Question

Ravi starts for his school at $8:20$ am on his bicycle. If he travels at a speed of $10$ km/h then he reaches his school late by $8$ minutes but on traveling at $16$ km/h, he reaches the school $10$ minutes early. At what time does school start?

Answer 1

Let the total distance $=x$ km
Let the time taken by Ravi to reach the school at sharp time $=t$ min.

We know that Speed $=\frac{\text{Distance}}{\text{time}}$

If the speed of the bicycle is $10$ km/ h then he reaches his school late by $8$ min.

$\therefore \frac{x}{10}=t+\frac{8}{60}$ [ Late time $=$ sharp time $+$ 8 min]

$\Rightarrow \frac{x}{10}=t+\frac{2}{15}...\left(i\right)$

If the speed of the bicycle is $16$ km/ h then he reach his school $10$ min early.

$\therefore \frac{x}{16}=t-\frac{10}{60}$ [ Early time $=$ sharp time $-10$ min]

$\Rightarrow \frac{x}{16}=t-\frac{1}{6}...\left(ii\right)$

On subtracting $Eq.\left(ii\right)$ from $\left(i\right),$ we get

$\frac{x}{10}-\frac{x}{16}=\frac{2}{15}+\frac{1}{6}$

$\Rightarrow \frac{8x-5x}{80}=\frac{4+5}{30}$

$\Rightarrow \frac{3x}{80}=\frac{9}{30}$

$\Rightarrow x=\frac{9\times 80}{30\times 3}=8$ km

Now, put $x=8inEq.\left(i\right)$, we get

$\frac{8}{10}=t+\frac{2}{15}$

$\Rightarrow t=\frac{8}{10}-\frac{2}{15}=\frac{24-4}{30}$

$\Rightarrow t=\frac{20}{30}=\frac{2}{3}$ hours

$=\frac{2}{3}\times 60=40$ min

So the time of school is $40$ mins after $8:20$ am $=9:00$ am.