BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

DS: Absolute Value again!

Expert replies
Source: — Data Sufficiency |

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.
Join the discussion

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.

--

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi

--

Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.

Yves Saint-Laurent

--

Learn more about ron
Join the discussion

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!
Join the discussion

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!!!
Join the discussion

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.
Join the discussion

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

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

Thanks

Puneet
Join the discussion