OG Diagnostice DS 41

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OG Diagnostice DS 41

by navalpike » Tue Jul 14, 2009 11:52 am

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If (x +y)/ z > 0, is x < 0?

1. x < y
2. z < 0

Looking for different ways to go about this one. My solution is taking far too long.

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Source: — Data Sufficiency |

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by yogami » Tue Jul 14, 2009 12:03 pm

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(1) says x < y. but this is insuff to say that x < 0 cos plug in few numbers if x = 4 y is 5 and z is positive then the condition is satisified. If x is -2 and y is -1 and z is negative then too the condition is satisfied
Hence INSUFF
(2) z < 0. WHich means that x + y has to be less than 0 but again pick x = 1 and y = -5. Here x > 0 aNd yet x + y < 0. Or pick x = -3 and y = -7. again x + y <0 but x < 0 so this is insuff

Combine
z < 0 and x < y means that x + y has to be < 0 and x has to be less than y
This means that x has to be negative cos if x is 0 or positive then y has to be more than x and x + y is not < 0
So combining is suff
hence (C)
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by Brent@GMATPrepNow » Tue Sep 17, 2019 8:19 am

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navalpike wrote:If (x +y)/ z > 0, is x < 0?

1. x < y
2. z < 0
Target question: Is x NEGATIVE?

Given: (x+y)/z > 0
What 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 NEGATIVE
Case b: x = 1, y = 2, and z = 1. In this case x is POSITIVE
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: z < 0
There 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 NEGATIVE
Case b: x = 1, y = -2, and z = -1. In this case x is POSITIVE
Since 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 < 0

Statement 1 tells us that x < y
If we subtract y from both sides of the inequality, we get: x - y < 0

So, we now have the following two inequalities:
x + y < 0
x - y < 0

When we ADD the two inequalities, we get: 2x < 0
Divide both sides by 2 to get: x < 0
In other words, x is NEGATIVE
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

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Brent
Brent Hanneson - Creator of GMATPrepNow.com
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