Are x and y both positive?


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

(2) \(\frac{x}{y} > 1\)

Are x and y both positive?


(1) \(2x-2y=1\). Well this one is clearly insufficient. You can do it with number plugging OR consider the following: both x and y are positive means that point (x,y) is in the I quadrant. \(2x-2y=1\) --> \(y=x-\frac{1}{2}\). We know it's an equation of a line and basically the question asks whether this line (all (x,y) points of this line) is only in I quadrant. This line for sure passes I quadrant (for example, x = 1.5 and y = 1) but it cannot entirely be only in I quadrant, so there must be some (x, y) points whose coordinates are not both positive. Not sufficient.


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

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

From (2): \(\frac{x}{y}>1\);

Substitute x into (2): \(\frac{y + \frac{1}{2}}{y}>1\);

\(1 + \frac{1}{2y}>1\);

\(\frac{1}{2y}>0\);

\(y > 0\). \(y\) is positive. Thus, \(x=y+\frac{1}{2}=positive+positive=positive\). So, \(x\) is positive too. Sufficient.


Answer: C.


You can also check GRAPHIC APPROACH below.
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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?

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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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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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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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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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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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

Thus, SUFFICIENT that x and y are both positive.

Answer: C
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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.
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