I saw this problem a few days ago and it got me thinking how I would present it in my algebra class:
“An airplane can fly 550 miles with the wind in the same amount of time it takes to fly 425 miles against the wind. Find the speed of the wind, if the airplane is flying at a constant speed of 195 miles per hour.”
First I would ask my students what are we equating? Is it a distance=distance problem or time=time? You could make it into distance = distance, but the time=time is simpler to deal with. Time is expressed as distance over rate. We know the distances, and if we let w = the wind speed, then we can express the rates as 195+w and 195-w.
We get this equation: 550/(195+w) = 425/(195-w), which we can simplify into
550(195-w) = 425(195+w).
Big numbers, so before they can ask whether they can use calculators, the first thing I ask the students is whether we can whittle down 550 and 425 by dividing both sides by a common factor? Dividing by 5 two times in a row (or 25 once) yields: 22/(195+w) = 17(195-w).
They still want to use their calculators. I forbid their use for this problem.
“Let’s just write it out without doing any multiplying,” I tell them.
17(195) + 17w = 22(195) – 22w
Let’s combine like terms: 39w = 22(195) – 17(195)
Is there a common factor on the right hand side? Yes; so now the equation looks like 39w = (5)(195) and w = 195(5)/39
Do you think 195 can be divided by 39? Let’s try. In fact it can; it’s 5. Now we have w = 5 x 5 = 25 mph. The only big number calculation we did was dividing 195 by 39.
This type of approach helps develop number sense, which some claim traditionally taught math doesn’t do a good job developing. I present it here as an example of what I consider a rich problem. It’s not the problem that is rich so much as the way it can be used to address different strategies that may be employed within the same problem.