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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, 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.

[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, 10:32
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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, 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

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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, 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?

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, 01:34
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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, 09:59
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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, 13:13
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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22 Dec 2013, 00:18
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, 04:48
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, 20:52
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 Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Kudos [?]: 3407 [0], given: 172 Intern Joined: 23 Sep 2014 Posts: 37 Kudos [?]: 25 [0], given: 210 Location: India Concentration: Marketing, Finance GMAT 1: 670 Q48 V34 Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink] ### Show Tags 28 Apr 2015, 00:38 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 Kudos [?]: 25 [0], given: 210 Math Forum Moderator Joined: 06 Jul 2014 Posts: 1272 Kudos [?]: 2312 [2], given: 178 Location: Ukraine Concentration: Entrepreneurship, Technology GMAT 1: 660 Q48 V33 GMAT 2: 740 Q50 V40 Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink] ### Show Tags 28 Apr 2015, 00:51 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 $$3$$ _________________ Kudos [?]: 2312 [2], given: 178 e-GMAT Representative Joined: 04 Jan 2015 Posts: 746 Kudos [?]: 2079 [1], given: 123 Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink] ### Show Tags 28 Apr 2015, 01:04 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 Kudos [?]: 2079 [1], given: 123 Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 1634 Kudos [?]: 837 [0], given: 2 Location: United States (CA) Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink] ### Show Tags 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 _________________ Scott Woodbury-Stewart Founder and CEO GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Kudos [?]: 837 [0], given: 2 Manager Joined: 28 Apr 2016 Posts: 100 Kudos [?]: 7 [0], given: 79 Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink] ### Show Tags 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: $$10<10^{n-1}$$ --> $$1<n-1$$ --> $$n>2$$. 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. Kudos [?]: 7 [0], given: 79 EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 9980 Kudos [?]: 3407 [0], given: 172 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink] ### Show Tags 24 Aug 2016, 14:11 Hi ameyaprabhu, 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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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]

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23 Sep 2016, 11:03
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.

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

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04 Jan 2017, 06:20
Bunuel wrote:
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.

Hi Bunuel

Is this the an efficient way to solve:

n>0

To find: (0.1)^n<0.01

This is only possible if n >2, otherwise (0.1)^n would be equal to or greater than 0.01.

Statement 1: n>2
Sufficient

Statement 2:
(0.1)^n-1 < 0.1

= ((0.1)^n)/(0.1) <0.1

=0.1^n< 0.1^2

= n<2

Sufficient. Same information as A.

Therefore D.

Is this correct?

Thanks.

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

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22 Jan 2017, 09:00
Hello All,

I did it as below:

S-1 - n>2 means (1/10)^n is atleast 0.0001 which is less than 0.01. Sufficient
S-2 - Separating the n-1 in exponents as => (1/10)^n*(1/10)^-1. This is (1/10)^n* 10 < 0.1. Divinding both sides by 10.
Statement itself gives (1/10)^n < 0.01. Sufficient
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If n is a positive integer, is (1/10)^n < 0.01? [#permalink]

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14 Feb 2017, 09:07
tkaelle wrote:

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

Please, CAN SOMEBODY EXPLAIN how ALL of you interpreted

$$\frac{1}{10}^n< 0.01$$ AS $$(\frac{1}{10})^n< 0.01$$

the original question implies that only the numerator (the value 1) of the fraction is raised to the power of "n" and NOT THE WHOLE FRACTION

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If n is a positive integer, is (1/10)^n < 0.01?   [#permalink] 14 Feb 2017, 09:07

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