Question

A bus was scheduled to travel from Town X to Town Y at constant speed V km/h. If the speed of the bus was increased by 20%, it could arrive at Town Y 1 hour ahead of schedule. Instead, if the bus travelled the first 120 km at V km/h and then increased its speed by 25%, it could arrive at Town Y 4/5 of an hour ahead of schedule. Find the distance between the 2 towns.

Answer 1

P.S. Would appreciate if you showed working as to how you arrived at the answer. Thanks.

Answer 2

let x be the distance between two towns and y be our speed.

time taken => distance/speed => x/y -------1

but if speed is increased by 20% => speed = y+0.2 y => 1.2 y
then, time taken - 1 => x/1.2y -------------2

or, from 1) and 2) we have x/1.2y+1 = x/y -------3

Answer 3

on the second condition, you have not specified at what point speed was increased by 25%.

if it was travelling with 25% extra speed ... then there's no way that it will have solution

Answer 4

experimentX, the speed was increased by 25% after travelling the 120 km. :)

Answer 5

increaseent by 25% i.e. aftertravelling 120. k.m.s

Answer 6

\[t = \frac{d}{V}\]
\[t-1 = \frac{d}{1.2V}\]
\[t-\frac{4}{5} = \frac{120}{V}+\frac{d-120}{1.25V}\]
Replace t with (d/V) in 2nd and 3rd equations
solve each equation for V in terms of d
set them equal to each other
\[\rightarrow \frac{d}{6}=\frac{d-120}{4}\]
solve for d
\[d = 360 \]
by substituting back in, V and t can be found as well
\[V = 60\]
\[t = 6\]