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Bunuel
If 0 < k < 1, then which of the following must be less than k?

(A) 3k/2

(B) 1/k

(C) |k|

(D) \(\sqrt{k}\)

(E) k^2

Let k = 0.50

(A) \(\frac{3k}{2} = \frac{3*0.50}{2} = 0.75\) ( Not Possible )

(B) \(\frac{1}{k} = \frac{1}{0.50} = 2\) ( Not Possible )

(C) \(|k| = |0.50|\) ( Not Possible )

(D) \(\sqrt{k}\) = \(\sqrt{0.50}\) = \(0.70\) ( Not Possible )

(E) \(k^2 = 0.50^2 = 0.25\) ( Possible )

Answer will hence be (E)
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Bunuel
If 0 < k < 1, then which of the following must be less than k?

(A) 3k/2

(B) 1/k

(C) |k|

(D) \(\sqrt{k}\)

(E) k^2


The question basically indicates that the value of k is a positive fraction. In such cases, the greater the value of denominator, the smaller the overall value of the fraction.
Out of all the options, \(k^2\) would increase the value of the denominator the most and thus E should be the correct answer.
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0<k<1
So k= proper fraction=1/2, 1/3, 1/4
Sub any of the above value in the options
Avoid decimals, as it may cause some confusions.

Option E alone will be less than k
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Bunuel
If 0 < k < 1, then which of the following must be less than k?

(A) 3k/2

(B) 1/k

(C) |k|

(D) \(\sqrt{k}\)

(E) k^2

0<k<1...

A.) 3k/2 > k

B.) 1/k > k

C.) |k| = k

D.) \(\sqrt{k}\) > k

E.) k^2 < k

Answer E.
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Bunuel
If 0 < k < 1, then which of the following must be less than k?

(A) 3k/2

(B) 1/k

(C) |k|

(D) \(\sqrt{k}\)

(E) k^2

We are given that 0 < k < 1, so k is a positive proper fraction. We may recall that when a positive proper fraction is raised to an exponent greater than 1, it will result in a lesser value. Thus, k^2 < k.

Answer: E
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