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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.

(2) a=-3 --> b=3 --> a+b=0>ab=-9. Sufficient.

Answer: B.

Hope it's clear.

Bravo. This is the very reason I marked B but was shocked to see that MR people had marked it as D. Thanks Bunuel. You made my day. _________________

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.

(2) a=-3 --> b=3 --> a+b=0>ab=-9. Sufficient.

Answer: B.

Hope it's clear.

Bravo. This is the very reason I marked B but was shocked to see that MR people had marked it as D. Thanks Bunuel. You made my day.

So the trick here is not to consider zero. Probably the owner of the question forgot to say "a and b are both non-zero integers".

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.

(2) a=-3 --> b=3 --> a+b=0>ab=-9. Sufficient.

Answer: B.

Hope it's clear.

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.

(2) a=-3 --> b=3 --> a+b=0>ab=-9. Sufficient.

Answer: B.

Hope it's clear.

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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