# By increasing her usual speed by 20 kilometers per hour, a bus driver decreases the time on a 20-kilometer trip by 5 minutes. How do you find the usual speed?

Apr 21, 2018

Usual speed of the bus was $60$ km/hour.

#### Explanation:

Let the usual speed be $x$ km/hour and the increased speed

is $\left(x + 20\right)$ km/hour. Time taken to cover $20$ km trip at

$x$ km/hour speed was ${t}_{1} = \frac{20}{x}$ hour and time taken to cover

$20$ km trip at $\left(x + 20\right)$ km/hour speed was ${t}_{2} = \frac{20}{x + 20}$

By given condition ${t}_{1} - {t}_{2} = \frac{5}{60} = \frac{1}{12}$ hour

$\therefore \frac{20}{x} - \frac{20}{20 + x} = \frac{1}{12}$ . Multiplying by $12 x \left(x + 20\right)$ on both

sides we get, $240 \left(20 + x\right) - 240 x = x \left(x + 20\right)$ or

$4800 + \cancel{240 x} - \cancel{240 x} = {x}^{2} + 20 x$ or

${x}^{2} + 20 x - 4800 = 0 \mathmr{and} {x}^{2} + 80 x - 60 x - 4800 = 0$ or

$x \left(x + 80\right) - 60 \left(x + 80\right) = 0 \mathmr{and} \left(x + 80\right) \left(x - 60\right) = 0 \therefore$

Either  x+80=0 :. x =-80 or x-60=0 :. x=60 ; x

can not be negative $\therefore x = 60$ km .

Usual speed of the bus was $60$ km/hour [Ans]