# Christmas in July Math Problem [Photo by Reenie-Just Reenie.]

In honor of my Google searchers, to demonstrate the power of bar diagrams to model ratio problems, and just because math is fun…

Eccentric Aunt Ethel leaves her Christmas tree up year ’round, but she changes the decorations for each passing season. This July, Ethel wanted a patriotic theme of flowers, ribbons, and colored lights.

When she stretched out her three light strings (100 lights each) to check the bulbs, she discovered that several were broken or burned-out. Of the lights that still worked, the ratio of red bulbs to white ones was 7:3. She had half as many good blue bulbs as red ones. But overall, she had to throw away one out of every 10 bulbs.

How many of each color light bulb did Ethel have?

Before reading further, pull out some scratch paper. How would you solve this problem? How would you teach it to a middle school student?

## Do It the Hard Way: Algebra with Words

One way to approach the problem is to read each line and translate the story information into algebra. For middle school students, algebra is easier to understand if we use words as variables, rather than letters. We will use the color names to represent the number of good bulbs in each category.

…she stretched out her three light strings (100 lights each)…

That part is simple: $total = 3 \times 100 = 300 \; bulbs$.

…the ratio of red bulbs to white ones was 7:3…

Be careful! Many students have trouble with ratios. The most common mistake is to write: $7 \; red = 3 \; white$.

No!!

Remember that ratios are fractions. Start with the fraction relationship, and then manipulate that to get any of the following… $\frac {red}{white} = \frac 73$, which means that $3 \; red = 7 \; white$, $red = \frac 73 \; white$, or $white = \frac 37 \; red$.

All of these statements are equivalent. Which one is the most useful for our problem remains to be seen.

## Now That We Have the Ratio Under Control

Let’s go back and find out what else our problem has to say:

…She had half as many good blue bulbs as red ones…

Be careful again! Make sure you write down the correct relationship: $blue = \frac 12 \; red$, or $red = 2 \; blue$.

…she had to throw away one out of every 10 bulbs…

This is another ratio, but this one is easy to figure mentally: $trash = \frac {1}{10} \; of \; total = \frac {1}{10} \times 300 = 30$.

…How many of each color light did Ethel have? $red + white + blue + trash = total$.

Since the white and blue bulbs were both compared to the red, it will be easiest to put everything in terms of red bulbs. And we can put in the numbers we know for trash and total: $red + \frac 37 \; red + \frac 12 \; red + 30 = 300$, $\frac {27}{14} \; red = 270$, and $red = 270 \times \frac {14}{27} = 140$.

Then we use substitution to figure out the number of white and blue bulbs.

Working with algebra leads us to some moderately complicated fraction calculations. I suspect that many middle school students would give up on the problem before reaching the end.

## Is There a Better Way? Bar Diagram Ratios

Students who work with bar diagrams have little trouble with ratios. They learn that a ratio of a:b means we have a multiplication problem: there are a units of the first quantity and b units of the second quantity, with the size of the unit to be determined by the rest of the data in the problem.

…the ratio of red bulbs to white ones was 7:3…

So there are 7 units of red bulbs and 3 units of white ones: …She had half as many good blue bulbs as red ones…

Now we add the blue bulbs to our diagram: In order to solve our problem, we need to have all bars in terms of the same size unit. (In algebra, this would be putting everything in terms of a single variable.) Since the middle unit of the red bulbs has been cut in half by the dotted line, let’s cut all the other units in half, too. And we’ll cut the blue bar the same way: That gives us 27 units of good bulbs. Now that we have the ratio diagram under control, we need to look at the other numbers in our problem.

…she pulled off her three light strings (100 lights each)…she had to throw away one out of every 10 bulbs… $good \; bulbs = total - trash = 300 - \frac{1}{10} \; of \; 300 = 270$.

As soon as we can connect a unit (or a set of same-size units) with a number, the difficult task of thinking through the problem is almost over. From here on, all we need is simple arithmetic. 27 units are 270 bulbs. That means: $unit = 270 \div 27 = 10 \; bulbs$.

Now that we know the size of our unit, it’s easy to find out how many bulbs of each color Aunt Ethel has.

## 7 thoughts on “Christmas in July Math Problem”

1. jd2718 says:

I thought I was on another blog, and then I saw the bar diagrams…

7:3,
2:1
7:3.5:3
14:7:6
14+7+6 = 27 (think, ratios coming back in a bit)
3×100 = 300
27 goes into 300 11 times, with remainder,
oops, 10% out
300 – 30 = 270
14*(270/27),
ok, 140, 70, 60

Nothing wrong, but at a certain point I became unteachable…

2. jd2718 says:

Ah, teaching, let’s get them to 14:7:6 and give them some freedom… (as I wish I’d been taught… probably wouldn’t work in most upper elementary or middle school classrooms)

3. Denise says:

That is actually the way I worked the problem out in my head before writing it down, to figure out how many bulbs should be burned out in the story. But I wouldn’t teach it that way, because I think the 3-part ratio would be a stumbling block to most of my middle-grades students.

4. Maria Miller says:

I worked this problem in an identical manner… start out with the ratio 7 : 3.5 : 3 then change that to 14: 7 : 6, add those, get 27 parts, etc.

I would feel that if ALL of us solved it that way, couldn’t we classify that as “expert” behavior, and that we would want to teach it to kids, as well?

Bar diagrams, IMO, would be useful to illustrate the concept of ratios, but as students become more able with ratios, they can leave them behind and just work with the numbers, like we all did.

In other words, encourage them to “pass through” the diagram stage. Maybe it’s more adequate to expect high schoolers to do that, though.

5. Denise says:

You are probably right. This is an upper-end ratio problem, and by the time students have worked through enough ratios to be able to handle this, they should be able to work directly with the numbers. The length of the algebra section (and the fact that I ended up with division of fractions) should have warned me that I had made the problem too hard for the students I had in mind.

6. Kari says:

-300 total bulbs
-1 out of every 10 discarded–> 270 bulbs
-ratio of red:white –> 7:3 + half as many blue as red
-since 7 is odd, I multiplied everything by 10 to make it easy to get half as many blue –>70:30:35 =135
twice that is the solution
140 red: 60 white: 70 blue
No fractions were hurt in the working of this equation!

7. Denise says:

“No fractions were hurt in the working of this equation!”