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If n is a positive integer, is (1/10)^n < 0.01? [#permalink]

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16 Mar 2012, 10:04

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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?

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]

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16 Mar 2012, 17:07

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]

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09 Aug 2012, 01:08

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?

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: \(10<10^{n-1}\) --> \(1<n-1\) --> \(n>2\).

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

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13 Jan 2015, 20:40

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This DS question is essentially about arithmetic rules (decimals and exponents). It can be solved conceptually or you solve it by TESTing VALUES.

We're told that N is a POSITIVE INTEGER. We're asked (1/10)^N < 0.01 This is a YES/NO question

Fact 1: N > 2

IF.... N = 3 (1/10)^3 = 1/1000 = .001 and the answer to the question is YES

N = 4 (1/10)^4 = 1/10,000 = .0001 and the answer to the question is YES

As N gets bigger, the resulting decimal point gets smaller and the answer to the question is ALWAYS YES. Fact 1 is SUFFICIENT

Fact 2: (1/10)^(N-1) < 0.1

Here, we have an interesting "restriction" - we can only use certain values for N....

IF.... N = 1 (1/10)^0 = 1 BUT this does NOT fit with the given information in Fact 2, so we CANNOT use this TEST CASE.

IF.... N = 2 (1/10)^1 = .1 BUT this also does NOT fit with the given information in Fact 2 EITHER.

This means that N CANNOT be 1 or 2. Since it has to be a POSITIVE INTEGER and we already have proof of what happens when N > 2 (the answer to the question is ALWAYS YES), we can stop working. Fact 2 is SUFFICIENT

You should use information not only from task but from statement too. When you use \(n = 1\) you break condition of second statement \((\frac{1}{10})^{(n-1)} <0.1\) and this mean that you can't use such value for \(n\) So you should combine conditions from task and statement.

This mean that you can take only those values for \(n\) that will satisfy condition of task and statement.

So when \(n = 1\) then it will be \((\frac{1}{10})^{(1-1)} <0.1\); \(\frac{1}{1}<0.1\); not possible

when \(n = 2\) then it will be \((\frac{1}{10})^{(2-1)} <0.1\); \(\frac{1}{10}<0.1\); not possible

when \(n = 3\) then it will be \((\frac{1}{10})^{(3-1)} <0.1\); \(\frac{1}{10}<0.1\); possible So \(n\) should be at least \(3\) _________________

It's good that you're being inquisitive about your mistakes. Analyzing the mistake you make once ensures that you don't ever make it again.

The part that you did wrong here that you considered only one piece of information about n: that n is a positive integer (this is given in the ques statement)

So, you thought that you could take n = 1 or 2. And then, got confused when these values of n did not satisfy the information given in Statements 1 and 2.

The correct way of talking about n is:

n is a positive integersuch that (1/10)^(n-1) <0. (info in St. 2)

OR

n is a positive integer such that n > 2 (info in St. 1)

So, the possible values of n will be those that satisfy:

i) The information given in question statement AND ii) The information given in Statements 1 and 2

You should use information not only from task but from statement too. When you use \(n = 1\) you break condition of second statement \((\frac{1}{10})^{(n-1)} <0.1\) and this mean that you can't use such value for \(n\) So you should combine conditions from task and statement.

This mean that you can take only those values for \(n\) that will satisfy condition of task and statement.

So when \(n = 1\) then it will be \((\frac{1}{10})^{(1-1)} <0.1\); \(\frac{1}{1}<0.1\); not possible

when \(n = 2\) then it will be \((\frac{1}{10})^{(2-1)} <0.1\); \(\frac{1}{10}<0.1\); not possible

when \(n = 3\) then it will be \((\frac{1}{10})^{(3-1)} <0.1\); \(\frac{1}{10}<0.1\); possible So \(n\) should be at least \(3\)

Thanks for the Explanation!! Did not concentrate enough!

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

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30 Apr 2016, 13:45

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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

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14 May 2016, 07:47

Here is my take:

We are given that n is a positive integer and need to determine whether (1/10)^n < 0.01. We can convert 0.01 to a fraction and display the question as:

Is (1/10)^n < 1/100 ?

Is (1/10)^n < (1/10)^2 ?

Using the negative exponent rule, we can take the reciprocal of our bases and switch the signs of the exponents.

Is 10^-n < 10^-2 ?

Because the bases are now the same, we equate the exponents.

Is -n < -2 ?

Is n > 2 ?

Statement One Alone:

n > 2

We see that statement one directly answers the question. We can eliminate answer choices B, C, and E.

Statement Two Alone:

(1/10)^(n-1) < 0.1

We can simplify the inequality in statement two.

(1/10)^(n-1) < 0.1

(1/10)^(n-1) < (1/10)^1

Using the negative exponent rule, we can take the reciprocal of our bases and switch the signs of the exponents.

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

The bases are now equal, so we can equate the exponents.

-(n – 1) < -1

n – 1 > 1

n > 2

We see that n is greater than 2. Statement two alone is sufficient to answer the question.

Answer: D _________________

Jeffrey Miller Scott Woodbury-Stewart Founder and CEO

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

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24 Aug 2016, 09:26

I had a somewhat similar approach for statement 2, but I landed up with an answer contradicting 1.

I considered 0.1 as 1/10

so then the equation is: n-1 < 1 (because the base is the same i.e. 1/10)

therefore n < 2...where am I going wrong?

Bunuel wrote:

pavanpuneet wrote:

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: \(10<10^{n-1}\) --> \(1<n-1\) --> \(n>2\).

If you choose to 'convert' the numbers into decimals, then that's fine, but you still have to remember how the rules of math 'work'

Which number is bigger: (1/10)^1 or (1/10)^2? WHY is that the case?

Since we're working with an inequality, we're bound by the rule that one value MUST be bigger than the other. If you're ever unsure about an 'algebraic' approach that you're using, you can easily verify whether you're correct or not by TESTing VALUES.

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