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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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24 Aug 2012, 12:30
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Are x and y both positive?
1) 2x2y=1 2(xy)=1 xy=1/2 >3/41/4=1/2....YES >1/4(3/4)=1/2...NO INSUFFICIENT
2) x/y>1 This just means that x and y have the same sign. They're either both positive or both negative. INSUFFICIENT
1&2) x=1/2+y
(1/2+y)/y>1 y/2 + 1 > 1 y/2 > 0 which means that Y is greater than 0. And since both x and y have the same sign, both x and y are Positive. YES.
Answer is C.



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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Bunuel wrote: Are x and y both positive?(1) 2x2y=1 (2) x/y>1 (1) 2x2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x2y=1 > y=x1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient. (2) x/y>1 > x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient. (1)+(2) Again it can be done with different approaches. You should just find the one which is the less timeconsuming and comfortable for you personally. One of the approaches: \(2x2y=1\) > \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) > \(\frac{xy}{y}>0\) > substitute x > \(\frac{1}{y}>0\) > \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient. Answer: C. Discussed here: ds193964.html?hilit=number%20plugging%20consider%20approaches and also here along with other hard inequality problems: inequalityandabsolutevaluequestionsfrommycollection86939.htmlHope it helps. Bunuel i would like tto know how \(\frac{1}{y}>0\) have this : if I have ( y + 1  y / 2 ) / y > 0 the result should be \(\frac{1}{2y}>0\) and not \(\frac{1}{y}> 0\) can you please explain ??' thanks @edited ............I have seen the explanation in another answer by you ) Ok
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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17 Jan 2013, 03:55
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Manbehindthecurtain wrote: Are x and y both positive?
(1) 2x2y = 1 (2) x/y > 1 1. xy = 1/2 This means that the distance between x and y is 1/2 unit and that x is greater than y. But x and y could be positive such as x=5 and y=4.5, OR x and y could be both negative such as x=4 and y=4.5 INSUFFICIENT. 2. x/y > 1 This shows that x and y must be positive meaning they are either both (+) or both (). ex) x/y = 5/2 OR x/y = 5/2 = 5/2 still > 1 INSUFFICIENT. Combine. Let x = 5 and y=9/2: 5/(9/2) = 10/9 > 1  This means when x and y are both positive it could be a solution to x/y > 1 Let x = 4 and y=9/2: 4/(9/2) = 8/9 < 1  This means when x and y are negative it could not be a solution to x/y > 1 Thus, SUFFICIENT that x and y are both positive. Answer: C
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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chris558 wrote: Are x and y both positive?
1) 2x2y=1 2(xy)=1 xy=1/2 >3/41/4=1/2....YES >1/4(3/4)=1/2...NO INSUFFICIENT
2) x/y>1 This just means that x and y have the same sign. They're either both positive or both negative. INSUFFICIENT
1&2) x=1/2+y
(1/2+y)/y>1 y/2 + 1 > 1 y/2 > 0 which means that Y is greater than 0. And since both x and y have the same sign, both x and y are Positive. YES.
Answer is C. Shouldn't (1/2+y)/y>1 simplify to (1/2y) + 1 > 1 ? Or am I missing something? Still get the right answer following this logic but I believe this step is off.



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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11 Apr 2013, 09:06
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Bunuel wrote: Are x and y both positive?(1) 2x2y=1 (2) x/y>1 (1) 2x2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x2y=1 > y=x1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient. (2) x/y>1 > x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient. (1)+(2) Again it can be done with different approaches. You should just find the one which is the less timeconsuming and comfortable for you personally. One of the approaches: \(2x2y=1\) > \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) > \(\frac{xy}{y}>0\) > substitute x > \(\frac{1}{y}>0\) > \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient. Answer: C. Discussed here: ds193964.html?hilit=number%20plugging%20consider%20approaches and also here along with other hard inequality problems: inequalityandabsolutevaluequestionsfrommycollection86939.htmlHope it helps. From 1 X=Y+1/2. Divide both sides by Y you get X/Y=1+1/2Y > 1+1/2Y>1 > 1/2Y>0 then Y>0. Then consequently X>0. Is the reasoning sound?



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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score780 wrote: Bunuel wrote: Are x and y both positive?(1) 2x2y=1 (2) x/y>1 (1) 2x2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x2y=1 > y=x1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient. (2) x/y>1 > x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient. (1)+(2) Again it can be done with different approaches. You should just find the one which is the less timeconsuming and comfortable for you personally. One of the approaches: \(2x2y=1\) > \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) > \(\frac{xy}{y}>0\) > substitute x > \(\frac{1}{y}>0\) > \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient. Answer: C. Discussed here: ds193964.html?hilit=number%20plugging%20consider%20approaches and also here along with other hard inequality problems: inequalityandabsolutevaluequestionsfrommycollection86939.htmlHope it helps. From 1 X=Y+1/2. Divide both sides by Y you get X/Y=1+1/2Y > 1+1/2Y>1 > 1/2Y>0 then Y>0. Then consequently X>0. Is the reasoning sound? Yes, your approach is correct.
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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14 Sep 2013, 21:23
Hello Bunuel, Request you to please provide your comments on the doubt posted here Usually, whenever I see combining an inequality and equation, I substitute the value of one of the variable in the inequality and then analyze the effect. So, going by that approach; xy=1/2 (1) x/y>1 (2) Substituting the value of x in equation(2) (y+1/2)/y>1 Lets assume that y is positive (y+1/2) > y 1/2>0 This means that our assumption is true since 1/2 is greater than Zero. Hence, y > 0 Now, Lets assume that y is negative Now, here I'm stuck, I know that multiplying by a negative number changes the sign of the inequality. I'm sure that the sign will be changed but what would be the resulting equation. I mean, do we need to replace y with "y" in the whole equation. Please clarify. Which of the following would be correct then a) y+1/2 <y b) y+1/2 < y c) y+1/2 < y Please help. Thanks
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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imhimanshu wrote: Hello Bunuel, Request you to please provide your comments on the doubt posted here
Usually, whenever I see combining an inequality and equation, I substitute the value of one of the variable in the inequality and then analyze the effect. So, going by that approach;
xy=1/2 (1) x/y>1 (2) Substituting the value of x in equation(2)
(y+1/2)/y>1
Lets assume that y is positive
(y+1/2) > y
1/2>0 This means that our assumption is true since 1/2 is greater than Zero. Hence, y > 0
Now, Lets assume that y is negative
Now, here I'm stuck, I know that multiplying by a negative number changes the sign of the inequality. I'm sure that the sign will be changed but what would be the resulting equation. I mean, do we need to replace y with "y" in the whole equation. Please clarify. Which of the following would be correct then
a) y+1/2 <y b) y+1/2 < y c) y+1/2 < y
Please help. Thanks Refer to the highlighted portion : Actually you don't have to take 2 cases at this point: The expression you have is : \(\frac{y+0.5}{y}>1 \to 1+\frac{0.5}{y}>1 \to \frac{1}{y}>0\)> Hence, y>0. As for your doubt, if y is negative, we crossmultiply it and get : \(y+0.5<y \to 0>0.5\), which is absurd. If y is negative, then y would be positive, and for multiplying a positive quantity, you don't need to flip signs. So , yes expression a is correct.
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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19 Nov 2013, 10:29
Hi I get confused in this question. I understand A,B,D are not answer bu confuse in C and E.However,official answer is C.
My approach 1) x=y+(1/2) Not sufficient 2) x/y>1 Not sufficient 1+2) x=y+(1/2) So plugging in a value of y which makes x>y by statement 2 . So If,y=2.5 which gives x=2 then No If, y=2 x=2.5 then Yes So answer is E. Please correct me where I am wrong.



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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St 1) 2x2y = 1 => 2 (xy) = 1 => xy =1/2 => all this tells us is that x > y (could be positive or negative) == hence INSUFF
St 2) x/y > 1 => we don't know if y is (+) or () . So we have two cases:
if y positive, then x>y; if y negative, then x<y (again INSUFF)
Combining 1) and 2) we get x>y (from 1) ...which means y is positive (from 2)
Hence, if y is positive, and x >y, then x is also positive. SUFF!!
Hope this was reasoned properly.



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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21 Dec 2013, 23:27
Bunuel wrote: Are x and y both positive?
\(2x2y=1\) > \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) > \(\frac{xy}{y}>0\) > substitute x > \(\frac{1}{y}>0\) > \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.
Hope it helps. Sorry for the bump but could you elaborate on the last part where you go from x/y>1 to (xy)/y>0 to 1/y>0 ..? I don't quite follow this algebra



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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Manbehindthecurtain wrote: Are x and y both positive?
(1) 2x2y = 1 (2) x/y > 1 Plug in approach that can be used without thinking much and very likely arrive at the correct answer. Values to be taken: x positive and negative and find the corresponding values for y based on the statements Note: x and y cannot be of different signs and also x cannot be zero as they will not satisfy (ii) (i) x=10, we have y =9.5 .Both positive satisfied And now x=10, we have y=9.5. Both negative also satisfied .Different results. So (i) alone not sufficient (ii) x=10, y can be positive. Both positive satisfied . And now x=10, y can be negative. Both negative also satisfied. So (ii) alone not sufficient (i) + (ii) x=10, y=9.5 satisfies both the statements . Both positive satisfied . And now x=10. Value of y is found from (i) and is negative , but we see it does not satisfy (ii). So both cannot be negative . So we can answer the question using (i) and (ii) together
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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29 Jan 2014, 21:28
Is it safe to solve this kind of questions based on logic?
I didn't jump into calculations/plugins, since statement (1) is clearly insufficient. And statement (2) states that x & y both have the same sign, so combining them together, the result of subtraction is a positive number, and given from (2) that they have the same sign, then they both must be positive.



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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09 Jun 2015, 22:22
Bunuel wrote: Are x and y both positive?(1) 2x2y=1 (2) x/y>1 (1) 2x2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x2y=1 > y=x1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient. (2) x/y>1 > x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient. (1)+(2) Again it can be done with different approaches. You should just find the one which is the less timeconsuming and comfortable for you personally. One of the approaches: \(2x2y=1\) > \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) > \(\frac{xy}{y}>0\) > substitute x > \(\frac{1}{y}>0\) > \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient. Answer: C. Discussed here: ds193964.html?hilit=number%20plugging%20consider%20approaches and also here along with other hard inequality problems: inequalityandabsolutevaluequestionsfrommycollection86939.htmlHope it helps. Question: Whenever I graph it, and get to statement two that says "\(\frac{x}{y}> 1\)" I can make \(x > y\) and \(x < y\) depending on if they are both positive or both negative. Is there some connection to absolute values here? If they are both positive, say: \(\frac{10}{5}\) Then 10 > 5. But if x = 10 and y = 5. 5 > 10. So, without absolue values, they can be in either the first quadrant or third quadrant



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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14 Jun 2015, 15:46
Bunuel, what if x=3/8 and y=1/8
It satisfies, xy=1/2> 3/8+1/8=4/8>1/2 And also satisfies x>y>3/8>1/8,
Could someone tell me what I am doing wrong??
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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14 Jun 2015, 22:06
fcojcruz wrote: Bunuel, what if x=3/8 and y=1/8
It satisfies, xy=1/2> 3/8+1/8=4/8>1/2 And also satisfies x>y>3/8>1/8,
Could someone tell me what I am doing wrong??
best regards, Hello fcojcruzHere is mistake: "And also satisfies x>y>3/8>1/8" if \(x=\frac{3}{8}\) and \(y = \frac{1}{8}\) than inequality \(\frac{x}{y}> 1\) is wrong \(\frac{3}{8}\) divided by \(\frac{1}{8}\) can't be bigger than \(1\) if \(y < 0\) than you should change sign of inequality then you multiply or divide inequality on \(y\) so if \(x>y\)>\(\frac{3}{8}>\frac{1}{8}\) than \(\frac{x}{y} < 1\) and this enaqulity contradicts to second statement
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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23 Jul 2015, 03:40
dauntingmcgee wrote: I found this one easiest to solve by drawing a graph. Clearly 1) and 2) alone are not sufficient as discussed, so what remains to be seen is if 2) adds enough information to 1) to determine if both x and y are positive.
Drawing a quick graph of the line y=x1/2 we find that the xintercept of the line is (0.5,0) and the yintercept is (0,0.5). From this graph we can clearly see that we don't need to worry about anything in the 4th quadrant (+x/y is not >1)or the 3rd quadrant (x<y, therefore x/y is not >1). All that is left is the 1st quadrant, in which x and y are both positive.
Sufficient. i did not understand the highlighted portion. why is that we dont have to worry about 3 rd qdrt. the line passes thr it. pls help to understand..?



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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23 Jul 2015, 04:35
riyazgilani wrote: dauntingmcgee wrote: I found this one easiest to solve by drawing a graph. Clearly 1) and 2) alone are not sufficient as discussed, so what remains to be seen is if 2) adds enough information to 1) to determine if both x and y are positive.
Drawing a quick graph of the line y=x1/2 we find that the xintercept of the line is (0.5,0) and the yintercept is (0,0.5). From this graph we can clearly see that we don't need to worry about anything in the 4th quadrant (+x/y is not >1)or the 3rd quadrant (x<y, therefore x/y is not >1). All that is left is the 1st quadrant, in which x and y are both positive.
Sufficient. i did not understand the highlighted portion. why is that we dont have to worry about 3 rd qdrt. the line passes thr it. pls help to understand..? What this means is that when we combine the statements for the 3rd quadrant with y<0 , x= y+0.5 , we get x/y <1 (goes against statement 2) . You can see that , when both x,y <0, x < y and this will give you x/y < 1 Consider 2 cases: y = 2.5 , x = y+0.5 = 2, but x/y < 1 (goes against statement 2) or y = 0.5, x = 0.5+0.5 = 0 , but x/y = 0/0.5 = 0 < 1 (goes against statement 2) Thus 3rd quadrant values are not allowed/possible. Hope this helps.




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