## Blog

### The Quest for 700: Weekly GMAT Challenge (Answer)

Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer:

We can attack this problem by doing Direct Algebra. First, carry out the replacement. That is, literally replace every x in the expression with 1 – x, putting parentheses around the 1 – x in order to preserve proper order of operations:

Original: 1/x – 1/(1 – x)

Replacement:

1/(1 – x) – 1/(1 – (1 – x))

Now simplify the second denominator: (1 – (1 – x)) = (1 – 1 + x) = x

So the replacement expression becomes this:

1/(1 – x) – 1/x

This should make sense. If we replace x by 1 – x, then it turns out that we are also replacing 1 – x by x (since 1 – (1 – x) = x). Thus, the denominators of the original expression are simply swapped.

Now we can either combine these fractions first (by finding a common denominator) or go ahead & multiply by x2x, as we are instructed to. Let’s take the latter approach.

[1/(1 – x) – 1/x] (x2x)

Instead of FOILing this product right away, we should factor the expression x2x first. If we do so, we will be able to cancel denominators quickly.

x2x factors into (x – 1)x. We can now rewrite the product:

[1/(1 – x) – 1/x] (x – 1)x

= (x – 1)x/(1 – x) – (x – 1)x/x

The second term, (x – 1)x/x, becomes just x – 1 after we cancel the x’s.

Since (x – 1) = –(1 – x), we can rewrite the first term as –(1 – x)x/(1 – x) and then cancel the (1 – x)’s, leaving –x.

So, the final result is

x – (x – 1) = –xx + 1 = 1 – 2x

Separately, since this is a Variables In Choices problem, we could instead pick a number and calculate a target. Since 0 and 1 are disallowed, let’s pick x = 2. We are told that x should be replaced by 1 – x, so we calculate 1 – x = –1 and put in –1 wherever x is in the original expression.

1/x – 1/(1 – x) = 1/(–1) – 1/(1 – (–1))

= –1 – ½

= –3/2

Now multiply this number by x2x = 22 – 2 = 2. We get –3 as our target number.

Finally, we plug x = 2 into the answer choices and look for –3:

(A) x + 1 = 2 + 1 = 3

(B) x – 1 = 2 – 1 = 1

(C) 1 – x2 = 1 – 22 = –3

(D) 2x – 1 = 2(2) – 1 = 3

(E) 1 – 2x = 1 – 2(2) = –3

We can eliminate choices A, B, and D, but to choose between C and E, we would need to pick another number. For instance, if we pick x = 3, we get a target of –5. Only E fits this target.

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