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

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If 0 < k < 1, then which of the following must be less than k?  [#permalink]

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13 Jun 2017, 07:50
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82% (00:54) correct 18% (01:01) wrong based on 131 sessions

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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

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If 0 < k < 1, then which of the following must be less than k?  [#permalink]

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13 Jun 2017, 08:14
1
Bunuel wrote:
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 = $$\frac{1}{2}$$

(A) $$\frac{3k}{2} = \frac{3}{2} * \frac{1}{2} = \frac{3}{4}$$ ------- Greater than k

(B) $$\frac{1}{k} = \frac{1}{(1/2)}$$ = 2 ------------ Greater than k

(C) |k| = $$\frac{1}{2}$$ ----------- Equal to k

(D) $$\sqrt{k}$$ = $$\sqrt{\frac{1}{2}}$$ = 0.707 ------------- Greater than k

(E) $$k^2$$ = $$(\frac{1}{2})^2$$ = $$\frac{1}{4}$$ -------- Less than k
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Re: If 0 < k < 1, then which of the following must be less than k?  [#permalink]

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13 Jun 2017, 08:48
Bunuel wrote:
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

Higher powers result in a smaller number when the number is between 0 and 1

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Re: If 0 < k < 1, then which of the following must be less than k?  [#permalink]

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13 Jun 2017, 08:54
Bunuel wrote:
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 )

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Re: If 0 < k < 1, then which of the following must be less than k?  [#permalink]

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13 Jun 2017, 21:23
Bunuel wrote:
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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Re: If 0 < k < 1, then which of the following must be less than k?  [#permalink]

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14 Jun 2017, 04:55
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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Re: If 0 < k < 1, then which of the following must be less than k?  [#permalink]

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14 Jun 2017, 05:11
Bunuel wrote:
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

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Re: If 0 < k < 1, then which of the following must be less than k?  [#permalink]

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15 Jun 2017, 15:21
1
Bunuel wrote:
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.

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Re: If 0 < k < 1, then which of the following must be less than k?  [#permalink]

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15 Aug 2018, 04:45
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Re: If 0 < k < 1, then which of the following must be less than k? &nbs [#permalink] 15 Aug 2018, 04:45
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