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 t square by 4 minutes. Find t.

Answer 1

$\underline{\underline{\Huge{\textbf{Solution:}}}}$


$\underline{\Huge{\textsf{Step-1:}}}$

$\sf\textsf{Rewrite the given data}$

$\sf\textsf{If the clock shows t min past 2 p.m,}\\\sf\textsf{the minute hand needs 60-t minutes}\\\sf\textsf{to show 3 p.m}$

$\begin{array}{c@{\implies}c@{\implies}c}3 \: pm& 15:00& 14: 60 \\2 \: pm& 14:\; t &\underline{14 : \; t }\\&&60-t \end{array}$

$\sf\textsf{3 minutes less than t square by 4 min = \dfrac{t^2}{4}-3 }$

$\sf\textsf{As per the given data,}$

$\implies \mathsf{\dfrac{t^2}{4}-3= 60-t}$ ———[1]





$\underline{\Huge{\textsf{Step-2:}}}$

$\sf\textsf{Simplify [1] to form Quadratic equation}$

$\implies \mathsf{240-4t = t^2 -12}$

$\implies \mathsf{t^2+4t-12-240}$

$\implies \mathsf{t^2+4t-252=0}$





$\underline{\Huge{\textsf{Step-3:}}}$

$\sf\textsf{Factorise the Quadratic equation}$

$\begin{array}{crclc}&&-252t^2&&\\&&\swarrow\searrow&&\\&18t&&-14t&\end{array}\quad\boxed{\begin{minipage}{6cm}\item\bigstar \quad \; \, t^2+4t-252=0 \\\begin{itemize} \item t^2\times(-252)\;\,=-252t^2 \item -252t^2\qquad\;\, = 18t\times-14t \item 18t+(-14t)\, =4t \end{itemize}\end{minipage}}$

$\begin{aligned}\implies \sf{t^2+18t-14t-252}&=0\\\implies \sf{t(t+18t)-14(t+18)}&=0\\\implies \mathsf{(t+18)(t-14)}&=0\\\implies \sf{t-14}&=0\\\implies \sf{t}&=0\end{aligned}$

$\sf\textsf{While equating the term (t+18) to 0, }\\\sf\textsf{a negative value is obtained.}$

$\sf\textsf{As t (time in min.) can't be negative,}\\\sf\textsf{the value t = -18 is ignored.}$

$\therefore\; \boxed{\LARGE{\boxed{\mathsf{t = 14}}}}$





$\underline{\underline{\Huge{\textbf{Verification :}}}}$

$\sf\textsf{Substitute t = 14 in [1]}$

$\begin{aligned}\implies \dfrac{14^2}{4}-3&= 60-14\\\implies\dfrac{196}{4}-3&= 46\\\implies 49-3&=46\\\implies 46&=46\end{aligned}$

$\sf\textsf{Given data satisfied accordingly.}$