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Is x < y ? [#permalink]
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Updated on: 23 Apr 2013, 05:46
Question Stats:
73% (00:55) correct 27% (00:42) wrong based on 368 sessions
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Is x < y ? (1) 2x < 3y (2) xy > 0
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Originally posted by kabilank87 on 23 Apr 2013, 01:23.
Last edited by Bunuel on 23 Apr 2013, 05:46, edited 1 time in total.
Renamed the topic and edited the question.



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Re: Inequalities  is x < y ? [#permalink]
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23 Apr 2013, 03:06
kabilank87 wrote: Is X < Y ?
1.2x < 3y 2. XY > 0 From F.S 1, for x=y=1, we have a NO, for x=1,y=2, we have a YES. Insufficient. From F.S 2, all we know that is Y and X are of the same sign. Insufficient. From both the statements together, use the same numbers used for F.S 1,again Insufficient. E.
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Re: Inequalities  is x < y ? [#permalink]
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23 Apr 2013, 03:42
Nothing to add here, just I wanna give you an other approach. Is X < Y ? is the red line, so the question now is : is "y" in the higher region defined by the red line? 1.2x < 3y this is the gree line \(y>\frac{2}{3}x\). x will be in the green area. Is this sufficient to say that x is also in the red area? As you see it's not. 2. XY > 0\((+,+) (,)\) are the combinations , so x and y will be in the I or in the III quadrant. Is this sufficient to say that x is in the red area? As you see it's not. 1+2 Since 1 and 2 give us no other info (the graph is key here), the answer is E. Hope you like my graphapproach, it helps me in many occasions
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Re: Inequalities  is x < y ? [#permalink]
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24 Apr 2013, 12:00
vinaymimani wrote: kabilank87 wrote: Is X < Y ?
1.2x < 3y 2. XY > 0 From F.S 1, for x=y=1, we have a NO, for x=1,y=2, we have a YES. Insufficient. From F.S 2, all we know that is Y and X are of the same sign. Insufficient. From both the statements together, use the same numbers used for F.S 1,again Insufficient. E. Hi Vinay , please explain F.S 1. i don't understand how you conclude F.S 1. Regards
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Re: Inequalities  is x < y ? [#permalink]
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24 Apr 2013, 12:19
kabilank87 wrote: vinaymimani wrote: kabilank87 wrote: Is X < Y ?
1.2x < 3y 2. XY > 0 From F.S 1, for x=y=1, we have a NO, for x=1,y=2, we have a YES. Insufficient. From F.S 2, all we know that is Y and X are of the same sign. Insufficient. From both the statements together, use the same numbers used for F.S 1,again Insufficient. E. Hi Vinay , please explain F.S 1. i don't understand how you conclude F.S 1. Regards The question is asking whether X<Y. The possible answer for this question can only be a YES/NO. Our job is to find out, whether the given Fact Statements are sufficient to answer the question stem with a YES or a NO. Thus, to conclusively say that a F.S is Suficient/Insufficient, we WILL have to get either a YES / NO;ALL THE TIME, for any value of the variable(s), which adhere to the given conditions in the Fact Statements/Problem/mathematical domain. For example, Is x>x^2 I. x>1 II.x is a nonnegative number From F.S 1, we know that x>1. Also, the question is asking whether x(1x)>0> x(x1)<0. Now, (x1) and x are both positive for the given F.S. Thus, for answering the Question Stem, we have a NO, which is a negative answer in the tradional sense, yet SUFFICIENT. Thus, armed with the knowledge that x>1[which is a fact statement], can we answer whether x>x^2[for all values of x, adhering to the giving fact]. As it turns out, we can and hence this statement is SUFFICIENt. From F.S 2, all we know about x is that it is not negative. Thus, x is greater than or equal to zero. Now, for x=0.5, we will get a YES for the question asked. However, for x=1, we will get a NO. The presence of a YES AND a NO renders this information to be INSUFFICIENT.So, this F.S is NOT Sufficient. Answer : A. Also, noteworthy point that the two F.S don't contradict each other.
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Re: Is x < y ? [#permalink]
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24 Apr 2013, 17:45
Is x < y ?
(1) 2x < 3y (2) xy > 0
1 just tells us that y>2/3x. example : lets say x=3, now y is greater that 2 . It can be 2.5 or 4. Insufficient.
2 tells us that both have same signs. hence Insufficient.
E wins.



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Re: Is x < y ? [#permalink]
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01 Oct 2013, 08:51
Hi Bunuel,
How would do this problem algebraically. Any suggestions?
Thanks



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Re: Is x < y ? [#permalink]
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02 Oct 2013, 02:32



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Re: Is x < y ? [#permalink]
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02 Oct 2013, 15:42
Thanks Bunuel. Do you have any link to effective number picking. I am not good at many sections in quant, however surely worse at number picking. If you can recall any such guidance, please provide  might be somewhere you have discussed.



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Re: Is x < y ? [#permalink]
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23 Jan 2016, 10:27
kabilank87 wrote: Is x < y ?
(1) 2x < 3y (2) xy > 0 (1) x < 1.5y, clearly not sufficient (for example x can be < 1,5y but =1,2y which is > y) (2) xy > 0, ok, same signs, gives us "zero" info to test x<y (1)+(2) we know that both a +ve and we are still by this statement x < 1.5y. So not sufficient. Answer E
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Re: Is x < y ? [#permalink]
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23 Jan 2016, 23:47
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) 2x < 3y (2) xy > 0 In the original condition, there are 2 variables(x,y), which should match with the number of equations. So you need 2 equations. For 1), 1 equation, for 2) 1 equation, which is likely to make C the answer. When 1) & 2), if x<0 and y<0, they become 2x<3y<2y. However, they also can be x>0 and y>0, which is not sufficient. Therefore, the answer is E. > For cases where we need 2 more equations, such as original conditions with “2 variables”, or“3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: Is x < y ? [#permalink]
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04 Feb 2017, 04:22
The question is asking you whether one thing is smaller (or bigger) than another. (1) tells you that 2 of one thing is smaller than 3 of another. That's not enough to tell whether one thing is smaller (bigger) than the other. Insufficient. (2) tells you that x and y share the same sign. Well, that tells you nothing about their relative magnitude; (2) is completely useless. Combining, we have insufficient (1) with useless (2) so in combo we still have no sufficiency. Choose (E).
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Re: Is x < y ? [#permalink]
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13 Jun 2018, 12:21
kabilank87 wrote: Is x < y ?
(1) 2x < 3y (2) xy > 0 Statement 1: In this type of question, I test whether the variables have any ability to change the inequality sign. How i test: (a) Take a same positive integer for both variables. [*]If the inequality holds true (normal case), the variables have no ability to change the inequality sign [all credit goes the given constants] in that situation, three scenarios are possible. (i) both variables are equal, (ii) first variable is greater than second variable, (iii) second variable is greater than first variable [*]If the inequality does not hold true (abnormal case), the variables are playing the main role to change the inequality sign [ the given constants have no ability to change the inequality sign]. in that situation, only a fixed role is possible [either first variable is greater than second variable or second variable is greater than first variable] as stated in the inequality. (b) Take a same negative integer for both variables. [*]If the inequality holds true (normal case), the variables have no ability to change the inequality sign [all credit goes the given constants] in that situation, three scenarios are possible. (i) both variables are equal, (ii) first variable is greater than second variable, (iii) second variable is greater than first variable [*]If the inequality does not hold true (abnormal case), the variables are playing the main role to change the inequality sign [ the given constants have no ability to change the inequality sign]. in that situation, only a fixed role is possible [either first variable is greater than second variable or second variable is greater than first variable] as stated in the inequality. back to the main question: Take a same positive integer [2] for both variables. 4<6. Here, the inequality holds true (normal case). i.e. the variables have no ability to change the inequality sign [all credit goes the given constants] in that situation, three scenarios are possible. (i) both variables are equal [x=y], (ii) first variable is greater than second variable (x>y), (iii) second variable is greater than first variable (x<Y). So statement 1 is insufficient. However, if you take a same negative integer [2] for both variables, [4<6] the inequality would not hold true (abnormal case), i.e. the variables are playing the main role to change the inequality sign [ the given constants have no ability to change the inequality sign]. in that situation, only a fixed role is possible [either first variable is greater than second variable or second variable is greater than first variable] as stated in the inequality [x<y]. Statement 2: xy>0. in other words x and y are both positive or both negative. i.e. (2,3) or (3,2) or (2,3) or (3,2). Thus, So statement 2 is insufficient. Combining: if x and y are positives, three scenarios are possible. (i) both variables are equal [x=y], (ii) first variable is greater than second variable (x>y), (iii) second variable is greater than first variable (x<Y). So, C is cancelled out. answer E. However, if x and y are negatives, x<y. Since statement 2 says both positive and both negatives are possible to exist. C is cancelled out.










