Question

At t minutes past 2 p.m., the time needed by the minute hand of a clock to show 3 p.m. was found to be 3 minutes less than $\dfrac{{{t}^{2}}}{4}$ minutes. Find t.

Answer 1

Hint: There are 60 minutes in an hour; use this statement to write the remaining time for reaching the minute’s hand from t past 2 p.m. to 3 p.m. Now, use the given condition and get an equation in terms of t only, and hence solve the quadratic to get value of t.

Complete step-by-step answer:
We know that there are 60 minutes in one hour. So, time taken by minute hand from t minutes past 2 p.m. ( $2:t$ o’clock) is the remaining time of the minute hand to reach 3:00 clock.
So, total time from 2:00 pm to 3:00 pm is 60 minutes, hence, time remaining after t minutes from 2:00 pm would be $\left( 60-t \right)$ minutes.
Now, it is also given in the question that time remaining from t minutes past 2:00 pm is 3 minutes less than $\dfrac{{{t}^{2}}}{4}$ minutes i.e. $\left( \dfrac{{{t}^{2}}}{4}-3 \right)$ minutes is the remaining time.
Now, we can equate $\left( 60-t \right)$ minutes and $\left( \dfrac{{{t}^{2}}}{4}-3 \right)$ minutes as both are representing the time remaining from ‘t’ minutes past 2:00 pm to 3:00 pm. So, we get
$\left( 60-t \right)=\dfrac{{{t}^{2}}}{4}-\dfrac{3}{1}$
Now, take L.C.M with the denominator in Right hand side to simplify it. So, we get
$\dfrac{60-t}{1}=\dfrac{{{t}^{2}}-12}{4}$
Now, cross multiply the above equation
$4\left( 60-t \right)={{t}^{2}}-12$
$240-4t={{t}^{2}}-12$
${{t}^{2}}+4t-252=0$ ……………………………………………….(i)
Now, we can use the quadratic formula to get the values of ‘t’ from the above quadratic. So, we know the roots of any quadratic $a{{x}^{2}}+bx+c=0$ can be given as
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ ……………………………………………..(ii)
Now, compare the calculated equation (i) and $a{{x}^{2}}+bx+c=0$ put the values of a, b, c in the equation(ii) to get values of ‘t’. So, we get
 $t=\dfrac{-4\pm \sqrt{{{\left( 4 \right)}^{2}}-4\times 1\times \left( -252 \right)}}{2\times 1}$
$t=\dfrac{-4\pm \sqrt{16+1008}}{2}$
$t=\dfrac{-4\pm \sqrt{1024}}{2}$ $t=\dfrac{-4\pm \sqrt{1024}}{2}$
Now, put $\sqrt{1024}=32$ to the above equation and hence, we get
$t=\dfrac{-4\pm \sqrt{1024}}{2}=\dfrac{-4\pm 32}{2}$
Now, take ‘+’ and ‘-‘ signs separately. So, we get
$t=\dfrac{-4+32}{2}$ and $t=\dfrac{-4-32}{2}$
$t=\dfrac{28}{2}$ and $t=\dfrac{-36}{2}$
$t=14$ and $t=-18$
Now, we can ignore $t=-18$ as time can never be a negative value.
So, the value of t is 14 minutes.
Hence, the answer is 14 minutes.

Note: We can solve the quadratic ${{t}^{2}}+4t-252=0$ by using the factorization method as well. Here, we need to split the middle term in two parts such that the product of them will be equal to the product of first and last terms.
So, we can split the middle term as
${{t}^{2}}+18t-14t-252=0$
$t\left( t+18 \right)-14\left( t+18 \right)=0$
$\left( t-14 \right)\left( t+18 \right)=0$
$t=14$ and $t=-18$
 Equating the term $\dfrac{{{t}^{2}}}{4}-3$ to $60-t$ is the key point of the question.
Put $\left( a,b,c \right)$ in the quadratic formula very carefully to get accurate values of ‘t’.