Alternative Solutions: Practice Test #1

DITW = Do it their way

Trial and Error = pick an answer, play with it, see if it fits what they said in the problem…

Back Door = make up numbers for the variables in the problem, work out an answer that is based on the numbers you made up, then put those made up numbers back into the answer choices, ruling out any that don’t produce a matching answer.

More detailed explanations available in the book…


 1. You can just try the answer choices one at a time…

 2. DITW

 3. Back door!  Make up your own m and p

 4. Recognize that this is y = b + mx…the b is the starting value.

 5. This looks like a back door problem and I guess you could do that, but this time, I agree that it is actually much faster to do the algebra: distribute the minus sign and group the terms…so DITW

 6. Either recognize that y = b + mx, and m is the rate of change…or to see how much the boy grows each year, just make up a number for a, plug it in and then plug in a second number, one year older.  For example, try a=10 and then a=11.  The difference is how much he grew in a year.

 7. DITW

 8. If you see the algebra, it’s quick.  But making up numbers is even quicker. I used a=4, and b=2…but any numbers that fit will do.

 9.  Use trial and error!  But be lazy: check the easier equation first.  Only one choice works.

 10.  This is an example of “the case of the missing constant” (see page 126). But if you do notice that the function is an even function, you can go right to the answer: g(4) = g(-4) = 8.  So “their” way is quicker but requires more insight.

 11. Trial and error would work, but faster to DITW

 12. Draw a neat diagram with a line through the origin that rises one and runs seven.  It will be obvious that only one of the given answers could possibly lie on that line. 

 13. This is a gift to students who know the back door!  Say x = 6…work out the fraction, common denominator and all that…then put  x = 6 into each answer.

 14.  Faster if you know the laws of exponents.  But if you are shaky…back door! 

I tried x=4, y=0.  Took a little while to show that 212 = 84, but it works.

 15. This is a tricky one whether you DITW or mine…First notice that if you start to foil out the expression, you find that ab=15.  And they gave you a + b = 8.  You could solve that system algebraically, but I prefer to look at the equations and try to make up numbers.  Seriously, the first numbers I tried were 5 and 3!  (It can be a=5, b=3 or the other way around.)  If you put a=5 and b=3 into the original expression and then go ahead and foil it all the way, you get 41.  You don’t need the other possible value to know that the answer is D.

 16. Algebra or just a quick trial and error

 17. These are two similar triangles – it’s one of the classic examples.  Fill in the lengths they give you and you will see that the bigger triangle’s sides are twice the length of the smaller…

 18. DITW

 19.  The sine of an angle is also the cosine of the complement.  So the answer is 4/5!  And by the way, if you have a hard time remembering that rule, it may help to think of “cosine” as short for “complement’s sine”!  Works for tangent and cotangent as well…

 20.  DITW



 1. DITW

 2. Like a “missing constant” problem, you could plug in their values for x and y, solve for k and then plug in the new x value.  Or you could recognize that if y=kx, then x and y are directly proportional – set up ratios, cross multiply and divide.

 3. Know the rules about parallel lines and transversals: pairs of angles that are not congruent are supplementary.  180 – 35 = 145.  (See page 95 for a review)

 4. Algebra is quick here, but this can also be solved with trial and error:  Be lazy though – start with the choices that are easily divisible by 8.  D does not work, but check out C:  If 8x = 16, then x = 2…and 16 + 4x comes out to 24 which is 10 more than 14.  How about that.

 5. DITW – what else could it be?  There is only one data set with a clear downward trend.

 6. DITW

 7. This is not really algebra.  Just add up the heights of the bars on the graph.  It adds up to 27.5.  But they tell you that the total is actually 27,500.  So the numbers on the side of the graph must be “thousands.”  The “official” solution says the exact same thing but makes it look more algebraic.

 8. You can do this by trial and error:  carefully try 0, 1 and 2.  None of them work.  So it has to be D.  You can also do this by thinking:  the smallest value you can have for the part of the expression with absolute value bars is zero.  Then they add 1 to that value.  So it can never add up to zero.

 9. Back door!  Say t = 10  Plug in and get a=1062.8.  Put that into each answer – only A comes out to 10.

 10.  Trial and Error – put each answer into the equation they gave you…

 11. Trial and Error.

 12.  Really helps if you just list out all of the data:  3,3,5,5,5,5,6,7,7,9,9,9 – now finding the average is just a matter of adding them up and then dividing.  And by the way, had they asked for the median, a list like this would make that very easy too.

 13. DITW

 14. DITW

 15. This is another example of y = b + mx.  The only difference is that the vertical axis is labeled with a C instead of the customary y.  But the intercept is still the starting amount…

 16.  …and the equation is still y = starting amount + slope  × run.  That the slope is 3 can be determined just by counting rise over run on the graph.  But don’t just count boxes – look at the scale of the graph!  Otherwise, you get ¾ which is wrong.

 17.  They are just asking you to find the INPUT (x-value) that gives the lowest OUTPUT (y-value).

 18.  DITW

 19.  Yes, you could do algebra here.  Trial and Error also works:  for example, if they sold 77 salads, that would be 77 × $6.50 = $500.50.  So they would need to take in another $336 from drinks.  At $2 each, that means 168 drinks.  But 168 + 77 is more than the 209 items they sold.  Now you try it with choice B.  You are in for a happy surprise. 

 20. Classic back door.  And like most back door problems involving percents, it’s a good choice to make the starting value $100…The 20% discount knocks the price down to $80.  With 8% sales tax, what you pay is p = $86.40.  Put that value into each answer choice…wait for the magic.

 21. DITW

 22. DITW

 23. DITW – annoying and kind of tedious, but what are you gonna do?  Read the chart, find the ratios, pick the one closest to human resources.

 24.  OK, this one makes me laugh.  You are supposed to use the center and endpoint of the radius to find the length of the radius and then write the equation of the circle in standard form.  But there is a lazy work-around:  that endpoint they give you does have to be a point on the circle.  So you can just put that point’s x and y values into the equations given in each answer choice.  Only one choice works.

 25.  Trial and error – plug in the choices, pick the one that gives a height closest to zero. (When you are on the ground, your height above the ground is zero.  That’s the only physics knowledge you needed to answer this.)

 26.  If you don’t want to do the algebra, go to each answer choice, find 20% and add it on.  When you get to choice B, you will see that you end up with the correct number of pears.

 27.  This is like a scale-up probability problem.  The sample regions have an area of 1 square meter.  The full region is 100 square meters.  So you just need the answer that is 100 times the average of the data points.  That average is in the neighborhood of 150…

 28. DITW

 29. DITW – but review the section on the remainder theorem (page 227).

 30.  Well, in theory you are supposed to complete the square.  But look at the graph: the vertex is clearly in the neighborhood of (1,-16) and there is only one answer choice where both of those appear directly as constants. 

 31. Just pick a number between 12 and 18 – let’s say he does 15 per hour.  Divide 72 by 15 and move on.

 32. DITW

 33. DITW

 34. DITW

 35. DITW

 36. You are looking for the value that makes the denominator zero.  This is one of those questions where you can make use of your calculator.  If you have a TI89, just set the denominator to zero and use the F2 command to solve that equation.  Or with any graphing calculator at all, graph the equation and then trace along the graph to find where the function equals zero.

 If you decide to do this one by algebra, a neat trick is to replace x-5 with a new variable, say m.  If you do this, you get an expression that is easy to factor.  But remember to go back and solve for x.

 37.  Review page 74 to see the quick way to calculate percent changes:  the base in that expression is 1.02 – that gives you a 2% increase each year.

 38.  Review percents and exponential growth (page 181).  Then you will see that all you need here is to calculate (100)(1.02)10 and 100(1.025)10 and then subtract. 

OK, that’s all for the first practice test. If you are keeping score, I believe that for 36 out of the 58 problems, there was an alternative method available. 



Polar Graphs in Interactive Physics

This week, I find myself having to remember how to graph in polar coordinates.  I am talking about cardioids, rose petals, lemniscates and the like.  It’s been a while…

There are many websites that will draw these for you.  Desmos has polar mode.  But I have not yet found a website that will show the graphs being constructed.  But I remember reading a very nice explanation that (I think) was in one of Howard Anton’s calculus texts.  If I am remembering this right, he said something like:

Imagine a clock with one hand that rotates around, connected to the x-y plane at the origin. 

Then, imagine an ant (Anton’s Ant!) walking along that clock hand as it rotates.  The ant’s distance from the origin is given by the function r=f(θ). 

As the clock arm rotates, the ant moves in an out along the arm.  The location of the ant traces out the graph of the polar function r=f(θ).

And note that if the radius is negative, that just means that the ant has walked “the other way” along the clock arm.

For example, these are three  pictures showing the clock arm and the ant at successive times as we graph the cardioid r = f(θ) = 2 + 4 cos (θ):


I don’t know how to make my own applets, but I do know how to use Interactive Physics as a work-around.  It’s not as good as a real applet.  You can change parameters easily but if you want to change to a new function, that takes more work.  And if you want to share with students and colleagues, they need the software, which is kind of pricey.  (Though there is a way to get student versions quite reasonably.  Email me if interested…totally legal!)

In any case, one thing you can do is export your experiments as videos.  You lose the ability to vary parameters but at least you get the flavor.  So that’s what I did…

A CARDIOID with an inner loop

YouTube Preview Image

A ROSE PETAL with 5 loops

YouTube Preview Image

If you do happen to have the Interactive Physics software package and would like the source files for these demos, post a comment or send me an email.  That way, you can run these yourself on your own computer, changing the parameters to make discoveries.

Or just leave them to run for as long as you feel like.  They are very soothing to watch.


3D visualization, Geometric Probability and 538’s Riddler #2

Hello all…sorry the posts are coming at such long intervals.  I am teaching a new class this year — multivariable calculus! — and it is taking up lots of time as I try to remember/recreate/invent some level of understanding, having last taken the course 30 years ago.  But I am having a lot of fun with it, teaching a small group of fine (and patient) students, learning a lot.

And speaking of things that come at long intervals…here is a puzzle that was posted on in their new feature: “The Riddler”.  This was from week #2:

You arrive at the beautiful Three Geysers National Park. You read a placard explaining that the three eponymous geysers — creatively named A, B and C — erupt at intervals of precisely two hours, four hours and six hours, respectively. However, you just got there, so you have no idea how the three eruptions are staggered. Assuming they each started erupting at some independently random point in history, what are the probabilities that A, B and C, respectively, will be the first to erupt after your arrival?

The full post is at:

and a complete solution is in the next week’s post:

But what I want to discuss is a different way of thinking about the solution than the one posted at 538.  I want to use geometric probability.  [If you want to solve this on your own, spoilers ahead.]

Let’s start with an easier version of the question…

Suppose the only geysers were A and B.

If you have just arrived, Geyser A will erupt in x hours, where x is between 0 and 2. And Geyser B will erupt in y hours, where y is between 0 and 4.  Intuitively, it seems likely that y will be greater than x.  One way to see this is to use geometric probability.  Think of every possible outcome as an ordered pair.  Then, all possible outcomes can be represented as an area on the coordinate plane.  I used to graph the region and also the line y=x, or more precisely, the shaded half-plane y>x.

2d geyser

Geyser A will erupt first whenever y>x. So the probability we seek is the area of the part of the box above the x=y line divided by the total area of the box.  You can just about count boxes to determine that this occurs 6/8 (or 3/4) of the time.

OK then.  Now let’s bring Geyser C back into the problem…

From the moment you arrive, you know that Geyser C will erupt in z hours, where 0<z<6.  But you can still use geometric probability.  Each outcome can be thought of as an ordered triple (x,y,z).  The space of all outcomes is now a 2 x 4 x 6 rectangular solid.  And if you are looking for the probability that Geyser A erupts first, you want to slice that solid up with the planes x=y and x=z.

I don’t know if Desmos can do this kind of thing.  But fortunately, Paul Seeburger’s Multivariable Calculus Exploration applet can be used to visualize this problem.  You can find the applet here:

Note: on my computer, I can only open this page in Firefox.  YMMV.

Here is the solid that represents the set of all possible outcomes: 3d geyser full space

And here is that same solid, cut by the planes x=y and x=z:3d geyser full space sliced

We need to find the volume of the part of this box “to the right” of the x=y plane and “above” the x=z plane. It’s easier to see what we are looking for if you rotate the diagram to view it “from behind”:

3d geyser full space sliced 2

In this diagram, the volume defined by points where y>x and z>x can be broken into four chunks.  I’ve added the x=2 plane and the z=2 plane to help you see them.

The rectangular solid on the upper left: V = 2x2x4=16

The triangular prism in the front on the lower left: V = (1/2)x2x2x2=4

The triangular prism in the front on the upper right: V = (1/2)x2x2x4 = 8

The front half of the pyramid on the bottom right: V = (1/3)(2)(2)(2) = 8/3

(If you are having trouble seeing the pyramid: in this view, we are looking through its 2×2 base, looking toward its vertex which is 2 units away.)

The probability of Geyser A erupting first is the sum of these volumes divided by the total volume of the box, or:

(16 + 4+ 8 + 8/3)/(2x4x6) which reduces to 23/36.

Using a similar method but with different cutting planes, you can find the pictures that will give you the probabilities of Geyser B or Geyser C erupting first.  And you do get the same answer that is posted on the Riddler’s webpage.

You can post questions (or ask me in class) if you need help with those last two…



“You keep using that word…”

The word is “exponential”.  It’s been showing up on my students’ tests and in their lab reports.  And, like Inigo Montaya, I have to say:  I do not think it means what you think it means.

In particular, it does NOT just mean “curving upward”.  When you state that a graph is “exponential”, you are making a very specific claim about the structure of that graph and the mathematical relationship it depicts.  Some curved graphs turn out to be exponential.  But not all of them.  (And in fact, not this next one.)

What got me started on this year’s version of this rant:

Here is a position vs. time graph from my student’s first test of the year.


One of the questions that I ask about the graph is:

“Between t = 0 seconds and t = 6 seconds, was the car accelerating?  How do you know?”

I am hoping for (and giving full credit for) answers like:

Yes.  The slope tangent line is getting steeper.


Yes.  If the velocity were constant, the graph would have been linear.

But I am also seeing a surprising number of papers that say:

It’s accelerating because the graph is increasing exponentially

I don’t take off credit (yet) but I do wish we could sort this one out.  So let’s try.

A Function is Exponential…

only if it can be written in the form: y=a·bx

where a and b are constants and x is the exponent.. And as long as b is greater than 1, the graph of the function will in fact curve upward.  For example, here is the graph of a function that is in fact exponential.  I don’t want to reveal what function it is just yet.


And here is a quick way to see if the function is exponential:  look at how much it increases in equal x-intervals.  A linear function will increase (or decrease) by the same amount in each interval. But an exponential function will increase (or decrease) by the same factor in each interval!

So now let’s take a look at the values of this function at regular intervals.  In this case, I’ll just use Δx =1 and I’ll read the values off the graph as best as I can.



As you can see, as x increases, the function increases, but not by equal amounts. That’s ok — we weren’t expecting equal increases.  The function is not linear.

But if you calculate the ratio of any two successive values, you do see that the function is increasing by (nearly) the same factor every time!  (You can trust me, or you can actually do the math.) This means that you are justified in suspecting that this curve really is “exponential”.  And you can figure out the equation by inspecting the graph and chart.  I’ll call that a puzzle — leave your answer in a comment (or if you are my current student, you can also post your answer in our class stream).

So let’s take another look at the first graph:



Let’s check on how much this graph increases each interval.  Here’s the data:


Some things to notice:

1. The values increase, but once again, not by the same amount each interval.  This one is not linear either.

2. The ratio of successive terms is NOT constant.  Try it and you will see.  This graph is NOT exponential.

But there is a pattern to be discovered here.  Instead of finding the ratio of successive terms, try calculating the difference.  You will discover a pattern, one that is true for “quadratic” functions.  Tell me about it in the comments if you like.


Bottom line: curved does not automatically mean exponential.  Exponential means increasing by equal factors in equal time intervals.

Another look at “Logarithms!”

OK, even I will admit that the logarithm rules do not seem to be infused with joy.  But once you know them, there is a way to apply them that my physics students seem to get a kick out of.  It’s a little exercise I learned by reading Professor David Mermin’s essay, “Logarithms!” in his excellent book, “Boojums All the Way Through” (Cambridge University Press, 1990).

The goal of the exercise is to be able to determine common (base-10) logarithms without a calculator.  At the time Mermin wrote, scientific calculators were actually not so common.  Now, everyone has access to them and Google will calculate them for you as well.  But that’s not the point.  It feels almost magical to be able to do this stuff without a calculator.  I would say it’s a hoot, definitely worth of the exclamation point Professor Mermin put in the title of his original article.

Here’s the basic plan.  Start by observing some coincidences, some “almost” equalities involving powers of small prime numbers and small powers of 10. Then use the logarithm rules to deduce the values of the logarithms.

[All of the examples that follow are from Mermin.]

For example, it happens that…

210=1024 and 103=1000

So we would be almost correct saying 210   = 103.

Then we could take the log of both sides, applying the power rule for logs to get:

10log 2 = log 1000

but we know that log 1000 = 3.

So 10log 2 = 3 and that means log 2 = 3/10 =.3

I have never presented this to a class without at least a handful of kids grabbing their calculators to see if we are close.  Their calculators tell them that log 2 = .3010 and I hear “Whoaaa!” (but I teach an impressively nerdy group).

Now what about log 3?  We need another coincidence, another “almost” equality.

While it is true that 32=9 which is almost 10, we can come closer with

34 = 81 which is almost 80.  This is helpful because 80 is also 10×2.

Starting from 34 = 10 × 23

we can take the log of both sides and actually enjoy ourselves as we apply the log rules to get:

4 log 3 = 1 + 3 log 2

And we already know log 2 — it’s (roughly) .3, which we can substitute in to get

4 log 3 = 1 + 3 × .3 = 1.9

log 3 = 1.9 ÷ 4 = .475

And when check out log 3 on your calculator, you see .4771….whoaaa.

We can get log 4 immediately from the fact that

Log 4 = log 22 = 2 log 2  ≈ 2 × .3 = .6  (whoaaa)

and log 5 = log (10 ÷ 2) = log 10 – log 2 ≈ 1 – .3 = .7 (Can I get another whoaaa?)

OK, I think you see the flavor of how this works now.  We just need to keep finding helpful coincidences…

Like 72 = 49 ≈ 50 = 100 ÷ 2.  Or  112 = 121 ≈ 120 = 10 × 2× 3.

These starting points expand our collection of approximate logs.  But you have to know your log rules to build the collection.

Mermin does not stop here.  He writes:

“Those whom the schools have taught to hate numbers will probably want to stop after this preliminary warmup, but the real fun is only beginning.”

He then goes on to improve on all of these estimated logs by constructing systems of equations built on closer, dazzling coincidences.  I recommend the essay and in fact the whole collection.




Lost In Translation — At Labyrinth Books in Princeton

In Advanced Math for Young Students, there is one section where I introduce a method to help students when they are stuck at the beginning of a word problem, not yet seeing how to translate the problem into the language of algebra.  Well, I have developed a workshop about this method and have been invited to present it at Labyrinth Books on Nassau Street in Princeton.

The presentation will be on June 11, 2015 at 5:30 pm.  If you are in the area, stop on in. But if not, check back here in mid-June.  I will post more about the method and include a link to the PowerPoint slides.

For more on the presentation:


For Small Angles Only

Two Surprising Trigonometric Almost-Equalities

Let’s warm up with an old math joke I first heard back in my own freshman year.  I have always thought it was funny even though I am not sure I get it.  But my students all LOVE my math jokes, so here goes:


I think it is meant to make fun of the way we sometimes do math in physics. For example…

When we are studying Young’s two-slit interference experiment, there comes a point when my students are asked to accept that for small angles, two things that they know are NOT equal can be sort-of, kind-of treated as equal:

sin(θ) ≈ tan(θ)

And when we are studying simple harmonic motion, I ask them to accept another one:

sin(θ) ≈ θ

That one is also only “true” when the small angles are measured in RADIANS (as you will see if you bravely read on).

Once you accept these two things, you can go on to derive other interesting rules about interference patterns and about the behavior of pendulums.  But why should anyone accept these things?  I’m going to give you a collection of reasons.

1. Look at the numbers.post30pic1

The chart above shows you values of the angle in degrees and then the sine, the radian value and the tangent listed in that order.  You can see that the values are close to each other.  The sine is a little less than the radian measure and the tangent is a little greater.  But even for angles as big as 20 degrees, the differences are less than 5%.  So just looking at the numbers, you have to accept that for small angles, these three functions — sin(θ), tan(θ) and θ — are equal-ish. 

2. Look at the Graphs

Here you see the graphs of the three functions graphed using the Desmos website:

post30pic2OK, clearly these are three very different functions.  But we are not interested in their entire domains.  We have only claimed that the functions are nearly equal for “small” angles.  So let’s just zoom in on the section of the graph from 0 to .4 radians…


Again, if we stay where the angles are small, the three graphs are “equal-ish.”

 3. Look at their Taylor Series Representations

This is something that students learn later in their first year of calculus.  These non-polynomial functions can each be expressed as a polynomials with an infinite number of terms.  And they can be approximated with polynomials of finite length.

How these polynomial formulas are developed and how much error you introduce by using only a finite number of terms are topics for several weeks of discussion in calc class.  But for now, let’s look at the first few terms of the Taylor polynomials for the sine function and the tangent function:


Notice that the first term is the same: it’s just ‘x’.  Then, after that, the next term has x-cubed.  But if x is already small, then x-cubed is going to be REALLY small.  Negligibly small.  As we already expect from looking at the charts and graphs, sin(x) is a little bit smaller than x and then tan(x) is a little bigger than x.  But for small values, the differences are negligible.

I know you are probably already convinced. But I have one more reason for you to believe this and it gives me an excuse to re-use one of my favorite diagrams…

4. Look at the Geometry

Here it is again:


This was the diagram we used to learn the geometric meaning of all of the trig ratios.  Now I am going to make some changes to the diagram:  I am going to make the angle smaller and also I am going to delete any of the segments we don’t need right now.  Then I am going to zoom in on the part we care about:


The length of the blue segment is sin(θ).  The length of the orange segment is tan(θ).  And because it is a unit circle, the length of that intercepted arc is just θ.  (If you are not sure why that is true, think about the definition of radian measure — or read about it in this earlier post here.)

Now look at how the lengths of those two segments compare to the length of the arc.  One of them (the sine) is a little shorter.  One of them (the tangent) is a little longer.  But as the angle gets smaller and smaller, the differences become negligible!

Well, I hope one of these four methods have convinced you that this is legal.  If not, I hope at least that you enjoyed the joke at the beginning.





Invert and Multiply…but WHY?

Most of my posts so far have addressed high school level math.  This one is for younger students (and their teachers).

At another blog that I follow, Kitchen Table Math, there has been recent discussion about the “invert and multiply” rule for dividing fractions and whether it is important and/or appropriate to teach the reasoning behind this rule.

You know I am going to vote “yes”.  But what kind of derivation should we use here?

The rule can be derived algebraically.  But that doesn’t really help the grade-school students who are just learning it.  It can also be derived in a algebra-ish way.  A commenter at Kitchen Table Math pointed to the explanation found here:

Mr. Novak’s explanation is elegant and concise. And it certainly answers the question: why do we invert and multiply in order to divide fractions?  But it would be nice if students also had an intuitive feel for why this works.

We can build the rule step-by-step in a manner that helps that intuition to develop.

Step 1: Division is for Sharing

You have 12 cookies – each a pretty good size —  and you need to put them in individual serving bags to hand out at a party.  By deciding how many to put in each bag, you also determine how many serving bags you can make.  For example, you could put 12 cookies in each bag, but then you would only have enough for 1 bag.  On the other hand, it you put 6 cookies in each bag, then you can fill 2 bags because 12/6 = 2. That seems obvious, but it establishes a method: we are going to divide the total number of cookies by the amount in each bag to find how many bags we can fill.  You soon notice that if you put fewer cookies in each bag, you can fill more bags.  For example:

If we take 12 cookies… And put 12 in each bag… 12 ÷ 12 = 1 We can fill  1 bag
If we take 12 cookies… And put 6 in each bag… 12 ÷ 6 = 2 We can fill  2 bags
If we take 12 cookies… And put 4 in each bag… 12 ÷ 4 = 3 We can fill  3 bags
If we take 12 cookies… And put 2 in each bag… 12 ÷ 2 = 6 We can fill  6 bags
If we take 12 cookies… And put 1 in each bag… 12 ÷ 1 = 12 We can fill  12 bags

So yes, dividing by a smaller number gives a bigger answer.  And with 12 cookies, 12 bags is the most you can fill…or is it?  I mentioned that these were pretty big cookies.  Hmmm…

 Step 2: Dividing by Fractions (of the “one over” variety)

You decide to put make the serving size a ½-cookie.  Since that serving size is even smaller, you expect to fill more bags.  Though you have not been taught how to divide fractions, you can do this one by thinking:

If I am putting ½ in each bag, then each cookie is enough for 2 bags.  So 12 cookies fill 24 bags.   I guess that must mean 12 ÷ ½ = 24.

Need to fill more bags?  Use smaller portions!  Give everyone 1/3 of a cookie…

You can see where this is going.  Let’s continue the chart…

If we take 12 cookies… And put 1/2 in each bag… 12 ÷ (1/2) = 24 We can fill  24 bags
If we take 12 cookies… And put 1/3 in each bag… 12 ÷ (1/3) = 36 We can fill  36 bags
If we take 12 cookies… And put 1/4 in each bag… 12 ÷ (1/4) = 48 We can fill  48 bags
If we take 12 cookies… And put 1/5 in each bag… 12 ÷ (1/5) = 60 We can fill  60 bags

The trend has continued.  Dividing by a smaller number gives a bigger answer.  And even if you don’t know the rule for dividing by fractions, you can certainly divide by these fractions!

To divide by “one over anything”, just multiply by the denominator.

 STEP 3: Any fraction at all!

Suppose you have those 12 big cookies and you decide to give everyone (2/3) of a cookie.  How many servings can you make?

We can figure this out using the reasoning we followed above, with one extra twist.

We have already seen that if you want to give everyone just (1/3) of a cookie, you will have enough to fill 12 × 3 = 36 bags.  But now you decide to give everyone TWO thirds of a cookie.  But you have already filled all those bags – why not just give everyone two of the bags?  You will have enough for 36 ÷ 2 = 18 servings.

Now go back and look at how we got that answer.  We don’t have to know an official rule to divide the 12 cookies into servings that each have 2/3 of a cookie.  We can just multiply by 3 and then divide by 2, which is the same as multiplying by 3/2.

Let’s try another:

Suppose you have 36 pizza pies to sell and you want to give 3/5 of a pizza to each customer.  (No one cuts pizzas into fifths, but don’t worry about that.)  How many customers can you serve before running out of pizza?

Well, if we give everyone ONE fifth of a pie, we can feed 36 × 5 = 180 customers.  Giving everyone THREE fifths means that we can only serve 180 ÷ 3 = 60 customers.

So to find 36 ÷ (3/5) we multiply by 5 and then divide by 3 which is the same as multiplying by 5/3.

Which derivation is “better”?

If I could only show students one explanation, I would go with Mr. Novak’s.  But if I had time to explore and play, I would start with what you see above.  I like that it encourages number sense and intuition.  Also, it introduces an idea that students will see again many times: that existing patterns can be extended to cover new situations.


Still Feeling Mean

Here’s a quick addition to that last post on various types of means.  By coincidence, yesterday, a question came up in my AP physics class about the meaning of the average radius of an ellipse.

We were looking at the laws that govern circular orbits.  But the orbits of planets are actually elliptical with the Sun at one focus.   The average orbit radius is the length of the semi-major axis. [When I first read that, it seemed wrong to me: how can the semi-major axis be the average distance — shouldn’t it be the greatest distance? But average orbit radius is average distance to the Sun, not to the center of the ellipse.  The Sun is at one focus so the planet distance is sometimes more than the length of the semi-major axis and sometimes less.  But that’s not the same thing as the average radius of the ellipse, which is the question that was asked.]

I’m not sure how useful it is to find the average radius of an ellipse, but it is interesting — and better with a picture!


The distance from the center of a circle to any point on the circle is a constant — that’s the length of the radius.

The distance from the center of an ellipse to a point on the ellipse is not constant.  The longest it can be is the semi-major axis, a.  The shortest it can me is the semi-minor axis, b.  But what do we mean by the average radius?  One possible interpretation is that the average radius is the arithmetic mean of the two semi-axes.  But here’s another interpretation:

What is the radius of the circle that would have the same area as the ellipse?

Clearly, that radius would have to be more than b and less than a.  So maybe it is their mean…but which kind of mean?

To answer this, you need to know two formulas:

Acircle = πr2

Aellipse = πab

So if the two areas are two be equal, we need r2 = ab.

In other words…

For a circle to have the same area as an ellipse, the length of its radius must be the geometric mean of the lengths of the ellipse’s semi-major and semi-minor axes! 

I didn’t know that.  And I was surprised to see a geometric mean pop up in class so shortly after my last post.  Also, I should add: I don’t know how to show that the semi-major axis is the average orbit radius.  If any of my fine students can come explain that to me, it is certainly worth a little extra credit.


One of my very favorite students greeted me this new year with a question:  what are the different kinds of means?  This is a fun question to explore.  We are going to look at three kinds of means, each associated with a type of sequence.

Suppose we start with two numbers.   Their mean is a number that lies between them.  But exactly where it lies depends on what type of mean you are talking about.

If left unspecified, the term “mean” probably refers to the “arithmetic” mean.  So let’s start with that one.  (And by the way, helpful note for beginners:  “arithmetic” here is an adjective.  Read with emphasis on 3rd syllable.)


A sequence of numbers is called arithmetic if the difference between any two consecutive terms is a constant.  For example:

2, 5, 8, 11, 14… has a common difference, d = 3.  (We calculate the current term minus the previous term.)

Say we have been given two numbers A and B.  We want to insert a number between them so that the three of them together form one of these arithmetic sequences. We are looking for the “arithmetic mean.”

We have A, x, B and we want the differences to be constant. 

That means we can write: x – A = B – x

From that we get: 2x = A + B

And then: x = (A + B)/2

This probably does not surprise you.  The arithmetic mean is also just the average. 

If A = 30 and B = 60 we get an arithmetic mean of 45, thus forming the sequence:

30, 45, 60

Then, if you check, you find the difference between consecutive terms is constant:

45 – 30 = 60 – 45 = 15. So the sequence is arithmetic.  45 is the arithmetic mean between 30 and 60.

Sometimes, this is the kind of mean we care about.  For example…

You drive for two hours at 30 mph and for two hours at 60 mph.  What is your average speed for the trip?   

When you solve this, you will discover that when you drive at different speeds for equal time periods, your average for the trip is the arithmetic mean of the two speeds.


A geometric sequence is one where the ratio between every two consecutive terms is constant.  For example:

1, 3, 9, 27, 81… is a geometric sequence with a common ratio of 3. (Now we calculate the current term divided by the previous term.)

So again we have been given two numbers A and B.  And now we want to insert a number between them so that the three of them together form one of these geometric sequences.  We are looking for the “geometric mean.”

We have A, x, B and we want the RATIOS to be constant. 

That means we can write: x /A = B/ x

Cross multiply to get: x2 = AB

And then: x = √(AB)

For example, if A=9 and B=16 then the geometric mean x = √(9×16) =√144  = 12.

We should check our work to see if the ratios are constant:

12/9 = 16/12 = 1.333… so yes, the sequence is geometric.  12 is the geometric mean of 9 and 16.

Sometimes, this is the kind of mean we care about.  For example, here is a problem you may have seen in geometry:


In the triangle above, if x = 9 and y = 16 determine the height, h.

You can use similar triangles to show that the height of that altitude is equal to the geometric mean of the lengths of the two segments on the hypotenuse.


A sequence is called “harmonic” if there is a common difference between the reciprocals of consecutive terms.  For example  (in fact, the classic example):

1, 1/2 , 1/3, 1/4, 1/5… is harmonic.  Its reciprocals are 1, 2, 3, 4, 5… which is clearly arithmetic with a common difference of 1.

And one more time, we have been given two numbers A and B.  Now we want to insert a number between them so that the three of them together form one of these harmonic sequences.  We are looking for the “harmonic mean.”

We  have A, x, B and we want the difference of the reciprocals to be constant. 

That means we can write:


So suppose A = 40 and B = 60. Their harmonic mean is:


 To check our work, we have to look at the sequence of the reciprocals:

1/40, 1/48, 1/60… or, expressed over a common denominator,

6/240, 5/240, 4/240…sure enough, the reciprocals form an arithmetic sequence.


Sometimes, this is the kind of mean we care about.  For example, here is an SAT classic:

You drive to work at 40 mph and then you drive back home at 60 mph. What is your average speed for the trip?

You can solve this by calling the length of each leg d, finding the time for each leg of the journey, adding to get the total time and then using total time and total distance to find average speed.  When you are done, you are in for a surprise.  When you travel equal distances at two different speeds, your average speed for the trip is the HARMONIC mean of the separate speeds.  

I’ll close with a question: why is this called “harmonic”?  Does it have something to do with music and harmony? 

Happy New Year!