2015 AMC 10A Problem 8

Attempt Problem 8 of the 2015 AMC 10A 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 2015 AMC 10A solutions, or check the answer key.

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

8.

Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be 2:1?2:1?
22
44
55
66
88
Answer: B
Concepts:agessystem of equations
Small hint:
Translate the two age statements using current ages
Big hint:
Solve for current ages before setting the future ratio
Solution:
Let pp and cc be Pete’s and Claire’s current ages respectively.
Then we have that p−2=3(c−2) p - 2 = 3(c - 2) and p−4=4(c−4). p - 4 = 4(c - 4). Simplifying both equations gives us p=3c−4 p = 3c - 4 and p=4c−12. p = 4c - 12. Setting them equal, we have 3c−4=4c−12 3c - 4 = 4c - 12 c=8. c = 8.
This means that p=3⋅8−4=20. p = 3 \cdot 8 - 4 = 20.
Now, we need to find the number of years (yy) until 20+y=2(8+y), 20 + y = 2(8 + y), which gives us 20+y=16+2y 20 + y = 16 + 2y y=4. y = 4. Thus, B is the correct answer.
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