2003 AMC 10A Problems/Problem 13

Problem

The sum of three numbers is $20$ . The first is four times the sum of the other two. The second is seven times the third. What is the product of all three?

$\mathrm{(A) \ } 28\qquad \mathrm{(B) \ } 40\qquad \mathrm{(C) \ } 100\qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 800$

Solution

Let the numbers be $x$ , $y$ , and $z$ in that order.

$y=7z$

$x=4(y+z)=4(7z+z)=4(8z)=32z$

$x+y+z=32z+7z+z=40z=20$

$z=\frac{20}{40}=\frac{1}{2}$

$y=7z=7\cdot\frac{1}{2}=\frac{7}{2}$

$x=32z=32\cdot\frac{1}{2}=16$

Therefore, the product of all three numbers is $xyz=16\cdot\frac{7}{2}\cdot\frac{1}{2}=28 \Rightarrow A$

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2003 AMC 10A Problems/Problem 13

Problem

Solution

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