2003 AMC 10A Problem 13

Question

The sum of three numbers is

20.

The first is

4

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

28

40

100

400

800

Let the third number be

c;

then the second is

7c

The first number equals

4(b + c),

so

4(b + c) + (b + c) = 20

Let the numbers be

a,

b,

c.

Since

a = 4(b + c),

we get

4(b + c) + (b + c) = 20,

so

b + c = 4

and

a = 16.

With

b = 7c,

we have

7c + c = 4,

so

c = \dfrac{1}{2}

and

b = \dfrac{7}{2}.

The product is

16 \cdot \dfrac{7}{2} \cdot \dfrac{1}{2} = 28.

Thus, the correct answer is A.

Problem 13 in other years