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OG 2016 DS 170

by nsuen » Fri Apr 29, 2016 6:20 am
If n is a positive integer, is ( 1/10) ^n <0.01
1) n>2
2) (1/10)^(n-1) <0.1

I got the answer correct, which is D, but my way is different and the greater than or less than sign is opposite of the official way ( underlined) so I want to make sure my logic is correct in order to master other similar questions. Thank you.

My way:
1) (1/10)^n < 1/100
(1/10)^n <(1/10)^2
n<2

2) (1/10)^(n-1) <(1/10)^1
n-1<1
n<2

Official answer:
1) (10^-1)^n <10^(-2)
-n<-2
n>2

2)(1/10)^(n-1) <0.1
(10^-1) ^(n-1)<10^(-1)
10^(-n+1)<10(-1)
-n+1<-1
n>2
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Source: — Data Sufficiency |
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by DavidG@VeritasPrep » Fri Apr 29, 2016 7:05 am
(1/10)^n <(1/10)^2
n<2
This move isn't valid. To see why, take a simple example: we know that (1/2)^3 < (1/2)^2, right? But obviously 3 < 2 isn't true! So we need to be careful with exponents in inequalities in a way we don't have to be careful with exponents in equations.
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by [email protected] » Fri Apr 29, 2016 9:11 am
Hi nsuen,

Sometimes the best way to tackle a Quant question (whether it's PS or DS) is to just write down all of the possibilities so that you can see the entire situation (and deduce whatever pattern might be hidden within the "math").

Here, we're asked "is (1/10)^N < .01?" This is YES/NO question. We're told that N is a POSITIVE INTEGER

Before dealing with the information in Fact 1, I'm going to do a quick series of minor calculations to see what pattern exists in this situation:

If N = 1, then (1/10)^1 = .1
If N = 2, then (1/10)^2 = 1/100 = .01
If N = 3, then (1/10)^3 = 1/1000 = .001
Etc.

So, the pattern is that as N gets bigger, more zeros appear between the decimal point and the 1.

In reference to the original question, we're essentially asked if N is less than, equal to, or greater than 2.

Fact 1: N > 2
So, N must be greater than 2 (it could be 3, 4, 5, etc.)

Each of those values would create more zeroes between the decimal point and the 1.
N = 3 gives us .001
N = 4 would be smaller and so on.
The answer to the question is ALWAYS YES.
Fact 1 is SUFFICIENT

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

While this might be "scary looking", it's based on the same pattern as before (we just need to adjust the math a bit)
N = 1 gives us 1
N = 2 gives us .1
N = 3 gives us .01
N = 4 gives us .001
Etc.

The information in Fact 2 tells us that N MUST be 3 or greater. This is the SAME information as what we had in Fact 1.
Fact 2 is SUFFICIENT.

Final Answer: D

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by Matt@VeritasPrep » Fri Apr 29, 2016 2:17 pm
If we have

Is (1/10)� < (1/10)² ?

we could approach it algebraically or conceptually.

Algebraically, I'd start with

Is 1/10� < 1/10² ?

Then, since you know both sides are positive, cross multiply

Is 10² < 10� ?

So the question becomes "Is n greater than 2?"

Conceptually, we could think of what happens with fractions between 0 and 1. Every time we multiply such fractions by themselves, they shrink! 1/10 > 1/100 > 1/1000 > ...

For 1/10� < 1/10² to hold, we'd need to multiply 1/10 by itself MORE than once, so we'd need n > 2.

Either way, same idea :)
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by nsuen » Fri Apr 29, 2016 11:15 pm
Thank you!

DavidG: Noticed, but then how come the official answer can get to n>2?

1) (10^-1)^n <10^(-2)
-n<-2
n>2
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by DavidG@VeritasPrep » Sun May 01, 2016 9:56 am
nsuen wrote:Thank you!

DavidG: Noticed, but then how come the official answer can get to n>2?

1) (10^-1)^n <10^(-2)
-n<-2
n>2
That's a valid move because the base, in this instance, is greater than 1. This is what makes inequalities with exponents tricky. If we have 2^x < 2^y, for example, we can say that x < y, but if we have (1/2)^x < (1/2)^y, then x > y. (This is the difference between the operation you performed and the Official Guide's solution.)
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