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 $t$.

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 $14$.

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Updated on: 10-Oct-2022

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