2003 AMC 10A Problem 13

Attempt Problem 13 of the 2003 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 2003 AMC 10A solutions, or check the answer key.

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13.

The sum of three numbers is 20.20. The first is 44 times the sum of the other two. The second is seven times the third. What is the product of all three?

2828

4040

100100

400400

800800

Answer: A
Concepts:system of equationssubstitution
Difficulty rating: 1310
Solution:

Let the numbers be a,a, b,b, c.c. Since a=4(b+c),a = 4(b + c), we get 4(b+c)+(b+c)=20,4(b + c) + (b + c) = 20, so b+c=4b + c = 4 and a=16.a = 16.

With b=7c,b = 7c, we have 7c+c=4,7c + c = 4, so c=12c = \dfrac{1}{2} and b=72.b = \dfrac{7}{2}.

The product is 167212=28.16 \cdot \dfrac{7}{2} \cdot \dfrac{1}{2} = 28.

Thus, the correct answer is A.

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