Bunuel
If \(\frac{x+y}{z}>0\), is x<0?
(1) x < y
(2) z < 0
Target question: Is x NEGATIVE? Given: (x+y)/z > 0What does tell us?
Not much.
It tells us that (x+y)/z is POSITIVE, which means EITHER (x+y) and z are both positive OR (x+y) and z are both negative
Statement 1: x < y This statement doesn't FEEL sufficient, so I'll TEST some values.
There are several values of x, y and z that satisfy statement 1 AND the given information. Here are two:
Case a: x = -2, y = -1, and z = -1. In this case
x is NEGATIVECase b: x = 1, y = 2, and z = 1. In this case
x is POSITIVESince we cannot answer the
target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: z < 0There are several values of x, y and z that satisfy statement 2 AND the given information. Here are two:
Case a: x = -2, y = -1, and z = -1. In this case
x is NEGATIVECase b: x = 1, y = -2, and z = -1. In this case
x is POSITIVESince we cannot answer the
target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined Statement 2 tells us that z is NEGATIVE
If z is NEGATIVE
and (x+y)/z > 0, then it must be the case that (x+y) is also NEGATIVE
In other words,
x + y < 0Statement 1 tells us that x < y
If we subtract y from both sides of the inequality, we get:
x - y < 0So, we now have the following two inequalities:
x + y < 0x - y < 0When we ADD the two inequalities, we get: 2x < 0
Divide both sides by 2 to get: x < 0
In other words,
x is NEGATIVESince we can answer the
target question with certainty, the combined statements are SUFFICIENT
Answer: C
Cheers,
Brent