Tuesday, March 4, 2014

Word Problems


Tips on Doing Word Problems

this lens' photo

One of the most common complaints I hear from my Math students is that word problems give them a headache. But really all you need to do is look for some keys words and phrases to translate the word problem into an equation.

(There are certain works that point you in the right direction as far as whether you use addition, subtraction, multiplication or division)

Here are some hints on dealing with them without an aspirin bottle.

First step: Read the problem and look for specific words

Many students have trouble with word problems. A word problems is a math problem that is described rather than written as a math problem would be. For example:

Jason, Nina and Susan all had raisins as a snack for lunch. They decided to count the raisins they each had. It turned out that Nina and Jason had the same number of raisins and Susan had three times as much as each of them. Together, they had 45 raisins. How many raisins did Susan have?

The first thing you need to do is put this word problem into numbers -- create an equation that says the same thing as the word problem. We notice there are three people with an undisclosed number of raisins. But there is a relationship between those numbers.

Let's call the number of raisins Nina has X. Since the number of raisins Nina has is equal to the number of raisins Jason has, Jason also has X raisins. Susan has three times as many raisins or 3X raisins. We are also told that the number of raisins they all had equaled 45. So we can now set up the problem:

X + X + 3X = 45 (the number of raisins Nina has plus the number of raisins Jason has plus the number of raisins Susan has equals 45)

This is now easier to solve -- 5X = 45; X = 9; 3X = 27

So Nina and Jason each have 9 raisins and Susan has 27.

(To check it, add the numbers together and see if it adds up to 45 -- 27+9=36+9=45)

That, of course, was a relatively simple example.

Next Step: Break it Down and Turn it into a Math Problem

The trick with word problems is to break them down to their component parts. Here is a problem:

At a vegetarian restaurant, the cost of two
veggie burgers and five orders of oven fried
potatoes is the same as the cost of four veggie burgers and two orders of oven fried potatoes. How many orders of oven fries could you buy for the same amount of money as two veggie burgers?

The first thing you have to do is figure out how to turn this problem written in English to a problem written as an algebraic equation.

Here's the breakdown:

At a vegetarian restaurant, this phrase is not part of the equation

the cost of two veggie burgers and five orders of oven fried potatoes is the same as the cost of four veggie burgers and two orders of oven fried potatoes. This is the main part of the equation. Let's call the price of veggie burgers X and the price of oven fries Y. This section can then be written as 2X + 5Y = 4X + 2Y
In this case, you need to solve for X in terms of Y or Y in terms of X (since there are two variables and the answer doesn't require you to find the price of either item). So... you first can subtract 2X from either side giving you 5Y = 2X + 2Y
then subtract 2Y from either side giving you 3Y = 2X
Since the question is:

How many orders of oven fries could you buy for the same amount of money as two veggie burgers? you have your answer -- the cost of two veggie burgers is 2X and we now have 3Y = 2X and Y is the cost of oven fries, therefore, if we say 3Y = 2X or 2X = 3Y, then the cost of two veggie burgers (2X) is the same as the cost of three orders of oven fries (3Y) so the answer is that you can buy three orders of oven fries for the same price as two veggie burgers.

Compugraph Designs' Printfection Store

Math (Fractions) Travel MugCompugraph Designs has a store on "Printfection" which includes cutting boards (good wedding or housewarming gifts), mugs and cups, tees, etc.

This apron is only one of several Math themed items at our store:

Compugraphd Printfection site

(Click on the picture to go directly to this product's page)

Sunday, March 2, 2014

Pluto: Planet or Dwarf?


Why Was Pluto Demoted?

When I was growing up, we were taught that there were 9 planets, Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and Pluto. Pluto, we were taught, was discovered in the 1930s. When I was going to school, Pluto was thought to have no moons, later they found a moon and named it Charon (after the "ferryman" in mythology who ferried the dead to Hades). Since then, Nix and Hydra and two other moons have been discovered. So how is it that something substantial enough to have at least 5 moons was demoted to a "dwarf planet"?

So What's the Issue?

So, as a good fan of the planet Pluto, I went to find out what the issue was. Why were they demoting my favorite little planet?

Well, apparently the scientists who work on these sorts of things (Astronomers) found some planetary type bodies in our Solar System. One of the planetary objects they found was named Eris. Eris, strangely enough, is larger than Pluto. But is that enough to change Pluto's status?

Why was Pluto demoted? Why didn't they promote Eris? Ceres, another planetary body, is now also considered a Dwarf Planet, but Ceres is a bit smaller than Pluto and Eris. When I was in school, I recall Ceres being considered an Asteroid. Asteroids are sometimes irregularly shaped, so perhaps that is why Ceres was promoted to a Dwarf Planet.

So where do we stand with planetary bodies? What constitutes a planet and what constitutes a Dwarf Planet? And, by the way, what constitutes an Asteroid? While many of the Asteroids are in the Asteroid Belt between Mars and Jupiter, there are some that are not.

Because space is so vast and there are so few objects within that vastness, I highly doubt that we are done finding planetary bodies or categorizing them. Like living organisms on the earth, planets and planetary bodies change over time; they change their position in the sky, they change their sizes and compositions, they change their place vis-a-vis the sun. We will continue learning about the solar system; we will continue to learn about the entire universe.

The Planets as I Learned Them

While I know Pluto is no longer a planet, this is how I learned it.
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Solar System Themed Designs from Compugraph Designs on Zazzle

Finding the Lowest Common Denominator


What's the "Lowest Common Denominator"?

this lens' photo

So you've heard the term "Lowest Common Denominator" but you're not quite sure what it is. It is a math term, but it's also used to mean "something of small intellectual content designed to appeal to a lowbrow audience ; also : such an audience" (according to http://www.merriam-webster.com/dictionary as of September 15, 2008).

But in math, it refers to a something needed for addition or subtraction of fractions.

When you need to add two fractions, the denominators (the bottom number in a fraction) need to be the same number. For example, you can add 3/5 to 4/5 because the denominator is the same (3/5 4/5 = 7/5).

But if the denominators don't match (like 3/5 and 1/6) then you have make them match.

Finding The Lowest Common Denominator, Part I

Factoring Numbers

When you need to find the Lowest Common Denominator, the first thing you need to do is factor the two denominators (You'll understand why later). Let's say we have two fractions, 3/78 and 5/98. We need to add these two fractions and they don't have the same denominator.

The first thing you need to do is factor 78 and 98 into their prime factors (Note: if a number in the denominator is already a prime number you would skip this step with that denominator number!). Let's start with 78. Since 78 is an even number, the first thing we would do is divide it by 2. 78/2= 39. Then we would divide 39 by another number that might go into it. Since 39 is odd, we know 2 doesn't go into it. So we try 3 (the next prime number after 2), and sure enough, 3 goes into 39 -- 39/3 = 13. 13 is a prime number (see below to find out how to check out whether a number is prime).

Next we do the same thing with 98. Since 98 is also an even number, the first thing we would do is divide it by 2. 98/2= 49. Then we would divide 49 by another number that might go into it. Since 49 is odd, we know 2 doesn't go into it. So we try 3 (the next prime number after 2), but that doesn't work. So we try 7 (since we know 5 won't work) and sure enough, 7 goes into 49 -- 49/7 = 7. 7 is a prime number.

Finding the Lowest Common Denominator, Part II

So now that we have the two denominators factored, we have to use that factoring to figure out what the Lowest Common Denominator is.

If you notice in the graphic, both factor lists have a 2 (circled in green). So when you multiply out to find the Lowest Common Denominator, you only include one 2. You then make sure you have at least one 3, two 7s and one 13 (see the factors to the side). When you multiply this all out, you get 3822. This would be the Lowest Common Denominator.

That is a bit of an extreme example, so I'll quickly show you another example. Let's say we have to add three fractions -- 5/12, 3/8, and 2/9. 12= 2 X 2 X 3 (two 2s and one 3), 8= 2 X 2 X 2 (three 2s) and 9= 3 X 3 (two 3s). We look at what we have and we take the largest number of each number that appears in any one of the factor lists (so we'd take three 2s to cover the 2s in 8; and two 3s to cover the 3s in 9; both will cover the factors in 12) -- this would give us 72 (2 X 2 X 2 X 3 X 3 = 72) -- so the 5/12 become 30/72 (12 X 6 = 72, 5 X 6 = 30), 3/8 becomes 27/72 (8 X 9 = 72, 3 X 9 = 27), and 2/9 become 16/72 (9 X 8 = 72, 2 X 8 = 16). Add them up and you get:
30/72 + 27/72 + 16/72 or 73/72.

How to find out if a number is prime

If you want to find out if a number is prime, you don't have to try to divide it by every number up to that number. (This would be really time consuming, even using [the bane of my existence] a calculator if you are trying to find out if a number such as 57,967 is prime!). First of all, 2 is the only even prime number, so other than 2, there are no even prime numbers. Second of all, any number ending in a "5" is divisible by 5 (15 is 3 X 5, 25 is 5 X 5, 35 is 7 X 5, etc.) so, other than 5 itself (which is prime) no number ending in a "5" is prime.

So, in order for a number to be prime (other than 2 and 5), it has to end in 1,3,7 or 9 (as you can see from the graphic on the side which shows the prime numbers from 0 to 100 -- 1 is not considered a prime number).

Another thing you have to understand is that composite numbers (that is numbers that are not prime) are divisible by at least two numbers (other than 1 and themselves). At least one of those number has to be either the square root of the number or less than the square root of the number.

[to illustrate this, let's take a few examples -- 39, for example, the square root of 39 would be greater than 6 (6X6=36) and less than 7 (7X7=49) -- so at least one of the numbers that 39 is divisible by, if it is composite, would be less than 7. So first you try 3, which 39 is divisible by, so 39 is not prime.

41, for example, the square root of 41 would be greater than 6 (6X6=36) and less than 7 (7X7=49) -- so at least one of the numbers that 41 is divisible by, if it is composite, would be less than 7. So first you try 3, which 41 is not divisible by (3X13=39, 3X14=42) then you try (since we know 41 is not divisible by 5 or any even number) 7, which 41 is notdivisible by (7X5=35, 7X6=42), so 41 is prime.]

Next, you only have to try dividing a number by prime numbers (since a number that is not divisible by 3, for example, won't be divisible by 6, 9, 12, 15, 18, 21 or any other number that 3 goes into) -- so that means you need to try dividing by 3, 7, 11, etc. (5 only goes into numbers that end in "0" and "5" like 55 or 110 or 9875 or 30670 and only even numbers are divisible by 2). This makes figuring out if something is prime a lot faster than trying every number.

By the way, 57,967 is not prime -- it equals 7 X 7 X 7 X 13 X 13.

Math Designs from Compugraph Designs' Shop on Zazzle

Pythagorean Theorem

More Math Hints

What was Pythagoras thinking?

One of the rules you learn in Geometry is the Pythagorean theorem. This states that the sum of the square of the two legs of a right triangle is equal to the square of the hypotenuse.

What exactly does that mean?

What is the Pythagorean Theorem?

Looking at the diagram on the right, you can see a right triangle. The Pythagorean theorem states that if you add the squares of the two legsequal the square of the hypotenuse (the side of the right triangle opposite the right angle -- this would necessarily be the largest side of the triangle).

This means that if you have the lengths of any two sides, you can find the third. For example, you have a right triangle and you know that one legs is 5 and the hypotenuse is 13. You take the formula and substitute in --
step 1 -- 5 squared + B squared = 13 squared
step 2 -- 25 + B squared = 169 --
step 3 -- B squared = 169 - 25
step 4 -- B squared = 144
step 5 -- B= the square root of 144 or 12

Pythagoras's Diagram

What this theorem represents

When Pythagoras developed his theorem, he was talking about the area of squares. The diagram at the right represents what Pythagoras meant. He meant that the area of a square with sides the length of one side of a triangle (side A, in magenta, length 3) added to the area of a square with sides the length of the second side of the triangle (side B, in blue, length 4) would add up to the area of a square with sides the length of the hypotenuse (side C, in purple, length 5).

What this means in our example, the area of the magenta square (with a side length of 3 and an area of 3X3 or 9) added to the area of the blue square (with a side length of 4 and an area of 4X4 or 16) equals the area of the hypotenuse (or purple square -- with a side length of 5 and an area of 5X5 or 25). 9+16=25.

This can help you find the length of the third side of a right triangle when you have the other two.

Compugraph Designs Arts Now Site

"Arts Now" is a "Print on Demand" site. They have a nice collection of clocks and watches, including the one pictured here (with math symbols on it). Click on the picture to see the entire math clock collection on the Arts Now site.

Math Themed Products from Compugraph Designs on Zazzle

Simultaneous Equations


Simultaneous Equations 

As my tutoring students know, I detest calculators. I think the current math curriculum encourages laziness. Working with the numbers themselves gives students more of an idea of how numbers fit together. For example, I often can solve math problems in my head. The reason I can do that is that I did the vast majority of my math problems with pencil and paper or in my head, not with a calculator.

The reason I mention this is that my main tutee (he's in high school and I've been working with him since he was in 2nd grade, so by now he knows how I feel about calculators) wanted to use his calculator, with a simultaneous equation function, to solve the problems. I told him if he wanted my help (which he did) he'd need to learn how to do it without the calculator.

Simultaneous Equations

This picture represents a set of simultaneous equations. These two equations represent two lines on a graph. The solution to the equations is also the point where the two lines intersect. The bottom equations shows how you can solve for x in terms of y)

The Next Step 

Multiply the entire top equation by 4 and the entire bottom equation by 3 (so that box equations now have 12x as the first part of thee equation) to get the results in this picture.

After That:

Because out goal is to get this equation down to a solvable one variable equation (and one equation), we will want to subtract the second equation (because the "x" part of both equations will "drop out" when you take 12x from 12x). So we multiply the lower equation by -1 (or, simply, change the plus signs to minus and the minus signs to plus).

Subtract and get this

As you can see from this picture, you subtract each column (x from x, y from y, etc.) and the x drops out leaving only the y and the numbers. Then you solve for y by dividing by 17 and you end up with y = -2

Once you solve for y, you need to find x:

Substitute -2 for y and this is what you get:

Do the same thing in the other equation:

You should get the same answer for x (this is a way to check your work -- if the answers are the same, then you probably did it right -- if not, you need to redo the problem)

One More Way to do Simultaneous Equations

There is another way to do Simultaneous Equations. That is by using the substitution method -- you solve for X in terms of Y or Y in terms of X.

In this example, we take the second equation (x-y=0) and move the y to the other side by adding it so -- X - Y + Y = 0 + Y. the -Y +Y on the left side cancel out (-Y+Y = 0) and you are left with X=Y.

In the OTHER equation (3x + 2y = -5) and substitute y for x (since y=x, as we just figured out) -- and we get 3y + 2y = -5. Add together and we get 5y=-5 or y = -1. Since x=y, then x=-1. So the answer it (-1, -1).

For the most part, it's harder to do this method, but occasionally, you'll see an equation that makes substitution easier. It's a matter of instinct. While this won't work well most of the time, it's nice to know how to do it when you need it