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Is x > y? (1) x= y+2 (2) x/2 = y1 [#permalink]
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02 Jul 2012, 02:39
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Is x > y? (1) x= y+2 (2) x/2 = y1 [#permalink]
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02 Jul 2012, 02:39



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Re: Is x > y? (1) x= y+2 (2) x/2 = y1 [#permalink]
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02 Jul 2012, 02:44
Hi,
Difficulty level: 600
Using (1), x = y + 2, so for any value of y, x > y. Sufficient.
Using (2), x/2 = y  1 for x = 0, y = 1 or x < y for x = 2, y = 2 or x = y for x = 4, y = 3 or x > y, Thus the relationship depends on value of x & y. Insufficient.
Thus, Answer is (A)
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Re: Is x > y? (1) x= y+2 (2) x/2 = y1 [#permalink]
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02 Jul 2012, 12:28
1) Sufficient . X is always 2 greater than Y. (+,ve and fraction). 2) Insufficient. For y=5, x = 8 so x> y but for y=5, x= 12 so y> x..
A)



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Re: Is x > y? (1) x= y+2 (2) x/2 = y1 [#permalink]
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03 Jul 2012, 09:53
(1) x = y+2 sufficient .try zero, negative, positive integers and nonintegers, and u will get answer YES, x>y. (2) x/2 = y1 means x=2(y1) if y=1 x =0 so, the answ is No, but if y=3 x=4 the answer is Yes
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Re: Is x > y? (1) x= y+2 (2) x/2 = y1 [#permalink]
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03 Jul 2012, 13:42
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Is x > y?
(1) x = y+2 Regardless of whether y is a positive or negative, if x=y+2, then y is greater than x. SUFFICIENT. (2) x/2 = y1 This kind of logic pops up frequently from the GMAT probelms I have done. x=2(y1) x=2y2 Whenever you see mult/div AND sub/add on the other side of x, you won't be able to tell whether whether one integer is greater than the other. It could be either of the case.
Answer: A



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Re: Is x > y? (1) x= y+2 (2) x/2 = y1 [#permalink]
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21 Nov 2015, 10:18
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Is x > y? (1) x = y+2 (2) x/2 = y1 We get xy>0? when we modify the questions. For condition 1, xy=2>0 This is a 'yes', and sufficient. For condition 2, x=2y2, xy=y2. We cannot determine the sign of xy, so this is insufficient. The answer (A). For (C), x=y+2, x=2y2 > y=2, x=4. This is insufficient as it is too trivial. Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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Re: Is x > y? (1) x= y+2 (2) x/2 = y1 [#permalink]
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13 Dec 2017, 21:25
Hi All, We're asked if X is greater than Y. This is a YES/NO question. We can solve it by TESTing VALUES. 1) X = Y +2 With this Fact, we can see that X is "2 greater" than Y, so the answer to the question is ALWAYS YES. You can see this pattern with a few TESTs: IF.... Y = 0, X = 2 Y = 5, X = 7 Y = 3, X = 1 Etc. Fact 1 is SUFFICIENT 2) X/2 = Y  1 IF.... Y = 1, X = 0 and the answer to the question is NO Y = 3, X = 4 and the answer to the question is YES Fact 2 is INSUFFICIENT Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: Is x > y? (1) x= y+2 (2) x/2 = y1 [#permalink]
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13 Dec 2017, 21:48
stmt 1: x= y +2 => xy = 2 => xy >0 => x > y (sufficient) stmt 2: x/2 = y1 => x = 2y 2 => x 2y <0 => x< 2y (not sufficient)



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Re: Is x > y? (1) x= y+2 (2) x/2 = y1 [#permalink]
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26 Jan 2018, 14:54
Bunuel VeritasPrepKarishmaCould you please provide graphical approach to solve Statement 2? I am able to draw the line on the graph. The line meets the x axis and y axis at points (2,0) and (0,1) respectively. From this point, how do we determine if there would be a case when y < x. Number plugging approach doesn't come very easily to me, so trying out other ways to solve. Thanks for your help!!
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Re: Is x > y? (1) x= y+2 (2) x/2 = y1 [#permalink]
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29 Jan 2018, 01:42
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sdlife wrote: Bunuel VeritasPrepKarishmaCould you please provide graphical approach to solve Statement 2? I am able to draw the line on the graph. The line meets the x axis and y axis at points (2,0) and (0,1) respectively. From this point, how do we determine if there would be a case when y < x. Number plugging approach doesn't come very easily to me, so trying out other ways to solve. Thanks for your help!! Hi They can provide better explanation but I will try. So you have drawn a line of x/2 = y  1. The other line you should also draw is x = y (or y = x). Why you may ask? Because the question asks you whether x is > y or not? Once you have the line y = x on the graph (it will be a line with positive slope of 1 passing through origin inclined at 45 degrees with x axis), you can see that: the area on graph below this line y=x will be the area where x > y and the area on the graph above this line y=x will be the area where x < y. So on one hand, you have this line y=x and on the other hand you have the line x/2 = y1. You will see that some part of line x/2 = y1 will fall below the y=x line and some part of line x/2 = y1 will be above the y=x line. So what does this tell us? This tells us that with the data x/2 = y1, we CANNOT be sure whether x > y or not.



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Re: Is x > y? (1) x= y+2 (2) x/2 = y1 [#permalink]
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30 Jan 2018, 03:00
sdlife wrote: Bunuel VeritasPrepKarishmaCould you please provide graphical approach to solve Statement 2? I am able to draw the line on the graph. The line meets the x axis and y axis at points (2,0) and (0,1) respectively. From this point, how do we determine if there would be a case when y < x. Number plugging approach doesn't come very easily to me, so trying out other ways to solve. Thanks for your help!! Yes, you have the line which depicts x/2 = y1. You see that this represents a line with slope 1/2. So y increases by 1/2 for every 1 unit increase in x. It passes through (2, 0) and (0, 1). So if x increases by 2, y increases by 1 so (2, 2) lies on this line. If x increases by another 2, y increases just by 1 so (4, 3) lies on this line too. In case of some points x > y and in other cases x < y. So this statement alone is not sufficient.
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Re: Is x > y? (1) x= y+2 (2) x/2 = y1 [#permalink]
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30 Jan 2018, 14:29
VeritasPrepKarishma wrote: Yes, you have the line which depicts x/2 = y1. You see that this represents a line with slope 1/2. So y increases by 1/2 for every 1 unit increase in x. It passes through (2, 0) and (0, 1). So if x increases by 2, y increases by 1 so (2, 2) lies on this line. If x increases by another 2, y increases just by 1 so (4, 3) lies on this line too. In case of some points x > y and in other cases x < y. So this statement alone is not sufficient.
Hi Karishma, Thank you very much for your response. Could you please explain the highlighted part? SD
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Is x > y? (1) x= y+2 (2) x/2 = y1 [#permalink]
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30 Jan 2018, 19:43
amanvermagmat wrote: They can provide better explanation but I will try. So you have drawn a line of x/2 = y  1. The other line you should also draw is x = y (or y = x). Why you may ask? Because the question asks you whether x is > y or not? Once you have the line y = x on the graph (it will be a line with positive slope of 1 passing through origin inclined at 45 degrees with x axis), you can see that: the area on graph below this line y=x will be the area where x > y and the area on the graph above this line y=x will be the area where x y or not. amanvermagmat Thanks a ton for your response. I do understand the solution now. A couple questions if you don't mind answering: 1) How did you think of drawing a line x=y for this question? One clue maybe since it was asked if x>y? But I couldn't even think of it. Any tips on how to solve these type of problems? 2) Probably a dumb question. How do decide that the side above the line y=x will have y>x, while below is x<y? Thank you very much for your help!
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Re: Is x > y? (1) x= y+2 (2) x/2 = y1 [#permalink]
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30 Jan 2018, 21:57
sdlife wrote: VeritasPrepKarishma wrote: Yes, you have the line which depicts x/2 = y1. You see that this represents a line with slope 1/2. So y increases by 1/2 for every 1 unit increase in x. It passes through (2, 0) and (0, 1). So if x increases by 2, y increases by 1 so (2, 2) lies on this line. If x increases by another 2, y increases just by 1 so (4, 3) lies on this line too. In case of some points x > y and in other cases x < y. So this statement alone is not sufficient.
Hi Karishma, Thank you very much for your response. Could you please explain the highlighted part? SD Check this post. It explains you the slope concept of the line. http://www.veritasprep.com/blog/2010/12 ... hegraphs/
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Re: Is x > y? (1) x= y+2 (2) x/2 = y1 [#permalink]
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30 Jan 2018, 22:08
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sdlife wrote: amanvermagmat wrote: They can provide better explanation but I will try. So you have drawn a line of x/2 = y  1. The other line you should also draw is x = y (or y = x). Why you may ask? Because the question asks you whether x is > y or not? Once you have the line y = x on the graph (it will be a line with positive slope of 1 passing through origin inclined at 45 degrees with x axis), you can see that: the area on graph below this line y=x will be the area where x > y and the area on the graph above this line y=x will be the area where x y or not. amanvermagmat Thanks a ton for your response. I do understand the solution now. A couple questions if you don't mind answering: 1) How did you think of drawing a line x=y for this question? One clue maybe since it was asked if x>y? But I couldn't even think of it. Any tips on how to solve these type of problems? 2) Probably a dumb question. How do decide that the side above the line y=x will have y>x, while below is x<y? Thank you very much for your help! Hi I dont mind answering any questions, as far as I know their answers 1) Yes, you are right. I thought of drawing x=y because I got the clue from x > y (which the question was asking). As you understand more about graphs, I am sure you will get better in these. 2) Once you plot the line y=x, I am sure you would agree that on one side of it, x > y and on another side of it x < y. How to check which is which? Just take any one point from any one side of the line, and see whether it fits in x > y or x < y. Eg., here you have the line y=x with you. Now on the right side of this line (or below) lets choose a point (3,1). The xcoordinate of the point is 3, and y coordinate is 1, and its clear that here x > y. So no need to check further, we can be sure that on the right side of the line x=y, all points will have x > y. This means automatically on the left side of the line (or you can say above the line), all points will have x < y (or you could check with any point above the line x=y, and you will find that x < y.




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