Financial math: equivalent annual rate

Tutoring financial math, this is an important concept.  The tutor introduces it.

Back in my September 18 post, I discussed the general case of compound interest.

Now we turn to the question: how do we compare 3% interest compounded annually with 3% interest compounded monthly? Perhaps the first problem to notice is that both are called 3%.

While there might be a few definitions, the nominal rate, to many, means the number the interest is called, as opposed to what it truly is. Therefore, 3%, compounded monthly, is a nominal rate (from my point of view, anyway).

To find the equivalent annual rate, we can imagine one dollar is invested in the account for 12 months. Subtracting one dollar off the end amount leaves the interest. Since the interest is on one dollar, it’s also the interest rate.

We proceed:

    \[A=1(1+\frac{0.03}{12})^{(12\times1)}=1.030415957\]

    \[interest=A-P=1.03041597-1=0.030415957\]

The calculations above point to an annual interest rate of 3.0415957%.

So, a nominal rate of 3%, compounded monthly, is really an annual rate of 3.0415957%.

Financial calculators often have a function to find the equivalent annual rate from a nominal one. I’ll be discussing the calculator aspect in future posts:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Math: the pigeonhole principle: another example

Tutoring high school math, the pigeonhole principle might not come up very often.  Because of its applications to everyday life, the tutor often thinks of it.

Back in my May 23 post, I introduced the pigeonhole prinicple. I pointed out that, although the theory seems straightforward, it can yield some surprising results. Here is an example for you sports fans:

Example:

Let’s imagine a sports team’s season is 35 games. How many must they win to guarantee a five-game streak?

Solution: Divide the 35 games into seven sets of five:

_ _ _ _ _|_ _ _ _ _|_ _ _ _ _|_ _ _ _ _|_ _ _ _ _|_ _ _ _ _|_ _ _ _ _

Now, imagine putting 4 Ws in each set of five games. Since there are seven sets, that makes 28 Ws. Notice that, by leaving an artful gap at the proper end of each set of games, you can place 4 Ws in each set of five so that you get no consecutive row of 5 Ws:

W W W W _|W W W W _|W W W W _|W W W W _|W W W W _|W W W W _|W W W W _ (for example)

However, if we need to insert one more W, we can’t escape getting a row of 5 consecutive Ws. Therefore, the solution is 28+1=29 wins. The team must win 29 games to guarantee a five-game winning streak.

Of course, most people might react that if you watch sports, you expect a team that wins even 25 of 35 games will likely have a 5 game streak. True, in all probability; probability, however, is a related but different branch of mathematics. (The problem above is a counting problem.) I’ll have more to say about topics from both those partitions of math in future posts:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Chemistry: stoichiometry

Tutoring chemistry, you field questions from stoichiometry.  The tutor starts with a simple example.

 
Back in my September 15 article, I opened the topic of what a mole means in chemistry. Specifically I pointed out that a mole is the number of atoms that gives the mass of each element as reported on the periodic table.

Molecules react on a molecule-to-molecule basis – or, on a larger scale, a mole-to-mole basis. However, chemists measure the chemicals in grams. Stoichiometry explores the quantitative amounts of chemicals needed – usually in grams – and the grams of products obtained from a chemical reaction.

Example: Consider the reaction

    \[C_3H_8 + 5O_2 \to 3CO_2 + 4H_2O\]

How many grams C_3H_8 (which is propane, as you’ll know from my post here) are required to produce 1 litre of water?

Solution:

(Before proceeding, you may want to take a look at my posts on significant figures. There are a number of them, but this one might be good to start with.)

First of all: a litre of water is 1000. grams at room temperature. Therefore, we need to produce 1000. grams of water.

First, we find the molar mass of water using the masses of H and O from the periodic table:

    \[mass\ H_2O=2(mass\ H)+mass\ O=2(1.0)+16.0=18.0\ grams\]

Similarly for propane:

    \[mass\ C_3H_8=3(mass\ C)+8(mass\ H)=3(12.0)+8(1.0)=44.0\ grams\]

Let’s examine the chemical equation again, in a slightly different way:

    \[1C_3H_8 + 5O_2\to 3CO_2 + 4H_2O\]

If not written, the coefficient of a term in a chemical reaction is 1. Notice the relationship implied:

    \[1\ mole\ C_3H_8\ yields\ 4\ moles\ H_2O\]

We need to figure out how many moles of water we need; we will need \frac{1}{4} that many moles propane.

Well, we know from above that a mole of water is 18.0 grams, and that we need 1000. grams of it. We can find the moles of water in 1000. grams as follows:

    \[1000.grams \times \frac{1mole}{18.0grams}=\frac{1000.mole}{18.0}=55.6moles\ H_2O\]

By the 1:4 ratio earlier observed between the moles of propane and the moles of water, we know we need \frac{1}{4}(55.6moles)=13.9moles propane. All we need now is to determine how many grams that is. From our earlier calculation, we know that the molar mass (mass of one mole) of propane is 44.0 grams. We proceed:

    \[13.9moles\times\frac{44.0grams}{1mole}=611.6\ grams\ C_3H_8\]

Of course, in significant figures, the answer is 612 grams propane.

I will have much more to say about stoichiometry in future posts. Cheers:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Halloween

As a tutor, you’re interested in the holidays and events that are relevant to your students.  An issue that hangs in the air at this time of year is the fascination about Halloween.

I’ve always loved Halloween. When I was a kid, living in farming places, it was a big deal. We’d make Halloween crafts in school and put them on the walls of the classroom. As you walked to gym or music class, you’d see Halloween adornments lining the corridors. Halloween was a kids’ celebration that was encouraged.

An interesting point about Halloween is that while it was promoted by the parents and teachers, the children agreed.  Rarely was the community so successful at directing the children’s interests.  Halloween was an occasion in which virtually everyone wanted to participate.  Even the “un-crafty” liked making Halloween crafts.  Halloween didn’t only resonate with children, either: even a “stick in the mud,” with no children, would buy a pumpkin and carve it into a convincing jack-o-lantern.  They’d have treats ready for the children whom, most other times, they’d rather not tolerate.

Perhaps one appeal of Halloween is its non-judgemental aspect.  You can’t choose the “wrong” costume or carve a jack-o-lantern “wrong”.  When people go out for Halloween – either trick-or-treating or supervising – they enjoy the variety of costumes and jack-o-lanterns they see, rather than just pointing out the best and worst.  Since Halloween has a grim dimension to it, you don’t have to seem happy while you’re celebrating it; you’re not expected to smile.  After all, with disquieted spirits everywhere about, doesn’t it make sense to be on guard?

Without a doubt, Halloween stokes the imagination like no other holiday.  It encourages people to imagine what they can’t see – and even encourages them to believe that it’s really there.  Perhaps above all, humans are imaginative creatures.  Halloween invites people to bring forth their spookiest conjurings.  It’s a labour of love for minds that are too often suppressed.

To be sure, Halloween’s connection with death makes it resonate with people.  Human beings constantly struggle with how meaningful life is in the moment, yet how fragile it is – and the certainty that it will end.  Yet, does it end entirely?  Halloween offers an endless array of ghouls who seem to relish being dead.  Although we might fear them, we find them reassuring as well:  it they have survived death, can we yet do the same?

Even without Halloween, the feelings would manifest.  Farmers’ fields, harvested and laid waste by frost, show the inevitability of death.  Yet, something seems to remain there, waiting to be visited.  People know they likely will not go there, but can’t help but wonder what they might find if they did….

Happy Halloween.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Calculator usage: Random numbers with the Sharp EL-520W

Tutoring math, you spend lots of time with calculators.  The tutor noticed a function the other day that’s convenient and fun.

My high school math students love the Sharp EL-520W – or its newer counterpart, which might be the EL-531. I see a few kinds. These calculators are “forward entry”, which is very popular nowadays. (Forward entry means that to enter, for example, square root of 26, you press the square root key, then key in 26, then equals. Though this seems obvious, calculators in the ’80s were commonly reverse entry: you entered first the number, then pressed square root.) Reverse entry calculators are still made, by the way; I have a number of them and some of my students still use them. They’re great, once you’re used to them – but that’s probably for another post.

Getting back to the Sharp EL-520W: the other day I noticed the RANDOM function. On mine, it’s 2ndF 7. If you press 2ndF 7, you’ll be greeted with a choice screen: RAND or R-DICE. To choose RAND (for random number), press 0. The screen then says RANDOM_; press = to receive a random number. (It’s between 0 and 1, as random number outputs commonly are.) You can keep pressing equals to get new random numbers.

If you press 2ndF 7, then key in 1, you choose R-DICE. Now, you see R.DICE_ across the screen. Pressing = gives you a number from 1 to 6: it’s a simulation of throwing a six-sided die. Not having dice is no longer an excuse to not play a board game;)

I seldom am asked about random numbers; yet, they are prominent in some university courses. I did my entire math degree with a calculator less powerful than the Sharp EL-520W:)

Have a great weekend.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Financial math, continued: calculation of mortgage payment using the TI BA-35 financial calculator

Tutoring math, you’re often asked, “Will I ever use this in life?”  About this particular topic, the math tutor responds, “I hope so.”

Back in my October 16 post on mortgage payments, I opened the topic with the use of mortgage tables. While I still prefer them, the world is moving towards calculators instead.

With so many financial calculators available, their use is an exciting topic to embark. Today I’ll look at my old TI BA-35; the model in front of me is, I believe, from the late 90s. Given this calculator’s commonality, plus the fact that calculator instructions usually remain fairly consistent from the old model to the newer one, I think it’s definitely worth talking about.

I’m going to give commentary on how to calculate the mortgage payment for the same situation posed in my October 16 post. Here are the parameters I was using:

loan 300 000
interest 6.5%
term 25 years
payment frequency monthly

With the TI BA-35, I suspect that changing modes to STAT, then back to FIN (by keying in 2nd FRQ, then 2nd N), will clear the old data. I don’t know this for sure, but it seems to work for me. At any rate, I make sure I’m in FIN mode (if so, you’ll see FIN at the bottom of the screen), then I key in the parameters like so:

300000 PV
0 FV
12×25=N
(the total number payments: monthly for 25 years)
Be careful here: 6.5÷12=%i

Now, to find the payment, I key in

CPT PMT

After a few seconds, the amount 2025.6215 shows on the display. Note this matches the monthly payment from my October 16 post.

Important to notice: with the TI BA-35, you need to enter the monthly interest rate (the interest rate divided by 12) to indicate monthly payments.

I’ll have much more to say about this in future posts; today was just to start us off with the TI BA-35:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Financial math: negative signs with the TI-83 Plus TVM solver

Tutoring math, the uptick in interest about financial math doesn’t go unnoticed.  The tutor responds.

The TI-83 Plus, aka the graphing calculator, has a time value of money solver.  When you press APPS, then select FINANCE, it’s the first choice (on my calculator, anyway).

The TI-83 Plus TVM solver is great, but it can be tricky. Depending on your point of view, understanding its use of negative signs might take some adaptation. I’ll open the issue with an example.

Example 1: Imagine a loan of 30 000, repayable by monthly payments over four years at nominal interest of 6.4%, compounded monthly. Payments are made at the end of the month. Find the monthly payment.

Solution:

Pressing APPS, then selecting FINANCE, you next select TVM Solver. A list of parameters appears wherein you can input your particular values:

N (the total number of loan payments): For our example, you can enter 48 or 4×12.

I%: the nominal interest: 6.4 for this example.

PV: the present value of the loan: 30 000

PMT: skip over this one for now.

FV: the future value of the loan. 0, since it will be paid in full.

P/Y: payments per year: 12 (monthly payments)

C/Y: compoundings per year: 12 (monthly compounding)

PMT: BEGIN END: means payments are made at the beginning or end of the month. For this example, select END.

Now, go back up to PMT, which you left blank before. Press ALPHA ENTER. You’ll see -706.29883…., which means that the monthly payment in this context is 706.30.

From the TI-83 Plus TVM solver’s point of view, if a loan has positive value, then payments against it are negative. Hence, the 30 000 loan generates a monthly payment of -706.30. If you enter -30 000 for the PV of the loan, then the payment will be 706.30.

The TVM solver’s handling of payments versus amounts – by use of negative signs – can be even more complicated in some cases. I’ll be looking at some in future posts:)

Jack of Oracle Tutoring of Jack and Diane, Campbell River, BC.

Math: memory calculations on the Sharp EL-520W

Tutoring math, you want to optimize your students’ calculator usage.  The tutor recalls his own use of the calculator memory back in high school; it was indispensable back then.

Let’s imagine you’re facing the following situation:

    \[(\sqrt[3]25+\sqrt14)^2 + (sin(0.27) +2)^3 - (\frac{1}{7} - \sqrt17)^{2/3}\]

True, you could just enter it, start to finish, into a forward entry calculator like the Sharp EL-520W. Assuming you enter everything perfectly, you’ll get the right answer: the scientific calculator will do the order of operations correctly.

With such a long and complex expression, however, the probability of entering something wrong is typically high. One wrong keystroke will result in the wrong answer. What a shame, following so much effort spent trying to enter the expression correctly!

The tutor admits that he’s as likely as anyone to make an entry error. How, then, do I increase the reliability of my entries?

For a long calculation like the one above, one strategy is to break it into parts, get a reliable answer for each part, then add or subtract them to arrive at the final answer. Let’s revisit our expression from above.

    \[(\sqrt[3]25+\sqrt14)^2 + (sin(0.27) +2)^3 - (\frac{1}{7} - \sqrt17)^{2/3}\]

One approach is to get the answer to (\sqrt[3]25+\sqrt14)^2. This expression, by itself, is short enough to enter reliably. Even so, I enter it a few times, until I get the same answer over again:

    \[(\sqrt[3]25+\sqrt14)^2=44.4312....\]

I do the same with the second part:

    \[(sin(0.27) +2)^3=11.6466....(calculator\ in\ rads)\]

Similarly with the third part:

    \[(\frac{1}{7} - \sqrt17)^{2/3}=2.5115....\]

Now, I add (or subtract) the three partial answers together to get the final one:

    \[44.4312....+11.6466....-2.5115....=53.56631271\]

At this point, some of my students will put the brakes on: “My teacher says I’m not allowed to round until the end. How do I get an exact answer if I have to do the computation in stages?”

While you could write each partial answer down from the calculator screen, keeping all ten digits each time, then re-enter them to get the final answer, I recognize that’s not practical. Worse yet, it’s downright error-prone: you could (and I likely would) make an error copying the answers down.

What you can do, instead, is use the calculator memory to store the partial answers, then recall them as needed to get the final answer. The key sequence to store an answer from the screen is STO letterkey. For instance, to store \sqrt18 in letter E, you’d key in square root 18, then the equals key, then STO, then finally the “log” key (just above the “log” key you’ll see a green ‘E’). To recall that answer later, key in RCL then the “log” key again.

To apply this memory feature to our procedure above: get the answer to (\sqrt[3]25+\sqrt14)^2, then key in STO CNST to store the answer in the ‘A’ variable. Continue, storing the next partial answer in ‘B’, the last one in C. Now, to get the final answer, key in RCL CNST + RCL y^x (for ‘B’) – RCL x^2 (for ‘C’). You, too, should arrive at 53.56631271.

You can store on top of old values as new calculations demand.

The memory feature can make long calculations much easier and safer. Enjoy playing with it:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Financial math: mortgage payments

The tutor notices that financial math is more prominent in high school than it was twenty years ago.   Probably the change is good; let’s embrace it with a first look at mortgage payments. This topic comes up annually in tutoring.

Even before the advent of financial calculators, people had to calculate annuity payments, mortgage payments, etc.  Such financial products have existed certainly since the industrial age – probably even before.

The formula to calculate a mortgage payment isn’t that difficult to use, but probably outside the comfort zone of most consumers.  The obvious question:  How, in the 70s, did people calculate mortgage payments, when financial calculators – if they existed at all – were rare?

The simple solution:  they used mortgage tables.  You looked up the interest rate and the term (in years) of repayment.  The amount you’d arrive at would be the payment per thousand dollars of the loan.  They were called mortgage tables, but you could use them for any loan.

I’ve found a table I like the look of over at realsavvyrealestate.com. Let’s put it to work:

Example: Calculate the mortgage payment for this case:

loan 300 000
interest 6.5%
term 25 years
payment frequency monthly

On the table from realsavvyrealestate.com., we look down the left column to find 6.5%. From there, we look across to the 25 year column. The number we arrive at is 6.75207, which means that the payment is $6.75207 per thousand dollars of loan. The loan amount in this case is $300 000. Therefore, the monthly payment will be

$6.75207×300=$2025.62

Mortgage tables enable easy loan payment calculations. Many people from an earlier generation – myself included – can’t help but prefer them even now.

Mortgage tables, of course, can’t handle the diversity of financial situations that a dedicated calculator can. In coming posts I’ll explore the use of a financial calculator – and perhaps even the option of doing the calculations using financial formulas.

Cheers:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Math: a sales tax problem feat. the “let statement”

Tutoring math, you know sales tax problems to be part of the elementary school repertoire.  The tutor brings forward a more subtle one that’s probably grade 10 or 11 level.

Sales tax problems were calculated by hand when I was in elementary school.  Two cultural changes have occurred since then:

1)  The widespread use of calculators, often even during elementary years.

2)  The use of cards instead of cash.

Both changes have diluted people’s consciousness of sales tax – but especially, I think, the second one.

In any case, sales tax word problems still occur now and then.  Here’s one that’s a little different from merely calculating the tax and adding it to the sticker price:

Example: If the total price paid, including tax, is a hundred dollars, what must the sticker price have been?

Solution: The first step is to invoke the all-powerful “let statement”:

Let\ x=price\ before\ tax

The tax, assuming 12 percent sales tax, will be 0.12x. Therefore,

    \[Total=price\ +\ tax\]

becomes, more specifically,

    \[100=x+0.12x\]

Since x=1x, this simplifies to

    \[100=1.12x\]

We now divide both sides by 1.12:

    \[\frac{100}{1.12}=\frac{1.12x}{1.12}\]

leading to

    \[89.29=x\]

So, the sticker price must have been $89.29 to give the final price, with tax, of $100.

The “let statement” is what makes this approach straightforward. It gives direction to the solver, and clarity to both solver and marker. When facing word problems, the importance of the “let statement” cannot be overemphasized:)

Hope you had a great Thanksgiving. Cheers!

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.