Problem Statement:

At 3.30 pm, Sajib goes to drawing class from home. The angle of the minute and the hour hand of the clock in this situation is $\theta _ 1$. When he returns from school, the minute hand of the clock is still in 6 hand side. The angle of the minute and the hour hand of the clock in this situation is $\theta_2.$ $\theta_1 - \theta_2 = 15^\circ$. When would Sajib return to home from school?

My trying:

The hour hand of the clock passes $6t$ degree angle in $t$ minutes. The minute hand passes $\frac{t}{2}$ degree angle in $t$ minutes. So after $t$ minutes, the angle between the minute and the hour hand is $\frac{11t}{2}.$

After then, what should I do?

• what does " the minute hand of clock is still in 6 hand side" mean? Dec 30 '17 at 14:41

The minute hand passes $6°$ per minute and the hour hand pass $\frac{1}{2}°$ per minute. Since the clock hands run clockwise, and from 3:30 ($\theta_1=75°$), the minute hand is in front of and runs faster than the hour hand, so we expect the minute hand to catch up next hour so we can have $\theta_1-\theta_2=15°$(which implies $\theta_1>\theta_2$).
The next time that minute and hour hands overlap should be approximately 4:21(calculate this by yourself). From then, since you know minute hand runs $5.5°$ per minute faster than the hour hand, you can solve for the $\Delta t$:
$\Delta t\cdot 5.5°/min=(75-15)°=60°$ -> $\Delta t=11min$. Then the time satisfying the requirement should be at approximately 4:33. And indeed, the minute hand of clock still is still in 6 hand side.
Note that the angle between the minute hand and the hour hand at $3:30$ is $75^\circ$. How? This is because the hour hand which is at $3$ would have moved $$\frac{30}{60}\times 30^\circ = 15^\circ$$ and the result follows.
As, $\theta_1=75^\circ$, it gives us: $\theta_2=60^\circ$.