Radioactive decay

$\frac{N_x}{N_y} = \frac{1}{7} \Rightarrow \frac{N_x}{N_x + N_y} = \frac{1}{8} = \left(\frac{1}{2}\right)^3$
So $t = 3T_{1/2} = 3\times1.4\times10^9$ years $= 4.2\times10^9$ years

$\frac{N_x}{N_y} = \frac{1}{7} \Rightarrow \frac{N_x}{N_x + N_y} = \frac{N}{N_0} = \frac{1}{8}$
Using $N = N_0e^{-\lambda t}$: $\frac{N_0}{8} = N_0e^{-\lambda t}$, i.e. three half-lives have passed.
$t = 3\times20$ years = 60 years

At $t_1$, $\frac{1}{3}$ has decayed, so $\frac{2}{3}$ remains: $\frac{2}{3}N_0 = N_0e^{-\lambda t_1}$ ...(i)
At $t_2$, $\frac{2}{3}$ has decayed, so $\frac{1}{3}$ remains: $\frac{1}{3}N_0 = N_0e^{-\lambda t_2}$ ...(ii)
Dividing (i) by (ii): $2 = e^{\lambda(t_2 - t_1)}$
$\ln2 = \lambda(t_2 - t_1) \Rightarrow t_2 - t_1 = \frac{\ln2}{\lambda} = t_{1/2}$
$t_2 - t_1 = 50$ days

$A_1$ (half-life 20 s): 40 g → 20 g (20 s) → 10 g (40 s)
$A_2$ (half-life 10 s): 160 g → 80 g (10 s) → 40 g (20 s) → 20 g (30 s) → 10 g (40 s)
Both become 10 g at t = 40 s.
Check: $40\left(\frac{1}{2}\right)^{t/20} = 160\left(\frac{1}{2}\right)^{t/10} \Rightarrow 2^{t/20} = 4 \Rightarrow t = 40$ s

After time t (in minutes): $N_P = 4N_0\left(\frac{1}{2}\right)^{t/1}$ and $N_Q = N_0\left(\frac{1}{2}\right)^{t/2}$
Setting $N_P = N_Q$: $4\times2^{-t} = 2^{-t/2}$
$2^{2 - t} = 2^{-t/2}$, so $2 - t = -\frac{t}{2}$
$t = 4$ minutes
At t = 4 min: $N_P = \frac{4N_0}{2^4} = \frac{N_0}{4}$ and $N_Q = \frac{N_0}{2^2} = \frac{N_0}{4}$
Nuclei of P decayed = $4N_0 - \frac{N_0}{4} = \frac{15N_0}{4}$; nuclei of Q decayed = $N_0 - \frac{N_0}{4} = \frac{3N_0}{4}$
Every decayed nucleus becomes R: $N_R = \frac{15N_0}{4} + \frac{3N_0}{4} = \frac{18N_0}{4}$
$N_R = \frac{9N_0}{2}$

X : Y = 1 : 15. If 1 part of X remains, 15 parts have decayed into Y, so the original amount of X was 1 + 15 = 16 parts.
Fraction of X remaining: $\frac{N}{N_0} = \frac{1}{16}$
$\frac{N}{N_0} = \left(\frac{1}{2}\right)^n$, where n is the number of half-lives.
$\frac{1}{16} = \left(\frac{1}{2}\right)^4$, so n = 4
Age of the rock $t = n\times T_{1/2} = 4\times50$
t = 200 years

Activity decays as $A = A_0e^{-\lambda t}$
At t = 5 min: $\frac{N_0}{e} = N_0e^{-5\lambda}$, so $e^{-1} = e^{-5\lambda}$
$5\lambda = 1$, so $\lambda = \frac{1}{5}\ min^{-1}$ (the mean life is 5 minutes)
Half-life: $T_{1/2} = \frac{\log_e2}{\lambda}$
$T_{1/2} = 5\log_e2$ minutes