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If x does not equal y, and xy does not equal 0, then when x
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02 Nov 2010, 00:48
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\(\frac{x+y}{xy}\) If x does not equal y, and xy does not equal 0, then when x is replaced by \(\frac{1}{x}\) and y is replaced by \(\frac{1}{y}\) everywhere in the expression above, the resulting expression is equivalent to: A) \(\frac{x+y}{xy}\) B) \(\frac{xy}{x+y}\) C) \(\frac{x+y}{xy}\) D) \(\frac{x^2y^2}{xy}\) E) \(\frac{yx}{x+y}\)
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Re: Variable Inversion
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02 Nov 2010, 02:24



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Re: Variable Inversion
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02 Nov 2010, 02:58
Bunuel wrote: Phaser wrote: \(\frac{x+y}{xy}\)
If x does not equal y, and xy does not equal 0, then when x is replaced by \(\frac{1}{x}\) and y is replaced by \(\frac{1}{y}\) everywhere in the expression above, the resulting expression is equivalent to: A) \(\frac{x+y}{xy}\) B) \(\frac{xy}{x+y}\) C) \(\frac{x+y}{xy}\)
D) \(\frac{x^2y^2}{xy}\)
E) \(\frac{yx}{x+y}\)
Please explain your answers if you can. OA will be provided in a few hours. When we replace we'll get: \(\frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x}\frac{1}{y}}=\frac{\frac{x+y}{xy}}{\frac{yx}{xy}}=\frac{x+y}{yx}=\frac{x+y}{xy}\). Answer: A. Yes, that is correct. But can you explain how the denominator ends up as yx instead of xy? Because that is what I did when I got this question on the MGMAT CAT and ended up with the wrong answer "A".



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Re: Variable Inversion
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02 Nov 2010, 03:04
Phaser wrote: Bunuel wrote: Phaser wrote: \(\frac{x+y}{xy}\)
If x does not equal y, and xy does not equal 0, then when x is replaced by \(\frac{1}{x}\) and y is replaced by \(\frac{1}{y}\) everywhere in the expression above, the resulting expression is equivalent to: A) \(\frac{x+y}{xy}\) B) \(\frac{xy}{x+y}\) C) \(\frac{x+y}{xy}\)
D) \(\frac{x^2y^2}{xy}\)
E) \(\frac{yx}{x+y}\)
Please explain your answers if you can. OA will be provided in a few hours. When we replace we'll get: \(\frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x}\frac{1}{y}}=\frac{\frac{x+y}{xy}}{\frac{yx}{xy}}=\frac{x+y}{yx}=\frac{x+y}{xy}\). Answer: A. Yes, that is correct. But can you explain how the denominator ends up as yx instead of xy? Because that is what I did when I got this question on the MGMAT CAT and ended up with the wrong answer "A". \(\frac{x+y}{yx}=\frac{x+y}{(xy)}=\frac{x+y}{xy}\)
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Re: Variable Inversion
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02 Nov 2010, 03:09
Ah, that explains. Thank you very much.



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Re: If x does not equal y...
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19 Sep 2012, 23:21
egiles wrote: x + y  x – y If x does not equal y, and xy does not equal 0, then when x is replaced by 1/x and y is replaced by 1/y everywhere in the expression above, the resulting expression is equivalent to... a) (x+y)  xy b) x  y  x + y c) x + y  x  y d) x^2  y^2  xy e) y  x  x + y When I do the algebra, I get d. Here are the steps I take. Where am I going wrong?...
1/x + 1/y (1/x + 1/y)(x  y) 1  y/x + x/y  1 x/y  y/x x^2  y^2  = = = =  1/x  1/y xy Hard to read your answer due to the formatting. \(\frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x}\frac{1}{y}}=\frac{\frac{y+x}{xy}}{\frac{yx}{xy}}=\frac{x+y}{xy}\cdot{\frac{xy}{yx}}=\frac{x+y}{yx}=\frac{x+y}{(xy)}=\frac{x+y}{xy}\) Answer A
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Re: If x does not equal y...
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19 Sep 2012, 23:32
I think the formatting of your solution (inside the spoiler tags) became illegible, but I may be able to guess at the mistake. When we replace 'x' with '1/x', and 'y' with '1/y', we have the following: \(\frac{ \frac{1}{x} + \frac{1}{y} }{ \frac{1}{x}  \frac{1}{y} }\) I think you then rewrote this as \(\left( \frac{1}{x} + \frac{1}{y} \right) (x  y)\), and if so, that isn't right. If that's what you did, you were probably trying to apply the rule "when you divide by a fraction, multiply by its reciprocal". There are two problems with trying to do that right away: we aren't dividing by a simple fraction here, so we can't use that rule (in the denominator, we're subtracting one fraction from another), and regardless, xy is not the reciprocal of (1/x)  (1/y). If those terms were reciprocals of each other, their product would be 1, and if you multiply them out, you'll see their product is not 1. As a first step, since we have fractions in both our numerator and denominator, I'd just multiply the top and bottom by xy. We then need to multiply by 1 on top and bottom only because of how the question designer chose to write the correct answer choice: \(\begin{align} \frac{ \frac{1}{x} + \frac{1}{y} }{ \frac{1}{x}  \frac{1}{y} } &= \left( \frac{ \frac{1}{x} + \frac{1}{y} }{ \frac{1}{x}  \frac{1}{y} } \right) \left( \frac{xy}{xy} \right) \\ &= \frac{y + x}{y  x} \\ &= \left( \frac{x + y}{x + y} \right) \left( \frac{1}{1} \right) \\ &= \frac{(x+y)}{xy} \end{align}\)
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Re: If x does not equal y, and xy does not equal 0, then when x
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10 Dec 2012, 02:59
\(=(\frac{1}{x}+\frac{1}{y})/(\frac{1}{x}\frac{1}{y})\) \(=(\frac{x+y}{xy})/(\frac{yx}{xy})\) \(=\frac{x+y}{yx}\) \(=\frac{{x+y}}{{(xy)}}\) Answer: A
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If x does not equal y, and xy does not equal 0, then when x
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30 Oct 2014, 00:32
\(\frac{x+y}{xy}\) Divide numerator & denominator by x \(\frac{1+\frac{y}{x}}{1\frac{y}{x}}\) Replacing x with \(\frac{1}{x}\) & y with \(\frac{1}{y}\) means replacing \(\frac{y}{x}\) with \(\frac{x}{y}\)\(\frac{1+\frac{x}{y}}{1 \frac{x}{y}} = \frac{\frac{y+x}{y}}{\frac{yx}{y}} =  \frac{x+y}{xy}\) Answer = A
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Re: If x does not equal y, and xy does not equal 0, then when x
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14 Aug 2016, 17:39
Take x = 2 and y = 1.
Value of original expression = (2 + 1)/(2  1) = 3
Value of original expression as per the suggested transformation = (0.5 + 1)/(0.5  1) = 3. This is value of original expression * 1.
Only choice (A) fits.



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Re: If x does not equal y, and xy does not equal 0, then when x
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18 Oct 2017, 10:38
Phaser wrote: \(\frac{x+y}{xy}\)
If x does not equal y, and xy does not equal 0, then when x is replaced by \(\frac{1}{x}\) and y is replaced by \(\frac{1}{y}\) everywhere in the expression above, the resulting expression is equivalent to: A) \(\frac{x+y}{xy}\) B) \(\frac{xy}{x+y}\) C) \(\frac{x+y}{xy}\)
D) \(\frac{x^2y^2}{xy}\)
E) \(\frac{yx}{x+y}\) (x+y)/(xy) ~ (1/x+1/y)/(1/x1/y) = [(x+y)/xy]/[(yx)/xy] = (x+y)/(yx) = (x+y)/(xy) Answer A
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If x does not equal y, and xy does not equal 0, then when x
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12 Dec 2017, 05:58
Bunuel wrote: Phaser wrote: \(\frac{x+y}{xy}\)
If x does not equal y, and xy does not equal 0, then when x is replaced by \(\frac{1}{x}\) and y is replaced by \(\frac{1}{y}\) everywhere in the expression above, the resulting expression is equivalent to: A) \(\frac{x+y}{xy}\) B) \(\frac{xy}{x+y}\) C) \(\frac{x+y}{xy}\)
D) \(\frac{x^2y^2}{xy}\)
E) \(\frac{yx}{x+y}\)
Please explain your answers if you can. OA will be provided in a few hours. When we replace we'll get: \(\frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x}\frac{1}{y}}=\frac{\frac{x+y}{xy}}{\frac{yx}{xy}}=\frac{x+y}{yx}=\frac{x+y}{xy}\). Answer: A. Would you be able to explain why the "" only affects the denominator and not the numerator? I was able to get the answer up until that point I'm just confused as to why the signs change in the denominator but not the numerator. Thank you for your help.



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Re: If x does not equal y, and xy does not equal 0, then when x
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12 Dec 2017, 06:19
imranmah wrote: Bunuel wrote: Phaser wrote: \(\frac{x+y}{xy}\)
If x does not equal y, and xy does not equal 0, then when x is replaced by \(\frac{1}{x}\) and y is replaced by \(\frac{1}{y}\) everywhere in the expression above, the resulting expression is equivalent to: A) \(\frac{x+y}{xy}\) B) \(\frac{xy}{x+y}\) C) \(\frac{x+y}{xy}\)
D) \(\frac{x^2y^2}{xy}\)
E) \(\frac{yx}{x+y}\)
Please explain your answers if you can. OA will be provided in a few hours. When we replace we'll get: \(\frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x}\frac{1}{y}}=\frac{\frac{x+y}{xy}}{\frac{yx}{xy}}=\frac{x+y}{yx}=\frac{x+y}{xy}\). Answer: A. Would you be able to explain why the "" only affects the denominator and not the numerator? I was able to get the answer up until that point I'm just confused as to why the signs change in the denominator but not the numerator. Thank you for your help. We are factoring out 1 from the denominator: \(\frac{x+y}{yx}=\frac{x+y}{1*(xy)}=\frac{x+y}{xy}\)
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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.
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