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Sajjad1994
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BrentGMATPrepNow
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If 0 < x < 1, then 0 < √x < 1, so answer A, 1/√x, will be greater than 1. Every other answer is less than 1, so A is largest.
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Another method is to substitute values, which can take up some of your time, but will definitely get you there.

Since x is > 0 and x < 1, it has to be a decimal/fraction.

Take a value which is a perfect square, as some options have a root value. Let's take x = 0.25 = \(0.5^2\)


Option 1: \(\frac{1}{\sqrt{0.25}}= \frac{1}{0.5} = 2\)


Option 2: \(\sqrt{0.25} = 0.5\)


Option 3: \(\frac{x}{\sqrt{\pi}}= \frac{0.25}{\frac{22}{7}} = 0.08\)


Option 4: \(0.25^3 = 0.015\)


Option 5: \(0.25^4 = 0.015^2 = 0.000225\)



Option A

Arun Kumar


Key takeaways after doing this question.

(a) Squares, cubes or higher powers of a positive decimal/fraction, is lesser than the fraction. i.e \( x^n < x\)

(b) Square root of a fraction/decimal is larger than the fraction

(c) If n is a positive integer and n is divided by a fraction x, then the result is always greater than n i.e [m]\frac{n}{x}> n
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Rule: Given X = a (+)Positive Proper Fraction that falls in the Range of: 0 < X < 1

(X)^2 < X < sqrt(X)


ex: X = 1/4

1/16 < 1/4 < 1/2
(X)^2 < X < sqrt(X)


-A- would give the Max Value

1 divided by the Square Root of a Proper Fraction will result in an Integer or Compound Fraction > 1


the only one that is close is -B- and we will still retain a Proper Fraction when we apply the Square Root to the Proper Fraction


-A- Correct Answer
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Sajjad1994
If \(0 < x < 1\), which of the following expressions is greatest?

(A) \(\frac{1}{\sqrt{x}}\)

(B) \(\sqrt{x}\)

(C) \(\frac{1}{π}\) x

(D) \(x^3\)

(E) \(x^4\)

I would also suggest taking a value for x, if you are not sure about how numbers behave in the range 0<x<1.

You can take x=0.01, so \(\sqrt{0.01}=0.1=\frac{1}{10}\)


(A) \(\frac{1}{\sqrt{x}}=\frac{1}{0.1}=10\)

(B) \(\sqrt{x}=0.1\)

(C) \(\frac{1}{π}*x=0.01*\frac{7}{22}\)

(D) \(x^3=0.000001\)

(E) \(x^4=0.00000001\)

Also, you can note the following
1) As the power increases in the range 0<x<1, the number becomes smaller...\(x^4<x^3<x^2<x<\sqrt{x}\)
2) x in the denominator will give you a bigger value than x in the numerator...\(\frac{1}{x}>x\)
If you know above two points, you can mark A as the answer.

A
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Sajjad1994
If \(0 < x < 1\), which of the following expressions is greatest?

(A) \(\frac{1}{\sqrt{x}}\)

(B) \(\sqrt{x}\)

(C) \(\frac{1}{π}\) x

(D) \(x^3\)

(E) \(x^4\)

Source: Nova GMAT

The rule that if \(0 < x < 1\), \(x^2 < x < \sqrt{x}\) helps us here to eliminate D and E.

Looking at \(\frac{1}{π}x\), we can sub in \(x=\frac{1}{4}\) to quickly compare: \( \frac{1}{π} * \frac{1}{4} = \frac{1}{4 π } \). We've made the fraction smaller, not larger. So this will be less than \(\sqrt{x}\).

That leaves \(\sqrt{x}\) and \(\frac{1}{\sqrt{x}}\). From the stem we can identify that \(\sqrt{x}\) will be a proper fraction, because x originally had a value between 0 and 1. Taking the reciprocal of a proper fraction will result in a value greater than itself. This means that \(\frac{1}{\sqrt{x}}> \sqrt{x} \). Thus, our answer is A.
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