Difference between revisions of "1962 AHSME Problems/Problem 10"
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==Solution== | ==Solution== | ||
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> | 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> | ||
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| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1962|before=Problem 9|num-a=11}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 22:14, 3 October 2014
Problem
A man drives 
miles to the seashore in 
hours and 
minutes. He returns from the shore to the starting point in 
hours and 
minutes. Let 
be the average rate for the entire trip. Then the average rate for the trip going exceeds in miles per hour, by:
Solution
Since the equation for rate is 
, you only need to find 
and 
. The distance from the seashore to the starting point is miles, and since he makes a round trip, 
. Also, you know the times it took him to go both directions, so when you add them up ( hours and minutes 
hours and minutes), you get 
hours. Since , after plugging in the values of and , you get 
, or 
. But what the question asks is for the positive difference between the speed going to the seashore and the average speed. The equation is , so when you plug in and , you get 
. Simplifying, you get 
. 
See Also
| 1962 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
