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If 0 < k < 1, then which of the following must be less than k?
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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?
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13 Jun 2017, 08:14
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 Answer E...



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Re: If 0 < k < 1, then which of the following must be less than k?
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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 Answer option E
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Re: If 0 < k < 1, then which of the following must be less than k?
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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 )Answer will hence be (E)
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Re: If 0 < k < 1, then which of the following must be less than k?
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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?
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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?
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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?
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15 Jun 2017, 15:21
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. Answer: E
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Re: If 0 < k < 1, then which of the following must be less than k?
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15 Aug 2018, 04:45
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