SOLUTION: The distance between towns A and B is 39km. A cyclist and a bus moved from A to B at constant speeds. A bus left A 15 minutes after a cyclist did and caught up with the cyclist in
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Question 932993: The distance between towns A and B is 39km. A cyclist and a bus moved from A to B at constant speeds. A bus left A 15 minutes after a cyclist did and caught up with the cyclist in 10 minutes. After reaching B, the bus immediately turned around, and 30 minutes after the first meeting met the cyclist the second time. Find the speeds of the car and the cyclist. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! The distance between towns A and B is 39km.
A cyclist and a bus moved from A to B at constant speeds.
A bus left A 15 minutes after a cyclist did and caught up with the cyclist in 10 minutes.
After reaching B, the bus immediately turned around, and 30 minutes after the first meeting met the cyclist the second time.
Find the speeds of the car and the cyclist.
:
A----x------*1st----------------*2nd------y------B
:
let x = distance from A to the 1st meeting point
Travel time of the cycle to x: 15 + 10 = 25 min
travel time of the bus to x: 10 min
therefore = speed of the cyle = speed of the bus
:
let y = distance from B to the 2nd meeting point
The total travel time of cycle: 25 + 30 = 55 min
The total travel time of the bus: 10 + 30 = 40 min = speed of the cycle = speed of the bus
:
Speed of cycle equation =
cross multiply
55x = 25(39-y)
55x = 975 - 25y
55x + 25y = 975
simplify divide by 5
11x + 5y = 195
Speed of bus equation =
Cross multiply
40x = 10(39+y)
40x = 390 + 10y
40x - 10y = 390
simplify divide by 2
20x - 5y = 195
:
Use elimination on these two equations
11x + 5y = 195
20x - 5y = 195
----------------adding eliminates y find x
31x = 390
x = 390/31
x = 12.58 km from A to the 1st meeting place
:
Find the cycle speed in km/hr = 30.2 km/hr speed of the cycle
find the bus speed = 75.5 km/hr speed of the bus
:
;
To check this find y using 11x + 5y = 195
11(12.58) + 5y = 195
5y = 195 - 138.4
5y = 56.6
y = 56.6/5
y = 11.3 km dist from B to the 2nd meeting point
We know the bus traveled 11.3 + 39 = 50.3 km in 40 min = 75.45, agrees with our previous calc for the bus speed
You can check it by finding the cycle speed to the 2nd meeting (39-11.3)
