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Printimc
algebra intermediate
Problem
The sum of three numbers is . 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?
(A)
(B)
(C)
(D)
Solution
Let the numbers be , , and in that order. The given tells us that \begin{eqnarray}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 \end{eqnarray} Therefore, the product of all three numbers is .
Final answer
A