$(a)$ Two stable isotopes of lithium $_{3}^{6} Li$ and $_{3}^{7} Li$ have respective abundances of $7.5 \%$ and $92.5 \%$. These isotopes have masses $6.01512 \; u$ and $7.01600 \; u$,respectively. Find the atomic mass of lithium.
$(b)$ Boron has two stable isotopes,$_{5}^{10} B$ and $_{5}^{11} B$. Their respective masses are $10.01294 \; u$ and $11.00931 \; u$,and the atomic mass of boron is $10.811 \; u$. Find the abundances of $_{5}^{10} B$ and $_{5}^{11} B$.

(N/A) Mass of $_{3}^{6} Li$ isotope,$m_{1} = 6.01512 \; u$
Mass of $_{3}^{7} Li$ isotope,$m_{2} = 7.01600 \; u$
Abundance of $_{3}^{6} Li$,$n_{1} = 7.5 \%$
Abundance of $_{3}^{7} Li$,$n_{2} = 92.5 \%$
The atomic mass of lithium is given by:
$m = \frac{m_{1} n_{1} + m_{2} n_{2}}{n_{1} + n_{2}}$
$m = \frac{6.01512 \times 7.5 + 7.01600 \times 92.5}{100} = 6.940934 \; u$
$(b)$ Mass of $_{5}^{10} B$ isotope,$m_{1} = 10.01294 \; u$
Mass of $_{5}^{11} B$ isotope,$m_{2} = 11.00931 \; u$
Let abundance of $_{5}^{10} B$ be $x \%$,then abundance of $_{5}^{11} B$ is $(100 - x) \%$.
Atomic mass of boron,$m = 10.811 \; u$
$10.811 = \frac{10.01294 \times x + 11.00931 \times (100 - x)}{100}$
$1081.1 = 10.01294 x + 1100.931 - 11.00931 x$
$0.99637 x = 19.831$
$x \approx 19.9 \%$
Abundance of $_{5}^{10} B \approx 19.9 \%$ and abundance of $_{5}^{11} B \approx 80.1 \%$.