Question

1. a cyclist leaves home at 7:30 am to cycle to school 7 km away. he cycles at 10 km/h until he has a puncture; then he has to push his bicycle the rest of the way at 3 km/h. he arrives at school at 8:40 am. how far did he have to push his bicycle?

Explanation:

Step1: Calculate total time taken

The cyclist leaves at 7:30 am and arrives at 8:40 am. So the total time taken is 1 hour and 10 minutes, which is $1 + \frac{10}{60} = \frac{7}{6}$ hours.

Step2: Let the time spent cycling be \( t \) hours, then time spent pushing is \( \frac{7}{6} - t \) hours.

The distance cycled is \( 10t \) km (since speed is 10 km/h), and the distance pushed is \( 3(\frac{7}{6} - t) \) km (since speed is 3 km/h). The total distance is 7 km. So we set up the equation:
\( 10t + 3(\frac{7}{6} - t) = 7 \)

Step3: Solve the equation for \( t \)

First, expand the equation: \( 10t + \frac{7}{2} - 3t = 7 \)
Combine like terms: \( 7t + \frac{7}{2} = 7 \)
Subtract \( \frac{7}{2} \) from both sides: \( 7t = 7 - \frac{7}{2} = \frac{7}{2} \)
Divide both sides by 7: \( t = \frac{7}{2} \div 7 = \frac{1}{2} \) hours.

Step4: Calculate distance pushed

Time spent pushing is \( \frac{7}{6} - \frac{1}{2} = \frac{7 - 3}{6} = \frac{4}{6} = \frac{2}{3} \) hours.
Distance pushed is speed (3 km/h) times time (\( \frac{2}{3} \) hours), so \( 3\times\frac{2}{3} = 2 \) km.

Answer:

2 km