Difference between revisions of "1966 AHSME Problems/Problem 27"
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== Solution == | == Solution == | ||
| − | <math>\ | + | Let the speed of rowing in still water and water speed be <math>r, w</math>. |
| + | |||
| + | Then: | ||
| + | |||
| + | <math>\frac{15}{r-w}-\frac{15}{r+w}=5</math>. | ||
| + | |||
| + | <math>\frac{15}{2r-w}-\frac{15}{2r+w=1</math>. | ||
| + | |||
| + | Solving (remember: both variables must be positive!), we have <math>r=4, w=2</math>. Select <math>\boxed{A}</math>. | ||
| + | |||
| + | ~hastapasta | ||
== See also == | == See also == | ||
Revision as of 15:46, 17 May 2022
Problem
At his usual rate a man rows 15 miles downstream in five hours less time than it takes him to return. If he doubles his usual rate, the time downstream is only one hour less than the time upstream. In miles per hour, the rate of the stream's current is:
Solution
Let the speed of rowing in still water and water speed be 
.
Then:
.
$\frac{15}{2r-w}-\frac{15}{2r+w=1$ (Error compiling LaTeX. Unknown error_msg).
Solving (remember: both variables must be positive!), we have 
. Select 
.
~hastapasta
See also
| 1966 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 26 |
Followed by Problem 28 | |
| 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. 
