Solving Distance, Speed, and Time Problems

This problem involves a man travelling a total distance over a total time, using two different modes of transport with different speeds. We need to find the specific distance travelled using one of the modes.

Understanding the Given Information

• Total distance travelled = 42 km
• Total time taken = 5 hours
• Speed on foot = 6 km/h
• Speed on bicycle = 10 km/h

We need to find the distance travelled on foot.

Setting up the Equations

Let's define variables for the unknown quantities:

• Let \(d_f\) be the distance travelled on foot (in km).
• Let \(d_b\) be the distance travelled on bicycle (in km).
• Let \(t_f\) be the time spent travelling on foot (in hours).
• Let \(t_b\) be the time spent travelling on bicycle (in hours).

From the given information, we can form equations based on total distance and total time:

Total distance:

\(d_f + d_b = 42 \quad (Equation \, 1)\)

Total time:

\(t_f + t_b = 5 \quad (Equation \, 2)\)

We know the relationship between distance, speed, and time: Distance = Speed × Time, which can be rearranged to Time = Distance / Speed.

Using this, we can express \(t_f\) and \(t_b\) in terms of distances and speeds:

Time spent on foot:

\(t_f = \frac{d_f}{\text{Speed on foot}} = \frac{d_f}{6}\)

Time spent on bicycle:

\(t_b = \frac{d_b}{\text{Speed on bicycle}} = \frac{d_b}{10}\)

Solving the System of Equations

Now substitute the expressions for \(t_f\) and \(t_b\) into Equation 2:

\(\frac{d_f}{6} + \frac{d_b}{10} = 5 \quad (Equation \, 3)\)

We now have a system of two linear equations with two variables (\(d_f\) and \(d_b\)):

1. \(d_f + d_b = 42\)
2. \(\frac{d_f}{6} + \frac{d_b}{10} = 5\)

We can solve this system using the substitution method. From Equation 1, express \(d_b\) in terms of \(d_f\):

\(d_b = 42 - d_f\)

Substitute this expression for \(d_b\) into Equation 3:

\(\frac{d_f}{6} + \frac{42 - d_f}{10} = 5\)

To eliminate the denominators, multiply the entire equation by the least common multiple (LCM) of 6 and 10, which is 30:

\(30 \left( \frac{d_f}{6} \right) + 30 \left( \frac{42 - d_f}{10} \right) = 30 \times 5\)

\(5 d_f + 3 (42 - d_f) = 150\)

Distribute the 3:

\(5 d_f + 126 - 3 d_f = 150\)

Combine like terms:

\((5 d_f - 3 d_f) + 126 = 150\)

\(2 d_f + 126 = 150\)

Subtract 126 from both sides:

\(2 d_f = 150 - 126\)

\(2 d_f = 24\)

Divide by 2:

\(d_f = \frac{24}{2}\)

\(d_f = 12\)

The distance travelled on foot is 12 km.

We can also find the distance travelled by bicycle to verify:

\(d_b = 42 - d_f = 42 - 12 = 30\) km.

Check the times:

Time on foot: \(t_f = \frac{12 \, \text{km}}{6 \, \text{km/h}} = 2\) hours

Time on bicycle: \(t_b = \frac{30 \, \text{km}}{10 \, \text{km/h}} = 3\) hours

Total time = \(t_f + t_b = 2 + 3 = 5\) hours. This matches the given total time.

Conclusion

The distance travelled on foot is 12 km.

Revision Table: Distance, Speed, Time Concepts

Additional Information: Alternative Approach (Using Time as Variable)

Instead of using distance as the main variable, we could have used time. Let \(t_f\) be the time spent on foot and \(t_b\) be the time spent on bicycle.

Given:

• Total time: \(t_f + t_b = 5\) hours

Distance travelled on foot = Speed on foot × Time on foot = \(6 t_f\)

Distance travelled on bicycle = Speed on bicycle × Time on bicycle = \(10 t_b\)

Total distance:

\(6 t_f + 10 t_b = 42\)

Now we have a system with \(t_f\) and \(t_b\):

1. \(t_f + t_b = 5\)
2. \(6 t_f + 10 t_b = 42\)

From equation 1, \(t_b = 5 - t_f\). Substitute this into equation 2:

\(6 t_f + 10 (5 - t_f) = 42\)

\(6 t_f + 50 - 10 t_f = 42\)

\(50 - 4 t_f = 42\)

\(50 - 42 = 4 t_f\)

\(8 = 4 t_f\)

\(t_f = \frac{8}{4} = 2\) hours

The time spent on foot is 2 hours. The distance travelled on foot is:

Distance on foot = Speed on foot × Time on foot = \(6 \, \text{km/h} \times 2 \, \text{hours} = 12\) km.

This confirms the result obtained using distance as the variable.