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Difference between revisions of "1987 AHSME Problems/Problem 18"

(Created page with "==Problem== It takes <math>A</math> algebra books (all the same thickness) and <math>H</math> geometry books (all the same thickness, which is greater than that of an algebra b...")
 
(→‎Problem)
Line 12: Line 12:
 
\textbf{(D)}\ \frac{AM-SH}{M-H}\qquad
 
\textbf{(D)}\ \frac{AM-SH}{M-H}\qquad
 
\textbf{(E)}\ \frac{AM^2-SH^2}{M^2-H^2}  </math>
 
\textbf{(E)}\ \frac{AM^2-SH^2}{M^2-H^2}  </math>
 +
 +
==Solution==
 +
 +
Let <math>x</math> and <math>y</math> be the thicknesses of an algebra book and geometry book, respectively, and let <math>z</math> be the length of the shelf. Then from the given information,
 +
\begin{align*}
 +
Ax + Hy &= z, \\
 +
Sx + My &= z, \\
 +
Ex &= z.
 +
\end{align*}
 +
From the third equation, <math>x = z/E</math>. Substituting into the first two equations, we get
 +
\begin{align*}
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\frac{A}{E} z + Hy &= z, \\
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\frac{S}{E} z + My &= z.
 +
\end{align*}
 +
 +
From the first equation,
 +
<cmath>Hy = z - \frac{A}{E} z = \frac{E - A}{E} z,</cmath>
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so
 +
<cmath>\frac{y}{z} = \frac{E - A}{EH}.</cmath>
 +
From the second equation,
 +
<cmath>My = z - \frac{S}{E} z = \frac{E - S}{E} z,</cmath>
 +
so
 +
<cmath>\frac{y}{z} = \frac{E - S}{ME}.</cmath>
 +
Hence,
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<cmath>\frac{E - A}{EH} = \frac{E - S}{ME}.</cmath>
 +
Multiplying both sides by <math>HME</math>, we get <math>ME - AM = HE - HS</math>. Then <math>(M - H)E = AM - HS</math>, so
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<cmath>E = \boxed{\frac{AM - HS}{M - H}}.</cmath>
 +
The answer is (D).
  
 
== See also ==
 
== See also ==

Revision as of 17:16, 28 June 2015

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, \begin{align*} Ax + Hy &= z, \\ Sx + My &= z, \\ Ex &= z. \end{align*} From the third equation, $x = z/E$. Substituting into the first two equations, we get \begin{align*} \frac{A}{E} z + Hy &= z, \\ \frac{S}{E} z + My &= z. \end{align*}

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).

See also

1987 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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