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First of all as \(|a|=b\) (\(b\) equals to absolute value of some number) then \(b\geq{0}\), as absolute value is always non-negative.

(1) \(a=-b\) (\(b=-a\)) --> so \(a\leq{0}\) and \(LHS=a+b=0\). But \(RHS=ab\leq{0}\) thus we can not say for sure that \(a+b>ab\), because if \(a=b=0\) then \(a+b=0=ab\) (so in case \(a=b=0\), \(a+b\) is not more than \(ab\) it equals to \(ab\)). Not sufficient.

Put that value in the equation a+b >ab then (-b) +b > (-b) (b) i.e. 0> -b^2 As b^2 is always +ve this equaliton will hold true. Hence statement 1 is sufficient.

As above explanations say statement B is also sufficient.

Hi Yogesh .. can you please share the access codes for these tests. It will be very nice of you. My email id is sandeepuc@gmail.com. Thanks in advance.

First of all as \(|a|=b\) (\(b\) equals to absolute value of some number) then \(b\geq{0}\), as absolute value is always non-negative.

(1) \(a=-b\) (\(b=-a\)) --> so \(a\leq{0}\) and \(LHS=a+b=0\). But \(RHS=ab\leq{0}\) thus we can not say for sure that \(a+b>ab\), because if \(a=b=0\) then \(a+b=0=ab\) (so in case \(a=b=0\), \(a+b\) is not more than \(ab\) it equals to \(ab\)). Not sufficient.

First of all as \(|a|=b\) (\(b\) equals to absolute value of some number) then \(b\geq{0}\), as absolute value is always non-negative.

(1) \(a=-b\) (\(b=-a\)) --> so \(a\leq{0}\) and \(LHS=a+b=0\). But \(RHS=ab\leq{0}\) thus we can not say for sure that \(a+b>ab\), because if \(a=b=0\) then \(a+b=0=ab\) (so in case \(a=b=0\), \(a+b\) is not more than \(ab\) it equals to \(ab\)). Not sufficient.

First of all as \(|a|=b\) (\(b\) equals to absolute value of some number) then \(b\geq{0}\), as absolute value is always non-negative.

(1) \(a=-b\) (\(b=-a\)) --> so \(a\leq{0}\) and \(LHS=a+b=0\). But \(RHS=ab\leq{0}\) thus we can not say for sure that \(a+b>ab\), because if \(a=b=0\) then \(a+b=0=ab\) (so in case \(a=b=0\), \(a+b\) is not more than \(ab\) it equals to \(ab\)). Not sufficient.

Thanks for the solution. But I still couldnt understand why are we taking a=0 and b=0 in the 1 st statement. It states a=-b, is it ok to write 0=-0..... It would be of great help if you can explain. It will help in clearing my doubts.

First of all as \(|a|=b\) (\(b\) equals to absolute value of some number) then \(b\geq{0}\), as absolute value is always non-negative.

(1) \(a=-b\) (\(b=-a\)) --> so \(a\leq{0}\) and \(LHS=a+b=0\). But \(RHS=ab\leq{0}\) thus we can not say for sure that \(a+b>ab\), because if \(a=b=0\) then \(a+b=0=ab\) (so in case \(a=b=0\), \(a+b\) is not more than \(ab\) it equals to \(ab\)). Not sufficient.

Thanks for the solution. But I still couldnt understand why are we taking a=0 and b=0 in the 1 st statement. It states a=-b, is it ok to write 0=-0..... It would be of great help if you can explain. It will help in clearing my doubts.

Re: If |a| = b is a + b > ab ? [#permalink]
25 Jun 2013, 15:33

1

This post was BOOKMARKED

If |a| = b is a + b > ab

(1) a = -b (2) a = -3

|a| = b so b MUST be ≥ 0

1.) a = -b b=-a

So what do we know?

|a|=|b| b is positive a is the negative value of b

So, a + b > ab a+b = 0 HOWEVER we are not sure what values a and b are. For example, a and b could be -3 and 3 or a and b could be 0 and 0. INSUFFICIENT

(2) a = -3 We know that |a|=b, so if a = -3 then b must = 3 a + b > ab -3+3 > (-3)(3) 0>-9 TRUE

Re: If |a| = b is a + b > ab ? [#permalink]
26 Sep 2014, 05:52

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