Carcass wrote:
\(|x| > |y|\) and \(x + y > 0\)
Quantity A |
Quantity B |
y |
x |
The approach used by @kaziselim is great.
However, if you missed that, here's a different approach:
GIVEN: |x| > |y|
This inequality tells us that the MAGNITUDE of x
is greater than the MAGNITUDE of y
In other words, on the number line, the distance from x to 0
is greater than the distance from y to 0
This gives us 4 possible cases:
case i: ----x----y----0------ (x and y are both negative)
case ii: ----x--------0--y---- (x is negative and y is positive)
case iii: ---y--0-------x-- (y is negative and x is positive)
case iv: --0---y----x-- (x and y are both positive)
Notice that, for all 4 cases, I've made the distance from x to 0
greater than the distance from y to 0
Now examine the second piece of given information...
GIVEN: x + y > 0
When we examine the 4 cases above,
only cases iii and iv satisfy the condition that x + y > 0
In case iii, x > y
In case iv, x > y
So, it MUST be the case that x > y
Answer: B
Cheers,
Brent
_________________
Brent Hanneson – Creator of greenlighttestprep.com
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