Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 3"
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==Problem== | ==Problem== | ||
− | A cyclist drives | + | A cyclist drives from town A to town B with velocity <math>40 \frac{\mathrm{km}}{\mathrm{h}}</math> and comes back with velocity <math>60 \frac{\mathrm{km}}{\mathrm{h}}</math>. The [[mean]] velocity in <math>\frac{\mathrm{km}}{\mathrm{h}}</math> for the total distance is |
<math> \mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 48\qquad \mathrm{(C) \ } 50\qquad \mathrm{(D) \ } 55\qquad \mathrm{(E) \ } 100</math> | <math> \mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 48\qquad \mathrm{(C) \ } 50\qquad \mathrm{(D) \ } 55\qquad \mathrm{(E) \ } 100</math> | ||
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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}} |
Latest revision as of 16:01, 24 May 2009
Problem
A cyclist drives from 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 | |
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