Question

Tim cycled from his house to his friend John’s house and then on to his (Tim’s) school without stopping. The average speed for his entire journey was 26 km/hr. The distance from John’s house to Tim’s school is 0.3 times the distance from Tim’s house to John’s house. Tim’s speed from John’s house to Tim’s school was twice that of Tim’s speed from Tim’s house to John’s house. What was Tim’s average speed from John’s house to Tim’s school in km/hr?

Answer 1

Correct Answer: 46

Let the distance from Tim's house to John's house be \(d\) km. Then, the distance from John's house to the school is \(0.3d\) km.
The total distance for the entire journey is \(d + 0.3d = 1.3d\) km.
Given the average speed for the entire journey is 26 km/hr, we have:

\[\frac{1.3d}{t} = 26\]

where \(t\) is the total time taken. Solving for \(t\), we get:

\[ t = \frac{1.3d}{26} \]

Let \(v_1\) be the speed from Tim's house to John's house, and \(v_2 = 2v_1\) be the speed from John's house to the school. The time taken from Tim's house to John's house is:

\[\frac{d}{v_1}\]

And the time taken from John's house to the school is:

\[\frac{0.3d}{2v_1}\]

The total time \(t\) can also be expressed as:

\[ t = \frac{d}{v_1} + \frac{0.3d}{2v_1} \]

Substitute \(t = \frac{1.3d}{26}\) into the equation:

\[\frac{1.3d}{26} = \frac{d}{v_1} + \frac{0.3d}{2v_1}\]

Simplifying, we find:

\[\frac{1.3}{26} = \frac{1}{v_1} + \frac{0.3}{2v_1}\]

Multiply the entire equation by \(2v_1\) to clear the denominators:

\[\frac{1.3 \cdot 2v_1}{26} = 2 + 0.3\]

\[\frac{2.6v_1}{26} = 2.3\]

Solve for \(v_1\):

\[2.6v_1 = 59.8\]

\[v_1 = \frac{59.8}{2.6}\]

Calculate \(v_1\):

\[v_1 = 23 \, \text{km/hr}\]

Therefore, \(v_2 = 2v_1 = 46 \, \text{km/hr}\).

The average speed from John's house to the school is thus 46 km/hr, which fits within the expected range of 46 to 46 km/hr.