pythagoras123
  • pythagoras123
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.
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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pythagoras123
  • pythagoras123
P.S. Would appreciate if you showed working as to how you arrived at the answer. Thanks.
experimentX
  • experimentX
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
experimentX
  • experimentX
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

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pythagoras123
  • pythagoras123
experimentX, the speed was increased by 25% after travelling the 120 km. :)
anonymous
  • anonymous
increaseent by 25% i.e. aftertravelling 120. k.m.s
dumbcow
  • dumbcow
\[t = \frac{d}{V}\] \[t-1 = \frac{d}{1.2V}\] \[t-\frac{4}{5} = \frac{120}{V}+\frac{d-120}{1.25V}\] 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 \[V = 60\] \[t = 6\]

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