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Difference between revisions of "1962 AHSME Problems/Problem 10"

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Since the equation for rate is <math>r=\frac{d}{t}</math>, you only need to find <math>d</math> and <math>t</math>. The distance from the seashore to the starting point is <math>150</math> miles, and since he makes a round trip, <math>d=300</math>. Also, you know the times it took him to go both directions, so when you add them up (<math>3</math> hours and <math>20</math> minutes <math>+</math> <math>4</math> hours and <math>10</math> minutes), you get <math>7\frac{1}{2}</math> hours. Since <math>r=\frac{d}{t}</math>, after plugging in the values of <math>d</math> and <math>t</math>, you get <math>r=\frac{300}{7\frac{1}{2}}</math>, or <math>r=40</math>. But what the question asks is for the positive difference between the speed going to the seashore and the average speed. The equation is <math>r=\frac{d}{t}</math>, so when you plug in <math>d</math> and <math>t</math>, you get <math>r=\frac{150}{3\frac{1}{3}}</math>. Simplifying, you get <math>r=45</math>. <math>45-40=5 \rightarrow \boxed{\text{A}}</math>

Revision as of 21:51, 21 February 2014

Problem

A man drives $150$ miles to the seashore in $3$ hours and $20$ minutes. He returns from the shore to the starting point in $4$ hours and $10$ minutes. Let $r$ be the average rate for the entire trip. Then the average rate for the trip going exceeds $r$ in miles per hour, by:

$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 4\frac{1}{2}\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 1$

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it. Since the equation for rate is $r=\frac{d}{t}$, you only need to find $d$ and $t$. The distance from the seashore to the starting point is $150$ miles, and since he makes a round trip, $d=300$. Also, you know the times it took him to go both directions, so when you add them up ($3$ hours and $20$ minutes $+$ $4$ hours and $10$ minutes), you get $7\frac{1}{2}$ hours. Since $r=\frac{d}{t}$, after plugging in the values of $d$ and $t$, you get $r=\frac{300}{7\frac{1}{2}}$, or $r=40$. But what the question asks is for the positive difference between the speed going to the seashore and the average speed. The equation is $r=\frac{d}{t}$, so when you plug in $d$ and $t$, you get $r=\frac{150}{3\frac{1}{3}}$. Simplifying, you get $r=45$. $45-40=5 \rightarrow \boxed{\text{A}}$

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