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At \( t \) minutes past \( 2 \mathrm{pm} \), the time needed by the minutes hand of a clock to show 3 pm was found to be 3 minutes less than \( \frac{t^{2}}{4} \) minutes. Find \( t \).
Given:
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.
To do:
Here, we have to find the value of $t$.
Solution:
Time needed by the minutes hand to show 3 pm $=60-t$ minutes.
According to the question,
$60-t = \frac{t^2}{4}-3$
$60-t = \frac{t^2-3\times4}{4}$
$60-t = \frac{t^2-12}{4}$
$4(60-t) = t^2-12$ (On cross multiplication)
$240-4t = t^2-12$
$t^2+4t-240-12=0$
$t^2+4t-252=0$
Solving for $t$ by factorization method, we get,
$t^2+18t-14t-252=0$
$t(t+18)-14(t+18)=0$
$(t-14)(t+18)=0$
$t-14=0$ or $t+18=0$
$t=14$ or $t=-18$
Therefore, the value of $t$ is $14$. (Time cannot be negative)
The value of $t$ is $14$.