# Manhattan GMAT Challenge Problem of the Week – 14 Jan 10

by on January 14th, 2010

Welcome back to this week’s Challenge Problem! As always, the problem and solution below were written by one of our fantastic instructors. Each challenge problem represents a 700+ level question. If you are up for the challenge, however, set your timer for 2 minutes and go!

## Question

If 1/(x – 2) = 1/(x + 2) + 1/(x – 1), which of the following is a possible value of x?

(A) –2
(B) –1
(C) 0
(D) 1
(E) 2

## Solution

The fastest way to solve this problem is first to recognize that an algebraic approach will take a little time. Essentially, we will have to multiply through by the product (x – 2)(x + 2)(x – 1), then simplify.

If, instead, we glance at the answer choices, we see that 3 of them make one of the denominators zero, a result that is not allowed (we cannot divide by zero). Specifically, x cannot be –2 because one denominator is x + 2; likewise, x cannot be 1 or 2, since we have x – 1 and x – 2 as denominators as well.

Thus, the only two possible answers are –1 and 0. We try each in turn.

If x = –1, then we have the following:
1/(–3) = 1/(1) + 1/(–2)?
–1/3 = 1 – 1/2?

This is not true.

However, if x = 0, then we have the following:
1/(–2) = 1/(2) + 1/(–1)?
–1/2 = 1/2 – 1?
–1/2 = –1/2?
This is true, so x can be equal to 0.

Alternatively, we could take the algebraic approach.
First, we multiply through by the product (x – 2)(x + 2)(x – 1) to eliminate denominators.

(x – 1)(x + 2) = (x – 2)(x – 1) + (x – 2)(x + 2)
+ x – 2 = – 3x + 2 + – 4
0 = – 4x
0 = x(x – 4)
x = 0 or x = 4

The correct answer is (C) 0.

To view the current Challenge Problem, simply visit the Challenge Problem page on Manhattan GMAT’s website.

• well easy one

• 1/(x-2) = 1/(x+2) + 1/(x-1)

=> 1/(x-2) = (2x+1)/(x+2)(x-1)
=> (x+2)/(x-2) = (2x+1)/(x-1)

By componendo-Dividendo, we have

2x/4 = 3x/(x+2)
x^2 + 2x = 6x
x^2 - 4x = 0 x=0 or 4.