Turn Algebra into Arithmatics

Algebra and arithmetic are very similar, but algebra uses variables where arithmetic would use real numbers. On certain CAT problems, we can taken a problem in which we were given variables and use real numbers instead “ we’re turning algebra into arithmetic! Recognize when you can turn algebra into arithmetic First, we need to know when we can use this technique. It’s possible to use this technique when there are variable expressions in the answers. Variable expressions do use variables but do not contain an equals sign or an inequality sign. For example, x+3 is a variable expression; x+y = 3 is not a variable expression but a variable equation. You can sometimes pick your own numbers even when a variable isn’t given in the problem “ for example, when the answers are in the form of fractions or percentages of something. In these problems, you’d have to introduce a variable (or variables) yourself in order to solve or you could just pick a real number. For instance, they could talk about the changing price of a TV set: first it goes up by 5% and then it goes down by 10%. And they might ask: what fraction of the original price is the new price? You can tell me that without knowing any real numbers. If they give you variable expressions in the answers, or if the problem keeps talking about some number but never gives you a real number for it, and the answers are in the form of a fraction or percentage of the original, then you may want to turn this problem into arithmetic. Deciding whether to turn algebra into arithmetic After we determine that we can use this method, we next have to decide whether we want to do so on this particular problem. There isn’t a one-size-fits-all answer to this question; different students will prefer different things. The very general rule is that you use algebra on problems that are easier (for you) and you use arithmetic on problems that are harder (for you). The particular line, though, is different for every student. So how do you decide? You figure it out while you’re studying. From now on, every time you see a problem that can be done using this method, try it both ways “ the algebra and the arithmetic. Then decide which way was easier for you on this problem and, as specifically as possible, why. And how are you going to remember that for future? Then, when you see a similar future problem, you’ll already have made the decision as to which method to use. Turning algebra into arithmetic First, we see how many variables we have and whether we need to pick for all of them or whether some will be determined by others. For example, if a problem talks about a cell phone plan costing x cents per minute and Jamie talked for y minutes last month, then you are going to pick for both of the variables in the problem. If, on the other hand, the problem tells me that x+y = a, then you know you are only going to pick for x and y; after that, a will be determined. Picking Numbers Next, you are going to be smart about picking my numbers. Full disclosure: the main drawback to this turn algebra into arithmetic method is that, if you are very unlucky, you could happen to pick some number that will work for more than one answer choice. This can’t happen on every problem, but it can happen sometimes.You can reduce the chances to almost zero, though, if you are smart about picking my numbers. Don’t pick zero. Don’t pick one. Don’t pick a number that already shows up in the problem. Don’t pick a number that would result in a calculation that gives you that same number again. For example, a percentage problem asks you to reduce the price of a TV set by a certain percentage and then do something with the new price relative to the amount saved off of the original price. You probably don’t want to discount that TV set by 50%, because then both the amount saved and the new price are the same number! Other than that, though, make your life as easy as you can. Pick small integers. If you need to pick more than one, choose numbers with slightly different characteristics “ an even and an odd, for instance. If a number needs to be divided at some point, pick something that will still give you an integer after you divide. When you’re studying, go back over what numbers really worked well or really didn’t. How could you have known, before you started solving, that picking 2 on this one was probably not a great idea? Or, conversely, that picking 2 on this other one was a great idea? Doing the Arithmetic Okay, now you’ve got your real number or numbers. (You’ve written them down clearly on your scrap paper, along with your variables, right?) Now, do the arithmetic. Wherever the problem says x, you now use 2, and wherever the problem say y, you now use 3. When you’re done, you’ve got a numerical answer. But this is all too abstract. Take a look at this problem: *Before being simplified, the instructions for computing income tax in country R were to add 2 percent of one’s annual income to the average (arithmetic mean) of 100 units of country R’s currency and 1 percent of one’s annual income. Which of the following represents the simplified formula for computing the income tax, in country R’s currency, for a person in that country whose annual income is I? (A) 50 + I/200 (B) 50 + 3I/100 (C) 50 + I/40 (D) 100 + I/50 You have one variable, I, and this is a percent problem. 100 is usually a nice number in a percent problem, but this problem uses the number 100 in this formula you are supposed to find so you are going to use something different on this one. You will also have to calculate 1% and 2%, so you don’t want to go smaller than 100 or you are going to have fractions; let’s go bigger instead. And, hey, you are eventually going to have to divide this I variable by 200, 100, 40, and 50 (look at the answers!), so let’s use 200. Now, 200 = one’s annual income. add 2% of one’s annual income okay, so that’s 200*2% = 4. Next, take the average (arithmetic mean) of 100 units of country R’s currency and 1 percent of one’s annual income okay, so 1% of one’s annual income is 200*1% = 2, and you are going to average 2 and 100. That’s (100+2)/2 = 51. Now add those two together: 4 + 51 = 55. That’s your target answer. You have to find which answer choice equals 55, and do that by substituting 200 in wherever the answer says I. (A) 50 + I/200: 50 + 200/200 = 50 + 1 = 51. Nope. (B) 50 + 3I/100: 50 + 3(200)/100 = 50+6 = 56. So close! But this isn’t right either. (C) 50 + I/40: 50 + 200/40 = 50 + 5. Bingo! (D) 100 + I/50: Let’s check just in case 100 + something? No. Too big. The correct answer is C. Notice that we did test D, even after finding that C worked. Here, we have to make a choice. Do we want to be thorough and make sure we didn’t hit the rare case where more than one answer works? Or are we satisfied with C and want to move on? Generally, if you feel as though you haven’t used up my time yet, you’ll be thorough; this is the same as quickly checking my work on another problem when you feel you have the time to do so. If, however, you feel as though you are already over my time limit and you are not 100% confident anyway, then  just stop as soon as you find one that works “ that’s good enough and I’m almost certain to be right. Take-Aways 1) You can turn algebra problems into arithmetic problems. The process is straightforward, though it does take practice to become efficient. 2) Know how to choose between algebra and arithmetic “ and know that this is based on your individual level. When studying, try problems both ways to determine which way is best for you in different situations. p.s. The answer to that TV set question is 94.5%. Pick the number 100 as your starting point for the price of the TV set. 5% of 100 is 5, so first the price rises to 100+5 = 105. Next, the price falls by 10%. 10% of 105 is 10.5, so the new price of the TV set is 105 “ 10.5 = 94.5. My questions was what fraction of the original price is the new price? My new price is 94.5 and my original price was 100. 94.5 / 100 = 94.5%.