Difference between revisions of "1987 AHSME Problems/Problem 18"
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Technodoggo (talk | contribs) (S2) |
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<cmath>E = \boxed{\frac{AM - HS}{M - H}}.</cmath> | <cmath>E = \boxed{\frac{AM - HS}{M - H}}.</cmath> | ||
The answer is (D). | The answer is (D). | ||
+ | |||
+ | ==Solution 2 (a little faster)== | ||
+ | |||
+ | Let <math>a</math> and <math>g</math> denote the widths of an algebra and geometry textbook, respectively. As in Solution 1, observe that the width of the shelf is equal to the following: | ||
+ | |||
+ | <cmath>Ea=Aa+Hg=Sa+Mg.</cmath> | ||
+ | |||
+ | The rest is simply algebra: | ||
+ | |||
+ | \begin{align*} | ||
+ | (E-A)a&=Hg \ | ||
+ | g&=\dfrac{(E-A)a}H \ | ||
+ | Ea&=Sa+M\left(\dfrac{(E-A)a}H\right) \ | ||
+ | &=Sa+\dfrac{M(E-A)a}H \ | ||
+ | E&=S+\dfrac{M(E-A)}H \ | ||
+ | &=\dfrac{SH+ME-MA}H \ | ||
+ | EH&=SH+ME-MA \ | ||
+ | E(H-M)&=SH-MA \ | ||
+ | E&=\dfrac{SH-MA}{H-M} \ | ||
+ | &=\dfrac{AM-SH}{M-H}. \ | ||
+ | \end{align*} | ||
+ | |||
+ | This is answer choice <math>\boxed{\textbf{(D)}}</math>. QED. <math>\Box</math> | ||
+ | |||
+ | ~Technodoggo | ||
== See also == | == See also == |
Latest revision as of 00:02, 8 September 2024
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).
Solution 2 (a little faster)
Let and denote the widths of an algebra and geometry textbook, respectively. As in Solution 1, observe that the width of the shelf is equal to the following:
The rest is simply algebra:
This is answer choice . QED.
~Technodoggo
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.