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Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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Are x and y both positive? (1) \(2x2y = 1\) (2) \(\frac{x}{y} > 1\)
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Last edited by Bunuel on 16 Aug 2017, 21:14, edited 3 times in total.
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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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.
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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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.
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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
I guessed and got it right with a 50/50 guess at the end. What I have done here is this 1) 2x  2y = 1 hence x  y = \frac{1}{2} {Dividing both side by 2} In sufficient 2) [fraction]x > y[/fraction] Alone in sufficient When (1) + (2) We can say that if X is greater than y than xy will yield a positive result. Please correct me if 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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zerotoinfinite2006 wrote: Manbehindthecurtain wrote: Are x and Y both positive?
1) 2X2Y = 1 2) (x/y) > 1
I guessed and got it right with a 50/50 guess at the end. What I have done here is this 1) 2x  2y = 1 hence x  y = \frac{1}{2} {Dividing both side by 2} In sufficient 2) x > y Alone in sufficient When (1) + (2) We can say that if X is greater than y than xy will yield a positive result.Please correct me if I am wrong First of all: the question is "are x and Y both positive?" not whether "xy will yield a positive result". Next, the red part is not correct. \(\frac{x}{y}>1\) does not mean that \(x>y\). If both \(x\) and \(y\) are positive, then \(x>y\), BUT if both are negative, then \(x<y\). What you are actually doing when writing \(x>y\) from \(\frac{x}{y}>1\) is multiplying both parts of inequality by \(y\): never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.So from (2) \(\frac{x}{y}>1\), we can only deduce that \(x\) and \(y\) have the same sigh (either both positive or both negative). See the complete solution of this problem in my previous post. Hope it helps.
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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: zerotoinfinite2006 wrote: Manbehindthecurtain wrote: Are x and Y both positive?
1) 2X2Y = 1 2) (x/y) > 1
I guessed and got it right with a 50/50 guess at the end. What I have done here is this 1) 2x  2y = 1 hence x  y = \frac{1}{2} {Dividing both side by 2} In sufficient 2) x > y Alone in sufficient When (1) + (2) We can say that if X is greater than y than xy will yield a positive result.Please correct me if I am wrong First of all: the question is "are x and Y both positive?" not whether "xy will yield a positive result". Next, the red part is not correct. \(\frac{x}{y}>1\) does not mean that \(x>y\). If both \(x\) and \(y\) are positive, then \(x>y\), BUT if both are negative, then \(x<y\). What you are actually doing when writing \(x>y\) from \(\frac{x}{y}>1\) is multiplying both parts of inequality by \(y\): never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.So from (2) \(\frac{x}{y}>1\), we can only deduce that \(x\) and \(y\) have the same sigh (either both positive or both negative). See the complete solution of this problem in my previous post. Hope it helps. I can clearly see how much weak I am in DS . I have no idea how to improve it. I am extremely weak in number system , including these kind of question. And day by day I am getting demoralize that I can't solve these kind of questions. Anyways, Thanks a lot for your explanation Bunuel. You are genius as always. +1 more .
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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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1 is not suff, x = 0, y = 1/2 2 is not suff,x and y can be both ve Combining both : x  y = 1/2 and (x  y)/y > 0 so 1/2/y > 0 => y is +ve and because x  y is +ve, x is +ve as well. So 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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i did same u subhashghosh (1) 2x2y=1 xy= 1/2 so y could be +ve or ve insuff. (2) x/y>1 here x and y both could be +ve or ve. so insuff. Considering C from (1) x is positive so from (2) y must be positive. Ans. C.
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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C is the answer. Question: Is x > 0 AND y > 0?Statement 1: 2x  2y = 1 => 2(x  y) = 1 => x  y = 1/2This just tells us that the difference is positive. But this can be true for cases when both x and y are positive, and when both x and y are negative. For instance, x = 1.5, y = 1 => x  y = 0.5; also, x = 1, y = 1.5 => x  y = 0.5. Thus, INSUFFICIENT. Statement 2: x/y > 1 This just tells us that x and y have the same sign. That is, both are positive or both are negative. INSUFFICIENT. Combining these statements, we can use the same numbers used in Statement 1 to find out that both the cases together do not work for negative numbers. For instance, x = 1, y = 1.5 => x  y = 0.5. However, x/y < 1. This violates statement 2. Thus, the combination of the given statements tells us that x and y both have to be positive. => x > 0 AND y > 0. SUFFICIENT.
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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17 Mar 2012, 18:54
Well, already proved by so many members in different ways, I will just share mine. Clearly 1) 2x2y =1, does not say much except simplifying it to xy = 1/2 , 2) x/y>1 , simplifying it to x>y => xy>0 I start with 2nd , x>y => X, Y positive or x,y negative. x/y +/ (Not an option since x/y>1) not I pick back 1st, xy>1/2, if x is ve, the y > x and +ve, but from 2nd we know that is not true. x>y. so, we know x is +ve. Now, if x is +ve, then y cannot be ve because x/y is +ve. So x,y are +ve and hence we need both statements to answer it.
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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18 Mar 2012, 07:31
Bunuel wrote: zerotoinfinite2006 wrote: Manbehindthecurtain wrote: Are x and Y both positive?
1) 2X2Y = 1 2) (x/y) > 1
I guessed and got it right with a 50/50 guess at the end. What I have done here is this 1) 2x  2y = 1 hence x  y = \frac{1}{2} {Dividing both side by 2} In sufficient 2) x > y Alone in sufficient When (1) + (2) We can say that if X is greater than y than xy will yield a positive result.Please correct me if I am wrong First of all: the question is "are x and Y both positive?" not whether "xy will yield a positive result". Next, the red part is not correct. \(\frac{x}{y}>1\) does not mean that \(x>y\). If both \(x\) and \(y\) are positive, then \(x>y\), BUT if both are negative, then \(x<y\). What you are actually doing when writing \(x>y\) from \(\frac{x}{y}>1\) is multiplying both parts of inequality by \(y\): never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.So from (2) \(\frac{x}{y}>1\), we can only deduce that \(x\) and \(y\) have the same sigh (either both positive or both negative). See the complete solution of this problem in my previous post. Hope it helps. Just to add we can multiply y to both numerator and denominator of x/y the advantage is that the denominator becomes a square i.e in this case \(y^2\) so now we can safely cross multiply in \(xy/y^2>1\) since square of a no. is always +ve \(xy>y^2\) or \(y(xy)>0\) This is a general method.



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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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09 May 2012, 10:15
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. Bunnel: I was trying to solve this question by plugging in numbers. I too agree that staements A and B both alone are insufficient. SO now by taking both the statements together x>y so let us take x=3/2 and y=1 and plugging this value we can satisfy the equation 2x2y = 1. Now let us take x=1 and y=3/2 and again plugging this value we can satisfy the equation 2x2y = 1. So the answer must be E. Please correct me where I am going wrong.



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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subhajeet 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. Bunnel: I was trying to solve this question by plugging in numbers. I too agree that staements A and B both alone are insufficient. SO now by taking both the statements together x>y so let us take x=3/2 and y=1 and plugging this value we can satisfy the equation 2x2y = 1. Now let us take x=1 and y=3/2 and again plugging this value we can satisfy the equation 2x2y = 1. So the answer must be E. Please correct me where I am going wrong. x=1 and y=3/2 don't satisfy the second statement: x/y=(1)/(3/2)=2/3 <1.
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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Statement (1): xy = 1/2. We can have x=1,y=1/2. Can also have x=0,y=1/2. Insufficient. Statement (2): x/y>1. We can have x=3,y=2. Can also have x=3,y=2. Insufficient. Combining both, (y+1/2)/y > 1 => 1/2y>0 => y>0 Also as x/y>1, x must be>0. Sufficient. C it is.
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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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02 Aug 2012, 09:10
Hi Bunuel please help me out...
I used the numbers x = 1, y = 1/2 and x = 1/2 and y = 1 while combining both the statements and both these sets satisfy, hence I get an E. Is this wrong? If so, why?



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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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I got it correct but took approx 2.5 minutes.
stmnt 1 : insufficient by plugging numbers stmnt 2 : x/y >1 => not suff as bot x and y can be ve or both +ve.
combined :
from stmnt 1 we have
x=y+1/2 => x/y = 1+1/2y => suppose x/y is 2 as x/y>1 => 2=1+1/2y = y=1/2 henc x=1
both positive.
Ans C
Am I right in my approach?



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Re: Are x and y both positive? (1) 2x2y = 1 (2) x/y > 1 [#permalink]
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Bunuel... For sure, this problem can be easily solved using algebra, just as you have explained before. However, I was trying to use geometry to solve this problem. The equation of the line is x/0.5 + y/0.5= 1, which means that the x and y intercepts are 0.5 and 0.5 respectively. Hence, the line passes through 1st, 3rd, and 4th quadrants.
In Q1, both x and y are +ve. In Q4, y is negative and hence this is out. Where I am getting confused is Q3: how do I verify whether x/y >1?




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