 ###### Eva Holtz

Harvard University
Perfect scores on the SAT and 4 SATIIs

Eva is a certified admissions counselor and the founder of PrepPoint, a premier test prep company in the San Francisco Bay Area.

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Turning Variables into Numbers, Part II

# Turning Variables into Numbers, Part I

Eva Holtz ###### Eva Holtz

Harvard University
Perfect scores on the SAT and 4 SATIIs

Eva is a certified admissions counselor and the founder of PrepPoint, a premier test prep company in the San Francisco Bay Area.

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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?