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

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commdiver
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If y ≠ x, then (x^3 + (x^2 + x)(1-y) - y)/(x - y) = ? [#permalink] If y ≠ x, then (x^3 + (x^2 + x)(1-y) - y)/(x - y) = ?   Updated on: 17 Mar 2020, 07:39
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If \(y ≠ x\), then \(\frac{x^3 + (x^2 + x)(1-y) - y}{x - y} =\) ?

(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:

Show 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.
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C

Originally posted by commdiver on 22 Nov 2012, 23:06.
Last edited by Bunuel on 17 Mar 2020, 07:39, edited 2 times 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] If y ≠ x, then (x^3 + (x^2 + x)(1-y) - y)/(x - y) = ?   24 Nov 2012, 06:16
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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:

Show 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:

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

Answer: C.
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Re: If y ≠ x, then (x^3 + (x^2 + x)(1-y) - y)/(x - y) = ? [#permalink] If y ≠ x, then (x^3 + (x^2 + x)(1-y) - y)/(x - y) = ?   23 Nov 2012, 20:40
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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:

Show 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

Answer must be (C)
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Re: If y ≠ x, then (x^3 + (x^2 + x)(1-y) - y)/(x - y) = ? [#permalink] If y ≠ x, then (x^3 + (x^2 + x)(1-y) - y)/(x - y) = ?   24 Nov 2012, 06:58
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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:



Show 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)

Answer : C
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Re: If y ≠ x, then (x^3 + (x^2 + x)(1-y) - y)/(x - y) = ? [#permalink] If y ≠ x, then (x^3 + (x^2 + x)(1-y) - y)/(x - y) = ?   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) \((2^2 + 2 +1)(2-3)=-7\)

Answer: C
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Re: If y ≠ x, then (x^3 + (x^2 + x)(1-y) - y)/(x - y) = ? [#permalink] If y ≠ x, then (x^3 + (x^2 + x)(1-y) - y)/(x - y) = ?   02 Dec 2020, 13:57
I've used X=1 and Y=2:

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

a = 0
b = 4
c = 3
d = 6
e = -3

I Think that this is FASTER
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Re: If y x, then (x^3 + (x^2 + x)(1-y) - y)/(x - y) = ? [#permalink] If y ≠ x, then (x^3 + (x^2 + x)(1-y) - y)/(x - y) = ?   13 Mar 2024, 22:21
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Re: If y x, then (x^3 + (x^2 + x)(1-y) - y)/(x - y) = ? [#permalink]
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