A sample of Boron has Boron 10 and Boron 11. The relative atomic mass of the sample is 10.8, therefore you should recognise by simple mathematical reasoning that this is likely to be 80% of Boron 11 and 20% of Boron 10. (0.8 x 11 + 0.2 x 10 = 10.8). Therefore the percentage of Boron 11 in the sample is 80%, which is option C.
A sample of Boron has Boron 10 and Boron 11. The relative atomic mass of the sample is 10.8, therefore you should recognise by simple mathematical reasoning that this is likely to be 80% of Boron 11 and 20% of Boron 10. (0.8 x 11 + 0.2 x 10 = 10.8). Therefore the percentage of Boron 11 in the sample if 80%, which is option C.
So x= 20% x is the percentage abundance for isotope 1 i.e. B10 Thus the percentage abundance for B11 will be 80%
Which is basically what the above comment did, sometimes having a method is easier than spotting relationships between numbers even if it does take longer- at least that's what i find lol!