Manhattan GMAT Challenge Problem of the Week - 25 Mar 10

by , Mar 25, 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 mins and go!

Question

[pmath]x[/pmath] is replaced by [pmath]1-x[/pmath] everywhere in the expression [pmath]{1/x}-{1/(1-x)}[/pmath], with x<>0 and x<>1. If the result is then multiplied by [pmath]{x^2}-x[/pmath], the outcome equals

(A) [pmath]x+1[/pmath]

(B) [pmath]x-1[/pmath]

(C) [pmath]1-{x^2}[/pmath]

(D) [pmath]2x-1[/pmath]

(E) [pmath]1-2x[/pmath]

Solution

First, carry out the replacement. Literally replace every [pmath]x[/pmath] in the expression with [pmath]1-x[/pmath], putting parentheses around the [pmath]1-x[/pmath] in order to preserve proper order of operations:

Original: [pmath]{1/x}-{1/(1-x)}[/pmath]

Replacement: [pmath]{1/(1-x)}-{1/(1-(1-x))}[/pmath]

Now simplify the second denominator: [pmath](1-(1-x))=(1-1+x)=x[/pmath]

So the replacement expression becomes this:

[pmath]{1/(1-x)}-{1/x}[/pmath]

This should make sense. If we replace [pmath]x[/pmath] by [pmath]1-x[/pmath], then it turns out that we are also replacing [pmath]1-x[/pmath] by [pmath]x[/pmath] since [pmath]1-(1-x)=x[/pmath]. 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 [pmath]{x^2}-x[/pmath], as we are instructed to. Lets take the latter approach.

[pmath]({1/(1-x)}-{1/x})(x^2-x)[/pmath]

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

[pmath]x^2-x[/pmath] factors into [pmath](x-1)x[/pmath]. We can now rewrite the product:

[pmath]({1/(1-x)}-{1/x})((x-1)x) = {{(x-1)x}/{(1-x)}}-{{(x-1)x}/x}[/pmath]

The second term, [pmath]{(x-1)x}/x[/pmath], becomes just [pmath]x-1[/pmath] after we cancel the [pmath]x[/pmath]s.

Since [pmath](x-1) = -(1-x)[/pmath], we can rewrite the first term as [pmath]{-(1-x)x}/{(1-x)}[/pmath] and then cancel the [pmath](1-x)[/pmath]s, leaving [pmath]-x[/pmath].

So, the final result is:

[pmath]-x-(x-1)=-x-x+1=1-2x[/pmath]

This is the answer.

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

[pmath]{1/x}-{1/(1-x)}={1/(-1)}-{1/(1-(-1))}[/pmath]

=[pmath]-1-{1/2}[/pmath]

=[pmath]-{3/2}[/pmath]

Now multiply this number by [pmath]{x^2}-x={2^2}-2=2[/pmath]. We get [pmath]-3[/pmath] as our target number.

Finally, we plug [pmath]x=2[/pmath] into the answer choices and look for [pmath]-3[/pmath]:

(A) [pmath]x+1=2+1=3[/pmath]

(B) [pmath]x-1=2-1=1[/pmath]

(C) [pmath]1-{x^2}=1-{2^2}=-3[/pmath]

(D) [pmath]2x-1=2(2)-1=3[/pmath]

(E) [pmath]1-2x=1-2(2)=-3[/pmath]

We can eliminate choices A, B, and D, but to choose between C and E, we would need to pick another number.

The correct answer is (E).

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