If a, b, x, and y are all positive, is a/b x/y ?

Question

If a, b, x, and y are all positive, is  \frac{a}{b} > \frac{x}{y}  ?

(1)  ay + 1 > bx

(2)  (\frac{ay}{bx})^2 > \frac{ay}{bx}

14101992 · Answer

As a,b,x and y are positive, it makes our life easy while playing with inequalities.

we have to find is : a/b>x/y ---> ay>bx ---> ay-bx > 0

S1.
ay+1>bx ---> ay-bx>-1

So, we cannot say whether ay-bx is always > 0 (for ex. ay-bx could be -0.5)
 
Insufficient!

S2.

(ay/bx)^2 > (ay/bx) ---> ay/bx > 1 ---> ay>bx ---> ay-bx>0

Sufficient.

Hence, answer should be B.

skothaka · Answer

rewrite the question as (ay/bx)>1

statement 1
> divide the entire inequality by bx
we get
(ay/bx) + (1/bx) > 1
that is (positive number) + (positive Number) >1. This does not say anything about (ay/bx) as it can be 0.5 or 1.5 and so on

This is not sufficient as we need to know whether (ay/bx)>1

Statement 2
(ay/bx)^2 > (ay/bx)
>(ay/bx)^2 - (ay/bx)>0
> (ay/bx) * ((ay/bx)-1) >0

this means that either both the boldface terms are greater than zero or both are less than zero.
as a,y,b and x are positive (ay/bx) cannot be less than zero. Hence both cannot be less than zero.
Now both are greater than zero. which means,
(ay/bx)> 0  and  (ay/bx) - 1 > 0

Remember that this is an AND condition. Both have to be true.

therefore ay/bx >1
statement 2 alone is sufficient.
So answer should be B

mohshu · Answer

simplify the given stem

ay>bx 
ay-bx >0 ????
stat 1: ay-bx > -1,,not suff

stat 2 : divide the entire equation by ax/by

gives ax/by >1

ax>by
 ax-by >0.. suff

ans B

14101992 · Answer

The solution provided for S1 is not complete. We cannot conclude anything with the equation (ay/bx) + (1/bx) > 1. See the above solution for a better explanation.

skothaka · Answer

The solution for S1 says that ay/bx could be anything between 0 and infinite. So S1 is not sufficient. That is enough to rule it out. Your solution is much simpler btw.

OindrillaDe · Answer

Bunuel  - Could you help me understand that if the statement did not mention that a,b,x and y are all positive - would the answer then be E

HWPO · Answer

I hope my reasoning is ok, but this is how I interpreted statement 2:  Bunuel  approve?

For POSITIVE fractions only: a positive fraction squared will ALWAYS be smaller than the fraction itself.

Example:  \frac{1}{2} > \frac{1}{4}  ,  \frac{1}{3} > \frac{1}{9}  , and so on.

IF, as in statement 2, our fractions squared is actually greater than our fraction itself, then the numerator had to be greater than the denominator to begin with.

Therefore, ay > bx ==== > Statement B is sufficient.

GMATCoachBen · Answer

HWPO : yes, that is solid reasoning and an interesting solution! It's great that you have a strong grasp of the important rule above.

My personal approach to statement (2) is to immediately cross off the exponent on the left side and put "1" on the right side, then multiply the "bx" over to the right side:

(\frac{ay}{bx})^2>\frac{ay}{bx} 
 (\frac{ay}{bx})>1 
 ay>bx

As others have noted above, we have to be careful with inequalities — we can only safely do this because the stem says all numbers are POSITIVE.

If a, b, x, and y are all positive, is a/b > x/y ?

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