2007 Cyprus MO/Lyceum/Problem 3

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 $A$ to $B$ 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 $\frac{2d}{\frac{d}{40}+\frac{d}{60}}=\frac{2d}{\frac{100d}{2400}}=\frac{4800d}{100d}=48\frac{\mathrm{km}}{\mathrm{h}}\Rightarrow\mathrm{ B}$

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2007 Cyprus MO/Lyceum/Problem 3

Problem

Solution

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