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integers

by naaga » Sat Feb 21, 2009 12:45 am
. If a,b, and c are integers, is a – b + c greater than a + b – c ?
(1) b is negative.
(2) c is positive.

OA is C

folks plz explain how the given answer satisfies regardless of the signs of individual integers, thanks in advance
Last edited by naaga on Mon Feb 23, 2009 3:50 am, edited 1 time in total.
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Re: integers

by Vemuri » Sat Feb 21, 2009 2:02 am
Well, I am getting C as the answer. Here's how:

1) b --> -ve; a & c can be +ve or -ve (i.e the various combinations can be a,c or -a,c or -a,-c or a,-c). When the combination -a,-c is used, the inequality looks like -a+b-c>-a-b+c ==> b-c>c-b (Insuff)

2) c --> +ve; a & b can ve +ve or -ve (i.e the various combinations can be a,b or -a,b or -a,-b or a,-b). When the combination -a,b is used, the inequality looks like -a-b+c>-a+b-c ==> c-b>b-c (Insuff)

Combing both the statements, we know b --> -ve & c --> +ve, so a can be either +ve or -ve. So, the various combinations possible are a,-b,c & -a,-b,c. With the combination a,-b,c the inequality looks like a+b+c>a-b-c ==> b+c > -b-c (True). With the combination -a,-b,c the inequality looks like -a+b+c>-a-b-c ==> b+c > -b-c (True).

So, combining both the statements we can say that a-b+c>a+b-c. So, according to me the answer should be C

Can someone explain why the OA is B?
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by Bidisha_800 » Sat Feb 21, 2009 8:20 pm
we have to prove a-b+c >a+b-c
or 2c > 2b
c > b

if b is negative c is positive c > b

(C)
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by cramya » Sun Feb 22, 2009 5:43 am
Agree with C

Proving its not b

Stmt II
c = 1
b=1000
a=0

a+c-b < a+b-c

a=0
b=-1000
c=1

a+c-b > a+b-c

Hence INSUFF or simply we have no idea about signs of a and b and their magnitude
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rep

by naaga » Mon Feb 23, 2009 3:54 am
friends, I am sorry , I rechecked the answer after the discussion , it is C only not B, I think I was confused while posting the answer, please apologize my mistake. I never repeat this in future posts.

Thanks friends,


Regarding,
Naag.
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