Jamalâ€™s trip requires him to drive 20mi to the airport, fly 2,300mi, and then take a bus 6mi to his hotel. His average speed driving to the airport is 40mph, the average speed of an airplane is 575mph, and the average speed of his bus is 15mph.

Aside from time waiting for the plane or bus, how long is Jamal in transit?

Bear’s Strategy: Lion’s Strategy:

Drive Time = 20 miles × 1hour/( 40 miles ) = 0.5 hours$\displaystyle Drive Time = 20 miles \times \frac{1hour}{40 miles} = 0.5 hours$

Fly Time = 2300 miles × 1hour/( 575 miles ) = 4 hours$\displaystyle Fly Time = 2300 miles \times \frac{1hour}{575 miles} = 4 hours$

Bus Time = 6 miles × 1hour/( 15 miles ) = 0.4 hours$\displaystyle Bus Time = 6 miles \times \frac{1hour}{15 miles} = 0.4 hours$

In Transit Time = Drive Time + Fly Time + Bus Time$\displaystyle In Transit Time = Drive Time + Fly Time + Bus Time$

0.5 + 4 + 0.4 = 4.9 hours$\displaystyle 0.5 + 4 + 0.4 = 4.9 hours$

In Transit Time = Drive Time + Fly Time + Bus Time$\displaystyle In Transit Time = Drive Time + Fly Time + Bus Time$

Drive Time = 20 miles × 1hour/( 40 miles ) = 0.5 hours$\displaystyle Drive Time = 20 miles \times \frac{1hour}{40 miles} = 0.5 hours$

Fly Time = 2300 miles × 1hour/( 575 miles ) = 4 hours$\displaystyle Fly Time = 2300 miles \times \frac{1hour}{575 miles} = 4 hours$

Bus Time = 6 miles × 1hour/( 15 miles ) = 0.4 hours$\displaystyle Bus Time = 6 miles \times \frac{1hour}{15 miles} = 0.4 hours$

0.5 + 4 + 0.4 = 4.9 hours$\displaystyle 0.5 + 4 + 0.4 = 4.9 hours$

1 Whose Strategy was Top Down? How do you know?

2 Whose Strategy was Bottom Up? How do you know?

3 Which way of thinking about the problem makes more sense to you?

 What’s happening with that Math?! When calculating Jamal’s drive time, we multiplied distance by speed. More specifically, we multiplied the starting value (20 miles$\displaystyle 20 miles$) by ( 1 hour )/( 40 miles )$\displaystyle \frac{1 hour}{40 miles}$. Why? Why not reverse it, to use ( 40 miles )/( 1 hour )$\displaystyle \frac{40 miles}{1 hour}$, as stated in the problem? Time$\displaystyle Time$ is the desired outcome. Looking at the units, we can see that speed must have miles$\displaystyle miles$ as its denominator to cancel out the miles$\displaystyle miles$ in the starting value. ( 20 mi )/1 × 1hour/40miles = ( 20\requireenclose/ \enclosehorizontalstrikemiles ×1hour )( 40 \requireenclose \enclosehorizontalstrikemiles ) = 20/40 hour = 1/2 hour$\displaystyle \frac{20 mi}{1} \times \frac{1hour}{40miles} = \frac{20\require{enclose} \enclose{horizontalstrike}{miles} \times1hour}{40 \require{enclose} \enclose{horizontalstrike}{miles}} = \frac{20}{40} hour = \frac{1}{2} hour$

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