Boron has 2 isotopes B-10 B-11. Th average atomic mass of boron is found to be 10.80u. Calculate the percentage abundance of these isotopes.

Given Average Atomic Mass of B = 10.80 amu
Let the percentage of ¹⁰B isotope = x
Then the percentage of ¹¹B isotope = 100 - x

Average Atomic Mass Of B = $\frac{\mathrm{X}\times 10+\hspace{0.17em}(100-\mathrm{x})\times 11}{100}$
Thus
$$\begin{gathered} \frac{\mathrm{X}\times 10+\hspace{0.17em}(100-\mathrm{x})\times 11}{100}=10.80 \\ 10x+1100-11x=10.8\times 100 \\ 1100-x=1080 \\ -x=1080-1100 \\ -x=-20 \\ x=20 \end{gathered}$$

Thus percentage abundance of ¹⁰B = 20 %
and percentage abundance of ¹¹B = 100 - 20 % = 80 %