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by puneetkhurana2000 » Sat Dec 22, 2012 6:24 pm
Statement 1) (x+1)(|x|-1) > 0

We have two cases when
1)x > 0 , |x| = x
(x+1)(|x|-1)>0 becomes (x+1)(x-1)>0
x^2 -1 > 0 i.e. x > 1 or x < -1 but we have assumed x > 0 so x > 1.

2) x < 0 , |x| = -x
(x+1)(|x|-1)>0 becomes (x+1)(-x-1)>0
-(x + 1 )^2 > 0 i.e. (x + 1 )^2 < 0 but square of any number can never be negative. So the case fails.

From Cases above we have x > 1. Sufficient!!!

Statement 2) |x|<5 means -5 < x < 5. Not Sufficient!!!

Answer A.
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by lunarpower » Sat Dec 22, 2012 7:14 pm
la1214 wrote:Hello,

Experts please help!!

Is x>1?
1. (x+1)(|x|-1)>0
2.|x|<5

Thanks
the solution posted above is quite good.

if you would rather not take such an algebraically intense approach, you can also try to find individual cases of "yes" and "no".
for either of the statements, if you can find both "yes" and "no", then the statement is insufficient; if you can't, then it's sufficient.

--

for statement 1:

Getting "yes"
if you plug in a really big number for x (say, x = 1000), then that number will definitely satisfy statement 1, and it will give a "yes" answer to the question.

Getting "no"
now, the goal is to get a number that actually satisfies statement 1 but isn't greater than 1.
but ...
* x = 1 doesn't work in statement 1, because the product will be 0.
* if x is anything between -1 and 1, then (x + 1) will be positive but (|x| - 1) will be negative, so statement 1 won't work.
* x = -1 doesn't work in statement 1 either; it makes the product 0 again.
* finally, no value of x below -1 will work in statement 1, because, for such values, (x + 1) is positive but (|x| - 1) is negative.

so, it's not possible to get a "no" answer with statement 1, so statement 1 is sufficient.

--

statement 2:

* if you let x = 0 (which satisfies statement 2), then you get a "no" answer to the question.

* if you let x = 2 (which also satisfies statement 2), then you get a "yes" answer to the question.

so, statement 2 is insufficient.
Ron has been teaching various standardized tests for 20 years.

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by la1214 » Sat Dec 22, 2012 7:23 pm
Puneet,

You consider two cases for x- i.e x >0 and x<0, what about x=0?

Thank you!
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by puneetkhurana2000 » Sat Dec 22, 2012 8:49 pm
In the first case we have proven that x > 1 as one solution.

Hope this helps!!!
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by viveksingh222 » Sun Dec 23, 2012 9:45 am
I have problem with inequalities can some one provide me a link for this topic, your help is appreciated.

Thank you.
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by puneetkhurana2000 » Sun Dec 23, 2012 3:47 pm
I find this really helpful.

https://www.manhattangmat.com/strategy-s ... -value.cfm

Thanks

Puneet
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