A Rate Problem: Two Speeds, Two Ways

(A new question of the week)

A question we got at the end of March asked about a standard kind of algebra word problem that can be solved in a couple very different ways. It illustrates several choices that can be made (both about the meaning of the problem and how to solve it), as well as why we ask students to show what they have tried when they ask a question! It’s also a good reminder that sometimes a nudge is all you need.

The word problem

Here is the question, from Elise:

How do I solve this question:

Damien drives his car at an average that is 40km/h faster than his average speed on a bicycle.  He drove his car to his local maintenance shop, because it needed a tune-up.  He had his bike on the roof of the car so he could ride his bike home while the car was being worked on.  The total distance to the garage and back home is 30km and the total trip to drive to the garage and ride his bike home took 2 h. Determine his average speed on the bike from the garage to the home.

Normally, when a student shows no work, I ask to see what they have tried, and perhaps give a hint to get them started in case they have no idea. This time I chose to go a little further in my hint than usual, in part because of an interpretational issue I saw. I first restated the problem in a more readable form to help me find the information, and then started the process of turning it into equations:

A procedure for setting up equations

Hi, Elise.

I presume you are taking algebra; if not, let me know what is the context of this problem.

The problem is:

Damien drives his car at an average that is 40 km/h faster than his average speed on a bicycle. 

He drove his car to his local maintenance shop, because it needed a tune-up. 

He had his bike on the roof of the car so he could ride his bike home while the car was being worked on. 

The total distance to the garage and back home is 30 km and the total trip to drive to the garage and ride his bike home took 2 h.

Determine his average speed on the bike from the garage to the home.

I would start by thinking about what we know, and what we want to find out.

We know how the two speeds are related, but not what they are.

We know the total distance of the trip, and the total time.

We know how speed, distance, and time are related: d = rt (where d is distance, r is rate (speed), and t is time.

We aren’t told that the trips each way follow the same route and have the same distance, though that seems likely; so we probably shouldn’t assume the trip is 15 km each way … but we might change our mind on that!

So it looks like we can either define a variable for each speed, or just one variable for one of the speeds.

This process of taking inventory is an important first step, as we focus our attention on what is known and what is unknown. Sometimes two variables are needed, and sometimes we could use only one. But for students familiar with using more than one variable, doing so makes the initial translation step easier. The most important thing at this step is clarifying what the problem means, and revealing a possible ambiguity to keep in mind.

I continued,

Now, let’s remove unnecessary information to reduce the problem to its essentials:

The car’s speed is 40 km/h faster than the bicycle’s speed

The total distance of the combined trip is 30 km.

The total time is 2 h.

We want the speed of the bike.

I see two reasons there for defining a variable like this:

R = (average) speed of bike

One reason is that this is what we want to find out; the other is that if we know this, we can easily write an expression for the speed of the car. What would that be?

___ = (average) speed of car (in km per hour)

Sometimes “what we want to find” and “what we can use to find other quantities in the problem” are not the same thing, and we have to choose the latter in order to proceed. But when the two criteria coincide, I would go for it!

But we have to say something about time, too:

We could write an expression for the distance each vehicle goes if we knew how much of the 2 hours was spent on each; so we might define another variable, and write another expression:

T = time spent on bike (in hours)

___ = time spent in car (in hours)

Or, we could use a variable for the time spent in the car, or for the distance the bike is ridden, …

So you have some choices to make; I think my suggestions are reasonable ones, but not the only good ones.

Let’s see you take it from here. Try writing expressions for other quantities you need, then use those to write two equations.

As I often do, I wrote those general thoughts before actually working on the problem, in order to show my initial thinking without bias. But …

Having written all that, I tried solving the problem, and found that I could only write one equation with two variables. So I’m guessing that we might be required to assume the same distance each way.

Here’s what we need now: Please tell me what topics you have been learning, assuming this problem is for a class; then show me whatever work you tried, and where you got stuck. There are several places I can see where you might have stopped because you weren’t sure of your work!

I had recognized from the start, without mentioning it, that many problems like this specify that the return trip is along the same route, but this one doesn’t say that. I didn’t want to make that assumption initially; but when I realized there was not enough information to solve the problem as I was interpreting it, I had to change my interpretation. That sometimes happens!

Solving the problem with two variables

The problem as now defined is:

Damien drives his car at an average that is 40 km/h faster than his average speed on a bicycle. 
The total distance to the garage and back home (on the same route) is 30 km (so each leg is 15 km), and the total trip to drive to the garage and ride his bike home took 2 h.
Determine his average speed on the bike from the garage to the home.

We’ll be using a different approach below, so let’s finish my method now.

I defined

\(R\) = (average) speed of bike
\(T\) = time spent on bike (in hours)

so that

\(R + 40\) = (average) speed of car (in km/h)
\(2 – T\) = time spent in car (in hours)

Applying the formula d = rt to each leg of the trip, I get the equations

\(RT = 15\)
\((R + 40)(2 – T) = 15\)

This is a nonlinear system of equations, which can be solved by substitution; one way to do this is to solve the first equation for T (since we want to solve for R):

\(\displaystyle\left(R + 40\right)\left(2 – \frac{15}{R}\right) = 15\)

Expanding this and then clearing fractions, we get

\(\displaystyle2R – 15 + 80 – \frac{600}{R} = 15\)

\(2R^2 + 50R – 600 = 0\)

\(R^2 + 25R – 300 = 0\)

We’d like to factor this, but can’t. I’ll finish by completing the square:

\(R^2 + 25R + 156.25 = 300 + 156.25\)

\((R + 12.5)^2 = 456.25\)

\(R = -12.5\pm\sqrt{456.25} \approx 8.86,-33.86\)

That was a little harder than I was originally expecting. The speed of the bike (which has to be positive) is 8.86 km/h.

A wrong way and an almost right way

Elise replied, showing two attempts, one very wrong and the other nearly right though quite different from mine:

Averaging speeds: no

I’m actually in grade 11 and we are covering Rational Expressions.  Here are some attempts I made but none of the answers I came up with seem correct. Please help!

First way:

Car speed (C) = v+40
Bike speed (B) = v
Distance (D) = 30 km
Total time (T) = 2 hours
Time it took the car to get to the garage: 15/(v+40)
Time it took to ride the bike: 15/v

avg = t + t + 40

30 km/2 h = 15 km/h

So

15 = 2t + 40

15 – 40 = 2t + 40 – 40

-25 = 2t

-12.5 km/h = speed of bike

She nicely listed various quantities, defining them either as known constants or using the one variable v, and used the formula t = d/r (derived from d = rt) to find the time for each part of the trip. Then she added the two speeds (evidently intending to average them but not dividing by 2) and set that equal to the average speed calculated from total time and distance. This gave an invalid equation, whose solution was clearly wrong.

It wouldn’t have worked even if she had divided by 2. This is a common error, due to the fact that averaging speeds over different parts of a trip weights them improperly (because the faster lap takes less time). For a good explanation, see

Average Speed of a Caterpillar

A rational equation: yes

Then she tried again:

Second way:

Car speed (C) = v+40
Bike speed (B) = v
Distance (D) = 30 km
Total time (T) = 2 hours
Time it took the car to get to the garage: 15/(v+40)
Time it took to ride the bike: 15/v

T = C + B

2 = 15/(v+40) + 15/v

2[v(v+40)] = 15/(v+40)[v(v+40)] + 15/v [v(v+40)]

2v^2 + 80v = 15v + 15v + 600

2v^2 + 50v – 600 = 0

This time the equation was good, formed by setting the known total time (in hours) equal to the sum of the times for each part. She started the work by multiplying by the LCD to clear fractions, leading to a quadratic equation. She offered two ways to continue from there:

To finish the equation, it could be this?

v^2 + 25v – 300 = 0

(v + 5)(v – 5) = 300

v + 5 = 300 or v – 5 = 300

v = 295 km/h or 305 km/h ?

Or it could be this?

v^2 + 25v – 300 = 0

v(v + 25) = 300

v = 300 or v = 325 ?

These are, respectively, a wrong way and a right way to factor the wrong expression. Now we know exactly where help is needed.

How not to solve a quadratic equation

I replied:

Thanks. I should have just asked for your work and context from the start, as we usually do.

Knowing you are studying rational expressions changes my approach a little; my method used my variables R and T largely to avoid rational expressions! The points at which I saw potential difficulties included a brief use of a rational equation anyway, a resulting quadratic equation, and a resulting irrational solution. But my solution is a reasonable speed for a bike, as most of yours are not …

I see that you did assume the same distance each way, as I had concluded we must. My equations were RT = 15 and (R+40)(2-T) = 15.

We saw my work with these equations above. I mentioned them just in case Elise later wanted to try my ideas.

Now I had to deal with the main issue, the quadratic equation:

Looking at your solutions, the first makes no sense to me, but the second is very good; you end up with the same equation I got, v^2 + 25v – 300 = 0.

But it isn’t useful to factor a quadratic equation unless the right-hand side is 0; when you get (v+5)(v-5) = 300, you can’t conclude that v+5 = 300 or v-5 = 300! Moreover, that isn’t the factorization of v^2 + 25v! In your last attempt you factored correctly, but again that is useless.

It turns out that v^2 + 25v – 300 can’t be factored; so the appropriate thing to do is to complete the square, or use the quadratic formula. That will lead to the correct answer!

So your error is not in rational equations, but in solving quadratic equations. That is not uncommon.

I look forward to seeing your finished work soon.

Getting it right

Elise answered with good work:

Okay sorry about that.  I didn’t realize that I just needed a quadratic formula to solve it.  It’s been a while since I’ve done quadratic equations.  Thanks for helping me with this. You’ve been a big help!  Okay here’s what I think is the answer. Let me know your thoughts!

Car speed (c) = v+40
Bike speed (b) = v
Distance (d) = 30 km
Time it took the car to get to the garage: 15/v+40
Total time (T) = 2 hours
Time it took to ride the bike: 15/v

T = C + B
2 = 15/(v + 40) + 15/v
2[v(v + 40)] = 15/(v + 40)[v(v + 40)] + 15/v [v(v + 40)]
2v^2 + 80v = 15v + 15v + 600
2v^2 + 50v – 600 = 0
v^2 + 25v – 300 = 0
v = [-b +- sqrt(b^2 – 4ac)]/[2a]
v = [-25 +- sqrt(25^2 – 4(2)(-300))]/2
v_a = 8.86
v_b = -33.86

Bike’s average speed is 8.86 km/h

The first part of this is a quicker way to get to the quadratic equation than mine; the rest is equivalent to my work apart from using the formula.

I replied:

Yes, you solved it – exactly as I did.

Of course, the next thing to do would be to check that your answer fits the story, by finding the other speed and the two times, which do add up to 2 hours.

As I mentioned, it is common for students working with rational equations to need a nudge when a quadratic equation arises! And when the solution is irrational … well, that’s how real life works, but we protect students from that reality a little too much!

Good work.

Elise closed with this:

Thanks so much for your help.  Yes, I totally needed a nudge.  If I had known I just needed a quadratic equation man I wouldn’t have been so confused for so long.

And if I’d just asked for her work (or been given it initially), we would have gotten to the answer more quickly – but I wouldn’t have shown an alternative method! Only a little help was ultimately needed, yet all of this was worth discussing.

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