If xy 0, is a y/x?

Question

If xy ≠ 0, is a > y/x?

(1) a = (1/x + 1/y)/(1/y)

(2) x and y are positive integers

aj13783 · Answer

xy <> 0 and is a > y/x?

1.) a = (1/x + 1/y)/(1/y)

a = ((y+x)/xy)*y

a = (y + x)/x

a = y/x + 1 > y/x

2.) X & Y are positive. Nothing substantial can be concluded from this.

Thus Option A

SavageBrother · Answer

If xy ≠ 0, is a > y/x?

(1) a = (1/x + 1/y)/(1/y)

(2) x and y are positive integers

Nice one.

1) A = (1/x + 1/y) / (1/y) 
(1/x) / (1/y) + (1/y) / (1/y) = (1/x) / (1/y) + 1
(1/x) / (1/y) = y / x, 
a = y / x + 1

S.

2) I.

A.

Bowtye · Answer

If xy ≠ 0, is a > y/x?

(1) a = (1/x + 1/y)/(1/y)

(2) x and y are positive integers

stmt 1: a = ((x+y)/xy)*y
 -> (x+y)/x 
-> y/x + x/x 
-> y/x +1=a >y/x

suff

stmt 2:nothing about a insuff

answer should be A

shriramvelamuri · Answer

A for me too. A tough one though.

Divyadisha · Answer

Is answer C?

We have to prove if ax>y
with 1st statement we get ax=y+x; However we do not know if X and Y are +ve or -ve.

With second statement we only know that both are +ve.
combining both statements we will arrive at the solution ax>y
Please correct me if I am wrong.

Lucky2783 · Answer

No divyadisha, we cannot multipky x to a the way you did (ax>y ). 
for eg :

4 > 6/(-2) ----> does this mean 4*(-2)>6? No. 
unless, you know the sign of x and y you cannot multiply/divide them with a.

mariannetremblay · Answer

If xy≠0 , is a>y/x ?

(1) a= 1/x + 1/y
1/y

(2) x and y are positive integers

The OS is A (statement 1 is sufficient alone).

Veritas uses algebraic manipulation, but I tried plugging in, and I thought it was insufficient.

here is what I did:

step 1: use plug-ins x=2 and y=4 in the equation given in statement 1 to find a:
(1/2 + 1/4 ) / (1/4) = ( 3/4 ) / (1/4) = 12/4 = 3 --> so a = 3

ans y/x = 4/2 = 2

a> (y/x) ? NO

2< 3, so (y/x)< a

Now let's try to get a YES:

use x=4 and y=2 this time. Plug those in to get a = [(1/4) + (1/2)] / (1/2) = (3/4) / (1/2) = 6/4 = 3/2 --> a = 3/2 or 1.5

y/x = 2/4 = 1/2 (or 0.5)

a> (y/x) ? YES 1.5> 0.5

Since I get a NO and a YES answer by using different values for x and y, this statement should be insufficient. What am I not seeing? Where did I mess up?

Thanks

yezz · Answer

From 1

A = y/x +1

Thus question becomes

Is y/x+1-y/x >0 , always true

2 insuff

A

Sent from my iPhone using  GMAT Club Forum mobile app

udaypratapsingh99 · Answer

1)) a = (1/x + 1/y)/(1/y)
 = (1/x + 1/y)×y
 = (1/x)×y + (1/y)×y
 = y/x + 1

=> a= y/x + 1 > y/x

Therefore A is correct

Posted from my mobile device

vivek4141 · Answer

Very short answer. 
As xy≠0, it means both x and y are non-zero (but they could be any other number, either fraction, integer).

(1) Given a = (1/x + 1/y)/(1/y)--> a = y/x +1. It means "a is 'something' (here y/x) plus 1". So, we dont need to check any sign or nature of y/x. Clearly, a > y/x.

(2) Given: x and y are positive integers. We don't even know their relation with a. No need to even check what y/x could be. So, Answer: A.

If xy 0, is a > y/x?

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