Statement 1: a = -y = | x |
If a = | x |, then a is positive.
If -y = | x |, then y is negative.
x could be negative or positive.
If x is positive, then ax > y.
If x = -1 then ax = y
If x < -1 then ax < y
Insufficient.
Statement 2: a < 1
Tells us nothing about x or y.
Insufficient.
Statements Combined:
The combined statements indicate the following.
-1 < y < 0
0 < a < 1
-1 < x < 0 or 0 < x < 1
Case 1: a and x are both positive, ax > y
Case 2: a and x are both fractions and x is negative, then ax > y.
Sufficient.
The correct answer is C.
(Did you provide the correct OA? If so, if you give me the question #, I will report the error to BellCurves.)
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DS: absolute value
Source: Beat The GMAT — Data Sufficiency |
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Statement 1:manik11 wrote:If x ≠0, is ax > y?
(1) a = -y = | x |
(2) a < 1
Since |x| > 0, y must be a NEGATIVE VALUE.
Case 1: y=-1, implying that a=1 and x=±1
If a=1 and x=1, then ax = 1.
Here, ax > y.
If a=1 and x=-1, then ax = -1.
Here, ax = y.
INSUFFICIENT.
Statement 2:
No information about x or y.
INSUFFICIENT.
Statements combined:
To satisfy both statements, y must be a NEGATIVE FRACTION.
Case 2: y=-1/2, implying that a=1/2 and x= ±1/2.
If a=1/2 and x=1/2, then ax = 1/4.
If a=1/2 and x=-1/2, then ax = -1/4.
In each case, ax > y.
Implication of Case 2:
Since y must be a NEGATIVE FRACTION, ax must be either a positive fraction or a negative fraction greater than y.
Thus, ax > y.
SUFFICIENT.
The correct answer is C.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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Let's write the question as
Is ax - y > 0 ?
S1::
a = |x|, so a ≥ 0
-y = |x|, so 0 ≥ y
We still don't know the sign of x, so this isn't sufficient.
S2::
1 > a
We don't know anything about x and y, so this isn't sufficient.
S1+S2::
We know that 1 > a ≥ 0, that y = -a, and that 1 > x > -1. So the question becomes
Is ax > -a ?
Let's consider two cases: x > 0 and 0 > x.
If x > 0, then we know ax > 0 and 0 > -a, so yes, ax > -a.
If 0 > x, then we have two possibilities:
0 > ax > -a
0 > -a > ax
But if the second case is true, we have
-a > ax
Dividing by a (which is positive), we have
-1 > x
But this contradicts what we learned about x, so it is impossible. Hence we must have the other case, ax > -a, and in either case (x > 0 or 0 > x), we're left with only ax > -a, or ax > y.
Is ax - y > 0 ?
S1::
a = |x|, so a ≥ 0
-y = |x|, so 0 ≥ y
We still don't know the sign of x, so this isn't sufficient.
S2::
1 > a
We don't know anything about x and y, so this isn't sufficient.
S1+S2::
We know that 1 > a ≥ 0, that y = -a, and that 1 > x > -1. So the question becomes
Is ax > -a ?
Let's consider two cases: x > 0 and 0 > x.
If x > 0, then we know ax > 0 and 0 > -a, so yes, ax > -a.
If 0 > x, then we have two possibilities:
0 > ax > -a
0 > -a > ax
But if the second case is true, we have
-a > ax
Dividing by a (which is positive), we have
-1 > x
But this contradicts what we learned about x, so it is impossible. Hence we must have the other case, ax > -a, and in either case (x > 0 or 0 > x), we're left with only ax > -a, or ax > y.
