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

(→‎Solution)
(S2)
 
Line 36: Line 36:
 
<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 \\
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&=\dfrac{SH+ME-MA}H \\
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EH&=SH+ME-MA \\
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E(H-M)&=SH-MA \\
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E&=\dfrac{SH-MA}{H-M} \\
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&=\dfrac{AM-SH}{M-H}. \\
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\end{align*}
 +
 +
This is answer choice <math>\boxed{\textbf{(D)}}</math>. QED. <math>\Box</math>
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 +
~Technodoggo
  
 
== See also ==
 
== See also ==

Latest revision as of 00:02, 8 September 2024

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).

Solution 2 (a little faster)

Let $a$ and $g$ 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:

\[Ea=Aa+Hg=Sa+Mg.\]

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 $\boxed{\textbf{(D)}}$. QED. $\Box$

~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

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