Author 
Message 
TAGS:

Hide Tags

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6385
GPA: 3.82

If xy=y, x+y=?
[#permalink]
Show Tags
05 Oct 2018, 01:36
Question Stats:
54% (00:56) correct 46% (00:50) wrong based on 123 sessions
HideShow timer Statistics
[ Math Revolution GMAT math practice question] If \(xy=y, x+y\)=? \(1) x=1\) \(2) y=0\)
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The oneandonly World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only $99 for 3 month Online Course" "Free Resources30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons  try it yourself"





Math Expert
Joined: 02 Aug 2009
Posts: 6958

Re: If xy=y, x+y=?
[#permalink]
Show Tags
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
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor



GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 374

Re: If xy=y, x+y=?
[#permalink]
Show Tags
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.
_________________
Fabio Skilnik :: http://www.GMATH.net (Math for the GMAT) Course release PROMO : finish our test drive till 31/Oct with (at least) 60 correct answers out of 92 (12questions Mock included) to gain a 60% discount!



Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6385
GPA: 3.82

Re: If xy=y, x+y=?
[#permalink]
Show Tags
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.
_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The oneandonly World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only $99 for 3 month Online Course" "Free Resources30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons  try it yourself"



Intern
Joined: 14 Jun 2018
Posts: 47
Location: India
Concentration: International Business, Marketing

Re: If xy=y, x+y=?
[#permalink]
Show Tags
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.



Math Expert
Joined: 02 Aug 2009
Posts: 6958

Re: If xy=y, x+y=?
[#permalink]
Show Tags
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
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor



Intern
Joined: 14 Jun 2018
Posts: 47
Location: India
Concentration: International Business, Marketing

Re: If xy=y, x+y=?
[#permalink]
Show Tags
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=? &nbs
[#permalink]
11 Oct 2018, 03:43






