SOLUTION: at 9 am, mark and tom get on their motor bikes to meet each other from towns located 45 km apart. mark travels 5 km/h faster than tom. at what must each travel if they were to meet
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: at 9 am, mark and tom get on their motor bikes to meet each other from towns located 45 km apart. mark travels 5 km/h faster than tom. at what must each travel if they were to meet
Log On
Question 1012536: at 9 am, mark and tom get on their motor bikes to meet each other from towns located 45 km apart. mark travels 5 km/h faster than tom. at what must each travel if they were to meet at 10:12 am? Found 2 solutions by ikleyn, fractalier:Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
at 9 am, Mark and Tom get on their motor bikes to meet each other from towns located 45 km apart.
Mark travels 5 km/h faster than Tom. At what must each travel if they were to meet at 10:12 am?
----------------------------------------------------------------
Let x be Mark' speed in . Then tom' speed is x-5
Since they move toward each other, the rate of decreasing the distance between them is x + (x-5) = 2x-5 .
The time they spent for their trip is 1 hour 12 minutes, or of an hour.
Your equation is Time = , or
= .
To solve it, multiply both sides by 2x-5. You will get
6*(2x-5) = 5*45, or
12x - 30 = 225,
12x = 225 + 30 = 255.
x = = 21.25 .
You can put this solution on YOUR website! In general, remember rate times time equals distance.
Let us call Tom's rate, r.
Then Mark's rate is r+5.
The time they travel is an hour and twelve minutes...that's an hour and a fifth, or 6/5 hours.
The total distance they cover is 45 km.
Thus we can write
(6/5)r + (6/5)(r+5) = 45
Now solve this...maybe multiply everything by 5 to clear fractions...we get
6r + 6(r+5) = 225
6r + 6r + 30 = 225
12r + 30 = 225
12r = 195
r = 195/12 = 16.25 km/h
r+5 = 21.25 km/h
