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.

9 thoughts on “A traditional approach to a traditional word problem in algebra”

1. Wayne Bishop

My only criticism/suggestions would be on the integer simplifications. Simple tests for common factors (the divisors of 10) and the sum of the digits divisible by 3 should be known and is usually enough (it makes the 39 much more obvious in this example) and reduction by multiples of 1001 for 7, 11, and 13 simultaneously for big numbers is kinda cute.

Beyond knowing and using these “tricks”, the “whys” of these simple tests are interesting and worth exploring in an algebra class. Even more, examining their equivalents in other base notation adds another level of understanding.

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• Thanks; good suggestions.

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2. Anna

Excellent question, Mr. Garelick. My daughter actually had a question similar to this for her algebra class this year. Once your students get to ratio and rates, they could tackle one like this:

“A plane flies from Alphaville to Betaville and then back to Alphaville. When there is no wind, the round trip takes 4 hours and 48 minutes, but when there is a wind blowing from Alphaville to Betaville at 100 miles per hour, the trip takes 5 hours. How many miles is the distance from Alphaville to Betaville?”

Solution:
Let d be the distance from Alphaville to Betaville in miles, and let r be the speed of the plane in miles per hour. When there is no wind, the time it takes to fly from Alphaville to Betaville in hours is d/r, and 48 minutes is 0.8 hours, so (2d)/r = 4.8.
Solving for d, we get d = 2.4r.

When there is a wind blowing from Alphaville to Betaville at 100 miles per hour, the plane’s speed going from Alphaville to Betaville is r + 100, and the plane’s speed going back from Betaville to Alphaville is r – 100, so the time it takes for the plane to complete the round trip, in hours, is d/(r + 100) + d/(r – 100). Combining the fractions, we get

d/(r + 100) + d/(r – 100) = [d(r – 100) + d(r + 100)] / [(r + 100)(r – 100)]
= (dr – 100d + dr + 100d) / (r^2 – 10000)
= (2dr)/(r^2 – 10000)

We are given that the round trip takes 5 hours when there is a wind blowing, so (2dr)/(r^2 – 10000) = 5. Cross-multiplying, we get 2dr = 5r^2 – 50000. Substituting d = 2.4r, we get 4.8r^2 = 5r^2 – 50000, so 0.2r^2 = 50000, which means r^2 = 250000. Taking the (positive) square root, we find r = 500, so d = 2.4r = 1200. Therefore, the distance from Alphaville to Betaville is 1200 miles.

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• Anna

It was fun to listen to my son (who is in pre-Algebra) discuss and argue concepts with her on this question. He couldn’t see why the tail wind/head wind didn’t cancel each other out and my daughter had to reason through the problem herself to explain it.

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• Nice problem. What algebra book does your daughter use? And in what city are you located? I’m always curious about what books are being used, and where.

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3. Anna

We finally pulled the plug on the public school and decided to go the homeschool route. They jettisoned Geometry in 8th grade 2-3 years ago and in the fall, they will no longer offer separate math courses. It will be 7th grade math, 8th grade math. Middle schoolers will be bussed to the HS if they want Algebra.

We live about 75 miles north of Seattle in a small town. The schools used to used TERC for K-6, Eureka now (both are horrible!). For middle school, they’ve just adopted Big Ideas by Larson and will change the HS curriculum to that series I think. I previewed the text and it is very heavy on linear algebra to the detriment of other concepts and has a strong focus on discovery. No partial fraction decomposition is presented, it is weak on ratios and rates, barely covers completing squares and factoring, and doesn’t touch optimization. It’s not the worst new textbook out there but not anywhere near the quality of Dolciani.

My oldest started with the 1988 Dolciani Algebra text and then moved to Art of Problem Solving Algebra. I think the AoPS material is great BUT students may need extra psets for review and practice. The online AoPS courses moved very fast. They covered the basics plus complex numbers, exponentials and logs, arithmetic and geometric series (finite and infinite), optimization with linear and quadratic inequalities, and special functions. The material requires a tremendous amount of effort and work from the student (and parent!).

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• I know about TERC; very bad. I also know about Eureka (nee EngageNY). And the school where I was teaching used the Big Ideas series. I didn’t care for the Big Ideas series; I skipped the discovery lessons and just focused on the direct instruction lessons. So I gave up on it and use Dolciani’s algebra textbook from 1962. I was able to get them for very low cost via Amazon. The students like them a lot better than Big Ideas, which I find interesting. Many word problems and fractions and quadratics are covered very well. (This Algebra I I’m talking about; I teach middle school math).

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4. SteveH

Back when I taught college math and computer science, I realized that when I prepared to explain how to do a problem, I consolidated all of my analysis into a final (and clear) process to explain. My error was thinking that my final solution process was helpful to them when they approached a new problem – as if they should be able to get immediately to that level of problem understanding. They might have understood exactly what I did, but that didn’t mean that it would be helpful to them on a new problem. I evolved into giving them a more general process that starts with finding the governing equation, like D=RT, and then defining unique and simple equations, like

550 = R1 * T1
and
425 = R2 * T2

I told them to not worry about too many variables or figuring out the solution ahead of time. Students make errors when they try to directly figure out a final set of M=N equations. Next, it should be easy to see that T1 = T2. Then after that, they can see that

R1 = 195 + W
and
R2 = 195 – W

They then have one equation in one unknown. They would get almost full credit at that point. I told them to define proper variables and write down legal equations. Let the math do the thinking for them. This is also psychologically good because they can quickly get partial credit even if they somehow get confused about wind being completely additive and subtractive to rate. Is it exactly? I can imagine my son questioning that. They might cancel out, but what cancels out might not be the free wind speed. What you add and subtract might be (W – delta).

In Anna’s more difficult problem, a student should see the need to divide the problem into two legs:

Dab = Rab * Tab
Dba = Rba * Tba

This is the result of no thinking step 1.

Then, Dab = Dba

D = Rab * Tab
D = Rba * Tba

Then, if the plane flies at a constant rate R, then
Rab = R + W
Rba = R – W

So

D = (R+W) * Tab
D = (R-W) * Tba

OK. I’m still not thinking too hard yet.

The total time T = Tab + Tba, so this is my general equation for any wind speed

T = D/(R+W) + D/(R-W)

Case 1, I have, W = 0 and T = 4.8 hours

4.8 = D/R + D/R … or D = 2.4R

Case 2, I have W = 100 and T = 5.0 hours

5.0 = D/((R+100) + D/(R-100)

Two equations in two unknowns – turn the crank, as my teachers would say.

Well, turning the crank may not be easy, but that’s a different set of unit skills.

Ultimately, I told students to break problems down into simple proven equations and let the math do the thinking for them. If the math does not agree with your thinking, then let it explain it to you. Never try to think your way to a fast solution path. You will always get into trouble. Too many variables and equations might seem like a slow process, but it isn’t.

The math will show you that distance might vary proportionately with rate, but it’s inversely proportional to time, as in:

T = D/R

If distance is 100 miles and R = 20, T = 5.
But if you add and subtract 5 from R=20, you get T = 4 and T = 6.67

The time differences are not the same +/- from R = 20. T is not linear with R.

Math will explain it all to you, but that doesn’t seem to be the “understanding” many educators are looking for. They seem to think that it’s a concept/gut level approach – which can get you into big trouble.

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