Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 3"

 
(Solution)
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==Solution==
 
==Solution==
Let the distance from <math>A</math> to <math>B</math> be <math>d</math>.
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Let the distance from town A to town B, in kilometers, be <math>d</math>.
  
 
The time it took the cyclist to travel from <math>A</math> to <math>B</math> was <math>\frac{d}{40}</math> hours.
 
The time it took the cyclist to travel from <math>A</math> to <math>B</math> was <math>\frac{d}{40}</math> hours.

Revision as of 11:40, 6 May 2007

Problem

A cyclist drives form town A to town B with velocity $40 \frac{\mathrm{km}}{\mathrm{h}}$ and comes back with velocity $60 \frac{\mathrm{km}}{\mathrm{h}}$. The mean velocity in $\frac{\mathrm{km}}{\mathrm{h}}$ for the total distance is

$\mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 48\qquad \mathrm{(C) \ } 50\qquad \mathrm{(D) \ } 55\qquad \mathrm{(E) \ } 100$

Solution

Let the distance from town A to town B, in kilometers, be $d$.

The time it took the cyclist to travel from $A$ to $B$ was $\frac{d}{40}$ hours.

The time it took the cyclist to travel from $B$ to $A$ was $\frac{d}{60}$ hours.

The cyclist's mean velocity was $\frac{2d}{\frac{d}{40}+\frac{d}{60}}=\frac{2d}{\frac{100d}{2400}}=\frac{4800d}{100d}=48\frac{\mathrm{km}}{\mathrm{h}}\Rightarrow\mathrm{ B}$

See also