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If |a| = b is a + b > ab ? [#permalink]
 13 Jul 2010, 14:59
Question Stats:
76% (01:56) correct
23% (00:36) wrong based on 43 sessions
If |a| = b is a + b > ab ? (1) a = -b (2) a = -3 I believe its B but the MR says its D.
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Yogesh Agarwal
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Ms. Big Fat Panda
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Given: |a| = b
To find: Is a+b>ab
Statement 1: a = -b - Sufficient
=> a+b = 0
ab = (-b)(b) = -b^2 < 0
0 > ab since ab is negative.
Statement 2: a = -3 - Sufficient
b = |-3| = 3
=> a+b = 0
ab = (-3)(3) = -9 < 0
0 > ab since ab is negative.
So the answer is D.
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Waoo I even came with D, but after looking to "Bunuel" post; I'm not sure what's wrong ?
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I think it is D.
Lets take statement 1, a = -b.
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.
B is answer.
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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.
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Bunuel wrote: yogeshagarwala wrote: Please solve:
I believe its B but the MR says its D. If |a|=b is a+b>ab? 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.
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yogeshagarwala wrote: Bunuel wrote: yogeshagarwala wrote: Please solve:
I believe its B but the MR says its D. If |a|=b is a+b>ab? 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".
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Oops what a catch !!
Even I missed the 0.
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Bunuel wrote: yogeshagarwala wrote: Please solve:
I believe its B but the MR says its D. If |a|=b is a+b>ab? 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. Thanks in advance
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Richa16 wrote: Bunuel wrote: yogeshagarwala wrote: Please solve:
I believe its B but the MR says its D. If |a|=b is a+b>ab? 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. Thanks in advance Yes, it's ok to write 0=-0.
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