Question

A cyclist covered a distance of $96$ km two hours faster than he assumed. Every hour he travelled $1$ km more than he intended to cover in $1$ h $15$ min. What was his speed?

Answer 1

He planned to cover $x$ km in $1hr15min$ $\left(\frac{75}{60}m\right)\to \left(\frac{5}{4}hr\right)$

He actually covered $(x+1)$ km in every hour.

Planned speed $=\frac{x}{\frac{5}{4}}=\frac{4x}{5}km/h$

Actual speed $\frac{x+1}{1}=x+1$km/hr

He reached the $96$ km 2 hrs faster than his assumed speed.

$\therefore$ $\frac{96}{\frac{4x}{5}}-\frac{96}{x+1}=2$

$\frac{120}{x}-\frac{96}{x+1}=2$

$120(x+1)-96x=2x(x+1)$

$120x+120-96x=2x^2+2x$

$2x^2-22x-120=0$

$x^2-11x-60=0$

$x^2-15x+4x-60=0$

$x(x-15)+4(x-15)=0$

$x=15$km/hr

Actual speed $=x+1=16$km/hr