Relative Atomic Mass of Boron = [19.9% 10^{10}10B} + [80.1% 11^{11}11B]
= [ 19.9% of 10] + [80.1% of 11]
= [ 19.9100×10\;\frac{19.9}{100}\times 1010019.9×10 ] + [ 80.1100×11\;\frac{80.1}{100}\times 1110080.1×11]
= 1.99 + 8.811
= 10.801
