1988 AMC 12 Problem 8

Attempt Problem 8 of the 1988 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1988 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

8.

If ba=2\frac ba=2 and cb=3,\frac cb=3, what is the ratio of a+ba+b to b+c?b+c?
13\frac13
38\frac38
35\frac35
23\frac23
34\frac34
Answer: B
Concepts:ratiossubstitutionproportions
Small hint:
Write both bb and cc as multiples of aa
Big hint:
Substitute those expressions into a+bb+c\frac{a+b}{b+c}
Solution:
We have b=2ab=2a and c=3b=6a.c=3b=6a. Therefore a+bb+c=3a8a=38.\frac{a+b}{b+c}=\frac{3a}{8a}=\frac38.
Thus the correct answer is B.
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