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Combining ratios, sometimes between different subgroups in a larger collection, a problem will express separate ratios, and we have to relate those. We’ll get a couple ratios and we’ll have to relate them to figure out something about either the whole probability, something like that, or we want the count of some subgroup.
So before we even begin the discussion, let me give a few reminders here about the math we are going to be using in this lesson. So, first of all, if you have been watching the ratio lessons, it should be clear to you that we typically write ratios as fractions. And what does this mean? Among other things, it means all the laws that apply to fractions apply to ratios.
And so for example, one-third I can express that, I can multiply numerator or the denominator by anything. So one to three ratio is equivalent to a 2 to 6 ratio, or a 4 to 12 ratio, anything like that. That’s important point number 1. Important point number two, and this is something we discussed in the ratio and portioning lesson.
That if we have a ratio, say, A/B is 3/8, that means for some unknown n, we could say that A= the number of A = 3N, three parts, and the number B equals 8N, 8 parts. It also means that if A and B are the only two groups in this larger collection, the total collection has a size of three plus eight, or 11N. And so this is an unfamiliar idea to you, I suggest watching the ratio proportioning lesson before you watch this one.
Finally, I will say what’s very important is what we can and can’t cancel in proportions. And so of course, we could cancel up and down, we could cancel P and Q, we could cancel R and S. We could also cancel across in the numerator, cancel two things in the numerator, two things in the denominator.
What we cannot do is cross cancel, cancel something with the P, the P with the S, or the Q with the R. And so again, if this is a new idea to you, or this is an unfamiliar idea to you, we discuss this in much more depth in the lesson Operations With Proportions. And so I want everyone to be clear on these, because all of these points are gonna come up in this lesson.
Okay, so now let’s talk about this scenario. The combining ratio scenario occurs when we are give, two or more separate ratios, usually two. Maybe sometimes you’d be given three within a larger group. And so what does that mean? All right, so suppose the larger group has Type 1, Type 2, and Type 3 people.
We don’t know what Type 1, Type 2, and Type 3 are. But we might be given a ratio, the ratio of Type 1 to Type 2 is something. And then, separately, given another ratio, the ratio of Type 2 to Type 3 is something. So maybe Type 1 to Type 2 is 3 to 5. Type 2 to Type 3 is something like 4 to 7.
And then we’re gonna have to figure out how those ratios relate. And we have two basic strategies for approaching such questions. So one strategy would be to find equivalent fractions for each ratio. So again, if the ratio you are given is one-half, we can write one-half, as 1:2. We could write that as 7:14 or 13:26. We could multiply the one and the two by any number.
As long as they both get multiplied by the number, it’s like multiplying a fraction by the same number in the numerator and denominator. And so we get an equivalent fraction. And so, one way to do it is to change one ratio by multiplying and change the other ratio to multiply so that the common term comes to be represented by the same number in both ratios.
Okay, so that’s one strategy. Another strategy, which is not mutually exclusive, we might use both of these in a problem. But the other strategy is if we’re given some absolute quantity in the prompts. So we’re actually told the number of members in this group is blah blah blah, then it might be easier to work with the absolute quantity.
So now, instead of 1:2 we’re gonna talk about N and 2N. We’re gonna use the N, or the parts, so we can actually talk about the absolute quantities. And so we might use one or both of these in any problem. All right, let me show you a few example problems of this. Here’s practice problem number one.
Pause the video, and then we’ll talk about this. Okay, so on a certain high school team we have sophomores to juniors is 2:3, juniors to seniors is 5:6. Sophomores are what fraction of the whole team? We’re assuming that those are the only three types on that team. So notice that what we’re being given is ratio information, and what we’re being asked is ratio information.
There are no counts here, there are no numbers and members. We have no idea, we don’t really care how many sophomores, how many juniors are on this team. So it makes sense to keep everything in terms of ratios. And so what we’re gonna do is we know the common element is juniors. And so what we’re gonna do is find equivalent ratios so that that common element winds up being the same number.
And let’s think about this, Soph:Jr = 2:3, Jr:Sr = 5:6. And so the number that represents juniors, we want that to be the same number in both ratios. And so this is an analogous process to finding a common denominator. It’s not exactly the same because we want the denominator of one fraction to equal the numerator of the other.
But we do have to multiply both fractions by something. And so what we’re gonna do, we have a three and a five, they have no factors in common. So the least common multiple of 3 and 5 is 15. We’re gonna multiply the sophomore fraction, both terms by 5. And we’re gonna multiply the Jr:Sr fraction, both terms by three.
And when we do that, we get 10 over 15 sophomores are 10, juniors are 15. And then Jr:Sr, juniors are 15, to the seniors are 18. Well now because it’s the same term, we can combine these all into one big ratio. Sophomores:Jr:Sr = 10::15:18, 10 parts to 15 parts to 18 parts.
And what this means is that the total number of parts, remember the proportioning video, the total number of parts Is 10+15+18 which is 43. So that’s the whole, and so sophomore are what fraction of the whole? Well they would have to be 10 over 13. So that’s practice problem #1. Here’s practice problem #2.
Pause the video and then we’ll talk about this. Okay, in a certain company, the ratio of programmers to marketers is 3:8, and the ratio of customer service reps to marketers is 2:3. If there are 27 programmers, there are how many customer service reps? Okay, so notice that the common element here are marketers.
But here, we’re also given a count, we’re actually told a count of one of the groups. There are 27 programmers, there’s actually 27 human beings in that group. So here we can solve for the absolute quantity in each term. And so let’s think about this. So first we’re gonna set up a ratio of programmers to marketers.
Because we have, so we know that ratio is 3/8, and we have the 27, and we’ll leave the M as the variable for the number of marketers. All right, so we have our proportion here. What can we cancel in a proportion? Well of course we can cancel those two numbers in the numerator. We can cancel across the numerator, and so then it reduces.
Well now that we’ve reduced, it’s always good to reduce before we multiply. But now that we’ve reduced, we can cross multiply, and we see that there are 72 marketers. Okay, well now there are 72 marketers, now we can put this in the other fraction. Okay. So customer service reps to marketers 2:3.
C we don’t know, but now we know marketers are 72. Again, what can we cancel? We can’t cancel the 2 with the 72, that’s an illegal cancelling. But we can cancel the 3 with the 72. And so that cancels down to 24, cross multiply and we get 48. There are 48 customer service reps.
Here’s another practice problem. Pause this and then we’ll talk about this. Okay, one cup of butter is enough for 12 of Roger’s cookies, and one cup of sugar is enough for eight of these cookies. If he used five more cups of sugar than of butter, how many cookies did he make? Okay, so Robert loves to cook.
Always good to see a man in the kitchen, isn’t it? So how are we gonna do this? Well first of all, let’s think about these ratios, and we have a ratio of butter to cookies and sugar to cookies. And so butter to cookies is 1:12, sugar to cookies is 1:8. Well what are we gonna do with this?
All right, we’re gonna find a, we’re gonna make the common term cookies equal. And the least common multiple of 8 and 12 of course is 24, so the top ratio gets multiplied by 2, the bottom ratio gets multiplied by 3. Now we can put everything together. B:S:C = 2:3:24. Now we’re gonna represent everything by N because we have a count, and we actually want to count, we don’t want a probability.
We want number of cookies. Okay, so in other words, butter is two parts. Sugar is 3 parts and cookies is 24 parts. All right, so now we can do some algebra. 5 more cups of sugar than of butter. So if he uses 3N cups of sugar and 2N cups of butter, we have to subtract those.
Well, 3N- 2N is just N. N = 5, well, now we’re golden, we have the value of N. So now all we have to do is plug that into the equation for C. C is 24 5, we can use doubling and halving, we don’t even need calculator for that, that is 120. And so, he made 120 cookies.
In summary, one method for solving these problems is to find equivalents so that the common terms are equal in both ratios. Then we can combine the ratios into one big ratio. Another method, if possible, solve one of the ratios, solve it completely for individual values, especially if you’re given an individual value. You can solve for some individual values, and then you can use the ratios to solve for the additional values.
And I’ll say even if there’s two different ways to solve the problem, it’s probably worthwhile to use both ways. Because it’s always a good thing to have more then one way to solve a problem.
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