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# S97-16

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Math Expert
Joined: 02 Sep 2009
Posts: 58312

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16 Sep 2014, 01:52
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Difficulty:

25% (medium)

Question Stats:

81% (02:06) correct 19% (01:59) wrong based on 21 sessions

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$$x$$ is replaced by $$1 - x$$ everywhere in the expression $$\frac{1}{x} - \frac{1}{1 - x}$$, with $$x \neq 0$$ and $$x \neq 1$$. If the result is then multiplied by $$x^2 - x$$, the outcome equals

A. $$x + 1$$
B. $$x - 1$$
C. $$1 - x^2$$
D. $$2x - 1$$
E. $$1 - 2x$$

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Math Expert
Joined: 02 Sep 2009
Posts: 58312

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16 Sep 2014, 01:52
Official Solution:

$$x$$ is replaced by $$1 - x$$ everywhere in the expression $$\frac{1}{x} - \frac{1}{1 - x}$$, with $$x \neq 0$$ and $$x \neq 1$$. If the result is then multiplied by $$x^2 - x$$, the outcome equals

A. $$x + 1$$
B. $$x - 1$$
C. $$1 - x^2$$
D. $$2x - 1$$
E. $$1 - 2x$$

First, carry out the replacement. 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: $$\frac{1}{x} - \frac{1}{1 - x}$$

Replacement:
$$\frac{1}{(1 - x)} - \frac{1}{1 - (1 - x)}$$

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

So the replacement expression becomes this:
$$\frac{1}{(1 - x)} - \frac{1}{x}$$

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 and multiply by $$x^2 - x,$$ as we are instructed to. Let’s take the latter approach.
$$( \frac{1}{1 - x} - \frac{1}{x}) (x^2 - x)$$

Instead of FOILing this product right away, we should factor the expression $$x^2 - x$$ first. If we do so, we will be able to cancel denominators quickly.

$$x^2 - x$$ factors into $$(x - 1)x$$. We can now rewrite the product:
$$( \frac{1}{1 - x} - \frac{1}{x} ) (x - 1)x = \frac{(x - 1)x}{(1 - x)} - \frac{(x - 1)x}{x}$$

The second term, $$\frac{(x - 1)x}{x}$$, becomes just $$x - 1$$ after we cancel the $$x$$'s.

Since $$(x - 1) = -(1 - x)$$, we can rewrite the first term as $$\frac{-(1 - x)x}{(1 - x)}$$ 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.
$$\frac{1}{x} - \frac{1}{(1 - x)} = \frac{1}{(-1)} - \frac{1}{(1 - (-1))} = -1 - \frac{1}{2} = -\frac{3}{2}$$

Now multiply this number by $$x^2 - x = 2^2 - 2 = 2$$. 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) $$1 - x^2 = 1 - 2^2 = -3$$

(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.

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Joined: 30 Jan 2018
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31 Jan 2018, 11:37
Found an easier way to solve this problem.
Assume (1-x) to be equivalent to a variable r.
Now the solution becomes 1/r - 1/x * x*-r
=> (x-r)/xr * (-rx) (cancel the numerator and denominator to get -1(x-r) = r-x
=> Now substitute r = 1-x
=> 1-x-x = 1-2x
Solution = E.
Intern
Joined: 17 Mar 2019
Posts: 3

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12 May 2019, 11:10
I think this the explanation isn't clear enough, please elaborate. the last part of the solution is a little unclear. Can it be explained in a little more algebraic depth?

x2−xx2−x factors into (x−1)x(x−1)x. We can now rewrite the product:

(11−x−1x)(x−1)x=(x−1)x(1−x)−(x−1)xx(11−x−1x)(x−1)x=(x−1)x(1−x)−(x−1)xx$The second term, (x−1)xx(x−1)xx, becomes just x−1x−1 after we cancel the xx's. Since (x−1)=−(1−x)(x−1)=−(1−x), we can rewrite the first term as −(1−x)x(1−x)−(1−x)x(1−x) and then cancel the (1−x)(1−x)’s, leaving −x−x. So, the final result is −x−(x−1)=−x−x+1=1−2x−x−(x−1)=−x−x+1=1−2x. This is the answer Math Expert Joined: 02 Sep 2009 Posts: 58312 Re: S97-16 [#permalink] ### Show Tags 12 May 2019, 21:17 asfandkhan wrote: I think this the explanation isn't clear enough, please elaborate. the last part of the solution is a little unclear. Can it be explained in a little more algebraic depth? x2−xx2−x factors into (x−1)x(x−1)x. We can now rewrite the product: (11−x−1x)(x−1)x=(x−1)x(1−x)−(x−1)xx(11−x−1x)(x−1)x=(x−1)x(1−x)−(x−1)xx$
The second term, (x−1)xx(x−1)xx, becomes just x−1x−1 after we cancel the xx's.

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

So, the final result is −x−(x−1)=−x−x+1=1−2x−x−(x−1)=−x−x+1=1−2x. This is the answer

You can find more solutions here: https://gmatclub.com/forum/x-is-replace ... 88681.html
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Re: S97-16   [#permalink] 12 May 2019, 21:17
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# S97-16

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