Manhattan GMAT Challenge Problem of the Week - 14 Jan 10

by , Jan 14, 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)

[pmath]x^2[/pmath] + x 2 = [pmath]x^2[/pmath] 3x + 2 + [pmath]x^2[/pmath] 4

0 = [pmath]x^2[/pmath] 4x

0 = x(x 4)

x = 0 or x = 4

The correct answer is (C) 0.

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