Naturally occurring Boron consists of two isotopes, whose atomic weights are 10.01 and 11.01. The atomic weight of natural boron is 10.81. The % of each isotope in natural boron is: 

Relative abundance of two isotopes of an element is the presence of isotope relative to each other. It is represented in relative percentage, let 'A' is the atomic mass of an element, then;

$\mathrm{A}\times 100={\mathrm{m}}_1{\mathrm{p}}_1+{\mathrm{m}}_2{\mathrm{p}}_2$

We have given;

$\mathrm{A}=10.81 \mathrm{g}$

${\mathrm{m}}_1=10.01 \mathrm{g}$

${\mathrm{m}}_2=11.01 \mathrm{g}$

We need to find p₁ and p₂. Let p₁ be 'x' then p₂ will be (100-x).

$10.81\times 100=10.01\mathrm{x}+(11.01\times 100)-11.01\mathrm{x}$

$\mathrm{x}=1101-1081=20\%$

$100-\mathrm{x}=80\%$

The relative abundance of boron isotope with 10.01 g is 20% and that of 11.01 g isotope is 80%.