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If n is a positive integer, is (1/10)^n < .01?

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If n is a positive integer, is (1/10)^n < .01? [#permalink] New post 22 Jun 2008, 23:17
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OG C DS 144
If n is a positive integer, is (1/10)^n < .01?
a. n>2
b. (1/10)^(n-1) <0.1
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Re: OG C DS 144 [#permalink] New post 22 Jun 2008, 23:23
If n is a positive integer, is (1/10)^n < .01?
a. n>2
b. (1/10)^(n-1) <0.1


And: D

(1/10)^n < .01 if & only if (1/10)^n < (1/10)^2 if & only if n<2

St1: Is sufficient!

In st2: (1/10)^(n-1) <0.1 - multiply by a 10 on both sides we get St1. Hence sufficint
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Re: OG C DS 144 [#permalink] New post 23 Jun 2008, 04:07
D for me as well. simplify the stem to ask if n is greater than 2.

stat 1 tells you n>2. suff.

stat 2 can be simplified to (1/10)^n < 1/100 ... this is what the stem is asking. suff.
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Re: OG C DS 144 [#permalink] New post 23 Jun 2008, 04:37
A for me - By (2) could be n= 2, thus (1/10)^n=0.01
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Re: OG C DS 144 [#permalink] New post 23 Jun 2008, 05:16
OG C DS 144
If n is a positive integer, is (1/10)^n < .01?
a. n>2
b. (1/10)^(n-1) <0.1


First lets make the question simpler...
n is a postive integer....so n can be 1, 2, 3.....
(1/10)^n < .01 ? ===> 10^-n < .01? ===== > 10^-n < 10^-2 ?
so the question now becomes ----- > is 10 raised to -n less than 10 raised to -2

a. n > 2====> since n can only be an integer, lets take the value of n as 3.
so, substitiuting n = 3,

10^-3 < 10^-2 i.e. 10^-3 IS less than 10^-2 .... so statement 1 is sufficient...

b. (1/10)^(n-1) <0.1
simplying statement 2 goes like this..:

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

(10)^-n+1 <0.1 =====> (10)^-n <0.01 ..ie. this is what the question was asking for...

so statement 2 is alos sufficient...

so answer is D........
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Re: OG C DS 144 [#permalink] New post 20 Mar 2011, 13:03
Hoping someone can please shed some light on this as I'm seeing conflicting answers in different threads.

OG C DS 144
If n is a positive integer, is \frac{1}{10}^n < 0.01?
a. n>2
b. \frac{1}{10}^{n-1} < 0.1

Taking the stem, I change it to \frac{1}{10}^n < \frac{1}{100}. For this to be true, n > 2. So the question is asking... is n > 2?

A. n > 2
Sufficient based on manipulation of stem.

B. \frac{1}{10}^{n-1} < 0.1
This is where it gets really confusing. There are two ways I can go about doing this resulting in different answers...

Method 1
\frac{1}{10}^{n-1} < 0.1
Change the bases into decimals.
0.1^{n-1} < 0.1^1
The bases on both sides of the inequality match, so I focus on the exponents:
n-1 < 1
n < 2
This is sufficient, but this answer conflicts with A. From what I understand, this should never happen on the GMAT.

Method 2
\frac{1}{10}^{n-1} < 0.1
Change the bases into whole numbers:
(10^{-1})^{n-1} < 10^{-1}
Now that the bases are the same on both sides, I focus on the exponents:
-1(n-1) < -1
-n+1 < -1
-n < -2
n > 2
This is sufficient and this answer agrees with A.

Method 2 is the explanation given in the OG. But why does Method 1 (my approach) give a different answer? What am I doing wrong?
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Re: OG C DS 144 [#permalink] New post 20 Mar 2011, 16:57
why it is not A? St1 is sufficient, but for St2, it can be 1 or .1 or .01 and so on.
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Re: OG C DS 144 [#permalink] New post 21 Mar 2011, 23:04
brandy96 wrote:
why it is not A? St1 is sufficient, but for St2, it can be 1 or .1 or .01 and so on.


(1/10)^(n-1) <0.1
=>(1/10)^n *10<0.1
=>(1/10)^n<0.01
=>(1/10)^n is always less than 0.01 and cant be 1 or .1 or .01 and so on.
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Re: OG C DS 144 [#permalink] New post 22 Mar 2011, 18:44
By Rephrasing the given expression , we now have to see if n>2?


1. Sufficient . Exactly we had to find
2. Sufficient. By rephrasing we see its the same expression.

Answer is D.
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Re: OG C DS 144 [#permalink] New post 24 Mar 2011, 11:05
GSDster wrote:
Hoping someone can please shed some light on this as I'm seeing conflicting answers in different threads.

OG C DS 144
If n is a positive integer, is \frac{1}{10}^n < 0.01?
a. n>2
b. \frac{1}{10}^{n-1} < 0.1

Taking the stem, I change it to \frac{1}{10}^n < \frac{1}{100}. For this to be true, n > 2. So the question is asking... is n > 2?

A. n > 2
Sufficient based on manipulation of stem.

B. \frac{1}{10}^{n-1} < 0.1
This is where it gets really confusing. There are two ways I can go about doing this resulting in different answers...

Method 1
\frac{1}{10}^{n-1} < 0.1
Change the bases into decimals.
0.1^{n-1} < 0.1^1
The bases on both sides of the inequality match, so I focus on the exponents:
n-1 < 1
n < 2
This is sufficient, but this answer conflicts with A. From what I understand, this should never happen on the GMAT.

Method 2
\frac{1}{10}^{n-1} < 0.1
Change the bases into whole numbers:
(10^{-1})^{n-1} < 10^{-1}
Now that the bases are the same on both sides, I focus on the exponents:
-1(n-1) < -1
-n+1 < -1
-n < -2
n > 2
This is sufficient and this answer agrees with A.

Method 2 is the explanation given in the OG. But why does Method 1 (my approach) give a different answer? What am I doing wrong?


Anyone? :( I'd really appreciate an explanation because clearly something is wrong with my logic.
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Re: OG C DS 144 [#permalink] New post 24 Mar 2011, 15:27
GSDster wrote:
GSDster wrote:
Hoping someone can please shed some light on this as I'm seeing conflicting answers in different threads.

OG C DS 144
If n is a positive integer, is \frac{1}{10}^n < 0.01?
a. n>2
b. \frac{1}{10}^{n-1} < 0.1

Taking the stem, I change it to \frac{1}{10}^n < \frac{1}{100}. For this to be true, n > 2. So the question is asking... is n > 2?

A. n > 2
Sufficient based on manipulation of stem.

B. \frac{1}{10}^{n-1} < 0.1
This is where it gets really confusing. There are two ways I can go about doing this resulting in different answers...

Method 1
\frac{1}{10}^{n-1} < 0.1
Change the bases into decimals.
0.1^{n-1} < 0.1^1
The bases on both sides of the inequality match, so I focus on the exponents:
n-1 < 1
n < 2
This is sufficient, but this answer conflicts with A. From what I understand, this should never happen on the GMAT.

Method 2
\frac{1}{10}^{n-1} < 0.1
Change the bases into whole numbers:
(10^{-1})^{n-1} < 10^{-1}
Now that the bases are the same on both sides, I focus on the exponents:
-1(n-1) < -1
-n+1 < -1
-n < -2
n > 2
This is sufficient and this answer agrees with A.

Method 2 is the explanation given in the OG. But why does Method 1 (my approach) give a different answer? What am I doing wrong?


Anyone? :( I'd really appreciate an explanation because clearly something is wrong with my logic.


GSD, the Method 2 deals with whole numbers, Method 1 is deals with decimals/fractions. A fraction raised to the (n+1) power is smaller than that same fraction raised to the n power (assuming n is positive), whereas a whole number raised to the (n+1) power is larger than that same number raised to the n power. That's why you're seeing the conflicting answers.
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Re: OG C DS 144 [#permalink] New post 24 Mar 2011, 16:49
jko wrote:
GSD, the Method 2 deals with whole numbers, Method 1 is deals with decimals/fractions. A fraction raised to the (n+1) power is smaller than that same fraction raised to the n power (assuming n is positive), whereas a whole number raised to the (n+1) power is larger than that same number raised to the n power. That's why you're seeing the conflicting answers.


Thank you for the response. Do you mind elaborating on this? I understand the concept that when you take any number between 0 and 1 and raise it to a power, the value will decrease. But I still don't understand how it matters in this particular context. The stem was not manipulated to use whole numbers as the base, so why does Method 2 (using whole numbers) arrive at the correct conflusion? Are there any general rules about flipping inequalities when dealing with numbers between 0 and 1?
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Re: OG C DS 144 [#permalink] New post 24 Mar 2011, 17:35
Sure, so look at your process here:

(1) \frac{1}{10}^{n-1} < 0.1
Change the bases into decimals.
(2) 0.1^{n-1} < 0.1^1

The bases on both sides of the inequality match, so I focus on the exponents:
(3) n-1 < 1


Going from step (2) to step (3) is not logically or mathematically correct.

If: 0.1^{n-1} < 0.1^1
Then: n-1 > 1, since the higher the exponent of 0.1, the lower the resulting decimal.

For example, if n-1 WAS less than 1, you can see that 0.1^{n-1} > 0.1^1
If n-1 = 0, which is less than 1, then 0.1^{0} = 1 > 0.1^1
If n-1 = -1, which is less than 1, then 0.1^{-1} = 10 > 0.1^1
If n-1 = 1, which is greater than 1, then 0.1^{-1} = .01 < 0.1^1
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Re: OG C DS 144 [#permalink] New post 25 Mar 2011, 04:24
redbeanaddict wrote:
OG C DS 144
If n is a positive integer, is (1/10)^n < .01?
a. n>2
b. (1/10)^(n-1) <0.1


(1) n is greater than 2 that is at least 3 so 1 is divided by 1000 (1/1000=.001) less than 0.01 suff.
(2) if n=2 then left=right part, if n=3 then left part (.01) <0.1, n>2 suff.

Ans D.
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Re: OG C DS 144 [#permalink] New post 25 Mar 2011, 19:18
Thanks Onell, I did a quick check and didn't read in detail.
Re: OG C DS 144   [#permalink] 25 Mar 2011, 19:18
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