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17 Nov 2017, 09:29
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
21% (02:34) correct
79%
(02:02)
wrong
based on 154
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If x and y are integers, is \(\frac{|xy|}{y}\geq{0}\) ?
(1) \(y^3<|y|\)
(2) \(|xy|>|y|\)
self made - a tricky one
(1) \(y^3<|y|\)
(2) \(|xy|>|y|\)
self made - a tricky one
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
17 Nov 2017, 11:20
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We know that |xy| will always be >=0, so the expression |xy|/y will be >=0 if y > 0 and it will be < 0 if y < 0 (we cannot take y=0 as that would make the question undefined).
(1) y^3 < |y|
This expression will be true for ALL negative values of y. Thats because |y| will be positive but y^3 will be negative.
Now this will also be true for 0<y<1. If y lies between 0 and 1, y^3 will be less than y or |y|. BUT the catch here is that we are given y is an integer. So it cannot lie between 0 and 1.
So from this statement we can safely say that y can only be negative. So |xy|/y will be negative. We got NO as our answer. Sufficient.
(2) |xy| > |y|
LHS can also be written as |x|*|y|. So |x|*|y| > |y| or |x| > |y|/|y| or |x| > 1. So either x > 1 or x < -1.
But this doesnt tell us anything about whether y is positive or negative. Insufficient.
Hence A answer.
(1) y^3 < |y|
This expression will be true for ALL negative values of y. Thats because |y| will be positive but y^3 will be negative.
Now this will also be true for 0<y<1. If y lies between 0 and 1, y^3 will be less than y or |y|. BUT the catch here is that we are given y is an integer. So it cannot lie between 0 and 1.
So from this statement we can safely say that y can only be negative. So |xy|/y will be negative. We got NO as our answer. Sufficient.
(2) |xy| > |y|
LHS can also be written as |x|*|y|. So |x|*|y| > |y| or |x| > |y|/|y| or |x| > 1. So either x > 1 or x < -1.
But this doesnt tell us anything about whether y is positive or negative. Insufficient.
Hence A answer.
18 Nov 2017, 20:22
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chetan2u
Hi amanvermagmat,
your thinking has been correct but you have neglected the role of second variable x..
lets solve it..
\(\frac{|xy|}{y}\geq{0}\)
the numerator will be 0 if x is 0 OR it will be POSITIVE as the value is in MOD
(1) \(y^3<|y|\)
since y is an integer, y will be a negative integer..
BUT we have x too..
if x is 0, \(\frac{|xy|}{y}={0}\), and ans is YES
If x is an integer, \(\frac{|xy|}{y}<{0}\)... ans is NO
insuff
(2) \(|xy|>|y|\)
square both sides...
\((xy)^2>y^2......y^2(x^2-1)>0\)
although it does not give anything to solve but tells us that both x and y are non zero
insuff
combined-
y is negative and x is non zero
so \(\frac{|xy|}{y}\geq{0}\) is always NO.
C
