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

Question

If  y ≠ x , then  \frac{x^3 + (x^2 + x)(1-y) - y}{x - y} =  ?

(A)  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:

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.

Bunuel · Accepted Answer

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.

KarishmaB · Answer

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-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-y) = 14/2 = 7
Option (C) gives 7 while option (E) will give 7/2

Answer must be (C)

Arvars · Answer

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

mbaiseasy · Answer

Let x=2 and y=3
=  \frac{ }{(2-3)} 
=  \frac{(8 + (6)(-2) - 3)}{-1} 
=  \frac{-7}{-1} 
=  7

(A)    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

plaverbach · Answer

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

GMAT1034 · Answer

Are these kind of pure math questions relevant for focus edn ?  Bunuel

Bunuel · Answer

Yes. Pure algebra questions are still part of GMAT Focus Edition in the  Problem Solving  section. However, pure algebra questions are no longer part of  Data Sufficiency .

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

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GMAT Problem Solving (PS) Questions

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

Prep Toolkit

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My Rewards