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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) x/y>1 [#permalink] New post 10 May 2010, 08:31
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Are x and y both positive?

(1) 2x-2y=1
(2) x/y>1
[Reveal] Spoiler: OA

Last edited by Bunuel on 28 Apr 2012, 17:17, edited 1 time in total.
Edited the question and added the OA
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Re: DS1 [#permalink] New post 10 May 2010, 11:47
Opt- A)
2x-2y = 1
x-y = 0.5
let x= 5 , -5
y= 4.5 , -4.5

5-4.5 = 0.5 = +ve
-5-(-4.5)=-0.5 = -ve

So we can't tell with option A) -

Opt - B)
x/y > 1
+ve/+ve > 1 -
-ve/-ve > 1
So we can't tell with this option also independently.

Let us combine both,
Since x/y > 1, x>y when both x&y are +ve.
if x= -7 & y= -6.5 = - 0.5 which is -ve

So we can't tell with the help of A & B also..
Both A& B are bot sufficient to find the answer
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Re: DS1 [#permalink] New post 10 May 2010, 13:07
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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.
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Re: DS1 [#permalink] New post 11 May 2010, 03:40
I'm still not clear why X and Y has to be positive when X/Y > 1. Can you please explain the way you combined taking both X and Y to be positive and also X and Y as negative. Since in either case X/Y will be > 1.

Thanks
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Re: DS1 [#permalink] New post 11 May 2010, 03:59
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harikattamudi wrote:
I'm still not clear why X and Y has to be positive when X/Y > 1. Can you please explain the way you combined taking both X and Y to be positive and also X and Y as negative. Since in either case X/Y will be > 1.

Thanks
-H


From (2) \frac{x}{y}>1, we can only deduce that x and y have the same sigh (either both positive or both negative).

When we consider two statement together:

From (1): 2x-2y=1 --> x=y+\frac{1}{2}

From (2): \frac{x}{y}>1 --> \frac{x}{y}-1>0 --> \frac{x-y}{y}>0 --> substitute x from (1) --> \frac{y+\frac{1}{2}-y}{y}>0--> \frac{1}{2y}>0 (we can drop 2 as it won't affect anything here and write as I wrote \frac{1}{y}>0, but basically it's the same) --> \frac{1}{2y}>0 means y is positive, and from (2) we know that if y is positive x must also be positive.

OR: as y is positive and as from (1) x=y+\frac{1}{2}, x=positive+\frac{1}{2}=positive, hence x is positive too.

Hope it helps.
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Algebra DS. [#permalink] New post 12 Jun 2010, 08:44
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Guys,

Another question and here is my answer. Can someone tell why its not correct?

Are x and y both positive?

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

I have gone for E, neither is sufficient. But its an incorrect answer and the correct answer is C.

My explanation for E

Question is indirectly asking whether x and y are both greater than zero?

Cosidering choice 1.

Let say x=1 and y=1/2 --> Then yes both x and y are +ve. But if x=-1/2 and y=-1 then both are negative. So this answer choice is insufficient.

Consdering answer choice 2

x > y

if x=2 , y=1 --> Both are positive.

if x=-2 and y=-3 then both at negative. Therefore, this answer choice is insufficient.

Cosidering both options:

x>y and 2x-2y=1 we can still have both positive and negative answers and therefore I have gone for E.

Can someone explain where I have gone wrong and what's the best way and numbers to pick for these types of questions?

Thanks again for all your valuable help.
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Re: DS1 [#permalink] New post 12 Jun 2010, 09:07
I am still struggling to understand how come both together are sufficient? What is common in both the answer choices that makes c a correct choice?

:?
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Re: DS1 [#permalink] New post 12 Jun 2010, 09:15
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gsaxena26 wrote:
I am still struggling to understand how come both together are sufficient? What is common in both the answer choices that makes c a correct choice?

:?


Here is the logic for C:

When we consider two statement together:

From (1): 2x-2y=1 --> x=y+\frac{1}{2}

From (2): \frac{x}{y}>1 --> \frac{x}{y}-1>0 --> \frac{x-y}{y}>0 --> substitute x from (1) --> \frac{y+\frac{1}{2}-y}{y}>0--> \frac{1}{2y}>0 (we can drop 2 as it won't affect anything here and write as I wrote \frac{1}{y}>0, but basically it's the same) --> \frac{1}{2y}>0 means y is positive, and from (2) we know that if y is positive x must also be positive.

OR: as y is positive and as from (1) x=y+\frac{1}{2}, x=positive+\frac{1}{2}=positive, hence x is positive too.

Does it make sense now?
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Re: DS1 [#permalink] New post 12 Jun 2010, 09:33
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Thanks Bunuel. I got it why the answer is C and here is my explanation:

Considering choice 1.

Let say x=1 and y=1/2 --> Then yes both x and y are +ve. But if x=-1/2 and y=-1 then both are negative. So this answer choice is insufficient.

Considering choice 2.

x/y>1

But x can be both +ve and negative. But if x>y then answer choice will be true as it can only be positive when x >y. Is my understanding correct?
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Re: DS1 [#permalink] New post 13 Jun 2010, 10:41
gsaxena26 wrote:
x/y>1

But x can be both +ve and negative. But if x>y then answer choice will be true as it can only be positive when x >y. Is my understanding correct?


Your reasoning is almost correct.
Statement II doesn't say x>y, it only says x/y>1... Be careful there is a difference, this is the mistake you made initially

Taking this statement alone, it means x and y are both same sign (either both + or both -) and |x|>|y|. Thus, it is insufficient and you need to combine it with Statement I.

From statement I, you know x=y+1/2, hence x>y. If you know x>y and also |x|>|y| this means both are positive.
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Re: DS1 [#permalink] New post 24 Jun 2010, 20:03
Here is my confusion.
Here is how I approached the question
1. 2x-2y=1
so x-y=.5

now x=1, y=.5
or x=1/4, y=-1/4
so can't tell

2. x/y>1
x>y so again can't tell.

Now if we combine both
still the options x=1, x=.5 is true
and so is the option x=1/4, y=-1/4 true

So can't tell hence E. I know this is not the correct answer but what am I missing?
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Re: DS1 [#permalink] New post 25 Jun 2010, 03:18
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sam2010 wrote:
Here is my confusion.
Here is how I approached the question
1. 2x-2y=1
so x-y=.5

now x=1, y=.5
or x=1/4, y=-1/4
so can't tell

2. x/y>1
x>y so again can't tell.


Now if we combine both
still the options x=1, x=.5 is true
and so is the option x=1/4, y=-1/4 true

So can't tell hence E. I know this is not the correct answer but what am I missing?


Problem with your solution is that 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.

From (2) \frac{x}{y}>1, we can only deduce that x and y have the same sigh (either both positive or both negative).
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Re: DS1 [#permalink] New post 25 Jun 2010, 03:45
got it. Thanks :)
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Re: DS1 [#permalink] New post 05 Sep 2010, 11:35
Excellent explanation Bunuel. Its damn clear to me now.
Many thanks.
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Re: DS1 [#permalink] New post 07 Sep 2010, 13:15
Bunuel,

From the 1/y>0 case what happens when y=0?, how goes GMAT treat that case? I notice that the question is asking if they are positive, not nonnegative, this case would have accounted for 0?
Thanks
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.


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Re: DS1 [#permalink] New post 07 Sep 2010, 14:11
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mainhoon wrote:
Bunuel,

From the 1/y>0 case what happens when y=0?, how goes GMAT treat that case? I notice that the question is asking if they are positive, not nonnegative, this case would have accounted for 0?
Thanks
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.


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\frac{x}{y}>1 means that y\neq{0} (\frac{1}{y} means y>0). Also as division by zero is undefined then in cases when denominator could be zero GMAT would most likely state that denominator does not not equal to zero.
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Re: DS1 [#permalink] New post 15 Oct 2010, 10:04
Bunuel...great explanation..
I tht of doing it geometrically but missed something and came to E.
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Re: DS1 [#permalink] New post 17 Oct 2010, 00:56
Statement 1: Apparently not sufficient.
Statement 2: x,y either both +ve or both negative. Not sufficient.

Using 2:

\frac{x}{y} > 1 , \frac{(x-y)}{y} > 0

=> x>y and y> 0 OR y<0 and y<x

using 1; we get x-y = 1/2>0 => x>y

Hence y>0 using 2

Since x>y and y>0 => x>y>0 Hence both positive.

Hence C.
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Re: DS1 [#permalink] New post 18 Jun 2011, 10:17
LM wrote:
Please explain....


(1) x-y = 1/2
so 1/4 - (-1/4) = 1/2 [y could be -ve]
again, 1 - 1/2 = 1/2 [y could be +ve]
Insufficient.
(2) This option refers both x and y are -ves or +ves. insufficient.

For 1+2:
x/y>1 and in option 1 x is positive so y is also positive.

Ans. C
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Re: DS1 [#permalink] New post 23 Jun 2011, 03:21
a+b

x,y can be either + or -. checking for - values where |x| > |y|.
thus x,y both positive.

C it is.
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Re: DS1   [#permalink] 23 Jun 2011, 03:21
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