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# If y≠x, then [x^3 + (x^2 + x)(1-y) - y] / (x-y) = ?

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Manager
Joined: 09 Nov 2012
Posts: 170
GMAT 1: 700 Q43 V42
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Kudos [?]: 59 [0], given: 29

If y≠x, then [x^3 + (x^2 + x)(1-y) - y] / (x-y) = ? [#permalink]

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22 Nov 2012, 23:06
00:00

Difficulty:

45% (medium)

Question Stats:

74% (03:24) correct 26% (02:54) wrong based on 66 sessions

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This is problem #34 on page 238 of Manhattan GMAT's Advanced GMAT Quant book.

If y≠x, then [x^3 + (x^2 + x)(1-y) - y] / (x-y) = ?

(A) [(x-1)^2]y

(B) (x+1)^2

(C) (x^2 + x +1)

(D) (x^2 + x +1)y

(E) (x^2 + x +1)(x-y)

Please see the spoiler below for my question:

[Reveal] Spoiler:
I picked 3 for x and 2 for y, but both C and E are correct for that. I redid the problem and picked 4 for x and 2 for y. That makes C the only correct answer. How can I avoid this problem in the future? If this were the real GMAT, I would have wasted a 30 seconds to a minute on this problem because I would have had to do it twice.
[Reveal] Spoiler: OA

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If my post helped you, please consider giving me kudos.

Last edited by Bunuel on 23 Nov 2012, 02:00, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If y≠x, then [x^3 + (x^2 + x)(1-y) - y] / (x-y) = ? [#permalink]

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23 Nov 2012, 20:40
1
KUDOS
Expert's post
commdiver wrote:
This is problem #34 on page 238 of Manhattan GMAT's Advanced GMAT Quant book.

If y≠x, then [x^3 + (x^2 + x)(1-y) - y] / (x-y) = ?

(A) [(x-1)^2]y

(B) (x+1)^2

(C) (x^2 + x +1)

(D) (x^2 + x +1)y

(E) (x^2 + x +1)(x-y)

Please see the spoiler below for my question:

[Reveal] Spoiler:
I picked 3 for x and 2 for y, but both C and E are correct for that. I redid the problem and picked 4 for x and 2 for y. That makes C the only correct answer. How can I avoid this problem in the future? If this were the real GMAT, I would have wasted a 30 seconds to a minute on this problem because I would have had to do it twice.

When you pick numbers, it is normal for you to get 2 or even 3 options that work out. The reason for this is that we tend to pick really easy numbers so that the calculation does not get cumbersome. I do not suggest you to pick harder numbers of course; I suggest you to pick even easier numbers so that the iterations don't take time.

I picked numbers to solve this too but I took numbers in which it took me a few secs to get to the correct option.

I said to myself, 'nothing says one of them can't be 0. Let x = 1 and y = 0'
Then [x^3 + (x^2 + x)(1-y) - y] / (x-y) = 3
(A) and (D) are outright out since they equal 0 because y is a factor in them. (B) gives 4 so out. (C) and (E) are the only possible options since they both give 3.

Now I notice that (C) and (E) differ in the product (x - y). So I want that the difference between them should not be 1 (I think you noticed this too). So now, I pick x = 2 and y = 0 (why to give up a good thing? y = 0 makes life easy)
[x^3 + (x^2 + x)(1-y) - y] / (x-y) = 14/2 = 7
Option (C) gives 7 while option (E) will give 7/2

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Kudos [?]: 79582 [2] , given: 10022

Re: If y≠x, then [x^3 + (x^2 + x)(1-y) - y] / (x-y) = ? [#permalink]

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24 Nov 2012, 06:16
2
KUDOS
Expert's post
commdiver wrote:
This is problem #34 on page 238 of Manhattan GMAT's Advanced GMAT Quant book.

If y≠x, then [x^3 + (x^2 + x)(1-y) - y] / (x-y) = ?

(A) [(x-1)^2]y

(B) (x+1)^2

(C) (x^2 + x +1)

(D) (x^2 + x +1)y

(E) (x^2 + x +1)(x-y)

Please see the spoiler below for my question:

[Reveal] Spoiler:
I picked 3 for x and 2 for y, but both C and E are correct for that. I redid the problem and picked 4 for x and 2 for y. That makes C the only correct answer. How can I avoid this problem in the future? If this were the real GMAT, I would have wasted a 30 seconds to a minute on this problem because I would have had to do it twice.

ALGEBRAIC APPROACH:

$$\frac{x^3 + (x^2 + x)(1-y) - y}{x-y}=\frac{x^3 +x^2-x^2y+x-xy-y}{x-y}=\frac{(x^3-x^2y)+(x^2-xy)+(x-y)}{x-y}=$$
$$\frac{x^2(x-y)+x(x-y)+(x-y)}{x-y}=\frac{(x-y)(x^2+x+1)}{x-y}=x^2+x+1$$.

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Re: If y≠x, then [x^3 + (x^2 + x)(1-y) - y] / (x-y) = ? [#permalink]

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24 Nov 2012, 06:58
3
KUDOS
commdiver wrote:
This is problem #34 on page 238 of Manhattan GMAT's Advanced GMAT Quant book.

If y≠x, then [x^3 + (x^2 + x)(1-y) - y] / (x-y) = ?

(A) [(x-1)^2]y

(B) (x+1)^2

(C) (x^2 + x +1)

(D) (x^2 + x +1)y

(E) (x^2 + x +1)(x-y)

Please see the spoiler below for my question:

[Reveal] Spoiler:
I picked 3 for x and 2 for y, but both C and E are correct for that. I redid the problem and picked 4 for x and 2 for y. That makes C the only correct answer. How can I avoid this problem in the future? If this were the real GMAT, I would have wasted a 30 seconds to a minute on this problem because I would have had to do it twice.

This can be solved quickly, if we look for cancellations in the numerator on the basis of options. The options suggests that y gets eliminated, means a rearrangement of the numerator will help us in eliminating y.

When we open up the brackets we will find that there are factors of (x-y)

x^3+x^2+x-x^2*y- xy-y (Numerator)

lets combine in factors of x-y (Since it is in denominator)
x^2 (x-y) + x(x-y) +1 (x-y)

so we get (x^2+x+1) as the final answer (because x-y cancels out in the numerator and denominator)

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Re: If y≠x, then [x^3 + (x^2 + x)(1-y) - y] / (x-y) = ? [#permalink]

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08 Dec 2012, 04:36
Let x=2 and y=3
= $$\frac{[(2)^3 + ((2)^2 + (2))(1-(3)) - (3)]}{(2-3)}$$
= $$\frac{(8 + (6)(-2) - 3)}{-1}$$
= $$\frac{-7}{-1}$$
= $$7$$

(A) $$[(2-1)^2]3=3$$
(B) $$(2+1)^2=9$$
(C) $$(2^2 + 2 +1)=7$$
(D) $$(2^2 + 2 +1)(3)=21$$
(E) $$(2^2 + 2 +1)(2-3)=-7$$

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Re: If y≠x, then [x^3 + (x^2 + x)(1-y) - y] / (x-y) = ?   [#permalink] 08 Dec 2012, 04:36
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