**Question:**

At t min past $2 \mathrm{pm}$, the time needed by the minute hand of a clock to show $3 \mathrm{pm}$ was found to be $3 \mathrm{~min}$ less

than $\frac{t^{2}}{4}$ min. Find $t$.

**Solution:**

We know that, the time between 2 pm to 3 pm = 1 h = 60 min

Given that, at f min past 2pm, the time needed by the min hand of a clock to show 3 pm was found to be 3

min less than $\frac{t^{2}}{4}$ min $i, e$

$t+\left(\frac{t^{2}}{4}-3\right)=60$

$\Rightarrow \quad 4 t+t^{2}-12=240$

$\Rightarrow \quad t^{2}+4 t-252=0$

$\Rightarrow \quad t^{2}+18 t-14 t-252=0 \quad$ [by splitting the middle term]

$\Rightarrow \quad t(t+18)-14(t+18)=0 \quad$ [since, time cannot be negative, so $t \neq-18]$

$\Rightarrow \quad(t+18)(t-14)=0$

$\therefore \quad t=14 \mathrm{~min}$

Hence, the required value off is 14 min