If x is Positive, Which of the Following Could be the Correct Ordering GMAT Problem Solving

Question: If x is positive, which of the following could be the correct ordering of 1/x, 2x, and x^2?

I) \(x^2 < 2x < \frac{1}{x}\)

II)\(x^2 < \frac{1}{x} < 2x \)

III) \( 2x < x^2 < \frac{1}{x}\)

(a) None

(b) I only

(c) III only

(d) I and II only

(e) I, II and III

Correct Answer: D

Solution and Explanation:
We are given that x is a positive number. The problem statement asks to find out which of the given ordering could be correct.

Approach Solution 1:

It should be noted that it is asking which of these could be correct and not which of these must be correct.

Hence, we are required to determine the relationship between the three values 1/x, x^2, 2x in three regions.

1. x > 2

2. 1 < x < 2

3. 0 < x < 1

When x is greater than 2
x^2 always has the greatest value.
But in none of the options, x^2 is greatest, therefore, we know that x < 2

In the region 1 < x < 2
2x has the greatest value and after that x^2 and at last 1/x
2x < x^2 < 1/x
But it is also not given in any option So x <=1

In the case of 0 < x < 1,
x^2 has the least value, so the two expressions having x^2 as the least value is

I) \(x^2 < 2x < \frac{1}{x}\)

=> 2x < \(\frac{1}{x} \) ?

=> \(\frac{(2x^2-1)}{x}\)< 0 ?

The expression \(\frac{(2x^2-1)}{x}\) can be negative as well as positive for the values in 0 < x< 1

It can be checked by taking different values of x.

II)\(x^2 < \frac{1}{x} < 2x \)

=> \(\frac{1}{x} \) < 2x?

=> (\(2x^2\)- 1) > 0 ?

The expression \(\frac{(2x^2-1)}{x}\) can be negative as well as positive for the values in 0 < x< 1
Therefore both cases I and II could be the possibility.

Second condition :

\(x^2 < \frac{1}{x} < 2x \)

Let x = 0.9
Note that 0 < x < 1
Now x^2 = 0.81
2x = 1.8

\(\frac{1}{x} \) = \(​​\frac{1}{0.9} \) = 1.11

0.81 < 1.11 < 1.8

\(x^2 < \frac{1}{x} < 2x \)

So verified that this could be the correct ordering.
Therefore the correct answer is option D which says I and II could be the possible answer.

Approach Solution 2:
The problem statement states that:

Given:

• x is a positive number.

Asked: 

• Find out which of the given ordering of 1/x, 2x, and x^2 could be correct.

We could use the algebraic approach to solve the problem. Here is number picking:

1. \(x^2 < 2x < \frac{1}{x}\)
   --> x = ½
   --> x²=¼, 
   2x=1,
   1/x =2
   --> 1/4<1<2.
   Hence this COULD be the correct ordering.
2. \(x^2 < \frac{1}{x} < 2x \)
   --> x = 0.9
   --> x² = 0.81
   1/x=1.11, 
   2x=1.8
   --> 0.81<1.11<1.8. Hence this COULD be the correct ordering.
3. \( 2x < x^2 < \frac{1}{x}\)
   --> x² to be more than 2x, x must be more than 2 (for positive x−es).
   But if x>2, then 1/x is the least value from these three and can not be more than 2x and x².
   So III can not be true.

Thus I and II could be correct ordering and III can not.

Approach Solution 3:
The problem statement implies that:

Given:

• x is a positive number.

Asked: 

• Find out which of the given ordering of 1/x, 2x, and x^2 could be correct.

Let’s solve the problem by stating any example that satisfies the inequation, that statement will be correct.
I. \(x^2 < 2x < \frac{1}{x}\)
If x = 0.1, this implies 0.01 < 0.2 < 10

II. \(x^2 < \frac{1}{x} < 2x \) 
If x= ½, this implies 1/4 < 1/2 < 1

III. \( 2x < x^2 < \frac{1}{x}\)
If 2x < x^2 <=> x ( 2 -x) < 0, x > 0 then x > 2
with x > 2 ==> x^2 < 1/x <=> x^3 < 1 <=> x < 1
Therefore, (III) can't happen

Thus I and II could be correct ordering and III can not.

“If x is positive, which of the following could be the correct ordering”- is a topic of the GMAT Quantitative reasoning section of the GMAT exam. This topic has been taken from the book “GMAT Official Guide Quantitative Review 2022”. GMAT Problem Solving questions enable the candidates to consider facts and solve numerical problems. GMAT Quant practice papers help the candidates to get familiar with different sorts of questions that will improve their mathematical knowledge.

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