If n is a positive integer, is (1/10)^n < 0.01? : GMAT Data Sufficiency (DS)
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# If n is a positive integer, is (1/10)^n < 0.01?

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

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16 Mar 2012, 09:04
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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.

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

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16 Mar 2012, 09:32
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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?

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

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

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

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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?

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

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09 Aug 2012, 00:34
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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.

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: $$10<10^{n-1}$$ --> $$1<n-1$$ --> $$n>2$$.

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

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

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15 Mar 2013, 08:59
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given 10^(-n)<10^(-2)
n>2 ?

1) n>2 suff

2) 10^(1-n)<10^(-1)
1-n<-1
n>2 suff

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

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18 Mar 2013, 12:13
Great explanation Bunuel, thanks.
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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]

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21 Dec 2013, 23:18
Hi !

I have an issue with the 2nd equation :

(1/10)^n-1 < 0.1

What I would do is (1/10)^n-1 < 1/10)

n-1 < 1

My qn. is why is it necassary to change the sign to make n> 2??
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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]

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22 Dec 2013, 03:48
mba1010 wrote:
Hi !

I have an issue with the 2nd equation :

(1/10)^n-1 < 0.1

What I would do is (1/10)^n-1 < 1/10)

n-1 < 1

My qn. is why is it necassary to change the sign to make n> 2??

Please check here: if-n-is-a-positive-integer-is-1-10-n-129176.html#p1059737 and here: if-n-is-a-positive-integer-is-1-10-n-129176.html#p1111563

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

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13 Jan 2015, 19:40
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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]

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13 Jan 2015, 19:52
Hi All,

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

[Reveal] Spoiler:
D

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# Special Offer: Save $75 + GMAT Club Tests 60-point improvement guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Intern Joined: 23 Sep 2014 Posts: 39 Location: India Concentration: Marketing, Finance GMAT 1: 610 Q46 V29 GMAT 2: 560 Q44 V25 GMAT 3: 670 Q48 V34 Followers: 1 Kudos [?]: 22 [0], given: 210 Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink] ### Show Tags 27 Apr 2015, 23:38 The question says that n is any positive number. So when I use n=2 in the second statement then both the equations become equal and the equality fails i.e 0.1<0.1 doesn't hold, Similarly when i use n=1, then 1<0.1 again a different answer and when n=3, then it satisfies. So how come answer is D?? I know i am doing something wrong. Would hope to have a reasoning which can clarify where i am going wrong. TIA Math Forum Moderator Joined: 06 Jul 2014 Posts: 1274 Location: Ukraine Concentration: Entrepreneurship, Technology GMAT 1: 660 Q48 V33 GMAT 2: 740 Q50 V40 Followers: 153 Kudos [?]: 1797 [2] , given: 178 Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink] ### Show Tags 27 Apr 2015, 23:51 2 This post received KUDOS believer700 wrote: The question says that n is any positive number. So when I use n=1 in the second statement then both the equations become equal and the equality fails i.e 0.1<0.1 doesn't hold. So how come answer is D?? I know i am doing something wrong. Would hope to have a reasoning which can clarify where i am going wrong. TIA Hello believer700 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 $$3$$ _________________ e-GMAT Representative Joined: 04 Jan 2015 Posts: 513 Followers: 139 Kudos [?]: 1140 [1] , given: 90 Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink] ### Show Tags 28 Apr 2015, 00:04 1 This post received KUDOS Expert's post believer700 wrote: The question says that n is any positive number. So when I use n=2 in the second statement then both the equations become equal and the equality fails i.e 0.1<0.1 doesn't hold, Similarly when i use n=1, then 1<0.1 again a different answer and when n=3, then it satisfies. So how come answer is D?? I know i am doing something wrong. Would hope to have a reasoning which can clarify where i am going wrong. TIA Dear believer700 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 integer such 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 I hope this clarifies your doubt! - Japinder _________________ | '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com Intern Joined: 23 Sep 2014 Posts: 39 Location: India Concentration: Marketing, Finance GMAT 1: 610 Q46 V29 GMAT 2: 560 Q44 V25 GMAT 3: 670 Q48 V34 Followers: 1 Kudos [?]: 22 [0], given: 210 Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink] ### Show Tags 28 Apr 2015, 01:57 Harley1980 wrote: believer700 wrote: The question says that n is any positive number. So when I use n=1 in the second statement then both the equations become equal and the equality fails i.e 0.1<0.1 doesn't hold. So how come answer is D?? I know i am doing something wrong. Would hope to have a reasoning which can clarify where i am going wrong. TIA Hello believer700 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! GMAT Club Legend Joined: 09 Sep 2013 Posts: 13529 Followers: 577 Kudos [?]: 163 [0], given: 0 Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink] ### Show Tags 30 Apr 2016, 12:45 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. _________________ Director Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 516 Location: United States (CA) Followers: 21 Kudos [?]: 195 [0], given: 2 Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink] ### Show Tags 14 May 2016, 06:47 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 Intern Joined: 28 Apr 2016 Posts: 48 Followers: 0 Kudos [?]: 1 [0], given: 60 Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink] ### Show Tags 24 Aug 2016, 08:26 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: $$10<10^{n-1}$$ --> $$1<n-1$$ --> $$n>2$$. OR: $$(\frac{1}{10})^{n-1} < \frac{1}{10}$$ --> $$(10^{-1})^{n-1}<10^{-1}$$ --> $$10^{1-n}<10^{-1}$$ --> $$1-n<-1$$ --> $$n>2$$. Hope it helps. 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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]

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23 Sep 2016, 10:03
Here is how I solved this question
The question stem states that[b] (1/10))^n<0.01
(1/10)^n < (1/10)^2
Cross multiplying, we get
(10)^2< 10^n
The above expression is possible only when n is greater than 2
Therefore our question becomes: Is n>2 ?

Statement 1: n>2 Sufficient
Statement 2: (1/10)^n-1< 0.1
(1/10)^n-1< 1/10
10< 10^n-1
10< 10^n/10
Cross multiplying, we get
10^2< 10^n
Therefore, n>2

I hope my approach is correct.
Re: If n is a positive integer, is (1/10)^n < 0.01?   [#permalink] 23 Sep 2016, 10:03

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