Difference between revisions of "1987 AHSME Problems/Problem 18"
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\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, | ||
+ | <cmath> Ax + Hy = z </cmath> | ||
+ | <cmath> Sx + My = z </cmath> | ||
+ | <cmath> Ex = z. </cmath> | ||
+ | From the third equation, <math>x = z/E</math>. Substituting into the first two equations, we get | ||
+ | <cmath> \frac{A}{E} z + Hy = z, </cmath> | ||
+ | <cmath> \frac{S}{E} z + My = z. </cmath> | ||
+ | |||
+ | From the first equation, | ||
+ | <cmath>Hy = z - \frac{A}{E} z = \frac{E - A}{E} z,</cmath> | ||
+ | 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, | ||
+ | <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 | ||
+ | <cmath>E = \boxed{\frac{AM - HS}{M - H}}.</cmath> | ||
+ | The answer is (D). | ||
== See also == | == See also == |
Latest revision as of 18:23, 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,
From the third equation,
. Substituting into the first two equations, we get
From the first equation,
so
From the second equation,
so
Hence,
Multiplying both sides by
, we get
. Then
, so
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.