A car leaves city A to city B, the way's length is 210 km.
20 minutes after it, a courier on a bike leaves city A so he can pass a delivery to the car.His speed is 120 km/h.
Immmediately after he passes the delivery, he turns back to city A.The car keeps moving with the same speed, and at the moment it reaches city B, the courirer reaches city A.
Find car's speed.
Average speed = Distance travelled / Time
Let the distance of the meet-up = D
The distance from the meet-up to city B = 210 - D
Let speed of the car =S
Let the time taken by the car until they meet = T(hours)
The time taken by the biker to catch up to the car =T - 1/3(hours)
S*T = 120[T - 1/3], solve for S
S = 120 - 40/T - speed of the car
T*[120 -40/T] + 120[T - 1/3] = 2D,
[210 - D] / [120 - 40/T] =D/120, solve for D, T
D =120Km, and T =4/3 hours.
120/(4/3) =90 Km/h - The speed of the car.
Let R be the car's rate
Call the time for the car to travel (in hours) = 210/R
Call the total time for the bike to travel (in hours) = 210/R - 1/3
Now.....the distance bike travels = its rate * its time = 120 * [ (210/R) - 1/3 ]
And the distance the bike travels to meet the car = 120* [ (210/R - 1/3] / 2 = 60[ (210/R - 1/3 ] = the same distance the bike travels back to "A"
So.....the distance the bike travels back to "A" / ts rate = some time (1)
And the distance the car travels after meeting the bike / its rate = the same time (2)
So.......equating (1) and (2) in algebraic terms, we have
( 210 - 60 [ 210/R - 1/3] ) / R = 60[ 210/R- 1/3 ] / 120
(210 - 60 [ 210/R - 1/3 ] ) / R = [ 210/ R - 1/3 ] / 2 cross-multiply
2(210 - 60[ 210/R - 1/3 ] = R [ 210/R - 1/3 ]
420 - 120 [ 210/R - 1/3 ] = 210 - R /3
420 - 25200/ R + 40 = 210 - R/3
460 - 25200/R = 210 - R/3
R/3 + 250 - 25200/R = 0 multiply though by R
(1/3)R^2 + 250R - 25200 = 0 multiply through by 3
R^2 + 750R - 75600 = 0 factoring, we get
(R - 90) (R + 840) = 0
Only the first factor set to 0 will provide a positive solution for R = 90 kmh = the car's rate