Data Sufficiency

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

(1) n > 2
(2) (1/10)^(n-1) < 0.1[/img]

(1/10)^n < 1/100

meaning n > 2 as 1/1000 < 1/100 or so.

a sufficient.

b 10 * (1/10)^n < 0.1
means

(1/10)^n < 0.1/10

sufficient.

D it is.

sunilrawat wrote:If n is a positive integer, is (1/10)^n < 0.01 ?

(1) n > 2
(2) (1/10)^(n-1) < 0.1[/img]

Do some work on the question stem first. Turn (1/10)^n into 1^n / 10^n, which is basically 1/10^n
Turn 0.01 into 1/100, which is the equivalent of 1/10^2.

So now the question stem says "is 1/10^n < 1/10^2 ?

For the fraction on the left to be smaller than the fraction on the right, the denominator needs to be greater on the left than on the right; the question stem is really asking "is 10^n > 10^2?, or in other words, "is n greater than 2?"

Stat. (1) is therefore a clean "yes" answer, and is sufficient.

Stat. (2) a minus in the exponent means translates into a division: 10^n-1 is equal to 10^n/10^1, which in turn is equal to 10^n/10.

So (1/10)^n-1 is equal to 1/10^(n-1) = 1/ 10^n/10. Multiply the numerator 1 by the reciprocal of the denominator to get 1/1 * 10/10^n.

So stat. (2) says
10/10^n < 1/10.
Divide by 10 on both sides to get

10/10*10^n < 1/100

Reduce the 10s on the left side to get
1/10^n < 1/100 - which is the same as the question stem, privng that the answer is yes.

sunilrawat wrote:If n is a positive integer, is (1/10)^n < 0.01 ?

(1) n > 2
(2) (1/10)^(n-1) < 0.1[/img]

1) is straight forward
2) same as is requested just a small resemblence change in power.

in (2), why 0.1 cant be read as (1/10)^1 instead of (1/10)^-1
tht way we get n<2

awesome! thank you!

Geva@MasterGMAT wrote:

sunilrawat wrote:If n is a positive integer, is (1/10)^n < 0.01 ?

(1) n > 2
(2) (1/10)^(n-1) < 0.1[/img]

Do some work on the question stem first. Turn (1/10)^n into 1^n / 10^n, which is basically 1/10^n
Turn 0.01 into 1/100, which is the equivalent of 1/10^2.

So now the question stem says "is 1/10^n < 1/10^2 ?

For the fraction on the left to be smaller than the fraction on the right, the denominator needs to be greater on the left than on the right; the question stem is really asking "is 10^n > 10^2?, or in other words, "is n greater than 2?"

Stat. (1) is therefore a clean "yes" answer, and is sufficient.

Stat. (2) a minus in the exponent means translates into a division: 10^n-1 is equal to 10^n/10^1, which in turn is equal to 10^n/10.

So (1/10)^n-1 is equal to 1/10^(n-1) = 1/ 10^n/10. Multiply the numerator 1 by the reciprocal of the denominator to get 1/1 * 10/10^n.

So stat. (2) says
10/10^n < 1/10.
Divide by 10 on both sides to get

10/10*10^n < 1/100

Reduce the 10s on the left side to get
1/10^n < 1/100 - which is the same as the question stem, privng that the answer is yes.

sunilrawat wrote:in (2), why 0.1 cant be read as (1/10)^1 instead of (1/10)^-1
tht way we get n<2

You're right in your translation into 1/10^1, but wrong about the interpretation. Read my post - for a fraction to be smaller than another fraction, the denominator of the left hand side must be GREATER than the denominator on the right hand side. you get that (n-1) must be >1, so n is >2.

.01 = 1/100

1) 1/10^2 = 1/100. 1/10^3 = 1/1000 1/1000<1/100. (Sufficient)

2) 1/10^4 = 1/10000<1/100, 1/10^4-1=>1/10^3<1/100. (Sufficient)

Answer is D.