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Is xy>ab ?

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Is xy>ab ? [#permalink]

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New post 06 Feb 2015, 06:16
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A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

58% (01:35) correct 42% (01:20) wrong based on 160 sessions

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Re: Is xy>ab ? [#permalink]

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New post 09 Feb 2015, 02:55
Hey,

I would say D.

[1] says that \(\frac{b}{x}\) > \(\frac{y}{a}\)
If we cross multiply we get that ab>xy. So the answer is No, so it is sufficient.

[2] gives the same relationship we found for [1], but squared. I think it means the same, because if I remember correctly, we only take the positive value of a radical in GMAT.

So, it is also sufficient.

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Re: Is xy>ab ? [#permalink]

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New post 09 Feb 2015, 04:49
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Bunuel wrote:
Is xy > ab ?

(1) b/x > y/a
(2) (ab)^2 > (xy)^2

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Solution: E

Statement (1) tells us nothing about whether x or a are positive, so we aren’t able to get a definitive answer by multiplying both sides of the equation by x and a, respectively, as we don’t know whether to flip the sign; INSUFFICIENT. Statement (2) is likewise INSUFFICIENT because we still don’t know whether any of the variables are negative. Combined, we still don’t know about what’s positive and what’s negative, so xy could be negative and ab positive or vice versa; (E).
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Re: Is xy>ab ? [#permalink]

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New post 27 May 2017, 15:08
Is xy > ab ?

(1) b/x > y/a
(2) (ab)^2 > (xy)^2

Kudos for a correct solution.

1) b/x > y/a
variable involved and we dont know the sign . therefore cant multiply . not sufficient.

2)(ab)^2 > (xy)^2
since both side are raised to the power 2 .
both ab and xy , can be positive or negative .

(ab)^2 > (xy)^2
let ab =2 and xy = 1
then (ab)^2 > (xy)^2 satisfied
but
Is xy > ab ? No



let ab = -1 and xy =1/2
then (ab)^2 > (xy)^2 satisfied
but
Is xy > ab ? yes

st .2 is not sufficient

together both 1 and 2 not sufficient . Since dont give any information that is helpful to solve the question
Hence E


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Re: Is xy>ab ? [#permalink]

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New post 31 Aug 2017, 06:46
Bunuel wrote:
Is xy > ab ?

(1) b/x > y/a
(2) (ab)^2 > (xy)^2

Kudos for a correct solution.


--------------------
Is xy>ab

1) b/x > y/a : b=2 ,a=1, y=1, x =1 yields 2>1 so we can use these values ----> xy=1 ab =2 xy<ab
now use b=2,a=-1,x=1,y=1 yields 2>-1 so we can use these values-----------> xy=1 ab =-2 xy>ab

A and D eliminated

2) (ab)^2 > (xy)^2 : same values as above same set as above in one case take a=1 and other a=-1
using same values as statement 1) where a=1 we get xy>ab
using same values as statement 1) where a=-1 we get xy<ab

B eliminated

combine 1&2

use same values as statement 1 with a = +1 get xy<ab
use same values as statement 1 with a = -1 get xy >ab

Insufficient C eliminated


E is the Answer
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Re: Is xy>ab ? [#permalink]

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New post 31 Aug 2017, 07:47
Bunuel wrote:
Is xy > ab ?

(1) b/x > y/a
(2) (ab)^2 > (xy)^2

Kudos for a correct solution.



(2) (ab)^2 > (xy)^2

It means |ab| > |xy|......We do not know if ab > xy. For example

ab = -6 & xy =-5...........Answer to question is Yes

ab = 6 & xy =-5...........Answer to question is No

Insufficient

(1) b/x > y/a

If x>0 & a>0.......ab > xy ...........Answer is No (Note: it is the same as x<0 & a<0)

If x>0 & a<0.......ab < xy ...........Answer is Yes (Note: it is the same as x<0 & a>0)

Insufficient

combine 1 & 2

Take same examples above...No clear cut

Insufficient

Answer: E

Kudos [?]: 302 [0], given: 166

Re: Is xy>ab ?   [#permalink] 31 Aug 2017, 07:47
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