ShmoopTube

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And here's your shmoop du jour...

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Justin rode his bike to school, pedaling at an average speed of 12 miles/hour.

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When he rode his bike back home at a leisurely pace of 10 miles/hour, it took him an extra 15 min.

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How far is his school from his home?

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And here are the potential answers...

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OK, we want to find the distance between Justin's home and his school.

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We know that he biked to school at 12 miles per hour, but we don't know how long it

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took him to get there. We'll call the time it takes him to bike there, x.

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Next, we know that when coming home, he biked at only 10 miles per hour.

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And it took him 15 minutes longer than before.

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We can take the old time, add 15 minutes to it, and get the new time.

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But wait.

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His velocity is given in miles per hour, and the extra time is in minutes.

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The units don't match up.

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Remember that 60 minutes are in an hour, and 15 minutes is .25, or a quarter of an hour.

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Now, we can set up equations that help us solve this problem.

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We'll start with biking to school.

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We know that velocity is equal to the distance Justin travels divided by

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the time it takes him to travel that distance. His velocity is 12 miles per hour.

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We get the equation 12 = d (distance) / x (time).

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Okay, now biking home.

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We get 10 = d over the quantity x + .25.

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We have two equations, and two variables.

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Sounds a little like we have to solve a system of equations.

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By isolating x in the first equation, we get x = d/12.

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We can then plug x into our second equation, like this.

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We multiply both sides by d/12 + .25, and we get 10 times d/12 + .25 = d.

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Distributing in the 10 we get 10/12 d + 2.5 is equal to d.

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We then subtract 10/12d from both sides to get 2/12 d,

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which we can simplify to 1/6 d is equal to 2.5.

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Finally, multiplying both sides by 6, and we get that d is equal to 15 miles.

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So our answer is (E).

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