# Quantitative Challenge Question week-03

Quantitative Challenge Question week-03

# Challenge Question of the Week

Welcome to Quantitative Challenge Question week-03, here’s the weekly challenge question of GMAT, GRE and SAT exams. You can see the answer explanation below the question.

## Question:

If x is a number, and x2 > x3; which of the following MUST be true?

I. x > x2
II. x2 > x
III. √x > x

A) I only
B) II only
C) III only
D) I & III only
E) None of the above

### Explanation:

First of all, it is given that: x2 > x3, which is only possible when either value of x exists between 0 and 1, or its negative.

Now, we need to see whether the three conditions given are MUST be true? Remember that MUST be true means its true for all values that can be possibly used. If the condition is true in all values except for only 1 value, then that condition COULD be true, but no more MUST be true.

Condition I: x > x2

This condition is only true if value of x exists between 0 and 1, because higher the exponent, the lower is the value of number that exists between 0 and 1. But for negative values of x, this condition becomes false. As it is given that x can be negative, so this condition can be stated as COULD be true statement but not a MUST be true statement.

Condition II: x2 > x

This condition is true if value of x is negative. On other hand, for any value of x between 0 and 1, this condition becomes false. Thus, this condition can be stated as COULD be true statement but not a MUST be true statement.

Condition III:x > x

This condition is also clearly COULD be true statement , but not a MUST be true statement. As the square root of a negative number is undefined. However, if value of x exists between 0 and 1,this statement holds true. As the square root of x means that power of x is 12, thus the value of square root of x would be greater than x when x exists between 0 and 1.

Therefore, choice E is correct answer.

## Takeaways:

→ If x > x2,     ⇒ x only exists between 0 and 1.

→ If x2 > x,     ⇒ either x is negative or x is greater than 1.

→ If √x > x,     ⇒ x is positive, but less than 1.

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