Question

A few days ago, Kevin bought a textbook from a bookstore. However, he accidentally bought an Algebra 2 textbook instead of the Algebra 1 book that he needed. He got a ride to the bookstore from Harry, but he had to walk home at a speed of 0.05 miles/min. If it took a total of 38 minutes from Kevin's house to the bookstore and back, and Harry drove at a speed of 0.9 miles/min, how long did it take Harry to drive Kevin to the bookstore? (Assume that no time was wasted at the bookstore exchanging the book).

Answer 1

2???

Answer 2

You may see that they give you a rate: .05mi/min, and time, 38 mins. By multiplying how fast he went by the amount of time it took, we may find the distance he covered in that time frame.
This answer is misleading; this takes in account the distance to his house to the bookstore AND back; we just need from his house to the bookstore. To clear this problem, just divide the answer by 2 (if you want more elaboration on that, feel free to ask).
After we found the distance, and half'ed that, we simply divide the rate of the car by the distance traveled. Lemme see...

Answer 3

So:\[.05mi/\min*38\min=1.9mi \rightarrow1.9mi/2=.95mi\]
\[\frac{ .9mi/\min }{ .95mi }=.947368\min\]
To know whether or not to multiply or divide to get the unit you want (mi, min ect), just use the formula\[velocity=\frac{ distance }{ time }\] In this case, it would be\[mi/\min=\frac{ mi }{ \min }\]

Answer 4

kevin wouldn't know cuzz he got the wrong book that shows him this problem :P