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Combining Ratios


Combining Ratios. Sometimes between different subgroups in a problem, a problem will give separately expressed ratios, and we will have to relate these different ratios either the whole or to absolute quantities. So for example, there might be sets a, b, c, and d. And instead of giving us an overall ratio, it might tell us one ratio from a to b, another ratio from b to c, another ratio from c to d.

And then we have to combine those ratios to understand what's going on with the whole group. So before we begin this discussion, a few reminders. Number 1, of course, we write ratios as fraction. And what that means is we can find the equivalent fractions with all the laws of fractions we can use for ratios.

And so that's a very helpful thing to keep in mind. Also, we can write any ratio in what I would call algebraic terms. So if you have a ratio of A/B = 3/8, that means for some n, we can write 3n, A = 3n and B = 8n. And sometimes setting things in terms of the algebra, then that allows us to setup algebra and solve.

So that's one possible strategy for ratios also. Finally, a very important point, what we can and can't cancel in proportions. This is just a brief review. Of course, we can cancel up and down in any proportion. We also can cancel across two numerators, across two denominators, all those are fine.

The one thing we're not allowed to do in a proportion is cross-cancel. Now this is something absolutely new to you. I strongly suggest go back and watch the lesson Operations with Proportions before watching this lesson. Because if you're unclear in the math of proportions, there's a lot in this lesson that could confuse you.

It's very important to be very clear on what you can and can't do with proportions before you watch this lesson. Okay, let's start talking about these ratios. So, the combining ratio scenario, what we're talking about in this lesson, occurs when we're given two or more separate ratios within a larger group. Say, in a group with Type 1, Type 2, and Type 3, we might be given one ratio for Type 1 to Type 2, and a whole other ratio for Type 2 to Type 3.

So if we wanna understand what's going on in the whole group, we'd somehow have to combine those ratios. We have two basic strategies for approaching such questions. One strategy would be to find equivalent fractions for each ratio so that the common term comes to be represented by the same number in both ratios. So if we have one ratio from Type 1 to Type 2, and another ratio from Type 2 to Type 3, we'd wanna find equivalent expressions for both ratios so the type 2 number is the same in both ratios.

And that would allow us to put it together in one big ratio, Type 1 to Type 2 to Type 3. We'll have some examples of this in this video. The other strategy, and this is especially true if we have to find individual values, individual quantities. How many people in Type 1 were in Type 3, something like that.

Then it's much more advantageous to use what I was calling the algebraic approach where we actually represent the different terms into fraction. If we're given for example, three-sevenths, we represent that as 3 over n and 7 over n, that sort of thing. So here's a practice problem. Pause the video, work on this on your own, and then we'll talk about how to solve this.

Okay, so notice we're given two different ratios and all we're being asked is fraction information. In other words, there's no concern here about total number. How many sophomores? How many juniors? We're only concerned with fractional information.

So that means we can stick entirely with ratios. So it makes sense to use the first strategy. We're just gonna identify the common element and find equivalent ratios so that the common element is equal. So right now, we have two ratios, sophomores to juniors, juniors to seniors. Obviously, juniors are the common element.

And so we'd like to find equivalents so that the 3 in the first ratio and the 5 in the second ratio are the same. And of course, the easiest way to do this would be to multiply the first ratio by 5 over 5, multiply the second ratio by 3 over 3. Then we get these equivalent fractions. Juniors are now represented in both by 15.

So now we can put everything together in one big ratio, sophomores to juniors to seniors, 10:15:18. Now thats a ratio. It looks like you might be able to cancel something, but theres actually no way to reduce that. Theres no known number that is a common factor of all three of those numbers.

And so, that is our combined ratio. And of course, if sophomores are 10 parts, juniors are 15 parts, and seniors are 18 parts, we can find the whole just by adding the parts. The whole is 43, and so this means that sophomores to the whole would be 10/43. And so that one we can find purely from strategy one, just combining the two ratios, finding equivalent fractions.

Here's problem number 2, pause the video, and then we'll talk about this. Okay, in this one, we're given two different ratios. We have three groups, we're given two different ratios. And we're actually given an absolute count. We're told there are 27 programmers, and we want an absolute count. How many customer service reps?

And so here, because we're given an absolute count and we want an absolute count, this makes it ideal for solving in terms of the absent quantities. In other words, the algebraic approach where we're gonna use n. So the first thing we're gonna do is Let's say we have P/M, and in fact, we don't even need the M. We can solve directly for the numbers here.

Programmers to marketers is 3/8. We know the number of programmers is 27. So we can simply find the number of marketers. Marketers, remember, is the in-between term, the term that we need to get from one ratio to the other, that's the common term. So we need to solve for marketers.

And here, this is where operations with proportions come in. Before we cross multiply, notice that we can cancel a factor of 3 in the numerator. So we cancel that, we get it down to this fraction. Then we can cross multiple very easily, M = 72. There are 72 marketers. Now we take that number and use it in the other ratio.

The ratio customer service reps to marketers is 2/3, and we know marketers is 72. So C is gonna be the number of customer service rep. That's the answer we're looking for. Again, we can cancel, we cancel now and the denominator by a factor of 3. It simplifies to this, we cross multiply, and we get that there are 48 customer service reps.

And so that's the answer. This is a slightly harder problem. Pause the video and then we'll talk about this. Okay, so what we have here are few different ratios. We have cup of butter for a number of cookies, cup of sugar for a number of cookies, and then what we're looking for is a difference.

So we're not looking for how many cups of butter, or how many cups of sugar. We're given this odd piece of information, five more cups of sugar than of butter. So this is a tricky one, this is actually going to require us to use both strategies. So first, we're gonna have to come up with the ratio, the combined ratio for the whole thing.

So we're gonna start out, so here we're gonna use both strategies. First, we're gonna combine the ratios. So these are the separate ratios, butter to cookies 1:12, sugar to cookies 1:8. The common factor here are the cookies. And so we have to figure out what can we multiply by. And so what is the least common multiple of 8 and 12, of course, it's 24.

And so we have to multiply the first one by 2 over 2 and the second one by 3 over 3. And this allows us to get this combined ratio 2:3:24, all right? Now that we have the combined ratio, we're gonna use algebra. And so this means that the cups of butter are 2n, the cups of milk are 3n, and then the number of cookies is 24n.

So if we found the value of n, we'd be all set. In other words, if we found n, we plug in to that last equation and we're done. So now we can user this very strange piece of information, he used five more cups of sugar than of butter. So, 5 = (cups of sugar)- (cups of butter), that's the basic equation we have. Well, now we have algebraic expressions for both of those expressions.

So it's gonna equal 3n- 2n. Well then this becomes very simple, 3n- 2n is just n. Now we have the value of n. We plug that into the expression for the number of cookies, 24 times 5. Of course, we can use doubling and halving, that becomes 12 times 10, which is 120, and that's the answer.

He made 120 cookies. In summary, one method is to find equivalent ratios so that the common terms are equal in both expressions, and then combine them. And this is most useful if what we're looking for is purely ratio information. Another method, if we're given an actual count, then we can solve our individual values.

And then we can either use the algebra or solve directly for the values. And sometimes we have to combine both of these methods, especially if you're given information about a sum or a difference of two different groups. And if you can solve it in more than one way, it's always good to practice and make sure you can do that. Because it's always a good thing to have more than one option for solving a problem.

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