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Which of the answer choices properly lists the following in increasing [#permalink]
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Updated on: 12 Apr 2018, 10:26
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66% (01:38) correct 34% (02:03) wrong based on 180 sessions
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Which of the answer choices properly lists the following in increasing order from left to right (all roots are positive): \(2^{(\frac{1}{3})}\) \(5^{(\frac{1}{6})}\) \(10^{(\frac{1}{10})}\) \(30^{(\frac{1}{15})}\) A \((2)^{(\frac{1}{3})}\), \((5)^{(\frac{1}{6})}\), \((10)^{(\frac{1}{10})}\), \((30)^{(\frac{1}{15})}\) B. \((10)^{(\frac{1}{10})}\), \((30)^{(\frac{1}{15})}\), \((2)^{(\frac{1}{3})}\), \((5)^{(\frac{1}{6)}}\) C. \((10)^{(\frac{1}{10})}\), \((2)^{(\frac{1}{3})}\), \((5)^{(\frac{1}{6})}\), \((30)^{(\frac{1}{15})}\) D. \((30)^{(\frac{1}{15})}\), \((2)^{(\frac{1}{3})}\), \((10)^{(\frac{1}{10})}\), \((5)^{(\frac{1}{6})}\) E. \((30)^{(\frac{1}{15})}\), \((10)^{(\frac{1}{10})}\), \((2)^{(\frac{1}{3})}\), \((5)^{(\frac{1}{6})}\)
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Originally posted by imhimanshu on 11 Sep 2013, 17:59.
Last edited by Bunuel on 12 Apr 2018, 10:26, edited 1 time in total.
Edited the options.



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Re: Which of the answer choices properly lists the following in increasing [#permalink]
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11 Sep 2013, 19:32
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imhimanshu wrote: Which of the answer choices properly lists the following in increasing order from left to right (all roots are positive):
(2)^(1/3) 5^(1/6) (10)^(1/10) 30^(1/15)
a) (2)^(1/3) 5^(1/6) (10)^(1/10) 30^(1/15) b) (10)^(1/10) 30^(1/15) (2)^(1/3) 5^(1/6) c)(10)^(1/10) (2)^(1/3) 5^(1/6) 30^(1/15) d) 30^(1/15) (2)^(1/3) (10)^(1/10) 5^(1/6) e) 30^(1/15) (10)^(1/10) (2)^(1/3) 5^(1/6)
Please explain how to go about this. himanshu, My reasoning, take lcm of 1/3, 1/6, 1/10, 1/15  LCM is 1/30 Now power each digit by 30 and the result is 2^10, 5^5, 10^3, 30^2 so order is 30^2, 10^3, 2^10, 5^5 > E
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Re: Which of the answer choices properly lists the following in increasing [#permalink]
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12 Sep 2013, 17:06
maaadhu wrote: imhimanshu wrote: Which of the answer choices properly lists the following in increasing order from left to right (all roots are positive):
(2)^(1/3) 5^(1/6) (10)^(1/10) 30^(1/15)
a) (2)^(1/3) 5^(1/6) (10)^(1/10) 30^(1/15) b) (10)^(1/10) 30^(1/15) (2)^(1/3) 5^(1/6) c)(10)^(1/10) (2)^(1/3) 5^(1/6) 30^(1/15) d) 30^(1/15) (2)^(1/3) (10)^(1/10) 5^(1/6) e) 30^(1/15) (10)^(1/10) (2)^(1/3) 5^(1/6)
Please explain how to go about this. himanshu, My reasoning, take lcm of 1/3, 1/6, 1/10, 1/15  LCM is 1/30 Now power each digit by 30 and the result is 2^10, 5^5, 10^3, 30^2 so order is 30^2, 10^3, 2^10, 5^5 > E Thanks Madhu, However, my doubt is Is it possible to approach this problem by making bases same instead of exponents. I believe, that is more natural to me. Please suggest. Thanks



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Re: Which of the answer choices properly lists the following in increasing [#permalink]
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12 Sep 2013, 17:17
imhimanshu wrote: maaadhu wrote: imhimanshu wrote: Which of the answer choices properly lists the following in increasing order from left to right (all roots are positive):
(2)^(1/3) 5^(1/6) (10)^(1/10) 30^(1/15)
a) (2)^(1/3) 5^(1/6) (10)^(1/10) 30^(1/15) b) (10)^(1/10) 30^(1/15) (2)^(1/3) 5^(1/6) c)(10)^(1/10) (2)^(1/3) 5^(1/6) 30^(1/15) d) 30^(1/15) (2)^(1/3) (10)^(1/10) 5^(1/6) e) 30^(1/15) (10)^(1/10) (2)^(1/3) 5^(1/6)
Please explain how to go about this. himanshu, My reasoning, take lcm of 1/3, 1/6, 1/10, 1/15  LCM is 1/30 Now power each digit by 30 and the result is 2^10, 5^5, 10^3, 30^2 so order is 30^2, 10^3, 2^10, 5^5 > E Thanks Madhu, However, my doubt is Is it possible to approach this problem by making bases same instead of exponents. I believe, that is more natural to me. Please suggest. Thanks Himanshu, When comparing numbers, you have to multiply with the same number or divide by same number. In this case, if you multiply or divide by the same number, your bases will not be equal. Even if you try to make the base a multiple of 10, its hard to figure out because you still have fraction exponents.
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Re: Which of the answer choices properly lists the following in increasing [#permalink]
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12 Sep 2013, 19:37
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imhimanshu wrote: Which of the answer choices properly lists the following in increasing order from left to right (all roots are positive):
(2)^(1/3) 5^(1/6) (10)^(1/10) 30^(1/15)
a) (2)^(1/3) 5^(1/6) (10)^(1/10) 30^(1/15) b) (10)^(1/10) 30^(1/15) (2)^(1/3) 5^(1/6) c)(10)^(1/10) (2)^(1/3) 5^(1/6) 30^(1/15) d) 30^(1/15) (2)^(1/3) (10)^(1/10) 5^(1/6) e) 30^(1/15) (10)^(1/10) (2)^(1/3) 5^(1/6)
Please explain how to go about this. Responding to a pm: Here your first problem is the fractional powers. We cannot compare them in anyway until and unless we get rid of those so you raise each term to a power of 30.. You get rid of all fractional powers. Now you don't need to make anything same  base or power. The numbers are quite comparable without doing anything: 2^10 = 1024, 5^5 = 625*5 which is more than 3000, 10^3 = 1000 and 30^2 = 900
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Re: Which of the answer choices properly lists the following in increasing [#permalink]
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16 Nov 2016, 09:58
imhimanshu wrote: Which of the answer choices properly lists the following in increasing order from left to right (all roots are positive):
2^(1/3) 5^(1/6) 10^(1/10) 30^(1/15)
A (2)^(1/3) 5^(1/6) (10)^(1/10) 30^(1/15) B. (10)^(1/10) 30^(1/15) (2)^(1/3) 5^(1/6) C. (10)^(1/10) (2)^(1/3) 5^(1/6) 30^(1/15) D. 30^(1/15) (2)^(1/3) (10)^(1/10) 5^(1/6) E. 30^(1/15) (10)^(1/10) (2)^(1/3) 5^(1/6) I didn't think much clearly 30^(1/15) is the smallest one 10^(1/10) will be the second one so all but E are eliminated.



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Re: Which of the answer choices properly lists the following in increasing [#permalink]
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12 Apr 2018, 09:06
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