# Turning Variables into Numbers, Part I

###### Transcript

In this episode, I'm going to introduce you to the strategy of turning variables into numbers, probably the most important strategy for the SAT Math section. So listen up and if you want to revisit this episode until you've really, really got down what's going on, I highly recommend it. Now, turning variables into numbers is going to make many Math problems easier. It's not just one or two per test, it's several per section so you can this with a lot of frequency and in addition, the strategy is great because it works for everyone. If you're a total Math rock star, it will make the most difficult questions only moderately difficult and if you really struggling with Math, it will make the moderately difficult questions sort of easy to medium so it's helpful across the board and the strategy has three basic steps. First of all, you're going to assign numbers to your variables so whatever's a variable, you're going to make it a number. Next step, you're going to solve your problem using those numbers that you chose and then lastly, simplifying the answer choice that matches the solution you came up with. Let's look at these three steps in lots more detail with tones of examples.

So, I'm going to show you several examples where we're going to turn variables into numbers and if you can do these problems using "real Math" that's great, however, I want you to practice following these steps because you need to get used to them, sometimes on easier problems in order to be fluent when you actually have to tackle harder problems. So even if you can do this the right way, I want you to follow along to understand the strategy and be able to apply it.

Let's start with this question; the average or arithmetic mean of w and x is six and the average of y and z is 10. What is the average of w, x, y and x? So, the first step is to turn our variables into numbers and I should that you usually want to avoid the numbers one and two and you usually want to avoid repeating numbers. So, for instance, let's see, w and x, I need to pick two values of w and x but I should avoid one and two if possible and I shouldn't repeat them and they need to average to six. So, I can use a wide variety of numbers but let's try five and seven because I know that five and seven average to six and then also the average of y and z is ten. Yet again, there are tones of numbers I could pick but I'll keep it simple and say nine and 11, because those average to 10 too. So I've already finished that point which is turning them into numbers and I'm ready for step two; solving the equation using the numbers I chose. So, the question is, what's the average of w, x, y and z. So, what's the average of w and x and y and z so I add them all up and divide by four because that's what the average is and a calculator might help but the answer turns out to be eight. So that's pretty easy. I'm ready for step three which is just finding the match. I got eight, c is eight. The answer is c.

Let's look at another example. If a, b, c and d are consecutive even integers and 'a' is less than 'b' is less than 'c' is less than 'd' then b + d is how much greater than a plus c. So, I start by turning those variables into numbers and I have to follow any restrictions in the problem. So, I'm told a, b, c and d are consecutive even integers, that means even integers that go in order like two, four, six, eight, 10 et cetera and I'm told that 'a' is smallest and it goes up to 'd'. So, I'm going to go ahead and assign two, four, six and eight and I have to keep them in that order because it says that two 'a' is smallest and it goes up from there. I guess I'm pretty much done with the first step which is assigning numbers. Ready for the second step which is to answer the question. The question is b + d is how much greater than a plus c? So let's figure that out, b + d is well, what's 'b'? 'B' is four, d is eight and that comes out to be 12 and that's how much greater than a + c. So what's a? We decided a is two, c is 6 and that comes out to be eight. So the question is, this is how much greater than this and the answer is that's it's four greater so that's the answer. Now that we've gotten the answer to the question we're ready for the final step which is go find the answer choice that matches. We said it's four greater and that's b.

Let's look at another example; 'After the first term in a sequence of positive integers, the ratio of each term to the term immediately preceding it, is three to one. What's the ratio of the fifth term to the second term?' So, one thing I want to point out here is this might remind you of certain chapters of algebra two and that could come in handy but what a lot of people do is they see this and they say, "I don't remember the formulas, I can't do it. I better give up." But so many SAT problems, if you just read them and think them through you'll be fine, the same thing here. We're not going to use a crazy, fancy formula, just think it through, so that's one point I wanted to make about this problem. Another point I wanted to make is if there is no variable showing up it's kind of hidden in the problem. They talk about the first term and then the other terms, so the hidden variable is the first term and we can make the first term whatever we want that's our variable. So let's start with the number one to keep it simple. The first term let's make it one and then the pattern is that the ratio of each term to the term immediately preceding it, is three to one. Ratio means there's a multiplication so the second term it's going to be multiplied by three. The third term it's going to be multiplied by three again so that's nine, the fourth term is going to be multiplied by three again so that's 27. The fifth term is going to be multiplied by three again so that's 81, so I think we've gotten enough progress here on picking our numbers and we're ready to actually answer the question. What's the ratio of the fifth term to the second term?

So we're dealing with this term and this term. What's the ratio of 81 to three? So that's the answer but as you can see, it's not reduced so it doesn't match the answer choices which is our final third step. So if we divide both of these by three it becomes 27 to one and sure enough, the final step is to find the match and our match is d. Time for another example. 'If 10 percent of w equals y and 40 percent of w equals z, what percent of the z is y?' Okay, this one's a little bit harder so let's see here. Certain numbers make more sense to pick than others, for instance, I noticed we're taking 10 percent of 'w' here and 40 percent of w here and a lot of times when you're dealing with percents, it's good to work with numbers like 100. It's easy to take to take a percent of 100. If I ask you to take ten percent of 11.54 it's not quite as much fun unless you're kind of into that thing. Alright, so let's say that w is 100 and that will be relatively easy to work with. So 10 percent of w, 10 percent of 100 is y, what's 10 percent of 100? Well, hopefully you know it's ten but if not, your calculator could help you with that and since that's 10 and it's y we'll write it there so we don't lose track and then 40 percent of w, 40 percent of that same 100 equals z, what's 40 percent of 100? It's 40. Now I should point out here that since w appears twice, it has to be the same number both times. You don't get to make up different numbers as you work through the problem. So we finished assigning numbers and we're ready to answer the question, the second step. What percent of z is y? What percent of... let's see, z is 40 and y is 10, so the question is, what percent of 40 is 10 or in other words 10 is what percent of 40? Now there are many ways to solve this but it's basically asking as a percentage, 10 is what compared to 40? Well, it's one fourth and you can ask your calculator if you need to but maybe you remember that one fourth is 25 percent. So we've got our answer and then step three is just to look for the answer right there, b, 25 percent.

Another example, 'if m is divided by two, the remainder is zero. What is the remainder if 4m + 1 is divided by two? Now, this is a little different. Before, when we picked numbers, it was a little obvious what the rules we had to follow were. This one's a little more settled. We don't quite know in advance what m is, we have to play around until we find the value of m which when we divide it by two, gives the remainder of zero. For instance, if we were to try the number three, it wouldn't work, let me show you. When the number m is divided by two, the remainder is zero. Well, let's try this. Two goes into three one time, subtract that two and you get one. So, three divided by two is one with a remainder of one and that violates the requirement. Our requirement is the remainder zero, this is not for that requirement. So that value of three for m is not going to work. Now, you could play around a little bit but one value that would work is four and I'll just demonstrate for you, so you feel comfortable with that. When four is divided by two, it goes in two times, two times two is four, subtract it and that's your remainder of zero. So sure enough in choosing a value of four for 'm' we've met the rules stated in the equation or rather stated in the question and now we're ready to actually answer the question being asked. What is the remainder if 4m + 1 is divided by two? So let's see 4m +1, what is that? Well we decided that m was four, so four m plus one is four times four, that's 16 plus one that's 17. So the question is, what's the remainder when 17 is divided by two? Let's do that. Not too complicated, divided by two. Eight times two is 16, subtract and it's eight remainder one. So the question is what's the remainder and the answer is one. Now we go look for our answer here and there it is, b.

Time for one last example. 'If y equals four times x cubed, what happens to the value of y when x is doubled?' Now this is a little different from the problems we've seen before because they're sort of before and after. There's an original value for x and that makes y be a certain value according to the equation and then x gets doubled and as a result, y changes again. So there is the original situation and the new situation. So let's do a before and after. So let's try that x equals three. So before we're going to have x equals three and y equals four times x which is three cubed. Three cubed is 27, 27 times four happens to be 108, so that's our original value for y. Now we repeat it in the after scenario where x has already been doubled. So instead of being three, it's now going to be twice as big which is six and we go for the Math again. Four times the new value of x which is six also cubed and we go through all this Math it's pretty complicated, we get 864. Now we're ready to answer the question, what happens to the value of y when x is doubled? So y went from being 108 to 864 and the answer is, it got multiplied by eight. It would be really easy to verify that by dividing this by this that would give you a result of eight, it's eight times as big. So it's step three which is basically nothing at all. Let's just say, oh, hey, look! That's the answer choice E, y has now been multiplied by eight. So let's tie together everything we've covered with these examples here.

So to recap, there are three basic steps for turning variables into numbers and it might take a little while to get used to but it is definitely worth the time that you invest in this strategy to get fluent with it. So first off, you're going to assign numbers to your variables. When you do, try to avoid one and two, try not to repeat numbers for different variables and also try to pick numbers that fit the rules, so if the problems says that two numbers multiplied together to equal another number, make sure that that statement is true. If you're told the number has to be consecutive even, make sure it is and so on. Next step, once you've got those numbers, solve the problem using those numbers and lastly, find the answer choice very simply that matches the answer that you determined and that's turning variables into numbers.