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Any typos?

I’ve found a few:

Page 112: 0.08 g/mL should be 0.08 g/100 mL

Page 120: “$120/$120/”?

If you don’t find typos of your own, you’re probably not reading the text carefully enough.


STEP 0: Don’t panic! “But I don’t know how to set this problem up; the teacher didn’t show us how to do problems like this.” Imagine yourself into the problem. This brings your intuitive, instinctive knowledge into play. (As opposed to your rule-following, pencil-pushing knowledge.)

Let’s apply the steps to a problem that came up in my life on Monday while I was doing laundry, using Tide liquid laundry detergent at a laundromat.

The washing machine said “Use only 1/4 of a cup.”

The liquid detergent had a measuring cup – let’s call it a measuring scoop – but the scoop was smaller than a standard cup. How much smaller, I wasn’t sure.

Step 1: Understand the problem.

1/4 cup = how many scoops?

Step 2: Devise a strategy.

Read the label!

The container gave me two clues: The label said “100 fluid ounces” and it said “32 uses”.

Is this enough information? …

How many fluid ounces in a cup? … 8.

(I used Google from my PDA to look this up!)

1/4 cup = how many scoops?

1 jug = 100 fluid ounces

1 jug = 32 scoops

1 cup = 8 fluid ounces

Step 3: Carry out the strategy.

Use the magic-of-units method.

(1/4 cup) (8 fl oz / cup) (1 jug / 100 fl oz) (32 scoops / jug)

= 0.64 scoops

Step 4: Check.

Compare with common sense:

Most people would normally put in a full scoop.

So we expect the answer to be less than 1 (otherwise they laundromat wouldn’t have put up the sign warning people to use less detergent than they might typically use).

But it shouldn’t be too much less than 1.

So 2/3 seems about right.

Compare with a different way of solving:

Number sense: 32 is about one-third of 100.

If 100 fl oz = 1 jug = 32 scoops, each scoop is about 3 fl oz.

On the other hand, 1/4 cup is exactly 2 fl oz.

So 1/4 cup is about 2/3 of a scoop.

Let’s discuss the examples in the chapter in more detail.

Example 1: I’m assuming you read this example.

Turn to page 120.

(It’s a good idea to bring your book to class on a day when I’ve assigned some reading!)

What’s the method used here? …

This example is about trial and error.

The book shows “trial”, but let’s see where “error” comes in.

Tickets for a fundraising event were priced at $10 for children and $20 for adults. Sally worked the first shift at the box office, selling a total of $130 worth of tickets. However, she did not keep careful count of how many tickets she sold for children and adults. How many tickets of each type (child and adult) did she sell?

Suppose Sally sold two $10 child tickets. In that case, she would have sold $130 – $20 = $110 worth of adult tickets. Because the adult tickets cost $20 apiece, this means she would have sold $110/($20 per adult ticket) = 5.5 adult tickets. But this makes no sense.

Does this mean the problem has no solution?

NO! It means the problem has no solution in which Sally sells exactly two child tickets.

(This is the “error” part of “trial and error”!)

Disadvantage of trial and error: It can be time-consuming.

Example 2: Jill and Jack’s Race (page 120)

What’s the method used here? …

The magic-of-units method, combined with the dangerous method of making up the numbers we don’t know.

Why is this dangerous? … Because making up different numbers might change the answer!

What if Jill DIDN’T run the first race in 20 seconds?

What if Jill and Jack are racing turtles, and Jill ran the first race in 20 minutes, rather than 20 seconds? How would that change the answer? …

Every “s” (for seconds) in the solution becomes “min” for minutes.

Jill still wins.

Implicit in Bennett and Brigg’s method is their belief (which they never explain) that the final answer won’t depend on what numbers they make up.

The reason THEY know this is that they know algebra.

How can WE understand it?

By looking at our TWO ways of plugging in reasonable numbers, and noticing that they give the same answer, and understanding why.

What if Jill ran the first race in 20 nanocenturies?

(A nanocentury, or a billionth of a century, is about 3 seconds, so 20 nanocenturies is about a minute.)

The analysis will be the same as in the book, but with “seconds” now replaced by “nanocenturies”.

Any other unit of time will do just as well!

So we see that the actual time it took Jill to run the race is irrelevant.

We could call it t and solve the problem letting t be an unknown instead of a specific number.

Here’s a different way to do Example 2:

Use ratios and proportions.

Make a table:

1st race 2nd race

Jill: 100 m 105 m

Jack: 95 m x m

(Here x represents how far Jack has run when Jill reaches the finish line. To avoid dealing with the two cases separately, we’ll assume that Jack keeps running until they’ve both crossed the finish line.)

If x < 100, … Jill wins.

If x > 100, … Jack wins.

If x < 100, … it’s a tie.

95 / 100 = Jack’s speed / Jill’s speed = x / 105

x = 105 ( 95 / 100

We don’t need to know x; we just want to know how it compares to 100.

Does anyone see how to do this mentally?

Most people with a good practical command of algebra can do this mentally, more or less as follows:

x / 100 = (105 / 100) ( (95 / 100)

= (1.05) (0.95)

= (1 + 0.05) ( (1 – 0.05)

= (1 + c) ( (1 – c) (with c = 0.05)

= (1 – c^2)

< 1,

therefore Jill wins.

(We see that one reason to get good at algebra is that it gives us shortcuts for avoiding arithmetic!)

Did anyone have a different approach?

Example 4 (on page 123)

That’s the one with the coffee and milk.

How many of you found the answer obvious?

How many of you found it surprising?

Let’s consider a related problem.

There’s a math class at a boy’s school, and a math class at a girl’s school. Both classes are the same size (100, say).

Five students from the all-girls class switch to the all-boy class.

The teacher at the all-boys school picks five students at random from the over-large class and transfers them to the all-girls school.

Which are there more of now: boys in the girls’ class, or girls in the boys’ class?

Is this easier to think about than the coffee-and-milk problem?

A variant of Example 6 (on page 126)

A very long rope is just long enough to go around the equator. (Ignore the fact that Earth has oceans; or, if you like, move the problem to Mars.)

Now suppose we make the rope one foot longer.

How high off the ground can the rope be lifted, if we have people circling the equator who all lift the rope at the same time?

Guesses? …

Most people are surprised by the answer.

We’ll discuss it on Thursday.

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