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 x2 – x, as we are instructed to. Let’s take the latter approach.
[1/(1 – x) – 1/x] (x2 – x)
Instead of FOILing this product right away, we should factor the expression x2 – x first. If we do so, we will be able to cancel denominators quickly.
x2 – x 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) = –x – x + 1 = 1 – 2x
This is the answer.
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 x2 – x = 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.
The correct answer is (E).
