Trouble with Percents

Can your students solve this problem?

There are 20% more girls than boys in the senior class.
What percent of the seniors are girls?

This is from a discussion of the semantics of percent problems and why students have trouble with them, going on over at MathNotations. (Follow-up post here.) Our pre-algebra class just finished a chapter on percents, so I thought Chickenfoot might have a chance at this one. Nope! He leapt without thought to the conclusion that 60% of the class must be girls. After I explained the significance of the word “than”, he solved the follow-up problem just fine.


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How Old Are You, in Nanoseconds?

Conversion factors are special fractions that contain problem-solving information. Why are they called conversion factors? “Conversion” means change, and conversion factors help you change the numbers and units in your problem. “Factors” are things you multiply with. So to use a conversion factor, you will multiply it by something.

For instance, if I am driving an average of 60 mph on the highway, I can use that rate as a conversion factor. I may use the fraction \frac{60 \: miles}{1 \: hour} , or I may flip it over to make \frac{1 \: hour}{60 \: miles} . It all depends on what problem I want to solve.

After driving two hours, I have traveled:

\left(2 \: hours \right) \times \frac{60 \: miles}{1 \: hour} = 120 miles so far.

But if I am planning to go 240 more miles, and I need to know when I will arrive:

\left(240 \: miles \right) \times \frac{1 \: hour}{60 \: miles} = 4 hours to go.

Any rate can be used as a conversion factor. You can recognize them by their form: this per that. Miles per hour, dollars per gallon, cm per meter, and many, many more.

Of course, you will need to use the rate that is relevant to the problem you are trying to solve. If I were trying to figure out how far a tank of gas would take me, it wouldn’t be any help to know that an M1A1 Abrams tank gets 1/3 mile per gallon. I won’t be driving one of those.

Using Conversion Factors Is Like Multiplying by One

If I am driving 65 mph on the interstate highway, then driving for one hour is exactly the same as driving 65 miles, and:

\frac{65 \: miles}{1 \: hour} = the \: same \: thing \: divided \: by \: itself = 1

This may be easier to see if you think of kitchen measurements. Two cups of sour cream are exactly the same as one pint of sour cream, so:

\frac{2 \: cups}{1 \: pint} = \left(2 \: cups \right) \div \left(1 \:pint \right) = 1

If I want to find out how many cups are in 3 pints of sour cream, I can multiply by the conversion factor:

\left(3 \: pints \right) \times \frac{2 \: cups}{1 \: pint} = 6 \: cups

Multiplying by one does not change the original number. In the same way, multiplying by a conversion factor does not change the original amount of stuff. It only changes the units that you measure the stuff in. When I multiplied 3 pints times the conversion factor, I did not change how much sour cream I had, only the way I was measuring it.

Conversion Factors Can Always Be Flipped Over

If there are \frac{60 \: minutes}{1 \: hour} , then there must also be \frac{1 \: hour}{60 \: minutes} .

If I draw house plans at a scale of \frac{4 \: feet}{1 \: inch} , that is the same as saying \frac{1 \: inch}{4 \: feet} .

If there are \frac{2\: cups}{1 \: pint} , then there is \frac{1\: pint}{2 \: cups} = 0.5 \: \frac{pints}{cup} .

Or if an airplane is burning fuel at \frac{8\: gallons}{1 \: hour} , then the pilot has only 1/8 hour left to fly for every gallon left in his tank.

This is true for all conversion factors, and it is an important part of what makes them so useful in solving problems. You can choose whichever form of the conversion factor seems most helpful in the problem at hand.

How can you know which form will help you solve the problem? Look at the units you have, and think about the units you need to end up with. In the sour cream measurement above, I started with pints and I wanted to end up with cups. That meant I needed a conversion factor with cups on top (so I would end up with that unit) and pints on bottom (to cancel out).

You Can String Conversion Factors Together

String several conversion factors together to solve more complicated problems. Just as numbers cancel out when the same number is on the top and bottom of a fraction (2/2 = 2 ÷ 2 = 1), so do units cancel out if you have the same unit in the numerator and denominator. In the following example, quarts/quarts = 1.

How many cups of milk are there in a gallon jug?

\left(1\: gallon \right) \times \frac{4\: quarts}{1\: gallon} \times \frac{2\: pints}{1\: quart} \times \frac{2\: cups}{1\: pint} = 16\: cups

As you write out your string of factors, you will want to draw a line through each unit as it cancels out, and then whatever is left will be the units of your answer. Notice that only the units cancel — not the numbers. Even after I canceled out the quarts, the 4 was still part of my calculation.

Let’s Try One More

The true power of conversion factors is their ability to change one piece of information into something that at first glance seems to be unrelated to the number with which you started.

Suppose I drove for 45 minutes at 55 mph in a pickup truck that gets 18 miles to the gallon, and I wanted to know how much gas I used. To find out, I start with a plain number that I know (in this case, the 45 miles) and use conversion factors to cancel out units until I get the units I want for my answer (gallons of gas). How can I change minutes into gallons? I need a string of conversion factors:

\left(45\: min. \right) \times \frac{1\: hour}{60\: min.} \times \frac{55\: miles}{1\: hour} \times \frac{1\: gallon}{18\: miles} = 2.3\: gallons

How Old Are You, Anyway?

If you want to find your exact age in nanoseconds, you need to know the exact moment at which you were born. But for a rough estimate, just knowing your birthday will do. First, find out how many days you have lived:

Days\: I\:have\: lived = \left(my\: age \right) \times \frac{365\: days}{year}

+ \left(number\: of\: leap\: years \right) \times \frac{1\: extra\: day}{leap\: year}

+ \left(days\: since\: my\: last\: birthday,\: inclusive \right)

Once you know how many days you have lived, you can use conversion factors to find out how many nanoseconds that would be. You know how many hours are in a day, minutes in an hour, and seconds in a minute. And just in case you weren’t quite sure:

One\: nanosecond = \frac{1}{1,000,000,000} \: of\: a\: second

Have fun playing around with conversion factors. You will be surprised how many problems these mathematical wonders can solve.


[Note: This article is adapted from my out-of-print book, Master the Math Monsters.]


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Bill Gates Proportions II

[Feature photo above by Remy Steinegger via Wikimedia Commons (CC BY 2.0).]

Another look at the Bill Gates proportion… Even though I couldn’t find any data on his real income, I did discover that the median American family’s net worth was $93,100 in 2004 (most of that is home equity) and that the figure has gone up a bit since then. This gives me another chance to play around with proportions.

So I wrote a sample problem for my Advanced Math Monsters workshop at the APACHE homeschool conference:

The median American family has a net worth of about $100 thousand. Bill Gates has a net worth of $56 billion. If Average Jane Homeschooler spends $100 in the vendor hall, what would be the equivalent expense for Gates?

Continue reading Bill Gates Proportions II

Putting Bill Gates in Proportion

[Feature photo above by Baluart.net.]

A friend gave me permission to turn our email discussion into an article…

Can you help us figure out how to figure out this problem? I think we have all the information we need, but I’m not sure:

The average household income in the United States is $60,000/year. And a man’s annual income is $56 billion. Is there a way to figure out what this man’s value of $1mil is, compared to the person who earns $60,000/year? In other words, I would like to say — $1,000,000 to us is like 10 cents to Bill Gates.

Continue reading Putting Bill Gates in Proportion

Percents: The Search for 100%

[Rescued from my old blog.]

Percents are one of the math monsters, the toughest topics of elementary and junior high school arithmetic. The most important step in solving any percent problem is to figure out what quantity is being treated as the basis, the whole thing that is 100%. The whole is whatever quantity to which the other things in the problem are being compared.

Continue reading Percents: The Search for 100%

Percents: Key Concepts and Connections

[Rescued from my old blog.]

Paraphrased from a homeschool math discussion forum:

“I am really struggling with percents right now, and feel I am in way over my head!”

Percents are one of the math monsters, the toughest topics of elementary and junior high school arithmetic. Here are a few tips to help you understand and teach percents.

Continue reading Percents: Key Concepts and Connections

Negative Numbers for Young Students

[Rescued from my old blog.]

Would you like to introduce your students to negative numbers before they study them in pre-algebra? With a whimsical number line, negative numbers are easy for children to understand.

Get a sheet of poster board, and paint a tree with roots — or a boat on the ocean, with water and fish below and bright sky above. Use big brushes and thick poster paint, so you are not tempted to put in too much detail. A thick, permanent marker works well to draw in your number line, with zero at ground (or sea) level and the negative numbers down below.

Continue reading Negative Numbers for Young Students

Order of Operations

[Rescued from my old blog.]

Marjorie in AZ asked a terrific question on the (now defunct) AHFH Math forum:

“…I have always been taught that the order of operations (Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction) means that you work a problem in that order. All parenthesis first, then all exponents, then all multiplication from left to right, then all division from left to right, etc. …”

Many people are confused with order of operations, and it is often poorly taught. I’m afraid that Marjorie has fallen victim to a poor teacher — or at least, to a teacher who didn’t fully understand math. Rather than thinking of a strict “PEMDAS” progression, think of a series of stair steps, with the inverse operations being on the same level.

Continue reading Order of Operations

Fraction Division — A Poem

[Rescued from my old blog.]

Division of fractions is surely one of the most difficult topic in elementary arithmetic. Very few students (or teachers) actually understand how and why it works. Most of us get by with memorized rules, such as:

Ours is not to reason why;
just invert and multiply!

Continue reading Fraction Division — A Poem