If a boy goes from his home to his school at a speed of 40 km/h, he reaches 10 minutes early. If he goes at a speed of 35 km/h, he reaches 5 minutes late. What will be the time taken for the journey at a speed of 35 km/h?

Let's assume the distance between his home to his school is 'd' km.

When the boy reaches on time then the time taken by him is 't' hours.

If a boy goes from his home to his school at a speed of 40 km/h, he reaches 10 minutes early.

$$\frac{d}{40}\ =\ t-\frac{10}{60}$$

$$d\ =40\ \left(t-\frac{1}{6}\right)$$    Eq.(i)

If he goes at a speed of 35 km/h, he reaches 5 minutes late.

$$\frac{d}{35}\ =\ t+\frac{5}{60}$$

$$d\ =\ 35\left(t+\frac{1}{12}\right)$$    Eq.(ii)

So Eq.(i) = Eq.(ii).

$$40\ \left(t-\frac{1}{6}\right) = 35\left(t+\frac{1}{12}\right)$$

$$8\ \left(t-\frac{1}{6}\right)=7\left(t+\frac{1}{12}\right)$$

$$\ 8t-\frac{8}{6}=7t+\frac{7}{12}$$

$$\ 8t-7t=\frac{4}{3}+\frac{7}{12}$$

$$t=\frac{16}{12}+\frac{7}{12}$$

$$t=\frac{23}{12}$$

Put the value of 't' in Eq.(i).

$$d\ =40\ \left(\frac{23}{12}-\frac{1}{6}\right)$$

$$d\ =40\ \left(\frac{23}{12}-\frac{2}{12}\right)$$

$$d\ =40\times\ \left(\frac{7}{4}\right)$$
d = 70 km

Time taken for the whole journey at a speed of 35 km/h = $$\frac{distance}{speed}$$

= $$\frac{70}{35}$$

= 2 hours

We know that 1 hour = 60 minutes. So 2 hours = $$2\times60$$ = 120 minutes