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655-705 (Hard)|   Exponents|   Roots|      
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imhimanshu
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Which of the answer choices properly lists the following in increasing Updated on: 12 Apr 2018, 10:26
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.
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68% (02:12) correct 32% (02:12) wrong based on 1165 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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Which of the answer choices properly lists the following in increasing 12 Sep 2013, 19:37
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imhimanshu
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.
Show more

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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Which of the answer choices properly lists the following in increasing 11 Sep 2013, 19:32
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imhimanshu
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.
Show more

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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Which of the answer choices properly lists the following in increasing 12 Sep 2013, 17:06
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maaadhu
imhimanshu
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
Show more

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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Which of the answer choices properly lists the following in increasing 12 Sep 2013, 17:17
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imhimanshu
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imhimanshu
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
Show more

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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Which of the answer choices properly lists the following in increasing 16 Nov 2016, 09:58
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imhimanshu
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)
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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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Which of the answer choices properly lists the following in increasing 24 Aug 2019, 07:36
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imhimanshu
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
Show more


Hi,

I want to know how can we raise the term by 30, just like that. Wouldn't raising the term by 30 make the different ? As we do in the case of multiplication, when we multiply the number we divide the same to keep the number same. I am not understanding. Please help.
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Which of the answer choices properly lists the following in increasing 25 Aug 2019, 03:13
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imhimanshu
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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Which of the answer choices properly lists the following in increasing order from left to right (all roots are positive):

Let us raise each number with ^30

\(2^{(\frac{1}{3})} = 2^{10} = 1024\) I

\(5^{(\frac{1}{6})} = 5^5 = 3125\) II

\(10^{(\frac{1}{10})} = 10^3 = 1000\). III

\(30^{(\frac{1}{15})} = 30^2 = 900\) IV

900 (IV) < 1000 (III) < 1024 (I) < 3125 (II)

IMO E
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Which of the answer choices properly lists the following in increasing 07 Apr 2026, 03:38
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