Naturally occuring boron has two isotopes of masses 10 amu and 11 amu. Calculate the percentage of each isotope if the average atomic mass of boron is `10.2`.

Explanation

Let x be the fraction of `""^10B` isotope and `(1-x)` that of `""^11B` isotope, then, `(x)10+(1-x)11=10.2`
Fraction of `""^10B` isotope, `x=0.8 (" or "80%)` and fraction of `"^11B` isotope, `(1-x)=0.2(" or "20%)`.