Forum Home > GMAT > Quantitative > Problem Solving (PS)
Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Dec 1511:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease.
Dec 1608:00 AM PST
-11:59 PM PST
GMAT Club 12 Days of Christmas is a 4th Annual GMAT Club Winter Competition based on solving questions. This is the Winter GMAT competition on GMAT Club with an amazing opportunity to win over $40,000 worth of prizes!- Dec 17
09:00 AM PST
-10:00 AM PST
Join Manhattan Prep instructor Whitney Garner for a fun—and thorough—review of logic-based (non-math) problems, with a particular emphasis on Data Sufficiency and Two-Parts. - Dec 18
10:00 AM PST
-11:00 AM PST
Here is the essential guide to securing scholarships as an MBA student! In this video, we explore the various types of scholarships available, including need-based and merit-based options. - Dec 21
11:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment
24 Sep 2015, 02:08
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
52% (01:44) correct
48%
(01:52)
wrong
based on 151
sessions
History
Date
Time
Result
Not Attempted Yet
01 Oct 2015, 21:07
Kudos
Bookmarks
rajarshee
Is x+y>0 where x and y are non zero integers?
(1) |x|>y
(2) |y|>x
(1) |x|>y
(2) |y|>x
Question: Is x + y > 0
When will the sum of x and y be positive?
When either both x and y are positive or only one of them is positive with higher absolute value than the other one (which could be 0 or negative).
In other cases, x+y will be negative or 0.
(1) |x|>y
This could hold when both x and y are positive (x = 4, y = 2) or when both x and y are negative (x = -4, y = -2) or other cases.
Hence this statement alone is not sufficient.
(2) |y|>x
This statement is symmetrical to statement 2 with just the variables reversed. IF statement 1 is not sufficient, statement 2 will also not be sufficient.
Using both,
|x| > y
|y| > x
Can both x and y be positive? No, because then only one's absolute value will be greater than the other.
Is it possible that one of them is positive with higher absolute value than the other one? No. Say the numbers are x = 5 and y = -3.
Here |y| cannot be greater than x.
Both the conditions are not met. So, we can say for sure that x+y is not greater than 0
Answer (C)
General Discussion
HardWorkBeatsAll
Joined: 17 Aug 2015
Last visit: 19 Jul 2020
Posts: 90
Own Kudos:
Given Kudos: 341
Location: India
Concentration: Strategy, General Management
Schools: Duke '19 (II)
GMAT 1: 750 Q49 V42
GPA: 4
WE:Information Technology (Finance: Investment Banking)
24 Sep 2015, 04:38
Kudos
Bookmarks
Nice question. Thanks for posting. +1 kudos to you! Looks like a 700+ question to me.
(1) |x| > y
So, |x|-y > 0
The modulus can be opened in two ways:
So, x - y >0 ... (a1)
OR
-x - y > 0 ... (b1)
(b1) can be converted as x+y<0 and is sufficient.
However, consider what happens if we take (a1). From (a1), x could be 4 and y could be 2 (x+y > 0) OR x could be -2 and y could be -4 (in which case x+y < 0). Thus, (a1) is not sufficient, although (b1) is sufficient. So, overall, (1) alone is not sufficient as we don't know which of (a1) and (b1) will come into effect.
(2) |y| > x
Opening the modulus,
y > x
so, y-x > 0 ... (a2)
OR
-y > x
so, -x -y > 0
or x+y < 0 ...(b2)
Again, (b2) is sufficient, but (a2) is not (since (a2) is true in both cases viz. x=2, y=4 (in which case x+y is > 0) as well as x = -4 and y = -2 (in which case x+y<0)). So, we can't determine if x+y>0 uniquely from it. Therefore, overall, (2) is not sufficient.
Now, consider both (1) and (2) together.
For both (1) and (2) to hold, we cannot have these two together:
(a1) x-y>0
(a2) y-x>0
That means, the only options left are (b1) and (b2) which give us the same result viz. x+y < 0. This is sufficient to determine whether x+y is > 0. So, both (1) and (2) together are sufficient.
Hence, (C). BOTH TOGETHER ARE SUFFICIENT.
(1) |x| > y
So, |x|-y > 0
The modulus can be opened in two ways:
So, x - y >0 ... (a1)
OR
-x - y > 0 ... (b1)
(b1) can be converted as x+y<0 and is sufficient.
However, consider what happens if we take (a1). From (a1), x could be 4 and y could be 2 (x+y > 0) OR x could be -2 and y could be -4 (in which case x+y < 0). Thus, (a1) is not sufficient, although (b1) is sufficient. So, overall, (1) alone is not sufficient as we don't know which of (a1) and (b1) will come into effect.
(2) |y| > x
Opening the modulus,
y > x
so, y-x > 0 ... (a2)
OR
-y > x
so, -x -y > 0
or x+y < 0 ...(b2)
Again, (b2) is sufficient, but (a2) is not (since (a2) is true in both cases viz. x=2, y=4 (in which case x+y is > 0) as well as x = -4 and y = -2 (in which case x+y<0)). So, we can't determine if x+y>0 uniquely from it. Therefore, overall, (2) is not sufficient.
Now, consider both (1) and (2) together.
For both (1) and (2) to hold, we cannot have these two together:
(a1) x-y>0
(a2) y-x>0
That means, the only options left are (b1) and (b2) which give us the same result viz. x+y < 0. This is sufficient to determine whether x+y is > 0. So, both (1) and (2) together are sufficient.
Hence, (C). BOTH TOGETHER ARE SUFFICIENT.
rajarshee
Is x+y>0 where x and y are non zero integers?
(1) |x|>y
(2) |y|>x
(1) |x|>y
(2) |y|>x
