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01 Jul 2019, 08:00
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D
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Difficulty:
85%
(hard)
Question Stats:
46% (02:03) correct
54%
(02:01)
wrong
based on 735
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If \(cb < ab < 0\), is \(|c - a| = |c| - |a|\)?
(1) \(c > a\)
(2) \(b < 0\)
(1) \(c > a\)
(2) \(b < 0\)
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01 Jul 2019, 10:36
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cb<ab<0 -- given
so now 2 cases:
1. b>0
2. b<0
taking 1 : b>0:
cb-ab<0 ---> b(c-a)<0
b>0 so c-a should be <0. So c<a
also since ab<0 and b>0 so a<0. Similarly c<0
taking 2 : b<0
here c-a>0, so c>a
also since b<0 and ab<0, so a>0 and similarly c>0
To check if |c-a|=|c|-|a|
lets take statement 1:
c>a
means case 2 from above. This case gives a,c>0 and c>a. So for all values |c-a|=|c|-|a| is true. 1 is sufficient.
statement 2:
b<0:
again case 2. SO sufficient:
Hence Ans D
so now 2 cases:
1. b>0
2. b<0
taking 1 : b>0:
cb-ab<0 ---> b(c-a)<0
b>0 so c-a should be <0. So c<a
also since ab<0 and b>0 so a<0. Similarly c<0
taking 2 : b<0
here c-a>0, so c>a
also since b<0 and ab<0, so a>0 and similarly c>0
To check if |c-a|=|c|-|a|
lets take statement 1:
c>a
means case 2 from above. This case gives a,c>0 and c>a. So for all values |c-a|=|c|-|a| is true. 1 is sufficient.
statement 2:
b<0:
again case 2. SO sufficient:
Hence Ans D
01 Jul 2019, 08:34
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It's given that cb<ab<0.
So both cb and ab are negative.
For this to be possible, there are only two scenarios.
C<a and b>0
Or
C>a and b< 0.
In the question, both the statements imply the same and point to scenario 2.
Which implies c>a and that implies the given statement will always never be true.
|C-a| is always positive whereas |c| - |a| will equal -(c-a).
Therefore both statements alone are sufficient and hence option D.
Posted from my mobile device
So both cb and ab are negative.
For this to be possible, there are only two scenarios.
C<a and b>0
Or
C>a and b< 0.
In the question, both the statements imply the same and point to scenario 2.
Which implies c>a and that implies the given statement will always never be true.
|C-a| is always positive whereas |c| - |a| will equal -(c-a).
Therefore both statements alone are sufficient and hence option D.
Posted from my mobile device
