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If n is a positive integer, is (1/10)^n < 0.01? [#permalink]
16 Mar 2012, 09:04

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Question Stats:

67% (01:50) correct
32% (01:16) wrong based on 193 sessions

If n is a positive integer, is \frac{1}{10}^n < 0.01?

(1) n > 2

(2) \frac{1}{10}^{n-1} < 0.1

First post here, so I hope I got the format right. I understand the OG explanation to this problem, but I tried taking a slightly alternate route and am coming up with the wrong answer. It's a rather simple one, but hope someone can shed some light on to where I've gone wrong. I included the OG explanation as well as my own in the spoiler.

OG Explanation: Manipulate both sides to be expressed as powers of 10.

\frac{1}{10}^n < 0.01

(10^{-1})^n < 10^{-2}

10^{-n} < 10^{-2}

n > 2

1)n > 2. SUFFICIENT

2)\frac{1}{10}^{n-1} < 0.1

(10^{-1})^{n-1} < 10^{-1}

10^{-n+1} < 10^{-1}

-n+1 < -1

n > 2 SUFFICIENT

My slightly modified solution for statement 2 was to first manipulate the 0.1 on the right side of the inequality to become a fraction and to leave the left side as a fraction (my first instinct is to see that 0.01 is the same as 1/10). You would have:

\frac{1}{10}^{n-1} < \frac{1}{10}^1

n-1 < 1

n < 2

As you can see, I get an opposite answer. I know this is super simple, but where am I going wrong?

Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]
16 Mar 2012, 09:32

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This post received KUDOS

Expert's post

tkaelle wrote:

First post here, so I hope I got the format right. I understand the OG explanation to this problem, but I tried taking a slightly alternate route and am coming up with the wrong answer. It's a rather simple one, but hope someone can shed some light on to where I've gone wrong. I included the OG explanation as well as my own in the spoiler.

If n is a positive integer, is \frac{1}{10}^n < 0.01?

1) n > 2

2) \frac{1}{10}^{n-1} < 0.1

OG Explanation: Manipulate both sides to be expressed as powers of 10.

\frac{1}{10}^n < 0.01

(10^{-1})^n < 10^{-2}

10^{-n} < 10^{-2}

n > 2

1)n > 2. SUFFICIENT

2)\frac{1}{10}^{n-1} < 0.1

(10^{-1})^{n-1} < 10^{-1}

10^{-n+1} < 10^{-1}

-n+1 < -1

n > 2 SUFFICIENT

My slightly modified solution for statement 2 was to first manipulate the 0.1 on the right side of the inequality to become a fraction and to leave the left side as a fraction (my first instinct is to see that 0.01 is the same as 1/10). You would have:

\frac{1}{10}^{n-1} < \frac{1}{10}^1

n-1 < 1

n < 2

As you can see, I get an opposite answer. I know this is super simple, but where am I going wrong?

Welcome to GMAT Club. Below is an answer to your question.

From (\frac{1}{10})^{n-1} < (\frac{1}{10})^1 since the base, 1/10, is a fraction in the range (0,1) then it should be n-1>1. For example: (\frac{1}{10})^{2} < (\frac{1}{10})^1 --> 2>1.

Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]
16 Mar 2012, 16:07

Thanks for the quick response. I knew that you had to switch the inequality sign if you were multiplying or dividing by a negative value, but the same is also true when working with a value 0 < x < 1?

I'm a little confused because in this case, we're not doing any multiplying or dividing to the equation, but just ignoring the common base and comparing their exponents. Not quite sure why we change the sign

Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]
09 Aug 2012, 00:08

Hi Bunuel,

I actually landed up doing in the exact same way as tkaelle did. I understand what you mentioned .. however, is there any rule because I am sure I might up land up doing the same in the exam if I do not understand why we switch the signs/ or why cant we manipulate the fraction 1/10 and continue?

Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]
09 Aug 2012, 00:34

Expert's post

pavanpuneet wrote:

Hi Bunuel,

I actually landed up doing in the exact same way as tkaelle did. I understand what you mentioned .. however, is there any rule because I am sure I might up land up doing the same in the exam if I do not understand why we switch the signs/ or why cant we manipulate the fraction 1/10 and continue?

Thank you.

It's not about switching sign.

If you have a problem with fractions in powers, then manipulate to get rid of the them: (\frac{1}{10})^{n-1} < \frac{1}{10} --> \frac{1}{10^{n-1}}< \frac{1}{10} --> cross-multiply: 10<10^{n-1} --> 1<n-1 --> n>2.