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

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Are x and y both positive?


(1) \(2x-2y = 1\)

(2) \(\frac{x}{y} > 1\)
[Reveal] Spoiler: OA

Last edited by Bunuel on 16 Aug 2017, 21:14, edited 3 times in total.
Edited the question and added the OA

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

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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.

Answer: C.

Discussed here: ds1-93964.html?hilit=number%20plugging%20consider%20approaches and also here along with other hard inequality problems: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope it helps.
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Re: Are x and y both positive? (1) 2x-2y = 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=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.
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Manbehindthecurtain wrote:
Are x and Y both positive?

1) 2X-2Y = 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 x-y will yield a positive result.

Please correct me if I am wrong
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Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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zerotoinfinite2006 wrote:
Manbehindthecurtain wrote:
Are x and Y both positive?

1) 2X-2Y = 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 x-y 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 "x-y 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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Bunuel wrote:
zerotoinfinite2006 wrote:
Manbehindthecurtain wrote:
Are x and Y both positive?

1) 2X-2Y = 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 x-y 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 "x-y 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 :roll: , including these kind of question. And day by day I am getting demoralize that I can't solve these kind of questions. :cry:

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) 2x-2y = 1 (2) x/y > 1 [#permalink]

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zerotoinfinite2006 wrote:
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 :roll: , including these kind of question. And day by day I am getting demoralize that I can't solve these kind of questions. :cry:

Anyways, Thanks a lot for your explanation Bunuel. You are genius as always.
+1 more .


Check Number Theory chapter of Math Book for more on number properties (link in my signature).
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Re: Are x and y both positive? (1) 2x-2y = 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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i did same u subhashghosh
(1) 2x-2y=1
x-y= 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) 2x-2y = 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 = 1/2
This 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) 2x-2y = 1 (2) x/y > 1 [#permalink]

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New post 17 Mar 2012, 18:54
Well, already proved by so many members in different ways, I will just share mine.

Clearly 1) 2x-2y =1, does not say much except simplifying it to x-y = 1/2 ,
2) x/y>1 , simplifying it to x>y => x-y>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, x-y>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) 2x-2y = 1 (2) x/y > 1 [#permalink]

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New post 18 Mar 2012, 07:31
Bunuel wrote:
zerotoinfinite2006 wrote:
Manbehindthecurtain wrote:
Are x and Y both positive?

1) 2X-2Y = 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 x-y 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 "x-y 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 \(y(x-y)>0\)
This is a general method.

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

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jach2012 wrote:
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 \(y(x-y)>0\)
This is a general method.


More usual way of doing this would be: \(\frac{x}{y}>1\) --> \(\frac{x}{y}-1>\) --> \(\frac{x-y}{y}>0\).
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Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

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New post 09 May 2012, 10:15
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.

Answer: C.

Discussed here: ds1-93964.html?hilit=number%20plugging%20consider%20approaches and also here along with other hard inequality problems: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope 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 2x-2y = 1.
Now let us take x=-1 and y=-3/2 and again plugging this value we can satisfy the equation 2x-2y = 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) 2x-2y = 1 (2) x/y > 1 [#permalink]

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subhajeet wrote:
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.

Answer: C.

Discussed here: ds1-93964.html?hilit=number%20plugging%20consider%20approaches and also here along with other hard inequality problems: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope 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 2x-2y = 1.
Now let us take x=-1 and y=-3/2 and again plugging this value we can satisfy the equation 2x-2y = 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) 2x-2y = 1 (2) x/y > 1 [#permalink]

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New post 12 May 2012, 01:36
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Statement (1): x-y = 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) 2x-2y = 1 (2) x/y > 1 [#permalink]

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New post 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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RohanNanda wrote:
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?


x = -1/2 and y = -1 do not satisfy the second statement: x/y=(-1/2)/(-1)=1/2<1.
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Re: Are x and y both positive? (1) 2x-2y = 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) 2x-2y = 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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Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1   [#permalink] 18 Aug 2012, 21:01

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