Question 1205336: problem 1:
A cyclist, riding steadily from his hometown, reached his destination in 4.5 h. He would have taken 45 min less time if he had ridden his bicycle at a speed faster by 3 km/h. Find the distance from his hometown to his destination.
problem 2:
At 9:30 am, car A started its journey and travelled at 65 km/h. At 11:00 am, car B started its journey from the same place and travelled on the same road as car A. If car B , travelling at a constant speed, took five hours to catch up with car A, find the speed of car B.
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Found 3 solutions by Theo, mccravyedwin, Edwin McCravy:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! problem 1:
A cyclist, riding steadily from his hometown, reached his destination in 4.5 h. He would have taken 45 min less time if he had ridden his bicycle at a speed faster by 3 km/h. Find the distance from his hometown to his destination.
r * t = d
r = rate
t = time
d = distance
your first equation is r * 4.5 = d
45 minutes is .75 of an hour.
your second equation is (r + 3) * (4.5 - .75) = d
simplify the second equation to get (r + 3) * 3.75 = d
simplify the second equation further to get 3.75 * r + 11.25 = d
your two equations are now:
4.5 * r = d
3.75 * r + 11.25 = d
subtract the second equation from the first to get:
.75 * r - 11.25 = 0
add 11.25 to both sides of the equation to get:
.75 * r = 11.25
solve for r to get:
r = 11.25 / .75 = 15
your two equations become:
4.5 * 15 = d = 67.5
3.75 * 18 = d = 67.5
the distance is the same as it should be at 67.5 kilometers.
that's your solution.
problem 11.
At 9:30 am, car A started its journey and travelled at 65 km/h. At 11:00 am, car B started its journey from the same place and travelled on the same road as car A. If car B , travelling at a constant speed, took five hours to catch up with car A, find the speed of car B.
car A started its journey at 9:30 am, travelling at a constant speed of 65 km per hour.
car B started at 11:00 am, traveling the same road at a constant unknown speed. catching up to car B in 5 hours.
if car B started at 11:00 am and took 5 hours to catch up to car A, then car B caught up to car A at at 4:00 pm.
at 4:00 pm, car B was traveling for 5 hours.
at 4:00 pm, car A was traveling for 6.5 hours
car A traveled 6.5 hours at a rate of 65 kmph for a distance of 422.5 km.
car B traveled 5 hours for a distance of 422.5 km.
car B rate was 84.5 kmph.
to confirm, your two equations were:
65 * t = 422.5 for car A.
solve for t to get t = 422.5 / 65 = 6.5 hours.
84.5 * t = 422.5 for car B.
solve for t to get t = 422.5 / 84.5 = 5 hours.
they started at the same place and traveled the same road.
car B caught up to car A 422.5 at a distance of 422.5 kilometers from the starting point.
since car B started at 11:00 am and caught up to car B in 5 hours, the time was 4:00 pm when car B caught up to car !.
at 4:00 pm, car A was traveling 4 + 2.5 = 6.5 hours.
6.5 * 65 = 422.5
5 * 84.5 = 422.5
your solution is that the speed of car B was 84.5 kilometers per hour.
Answer by mccravyedwin(409)  (Show Source):
You can put this solution on YOUR website! A cyclist, riding steadily from his hometown, reached his destination in 4.5 h.
He would have taken 45 min less time if he had ridden his bicycle at a speed
faster by 3 km/h. Find the distance from his hometown to his destination.
45 minutes = 3/4 hour = 0.75 hour. 0.75 less than 4.5 = 4.5-0.75 = 3.75 hours.
Rate | Time | Distance
At actual speed r | 4.5 | 4.5r
If faster r+3 | 3.75 | 3.75(r+3)
The distances are the same
4.5r = 3.75(r + 3)
450r = 375(r + 3)
450r = 375r + 1125
75r = 1125
r = 15
Substitute in 4.5r = 4.5(15) = 67.5 km.
Edwin
Answer by Edwin McCravy(20064)  (Show Source):
You can put this solution on YOUR website! At 9:30 am, car A started its journey and travelled at 65 km/h. At 11:00 am, car
B started its journey from the same place and travelled on the same road as car
A. If car B , travelling at a constant speed, took five hours to catch up with
car A, find the speed of car B.
Let r = the speed of car B
At 11 AM, car A had already been traveling 1.5 hours at 65 km/h, and therefore
was (1.5)(65) = 97.5 kilometers down the road.
The catch-up rate is the difference of their speeds, r-65 km/h
To reduce the 97.5 km to 0 in a 5 hour time period, we use
(rate)(time)=distance
(r-65)(5) = 97.5
5r - 325 = 97.5
5r = 422.5
r = 84.5 km/h
Edwin
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