Question

A cyclist drove one kilometer, with the wind in his back, in three minutes and drove the same way back, against the wind in four minutes. If we assume that the cyclist always puts constant force on the pedals, how much time would it take him to drive one kilometer without wind?

• A. \(2^1_3\)
• B. \(3^3_7\)
• C. \(2^3_7\)
• D. \(3^7_12\)

Answer 1

The Correct Option is B

The problem involves calculating the cyclist's time to travel one kilometer without wind based on the times with and against the wind. Let's denote:

• t₀ - time to drive 1 km without wind (in minutes)
• V - the cyclist's speed without wind (in km/min)
• W - wind speed (in km/min)

The cyclist traveled with the wind in 3 minutes, so his effective speed is \(V + W\). Thus, the distance-time relationship with the wind is:

\( \frac{1}{V+W} = 3 \)

The cyclist traveled against the wind in 4 minutes, so his effective speed is \(V - W\). Thus, the distance-time relationship against the wind is:

\( \frac{1}{V-W} = 4 \)

From these two equations, we have:

\( V + W = \frac{1}{3} \) and \( V - W = \frac{1}{4} \)

Solving these equations simultaneously, add both:

\( 2V = \frac{1}{3} + \frac{1}{4} \)

To add the fractions, find a common denominator:

\( 2V = \frac{4}{12} + \frac{3}{12} = \frac{7}{12} \)

Thus, \( V = \frac{7}{24} \)

Now, use either equation to solve for \(W\). Using \(V + W = \frac{1}{3}\):

\( \frac{7}{24} + W = \frac{1}{3} \)

Convert \(\frac{1}{3}\) to a common denominator:

\( W = \frac{8}{24} - \frac{7}{24} = \frac{1}{24} \)

With \(V = \frac{7}{24}\) and \(W = \frac{1}{24}\), we find t₀:

\( t_0 = \frac{1}{V} = \frac{1}{\frac{7}{24}} = \frac{24}{7} \) minutes

\(\frac{24}{7}\) minutes is equivalent to \(3\frac{3}{7}\) minutes. Thus, the correct answer is:

\(3^3_7\)