Question

A watch gains $5$ seconds in $3$ minutes was set right at $7\mathrm{AM}$ In the afternoon of the same day, when the watch shows a quarter past $4$ 'o'clock, the true time is ______________.

Answer 1

Calculate the true time of the clock.

Time in hours from $7:00\mathrm{AM}$ to $4:15\mathrm{PM}$$=9\mathrm{hours}15\mathrm{minutes}$

Convert $9$ hours $15$ minutes into hours.

$$\begin{gathered} \therefore 9\mathrm{hours}15\mathrm{minutes}=\left(9+\frac{15}{60}\right)\mathrm{hours}\left[\because 1\mathrm{hours}=60\mathrm{minutes}\right] \\ \Rightarrow =\frac{37}{4}\mathrm{hours} \end{gathered}$$

Now convert $3$ minutes $5$ seconds into hours.

$$\begin{gathered} \therefore 3\mathrm{minutes}5\mathrm{seconds}=\left(\frac{3}{60}+\frac{5}{3600}\right) \\ \Rightarrow =\frac{180+5}{3600} \\ \Rightarrow =\frac{185}{3600} \\ \Rightarrow =\frac{37}{720}\mathrm{hours} \end{gathered}$$

Since the given clock gains $5$ seconds in $3$minutes.

$\therefore$ $3$ minutes $5$ seconds of the given clock$=$$3$ minutes of the correct clock

$$\begin{gathered} \because \frac{37}{720}\mathrm{hrs}\mathrm{of}\mathrm{given}\mathrm{clock}=\frac{1}{20}\mathrm{hrs}\mathrm{of}\mathrm{the}\mathrm{correct}\mathrm{clock} \\ \therefore \frac{37}{4}\mathrm{hrs}\mathrm{of}\mathrm{given}\mathrm{clock}=\frac{1}{20}\times \frac{720}{37}\times \frac{37}{4}\mathrm{hrs}\mathrm{of}\mathrm{the}\mathrm{correct}\mathrm{clock} \\ =9\mathrm{hrs}\mathrm{of}\mathrm{the}\mathrm{correct}\mathrm{clock} \end{gathered}$$

So the correct time is, add $9$ hours in $7\mathrm{AM}$ i.e $4\mathrm{PM}$

Hence, the true time is $4\mathrm{PM}$.