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For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which

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For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which [#permalink]

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Tough and Tricky questions: Algebra.



For any numbers \(x\) and \(y\), \(x \diamond y = 2x - y - xy\). If \(x \diamond y = 0\), then which of the following CANNOT be \(y\)?

A. 3
B. 2
C. 0
D. \(-\frac{4}{3}\)
E. -2

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which [#permalink]

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New post 13 Nov 2014, 09:09
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For y=2 the equation gives 1=0 which is impossible.

Answer (B)
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Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which [#permalink]

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New post 13 Nov 2014, 10:52
A. plug in 3 for y to get 2x-3x-3=0 or -x=3. This could work
B. plug in 2 for y to get 2x-2-2x=0 or 0=2 and this is not true

Answer B!!
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Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which [#permalink]

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New post 13 Nov 2014, 19:25
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Answer = B = 2

Given that 2x - y - xy = 0

Bring "x" in terms of "y"

2x - xy = y

\(x = \frac{y}{2-y}\)

Denominator cannot be 0

y cannot be 2
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Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which [#permalink]

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New post 13 Nov 2014, 19:33
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Answer should be B)

We can reduce 2x - y - xy = 0 to y = 2x/1+x

All numbers apart from y = 2 can be equated correctly.
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Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which [#permalink]

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New post 14 Nov 2014, 04:34
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Given \(x \diamond y\) = 2x - y - xy.
And \(x \diamond y\) = 0

Let's Put this value in first equation. So, 2x-xy=y => x= y/2-y
Now we can check the given options one by one from the above equation.

a) y=3 => x = -3 Possible
b) y=2 => x= infinite, Y can not be 2.

hence Answer is B --> 2
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Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which [#permalink]

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Bunuel wrote:

Tough and Tricky questions: Algebra.



For any numbers \(x\) and \(y\), \(x \diamond y = 2x - y - xy\). If \(x \diamond y = 0\), then which of the following CANNOT be \(y\)?

A. 3
B. 2
C. 0
D. \(-\frac{4}{3}\)
E. -2

Kudos for a correct solution.


Official Solution:

For any numbers \(x\) and \(y\), \(x \diamond y = 2x - y - xy\). If \(x \diamond y = 0\), then which of the following CANNOT be \(y\)?

A. 3
B. 2
C. 0
D. \(-\frac{4}{3}\)
E. -2

We must determine which of the answer choices cannot be a value of \(y\).

If \(x \diamond y = 2x - y - xy\) and \(x \diamond y = 0\), we can combine equations to get: \(2x - y - xy = 0\). Because the problem tells us that one value for \(y\) is impossible, one answer choice will make the equation \(2x - y - xy = 0\) false.

Plug in each answer choice and see which makes the equation false.

A. If \(y = 3\), then \(2x - y - xy = 0\) becomes \(2x - 3 - x(3) = 0\). Combine like terms: \(-x - 3 = 0\), so \(x = -3\). Because \(x\) is a variable, it can be equal to any quantity. Therefore, this equation is not false.

B. If \(y = 2\), then \(2x - 2 - x(2) = 0\). The \(x\) terms cancel, leaving \(-2 = 0\). This is false, so B is correct. Double-check by making sure that no other answer yields a false equation.

C. If \(y = 0\), then \(2x - 0 - x(0) = 0\). This simplifies to \(2x = 0\), which is not false.

D. If \(y = -\frac{4}{3}\) then \(2x - (-\frac{4}{3}) - x(-\frac{4}{3}) = 0\). Combine like terms: \(\frac{10}{3}x + \frac{4}{3} = 0\). Isolate the \(x\) term: \(\frac{10}{3}x = -\frac{4}{3}\), so \(x = -\frac{2}{5}\). This is not false.

E. If \(y = -2\), then \(2x - (-2) - x(-2) = 0\). Combine like terms: \(4x +2 = 0\). Isolate the \(x\) term: \(4x = -2\), so \(x = -\frac{1}{2}\). This is not false, so only choice B yields a false equation.

Choice B is the correct answer.

Alternatively, we can solve this problem using algebra. Recall that a value can be impossible for a variable in two ways: if the value of the variable causes a division by 0 or if the value forces the equation to look for the square root of a negative number. Since there are no exponents or roots in this equation, the answer will likely involve dividing by zero. If plugging in a certain value for \(y\) causes a division by zero, we should look for \(y\) in the denominator of a fraction. Solving the equation for \(x\) may give us a fraction in terms of \(y\).

First factor out the \(x\) to get \(x(2 - y) - y = 0\).

Add \(y\) to both sides: \(x(2 - y) = y\).

Divide both sides by \(2 - y\) to get \(x = \frac{y}{2-y}\).

If \(y = 2\), the denominator becomes \(2 - 2 = 0\). Any value that makes a denominator equal to 0 cannot be a valid solution to an equation because dividing by 0 is undefined. Therefore, \(y\) cannot be equal to 2.

Answer: B.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
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For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which [#permalink]

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New post 14 Nov 2014, 22:49
Means that

2x-y-xy=0
x(2-y)-y=0
x(2-y)=y
so, test every option

A) x(2-3)=3 is possible when x=-3
B) x(2-2)=2 is not possible in any x
C) x(2-0)=0 is possible when x=0
D) x(2--4/3)=-4/3 is possible when x=-4/3:10/3
E) x(2--2)=-2 is possible when x=-1/2

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Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which [#permalink]

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Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which   [#permalink] 01 Apr 2017, 23:59
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