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Is |x|+x>|x|? 1) x<y^2 2) x+y^2<0 10 Mar 2019, 09:52
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

35% (medium)

Question Stats:

62% (01:12) correct 38% (01:23) wrong based on 168 sessions

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Is \(|x|+x>|x|\)?
1) \(x<y^2\)
2) \(x+y^2<0\)
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B
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m1033512
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Is |x|+x>|x|? 1) x<y^2 2) x+y^2<0 10 Mar 2019, 10:02
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statement 1

x is always positive so the the answer is yes .

statement 2 .

x is negative , then

the answer is always No .



IMO D .

each one is sufficient

Posted from my mobile device
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Is |x|+x>|x|? 1) x<y^2 2) x+y^2<0 10 Mar 2019, 10:06
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Is \(|x|+x>|x|\)?
Now |x| is not negative, so subtract it from both sides without changing the inequality sign..=> the question becomes ' Is x>0?'
1) \(x<y^2\)
x can be both negative and positive.. say y=2 and x = -1, or y=2 and x=1...Insufficient
2) \(x+y^2<0\)
\(y^2\) will be 0 at the least so x<0 OR \(x+y^2<0....x<-y^2...x<0\)
The answer is NO...Sufficient
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Is |x|+x>|x|? 1) x<y^2 2) x+y^2<0 10 Mar 2019, 10:06
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m1033512
statement 1

x is always positive so the the answer is yes .

statement 2 .

x is negative , then

the answer is always No .



IMO D .

each one is sufficient

Posted from my mobile device
Show more

Check statement 1....
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Is |x|+x>|x|? 1) x<y^2 2) x+y^2<0 10 Mar 2019, 10:13
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chetan2u
m1033512
statement 1

x is always positive so the the answer is yes .

statement 2 .

x is negative , then

the answer is always No .



IMO D .

each one is sufficient

Posted from my mobile device

Check statement 1....
Show more

Thanks Chetan ,

x < y^2 , so x can be negative also,

I wrongly assumed x can not be negative
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Is |x|+x>|x|? 1) x<y^2 2) x+y^2<0 10 Mar 2019, 10:19
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chetan2u
Is \(|x|+x>|x|\)?
Now |x| is not negative, so subtract it from both sides without changing the inequality sign..=> the question becomes ' Is x>0?'
1) \(x<y^2\)
x can be both negative and positive.. say y=2 and x = -1, or y=2 and x=1...Insufficient
2) \(x+y^2<0\)
\(y^2\) will be 0 at the least so x<0 OR \(x+y^2<0....x<-y^2...x<0\)
The answer is NO...Sufficient
Show more

In Statement 1, x could be zero as well.
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Is |x|+x>|x|? 1) x<y^2 2) x+y^2<0 20 Nov 2021, 02:30
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