Please help, how do i solve this chemistry question?

Question

K-40 decays to Ar-40 with a half life of 1.27 x 10^9 yrs. The ratio of Ar to K in the rock is 0.812. How old is the rock?

Answer 1

Ar:K:total = 0.812:1:1.812

Present K:Original K = 1:1.812

N = No(e^(-t(0.6931/(1.27 x 10^9))))

N/No = e^(-t(0.6931/(1.27 x 10^9)))

1/1.812 = e^(-t(0.6931/(1.27 x 10^9)))

ln(1/1.812) = ln{e^(-t(0.6931/(1.27 x 10^9)))}

-ln(1.812) = -(t)(0.6931)/ (1.27 x 10^9)

0.59443 = (t)(0.6931)/ (1.27 x 10^9)

t = (1.27 x 10^9)*0.59443/(0.6931)

= 1.0892 x 10^9 years

Notes: For every gram of K there are 0.812 grams of Ar.

All the Ar was originally K. The original K was 1 + 0.812 = 1.812.