If -2<a<11 and 3<b<12, then which of the folloin

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by neelgandham » Thu Oct 27, 2011 12:53 pm
Why do you think it is E ?

If -2<a<11 and 3<b<12, then -24<ab<132,
let us say a=-1.9 and b =11.9 , ab = -22.61 !
let a=10.9 and b =11.9 then ab =129.71 !
So, -24<ab<132 is correct !

C). -7<b-a<14 - Let a = 10.9 and b =3.1, b-a =3.1-10.9=-7.8 -7.8 is < -7 So, this inequation is NOT true!
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by GmatMathPro » Thu Oct 27, 2011 1:18 pm
Yeah, keep in mind that to make ab as small as possible, it doesn't necessarily mean you should make a and b both as small as possible. That would make sense if both a and b had to be positive, but since we have negative values to consider, the smallest value of ab comes when ab is most negative. In this case it means making b as big as possible, and making a as negative as possible.
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by Amiable Scholar » Thu Oct 27, 2011 6:21 pm
lenagmat wrote: [spoiler] for me the only question here is what to do with E. to be true, because for me it is not true
-2 < a < 11 ------------- (1)
3<b<12 ------------------ (2)
so for a consider 2 cases
(A)
-2< a < 0 in that case reverse the sign to make middle term positive
=> 0 < -a < 2
remember .. -a is positive here
now you can multiply it with second inequality to check the boundaries
=>
0< -ab < 24
now reverse the sign => -24 < ab < 0 --- (C)
(B)
0<a<11 here a is definitely positive ...
you can multiply it directly with 2
=>
0<ab <132 - ----(D)
from C and D and we also didn't include case a = 0 so ab can be 0 too if a is zero

-24 < ab <132
so E is true
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by VivianKerr » Thu Oct 27, 2011 8:22 pm
As long as we can pick two allowed values and make a statement TRUE, we know it cannot be correct.

For E, the biggest a must be smaller than 11 and the biggest b must be smaller than 12. 11 x 12 = 132, so the product MUST be smaller, true.

The smallest a is greater than -2, and the smallest b is greater than 3, so the smallest ab must be greater than -6. All values greater than -6 are certainly greater than -24.
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