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If xy=y, x+y=?
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05 Oct 2018, 01:36
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[ Math Revolution GMAT math practice question] If \(xy=y, x+y\)=? \(1) x=1\) \(2) y=0\)
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Re: If xy=y, x+y=?
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05 Oct 2018, 03:15
If \(xy=y, x+y\)=? \(xy=y.......xyy=0.......y(x1)=0\), so y=0 or x=1 or both \(1) x=1\) So \(x\neq{1}\), and therefore y=0 Thus x+y=1+0=1 Sufficient \(2) y=0\) Nothing about x Insuff A
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Re: If xy=y, x+y=?
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05 Oct 2018, 08:28
MathRevolution wrote: [ Math Revolution GMAT math practice question] If \(xy=y, x+y\)=? \(1) x=1\) \(2) y=0\) \(xy = y\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,y\left( {x  1} \right) = 0\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( * \right)\,\,\,\,\,\left\{ \matrix{ \,\,y = 0 \hfill \cr \,\,{\rm{OR}} \hfill \cr \,\,x = 1 \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\) \(? = \left x \right + \left y \right\) \(\left( 1 \right)\,\,\,x =  1\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,y = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = \left {  1} \right + \left 0 \right\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\rm{SUFF}}.\,\,\,\,\,\,\) \(\left( 2 \right)\,\,\,y = 0\,\,\,\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {0,0} \right)\,\,\,\, \Rightarrow \,\,\,{\rm{?}}\,\,{\rm{ = }}\,\,0\,\, \hfill \cr \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,0} \right)\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{?}}\,\,{\rm{ = }}\,\,{\rm{1}}\,\, \hfill \cr} \right.\,\) This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.
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Re: If xy=y, x+y=?
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07 Oct 2018, 18:35
=> Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. Modifying the original condition: The equality \(xy=y\) is equivalent to \(y = 0\) or \(x = 1\) as shown below: \(xy=y\) \(=> xyy=0\) \(=> y(x1)=0\) \(=> y = 0\) or \(x = 1\) Since we have \(2\) variables (\(x\) and \(y\)) and \(1\) equation (\(xy=y\)), D is most likely to be the answer. Condition 1) Since \(x = 1\) from condition 1) and \(y = 0\) or \(x = 1\) from the original condition, \(y = 0\). Thus, \(x + y = 1 + 0 = 1.\) Condition 1) is sufficient. Condition 2) If \(x = 1\) and \(y = 0\), then \(x+y = 1\). If \(x = 2\) and \(y = 0\), then \(x+y = 2\). Since we don’t have a unique solution, condition 2) is not sufficient. Therefore, A is the answer. Answer: A Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: If xy=y, x+y=?
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10 Oct 2018, 10:16
chetan2u wrote: If \(xy=y, x+y\)=?
\(xy=y.......xyy=0.......y(x1)=0\), so y=0 or x=1 or both
\(1) x=1\) So \(x\neq{1}\), and therefore y=0 Thus x+y=1+0=1 Sufficient
\(2) y=0\) Nothing about x Insuff
A why is x not equal to 1 in statement 2. why is it insufficient just because y = 0? in statement 1, only x =1 and we take y = 0 from the given. so why not the same in statement 2? please clear this doubt Sir.



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Re: If xy=y, x+y=?
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10 Oct 2018, 19:01
Shri15kumar wrote: chetan2u wrote: If \(xy=y, x+y\)=?
\(xy=y.......xyy=0.......y(x1)=0\), so y=0 or x=1 or both
\(1) x=1\) So \(x\neq{1}\), and therefore y=0 Thus x+y=1+0=1 Sufficient
\(2) y=0\) Nothing about x Insuff
A why is x not equal to 1 in statement 2. why is it insufficient just because y = 0? in statement 1, only x =1 and we take y = 0 from the given. so why not the same in statement 2? please clear this doubt Sir. Hi.. The equation is y(x1)=0 So three cases.. 1) y=0 2) X=1 3) both y=0 and x=1 Statement I gives value of X as 1 so y HAS to be 0, thus we have value of both X and y Statement II gives y as 0 so X can be 1 or it can be any other value as y=0 makes equation y(x1)=0(x1)=0 true .. here X can be 10 1,100 anything still equation will be true. Had it been given y is 5 or y is not equal to 0, X would have been 1
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Re: If xy=y, x+y=?
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11 Oct 2018, 03:43
chetan2u wrote: Shri15kumar wrote: chetan2u wrote: If \(xy=y, x+y\)=?
\(xy=y.......xyy=0.......y(x1)=0\), so y=0 or x=1 or both
\(1) x=1\) So \(x\neq{1}\), and therefore y=0 Thus x+y=1+0=1 Sufficient
\(2) y=0\) Nothing about x Insuff
A why is x not equal to 1 in statement 2. why is it insufficient just because y = 0? in statement 1, only x =1 and we take y = 0 from the given. so why not the same in statement 2? please clear this doubt Sir. Hi.. The equation is y(x1)=0 So three cases.. 1) y=0 2) X=1 3) both y=0 and x=1 Statement I gives value of X as 1 so y HAS to be 0, thus we have value of both X and y Statement II gives y as 0 so X can be 1 or it can be any other value as y=0 makes equation y(x1)=0(x1)=0 true .. here X can be 10 1,100 anything still equation will be true. Had it been given y is 5 or y is not equal to 0, X would have been 1 Got it! Thank you sir!




Re: If xy=y, x+y=?
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