Question

At t minutes past 2 pm the time needed by the minutes hand and a clock to show 3 pm was found to be 3 minutes less than $\frac{t^2}{4}$ minutes. Find t.

Answer 1

It is given that at t minutes past 2 pm the time needed by the minutes hand and a clock to show 3 pm was found to be 3 minutes less than $\frac{t^2}{4}$ minutes i.e. $\frac{t^2}{4}-3$.
There are 60 minutes in 1 hour, so at t minutes past 2, time left to 3 will be $60-t$ minutes.
$$\begin{gathered} \Rightarrow 60-t=\frac{t^2}{4}-3 \\ \Rightarrow 240-4t=t^2-12 \\ \Rightarrow t^2+4t-252=0 \\ \Rightarrow t^2+18t-14t-252=0 \\ \Rightarrow t\left(t+18\right)-14\left(t+18\right)=0 \\ \Rightarrow \left(t-14\right)\left(t+18\right)=0 \\ \Rightarrow t=14\mathrm{or}t=-18 \end{gathered}$$
We ignore the negative value because time cannot be negative.
Therefore, t = 14