A skier had to cover a distance of $30$ km. Having started $3$ minutes later than the fixed time, he glided with the speed which exceeded the assumed speed by $1$ kmph and reached the finish at the time he would reach it if he began to race strictly at the appointed time and ran with the assumed speed. Find the speed of the skier.
Total distance covered by the skier $= d = 30$ km
Let the time taken to cover $d = 30$ km be equal to $t$ hours.
$\therefore \;$ Assumed speed of skier $= s = \dfrac{30}{t}$ kmph
Actual time taken by the skier $= t$ hours $- 3$ min $= \left(t - \dfrac{3}{60}\right)$ hr
$= \left(t - \dfrac{1}{20}\right)$ hr $= \left(\dfrac{20 t - 1}{20}\right)$ hr
$\therefore \;$ Actual speed of skier $= \dfrac{30}{\left(20t - 1\right) / 20} = \dfrac{600}{20t - 1}$ kmph
Given: $\;$ Actual speed $=$ Assumed speed $+ 1$ kmph
i.e. $\;$ $\dfrac{600}{20 t - 1} = \dfrac{30}{t} + 1$
i.e. $\;$ $600 t = \left(30 + t\right) \left(20 t - 1\right)$
i.e. $\;$ $600t = 600t - 30 + 20 t^2 - t$
i.e. $\;$ $20 t^2 - t - 30 = 0$
i.e. $\;$ $\left(4t - 5\right) \left(5t + 6\right) = 0$
i.e. $\;$ $4t - 5 = 0$ $\;$ or $\;$ $5t + 6 = 0$
i.e. $\;$ $t = \dfrac{5}{4}$ $\;$ or $\;$ $t = \dfrac{-6}{5}$
Since time cannot be negative
$\therefore \;$ Time taken to cover $d = 30$ km is $t = \dfrac{5}{4}$ hour
$\therefore \;$ Actual speed of skier $= \dfrac{600}{20 \times \dfrac{5}{4} - 1} = \dfrac{600}{24} = 25$ kmph