b, and the reciprocal of the number y is the average (arithmetic mean) of the reciprocals of a and b. In terms of a and b, x - y = ?

(A)A^2 + b^2 / 2(A+B)
(B)(A-B)^2 / 2(A+B)
(C)A+B / 2
(D)A-B / 2
(E)0

I answered this question incorrectly on the test. I understand how to put this in equation format to a certain point, but after that my algebra skills are insufficient. The explanation involves a lot of tedious algebra. Someone please let me know if there is an easy way to do this without a lot of complicated factoring and other manipulative BS.
Pick numbers:

A = 2
B = 4

A+B/2 = X
2+4/2 = 3

1/a + 1/b = 1/y
1/2 + 1/4 = 3/4

y = 4/3

x-y
3- 4/3 = 5/3

Can eliminate E,D,C

Try A:

A^2 + b^2 / 2(A+B)
4 + 16 / 2(2+4)
20/12 = 5/3