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.
