Two cars starts out simultaneously from a point in the same direction, one of them going at a speed of `50 kmhr^-1` and the other at `40 kmhr^-1`. In half an hour a third car starts out from the same point and overtakes the first car `1.5` hours after catching up with the second car. The speed of the third car is.

Options

A. `55 kmhr^-1`
B. `60 kmhr^-1`
C. `72 kmhr^-1`
D. `90 kmhr^-1`

Answer

B

Explanation

Positions of the cars when third car starts is shown
Let third car catches `2nd` car in time `T`after it starts out, then
`vT = 40 T + 20` …(i)
Now third car catches `1 st` car after time `T + 1.5`, so
`v[T + 1.5] = 50[T + 1.5]+ 25` ....(ii)
Put the value of `T` from equation (i) into equation (ii),
we get `1.5 v^2 - 140 v + 3000 = 0`
Put the value of `T` from equation (i) into equation (ii), we get
`rArr 3v^2 - 280 v + 6000 = 0`
`rArrv=(280+-sqrt((280)^(2)-72000))/(6)`
`rArr v =(280 +- 80)/(6) = 60 km//h, (100)/(3) km//h`
Since `v` can't be less than `50 km//h`, we have
`v = 60 km//h`.