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If n is a positive integer, is (1/10)^n < .01? [#permalink]
23 Jun 2008, 00:17
Question Stats:
25% (00:00) correct
75% (00:47) wrong based on 8 sessions
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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Manager
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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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SVP
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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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Manager
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A for me - By (2) could be n= 2, thus (1/10)^n=0.01
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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
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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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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Manager
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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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Manager
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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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Director
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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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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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Ross Thread Master
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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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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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Ross Thread Master
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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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Director
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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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I am student of everyone-baten Collections:- PSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html DS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html 100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html
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Thanks Onell, I did a quick check and didn't read in detail.
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