Question 5689: A bicycle club left the campus to ride to a park 24 mi away for an outing. Lynn left from the same place by car with picnic supplies 1 1/2 hr later, taveled at a speed 4 times as fast, and arrived at the parl at the same time as the cyclists. How fast did Lynn drive?
Answer by prince_abubu(198)   (Show Source):
You can put this solution on YOUR website! This is quite a tricky problem that seems to not give enough information. We should probably start with gathering all the facts we can.
distance from campus to park = 24 miles.
bikers' speed = B (This, we don't know.).
Lynn's speed = 4B (All we know is that she was 4x faster than the bikers)
Any delays? Yes. According to the problem, Lynn only left campus when the bikers were already out there on their way for 1.5 hours.
Time it will take Lynn to drive to the park = T.
One useful trick in this problem is that we should focus on what was done by the working unit when the other unit was idle. In this case, we should focus on how much the bikers travelled while Lynn wasn't yet driving. Lynn wasn't driving for 1.5 hours. So in this case, the bikers' distance for the first 1.5 hours would be B*1.5.
After 1.5 hours that the bikers are on the road, Lynn starts driving. Now, the problem does say that they'll arrive at the park at the same time. This means that the time remaining for the bikers will be equal to the time it will take Lynn to drive the entire distance. So, the bikers' remaining distance is B*T.
Putting the previous two paragraphs together, the bikers total distance (24 miles as given) is actually the distance they travelled for the first 1.5 hours that Lynn wasn't driving, plus the remaining distance they'd travel when Lynn tries to catch up. Our equation (let's call it equation 1) is
 <---- The bikers.
Now we have to deal with Lynn. We already agreed that T will be the time it takes Lynn to drive, and we also agreed that her time delay would be "given" to the bikers as the time it took them to travel while she wasn't driving. We know that she drove 4x faster than the bikers. So, Lynn's equation is
 <---- Lynn. (Equation 2).
Since both the bikers and Lynn have to travel the same distance, we can set the bikers' and Lynn's equation (1 and 2) to each other to get
 <--- We just factored out the B
 <----- then we divided both sides by B (in this case, cancelling B from both sides.
 <---- subtract T from both sides.
 <---- So it took Lynn half an hour to drive from the campus to the park.
We're not done yet. The question asks for Lynn's actual speed. If it took Lynn half an hour to travel 24 miles, her speed MUST have been 48 miles per hour.
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