Think Before You Start

I ended the previous post by stating a problem from the 9th grade IMP math textbook called The Perils of Pauline. The girl is 3/8 of the way through a railroad tunnel when she hears a train from behind. She just has time to run back the way she came to reach the beginning of the tunnel when the train gets to the beginning, and she just has time to run forward to get to the end of the tnnnel when the train gets to the end. If the train goes 60 mph, how fast does Pauline run?

When a problem seems at all complicated, it is always a good idea to think about the problem before you start trying to express it mathemaically. In particular it is a good idea to make sure we understand what conditions we are given and what we are being asked to find

Condition A--She is 3/8 of the way into the tunnel when she hears the train.

Condition B--Running, she can just reach the beginning when the train does.

Condition C--Running, she can just reach the end of the tunnel when the train does.

Condition D--The train goes 60 mph

Find--How fast Pauline runs. From contest we may assume she will run the same speed in either direction.

Next we will see if we can re-express these conditions in a way might better help us solve the problem. Since the problem talks about her running in each direction we will re-express the first coondition as:


Condition A--When she hears the train, she is 3/8 of the way from the beginning and 5/8 of the way from the end of the tunnel.


It also makes sense to to look at the obvious mathematical implications of the next two conditions.


Condition B--Pauline can run through 3/8 of the tunnel in the time it takes the train to reach the beginning of the tunnel.


Condition C--Pauline can run through 5/8 of the tunnel in the time it takes the train to reach the end of the tunnel.

At this point what could have been a rather messy algebra problem can now be solved with just arithmetic. Looking at the difference between Conditions B and C we see that in C compared to to B, the train goes the extra distance represnted by the length of the tunnel, while the extra distance Pauline runs is 5/8 - 3/8 = 2/8 = 1/4 the length of the tunnel. Pauline runs 1/4 as fast as the train and 1/4 of 60 mph is 15 mph.

If we wish to solve this problem using algebra we must still think carefully about what we do.

We will need to express Condition B and Condition C algebraically. To do this we will not only need a variable such as P to represent Paula's running speed, we will need additional variables to represent the distances than Paula and the train travel. One way to do this would be to let L be the length of the tunnel and D be the distance the rain has to travel to reach the beginning of the tunnel.

With these variables we can set time equal to distance divided by rate--and if the train and Pauls could reach the beginning of the tunnel at the same time then

Condition B1 would be D/60 = (3/8)L/P

and if both could reache the end at the same time

Condition B2 would be (D+L)/60 = (5/8)L/P

At this point we have two equations and 3 unknowns which normally would not be enough to solve for any variables. But if we subtract B1 from B2 we can at least eliminate one variable and we get

L/60 = (2/8)L/P

Multiplying by 60P we are able to clear the fractions and we get

LP = (60)(2/8)L = 15L and dividing by L we get

P=15

An alternate approachh would involves clearing away the fraction in Conditions B1 and B2 first, and then subtracting one condition from the other and solving in much the same way that we did above.

Subtracting one condition from another is something we frequently find helpful in solving problems. We look at more at this technique in the next post.

Mathematics is a Language

The most important thing to recognize about mathematics is the fact that mathematics is a language. Indeed, mathematics is often called the languae of science. As with any language we can makke a distinction between how much of the language we know, and what we are able to do with that knowledge.

With an ordinary language, you are able to make good use of your knowledge if you are fluent in the language. A typical American high school student studying French will learn more French than the average French 4 year old knows. But French 4 year olds are fluent in French and can use their limited knowledge of French to communicate far more effectively than a typical American student.

I like to use the term math fluency to describe a similar ability to make use of our mathematical knowledge in flexible and creative ways. But whatever you wish to call it, this ability to make flexible and creative use of mathematics can be much more important than simply having a lot of knowledge that you can only use in very familar situations.

In all likelyhood, you would not consider yourself to be fluent in math. So how do you become fluent in mathematics? If you are studying calculus and you have never been fluent in math at any level, calculus is not the place to start to try to achieve fluency in math. Neither is algebra. The easiest way to develop fluency in math is to start at a level that is as close as possible to the ordinary language in which we are fluent--which in my case is English.

Mary is two years older than her sister. If Mary is 10, how old is her sister?

This problem can be solved with simple arithmetic. But you have to ne alert. It is a bit of a trick question. It sounds like it should be an addition problem--but it turns out to require subtraction. A large pat of what it means to be fluent in math is being able to notice when we are being given information in a form that is not the ideal form for us to use, and being able restate that information in a way we can use. In this case our restated condition is that her sister is two years younger than Mary.

Next we will consider some problems that are mathematically more advanced. A radio sells for $13 ar 35% off. What was the original price? A meal including a 15% tip comes to $46. What was the cost of the meal before the tip?

35% off mean 35% less than 100% or the amount we pay is 65% or 0.65 times the original price. To find the original price we must divide $13 by 0.65 which gives us $20 as the original price. You can check that 35% of $20 is $7 and $20 - $7 does equal $13.

For the meal we add 15% to our basic cost of 100% to get 115%. The $46 is 115% of the basic cost of 1.15 times the basic cost. To find the basic cost we divide $46 by 1.15 and get an answer of $40. You can check that 15% of $40 is $6 and 40 and 6 do add up to 46.


All of this can be expressed very efficently using algebra. As long as students are going to study algebra, why not wait and cover this material using the language of algebra? I would argue that a solution given in plain English will paint a clearer picture for most people than one expressed in the symbolic language of algebra. More importantly, when students do study algebra, what they study will be easier to understand if they have seen something similar before. Indeed, it can be very helpful begin the study of algebra by seeing how basically the same arguments can be made either with verbal reasoning or symbolically working with algebraic expressions. For most students, this will make what they do in algebra seem more understandable and meaningful.

I would like to conclude this post with a problem from the 9th grade IMP textbook called the Perils of Pauline. The girl is 3/8 of the way through when she hears a train whistle from behind her. She just has time to run back to the beginning of the tunnel and get there at the same time as the train. She also just has time to run forward and get to the end of the tunnel at the same time as the train. If the train goes 60 mph how fast doe Pauline run?

I will look at this problem in my next post--but I wanted you to see it first wiout mmy commentary. You may try to solve it if you like. Whether you try to solve it or not, I would like you to make an assessment of how difficult you think this problem would be for the average person.