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18 Jun 2007, 19:37
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Is xy > x/y?
(1) xy > 0
(2) y < 0
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
OA - 'E'
(1) xy > 0
(2) y < 0
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
OA - 'E'
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18 Jun 2007, 19:49
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1) any two positive numbers or any two negative numbers will give you >0. You can make X=2 and Y=1 and it doesnt work.
Since:
2 x 1 = 2 / 1
2=2 which doesnt meet the criteria of greater than
2) just use the same numbers as above (y=-1 which is <0) but make both negative and you get the same result.
The solution for 2) also proves that together they dont work since -2 x -1 = 2 which meets the requirements for 1) but is also is equal to -2 / -1 and not greater than.
Since:
2 x 1 = 2 / 1
2=2 which doesnt meet the criteria of greater than
2) just use the same numbers as above (y=-1 which is <0) but make both negative and you get the same result.
The solution for 2) also proves that together they dont work since -2 x -1 = 2 which meets the requirements for 1) but is also is equal to -2 / -1 and not greater than.
18 Jun 2007, 23:19
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(E) for me 
xy > x/y?
x*y - x/y > 0 ?
x*(y-1/y) > 0 ?
x*((y^2-1)/y) > 0 ?
x*(y+1)*(y-1)/y > 0 ?
From 1
xy > 0
sgn(x) = sgn(y)
Then,
sign(x*(y+1)*(y-1)/y)
= sign(x)*sign(y+1)*sign(y-1)/sign(y)
= sign(y+1)*sign(y-1)
So,
o If y > 0 then, y+1 > 1 > 0 and y-1 > -1... sign(x*(y+1)*(y-1)/y) = sign(y-1) which could be positive or negative
o If y < 0 then, y+1 < 1 and y-1 < -1 < 0... sign(x*(y+1)*(y-1)/y) = sign(y+1) which could be positive or negative
INSUFF.
From 2
y < 0
We have no clue about x.
INSUFF.
Both 1 & 2
We are in the second case of stat 1 : y < 0
Implying that:
sign(x*(y+1)*(y-1)/y) = sign(y+1) which could be positive or negative
INSUFF.
xy > x/y?
x*y - x/y > 0 ?
x*(y-1/y) > 0 ?
x*((y^2-1)/y) > 0 ?
x*(y+1)*(y-1)/y > 0 ?
From 1
xy > 0
sgn(x) = sgn(y)
Then,
sign(x*(y+1)*(y-1)/y)
= sign(x)*sign(y+1)*sign(y-1)/sign(y)
= sign(y+1)*sign(y-1)
So,
o If y > 0 then, y+1 > 1 > 0 and y-1 > -1... sign(x*(y+1)*(y-1)/y) = sign(y-1) which could be positive or negative
o If y < 0 then, y+1 < 1 and y-1 < -1 < 0... sign(x*(y+1)*(y-1)/y) = sign(y+1) which could be positive or negative
INSUFF.
From 2
y < 0
We have no clue about x.
INSUFF.
Both 1 & 2
We are in the second case of stat 1 : y < 0
Implying that:
sign(x*(y+1)*(y-1)/y) = sign(y+1) which could be positive or negative
INSUFF.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
