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S97-16

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S97-16  [#permalink]

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New post 16 Sep 2014, 01:52
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A
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C
D
E

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  25% (medium)

Question Stats:

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

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S97-16  [#permalink]

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New post 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.


Answer: E
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S97-16  [#permalink]

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New post 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.
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Re S97-16  [#permalink]

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New post 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
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Re: S97-16  [#permalink]

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New post 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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