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by GmatKiss » Sat Aug 06, 2011 11:02 am
Together, Mary and Joe have x dollars. If Mary has $80 less than Joe, how much money, in terms of x, will Joe have if he receives an additional x dollars?

X+40

2X+40

(3X/2)+40

(X/2 )+40

(X/2)+40
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by Touseef » Sat Aug 06, 2011 11:08 am
M+J=X
(J-80)+J=X
2J-80=X

2J-80+X=2X
2J-80=X
J=(80+X)/2

Ans:40+(X/2)
Last edited by Touseef on Sat Aug 06, 2011 11:39 am, edited 1 time in total.
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by goalevan » Sat Aug 06, 2011 11:19 am
We are given two equations to begin:

M + J = x, and
M = J - 80

We can calculate for Joe's current dollars in terms of x:

[J - 80] + J = x
2J - 80 = x
2J = x + 80
J = (x + 80)/2

If we add x to both sides, we will have the answer to the question in terms of x on the right side:

J + x = (x + 80)/2 + x
= (x + 80 + 2x)/2
= (3x + 80)/2
= 3x/2 + 40

C
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by Tani » Sat Aug 06, 2011 11:54 am
This question lends itself to picking numbers.
Say x = 200. Then Joe has 140 and Mary has 60.

Give Joe an additional 200, he now has 340.

Plugging 200 back into the answer choices, only C gives us 340.
Tani Wolff
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by MBA.Aspirant » Sat Aug 06, 2011 2:43 pm
M+J = x

m = x- j

m + 80 = j

x - j + 80 = J

x+80 = 2J

J = 1/2 x + 40

j = 1/2x + 40 + x = 3/2 x + 40
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