SOLUTION: The distance between city A and city B is 100 km. At 7 AM, a car carrying bananas left from city A to city B, and at the same time, a bike rider left from city B to A. They met at

Question 638023: The distance between city A and city B is 100 km. At 7 AM, a car carrying bananas left from city A to city B, and at the same time, a bike rider left from city B to A. They met at 8:40, and immediately after they both continued to their destination. The car arrived at city B, waited there for 55 minutes, and left back to city A. At 11 AM, the car passed the bike rider that was still on his way to city A. The bike and car's speeds did not change during the drive. What were their speeds?
Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
Basic formula d = rt, where d = distance, r = rate, t = time
Let c = car's rate of speed, km/min
Let b = bike's rate of speed, km/min
Let d1 = distance travelled by car during 7:00am-8:40am, 100 min
Let d2 = distance travelled by bike during 7:00am-8:40am, 100 min
Let d3 = Distance travelled by bike during 8:40am-11:00am, 140 min
First period of time, 7:00am-8:40am. 100 min
(1) d1 = c*100
(2) d2 = b*100
(3) d1 + d2 = 100
Add (1) and (2) yields
d1 + d2 = 100(b+c)
Equate to (3) yields
(4) b + c = 1
Second period of time 8:40am-11:00am, 140 min
The distance travelled by car is
(d2 + 0 + d2 + d3) = 2*d2 + d3
Using d=rt yields
(5) 2*d2 + d3 = c*140
The distance travelled by bike is
(6) d3 = b*140
Substitute (6) into (5) yields
(7) 2*d2 + 140b = 140c
Substitute (4) into (7) yields
(8) 200b + 140b = 140c or
(9) 340b = 140c
Substitute (4) into (9) gives
(10) 340b = 140(1-c), simplifying yields
b = 7/24
Using (4) yields
c = 17/24
Check your answers.
Is (d1 + d2 = 100)?
Is (100*17/24 + 100*7/24 = 100)?
Is (100*(17/24+7/24) = 100)?
Is (100*1 = 100)? Yes
Is ((2*d2+d3)/(17/24)= 140)?
Is ((200+140)*7/24/17/24 = 140)?
Is (340*7/17 = 140)?
Is (140 = 140)? Yes
Answer: The car 's speed is 17/24 km/min and the bike's speed is 7/24 km/min.
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