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ACT Pre-Algebra

Teacher/Instructor Devorah Goldblatt
Devorah Goldblatt

Case Western Univ., summa cum laude
Perfect scorer on the SAT & the ACT

Devorah is the founder of Advantage Point Test Prep and the author of the book “Boost Your Score” The Unofficial Guide to the Real ACT.

I'm planning a party after a football game. Now I know the players are going to be really, really hungry. So I figured out I need about two pies for every three players. But here is the thing there're 36 players coming, so how many pies of pizzas do I need? Well in this episode we're going to talk about propositions which can help us solve this. We're also going to talk about the other top pre-algebra concepts that you'll see on the ACT.
Let's get started talking about a concept you need to review for the math part of the ACT. Now PEMDAS, 'Please Excuse My Dear Aunt Sally' as your teacher probably told you. Order of operations, this is not tested directly on the ACT, there's not going to be a PEMDAS question for pre-algebra but this is going to show up in a lot of questions as you do the questions. You're going to need to make sure that you're doing the order of operations properly. So PEMDAS stands for Parentheses, which you'll do first, Exponents, Multiplication, Division, Addition and then Subtraction. You're going to want to do them in this order. Okay let me tell you something really important that I see students doing a lot. They know PEMDAS and they feel really good about it, they have seen it in thier pre-algebra but they don't use thier calculator correctly to help them to calculate it. This is an example one of my students just had trouble with. So you hand her something and she had done this whole complicated algebra problem she was just about finished. This was is what was left to calculate. Okay 12 plus 29 over 132. Right 29 divided by 132. So you know PEMDAS, you know okay 'division' where do you go before addition so you would do this first and then you would add 12. But when you put it in your calculator, you have to make sure to tell your calculator to do that. What she did was input her numbers just from left to right into her calculator. And she kind of thought her calculator would just know what to do. If you put it into your calculator, your calculator will think, 'Oh it's 12 plus 29, whatever that is and all of that divided by 132,' which is a completely a different number than 12 plus parentheses on your calculator 29 over 132, signifying to your calculator that's a separate expression, a separate fraction or division that needs to happen first. So really watch this you guys, watch your calculators and think about PEMDAS. Alright let's look at the first pre algebra concept that you need you need to know that's actually is going to be straight forwardly tested on the ACT.
Okay percents, students often ask me, you know I see a whole wordy percent problem, how do I change it into algebra so I can solve it. You know you'll see something like what is 50 percent or what is 12 percent of 29. So they got a lot of words there and it's not algebra. It's actually not that difficult as long as you remember these rules. When you see 'what' in a problem it's going to be your X, when you see is, that's going to be your equal sign in your algebraic equation. And this is key, 'of', the word 'of' always means multiplication and the word 'of' in any percent problem will always be between the two parts that need to be multiplied. So if you ever stuck in thinking, what do I need to multiply together here, the word 'of' is going to between them. Now this probably still sounds kind of confusing so let's try on a real problem. If 50 percent of X is 150 what is two percent of X? So you see we actually have a two part percent problem here. Let's take it piece by piece, first we need to find '50 percent of X'. For we have to find what 'X' is if '50 percent of X' is '150' and next we've to find out what's two percent of 'X', the number that we find first. Alright let's hoop right in and then you'll see how I use that knowledge about 'what', 'is' and 'of' that I talked about earlier. If 50 percent of X is 150, how is it algebra? Well remember 'of' is always between the two things we need to multiply together. So 50 percent of X is 150. 0.5 and remember 50 percent in algebraic terms is 0.5. And if you need a review there is a little sheet about that in your bonus material. So 50 is 0.50 times X right because you've got your 'of', is equals 150. Great and now we can just solve it. It's just a simple algebra problem, you know divide both sides by 0.5 you can cross it off, divide here by 0.50 and we know that then X is equal to 300. Okay so let me circle that. But we're not done right? We next need to know, what's two percent of X? What's two percent of 300? So let's do this again. Okay what is two percent of 300? So 'what' that's our X is, equal, two percent of X. Again 'of' between the two things that need to be multiplied so 0.2 times X I'm sorry times 300 because we already know, so times 300 the number we just found. Okay so use your calculator. 0.2 times 300 is equal to six. So now we know that our answer is B, six. And again if you need more practice there is a lot of practice problems in your bonus material for percents.
Let's keep going. So we talked about some basic percents, there also something also that will be tested involving percents which is percent increase and decrease. The ACT likes to test, how much did something increase by or decrease by with regards to percent. We'll talk about in a second exactly what that's going to look like. The formula for it though is the amount of change between the first amount and the second amount divided by what you started with. Let me show you what I mean here. 'If Sharon had 65 followers on twitter in March and 95 followers in April, by what percent did her followers increase?' So you see how this is a different kind of percent problem, before we were just finding the percent of something, now we want 'by what percent did you have a decrease or an increase between a first amount and a second amount'. Okay so remember the, remember the formula that we had, we had the amount of the change over the amount that he started with. Okay while we had 65, we ended up with 95, what's the amount that changed? Well 30 right, from 65 to 95 perfect. Very straight forward, so 30 is the amount of change and what did we start with? Well we started with 65, 30 over 65. And then when you use your calculator and you'll get something about 0.46. And remember how switch percents from algebra into percent terms, 0.46 is about 46 percent so B will be the right answer. Okay let's keep going.
Let's move on to one of my favorite pre-algebra topics, Scary Plug-in Problems. These are the ones you'll see. You'll probably remember them from the practice test you hopefully took. They look horrible, you'll see some really, really scary equation and you'll think, "Oh my God I didn't know I have to know this, something about volume, something about calculating temperature." And you think, "Oh my gosh, there is no way I can solve this." Actually you can. What's its going to be is a just a simple straight forward plug-in and problem where you plug-in a given amount into the scary equation. Let's take a look and see what I mean. A formula for the volume V over sphere with radius R is V equals four thirds pie R cubed. And already you are probably getting chest pain, "Oh my God, what is this." Right? If the radius of the pitch ball is 12 inches, what's the volume to the nearest cubic inch? Never be intimidated about these problems. You'll see three to four of them and they're really, really not bad because look we've got all the parts we need to easily solve his problem. We know that the radius is 12 and so we've got that the volume and they tell you the volume B when you have your radius R, so we would just plug-in our radius into R, no big deal. Let's go ahead and solve it. So we have four thirds times pie times 12 cube, easy okay. And so you see it's not intimidating after all. Now it's just a straight forward algebra problem. Remember PEMDAS, we're going to do exponents first. So we've four thirds times pie times 12 cubed, Let's figure out what that is. So I got my calculator here and 12 cubed is 1728 and hopefully you're doing this along with me at home. So we've got four thirds times pie times 1728. And now that we've got everything is multiplied across we can just multiplied across. By the way because the answer choices don't have pie in them, that's a sign to us that we actually have to calculate pie. We can just call it three point one four, okay. So I'll just multiply across. I'm just going to use my calculator and we'll see what we end up with. So we have four thirds or one point three, three about times three point one four, you can do this on a calculator, times 1728 okay. If you're doing this along with me hopefully you ended up with about 7234, let me write that down, 7234 and now if you're trying to worry oh my gosh nothing looks identical take a closer look at the problem it does tell you what's the volume to the nearest cubic inch so we just want the closest, the closest answer choice which it looks like it would be E, 7238 great. So that's it for scary looking plugging problems never be intimidated by these just look and see what do you have to plug in.
Okay let's take a look at proportions I've seen ACT tests where proportions will show three or four or sometimes even five times so I want you to feel really good about proportions. What are proportions? Well, proportions are an equation that states that two ratios are equal. This is kind of like a fraction and then the reduced form of the fraction and you know they're equivalent to each other, for example four eighths is the same as one half right 'cause if you reduced four eighths you'd end up with a half, the cool thing about proportion is you can cross multiply and they should be equal to each other. So four times two for example that's eight and so is eight times one and you can use this to solve ACT problems.
Let's go to the pizza problem and then we'll look at a real ACT problem testing proportions, okay if you'll remember I need two pizzas for every three football players coming to my party, so how many pizzas do I need for thirty six football players pay attention this really does help you in real life. So two pizzas for every three football players okay so we can write that out as a fraction or is it really a ratio, really same thing here, so two 'to' three right or two over three for every two pizzas well every two pizzas are going to feed three people okay and we know that we need an equivalent fraction just bigger for how many people we're going to feed, sorry for how many pizzas we're going to need to feed 36 people. Okay so if you need two pizzas for every three football players how many, we don't know so we'll call that X, do you need for 36 people so you see how we again, just have two equivalent fractions now like I said we can have two equivalent fractions. Now like I said we can cross multiply to actually solve for X so we have two times 36 equals three X let me write that down, two times 36 is equal to three X okay so two times 36 that's 72 so 72 equal to three X we're just going to divide both sides by three right, great so X is going to be equal to 24 good. So now let's take a look at this if I need two pizzas for every three football players I'm going to need 24 pizzas that's a lot of pizzas to feed 36 football players but now I know great.
Let's look at a real ACT problem testing proportions, Julie and Mike are on a carousel that rotates 30 degrees every five seconds. How many degrees will they rotate in a minute? Okay so we know we're going to have an equivalent ratio or an equivalent fraction here, we've got 30 degrees every five seconds, so 30 every five seconds and that's going to be equivalent to however many degrees they rotate in a minute right this is just going to be a simplified fraction. So how many degrees will they rotate in a minute well X degrees make sure though you've got your 60 seconds for a minute because when you do proportions your equivalent parts have to be in the same units. Okay now we can just go ahead and cross multiply so we've got 30 times 60 equals five X so 30 times 60, 1800 equals five X and therefore if you divide both sides by five to isolate your variable X is going to be equal to 360. Okay so we see if the carousal rotates 30 degrees every five seconds in an entire minute they're rotating an entire 360 degrees. And C is the correct answer choice here.
Okay let's move on to another concept taking a look at average problems these will always show up on the ACT and usually you'll see them two or three or sometimes even four times. The formula hopefully you're familiar with already but let's just go over it, sum of terms divided by the number of terms meaning adding up all the different parts you have that you want to take the average of and divided by how many different things there are. We're going to look at two example average problems because there are two types of average problems that show up, a really basic average problem and then a slightly more complicated one.
First let's look at the easier one, Tom scored an 89, 72, 54, 50, 80 and 69 on six high school biology tests. Approximately what's Tom average in the class? Hopefully this looks really familiar to you so we've got a bunch of different terms and we know that the formula is that you add up all the terms and divide by how many you have to find the average, let's do that. So we've got 89 plus 72 plus 54 plus 50 gosh this is a lot of tests this guy took, plus 80 plus 69 did I get that right, yes. So we've got everything added up divided by how many we have so let's divide by six, write six different tests that's going to equal out average which is X, okay. So let's go ahead and pick up our calculators because this is a lot to add, so we have got 89 plus 72 plus 54 plus 50, hopefully you're doing this along with me, plus 80, plus 69 okay that's equal to 414 so 414 over six is going to be our average so let's do that 414 divided by six and this guy's average is a 69, great so answer choice B.
Now let's take a look at a slightly more difficult average problem that you'll see, be on the lookout for this you will definitely see this problem or this problem type and you can totally nail it if you know what to do. Tom scored an 89, a 72, 54, 50, 80 and 69 same as before and six high school biology tests tomorrow he's going to take a seventh biology test if Tom needs an average of 65 to pass the class what's the lowest possible score he can get on the test and still pass the class. Okay so let's think about this for a second, we've got the scores for six things right, but we're missing a score we need the score for the seventh, this is like I said there is something you're going to see the ACT loves to test average problems where what you're missing is actually something at the top of your equation remember what our equation is everything added up divided by how many you have that gives you your average right. Here we know what we want the average to be we want him to get a 65 we also know how many tests there are, there are seven let's write that down so we've got our typical formula everything added up is going to go here divided by how many terms okay I mean there are going to be seven, seven total tests and we know his average needs to be a 65 so you see how interesting right, we actually have two of the parts but we're going to miss, we're missing part of the top. But we've got the rest of the test we know he got an 89 plus a 72 plus that 54 plus that what else 50 plus the 80 plus the 69, okay guys plus X.
Alright so just watch up for this trail over here, this is our whole top so you see how we're missing that seventh right but we know everything else added plus that seventh test whatever it is I'm calling X divided by seven the amount we have it can give you 65. A lot of students look at this and they think, "Oh my gosh what do I do", don't worry about this all you're going to do is treat it like you would a typical algebra problem remember if you have something going on in the bottom you have division and you have an equal sign with something on the other side you can multiply the seven to the other side you remember this if you think about algebra if I had multiplied both sides by the same thing which is allowed, multiply both sides by seven I can cross of my sevens here and this becomes 65 times seven.
Okay so now this whole chunk here is going to be equal to 65 times seven let me just figure out what 65 times seven which will make things a little bit clearer here. So 65 times seven is 455, okay so this whole chunk over here is going to equal 455 I don't want to rewrite this whole chunk because remember on the last problem we added this all up and they're equal to 414 right we did that for the last problem. So now we know 414 plus X is equal to 455 okay so you see what we did we just added up all the test scores for leaving our X our missing score, but it just turned into a really straight forward algebra problem all of this you move this to the other side so 414 plus X is 455 now we can just subtract to figure out what is, X is going to be 455 minus 414 that is 414 okay, let's use our calculators for that minus 414 and wow a 41. So lucky Tom just a pass if that's all he wants to do all he needs is a 41 on this test great and that looks like answer choice A.
Now i wanted to do it this way so you guys see how I set this equation but remember our pacing strategies and our back door strategies that we talked about keep in mind when you do these problems sometimes there's a shorter way to do it, you still need to know basically how the equation is going to be set up but remember you can also work backwards from the answer choices in a problem like this. So I wanted to do it straight through so you can see what it looks but another thing you could have done is once you realize it was everything added up divided by the seven gives you that 65 that you need, you could have played in through from the answer choices and said okay is X 41 there's this all added up you know with the 41 here divided by seven give you that 65 and so on and picking from the answers.
So it's just a cool thing to remember sometimes or actually pretty often there are a couple of different ways to solve a problem which is a really powerful feeling to have kind of a bunch of tools that you can pull out and use on a problem. Great so that's it for the top pre-algebra problems that you're going to see repeatedly on the ACT and I want you to feel great about these because pre-algebra shows up 14 times on the math that's almost a quarter, 60 questions on the math 14 out of the 60 questions are going to be pre-algebra. If you need some more practice then really make sure to head to your bonus materials and then with some practice I think you're going to do a great job.