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## 7.RP.A.3 Transcript – Part 1

This is Common Core State Standards support video in mathematics. The standard is 7.RP.A.3. This standard states: Use proportional relationships to solve multi-step ratio and percent problems. Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees, percent increase and decrease, percent error.

Let’s look at some of the related standards to this one. In the grade before, in Grade 6, there’s standard 6.RP.A.3c that talks about finding a percent of a quantity as a rate per hundred. Now, notice that they do use the idea of percent. Now, other than three high school standards, 7.RP.A.3 and 6.RP.A.3c are the only standards where the term percent is specifically addressed. Topics such as simple interest, tax, gratuity, and so forth imply the use of percentages even though the term percent is not explicitly stated. Back in Grade 4, students convert fractions with denominators of 10 and 100 to decimals. In Grade 5, the focus is mostly on using decimals with the four operations, and in the grade preceding this, Grade 6, the term decimal does not appear in any standard. The number system focus at that grade level is rational numbers including integers.

In order to succeed with the problems posed by this standard, students have to be proficient with conversions among fractions, decimals, and percents. Now, there is standard 7.NS.A.2d that talks about converting a rational number to a decimal using long division; and they need to know that the decimal form of a rational number terminates in zeros or eventually repeats. We have this idea of knowing how to convert from one form to another—decimals, fractions, and percents. When you look at converting decimals to percents or to fractions, what we’re dealing with here is computation. So, for computation purposes, it’s really about the relationship between fractions and decimals or percents and decimals. If you’re talking about the direct connection between fractions and percent, that’s really more about fluency. For example, students need to pretty much have ingrained that 1/4 is the same as 25% or that 3/4 is the same as 75%. So, students have to have a conceptual understanding of this relationship between fractions and percents. Again, they need to be able to recognize pretty quickly that 18% is the same thing as the fraction 18/100.

Let’s focus on this idea of using proportional relationships to solve multi-step ratio and percent problems. In sixth grade you had standard 6.RP.A.3, and in seventh grade you also have standard 7.EE.B.3 that talks about solving multi-step real-life and mathematical problems. This standard 7.EE.B.3, the big difference is that it deals with a variety of numbers. It deals with whole numbers, fractions, and decimals, again, numbers in any form, and students have to convert between forms as appropriate.

This idea of using proportional relationships—now it’s not necessary to set up a proportion to solve these problems. However, the general equations used to solve these problems can be derived from a proportion. Let’s start off with this ratio. You have some original amount and that’s 100%, and then we set that up equal to some new amount over 100% plus the percent change. Now, this is for contexts where there’s an increase from the original amount, and notice the plus sign. It’s 100% plus the percent change. So, for example, if we had a 12% increase, then it would be the new amount compared to 112%, which came from 100% plus the 12% increase. Now, if there’s a decrease from the original amount, then it would be 100% minus the percent change. So, for example, if we had a 9% decrease, then the proportion changes to the original amount compared to 100% would be equivalent to the new amount compared to 91%. It might be easier to think of the 100% as one. So that converts to the original amount compared to one would be equivalent to the new amount compared to one plus whatever the percentage of change was.

Now let’s take this proportion and do some manipulation with it. Let’s multiply both sides by one plus the percent of change. When we do that, this is what we get—the original amount times one plus the percent of change would be equivalent to the new amount. And if it’s for a context where there’s a decrease from the original amount, then it would be one minus the percent of change instead of one plus the percent of change. Now, when we do the computation, we’re going to have to make that percent change into a decimal. Again, for computational purposes, we need to have that percent change in decimal form, and in the case of a decrease, it would be one minus the percent change instead of one plus the percent change.

This equation, notice again, that it came from our proportional relationship. We derived it from that, so it didn’t come out of thin air. There is a justification for it, and we can show this. It’s best to do this because students will understand that again, this came from somewhere, and it came from this idea of a proportional relationship. Now, we can express this relationship the way we have it originally, but we can use our distributive property and tweak it just a bit. If we use our distribute property, we have this expanded version that will be the original amount times one plus the original amount times the second term in that parentheses, the percent of change, and that would be equal to the new amount.

Now, the original version of this that we came up with, this form has the advantage of less computation. Students don’t have to take as many steps, but there are some situations where the expanded version might make more sense. If you look at the second part of this, on the left side of the equal sign, the original amount times the percent of change, that actually gives us the amount of change, the actual amount that changed from the original. Again, although it involves additional computation, this second form of the equation provides the actual amount of change if it’s something that is asked for in the context.

Okay, let’s look at these different types of problems. Simple interest—let’s take this context. Monica has $500 in a savings account where she earns 3% interest per year. What’s her new balance after one year? Well, we have an option of which form of this relationship to use, and let’s go ahead and use the longer version, the expanded version where we use the distributive property. So, we’re going to have our original amount times one plus the original amount times the percent of change equal to the new amount. Now, we don’t really need the one because that doesn’t change anything, so we can get rid of that. It would be a good idea to use this as the template until students become proficient with it, then move on to the use of variables. But, if you leave it in this form, it’s a lot easier for students to figure out where it is that the different numbers should be placed.

In this context, the original amount was the $500 that she had in her savings plus the original amount, which is still 500, times the percent change, which was 3%, and that’s going to be our new amount. We need to change the percent to a decimal for computation purposes, so that becomes .03. So, now we have 500 + 15 is equal to the new amount. So, the new balance is $515. Well, what about that shorter version of this relationship, the original amount times one plus the percent of change is equal to the new amount? Let’s try that. So, 500, the original amount, times one plus 3% is equal to the new amount. We need to change the 3% to a decimal, then one plus .03 is 1.03 Do the computation, and we get the new balance directly being $515.

Notice that with this method using just one plus the percent of change, again it’s more direct. But students need to understand what’s going on here. What really happens when you have an increase of 3%? Well, we have one plus the percent of change, which in this case was 3%. We changed the one back to 100%. So we have 100% plus 3%, which is 103%, and as a decimal that converts to 1.03. So, 1.03 is a shorter way to represent an increase of 3%. Again, you get the new amount directly by multiplying, in this case by 1.03. Now, what if you had a decrease of 20%? So, now you would have one minus the percent of change. But let’s go ahead and change that to all percentages, 100% minus the percent of change, plug in our numbers, 100% minus 20% being 80%, which is .80 as a decimal. So, a nice easy way of representing a decrease of 20% is simply by multiplying by .80 because that is what you have left, 80%.

Let’s look at another example, this time using tax. Jesse bought a new car for $24,000 but must pay 6% sales tax. What’s the cost of the car with taxes included? Now, they’re not asking us for the actual amount of tax so we can actually go straight to the solution and use the shorter form of the relationship, the original amount times one plus the percent of change is equal to the new amount. So, we plug in our numbers. The original amount was 24,000, 100% plus 6% equals to the new amount. So we have 24,000 times 106%, which converts to 24,000 times 1.06. So, we get directly that the total cost of the car is $25,440.

Now, let’s say the context was a little bit different. Instead of asking directly for the cost of the car with the taxes included, what if the question asked for the amount of sales tax? Now, if we take the expanded version, the actual amount of sales tax is going to be given by this, by the original amount times the percent of change. So that’s really all we need. We don’t need all the rest of it because all they want is just the sales tax amount. So, the original amount times the percent of change would be the actual amount of change. So, here’s a slightly different form that we derived where if we’re looking for the actual amount of change, it’s going to be the original amount times the percent of change. In this case, it’s the sales tax. The original amount would be the price. So, we’re going to take our price of 24,000, multiply that times 6% because we had a 6% tax rate, and that’s going to be the actual amount of sales tax. So, change it to a decimal, so we have 24,000 times .06. So, we come out with $1,440 for the sales tax.

Let’s change the context one more time. What if they wanted to know both things? They want to know the sales tax and the total cost of the car with the taxes included. Well, we could do it with the shorter form of the equation. By doing it this way, we’re going to get just the total cost of the car, but all we’d have to do is take the total cost of the car minus the original price of 24,000, and that would give us our sales tax, $1,440.

### Part 2

Let’s look at contexts that involve markups and markdowns. Let’s say we had this situation. To increase sales, the store manager decides to mark down the price of all bicycles by 20%. If the price of a bike is $180, how much is the discount and what is the new sales price? Now, notice that they want two things here. They want the discount, and they want what the new sales price would be. So that being the case, we’re probably better off using the expanded equation, the original amount minus the original amount times the percent of change is equal to the new amount. So, let’s plug in our values. The original amount is $180 for the price of the bike minus the original amount times 20%, and that’s equal to the new price.

Now, notice that it’s subtraction because it’s a discount. We’re decreasing the original price. So, let’s crunch our numbers. Change the 20% to .20. Multiply 180 times .20. We get 36. Now, notice that this answers one of the questions already because $36 is the actual amount of discount. So we’ve already answered that question by doing the computation. Then if we continue on, subtract 36 from 180. That’s 144. So, that is our actual sales price, $144.

Now, let’s say they didn’t want the discount amount. All they wanted was the new sales price. Okay, so if we look at part of that equation, the one minus percent of change, well, the decrease of 20% would be 100% minus 20%, which is 80%, which is .80 as a decimal. So now, I can directly plug in my numbers. The original amount is 180. The percent change would be .80, and that’s going to give us the sales price directly, which is $144. Now, if we had done it this way and they also wanted to know the actual discount, well, we can just take this a step further. We know that if you take the difference between the original price that they had it marked at and subtract the actual selling price, that’s going to give us the amount of discount. So, the original price was 180. We got it for 144. Take the difference between the two. We get the $36 discount. That’s how much we saved.

Gratuities and commissions—let’s look at that. Now, that type of context is not going to be that much different than simple interest and tax and so forth. However, more than likely, you’re probably just going to need to find the actual amount of the gratuity or the actual amount of the commission. So, like in this case, typical scenario, the tip for a meal, the meal is $24.50, the gratuity rate is 15%. So, how much is the tip? So, all we really need is this part where the actual amount of change is going to be the original amount times the percent of change. Well, the original amount was $24.50. It’s a 15% gratuity, and when we multiply those two, that will give us the actual tip amount. So, we multiply $24.50 times .15, and we get 3.675...little bit of a problem because we don’t have parts of pennies in our money system. So, we have to round that up. So, the gratuity amount would be $3.68, and that’s it. We’re done with the problem. We’ve solved it.

Now fees, let’s go ahead and skip this one because again, those types of contexts are going to be very similar to some of those others like taxes and so forth. So, again, let’s go ahead and go directly to percent increase and decrease. Now, out of all of these contexts, percent increase and decrease, for whatever reason give people a lot of difficulty. There’s just something about it. Part of it really boils down to well, what is it that I am comparing to? Here’s what we did earlier. We took our original proportion, and we derived a formula or a relationship from it, where we used the distributive property and multiplied it out. So, it’s just a slightly expanded form.

Now, we can actually derive where the percent increase and decrease relationship comes from, and it’s best that we do that so students have a better understanding and will get a handle for what really goes on and remember the process a lot better. Now, when it comes to the percent increase and decrease, what we’re looking for there, that’s really your percent of change. So, if we take this relationship, and we treat it like an equation with variables, well, let’s solve this. Let’s isolate percent change. So, the first thing we have to do on both sides of the equal sign, we need to subtract the original amount, because again, we’re trying to get percent change by itself. So, here’s what we have now. We need to get rid of the original amount on the left-hand side of the equation. So, this time we need to divide by the original amount, and when we do that, this is what we have.

So, we actually derived the relationship, the equation. The percent of change would be the new amount, whatever that is, minus the original amount, the difference between the two, and you always compare to the original amount before the change happened. So, again, here’s our relationship. Again, the percent increase or decrease, which would be your percent change, is always going to be the difference of the two amounts compared to the original amount, and it’s always the new amount that includes whatever the change was, minus the original amount before the change happened. If it’s a percent increase, the difference between the new amount and the original amount will be positive. If the difference between the new amount and the original amount is negative, then that’s going to be a percent decrease.

So, let’s look at some examples. John’s rent is $400 per month, but it will go up to $500 next month. Ouch! What’s the percent increase in his rent? Okay, so we’ve already determined that to get the percent increase in this case, we’re going to take the new amount minus the original amount and compare it back to the original amount. We just have to be a little bit careful here and make sure that we understand which is which. The original amount was $400, so we plug 400 in place of the original amount, and then $500 is the new amount. That’s what we’re going to have to pay after the increase. So, we take the difference between the two. That’s 100 over the original amount of 400. Convert that to a decimal and then to a percent. So, we get 25%. So if you start off with a rent of $400 and it goes up to $500, that’s a 25% increase in the rent.

Let’s look at a second scenario. John’s health insurance is $500 a month. He found a better rate of $400 a month. What’s the percent decrease in his monthly health insurance? Okay, start with our basic fundamental equation; figure out the new amount and original amount. Well, let’s see. We were paying $500 a month, but we’re going to get a new policy that’s 400. So, that’s our new amount Take the difference between the two. Ah, but it’s negative. So, we have -100/500, which as a decimal would be -.20, which is -20%. So, by changing insurance, going from $500 a month to a new payment of $400 a month, again, that is a 20% decrease. And again, that’s what the negative sign indicates, that it’s a decrease rather than an increase.

Now, notice that even though the answer was negative, when we say decrease, well, that’s what we’re talking about. That explains what the negative sign is. Again, we have a decrease of 20%. Now, back in sixth grade there was standard 6.NS.C.7c that talks about understanding the absolute value of a rational number as a distance from zero on the number line. So, what we can do here to kind of understand the relationships a little better is to just look at this in terms of absolute value. So, when we get that -20%, we really just need the absolute value because that’s the actual change. It’s 20%, but it just happened to be a decrease rather than an increase.

Notice that in both these scenarios, we were talking about 400 and 500, but in one case, in the rent increase, compared to the insurance decrease, well, we came out with different percentages. For the increase in the rent, it was 25%, but the decrease in the insurance was 20% even though we were dealing with the same two quantities, 400 and 500. But again, it boils down to, well what was the original amount? For the rent increase, the comparison was back to the original rent of $400. That’s what the change is based on as opposed to the insurance where we started off paying $500 a month in insurance, and the change was based on 500.

So again, it’s real important that kids realize the distinction here. You’re always comparing to the original amount before the change occurred. Again, it’s very important here, even though we were dealing with the same numbers that we are always comparing back to the original amount. In this case, for the rent increase we were comparing to 400 as opposed to the insurance decrease where we started off with 500; again, very, very important that kids realize the distinction here. You always compare the difference of the two amounts to the original amount before the change occurred.

Let’s look at percent error, and these types of contexts will be very similar to the percent increase or decrease. It’s basically the same thing. So, percent error, percent change—the same idea, but there is some distinction. There might be a little bit more difficulty determining which is the new amount and which is the original amount. Again, you’re still going to take the difference between the two, and again, depending on the context, it’s going to always be a case where the original amount is going to be what’s being used for comparison. Again, the decision here is going to be a little bit more difficult, but again, in a percent error situation, what am I going to compare to?

Let’s look at an example of a percent error context. Let’s say there are 480 jelly beans in a jar. Marla guessed that there are 450 of them. What was the percent error? Okay, now, we’re going to use the same basic equation that we had for percent change. Now we have to figure out, well what are we going to use for the new amount and the original amount? Well, the amount that’s going to be used for comparison is the actual amount of jelly beans because we are going to see how far off that the guess was. So, the actual amount is going to be the original amount, and the new amount should be what was guessed.

Okay, so now, to get the percent error, we’re going to take the guessed amount and subtract the actual number of jelly beans and compare that to the actual number of jelly beans.

So we plug in our numbers. She guessed 450 and the actual amount of jelly beans is 480. So, it’s 450 minus 480 compared to 480. We get -30/480, which as a decimal would be -.0625. Convert that to a decimal. That’s 6.25%. Now, the percent error is 6.25%. But hey, wait a minute. There was a negative involved here. Well, okay. Go back again to that absolute value standard back in sixth grade, 6.NS.C.7c. So, let’s do this in terms of absolute value. So, we had a percent error of -.0625, but if we just look at the absolute value, the error is 6.25%. What happens here is that again, the negative just indicates where we were as far as the actual guess. She guessed 450, so she was actually low on her estimation. So, again, that’s how come it’s a negative. That shows that she was underestimating instead of overestimating.

One more example of percent error: the marketing department estimated $400,000 in sales for the year. The actual sales were $450,000. What was the percent error in the sales estimate? Again, let’s go ahead and do this directly with absolute value. Now, we need to decide what amount is which. Well, the amount that we’re going to compare to is the actual sales amount. So, the actual sales amount was $450,000. So, we’re going to take the estimated amount and subtract the actual amount and compare it to the actual amount to get our percent error. So, we’re going to have 400,000 minus 450,000, absolute value. Compare that to the actual sales of 450,000. We get -50,000/450,000, which is .1111 repeating, which is 11.11%. And again, we’re not worried about positive or negative because we just want the actual percent error. So, it’s a little over 11%. And again, the reason it was a negative difference was because again, we underestimated. We guessed $400,000 and the actual sales were $450,000.

Let’s look at our standards for mathematical practice. If we look at the first four, if we do some of these activities, we’re actually going to be doing all of these. Students will make sense of problems and persevere in solving them. They’ll reason abstractly and quantitatively. They should construct viable arguments and critique the reasoning of others, and they will be modeling the mathematics. If we look at the last four standards of mathematical practice, students will be attending to precision, and they will look for and express regularity in repeated reasoning.