Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 3"
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The time it took the cyclist to travel from <math>B</math> to <math>A</math> was <math>\frac{d}{60}</math> hours. | The time it took the cyclist to travel from <math>B</math> to <math>A</math> was <math>\frac{d}{60}</math> hours. | ||
| − | The cyclist's mean velocity was <math>\frac{2d}{\frac{d}{40}+\frac{d}{60}}=\frac{2d}{\frac{100d}{2400}}=\frac{4800d}{100d}=48\frac{\mathrm{km}}{\mathrm{h}}\ | + | The cyclist's mean velocity was <math>\frac{2d}{\frac{d}{40}+\frac{d}{60}}=\frac{2d}{\frac{100d}{2400}}= \frac{4800d}{100d}=48\frac{\mathrm{km}}{\mathrm{h}} \Longrightarrow\mathrm{ B}</math> |
==See also== | ==See also== | ||
{{CYMO box|year=2007|l=Lyceum|num-b=2|num-a=4}} | {{CYMO box|year=2007|l=Lyceum|num-b=2|num-a=4}} | ||
Revision as of 20:37, 6 May 2007
Problem
A cyclist drives form town A to town B with velocity 
and comes back with velocity 
. The mean velocity in 
for the total distance is
Solution
Let the distance from town A to town B, in kilometers, be 
.
The time it took the cyclist to travel from 
to 
was 
hours.
The time it took the cyclist to travel from to was 
hours.
The cyclist's mean velocity was 
See also
| 2007 Cyprus MO, Lyceum (Problems) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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 | ||
