Candy Dish Blog

I found an article online last Thursday about Green & Black’s chocolate going to Fairtrade completely by the end of 2011. I was pretty impressed by the move and even added it to our Amplify feed. “Wow,” I thought. “Good for them.” Then I read further and saw a statistic that said the Fairtrade market has grown from £22 million in 1999 to £635 million last year. It’s a staggering growth for any business, especially when your stock and trade is helping small businesses. That’s apparently a very lucrative business to be in!

Personally, I think of Fairtrade products in the same light as organic products, since it is the same type of store that carries the most variety of each, at least in my area. However, organic products have taken a hit in the current economy while Fairtrade products are seeing growth in sales. I attribute this to the differing reasons people buy and don’t buy each.

People buy organic products for a rational reason: they have the perception of better nutrition and safety. We are talking perception here. They are also more expensive than the non-organic alternatives. And when you are buying produce, for example, and the organic apples are smaller than or identical to the non-organic alternative and you have less money in your pocket, it becomes easy to choose a non-organic product. In other words, the choice comes down to price.
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Manipulatives are visual, tactile devices that are meant to be handled and are especially useful for teaching abstract topics like fractions and percentages. I pointed out earlier that one math teacher in Glendale, CA Chamlian Armenian School uses Hershey’s bars as a visual aid to teach math concepts.

The chocolate bars are divided into four rows of three squares each, totaling 12 squares in all. If a student breaks the four rows apart, it is easy and delicious to teach concepts like half and quarters. If you break the bar other ways, you can teach thirds and even twelfths. Other brands that are scored in different ways can help teach other fraction concepts. Let’s look at a simple math problem using a Hershey bar:

What is one-fourth of 12? Break the chocolate bar into squares. Next, divide the squares into four equal groupings. Each grouping has three pieces, so one-fourth of 12 is three.

As the students progress with the math, they can get to more advanced concepts like remainders. How many times does five go into 12? To solve this, start with 12 squares and make as many piles of five squares as possible. You end up with two piles of five squares with two squares leftover, so the answer is two remainder two.

This is not difficult math for most of us but getting kids to see a concrete visualization like this can be a big key in getting them to think abstractly.

Those of us who have been in the classroom know that there are different types of intelligence and different learning styles. As educators, we have a golden opportunity to be creative and implement some approaches we did not have ourselves so we can communicate ideas to young minds. To me that was where both the frustration and joy of teaching came from, changing from one to the next and ending in success for the kids I worked with.

Back then I never thought of using chocolate as a manipulative but I salute the teacher who thought of this. It’s a sweet ending to math class – a course often feared and dreaded.

Chocolate Squares by Siona Watson.

Candy Math Monday is all about demystifying math for the benefit of all, especially our young friends who struggle with math in school or while preparing for the SAT or a similar test. It’s a pain, to be sure, but that’s why we are working with candy – to make math a bit sweeter for you.

Today we will look at probabilities. Notice that the Mike and Ike candies in the photo have about twice as many green pieces than red. Let’s assume that is the exact proportion: two green to one red. If you have a bag of these you are sharing with a friend and you randomly pick one out, what is the probability that it will be a red one?

Break it down
First, what is probability? It is a term used quite a bit, so let’s come to an understood agreement. Basically put, probability is the quantifiable likelihood of something happening. In other words, it is the possibility of a particular outcome over the entire set of outcomes. Simple? I thought so.

How it all works
If we have two green candies for each red one, that is a ratio of 2:1. Add those numbers together and you have three candies total (2 green + 1 red = 3), or some multiple of three, assuming you have a big bag of them. One way to think of this is to imagine what you are trying to find (a red one) out of the total in the bag (including green and red).

One out of three of the candies is red, so you have a one-third chance of getting a red one. Simple answer, right? Well, this is an entry-level problem. Let’s kick it up a notch. If you have the same ratio and five red candies, what is the probability of getting two red ones in a row?

The first thing you have to do is calculate how many candies you have. If the ratio is 2:1 (G:R) and you know you have five reds, then you have ten green ones, right? That means your expanded ratio is 10:5.

The Answer
The first part is easy. We already did that. Five red candies, ten green ones means a 5 out of 15 (5/15) or one-third chance.

The second part is tricky. You have already eaten one red one, or maybe shared it with your friend, so how many are left? You have four left. This changes the ratio, so you have to do some more math. Your new ratio is 10:4. The new probability of getting a red one is 4 out of 14. 4/14 can be reduced to 2/7. It’s easier to work with smaller numbers, so let’s go with that.

So now we have the answers to the first and second parts of the problem. Now we have to bring them together. When dealing with probabilities like this, you have to multiply them to get the final answer.

1/3 * 2/7 = 2/21

So you have a 2/21 chance of getting two red candies in a row. Simple? I thought so.

The key to solving complicated math problems is to break them down into manageable, simpler problems. In addition, putting some candy in the mix adds some fun to the process.

If I based my life’s philosophy on candy poetry, and don’t think I haven’t, I would know in my heart of hearts that there just aren’t enough mellowcreme pumpkins in a bag of autumn mix. It’s a truth, you see, not a belief. Based on that, let’s do a little math to figure something out.

If a company wants to limit the exposure of mellowcreme pumpkins by mixing them in with regular candy corn, and thus dilute the goodness and love and enjoyment in the world, and has figured out the ideal ratio of 3:2:1 for regular candy corn : caramel candy corn : mellowcreme pumpkins, how many of each are there in a package, and how many candy pieces total, if there are 92 caramel pieces?

Step 1. Set up a table to organize your information:

Step 2. Figure out what’s missing. I would go for that right column first. If 92 is twice the number of pumpkins, then there are 46 pumpkins, right? You can test this by multiplying 46 by 2 to see if you get 92. Bingo. Then the left blank. If we know there are three times the number of candy corn kernels as there are pumpkins, then we just multiply 46 by 3 to get the answer. 138.

Step 3. Figure out the total number of pieces of candy. That’s easy. Just add it up:

Alternate method:
After finding the number of pumpkins, you have the basis for how to find the total number of pieces of candy in a different, even quicker, way. The ratio is 3:2:1. Adding those numbers up (3+2+1) you get 6. Multiply 6 times the base number, 46, and you get the same answer – 276.

Bada bing, bada boom. Here I am, keeping the world safe for math and candy. Huzzah!

Candy Pumpkin by adobemac

To continue the candy math series, today we will look at problems dealing with sets and overlapping sets. If you are studying for the PSAT or SAT, you should become very familiar with this concept. Let’s start by looking at a typical problem:

After school, kids gather to trade gum (unchewed, of course). If 25 students have bubble gum, 19 have chewing gum, 12 have both and 5 have neither (poor gumless kids), how many kids are there?

This type of problem requires more than simple adding because some kids fall into more than one category. There are two basic ways to solve this problem. One is to use an equation. For overlapping sets, you can always use this equation:

Total = Set 1 + Set 2 + Neither – Both

So to plug in these numbers you get:

x = 25 + 19 + 5 -12

and thus:

x = 37 kids

There is another way to solve this problem. Some people do just fine with this equation but some people think more visually and need to see things graphically. I am that type of person. Did I mention that I am a photographer? That’s part of me being a visual person. The second method of solving this requires Venn diagrams, which give a visual representation of the problem.

The problem lots of people have with these problems has to do with the overlap. One way to use the Venn diagram is just as a visual cue to remind yourself to subtract the overlap. Another is to help you determine precise numbers of how many of these wonderful young people have only chewing gum or only bubble gum.

So you can see that when adding these numbers together, you have to subtract out the 12 students who overlap. 13 students have only bubblegum, seven have only chewing gum, 12 have both and five have neither. 13 + 7 + 12 + 5 = 37.

The moral of the story is that math is fun and tastes like candy.

Candy Math Mondays are a biweekly feature here and will be around for as long as I can stretch this theme out. Are you a teacher wanting to incorporate candy into the classroom to get kids more interested in math? Let us know what problems you would like to see tackled with a candy twist. We will do our best to help.

Last week we looked at one type of combination problem, figuring out how many different possibilities of candy you can have mixed into your ice cream. This week we will look at a different combination problem, really a permutation problem – how many ways can you do something? Let’s look back at Sera’s photo of gummi bears from a couple weeks ago:

Ten different flavors of gummi bears are represented here. Yes, I know the two red ones look the same. Let’s assume one is cherry and the other is raspberry or some other red flavor. I want to eat them, one at a time, and want to know how many different possible ways I can do it if I have no order preference, choosing them randomly one at a time. This is another typical SAT problem with a simple solution. In fact, it is even easier than last week’s problem.

I choose my first gummi bear. How many options do I have? Ten. Then I have nine left so I can choose one out of nine. Then eight, seven, etc., all the way to when we have only one gummi bear left and thus have only one option. So we have worked out the logic of this problem, but how do we find the answer? The solution is to multiply these different possibilities together:

10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3,628,800

Simple, right? You probably had no idea that ten little gummi bears offered such a huge number of possibilities for your life! So what do you think? Is math more fun with candy or without?

Gummi Bears by princess_of_llyr.
Candy Math Mondays are a new feature here and will be around for as long as I can stretch this theme out. Are you a teacher wanting to incorporate candy into the classroom to get kids more interested in learning? Let us know what problems you would like to see tackled witha a candy twist. We will do our best to help.

On Wednesday Tori talked about candy in ice cream at the DQ and said, “the options are practically endless,” after listing fives different candies. Of course, she knew the options were not really endless. She’s a smart one, that Tori. This led me to remember the old SAT problem of how many combinations you can have from a given set of options. Ah, the useful stuff I picked up while teaching SAT prep classes.

In Tori’s example, she listed five candies that can be mixed into the ice cream, but let’s thin the herd a bit. To replicate a good SAT problem, it must be a bit shorter because each math problem on the SAT can be done in one minute or less. Believe me on that one. Every problem has a relatively simple, if not straightforward, solution. Our choices are:

Snickers (S)
Butterfinger (B)
Reese’s (R)

This is not a real equation and is just here for our amusement.

If we assume that the customer is going to have at least one candy and up to three, how many combinations of different candies could there be? What we have to do is separate this out into three easy problems, find each individual answer and then add them all together. It is three problems because the customer can have 1, 2 or 3 types of candy added.

1 Candy. This is easy. Pick one out of three. There are three options, so the answer is 3.

S
B
R

2 Candies. This gets a bit trickier. Choose your first candy. Let’s say you choose Butterfinger for your first candy. There are two options left for the second candy, right? To figure this out, make a list of your options:

SB
SR
BR

Note that I have not listed RB as an option because mixing Reese’s with Butterfinger is the same as mixing Butterfinger with Reese’s. This is one mistake people make with this problem. 3 Options total.

3 Candies. This is an easy step. If you have three candies and want to put in all three, how many options are there? 1 option.

SBR

The Answer. To get your final answer, simply add up your individual answers:

3+3+1=7

Easy as pie, right? Now, for those of you who are hardcore math lovers like our young friend Laura, let’s go back to Tori’s first example, which had five candy choices. How many combinations can you derive from that?

Update: It has been pointed out that the equation listed above is not the correct one for combinations. It is, in fact, a nonsense equation, chosen out of humor and a perceived complexity of math. THat was probably a bad choice on my part. Attentive reader Elise (aka my Lovely Wife) sent me the proper equation, which is this:

Nonsense Equation by Adam Coster.
Blizzards by digiputz.