2003 AMC 10A Problem 13

Below is the professionally curated solution for Problem 13 of the 2003 AMC 10A, 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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Concepts:system of equationssubstitution

Difficulty rating: 1310

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

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.

Problem 13 in Other Years