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If x > 0, is x^2 < x ?

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If x > 0, is x^2 < x ?

by Vincen » Mon Sep 25, 2017 6:53 pm
If x > 0, is x^2 < x ?

(1) 0.1 < x < 0.4
(2) x^3 < x^2

The OA is D.

Experts, should I have to make the graph of the options to see that both statements alone are sufficient? or can I solve it with some calculations?
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Source: — Data Sufficiency |

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by ErikaPrepScholar » Tue Sep 26, 2017 9:37 am
This problem is easiest to solve if you understand some number properties.

Here, we need to know that multiplying by a fraction (a number between 0 and 1) yields a smaller number. This means that if x is positive *and a fraction*, x^2 will be a smaller fraction. For example:

1/2 * 1/2 = 1/4
1/3 * 1/3 = 1/9
1/4 * 1/4 = 1/16
1/8 * 1/8 = 1/64

and so on. So we know that if x is a fraction, x > x^2.

Statement 1 tells us that x is a fraction. SUFFICIENT.

Statement 2 tells us that x^3 > x^2. The only ways for that to be possible are if x is negative or if x is a fraction. If x is 1, x^3 = x^2, and if x is > 1, x^3 > x^2. We already know that x >0, so x cannot be negative. This means that x must be a fraction. SUFFICIENT.
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If x > 0, is x^2 < x ?

by Brent@GMATPrepNow » Tue Sep 26, 2017 9:37 am
Vincen wrote:If x > 0, is x² < x ?

(1) 0.1 < x < 0.4
(2) x³ < x²
Target question: Is x² < x ?

Given: x > 0
So, x is POSITIVE

Statement 1: 0.1 < x < 0.4
There's a nice property that says: If 0 < k < 1, then k² < k
Statement essentially 1 tells us that 0 < x < 1 , which means we can be certain that x² < x
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x³ < x²
Since we're told that x is POSITIVE, we can safely take the inequality, x³ < x², and divide both sides by x
When we do so, we get: x² < x
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D

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