The atomic mass of element a is 63.6 atomic mass units. the only naturally occurring isotopes of element a are a-63 and a-65. the percent abundances in a naturally occurring sample of element a are closest to
(1) 31% a-63 and 69% a-65
(2) 50% a-63 and 50% a-65
(3) 69% a-63 and 31% a-65
(4) 100% a-63 and 0% a-65

Let x (%) be the abundance of A-63 and y (%) be the abundance of A-65.

x (%) + y (%) = 100 (%)
⇒ x = 100-y

We have: α= ( 63*x + 65*y) / 100 (in which α is the atomic mass of element A).
⇒ 63.6 = ( 63* (100-y) + 65*y) / 100 (replace x with 100-y).
⇒ 63.6*100 = ( 63* (100-y) + 65*y) * (100/100)
⇒ 6,360 = 6,300 - 63*y + 65*y (distributive property)
⇒ 6,360- 6,300= (6,300-6,300) + 2*y
⇒ 60 = 2*y
⇒ 60/2 = (2/2)*y
⇒ 30 = y
The abundance of A-65 is about 30%.

x = 100-30 = 70.
The abundance of A-63 is about 70%.

Conclusion: (3) 69% A-63 and 31% A-65 is the correct answer.

Hope this would help~

Your question is not clear but if you can explain yourself i would be glad to .  just trying to : ))

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