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Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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02 Oct 2012, 19:42

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Bunuel wrote:

Are x and y both positive? (1) 2x-2y=1 (2) x/y>1

(1) 2x-2y=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. 2x-2y=1 --> y=x-1/2, 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 time-consuming and comfortable for you personally.

One of the approaches: \(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

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
_________________

Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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17 Jan 2013, 04:55

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Manbehindthecurtain wrote:

Are x and y both positive?

(1) 2x-2y = 1 (2) x/y > 1

1. x-y = 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

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.

Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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11 Apr 2013, 10:06

Bunuel wrote:

Are x and y both positive? (1) 2x-2y=1 (2) x/y>1

(1) 2x-2y=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. 2x-2y=1 --> y=x-1/2, 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 time-consuming and comfortable for you personally.

One of the approaches: \(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

(1) 2x-2y=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. 2x-2y=1 --> y=x-1/2, 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 time-consuming and comfortable for you personally.

One of the approaches: \(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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14 Sep 2013, 22:23

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;

x-y=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

Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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14 Sep 2013, 22:39

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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;

x-y=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 cross-multiply 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.
_________________

Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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19 Nov 2013, 11:29

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.

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.

x=-2 and y=-2.5 do not satisfy x/y>1.
_________________

Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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22 Dec 2013, 00:27

Bunuel wrote:

Are x and y both positive?

\(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> 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 (x-y)/y>0 to 1/y>0 ..?

\(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> 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 (x-y)/y>0 to 1/y>0 ..?

I don't quite follow this algebra

\(\frac{x}{y}>1\) --> \(\frac{x}{y}-1>\) --> \(\frac{x-y}{y}>0\). Now, substitute \(x=y+\frac{1}{2}\) there to get \(\frac{1}{2y}>0\), which further simplifies to \(\frac{1}{y}>0\).

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
_________________

Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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29 Jan 2014, 22:28

Is it safe to solve this kind of questions based on logic?

I didn't jump into calculations/plug-ins, 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.

Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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09 Jun 2015, 23:22

Bunuel wrote:

Are x and y both positive? (1) 2x-2y=1 (2) x/y>1

(1) 2x-2y=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. 2x-2y=1 --> y=x-1/2, 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 time-consuming and comfortable for you personally.

One of the approaches: \(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

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

Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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14 Jun 2015, 23:06

fcojcruz wrote:

Bunuel, what if x=3/8 and y=-1/8

It satisfies, x-y=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 fcojcruz Here 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
_________________

Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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23 Jul 2015, 04:40

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=x-1/2 we find that the x-intercept of the line is (0.5,0) and the y-intercept 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..?

Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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23 Jul 2015, 05:35

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=x-1/2 we find that the x-intercept of the line is (0.5,0) and the y-intercept 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.

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