Question

⭐At t min past 2 pm, the time needed by the minute hand clock to show 3 pm was found to be 3 min less than t^2/4 min. find t. ⭐

CLASS 10 CHAPTER :QUADRATIC EQUATIONS

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Answer 1

Given that At t min past 2 pm, the time needed by the minute hand clock to show 3 pm was found to be 3 min less than t^2/4 min.

We know that 1 hour = 60 minutes.

According to the given condition:

⇒ t + (t^2/4) - 3 = 60

⇒ 4t + t^2 - 12 = 240

⇒ t^2 + 4t - 12 = 240

⇒ t^2 + 4t - 252 = 0

⇒ t^2 + 18t - 14t - 252 = 0

⇒ t(t + 18) - 14(t + 18) = 0

⇒ (t - 14)(t + 18) = 0

⇒ t = 14,-18{t cannot be negative}.

⇒ t = 14.

Therefore, the value of t = 14.

Hope it helps!

Answer 2

$<b><font color = "red"> Hello Friend$


Find your answer below


Since minute hand is t min more than 2 pm minutes hand has completed t min after 2 pm.


So, to show show 3 pm it will cover 60 - t minute more.


$\mathsf {\implies According \: to \: the \: Question }$


$\mathsf {\implies 60 - t = {\dfrac {t^{2}}{4} - 3}}$


$\mathsf {\implies 60 - t = {\dfrac {t^{2} - 12}{4}}}$


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


$\mathsf {\implies t^{2} + 4t - 240 - 12 = 0}$


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


$\mathsf {\implies t^{2} + 18t - 14t - 252 = 0}$


$\mathsf {\implies t ( t + 18 ) - 14 ( t + 18 ) = 0}$


$\mathsf {\implies ( t - 14 ) ( t + 18 ) = 0}$


$\mathsf{\implies t = 14 \: or - 18}$


$\mathsf {\implies Since \: time \: cannot \: be \: negative }$


$\mathsf {\implies t = 14 \: minutes}$


$\huge\bold\color {green}Hope \: It \: Helps$


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