Difference between revisions of "1957 AHSME Problems/Problem 39"

Latest revision as of 08:13, 27 July 2024

Problem

Two men set out at the same time to walk towards each other from $M$ and $N$ , $72$ miles apart. The first man walks at the rate of $4$ mph. The second man walks $2$ miles the first hour, $2\tfrac {1}{2}$ miles the second hour, $3$ miles the third hour, and so on in arithmetic progression. Then the men will meet:

$\textbf{(A)}\ \text{in 7 hours} \qquad \textbf{(B)}\ \text{in }{8\frac {1}{4}}\text{ hours}\qquad \textbf{(C)}\ \text{nearer }{M}\text{ than }{N}\qquad\\ \textbf{(D)}\ \text{nearer }{N}\text{ than }{M}\qquad \textbf{(E)}\ \text{midway between }{M}\text{ and }{N}$

Solution

Let $t$ be the time it takes (in hours) for the men to meet. Then, the distance the first man will have travelled is $4t$ . If $t$ is an integer, the second man will have travelled $\tfrac{2+2+0.5(t-1)}{2} \cdot t$ , the sum of the first $t$ terms of the arithmetic progression. If $t$ is not an integer, then we should have a good enough approximation to choose an answer choice (given the relatively large distance of $72$ miles compared to their speeds). Because the combined distance that the men travel must be $72$ miles, we can now solve for $t$ in the following formula: \begin{align*} 4t+\frac{2+2+0.5(t-1)}{2} \cdot t &= 72 \\ 4t+\frac{4+0.5t-0.5}{2} \cdot t &= 72 \\ 8t+3.5t+0.5t^2 &= 144 \\ t^2+23t &= 288 \\ t^2+23t-288 &= 0 \\ (t+32)(t-9) &= 0 \end{align*} Because $t>0$ , we are left with $t=9$ . At $t=9$ , the first man will have travelled $36$ miles, which is half of te distance between $M$ and $N$ . Thus, our answer is $\boxed{\textbf{(E)} \text{ midway between } M \text{ and } N}$ .

1957 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 38	Followed by Problem 40
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Difference between revisions of "1957 AHSME Problems/Problem 39"

Latest revision as of 08:13, 27 July 2024

Problem

Solution

See Also

@@ Line 11: / Line 11: @@
 == Solution ==
-<math>\boxed{\textbf{(E)} \text{ midway between } M \text{ and } N}</math>.
+Let <math>t</math> be the time it takes (in hours) for the men to meet. Then, the distance the first man will have travelled is <math>4t</math>. If <math>t</math> is an integer, the second man will have travelled <math>\tfrac{2+2+0.5(t-1)}{2} \cdot t</math>, the sum of the first <math>t</math> terms of the [[arithmetic progression]]. If <math>t</math> is not an integer, then we should have a good enough approximation to choose an answer choice (given the relatively large distance of <math>72</math> miles compared to their speeds). Because the combined distance that the men travel must be <math>72</math> miles, we can now solve for <math>t</math> in the following formula:
+\begin{align*}
+t+\frac{2+2+0.5(t-1)}{2} \cdot t &= 72 \\
+t+\frac{4+0.5t-0.5}{2} \cdot t &= 72 \\
+t+3.5t+0.5t^2 &= 144 \\
+t^2+23t &= 288 \\
+t^2+23t-288 &= 0 \\
+(t+32)(t-9) &= 0
+\end{align*}
+Because <math>t>0</math>, we are left with <math>t=9</math>. At <math>t=9</math>, the first man will have travelled <math>36</math> miles, which is half of te distance between <math>M</math> and <math>N</math>. Thus, our answer is <math>\boxed{\textbf{(E)} \text{ midway between } M \text{ and } N}</math>.
 == See Also ==