1987 AHSME Problems/Problem 18

Problem

It takes $A$ algebra books (all the same thickness) and $H$ geometry books (all the same thickness, which is greater than that of an algebra book) to completely fill a certain shelf. Also, $S$ of the algebra books and $M$ of the geometry books would fill the same shelf. Finally, $E$ of the algebra books alone would fill this shelf. Given that $A, H, S, M, E$ are distinct positive integers, it follows that $E$ is

$\textbf{(A)}\ \frac{AM+SH}{M+H} \qquad \textbf{(B)}\ \frac{AM^2+SH^2}{M^2+H^2} \qquad \textbf{(C)}\ \frac{AH-SM}{M-H}\qquad \textbf{(D)}\ \frac{AM-SH}{M-H}\qquad \textbf{(E)}\ \frac{AM^2-SH^2}{M^2-H^2}$

Solution

Let $x$ and $y$ be the thicknesses of an algebra book and geometry book, respectively, and let $z$ be the length of the shelf. Then from the given information, $Ax + Hy = z$ $Sx + My = z$ $Ex = z.$ From the third equation, $x = z/E$ . Substituting into the first two equations, we get $\frac{A}{E} z + Hy = z,$ $\frac{S}{E} z + My = z.$

From the first equation, $Hy = z - \frac{A}{E} z = \frac{E - A}{E} z,$ so $\frac{y}{z} = \frac{E - A}{EH}.$ From the second equation, $My = z - \frac{S}{E} z = \frac{E - S}{E} z,$ so $\frac{y}{z} = \frac{E - S}{ME}.$ Hence, $\frac{E - A}{EH} = \frac{E - S}{ME}.$ Multiplying both sides by $HME$ , we get $ME - AM = HE - HS$ . Then $(M - H)E = AM - HS$ , so $E = \boxed{\frac{AM - HS}{M - H}}.$ The answer is (D).

1987 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1987 AHSME Problems/Problem 18

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