Simple Atomic ratio question?

Question

The element indium has an atomic mass of 114.8 g and an atomic number of 49. Naturally occurring
indium contains a mixture of indium-112 and indium-115, respectively, in an atomic ratio of
approximately

(a) 6/94 (b) 25/75 (c) 50/50 (d) 75/25 (e) 94/6

I'd appreciate it if someone could show me how to do this.

Answer 1

We'll say there is x% of indium-112 is ocurring naturally.then(100-x)% of indium-115 is there.

114.8={112 * x%+ 115 * (100-x)%} / 100
when solve this equation you'll get 6.6. for X.
so (100-x)= 100- 6.6= 93.4
ratio=indium 112/ indium 115 = 6.6 / 93.4
approximately it is 6/94
answer is (a)

Answer 2

The atomic mass of 114.8 is much closer to 115 than to 112, so we know that the natural abundance of In-115 is much higher than that of In-112. By a quick calculation, 114.8 is 1/15 of the way from 115 down to 112, so the abundance of In-112 is 1/15 or 0.0667, which is 6.67%, leaving 93.33% for In-115, or a ratio of 7/93. Accounting for rounding, the clear anser is (a) 6/94, the only one close to this result.