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n and k are positive integers. If n=(22)(3)(53)/15k, is n<k?

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Math Revolution GMAT Instructor
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n and k are positive integers. If n=(22)(3)(53)/15k, is n<k?  [#permalink]

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New post Updated on: 13 Jun 2016, 01:09
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n and k are positive integers. If n=(2^2)(3)(5^3)/15k, is n<k?
1) n<10
2) k>4

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Originally posted by MathRevolution on 09 Jun 2016, 22:45.
Last edited by MathRevolution on 13 Jun 2016, 01:09, edited 1 time in total.
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n and k are positive integers. If n=(22)(3)(53)/15k, is n<k?  [#permalink]

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New post 13 Jun 2016, 01:08
If we modify the original condition and the question, we get n=100/k, k=100/n. If we substitute this into n<k?, the question becomes n<100/n?, n^2<100=10^2?, n<10?. Hence, the correct answer is A.

- Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
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Re: n and k are positive integers. If n=(22)(3)(53)/15k, is n<k?  [#permalink]

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New post 23 Sep 2017, 11:10
How did the substitution occur?
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Re: n and k are positive integers. If n=(22)(3)(53)/15k, is n<k?  [#permalink]

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New post 24 Sep 2017, 10:13
Madhavi1990 wrote:
How did the substitution occur?


simplify the eq: it comes down to n = 300/k

1. n<10 to make that happen k>30
2. k>4 so k=5 n=60... keep going and k=31 n<10 so we have multiple values
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Re: n and k are positive integers. If n=(22)(3)(53)/15k, is n<k?  [#permalink]

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New post 24 Sep 2017, 11:26
Simplifying question stem:
\(n=\frac{(4*3*5*5*5)}{(5*3)k}\)
\(n=\frac{100}{k}\)

Statement 1: n<10
Rewrite question stem as \(k=\frac{100}{n}\)
For 0<n<10, k will be more than n.-----------Sufficient

Statement 1: k>4
Rewrite question stem as \(n=\frac{100}{k}\)
For k=100, n=1.................Is n<k? Yes
For k=5, n=20................Is n<k? No----------Insufficient

Answer is A
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Re: n and k are positive integers. If n=(22)(3)(53)/15k, is n<k? &nbs [#permalink] 24 Sep 2017, 11:26
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