A cyclist covered a distance of $96$ km two hours faster than he assumed. Every hour he traveled $1$ km more than he intended to cover in $1$ hr $15$ min. What was his actual and assumed speed?
Total distance covered by the cyclist $= 96$ km
Let the assumed time taken $= t$ hours
Then, assumed speed of the cyclist $= \left(\dfrac{96}{t}\right)$ kmph
Actual time taken by the cyclist $= \left(t - 2\right)$ hours
$\therefore \;$ Actual speed of the cyclist $= \left(\dfrac{96}{t - 2}\right)$ kmph $\;\;\; \cdots \; (1)$
Let the distance covered by the cyclist in $1$ hr $15$ min $\left(\dfrac{5}{4} \; \text{hr}\right)$ be $= d$ km
Then the distance covered in $1$ hr by the cyclist $= \left(d + 1\right)$ km
Now, distance covered in $\left(\dfrac{5}{4}\right)$ hr is the assumed distance covered with an assumed speed of $\left(\dfrac{96}{t}\right)$ kmph
$\therefore \;$ Assumed distance covered $= d = \dfrac{96}{t} \times \dfrac{5}{4} = \dfrac{120}{t}$ km
Actual distance covered in $1$ hour $= d + 1 = \left(\dfrac{120}{t} + 1\right)$ km
$\therefore \;$ Actual speed in $1$ hr $= \left(\dfrac{120}{t} + 1\right)$ kmph $\;\;\; \cdots \; (2)$
$\therefore \;$ We have from equations $(1)$ and $(2)$
$\dfrac{96}{t - 2} = \dfrac{120}{t} + 1$
i.e. $\;$ $96 t = \left(120 + t\right) \left(t - 2\right)$
i.e. $\;$ $96 t = 120 t - 240 + t^2 - 2t$
i.e. $\;$ $t^2 + 22t - 240 = 0$
i.e. $\;$ $\left(t + 30\right) \left(t - 8\right) = 0$
i.e. $\;$ $t = -30$ $\;$ or $\;$ $t = 8$
$\because \;$ Time taken cannot be negative
$\implies$ Assumed time of the cyclist $= t = 8$ hr
$\therefore \;$ Assumed speed of the cyclist $= \dfrac{96}{t} = \dfrac{96}{8} = 12$ kmph
and actual speed of the cyclist $= \dfrac{96}{t - 2} = \dfrac{96}{8 - 2} = \dfrac{96}{6} = 16$ kmph