Is X^2 greater than X?

[its4christian]

Is X^2 greater than X?

1)X^2 is greater than 1
2) X is greater than -1


Rephrasing for the the question:

X^2>X
X(X-1)>0
X>0
X>1
so....
X>1 ?


1) It says that X is >1 and <-1. SUFF

2) X>-1 [0, 1, 2, 3, ...] SUFF

I chose D

The OA is

Code: Select All Code

A

Can anyone explain why B is not sufficient?

I'm posting this question because this is the only time that using the MGMAT techinique of finding the minimum needed in the DS problem get's me in trouble. What am I missing here??

Thank you very much!
Christian
Mgmat student

[nimish.tiwari]

X(X-1)>0

If product of two numbers is greater than zero, then either both of them are positive or both are negative (so not necessarily, X>0).

Answer to your Question lies within.
for (1), clearly this means X too is greater than 1 but less than X^2. So, Sufficient.

for (2), Consider X= 0.1 (which is still >-1). In this case X^2 < X.
and for any X>1, X^2> X. So, Insufficient.

Hence, OA is A.

The only thing you probably forgot to consider is that not necessarily X can take only integer values. Consider for fractional values as well.

Hope this helps.

[its4christian]

Hi Nimish,
yes you're right - I overlooked the fact that X can take non integer values. If X is -0.5, 0, 0.5 the answer is N, if X is 2 the answer is Y.
Thanks for the eye opener!

C-

[RonPurewal]

nicely done, posters.

[linkon.nafiz]

When I attempted this problem, this is the approach I took and would like to know if it is a good way of tackling these types:

When I see X^2 - X > 0, I always think of consecutive integers and signs. So this simplifies to X (X-1) >0?

(1) X^2 > 1

X^2-1 >0

(x+1) (x-1) >0

Both of these must have same signs, each is one number to the left and right of X. Therefore, if X+1 is positive, X-1 is positive, and X has to be positive. The same is true if positive is switched with negative.

Am I correct with this way of thinking?

[RonPurewal]

here's a more efficient way to handle things like this:

• find the values where the left-hand side IS zero
• these are the only places where the sign can change. (if the expression goes from positive to negative, or from negative to positive, then clearly it has to pass through zero!)
• observe what happens BETWEEN these places.
• since the sign can't change except at the points you found, you only need to test one value per 'region'.

[RonPurewal]

for the case in question we're considering x(x – 1).
this expression IS zero at x = 0 and x = 1.
thus there are three 'regions' of interest:
... from negative infinity to 0,
... from 0 to 1,
... from 1 to positive infinity.
for each of these 'regions' you only need to test one value, since, again, the sign of x(x – 1) can change ONLY at x = 0 and/or x = 1.

if you test these 'regions' you'll find that x(x – 1) is positive in the first and last ones, and negative in the middle one.

[RonPurewal]

nb: what i explained above ^^ is actually the 'standard' way of solving these sorts of things, from high-school algebra. if it's any clearer here than it was back then, that's only because i don't use as many gratuitous symbols and big words when i explain stuff.
(:

[sahilk47]

Ron

Referring to statement 1 here,

x^2 >1

If we take positive square root on both sides :

Mod x > 1

Isn't Mod x > 1 equivalent to:
(a) x > 1, and
(b) x < -1

Don't we have to prove that in this current question, case (b) is not applicable ?

Thank you

[tim]

Isn't Mod x > 1 equivalent to:
(a) x > 1, and
(b) x < -1

Don't we have to prove that in this current question, case (b) is not applicable ?

Thank you

Technically that "and" should be an "or", but what makes you think case b is not applicable or that we should care? As long as x^2>x for any value of x that is greater than 1 or less than -1, statement 1 is sufficient under either case.

[RonPurewal]

in fewer words:

• both cases are applicable

• for each case, you have to answer the question that's actually there

• the answer to the question that's actually there will be the same both times.