Data Sufficiency

hi... I am back after a gap (courtesy my tiring office work), trying to check how am I doing..

Ques: Are X and Y both positive?
1) 2X - 2Y =1
2) X/Y >1

A. 1 alone is sufficient
B. 2 alone is sufficient
C. Together sufficient
D. Each sufficient
E. Togther not sufficient

I chose E but Answer says: C.

My reasoning:
X= 1/4, Y= -1/4, both satisfies 1), 2)
X= 1, Y=1/2, both satisfies 1), 2)

so, cant say if X and Y both +ve

appreciate your input

gmattarget700 wrote:hi... I am back after a gap (courtesy my tiring office work), trying to check how am I doing..

Ques: Are X and Y both positive?
1) 2X - 2Y =1
2) X/Y >1

A. 1 alone is sufficient
B. 2 alone is sufficient
C. Together sufficient
D. Each sufficient
E. Togther not sufficient

I chose E but Answer says: C.

My reasoning:
X= 1/4, Y= -1/4, both satisfies 1), 2)
X= 1, Y=1/2, both satisfies 1), 2)

so, cant say if X and Y both +ve

appreciate your input

Hi,

the only way a fraction can be positive is if both the numerator and the denominator share the same sign.

So, x=1/4 y=-1/4 doesn't satisfy statement (2)

(if you do the math (1/4)/(-1/4) = (1/4)(-4) = -1, which isn't > 1).

thanks Stuart, appreciate it...

I was overlooking this part of the hidden trick.... my thinking direction was following:
X/Y >1
=> 2) as X > Y

and 1/4 > -1/4 and so was my answer...

Now going with your hint, so now I know that 2) => X and Y are either both positive or both negative

Now, how 1) helping to take it further to deduce that both are positive or both negative? As I understand, 1 and 2 together has to prove that Both are either positive or both negative...

may be a simple one, just not able to focus much....

gmattarget700 wrote:hi... I am back after a gap (courtesy my tiring office work), trying to check how am I doing..

Ques: Are X and Y both positive?
1) 2X - 2Y =1
2) X/Y >1

OK.. let's break it down.

(1) dividing both sides of the equation by 2, we get:

x - y = 1/2

So, we could put x and y anywhere on the number line as long as x is .5 to the right of y; they could both be +, both be - or could straddle 0. Accordingly, (1) is insufficient.

(2) x/y > 1

We have to be very careful with inequalities and variables; remember, whenever you multiply or divide both sides of an inequality by a negative, you have to swap the inequality's direction. So, if we want to solve algebraically, we have to consider two cases:

(a) y > 0

If y > 0, and we multiply both sides by y, then we get:

x > y

and since x > y > 0, we get a "yes" answer to the original question.

(b) y < 0

if y < 0, and we multiply both sides by y, then we get:

x < y

and since x < y < 0, we get a "no" answer to the original question.

Accordingly, (2) is insufficient.

Each statement is insufficient alone, so let's combine and see what happens.

From (1), we know that x is .5 to the right of y on the number line; in other words, x > y.

From (2), we know that if y < 0, x will end up being less than y. Since this violates (1), it's no longer a possibility.

Therefore, we now know that y > 0, which means that x is also > 0, so we get a definite YES answer to the question.

Together sufficient, choose C.