A radio isotope X with a half life 1.4 × 10^8 year, decays to Y,...

Let the initial number of atoms of X be $N_0$.

Since X decays to Y, the number of atoms of Y formed is $N_Y = N_0 - N_X$, where $N_X$ is the number of atoms of X remaining.

Given the ratio of X to Y is $1 : 7$, we have $\dfrac{N_X}{N_Y} = \dfrac{1}{7}$.

Substituting $N_Y = N_0 - N_X$, we get $\dfrac{N_X}{N_0 - N_X} = \dfrac{1}{7}$.

$7N_X = N_0 - N_X \Rightarrow 8N_X = N_0 \Rightarrow N_X = \dfrac{N_0}{8}$.

The amount of X remaining can be written as $N_X = N_0 \left(\dfrac{1}{2}\right)^n$, where $n$ is the number of half-lives.

$\dfrac{N_0}{8} = N_0 \left(\dfrac{1}{2}\right)^n \Rightarrow \left(\dfrac{1}{2}\right)^3 = \left(\dfrac{1}{2}\right)^n \Rightarrow n = 3$.

The age of the rock is $t = n \times T_{1/2}$.

$t = 3 \times 1.4 \times 10^8 = 4.20 \times 10^8$ years.

Answer: $4.20 \times 10^8$