Difference between revisions of "1987 AHSME Problems/Problem 18"
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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, | 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*} | + | \[ \begin{align*} |
Ax + Hy &= z, \\ | Ax + Hy &= z, \\ | ||
Sx + My &= z, \\ | Sx + My &= z, \\ | ||
Ex &= z. | Ex &= z. | ||
| − | \end{align*} | + | \end{align*} \] |
From the third equation, <math>x = z/E</math>. Substituting into the first two equations, we get | From the third equation, <math>x = z/E</math>. Substituting into the first two equations, we get | ||
\begin{align*} | \begin{align*} | ||
Revision as of 17:17, 28 June 2015
Problem
It takes 
algebra books (all the same thickness) and 
geometry books (all the same thickness,
which is greater than that of an algebra book) to completely fill a certain shelf.
Also, 
of the algebra books and 
of the geometry books would fill the same shelf.
Finally, 
of the algebra books alone would fill this shelf. Given that 
are distinct positive integers,
it follows that is
Solution
Let 
and 
be the thicknesses of an algebra book and geometry book, respectively, and let 
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, 
. 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,
![\[Hy = z - \frac{A}{E} z = \frac{E - A}{E} z,\]](http://latex.artofproblemsolving.com/5/e/d/5edcb2c71ddef42b3b5eac74d9ebed2bd4aa9beb.png)
so
![\[\frac{y}{z} = \frac{E - A}{EH}.\]](http://latex.artofproblemsolving.com/6/5/a/65a8e24e0bea18b7fd71c842c75a2f2e40ed23cb.png)
From the second equation,
![\[My = z - \frac{S}{E} z = \frac{E - S}{E} z,\]](http://latex.artofproblemsolving.com/8/c/4/8c4b4dde8d0c5e479fe3dfb651eacdc683a84993.png)
so
![\[\frac{y}{z} = \frac{E - S}{ME}.\]](http://latex.artofproblemsolving.com/4/0/a/40af77862a87fd9ddf8767df4d9205e0ed89296a.png)
Hence,
![\[\frac{E - A}{EH} = \frac{E - S}{ME}.\]](http://latex.artofproblemsolving.com/0/7/e/07e9910c60606ecd16e5899c4c569395f6446465.png)
Multiplying both sides by 
, we get 
. Then 
, so
![\[E = \boxed{\frac{AM - HS}{M - H}}.\]](http://latex.artofproblemsolving.com/7/4/3/74316e95c310c34e3fcc8a88d311334bfd0cddde.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
