2015 AMC 10A Problem 8

Below is the professionally curated solution for Problem 8 of the 2015 AMC 10A, 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.

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Concepts:agessystem of equations

Difficulty rating: 1280

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 22 : 11 ?

22

44

55

66

88

Solution:

Let pp and cc be Pete's and Claire's current ages respectively.

Then we have that p2=3(c2) p - 2 = 3(c - 2) and p4=4(c4). p - 4 = 4(c - 4). Simplifying both equations gives us p=3c4 p = 3c - 4 and p=4c12. p = 4c - 12. Setting them equal, we have 3c4=4c12 3c - 4 = 4c - 12 c=8. c = 8.

This means that p=384=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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