SOLUTION: A tour bus normally leaves for its destination at 5:00 p.m. for a 300 mile trip. This week however, the bus leaves at 6:00 p.m. To arrive on time, the driver drives 10 miles per

Question 1176546: A tour bus normally leaves for its destination at
5:00 p.m. for a 300 mile trip. This week however, the bus leaves at 6:00 p.m. To arrive on time, the driver drives 10 miles per hour faster than usual. What is the bus` usual speed?
Found 2 solutions by ikleyn, ewatrrr:
Answer by ikleyn(52776) (Show Source): You can put this solution on YOUR website!
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A tour bus normally leaves for its destination at 5:00 p.m. for a 300 mile trip. This week however, the bus leaves at 6:00 p.m.
To arrive on time, the driver drives 10 miles per hour faster than usual. What is the bus` usual speed?
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Let x be its regular speed, in miles per hour. Then the increased speed is (x+10) mph. The regular travel time is hours. The travel time with increased speed is hours. The difference of traveled time is (from the context) 1 hour - = 1 hour. It is your time equation to find x. You can guess the solution MENTALLY just from this point: it is 50 mph. ANSWER Indeed, - = - = 6 - 5 = 1 hour. Or, alternatively, you may reduce equation (*) to quadratic equation and solve it formally.

At this point, the solution is completed.

Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!

Hi d = rt 300 = rt 0r 300/r = t 300 = (r+10)(t-1) 300 = (r+10)(300/r -1 ) 300 = (r+10)((300-r)/r 300r = (r+10)((300-r) 300r = 300r + 3000 - r^2 - 10r r^2 + 10r + 3000 = 0 (r+60)(r-50) = 0 r = 50mph, the usual speed |tossing out negative number for speed 50mph*6hr = 60mph*5hr = 300mi checks Wish You the Best in your Studies.

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