SAT Math COMPLETE Test Review (50 Problems) || Algebra & Geometry, Calculator & Non-Calculator

by the end of this video you're going to be feeling much more confident with these sat math problems we're going to be going over 50 of these problems in this video and before we get into this here's everything that you need to know about this video in under a minute as the video goes on these problems will get more and more difficult so especially early on if there's a problem you feel comfortable with and you want to skip to the next problem i have the time stamps for every single problem right in the play bar also the calculator problems for this video will be marked with a little c early on we do a lot of no calculator problems with some calculator problems mixed in but the last 20 problems are almost all calculator problems now we do have a printable problem sheet and answer key for this video linked right in the description i know a lot of you have been asking me to do that for this new video so i made it happen and i also added a little qr code on the notes that can take you back to this video and of course you're going to get questions as you go throughout the video so ask whatever questions you have in the comments i'll get back to you when i can and if you're looking to review harder sat math problems only i have another video where i do 25 of those and lastly i'll talk about these guys later because i think it's about time we get into this video so let's try and get through these first five problems and then i'll give you a break so let's start off with problem one problem one says that kathy is a repair technician for a phone company each week she receives a batch of phones that need repairs the number of phones that she has left to fix at the end of each day can be estimated with the equation p equals 108-23d where p is the number of phones left and d is the number of days she has worked that week what is the meaning of the value 108 in this equation now something i like to do whenever i finish reading a word problem is just underlining what it's actually asking for and the reason for that is because like i don't know if this happens to you but for me especially after i read a word problem i just forget everything i just read i'm just like wait what so i think a great way to start like countering that is to underline what the problem's asking for and then you can ask some other questions from there like here it says what is the meaning of the value 108 in this equation well what equation is it referring to well the p equals 108 minus 23d and if you want to know what p and d mean well it tells you that later in the problem so that's how i like to start dissecting these problems and if you notice here the equation that it gives us is the equation of a line we can rewrite this as p equals negative 23d plus 108 and that's how you generally see this equation written right with your slope first and your y intercept second and you'll notice what i just said here the 108 is the y intercept so what does the y-intercept mean well the y-intercept of a line is your starting point and if you don't understand why it's the starting point i can talk about why that is in a second but the line also has another aspect of it has the slope because you not only need a place to start from but you also need a direction the slope tells you the rate of change it tells you whether your line is you know looking like this if it's going up really fast if it's going down kind of slow that's what your slope will tell you and just with this information right here if you understand that 108 here is your y intercept and you know that the y intercept is a starting point you already know that the answer to this problem is choice b and the reason for that is because we'll look at the other question there are the other answers choice a says kathy will complete the repairs within 108 days that says nothing about starting something so no choices c and d are talking about rates well the rates i mean that would be talking about the slope and if we wanted to talk about the slope we'd need to be talking about the negative 23. so it's not those answers here but choice b says kathy starts each week with 108 phones to fix and that works with what the y-intercept is so you might get the gist of what i'm saying here right the y-intercept is the starting point so we know how to solve this problem but why is the y-intercept the starting point that might be a question that you still have so let's talk about that in the context of this problem so we can see here at the beginning of the problem each week kathy is receiving a batch of phones that need repairs so at the start of each week right starting point at the start of each week zero days have passed that means that d is equal to zero and since 23 times 0 that's 0 p is equal to 108 so what we have just done here is we found how many phones that kathy has to fix at the start of each week that is the significance of the y-intercept it's telling you where you're starting from okay so if you understand that then i think it's clear for us to move on to problem two problem two says that a pediatrician uses the model above to estimate the height h of a boy in inches in terms of the boy's age a in years between the ages of two and five based on the model what is the estimated increase in inches of a boy's height each year so again what is that word problem actually asking for just because you know probably most of what you just read went right in one year and out the other this problem is saying based on the model what is the estimated increase in inches of the boy's height each year so we're looking for that keyword is increase and here since we have a line that we're starting with we know that the increase that rate of change is the slope right we're talking about the slope of this line and the slope here is 3. so that would get you to believe that the answer here is choice a but you might want to check the units on that just to make sure the units for slope or just slope in general is change in y over change in x here that y is an h and the x has an a there and what are the units on both of those well the h the height is measured in inches and the a is the age that's measured in years so it's being measured in inches per year and so this 3 right here right the slope here we know is 3. so the slope is three inches per year and that's exactly what we want here we want the change in the boy's height each year so the slope here is not representing years per inches so we're safe right we know that the answer is definitely 3. so moving on to problem 3 here we have that the function f is defined by a polynomial some values of x and f of x are shown in the table above which of the following must be a factor of f of x so based on this table here which just gives us a couple points on this function on this on this graph we're looking for which of these guys are a factor of f of x and so well what do the factors actually tell us about a graph well the factors like let's say we have just some quadratic or something like this the factors which in this case would be x plus 2 and x plus 3. the reason why these are useful is because it tells us where the y is equal to zero right so if we if we want to figure out where the y is equal to zero we can see right here that that's going to happen where x is equal to negative 2 because that's going to give us a 0 here if we plug in that x is negative 2 and 0 times anything is 0 and we also know that x equals negative 3 would work here because if you plug in a negative 3 here you're going to get a 0 and 0 times whatever this guy is is going to be 0. so factored form points out those places it points out the zeros which are the x-intercepts so you know you're looking at you know either this guy or this guy right where something is zero remember these guys are points where y is equal to zero i have that written right up here right it's where we set y equal to zero and if you think about that a little bit better graphically just so you don't have to you know go through all this conceptual mess that i just put you through think about your like x y axes right where are the zeros the zeros are right here on both at both of these points is y zero or is x zero well y controls how far up or down you are and here you're right in the middle so that means y is equal to zero okay so you can think about that either way that you want to but since you know that you're looking for where y is equal to zero you see that that happens when x is equal to four and what's the factor that corresponds to x equals four well it's going to be x not plus four it'll be minus four because remember we're trying to get this factor to like equal 0 right like here we want that x plus 2 to equal 0. so we're asking what value of x will make this thing 0 and that would be negative 2 in this case and in this case 4 would be the value of x because then you get a 4 minus 4 which is 0. so that's why we always flip that sign there i know a lot of people are always confused on why the heck we do that so x minus 4 is going to be the answer here all right for problem 4. this problem says if a over b is equal to 2 what is the value of 4b over a and so there's a lot of different ways that people like to solve this problem a lot of people like i've done this problem on tech talk before and so many people just argue about what way is the best to do it there's like three different ways i can think of to solve this problem i'm going to show you one different one just one way to solve it and if you have a different way that you like better let me know what it is in the comments but here's how i would go about solving this problem or at least the easiest way i would i could explain how to solve this problem you can just solve it like a substitution so i can solve for a here really quick i can just multiply b on both sides and that will give me that a is equal to 2b and if i have that i can plug that into 4b over a because that'll get everything in terms of one variable now that i have a in terms of b it's just like substitution method when we solve a system of linear equations so this becomes 4b over a is now 2b and from here the b's cancel you're left with 4 divided by 2 which is 2 and that is choice c all right so that's the answer there now moving on to problem 5. this is actually the first of our calculator problem so you see that c up in the top right this problem says for what value of n is the absolute value of n minus one plus one equal to zero and i think a lot of people they get confused when solving like absolute value equations because that's not something that you do very much of you like cover it once in algebra one and then never talk about it ever again so let's review a little bit of that here we have that the absolute value of n minus 1 plus 1 is equal to 0. and if we want to start solving this for n what we want to do is get this absolute value by itself so anything that is not in the absolute value has to go to the other side we can get it to the other side that plus 1 we can get that to the other side by subtracting it on both sides and that will give us that the absolute value of n minus 1 is equal to negative 1. but this right here poses a problem doesn't it because well the absolute value of something can't be negative that should be an issue that you see right away that's not a thing so no values of n can make this work because just by definition you can't have the absolute value of a number being negative and so the answer here would actually be d there's no value of n that could make this happen i know a lot of people like i get this way a lot i'm not sure if you get this way too but like i always am so hesitant to have the answer be there's no possible way this could happen right like i don't want to have that be the answer i want to figure out how to actually make it work but here there's actually no way to make it work so yeah that's the answer here but something you might still be wondering is okay well if this did work if i was able to solve this how would i go from there so let's talk about that let's uh i'll give you an example you could solve let's say that we had the absolute value of x minus 1 is equal to 3. okay how would we end up solving this well first off we do have that absolute value that is by itself that's exactly what we want everything that's not in the absolute value is over on the other side well what we would do at this point is well look at this equation whatever is inside the absolute value needs to be one of two values to make this equation true this could be a 3 because the absolute value of 3 is equal to 3 that would work but if this x minus 1 here was a negative 3 this equation would still be true because you again you have to take the absolute value of it and if you did that you would get 3. so that x minus 1 there the thing that is inside the absolute value could either equal 3 or it could equal negative 3 and you can solve both of those equations here just by adding 1 on both sides and get your values of x so here you would get x is equal to 4 and x is equal to negative 2. okay so that's all because like once you get to this point you look and you see oh well what's inside this absolute value here could be a negative 3 or a 3. because when i take an absolute value of both of those numbers it could be a 3 and you split that up you solve these two equations and that will give you your values of x so here's an example where you can't solve it here's an example where you can and so that's how to deal with absolute value equations on the sat so that does it for our first set of five problems here and so at this point i would just take a quick break i'm not sure how long we've been going here maybe like i don't know 12 minutes or something like that but uh take a quick like five minute break or something like that and get back into this next set of five so assuming that you've taken that break let's move on to our next set of five problems for this video so for our next problem here problem six this problem says the line y equals kx plus four where k is a constant is graphed in the x y plane if the line contains the point c comma d where c and d are both not equal to zero what is the slope of the line in terms of c and d so what are we trying to find here again let's underline that we're trying to find the slope of this line in terms of c and d this line is y equals kx plus 4. so how do we find the slope well the slope is found using the slope formula the y2 minus y1 over x2 minus x1 we need two points to find the slope in this problem though we're given the one point we're given if the line contains the point c comma d so we know that that's a point i'll write that up here we know that that's a point but what's our other point that's actually that's where the problem lies here we have to actually find that other point and it's a little bit tricky here but the other point comes from the equation itself y equals kx plus 4 because the y intercept gives you a point if i draw my xy axes here we know that this line crosses the point zero comma four we know it's got to cross this we don't know what the slope is but we know it at least crosses this point and the coordinates of this point are zero comma four not four comma zero and that is my other point so which one of these should i call my first point which one should i call my second point um it doesn't matter you'll get the right answer either way i can actually show you either way but i'm going to go and put this guy as the second point and this guy is the first point and you might be able to tell why i'm doing that if you look at these answers but you'll be able to see that as i go through here so again we know the slope is y2 minus y1 over x2 minus x1 i think i accidentally erased that and the slope is going to be well y2 the y coordinate of my second point is d the y coordinate of my first point is four the x coordinate of my second point is c and the x coordinate of my first point is zero so all in all i'm going to get a d minus four over c and that right there is equal to choice a so that's my slope now what would happen if you called this guy your first point though because some people might do that and you know it probably would make more sense that way since this is the first point that we found so if you did that i'll write this here so if you did this is your first point this is your second point what would happen is you would get well the y coordinate of your second point is now four you would get that minus the y coordinate of your first point that's d and then you would get the x coordinate of your second point that's zero and then the x coordinate of your first point is c so what you would end up with is a four minus d over negative c that's not yet the same thing as d minus 4 over c right it doesn't at least it doesn't look like it so what do we do how do we get this to look like this well what we can do is notice hey this denominator's negative and if you look at all our denominators here i mean they're all positive so why don't we just multiply by a negative one over negative one right we multiply by a negative one on the top and a negative 1 on the bottom if we do that that's going to give us a positive denominator and in the numerator if we multiply negative 1 to the 4 that'll give us a negative 4. multiply the negative 1 to the negative d that gives us a positive d and so now you get negative 4 plus d you're almost done just switch the order of these two and that gives you your answer so it's a little more work if you do it that way and the reason why i chose to pick it this way now you might be able to see this is because i knew that if i picked it this way i would get a negative in the denominator right you can kind of see how these would go if you get comfortable with this and so i'm like yeah i don't want to deal with that so let's make this guy the second point and that way my denominator ends up being positive so just something to watch out for so moving on to problem seven here this problem says what is the solution x comma y to the system of equations above so we've got this gross looking system of equations and they're already trying to trick you here they're already doing something to trick you which is kind of annoying to be honest what they're doing and by the way this is not a calculator problem sadly what they're doing is they've switched the x's and the y here so that they're assuming that you know you'll make the mistake of putting this as 2x instead of 2y they're cheap that way so what we're going to do is we're going to switch this order okay we're not going to get tricked by them the negative x is going to go first and then the positive 2y is going to go second and this is equal to negative 19. so we can solve this problem i mean there's a couple different ways that we could go about it there's substitution method there's elimination method and honestly what i would do here is just solve this using elimination method i like using elimination most of the time and unless there's like an easy substitution i can do so what i'll do is i can see that these y's can be canceled out pretty easily if i just multiply by a negative 2 on both sides on the bottom equation because if i do that that's going to give me a negative 4y here and that will cancel out with the 4y okay you'll see that in a second especially if you're rusty with systems of equations you might not understand what i'm really getting at yet and so the top equation stays exactly the same it's still the 3x plus 4y equals negative 23. the bottom equation is now negative 2 times negative x that's a positive 2x then we have a negative 2 times 2y which we was the negative 4y that we wanted and then this side is negative 19 times negative 2 that's going to be a positive 38. so now what i'm going to do is add these two equations together and the reason why i mean in case this is still random to you why can you add two equations together why does that even make sense this is just the same thing as you've always been doing like when you have an equation x plus 3 equals 5 and you solve it by subtracting the same thing on both sides right subtracting 3 in this case on both sides here i want you to realize that you were adding the same thing on both sides of this equation on the left hand side you were adding a 2x minus 4y and on the right hand side you're adding a 38. that's the same thing as adding something on both sides how do you know that though how do you know that this guy is the same thing as this well it's because they're equal that equation says that so you're quite literally still adding the same thing on both sides that's why we're able to add or subtract equations and i didn't know that for like the longest time when i was in algebra 1. i didn't really get that that's why we could do that so when we add the 3x to the 2x we get a 5x these guys cancel out because 4y minus 4y is 0. and so 5x is equal to this guy it's a 38 minus 23. that will be a 15. so we can just divide by 5 on both sides that gives us that x is equal to 3 and there's only one solution where x is equal to 3 so we already know what the answer is so that works and if you wanted to solve for y here you don't you shouldn't right you shouldn't you should at this point you should be done unless you have time to go back and check your answers but you can take one of these equations and just plug in what you have for x sorry this should be an x plug in the x is 3 so do 2 y minus 3 is equal to 19 and then solve for y you'll be able to see that it's negative 8. so that's the way that you can do it if you are struggling if you don't know how to solve something using system of equations if you forget all of a sudden that's you know it's fine what you can do is start plugging your answers in plug in negative 5 comma negative 2 into both of these equations and see if it works and then move on to your next point because if you plug in 3 comma negative 8 to both of these equations it will work meaning you'll get the same thing on both sides so there's there's that backup plan if you don't know how to solve system of equations like that you can always just plug in answers and honestly they both probably take like similar amounts of time so moving on to problem 8 this is another problem that involves a system of equations but this one is a little bit harder this problem says in the system of equations above k is a constant and x and y are variables for what value of k will the system of equations have no solution now the big key to this problem is that this system is going to have no solutions and i think when a lot of people you know they start deviating from like the regular one solution they start talking about no solutions and infinite solutions now they start to get in trouble they start to forget absolutely everything that they learned because it's been a while it's been you know maybe one or two years for you since you've actually talked about that stuff so as a quick recap of what systems of linear equations actually are here is an x y axis here okay here's your x y plane and usually what happens with a system of equations so like here's a system of equations these two are lines you can convert them both into y equals mx plus b form and let's say that you graph them they'll most likely intersect at a certain point and that point that they intersect at is called the solution you're trying to find where these two lines intersect that's what we do in this problem here in problem seven these two are both lines and we found where they intersected that's what goes on here but these lines don't have to intersect these lines can be parallel to each other and now you start to get in trouble because these two lines are parallel so they'll never intersect and this is an example of no solutions and the important thing to know about a system of equations which has no solutions is that these two lines have the same slope that's actually what we're going to use for this problem here but before we do that i should also note that if these two lines are right on top of each other they're still parallel right they're still the same slope but now they're right on top of each other they're now going to intersect at every single point right they're on top of each other they're always going to be on top of each other that is what infinite solutions is so now they instead of just having the same slope they have the same slope and the same y intercept and that is the difference between no solutions and infinite solutions and while i erase this here i should note if you want to learn more about like actually solving systems of equations which also can appear on the sat where you have no solutions and infinite solutions like you need to see the zero equals zero case and the zero equals two or something like that i have a full youtube video where i go over all of that so let's actually get back into this problem here remember for no solutions okay if we have no solutions that means that these two lines right here have the same slope that's what piece of information this gives us because without it there's no way that we can find k here so if we know these two lines have the same slope then we can find the slope of each of these lines and we can set them equal that's how we're going to be able to solve for k and you're going to see that so let's start off by finding the slope of kx minus 3y equals 4. we've got to get that into y equals mx plus b form so we can see what the slope is to find that slope let's subtract kx on both sides got to solve for y here so we're going to get a negative 3y is equal to a negative kx plus 4. i'll write that x term first like you're used to seeing it and then we're going to divide by negative 3 on both sides and that's going to give us that y is equal to this is they're both negative so this is going to be a positive k over 3. i'll put the x here you can leave the x up here but i like to put it separate it's still the same thing but the reason why i like to keep it separate is because now you can see clearly what your slope is and then we have a negative four thirds right here great we get that that works we know our slope is k over 3 for that line what about for the other line for the 4x minus 5y equals 7. well now we subtract 4x on both sides to solve for y we get negative 5y is equal to a negative 4x plus 7 and we divide by negative 5 on both sides to finally solve for y so we get that y is equal to this will be a four-fifths x and then this is a negative seven over five that we have that we again know that these slopes are equal to each other they are the same and so k over 3 must equal 4 over 5. so now that we know that let's just cross multiply and that'll allow us to solve for k we get the 5k actually no let's not cross multiply that's way too much effort let's multiply by 3 on both sides because if we multiply by 3 on both sides then the 3s go away we already have k solved for so we get that k is equal to a 3 times 4 which is 12 divided by 5. so that's going to be a 12 over 5. and that is choice a all right so that's the key when you think about no solutions they have the same slope but they have different y intercepts and you can actually see here even they do have different y intercepts negative four thirds and negative seven fifths if these two lines if we were looking for where they had infinite solutions then these two y intercepts would have to be equal that can't happen here but if you know there's problems with even more constants involved which i have seen before then just remember you also have to set the y intercepts equal if there is infinite solutions but yeah that's this problem so moving on to our next problem here problem nine this problem says for the function g defined above a is a constant and g of four is equal to eight what's the value of g of negative four so you might be able to solve this problem in like no time at all and like five seconds because you can see that this is an even function so g of four would need to be the same as g of negative four so since this guy's eight this needs to be eight and that means that your answer is choice a all right moving on no i'm kidding so yeah obviously some of you are going to be confused with first off what's an even function what does it mean to have an even function how do i know that i have an even function and all that stuff so we're going to go through that and if you don't even think about this whole even function thing at all i'm still going to show you how to solve the problem anyway without knowing anything about even functions so let's talk about this first off i can tell that this is an even function because it only has even powers of x this term also is an even power of x because with 24 right like i can write this as ax squared plus 24 times x to the 0 so x to the 0 is just 1. so this is completely fine already like this because this will be 24 times 1 which is still 24. and here we have an even power of x so again it's all even powers that's how i know we have an even function and an odd function would be only odd func only odd powers of x now an even function has symmetry about the y axis like this so if we know that the x the y value at x equals four is eight then if we want the x value sorry we want again we want the y value at x equals negative four it's got to be the exact same thing because it's got that symmetry so that's how i can show you that with the symmetry argument but what if you don't know that that's not something that comes to your mind right away well we can solve this algebraically too so what we can do here is say okay well i know that g of 4 is a times x squared which i can write as a 4 squared plus 24 and this guy's equal to 8. from here you can solve for a and by the way from here you can also see that if i was doing g of negative 4 and i plug in this negative 4 for x it gets treated just the same as four does because it gets squared and what that means is that the negative drops so this right here is a a times 16 and this guy is also an a times 16. so there's another clue that g of 4 is the same as g of negative 4. but if you don't if you miss that too if you don't see that you can still solve this problem okay so i'm giving you all the clues to solve this a lot faster but if you don't see it here's how to go you can solve for a you can say that this is 16 a plus 24 that's equal to 8. let's solve for a we subtract 24 on both sides we get that 16a is equal to a negative 16. divide by 16 on both sides and you get that a is equal to negative 1. so there you go you have your value of a and then plugging in to g of negative 4 you know that it's going to be ax squared plus 24 but here you know that a is a negative 1 and then x squared that'll be the negative 4 squared since you know you're plugging in a negative 4 for x here then we just add on 24 so this becomes negative 1 times 16 which is negative 16 plus 24 and that's equal to 8. so you see the g of 4 is 8 here so that is the very long way of solving this problem but again this work is not necessary if you realize this this is an even function so if you realize that g of x is even all right so that is problem nine now we're going to move on to our last problem for this set of five so we'll have 10 problems done that'll be great so this problem says in the x y plane the parabola with equation y equals x minus 11 squared intersects the line with equation y equals 25 at two points a and b what is the length of the segment a b so this problem is going to leave you clueless if you don't like if you can't visualize what's going on here so you're going to need to know a little bit about how graphs transform i'll show you this so if we want to visualize what's going on here we have two curves we have a parabola this guy and we have a line so how do i draw x minus 11 squared well if you just were to draw x squared that's a parabola just like normal you've got your vertex right at the origin but if what you do is you take this square or you take what's inside that square and you subtract 11. what that does is it shifts it either left or right here since you're subtracting it actually shifts it to the right that's what it will do and i'll just not exaggerate that as much so i'll draw it that way the reason why you have to do the opposite like you would think subtraction moves it to the left and addition would move it to the right but the opposite happens the reason for that let's think about this parabola here right let's think about where your y would equal zero that's this point right here where on the x axis would your y equal zero for x minus 11 squared well if you set this thing equal to zero you can see that the only x value that you could plug in here that would make this thing equal to 0 would be 11 because then you'd get 11 minus 11. so that's why this moves to the right not to the left that's how to check yourself on that but if you are let's say adding a number to this right so let's say that you have x squared plus three okay that is your same parabola but it is moved up by three so this adding will move it up subtracting will move it down that will do exactly what you expect it's only when you put things in the parentheses that they do the opposite so that's that little bit of explanation out of the way and now all we need to do is graph y equals 25 so i'll just graph that as some line coming through here what that will do is give us our points okay and b those are the two intersection points and what we're trying to do is figure out what the length is of this segment right here so how do we do that well if we find the x coordinates of this intersection then we can just subtract them right because the distance here is the same as the distance here so that's exactly what we want to do let's find the x coordinates of the intersection let's subtract them and to find the intersection all you have to do is set these two equations equal to each other we know that y is equal to this and y is equal to this therefore if we want to find where they intersect we want to know where they're equal to each other so to solve for x here let's just square root both sides when you square it this side it cancels out with the square so you just get x minus 11 and then square root of 25 is 5. and when you square root both sides of an equation you need to put a plus or minus on one of the sides so here i'll make it a plus or minus 5. okay so now i can just keep solving here by adding 11 on both sides and that will give me that x is equal to a 5 plus 11 and it also gives me that x is equal to a negative 5 plus 11 and so x is equal to 16 and x is also equal to that's a 6. so we know the x coordinate here is 16 the x coordinate here is six and therefore the distance between them would be 16 minus six and that's ten so there you go that is your next set of five problems we now have ten problems done and we've got a lot to go but at this point i think it's a good idea for you to take a just quick five minute break get some water go stretch do something because you're not going to want to sit and just try to watch all of these because you're just your mental capacity is just going to go down and down and down and just it's not a good idea okay so take a quick five minute break and then come back to the video we'll get another set of five done and assuming you've taken that break let's hop into the next set of five problems for this video so we're gonna start with problem 11 and this problem says in the equations above b and c represent the price per pound in dollars of beef and chicken respectively x weeks after july first during last summer what was the price per pound of beef when it was equal to the price per pound of chicken so this is what the word problem is asking for and all right i just want to like address a little pet peeve before we keep going on here i don't understand why the sat does this look at how this was a two sentence paragraph look how long that sentence was that is a run-on sentence that is the dictionary definition of a run-on sentence if the sat is going to assign an english section you would think that they would like you know follow those rules when making their math problems like just a thought right like that should probably happen but anyways when we're in this problem this problem is asking for the price per pound of beef when it is equal to the price per pound of chicken so we're saying b is equal to c here those are the price per pounds that it gives us earlier in the problem and we want to know well what's the value of these guys when they are equal so if we know the b is equal to c all right let's set these two equal to each other 2.35 x nope already wrong 2.35 plus 0.25 x is equal to 1.75 plus 0.40 x and now we just solve for x so i'm going to subtract 0.25 x on both sides get all my x's to the right-hand side you know some of you may not like that i'm going to the right hand side not the left hand side but i don't want negatives so sorry this right here is 0.6 this guy is 15 so it's right there and yeah this is not a calculator problem either so you got you got to do that and so now we just divided by 15 on both sides this is easy to do division i mean it's the same thing as 60 divided by 15 which is 4 so x is equal to 4 here now you'll notice that is not the answer here because what we're asking for is not the time when b is equal to c we're asking what the price is per pound when b is equal to c so we've got to plug this x equals 4 back into one of these two equations so i'm going to plug into i'm going to plug into the b equation actually because it's going to be really nice if we multiply this guy by 4. all right so b is 2.35 plus 0.25 times x if we plug in that x is 4 and i can write it like that and so this is 1 and this is 2.35 so that's 3.35 when you add those two together and that is going to be your answer so that's all you got to do for that problem it's actually one of the nicer system of equations problems that we've done in this video all right moving on to our next problem here problem 12. this problem is a little bit uh geometrical we got to deal with angles here it says in the figure above lines k l and m intersect at a point if x plus y equals u plus w which of the following must be true and we're given a bunch of statements that may or may not be true here so first off you might see that we have a bunch of vertical angles going on here so we know that y is equal to u we know that z would equal w we know that t would equal x and that's just to start those are the vertical angles now even though it looks like you know z and x would probably equal each other and you know y and t like just because they they look the same it looks like all these angles are exactly the same but we can't assume that it says figure not drawn to scale so where do we go from here well we also have this equation that they gave us in the problem so you know that can help as well we know that x plus y is equal to u plus w and you might be able to see here we already know that y is equal to u this is the same thing as this which means that x needs to equal w because these two things are not going to be equal if x doesn't equal w think about it if y is equal to u these two are the same exact thing like that's kind of the same thing as like x plus three equals three plus w right these two are the same thing like i was saying so if we subtract them off on each side we get that x is equal to w that's why that makes sense so if we know that x is equal to w well we also know that w is equal to z and we know that x is equal to t so now we know that all four of these guys are equal so where do we go from this well let's start looking at our statements x is equal to z do we have that well yeah we do right here so that's gonna work what about y equals w well i don't see that anywhere i don't see anything that would that would equate y to w so we can't really say that yet three z equals t we do have that so that works but what about this y equals w thing is there any way that we could figure this out i mean you could probably start saying oh well we know that x plus y plus z is equal to w plus u plus d because they're supplementary angles i mean but that's not going to get you anywhere right you're not going to be able to do anything else and you can try but i mean there's nothing else you can do and if you've tried for a while and there's nothing else that you could think of just call it quits all right it's probably not that as hard as you're making it out to be so one and three is going to be the answer here and that does it for problem 12. oh that's not nope one and three is here that's gonna do it for problem 12. so moving on to problem 13 this problem says a line in the xy plane passes through the origin and has a slope of one over seven which of the following points lies on the line so let's think about this here we have a line in the x y plane so let's bring out an x y plane and let's just draw a point at the origin we know that's a point that is on this line we also know this line has a slope of one over seven so does that mean that the you know if we do our rise over run thing we're going to end up at one comma seven well no not quite and that's kind of a little bit of a trick to this problem they would get you to assume oh well if it's a slope of 1 7 then we'll end up at 1 comma 7. but that's not quite right remember with the slope being rise over run that means we're going to go up by 1 and over by 7. the point that we end up at is not 1 comma seven it's actually seven comma one because the x coordinate is not a one it's a seven so that's how we get to the first point but i mean you can see here that doesn't match any of our answer choices here so where do we go from here might be a little confusing right well let's go one more over and see if we can get any of these answer choices so if we go up by one over by seven again we end up at 14 comma 2 this time so choice d is the answer here moving on to problem 14 this problem says based on the histogram above of the following which is closest to the average arithmetic mean number of seeds per apple so we're trying to do an average off of this histogram and that might be a little confusing to you because usually when you're trying to think about averages you're given some list like something like this and they're like okay find the average of this so how do we find the average of a histogram well we can turn the histogram into a list and if you think about what we're trying to average here it would make sense how we would turn it into a list what we're doing in this histogram if you look at it close we are taking 12 apples and we're finding the number of seeds in each apple the results are plotted in the histogram and what we're trying to do is average the number of seeds in each apple so what we'd want to do is we'd want to add up the total number of seeds and divide by the number of apples so that's how we would go about doing this but well how do we find that total number of seeds we can just form a list if you want to i'm going to show you a quicker way to do it but if you wanted to start making a list you can so we see that we have two apples where there are three seeds so we could write that as three comma three there you go you have two apples with three seeds you have four apples with five seeds so you would write four fives and you can make your list that way and then eventually when you're actually trying to compute the average you would add these all up and everything that comes after it but instead of doing that i'm gonna show you a little bit of a quicker way we have two threes so instead of writing that as three plus three let's write that as two times three and instead of writing four fives let's write it as four times five and if we keep going there there's one apple with six seeds so we write that as one times six we have a two times seven and a three times nine and we're putting that all over it's not five things that we're adding together this is a compact way of writing all of the different pieces in the list because there are they're going to be 12 of these how do i know because there's 12 apples that we were looking at so this is going to be divided by 12. and now you can put this in your calculator this is a calculator problem but what i'm going to do is just limit the amount of thumb work that i need to do i'm going to just do this numerator part in my head first we have 6 plus 20 is 26 plus another 6 would be 32 32 plus 14 is 46 and 46 plus 27 is not going to be a 63 it's going to be a 73. so it's going to be 73 divided by 12. and so now we just need to put that in our calculator if you do that you get 6.083 repeating i believe so you get this guy and that might make you think you did something wrong because it's not one of these numbers but reread the problem and it says closest to the average so we're not trying to find i wish i really wish i would not get emails while i'm recording my videos like i thought i thought that i turned it off it appears that i haven't so all right anyways so the average the the closest number to the average is going to be choice c which would just be six because this guy is about equal to six all right so that is it for problem 14. now moving on to problem 15 the last problem for this set of five this problem says in the quadratic equation above a is a non-zero constant the graph of the equation in the xy plane is a parabola with vertex c comma d which of the following is equal to d so what are we trying to do in this problem here we're trying to find the value of d and what is d being defined as it's saying that it is the y-coordinate of the vertex now it does give us the equation of this parabola which is really nice so i mean this equation does give us some pretty good information it tells us where the zeros are for this parabola so i can graph this and just give you a good understanding of what this thing is going to look like so we know that there's going to be a 0 at x equals 2 into 0 at x equals negative 4. so let's do that we'll put a zero this is going to be probably about where two is this is going to be about where negative four is so if i was to draw a parabola let's see if i can get this on the first try all right not good but um it'll work it'll work it'll work so great there's your parabola and i don't know if it's facing upwards or downwards because i don't know whether a is positive or negative the only thing it tells me about a is that it's non-zero which means that it's you know not zero so that's great that's a good start now we want to talk about the vertex and what i want you to understand about this is that parabolas are symmetrical that means if you want to find the vertex all you have to do is look in between the zeros look at the middle number in between the zeros the vertex is going to fall at x equals negative one because it is the middle number in between negative four and two if you think about this is negative three negative two you got zero and then you have one over here i know that this is negative one i know that's the middle number because well one two three one two three if you don't like doing that if you just want a better way of doing that you can average the two numbers so negative four and two i'll average them by adding them together and dividing by two and that'll give me negative two divided by two which is negative one so what that gives me the middle point there gives me the x coordinate of the vertex so that's nice if it gives me that but i want the y coordinate i don't want c i want d so well if i know the x coordinate of the vertex i can just plug that in to the equation for this parabola and that'll give me the y so let's do that y is equal to a times x minus two times x plus four plug in that x is equal to negative one and what we end up getting is a times negative 3 times 3 and multiplying those together we get that y is negative 9a and that right there is choice d oh sorry choice a what it's choice a so that's just something you got to understand about parabolas how they are symmetrical i think a lot of people get confused with that concept but that's just a property of parabolas so that is going to do it for our third set of five i believe yeah it's our third set of five so go take a break and while you're taking that break if you're looking for more help with the sat math section and you like how i'm explaining things so far then i have good news for you because i have my full sat course which goes over every single topic that you need to know for the sat and in that course i walk you through hundreds of practice problems to really get you feeling confident with each topic if you're interested in that the link is in the description and assuming that you've taken your five minute break let's hop into the next five problems for this video so we start with problem 16. this problem says if x is greater than 3 which of the following is equivalent to this gigantic fraction and i think you know before we even talk about this gigantic fraction where a lot of people will get confused is just with this inequality because i'm going to show you the solution to this problem and it's not going to involve this inequality at all so people are like well why is it even there it confuses them and the answer is this inequality is just stopping the fraction from being undefined it's stopping there from being a 0 in the denominator if you look at this fraction here you can't have a negative 3 for x you cannot have a negative 2 for x and so what this inequality does i don't really know why they chose 3 but it just keeps it so x can't be negative 2 and it can't be 3 because if x is greater than 3 it can't take either of those values so yeah basically you never have to worry about that inequality again it just stops the denominator from being 0. so let's actually get into this problem here we need to get this guy to look like one of these and how we need to start doing that if you want to clean up a just gross fraction inside a fraction whatever what you want to do is get a fraction in the numerator and a fraction in the denominator here we have two fractions in the denominator we need to combine those into one and so let's start doing that i'm going to take the one over x plus two i'm going to add that to the one over x plus three we won't worry that it's you know all below a one we'll take care of that later but let's start simplifying this here or not simplifying but finding our common denominators because that's how we're going to have to combine it we need to multiply the one over x plus 2 by x plus 3 over x plus 3 because that's going to be our common denominator we just need to multiply the two denominators together so then this guy is going to get multiplied by an x plus 2 over x plus 2. so this numerator here is going to be an x plus 3. the denominator here that's going to be the x plus 3 times x plus 2. you can multiply that out you might even be able to see it's exactly this x squared plus 5x plus 6 that it's showing you in the answer choices so the x times x gives you the x squared 3 times x gives you 3x 2 times x gives you 2x and then 3 times 2 gives you 6. and just so i mean i don't want to keep writing out a bunch of different work here i'm going to combine these already this is a 5x so that's our first fraction then the other one is going to be x plus 2 divided by the same x plus 3 times x plus 2. so it's going to again be x squared plus 5x plus 6. and so now you can combine these guys the denominator stays the same in the numerator you add them together so x plus 3 plus x plus 2 just combine like terms x plus x is 2x and then we have 3 plus 2 which is 5. and so your answer is choice a right well no we have to remember that this entire thing is beneath a one right we were just talking about the denominator part we have to now worry about this entire thing as a whole from all the work we've just done we know that 1 all over 1 over x plus 2 plus 1 over x plus 3 is the same thing as a 1 all over a 2x plus 5 over x squared plus 5x plus 6. and we know that because we just found that this is equal to this that's all the work we just did up here so now if you want to get this thing into the denominator because that's what you can do when you have a fraction in the numerator and a fraction in the denominator what you need to do is bring this guy up to the numerator and flip it the reason that you can do that by the way is the same exact thing it's literally stay change flip or keep change change or whatever you learned in like middle school of how to divide two fractions this is the same thing because remember like people get so disconnected from this that fraction bar that is division you're dividing there so this can be rewritten as a one divided by two x plus five over x squared plus five x plus six and if you rewrite the one as a one over one so it's a fraction and you can do that because 1 divided by 1 is 1. if you rewrite it like that then what you end up doing is keeping this fraction you change the sign to multiplication and you flip this guy this is exactly why you can take something from the denominator bring it up to the numerator and flip it it's the same thing okay so hopefully you see that and that means that this guy is the same thing as bring this up to the numerator and flip it this guy goes to the numerator so it ends up being x squared plus 5x plus 6 and then you have this guy in the denominator so that's it that's really what this entire thing reduces to and that is choice b so moving on to problem 17 this problem it looks really gross but really it's it's not as bad as you might think it says the equation blank i'm just going to say blank is true for all values of x not equal to 2 over a again why is it giving us this it's because if you plugged in that x was 2 over a here you would get a 0 in the denominator so it's saying it's true for all of those values of x that's not this where a is a constant what is the value of a so for these types of problems that you might not know what to do at all here but if we want to start making this look nicer making this look not so horrible i mean you can see that we almost have common denominators here we just need that this negative 8x minus 3 to have an ax minus 2 as the denominator so what we can do is take this negative 8x minus 3 and multiply it by an ax minus 2 over ax minus 2. that will give it the ax minus 2 in the denominator and in the numerator that's going to be a little grosser we're going to have negative 8x times ax that'll be negative eight a and then there's two x's so it's x squared next we'll do the negative eight x times negative two that will give us a positive sixteen x we have a negative 3 times ax we have negative 3 times negative 2 and that's it so now we have common denominators everywhere because this negative 8 x minus 3 is this thing i know it looks like we just messed this up even more but we haven't because now that we have common denominators everywhere the denominators go away and you might be like well why do they go away well there's a couple ways i can explain that i think the easiest way to explain that i mean first off if you have common denominators everywhere you could just multiply them on both sides so i get ax minus 2 that cancels out here the x minus 2 would cancel out with the x minus 2s in the denominator on the right hand side so that's a way that you can make the denominators go away but i also want you to see it from this angle so first off i know that 2 over 5 plus 3 over 5 is equal to 5 over 5 right that's just basic addition of fractions now i have common denominators everywhere here and so yeah you can think about it as multiplying by 5 on both sides and that gets rid of all the different denominators but how you can also think about it is that when all of my denominators are the same the numerators have to be the same as in 2 plus 3 needs to equal 5 and it does that would make sense 5 is equal to 5. that's how this works so when the denominators are equal the numerators have to be equal so that means that this 24 x squared plus 25x minus 47 needs to equal the numerator for this guy which is this mess which is negative 8x x squared plus 16 x minus 3 a x plus 6 and then we have a minus 53. so these two things have to be equal if we have the same denominators so we have that that's great we can combine a little bit of like terms here that'll give us a negative 47 so we can write all of this again but now this has a negative 47 in it but uh yeah that doesn't really it doesn't really help us much here it doesn't help us at all and actually what are we even trying to do here because this just looks really gross what are we doing we're finding the value of a and it's very it's very easy to do that actually if these two things are equal if they are equal they have to be the same thing that means that this thing needs to be a 24x squared plus 25x minus 47 because these things are the same for it's not just like one or two values of x it's all values of x that are just not this so you know x equals 1 2 3 4 5 whatever that all works just not this one and so if these are going to be the same for all values of x the only parabola that can be equal to this is itself that means that this negative 8a needs to be a 24 so that on this side we also get a 24x squared if you think about the x terms we don't need to to solve this problem you can already see that a is negative 3 but the 25 here needs to be equal to 16 minus 3a and you'll see that that works as well that'll be our way of checking but negative 8a just divide this by negative 8 on both sides you again see that a is negative 3 and that's all you need to do so that's our answer here but just going a little bit further making sure that this is our answer if you plug in an a equals negative 3 here then this negative 8 times negative 3 is a 24. this combines to be a 25x once you plug in the a is negative 3 and then you get that negative 47 there and that will make this thing equal to this so that's how these problems work and hopefully with that explanation that makes sense to you because i know a lot of people get tripped up with these types of problems because honestly did you ever learn these in high school i don't know if i ever learned how to do a problem like this so yeah that the sat kind of just throws that in there moving on to problem 18 this problem says the equation above gives the height h in feet of a ball t seconds after it is thrown straight up with an initial speed of vp v feet per second and from a height of k feet which of the following gives v in terms of h t and k so all we're trying to do here is we're trying to solve for v that's all this problem is asking for and so to start solving for v with this equation i'll write it out below here h equals negative 16 t squared plus v t plus k if we want to start solving for v we need to get all these terms that do not have v over on the other side so what i'll start by doing is adding 16 t squared on both sides that'll get that guy over to the left-hand side and then i'll subtract k on both sides so now i'm going to end up getting that h plus 16 t squared minus k is equal to the only thing left on the right hand side is the vt and now all we have to do is divide t on both sides doing that i'll switch the sides now so i'll put v on the left hand side v is going to equal h over t plus a well now we're going to have a t cancel off right one of these t's is going to cancel off with this t so we're going to be left with a 16t there and then we have that minus k over t on the end and we can simplify this a little bit more here we can write this as well since we have two fractions here with common denominators this can be written as h minus k divided by t plus 16t and that right there is our answer that's choice d now somewhere where you might have gotten caught up here when some people divide the t on both sides like this what they'll do and here's what i usually don't write it like this what i'll usually write it as is like the h plus 16 t squared minus k i'll write that all over t and that's the exact same thing as dividing each piece by t and if you don't believe me if you're having a hard time believing that this is just a bunch of fractions with a common denominator we can add and subtract all those together to get this so it's the exact same thing it's just in this problem especially it was easier to write it like this because then we were able to simplify a little bit and get something like one of our answers so just a little thing to watch out for there moving on to problem 19 this problem says the table above lists the lengths to the nearest inch of a random sample of 21 brown bullhead fish the outlier measurement of 24 inches is an error of the mean median and range of the values listed which will change the most if the 24 inch measurement is removed from the data so what this problem is saying is hey we're going to get rid of this 24 it's an error which of these guys is going to be affected the most and so let's think about it remember i mean we don't need to actually go and calculate how much the mean is going to be affected just get a decent idea for it you got to think these sat problems they're not supposed to take like 10 minutes to do or you don't have to be exact with this is multiple choice so if you think that doing all this computation is going to take like 10 minutes that probably tells you that you're thinking about this a little too hard if we just think about the mean here i mean we're going to take away that 24. yeah that'll affect the average a little bit not i mean not a crazy amount the median i mean take away that 24 and your center number your center number is probably what like a 12. if you take away that 24 now your center number might be like over here that doesn't change the median at all actually and what about the range well remember that the range is the largest number minus the smallest number so the range here is 24 minus 8 and that's 16. after we take away that 24 the largest number is a 16. and so we get 16 minus eight is the range the range drops by eight by eight inches that's huge is the mean gonna drop by eight inches is that is it going to change that much no there's no way the mean would drop by that much and so the answer here is going to be c the range all right moving on to the last problem for this fourth set of five problems we have this problem right here what are the solutions to this quadratic equation so do not here's one thing i'm going to say right off the bat do not waste your time trying to factor this there is a clear indication that you should use the quadratic formula here and what's telling you to do that is the fact that all of these answer choices have radicals in them they all have a square root so i mean you're not going to be able to factor that and get a square root so what you want to do is pull out the old quadratic formula and honestly before you go and start plugging in the a b and c i wouldn't even go that far yet what i would do is start off by simplifying this quadratic equation i'm really trying hard not to write slanted like this this is annoying like like my entire thing is just rotated all right so what i would do i mean i see that these terms are all multiples of three and so i'm just going to divide by 3 on both sides and remember if i divide this entire thing by three it's the same thing as dividing each piece by three and so what that gives me is x squared plus four x plus 2 equals 0. and these are going to be some much nicer numbers to plug in because in this quadratic a is equal to the number out in front of x squared that's a 1 here because you don't see a number there b would be equal to 4 and then c would be equal to 2. and that's much nicer than having a 12 in there which you know just makes things a little grosser and i should note too these are going to have the exact exact same solutions we haven't changed this equation at all because we divided the exact same thing on both sides that is what makes it not change so let's actually plug into the quadratic formula here i'll drag this guy this way and we know that b is 4 so this is going to be a negative 4. we have the square root of a b squared that's a 4 squared that's going to be 16 minus a 4 times a which is one times c which is two and that's all going to be over two times a a here is one now let's simplify a little bit here what we're going to do is write the negative 4 plus or minus this is the square root of well there's a 16 minus this is just an 8. 4 times 1 times 2 is 8. so you get a 16 minus 8 which is 8 and that's going to be divided by 2 times 1 which is 2. so we get this and that doesn't match any of our answer choices here so what do we do well we're going to have to simplify this radical but before we do that there is something i want to know here and that is that you can already see which of these answers it is without simplifying the radical in the first place you just might not see that right away so i'm gonna fully solve it anyway but you can see here that this first term the one before the plus or minus is going to be a negative two it's negative four divided by two and if you look at the answer choices choices c and d they both have a negative 6 so they can't be right and you can see if it's going to be a or b just because b is the only one with the square root of 30 and there's no way you're pulling a square root of 30 out of this you have a square root of 8. how are you just going to magically make that bigger so b is not the answer either and a would be right so if you can see that right away great if you can't here's how to simplify that square root of 8. so if we want to simplify a square root what we need to do is take out the largest perfect square that we can out of this eight and perfect squares are just all the numbers that you can square root and get something nice so for instance you know square root of all these numbers you get 1 2 3 4 5. perfect squares can also be variables like x squared is a perfect square because if you square root it you get something nice that just being x and that's why when you have a difference of perfect squares and you can factor it like this these are two perfect squares and they are being subtracted that's why they call it a difference of perfect squares so that's just something i just figured i would i would add on there but yeah we're trying to take out the largest one of these numbers that we can out of 8 and of course that's going to be 4 because we can divide in 8 up into a 4 and a 2 because the square root of 4 times the square root of 2 would be the square root of 8. and the reason why we want to take out the largest perfect squares because then we can just when we square root it the square root of 4 is 2 and so the square root of 8 can be written as 2 times the square root of 2. if we write this all out we're going to have a negative 4 divided by 2 plus or minus a 2 rad 2 divided by 2 and we can simplify that all at once to get negative 2 plus or minus the 2's cancel we get rad 2. and that was exactly the answer that we thought it was going to be so that is going to do it for our fourth set of five problems for this video so again go take a break just do something other than this video come back with a clear head and then we'll get into the next set of five problems which is honestly i think quicker than this one there's just like one problem that involves a little bit more work so assuming that you've taken that break let's hop in to the next set of five problems starting with problem 21. so problem 21 says that the cost of using a telephone in a hotel meeting room is 20 cents per minute which of the following equations represents the total cost c in dollars for h hours of phone use so what are we trying to find here we're trying to find the cost c in dollars for h hours of phone use so we're trying to find c in terms of h and you can see all those equations there with c in terms of h so what information does this problem give us well it says in the first sentence that the cost is 20 cents per minute so we can write that as an equation we can write that c is equal to 20 cents per minute we'll multiply it by m that way if we've been on call for 2 minutes then this is 20 cents times 2 it's 40 cents like we would expect it to be the problem is though we don't want this in terms of m we want it in terms of h so we need an equation for m in terms of h and no the equation that you want is is not 60 m equals h it is true that there is 60 minutes in one hour that is a conversion factor that is not an equation because as you can see here if we plug in that one minute has passed then it would be true that h is equal to 60. and one minute is not equal to 60 hours so the conversion is not the same thing as the equation here's how to write the actual equation part for the equation we need to write the 60 on the other side and that way you can even check this if we put the 60 on the other side then plug in that one hour has passed and it's true that 60 minutes have passed in one hour so that's just a double that's a way to double check that your equation is actually right and if we have that we can plug that in plug in that m is equal to 60h and now you have your answer which is choice a and i don't know if this got you tripped up at all if you try this on your own first but they are trying to bait you they're trying to assume that you'll say that 60 m is equal to h instead and if you did that then you would get one of these other answers and i think it would be let's see if you did that you would get you divide by 60 on each side and so you would get this guy right here and of course that would be wrong so you just got to watch out for that moving on to problem 22 this problem says if ax plus 2 times bx plus 7 is equal to 15x squared plus cx plus 14 for all values of x and a plus b is equal to 8. what are the two possible values for c so let's write out what we know here we know that ax plus 2 times bx plus 7 is equal to 15x squared plus cx plus 14. and we know this is true for all values of x this is another one of those correlating coefficient problems it's the same as a problem that we tackled earlier where was it i think it was the one right before this one it's the same as this guy right we were talking about how the 24 is equal to the negative 8a it's the same kind of problem so this is definitely going to be some good practice for you so what we need to do is expand this guy and if we do that we multiply the ax times the bx that gives us an a times b times x squared multiply the 7 by the ax next and then the 2 times the bx and then we can multiply the 2 by the 7 to end it off it's equal to 14 and then we get 15x squared plus cx plus 14. and hopefully you can see here why i've actually multiplied this out right we need to be able to compare the coefficients on this side with the coefficients on this side and now that we've multiplied it out we have those coefficients we have the a b is the coefficient on x squared and here it's 15. so we know if these two guys are the same for all values of x then this side needs to be a 15x squared plus cx plus 14 just like this side is so a b must be the same exact thing as 15. so we get a 15x squared here so we have that that's great we also can now look at the x terms we know that 7a plus 2b is equal to c and that gets the x coefficients equal to each other and of course we know that 14 is equal to 14. that's true that's good and i think you know that just tells us that hey we multiplied things out correctly and that's that's nice so we know that that is true but what you might see here is that well there's no way we could solve for c at this point the reason being we have three variables a b and c and well only two equations remember with a system of equations we can solve for like x and y if we have two equations that have x and y's in them but if we have three variables to solve for we can't just get away with two equations we need a third equation so where are we gonna get that from well the problem gives us another equation it gives us that a plus b is equal to eight and so if we combine all of these equations together we can actually solve for c so let's start solving this system of equations here and we're not going to be able to solve for c right away but we do have these nice two equations for a with a's and b's in them so we can solve for one of those variables i'll actually just solve for b in this first equation so i'll subtract a on both sides and so what i'll do is i will subtract a on both sides like that i will get b is equal to 8 minus a and if i've done that then i can plug that in to a b equals 15. so i know that a times b is now an 8 minus a we know that's equal to 15. so multiplying this out here i get that 8 a minus an a squared is equal to 15. and if i i mean this is a quadratic at this point we can solve that subtract 8a on both sides add a squared on both sides and what i've done here is now i have 0 on the left hand side and on the other side i have an a squared minus 8a plus 15. awesome so we can factor that what are two numbers that add to be negative eight and multiply to be fifteen well negative three and negative five and so that means that a is equal to 3 and a is equal to 5. those are the two possible values for a great so if we have that we can actually just go and solve for c now and i'll show you how take this equation the 7a plus 2b is equal to c and what you can do is i mean you can just go and use these a's to solve for b but you can do that as you plug in if i plug in a 3 here then we already know what the value of b is going to be since a plus b is equal to 8 i know that if a is equal to 3 like we just plugged in then b would need to be 5 so you get 3 plus 5 which is 8. so this is one possible value of c now another one would be if we plug 5 in for a because that's the other value it can take and if a is 5 then b has to be 3 so they add to b 8. and so right there you go so from here you're going to get a 21 plus 10 that's 31 and from here you get a 35 plus 6 that's going to be 41 and so that gives you the answer of choice d which i'll go back up and show you in a second so that is honestly that's the slow way to go about solving this okay i wanted to show you like the full way first because that's probably the default way that you'd solve this and this problem is one of the longest problems that i have in this video if not the longest okay it does take a lot of effort but there is a quicker way to solve this and you might have noticed this while i was solving it or you were just like oh my god there's so much and you got overwhelmed and you were like all right let's let's chill on this so so let me show you an easier way to go about solving this problem something i noticed while i was solving this was we could have found the values for a and b much easier because this is actually just the same thing that we do when we factor we find two numbers that add to be eight and we need those two numbers to multiply to be 15. so we could have found those numbers quick we would have been oh well first off we could have a equal to three and b equal to five and then we could also have a equal to five and a equal to three those two numbers would work in both of those equations and remember those are exactly the same values that we plugged down here so you could have skipped all of this math right here if you would have realized that but realizing that is the hard part and the reason why i didn't just show you the short way and was done with it is because well you're not always going to see the short way i mean you're not going to see the short way to solve these problems 100 of the time really no one is not not me either you think like with every single problem that i'm showing you i always just immediately know the quickest way to solve these problems no no not at all so we're going to move on now to problem 23 and problem 23 says in order to determine if treatment x is successful in improving eyesight a research study was conducted from a large population of people with poor eyesight 300 participants were selected at random half of the participants were randomly assigned to receive treatment x and the other half did not receive treatment x the resulting data showed that participants who received treatment x has significantly improved eyesight as compared to those who did not receive treatment x based on the design and results of the study which of the following is an appropriate conclusion yeah this is another one of those word problems where you read it and then just forget everything so let's go back into this word problem and just figure out what we're actually talking about here so yeah i think i honestly just even underlining as you go here is a is a good idea but we're trying to draw a conclusion for this study here and what did this study do well it took a large population of people with poor eyesight it took 300 people to be exact half of these people received treatment x and the other half they did not receive this treatment and so the resulting data showed that those who received that treatment had significantly improved eyesight as compared to those who did not receive it and so now we have to draw a conclusion based off of that so choice a says treatment x is likely to improve the eyesight of people who have poor eyesight and that's true yeah so i would probably go with a right off the bat because we took the people that had poor eyesight we gave half of them the treatment and those who got the treatment showed significantly improved eyesight compared to those who did not choice b says it improves eyesight better than all other available treatments and nowhere in the study did it talk about other treatments so no choice c says treatment x will improve the eyesight of anyone who takes it no the study talked about people who had poor eyesight and then choice d says treatment x will cause a substantial improvement in eyesight yet maybe for people who have poor eyesight like uh i these problems are so weird the conclusion problems like they're essentially just completely common sense problems you have to just make sure you're reading the word problem carefully if you do read it and you understand it then the like the answer choice is very easy to see from there at least that's what i take away from these types of problems all right problem 24 this one's a little bit of a break here it says if t is greater than zero and t squared minus four is equal to zero what is the value of t so let's think about this here we know that t squared minus four is equal to zero that's a difference of perfect squares so we can factor this as t minus two times t plus two and we can find the values of t so from this this this handwriting is bugging me t okay cool nope even worse all right anyways t is equal to two and t is equal to negative two however we can't just say either one of these for our answer it says in the problem that t is greater than zero and so since t is greater than zero t is equal to negative two doesn't work and so we're going to have to answer this with t is equal to 2. and so you're just going to you know bubble in a 2 for your answer so moving on to problem 25 here this is the last problem for set of five and this one's almost just as quick honestly as the last one this problem says graphs of the function f and g are shown in the x y plane above for which are the following values of x does f of x plus g of x equals zero so what does this mean again this notation i think a lot of people can get caught up with this f of x is talking about the y value of the f graph and the graph of f is this one right here it points to it this arrow right here the graph of g is this one right here so we're asking well when do the y values of each of these graphs cancel out so when do they add to be zero and if they add to be zero they need to be on opposite sides of the of the x-axis they need to be so if one is at four the other one needs to be at negative four that way when you add them together you get zero so where on the graph do you have that well it's kind of sticking out right in front of you right here they're on the opposite sides of that x-axis so here we have one at 2 1's at negative 2. and so at this point we're at let's see so this is where we're looking at this would be x equals negative 2. it's hard to see the the y-axis here so this is the y-axis and therefore this is negative one this would be negative two so and i can write this out with notation just to get you feeling more comfortable with it if we are looking at x equals negative two we know that f of negative two that's the y value for the f graph that's negative two and the y value for the g graph that's this one at x equals negative 2 is a positive 2. and so these indeed equal 0 when you add them together so remember what the problem's asking for it's asking for the value of x where this occurs and we already figured out that that happens at x equals negative 2. so the answer we circle is choice b and so guys the the time has come we're halfway done with this video at this point we've done the first 25 problems and now we have i mean it's all downhill from here and i'm also saying i'm saying that for you but i'm also saying that for me because like now i only have to record 25 more problems like i'm just sitting here in my room anyways so take your break here just especially i mean hey reward yourself take a 10 minute break this time and then let's get back in and finish this video up so we are going to start with problem 26 here this problem is pretty quick it says the complete graph of the function f is shown in the xy plane above for what value of x is the value of f of x at its minimum so we're looking at where f of x is at its minimum remember what f x actually means it's talking about the y value of the graph that we're calling f and that's this guy right here we're calling this f so we're asking where is this y value at its minimum where is it the lowest and that's happening right here the answer is not negative 2 because that's not what this problem is asking for the problem is asking for the value of x where this happens where it's at its minimum and that value of x is going to be at negative 3. so don't get caught up there you can see that they're baiting you for that negative 2. so that was pretty quick and moving on to problem 27 here this problem says according to the system of equations above what is the value of x and we've done a couple problems at this point involving a system of equations so if you're really getting comfortable with this if this video is actually helping you then you should be able to solve this problem you know pretty quickly you don't have to speed through it or anything but you should be able to kind of get the hang of this so let's look at this problem how would we solve it well i did show a couple ways to solve systems of equations using elimination method so maybe it's time that i talk about substitution method this is another way that people like especially when they're first learning how to solve systems of equations like i always went with substitution method so what we can do here if we're trying to solve for x we can find a substitution for y that way the y's cancel out when we substitute so what we can do is take this equation here this x plus y equals negative nine and we can solve for y and we can solve for y by subtracting x on both sides so if we do that here we get y is equal to a negative 9 minus x i'm going to write that as a negative x minus 9 just because that looks more normal to me and if i've done that i can substitute that y in for the second equation and doing that i get x plus a 2 times y is now negative x minus 9. that's equal to negative 25. so i can distribute that 2 through distributing the 2 through to the negative x i get negative 2x distributing that 2 3 to the negative 9 i get negative 18 and that's equal to negative 25. combining like terms here i'm going to get a negative x and then honestly i mean i don't want this negative 18 on this side so what i can do is add it over so these two you know these like terms still combine to be a negative x but now the 18 is gone and this is all just equal to whatever negative 25 plus 18 is which is negative 7. and how do i know that well just think about this as 25 minus 18 that's 7 but the larger number here was negative so we put a negative on it okay just making sure we're all on the same page with that so we can divide each side by negative 1 here and that's going to give us the value of x so x would equal 7. so 7 is going to be your answer for the second problem for this set of 5. so moving on to problem 28 here we have the in the xy plane if zero comma zero is a solution to the system of inequalities above which of the following relationships between a and b must be true so we're given the system of inequalities and we're given that zero zero works for it so if 0 0 works here let's plug it in and see what we get if we plug in 0 0 into the first inequality plug in the 0 for y and plug in the 0 for x well negative 0 that doesn't make sense i'm just going to put a 0 there and that's going to give me 0 is less than a now just do it for the second inequality y is greater than x plus b if i do that i'm going to get 0 is greater than 0 plus b and simplifying that 0 is greater than b so now i have two things here i have that 0 is less than a and i have the 0 is greater than b so how do i get a relationship here between a and b what connects these two inequalities well what connects them is the fact that they are both showing a relationship to zero so if they both have a relationship to zero then they have a relationship to each other i'll show you what i mean we know that 0 is greater than b that's the second inequality here but we also know that 0 is less than a and how do we write that here well we can write this inequality backwards because we can read inequalities backwards we can write this or we can read this as 0 is less than a or we can read this as a is greater than 0. so if we read it as a is greater than 0 then we can write it as a is greater than zero so now if a is greater than zero and that's greater than b well a has to be greater than b then right that it wouldn't work for a to be less than b because a is greater than zero and 0 is a number that's already greater than b and so here we get that a is greater than b which is choice a so that is our answer for problem 28 and now we can move on to problem 29 problem 29 we're going to get into some geometry here we have the two isosceles triangles are shown above if 180 minus z is 2y and y is equal to 75 what is the value of x so we know that y is 75 we also now have an equation to find z z is not also 75 because these are not vertical angles they would be vertical angles if we had two lines that were intersecting then this angle would be equal to this angle and this angle would be equal to this angle but that's not what we have here because this isn't just one line this is two lines that are intersecting they're different right you can see it's bent here so that's not just a straight line so we're going to have to use that equation to find z we can do that very quick so y is 75 we can plug that in you guys like how i'm writing my z's isn't it cute writing a little cursive z's nobody cares all right so this is 150 and then i can what am i going to do here let's uh let's just subtract 180 i guess i was thinking about doing something else but no i'm going to have to have the negative here anyways negative z is equal to negative 30 and we can divide by negative 1 on both sides and get that z is equal to 30. so now that we've found z we're getting closer to finding x now and there's one more thing we have to realize because we have to make a connection here between z and x and so what we need to realize is that this is an isosceles triangle and what that means i mean since these two sides are the same length it shows that right there with the two bars that means that their corresponding angles are the same length and these angles matter because this angle right here is supplementary to this angle right they're supplementary angles which means that they add to be 180 degrees so if we know this angle we can find x and that angle right there let's just give it a name we'll call it w and now i mean this angle would be w2 because it's exactly the same so all of these angles right here are going to add to be 180 degrees because well that's the interior angle sum of a triangle so we have two w's there plus z is equal to 180. so 2w plus z is equal to 180 and so we know z is 30 so i'll just substitute that in there and we can subtract that on both sides so you get that 2w is equal to 150 and dividing by 2 on both sides that gives you the w is 75. and now that we have the value of w which is this guy right here we can find x they are supplementary angles so that means 180 or well that means they add to be 180 so i'll write that w plus x is equal to 180 and we know the w 75 and we can subtract that on both sides the 75 180 minus 75 is 105 and that gives you the value of x this is yellow yeah okay now before we move on i gotta note one thing about this problem like i'm showing you all the work right now when you're on the sat you don't have to like show all this if you're comfortable not showing it like for example right here i would have done all this in my head i would have been like i if i would have seen this y is 75 right away i can already tell the z is 30 because this is going to be a 180 minus z equals 2 times 75 which is 150 you can already see here that z has to be 30 and you can do that little bit of work in your head so you don't have to show everything i just showed so you might be able to do that part in your head there's also other parts of this problem that you could do a little bit quicker and not have to show as much work as i did like you don't need to write out w plus w plus z if you already see that's going to be a 2w write the 2w you know what i'm saying so it all depends on the person i'm showing out everything you don't have to show out all of this if you can already do some of it in your head with that being said let's move on to the last problem for this set of five we'll move on to problem 30 and this one has to do with trig so this problem says in a right triangle one angle measures x where sine of x equals four over five what is cosine of 90 minus x so this whole sine of x cosine of 90 minus x thing this is something that they'd love this on the sat they'll put this like once per test it's pretty much about that frequent this isn't something that you really talk about much in class it's just something that you kind of got to figure out when you're on the test sine of x is actually equal to cosine of 90 minus x i want to prove that to you so let's just think about some of these trig stuff you if you don't know your unit circle it's fine like you'll get through this with me if i make x equal to 30 i get sine of 30. i'm going to show you that's equal to cosine of 90 minus that because this is sine of 30 this is cosine of 60 and sine of 30 is one half cosine of 60 is also one half so that works and that's one example of where sine of x is equal to cosine of 90 minus x you can also have sine of 45 that's equal to cosine of 45 or sorry 90 minus 45 because 90 minus 45 is 45 and we know that both of these are rad 2 over 2. so since you now know that sine of x is equal to cosine of 90 minus x if you're given that sine of x is four-fifths then cosine of 90 minus x is also four-fifths and that's going to be your answer for this problem and so that's going to do it for another set of five problems and now we've done 30 problems so far in this video we've got another 20 problems to go and these 20 problems are pretty much all going to be calculator problems in the beginning of this video we did a lot of no calculator stuff with some calculator problems sprinkled in but from here on out it's going to be solely calculated problems and so these problems they're going to have some more geometry in it there's going to be some graphs that we have to interpret and stuff like that so you're going to be seeing a lot of different problem types from here on out okay so assuming that you've taken your break remember to do that assuming that you've taken your break and you're in a good head space to get into these next set of five problems let's get into this so we are going to start off here with problem 31 and this problem says a food truck sells salads for 6.50 each and drinks for two dollars each the food trucks revenue from selling a total of 209 salads and drinks in one day was 836 dollars and 50 cents how many salads were sold that day so we're asking for how many salads were sold and we're given some information about salads earlier on if we're trying to find salads i'm going to give that a variable i'm going to call that x and the only reason why i'm trying to do this is because i don't see an obvious way to like solve this problem in five seconds right so i feel like we're going to have to go and work with some equations here so starting off i'm naming solids so i also just reading through this word problem i do see some information about drinks right with salads so i think it's only right that we give drinks its own variable as well we'll call the number of drinks that we sell why and now that we have that we can actually start organizing the information that it gave us in this problem first we have that a food truck sells a salad for 650 each and drinks for two dollars each so we know that 650 for every salad plus two dollars you can write as two with you know the decimal points and all that or you could just write it as 2y that's how i'm going to write it because i think it clears things up a little bit i know that that's equal to the total amount that they sold that day which is 836 dollars and 50 cents because this right here the 650x that gives me the amount that they sold from salads the 2y that gives me the amount that they earned from drinks so then in total if i add these two things together i get the total amount that they made off of salads and drinks so is there any other pieces of information in this problem well yeah because what about this 209 well that's how many salads and drinks they sold in that day and that's actually good that we get that other piece of information because right here with this equation we're not going to be able to solve for x remember if we have two variables in an equation we need two equations to be able to actually solve for one of those variables so what we can do is say well okay we know x plus y the number of salads that we sold plus the number of drinks that we sold is equal to 209 and now from here we can solve a system of equations so you can solve this with elimination method personally i'm going to solve this with substitution method here and i'm going to start by solving for y so i'm going to subtract x on both sides here i'll solve for y in this equation and then we'll plug that in to the y and the other equation so subtracting x on both sides here i'm going to get that y is equal to a you can write that as 209 minus x i'll write that as negative x plus 209 and now we have y solved for we can plug that into the top equation doing that we're going to get 650x plus 2 times y is now negative x plus 209 and that's equal to 836 dollars and 50 cents so now that we have that we can do a little bit of distributing with that 2. 2 times negative x is negative 2x and 2 times 209 is 418 and that's equal to 836 dollars and 50 cents so we can combine like terms here and also you know at the same time we want to get everything that doesn't have an x over to the other side and so i'm going to do all of this at once because i don't like writing that much so i'm going to subtract that 418 to the other side and i'm going to combine like terms so on the left hand side since i don't have that 418 anymore all i have left is the 650x minus 2x that's going to be 450x and i know that that is equal to 836.50 minus 418 which you can do in your calculator if you want you might also see right away though that that's 418. 50. so you know it's up to you if you want to use your calculator for some of this this is a calculator problem but at this point i'm going to divide each side by 450 and this i definitely am going to use my calculator for but i'm kind of cheating because i already know what the answer is here so i already know that if we divide these two in our calculator we're going to get 93. and 93 is what it's x but what is x x is the number of salads that were sold that day and that's exactly what we're asking for here and so we can safely say that the answer to this problem is choice b so moving on to problem 32 this is the start of those geometry problems i was talking about this problem says in triangle abc the measure of angle b is 90 degrees bc is 16 and ac is 20. triangle def is similar to triangle abc that's an important node right there where vertices d e f correspond to vertices a b c respectively and each side of triangle def is one-third the length of the corresponding side of triangle abc that's another really important node right there then it asks us what's the value of sine of f okay so i know that probably seems really overwhelming we were given a bunch of side links we said okay these triangles are similar and now we're asking for like trig stuff right let's unpack this for a second um first off i know that i have this right triangle here triangle abc and it tells me right in the problem that this is a right triangle b is the right angle so now that we have that triangle let's talk about the other triangle triangle def and we know what that triangle is similar that means it has all of the same angles but its sides are 1 3 the length of that of this triangle the triangle abc so what i'm going to do i'm going to make the exact same triangle but just a little bit smaller a little bit bigger than that honestly i don't want and i want you to be able to see it still and so this is going to be def so there you go great now we know a couple things about these the sides of these triangles we know that bc has a length of 16. we know that ac has a length of 20. and what we're trying to do is find sine of f so we're trying to find sine of this angle here and we know from sohcahtoa oh actually that's probably a great place to start in sohcahtoa the so part that tells you about sine sine is opposite over hypotenuse and well looking at the angle f here the triangle def what side is opposite to f well it's d e so this is the opposite side that we need and where's the hypotenuse of triangle def well the longest side is over here it's this guy right here so now we have the opposite side and the hypotenuse labeled and so all we need to do is find the lengths of these sides but how do we get there well first off it's pretty easy to find the hypotenuse of this side since we know that each side of triangle def is one-third the length of the corresponding side of triangle abc so that means the hypotenuse isn't 20 it's 20 divided by 3. and if you don't see that right away you're not sure where i got that from just multiply 20 by 1 3 and that's equal to 20 divided by 3. so if we're on the same page there then the only thing left to do is find the opposite side of triangle d e f right the opposite side of angle f rather and so for that we need the length of side a b we don't have that yet we can find that using the pythagorean theorem since this is a right triangle or if you want to be a little bit quicker if you know your pythagorean triples you know what the other side is because let me go through the pythagorean triples real quick so there's two main pythagorean triples that i would say to memorize for the sat and honestly the 345 triangle that is the most common that's what this one is it's a three four five and you i mean you don't see it yet i'll talk about that in a second another one to memorize is five twelve thirteens that's another one and so these are just common triples that you can see so for instance if this side is three and this one's four you know that this one's five if you didn't know one of these sides you could recognize oh well the other side must be a three then so we have a three four five triangle so that's what pythagorean triples are all about now if you know these you can do a lot of the pythagorean stuff on the sat in your head and you might be like though okay i can't draw the connection though how are you talking about three four five triangles and you're saying that's going to help me here when our side lengths are like 16 and 20. well that's because all the multiples of your pythagorean triples will also work so if i just multiply all these guys by 2 another pythagorean triple is 6 8 10. multiply these by three another pythagorean triple is 9 12 15. and if you multiply by 4 here you get the one that we have which is a 12 16 20. so these are all pythagorean triples if you see this three four five ratio which i kind of saw here i saw that we had you know the four and the five piece so right now we need the three piece we need the 12. so if you see that ratio then great you're you're skipping a little bit of pythagorean theorem and that does speed you up a little bit if you don't see that it's not the end of the world just calculate it using pythagorean theorem and you still are going to get the right answer so if we know that side a b is of length 12 then we know what the length of the opposite side is it would be this divided by 3 12 divided by 3 is 4. and now that we have that sine of f is easy to calculate sine of f is going to be your opposite side which is 4 divided by your hypotenuse which is 20 divided by 3. and remember how we do division of fractions that's all this is just division of fractions when we want to divide two fractions think about this as a four over one divided by 20 over three it's that stay change flip keep change change whatever you call it all we're gonna do is keep this one change the sign to your multiplication and flip this fraction and all that equates to you don't need to think about it like this every time all you need to do is take this fraction that's in the denominator bring it up to the numerator and just flip it that's all you have to do so if we do that we're going to get a 4 times flip this fraction it's going to be a 3 over 20. notice how that's the exact same thing as the whole state change flip keep change change thing i was talking about earlier so if you multiply this the 4 times 3 is 12 so you get 12 over 20. and if we keep going here sine of f is equal to 12 over 20 we can simplify that just divide by 4 over four right divide by a sorry probably should write it like this you're dividing four on top and bottom so what you get is a three over five so that is sine of f right there it's three over five and yeah that's that's just our answer here there's no multiple choices to circle so moving on now to problem 33 this problem we're given a graph here and it says michael swam 2 000 yards on each of 18 days the scatter plot above shows his swim time 4 and corresponding heart rate after each swim the line of best fit for the data is also shown for the swim that took 34 minutes michael's actual heart rate was about how many beats per minute less than the rate predicted by the line of best fit so i did take a little pause there i was going to say let's just find that already but that swim that took 34 minutes let's go find that on the graph so swimming time is right on the x-axis so we're looking at whatever point is right on this line and i'll zoom in so you can see it a little bit better we're talking about this point right here that's the swim that took 34 minutes and what are we asking about that swim i'll keep that highlighted we are asking how many beats per minute less than the rate predicted by the line of best fit so beats per minute is right here we're asking how much lower is it than the line of best fit the line of best fit would have said that the beats per minute would be at 150. well what is it at right now look at your y-axis don't make a silly mistake these increments are going up by twos so we're going like 142 144 146 148 and we're at 148. so we are two lower than what the line of best fit would have suggested we would be at and so b is going to be our answer here so moving on to problem 34 here we have a table and this problem says the data in the table above were produced by a sleep researcher studying the number of dreams people recall when asked to record their dreams for one week group x consisted of a hundred people who observed early bedtimes and group y consisted of a hundred people who observed later bedtimes if a person is chosen at random from those who recalled at least one dream what is the probability that the person belonged to group y so again this is another word problem where you could literally read the whole thing and then just forget anything that you just read so what we need to do is first off identify the actual question so let's read the whole last sentence because it's a little wordy if a person is chosen at random from those who recalled at least one dream so we are choosing a person at random from those who recalled at least one dream we're not talking about anyone who didn't record any dreams we're talking about just the people who recorded one or more dreams and we're saying well what's the probability that if we pick a random person out of that group that that person belonged to group y so i think it's very easy to get messed up with what this probability is actually going to be so i think it's best to just write it out the probability i'll call that p is going to be well first off what's the number of people that we're picking out of we are picking out of the total number of people with one or more dream recalled that's our denominator because that's the entire number of people that we're picking out of and well what's the numerator that's going to be the group that we're trying to look at here we're talking about group y and it's not going to be the total number of people in group y no that's not what we're talking about we're only saying we're saying hey we're picking out of the people who have one or more dream that is recorded so the numerator is going to be just the number of people in group y with one or more dream recorded okay so that's how this probability actually looks and if you don't get this part then you of course are not going to end up getting the answer and i think a lot of people end up making silly mistakes on problems like this so that's why i thought it'd be good to write this out so let's find these numbers now how many people in group y had one or more dream well in group y eleven recorded one to four in 68 recorded five or more so the numerator is going to be 11 plus 68. and the denominator that's the total number of people with one or more dream that's going to be 39 plus 125 okay we're not gonna count that 36 because they didn't record one or more dream that's the p that's the people that we're picking out of so again the denominator is going to be 39 plus 125 and so the probability here is going to be 11 plus 68 that's 79 divided by 39 plus 125 that right there is 164. and i didn't actually do that in my head i just looked at the answer because i knew it wasn't gonna be a hundred so i cheated i guess but yeah so that right there is the answer for problem 34. and now moving on to the last problem for this set of five this problem says of the following four types of savings account plans which option would yield exponential growth of the money in the account so this problem's asking which of these guys is demonstrating exponential growth so i think a good place to start for this problem would be well okay what's exponential growth we probably should talk about that first and well the best way that i can describe it is in in this case we're talking about money exponential growth is going to be when you are making money like let's say every year based off of how much money you currently have in your account so let's say that you've had the money in for three years well that account has grown by eight percent every single year and so now let's say you put in a hundred dollars it's not going to be 100 anymore maybe it's like 120 so you're gonna make eight percent off of 120 not a hundred so does that make sense that's what exponential growth really is you're gaining by a certain percentage every single year so let's look at these answer choices here choice a says each successive year two percent of the initial savings is added to the value of the account well that's exactly what i just said that exponential growth was not because we're talking we're not talking about a percentage of the initial savings we want a percentage of how much money we have in the account right now so choice a is not going to be it choice b says each successive year 1.5 percent of the initial savings and 100 is added to the value of the account again no we're not talking about a percentage of the initial savings here okay because if we have if we throw in a hundred dollars okay let's say that we make you know two percent of the initial amount every year well two percent of a hundred is two and so that means that every year we'd be making two dollars and the amount that we make won't increase it's just going to be two dollars every single time okay and even if we add in an extra 100 then we'd be increasing by 102 every year it's still the same exact amount so choice c says each successive year one percent of the current value is added to the value of the account so now we're taking a percentage of the current value and that is exactly what exponential growth is it's a percentage of the amount that we currently have in there and choice d says each successive year 100 is added to the value of the account that's just like linear growth okay that's another example of linear growth we're just adding 100 every year that's definitely not exponential so the answer to this problem is definitely going to be choice c so that does it for another set of five problems here and so i mean i don't know i think i've been recording for like 30 minutes here uh just on these like last five problems so i assume that we've been going here for like maybe 20-25 minutes so at this point definitely take another break take five minutes go stretch get some water do something and then i'll be here when you get back let's go through another five i always got to make sure i'm giving you these reminders because i don't want you guys to just binge this and i mean if you're starting to binge this you'll probably get to like problem 15 right and then just be like for the rest of the time you know so i don't want i want you to be refreshed after every set of five that's why i keep doing this i feel like i'm justifying why i'm giving you guys breaks every single time that there's a break i don't i don't know anyways let's move on to problem 36. so problem 36 says that the sum of three numbers is 855. one of the numbers x is 50 more than the sum of the other two numbers what is the value of x so first off we have three numbers that add to be 855. okay that's just three numbers i'm going to call that x y and z and what we know is that well one of the numbers this x here it's fifty percent more than the sum of the other two numbers so x is fifty percent more than y plus z so how do we write that how do we write this relationship between x and y plus z well what we can do is say that x is 150 of y plus z and i still have not given myself enough room here now you might be very confused why did we put 150 percent there where does that come from we're talking about 50 here why is it now 150 let's let's think so if i have a number w okay let's say i want a hundred percent of w that means that means i want all of it a hundred percent is just one and so this is the same thing as w it's the same thing as one times w which is w from here that's where i apply the increase because if i just said 50 then i would be cutting this in half that's not what's going on here i'm increasing this value by 50 which means i'm adding 50 percent to the one that's like technically out in front of the w you get what i'm saying so if i increase by 50 now all of a sudden we're not getting w we're getting 1.5 times w so that is why i'm putting 150 percent there this is the increase of 50 that's why you always see a 1 there that's why that one always magically gets added it's because we have to start from a hundred percent and increase or decrease from there so 150 percent you can rewrite that in decimal form as 1.5 and now you actually have a substitution because you know that if x oops yeah let's just erase the entire thing if x is 1.5 times y plus z then i can just replace that in here i know that 1.5 times y plus z plus y plus z is equal to 855. see i just plug that in for x because that's what this equation tells me to do okay so at this point i know you might be like all like okay we have so many y plus z's floating around here this is really gross i agree there's a way that we can make this like look a little nicer okay because we keep talking about the sum of these two numbers we don't care what the value of y or z is we care about what their sum is so if we only care about the sum then let's just call that a single variable that's a gross z gotta do better than that there we go let y plus z equal let's call it b because now things clean up and i think it becomes a lot clearer to you how this actually works and this is something you can do like if it just you know you see this y plus z repeating all right let's let's clean that up a little bit so instead of 1.5 times y plus z now we just have 1.5 times b and then this right here the y plus z again that's just b and that's equal to 855. so adding these two together a 1.5 b plus b that's going to be 2.5 times b because remember this is just a 1. and so to find the value of the sum just divide by 2.5 on each side and you're going to get that b is equal to actually i don't know what you're going to get where's my calculator where's my calculator i actually don't know where my calculator is all right let's cut guys i found my calculator it was just my room is a mess okay i'm not even gonna talk about it so if we do 855 divided by 2.5 we get that b is 342. all right great calculator part done and no it's not actually because this is not our answer here that's just the sum of the two numbers we know that x is 1.5 times the sum of the two numbers which remember we called b so we have to multiply this by 1.5 still x is 1.5 times the sum and so it's 1.5 times 342 and that will be equal to i don't know why i'm putting this in my calculator i actually know what the answer is this is going to be 513. there we go so yeah that's the answer for this problem and i should note just before i move on because i'm going to feel guilty if i don't mention it there was one guy when i made a a tick tock video on this problem who gave me just another way to think about this problem and i actually think i like it a lot so you might be able to see here that x is going to be 60 percent of a 55 and how would you be able to see that well because when you're talking about the sum of the numbers that's 100 of the sum here when we're talking about x that's 150 percent of the sum and you add those two things together as we went through the problem and that's how we got the 2.5 right so what we want there we don't want 2.5 we want 1.5 we want 60 i don't know if you can see that but if you divide like here this is equal to 60 so the value that we want we want x it's 60 of that 855. so if we just multiplied that 855 by three fifths or sixty percent you would end up getting 513. so i don't know if you can see that if not no big deal just make sure you understand how we solved it here and yeah let's move on to the next problem moving on to problem 37 here so in this problem we are dealing with some angles it says the angles shown above are acute and sine of a equals cosine of b if a is equal to 4 k minus 22 and b is equal to 6 k minus 13 what is the value of k so the big piece of information here that you might not know what to do with is that sine of a equals cosine of b what does that mean for us here well remember we talked about this earlier i think it was problem 29 maybe or no it was problem 30 we talked about how sine of x is equal to cosine of 90 minus x okay this is something that you need to know because again here's another example of how it can show up on the sat when you have sine equal to cosine that means that the angle measures will add to be 90. wow losing my voice there if you add these two angle measures together you get 90. okay and that should actually kind of make sense to you if you know your trig a little bit because remember how like sine of 30 that's a half and if you want to get another half i mean that's cosine of 60. right these are both a half and in this case again their angles add to be 90. so i mean if you want to think about it this way you can or just start writing some examples of where sine is equal to cosine you're quickly going to see like for another example sine of 45 is equal to cosine of 45. we could also have that sine of 0 is equal to cosine of 90. they're both zero right in all of these examples the two angles had to be 90 degrees so if you can see it this way that's great if you can see it this way that's great but what that means for us is that a and b they add to be 90 degrees so since we know that a plus b is equal to 90 we can write this as 4k minus 22 plus a 6k minus 13 is equal to 90. combining like terms here we're going to get a 10k uh do we have a 22 and a 13 that's going to give us a 35 and it's negative so this is a calculator problem here just so you know so we might need our calculator in a second here i think all the problems at this point are calculator going forward yeah 10k is equal to 125 and dividing by k on both sides i guess you don't need a calculator for this you're going to get the k is equal to 12.5 and that right there is choice c so great now that we have the answer to problem 37 we can move on to problem 38. so problem 38 says which of the following is an equation of a circle in the x y plane with center zero comma four and a radius with endpoint four over three comma five so first off let's talk about the equation of a circle because if you don't know the equation of a circle you're not going to know how to do this problem so the equation of a circle is x minus h squared plus y minus k squared is equal to r squared the important pieces that we need to fill in here are this h this k and we need to fill in this r the center of the circle is h comma k and it has a radius of r so i mean if you know that the problem's really not that bad to figure out from here we know what the center is it's zero comma four and so writing out the equation here we're going to get a x minus zero squared plus a y this is a going to be a minus right we still have that minus there that's not going to go away we have k is 4 and that's going to be equal to r squared so simplifying that a little bit this is just an x squared and then we have the y minus 4 squared and we don't know the radius yet so looking at this now and just comparing to our answer choices we can see that it's not choices b or d because they have a y plus 4 there and that's where they're trying to get you they really do try to get you with remembering this minus sign that minus sign is extremely important because like let's say our center was zero comma negative four instead a lot of people will still say it's y minus four squared and that's just not true you can think about it as always doing the opposite of the sign so in this case we'll get x squared plus and then we'll have a y minus k now is a negative 4. so we plug that in that doesn't count for this minus that minus is still there k is negative 4 that puts in another negative and now we get a positive right when you see two negatives like that you can just make them both positive and so you get a y plus 4 squared out of that so that's how we could possibly get a y plus 4 squared would be if k was negative 4 instead of a positive 4. but here it's a positive 4 so we got the y minus 4. so from here we need to find a radius but honestly you might be able to look at the answers and already guess what the answer is going to be the reason why is because a common mistake that a lot of people make is that they don't square the radius what goes out here is not the radius it's the radius squared so i'm willing to bet you that this is the radius and this is the radius squared so that the answer is definitely choice a and this is a trap so let's find out we could find out we know one of the endpoints of the circle so we can use the distance formula because i mean just drawing a picture here what they gave us they gave us the center of the circle and they give us a point on the circle so if we determine that distance which we'll call d here i will d is r that's the radius of the circle so actually i'll call it a i'll call it r so the radius is going to be the square root of remember that distance formula this is going to be x minus i don't know you might call this x2 minus x1 squared maybe that's what you call it in class it's the distance between the two x points squared plus the distance between the two y points squared and you have to square root that entire thing so the radius here is going to be let's pick what we're going to call our second point in our first point let's call this our second point and this our first point so for our x's we're going to have a four-thirds minus zero and then we're gonna have a five minus four there we go and so we can simplify that r is going to be the square root of this is going to be a four thirds squared which if you want to square a fraction you just square the numerator and square the denominator so it's going to be 16 over 9. right just 4 square root of 16 3 squared is 9. and then this is a 1 squared that's 1. so what we get under here is a 16 over nine plus one you can write that one with you know a common denominator you can write it as nine over nine just you know multiply this by a nine over nine and you end up getting that the radius is 25 over 9 and uh well square root of 25 over 9. so that gives you the radius if you square root that you can just as we we could square something a square fraction by just squaring the numerator and squaring the denominator if we want to square root a fraction we can square root the numerator and square root the denominator so this would be square root of 25 is 5 and the square root of 9 is 3. so it's equal to five thirds but remember i said this was a trap what we want here is not the radius we want the radius squared and you don't even need to find the five thirds in order to find the radius squared for the radius squared just square the square root and get rid of it so if you square that cancels out with the square root you get 25 over 9 is your r squared and that is what you're going to be using for the equation so you get your final equation as x squared plus y minus 4 squared remember that's what we had before is equal to the r squared and r squared is 25 over 9. so after all that we figured out that the answer was exactly what you would have expected it to be it's choice a so that is the answer for problem 38. so moving on to problem 39 this problem says mr cole has a beaker containing n milliliters of solution to distribute to the students in his chemistry class if he gives each student three milliliters of solution he will have five milliliters left over in order to give each student four milliliters of solution he will need an additional 21 milliliters how many students are in the class so aside from the fact that you read a word problem that you might have just forgotten everything about there is a quick way to just solve this problem in your head i will talk about that after i solve it the slower way because i think just if i solve it the slow way and everybody can see that way then it's going to make it very easy for you to do it in your head to think about how you would have done it in your head so let's talk about the i guess slower way first so the question here that we're asking that's the first thing we need to think about when we go through a word problem is how many students are in the class so we want to find that so i'm going to give that a variable we'll call it x x is the number of students and now that we have that written out let's just start going through the information in this problem first off we're given that his beaker contains n milliliters and it says if he gives each student three milliliters of solution that means he will have given out three milliliters for every student so that's three x because if he has two students he'll give out six milliliters if he has three students he'll give out three times three milliliters which is nine so three x represents the total amount of solution that he gives out and it says if he gives out that three milliliters of solution per student he's going to have five milliliters left over so that means that the amount that he has the amount of solution that he has in total is going to be the amount that he gives out plus the five that he has left over and now that we have that we can use that exact same method to make another equation for the next set of information the next set of information says in order to give each student four milliliters of solution so now he's giving out four x milliliters because it's four for every student in order for him to do that he's gonna need an additional 21 milliliters so let's think about how we would actually write that in an equation the total amount of solution that he has he doesn't have enough to give 4 milliliters for every student he doesn't have that much he needs an additional 21 milliliters so the total amount that he has isn't the 4x it's 4x minus 21 because after he gives out that 4x he still needs 21 to actually give everybody 4 milliliters a big mistake that i could see people making here is actually doing plus 21. but that's not true the total amount of milliliters he doesn't even have 4x so 4x cannot be equal to n it's got to be something less than 4x and it says here that he needs an additional 21 to do the 4x so if we subtract 4x from 20 or we subtract 21 from 4x we'll get the amount that he currently has so that's why it's minus 21 and not plus 21. but if you have that it's very easy to solve for x here which is what we want we know that both of these things are equal to n and if they're both equal to the same number that means they are equal to each other so from this got another email during class or during my during this youtube video that's great all right so we can set them equal to each other and now we just get all our x's to one side i'm going to get the x's to the right-hand side and i'm going to get everything that doesn't have an x to the left-hand side and you'll notice why i did it kind of opposite right we usually get all our x's to the left-hand side but i noticed if i get my x's to the right-hand side i'm going to have a 4x minus 3x that's just x so that makes my life a lot easier i'm not going to have to deal with any negative garbage or anything like that so then we get 5 plus 21 on the left hand side that's 26. so this tells us that he has 26 students okay not going to make a rectangle for me there we go thank you all right so with that that's choice d so that is a slower way to think about this problem but you honestly could have thought about this a little bit quicker it's not that hard to set up these equations but there is a way you can think about it mentally so think about it he's giving out three milliliters of solution and still having five left over and then he's giving out one more milliliter to every single student and now he's lacking 21. so that means he not only used up the five that he had extra to give one out to every student but he also now needs an extra 21. so in order to give that additional milliliter to every single student he needed these five milliliters and he needs 21 more and so just looking at that you can see that he wanted to give out 26 extra milliliters and if there's one milliliter for every student that means he has 26 students so that is a quicker way to think about this problem hopefully that makes sense just be talking about it mentally i know a lot of people um they like to see it rather than just me saying it but hopefully now that i've actually shown you some of the work maybe you can trace back what i said to the work and see how it's basically the same thing so that is problem 39 now moving on to problem 40. so this problem says the equation above expresses the approximate height h in meters of a ball t seconds after it is launched vertically upward from the ground with an initial velocity of 25 meters per second again that sentence literally went from here to here okay after approximately how many seconds will the ball hit the ground so again looking at this question here we are looking at after approximately how many seconds the ball hit the ground and this is a parabola this is a you know it's a quadratic equation that forms a parabola i don't know if you ever had solved these problems before where you know you throw a ball up in the air it essentially is a problem because that's what happens in real life you get a parabola so from that you can calculate where the ball landed and that's exactly what we're really doing here we're saying after how many seconds the ball hit the ground the path of this ball is going to look something like this right it's a parabola and so where it's where it started and where it landed are the zeros of the parabola so if we want to see after how many seconds the ball hit the ground let's just calculate the distance between the zeros and so how do we do that well the quadratic itself is actually very easy to factor so we want to set that h equal to zero because that's what we're trying to find this is the h axis this is the t axis and zero is going to be equal to negative 4.9 t squared plus 25 t so since both of these have a t in them we can bring a t out we can factor it out and write this as t times negative 4.9 t plus take a t away from here all you're left with is a 25. and this t is really gross so i'm going to replace it so from here there are two things that could happen to make this side equal zero there's two there's two things that could happen to make this equation right here true first t could equal zero that's one solution so this thing starts at t equals zero that's initially when it's thrown which kind of makes sense and why would t equal zero because if t is zero here it doesn't matter whatever this thing is zero times anything is zero so that would make the equation true now what else would make the equation true is if this thing was zero because again then you get zero times some number and that is zero so let's set that thing equal to zero and solve for t let's subtract 25 on each side and then we'll divide by negative 4.9 so here comes the dividing step and this is something you can do in your calculator or you can just see that it's saying after approximately you're trying to estimate here you can see that this is about negative 25 divided by negative 5 which is 5. so you can probably already see that t is about 5. and that is choice d okay so that does it for another five problems now we only have 10 problems left these are all calculator problems from here we've got 10 problems left so go take your break and then you're going to have problems 41 through 45 to get done and while you're taking that break i have a little proposition for you i'm really trying to get this video out there to as many people as i can and so if you can make a tick tock about this video and it gets let's say 25 000 likes then i'll give you three hours of one-on-one tutoring for free just tag me in the video so i can see it and speaking of tick tock i'll probably go try to make a tick tock about this youtube video tomorrow that's when i plan on releasing this video and yeah you're kind of seeing like a future me right now i've already recorded all 50 problems and i've edited them too so like i've been editing for like 10 hours now and so now i'm just recording a few clips that i just want to throw in the video so i'm going to try to make a tick talk about this video tomorrow and i'll try to like you know get some extra traction to it because again trying to help as many people as i possibly can and so assuming that you've taken your five minute break let's hop into the next five problems for this video so we're going to start off here with problem 41 and this problem says that a grain silo is built from two right circular cones and a right circular cylinder with internal measurements represented by the figure above of the following which is the closest to the volume of the grain silo in cubic feet so we're trying to find the volume of this object here and something you need to know about the sat is that it actually gives you the formulas for all of these things that you need to find the volume of so it's got a formula sheet at the front of the booklet and if you look at that you can see that the volume of a cone is 1 3 pi r squared h and the volume of a cylinder is pi r squared h and this one should make sense to you like the volume of a cylinder makes perfect sense because the area of a circle okay that's not a circle the area of a circle is pi r squared and so when you extend it into the third dimension the volume will be the area of the circle times the height at which you you know extended it into the third dimension anyways i'm kind of nerding out here let's actually solve the problem um so the volume here is going to be the volume of two cones because as you can see here we have two identical cones we have a cone here and a cone here so the volume is going to be two times the volume of a cone which is 1 3 pi r squared h and then we're going to add on the volume of the cylinder here which is pi r squared h and just doing that here we're not going to combine all this notation here because remember these are different variables here this r and h is not the same as the other r and h and i'll get through that confusion as i go through here but the volume here we can multiply the 2 in the one-third that's going to give us a 2 over 3. we'll have a pi times the radius of the cones is 5 feet so we'll write a 5 there that's being squared the height of the cones is 5 feet so that's going to be your h and then you have plus pi times r squared h for the cylinder now in the cylinder it has the same radius actually so we're going to keep that with five and then we'll multiply by the height the height of the cylinder is 10. so that means that the volume of this entire object we can simplify this a little bit honestly if you want to before you put it in your calculator it's going to be two-thirds pi this is three fives here so that's going to be 5 times 5 is 25 times another 5 is 125 so this is going to be 2 over 3 pi times 125 and then this is plus this is a 25 times 10 which is 250. so now it might make it a little bit quicker for you to put it in your calculator it's less thumb work that's why i like to simplify as much as possible to be honest for me i would simplify even more than this before plugging into my calculator but i'm a little weird we'll do the 125 times two-thirds first and so that is going to give us if you put that into a fraction this will be 250 over 3 pi actually i could have just told you that anyway and then you're adding on 250 pi here now we'll go and add these together so if you do that multiply your answer by pi fiddling on the calculator here you're going to get a volume of 1047 point which rounds up to a 0.2 and the units on that would be feet cubed so that is the first problem done of this next set of five so that's choice d so moving on to problem 42 this problem says that katarina is a botanist studying the production of pears by two types of pear trees she noticed the type a trees produced twenty percent more pears than type b trees did based on katarina's observation if the type a trees produced 144 pairs how many pairs did the type b trees produce so with this problem you might be able to do it in your head honestly and i can talk about how to do it in your head after i explain i guess more of the longer more algebraic way of of doing it like you know i'll show out all the algebra but what we're going to need to start off with here is we're going to need two variables we need one to represent the pairs in type a trees and we need one to represent the number of pairs and type b trees so i'm going to do p sub a for the pairs in type a trees you could honestly just put an a there and you'll know what you're talking about so i'll say number of pairs in type a trees and then p sub b would be the number of pairs in type b trees cool so if we have that i mean look it even tells us that type a trees are producing 20 more than type b trees did so how do we write that well there's going to be some equation here between p a and p b we just need to know where to put that 20 increase so type a trees are the ones that are producing 20 more and so that means that this one is 20 greater than pb so for these two things to be equal you need to raise the smaller one by twenty percent how do you raise it by twenty percent well you're not going to multiply it by twenty percent like this because then you're going to literally end up dividing this thing by five if you want to increase it by 20 you add it to a hundred percent right because 100 we talked about this in a problem before a hundred percent of the pairs in type b trees is going to be all of the pairs and type b trees what we're doing is we're trying to increase 20 from there so we are not multiplying this by 20 we're multiplying by 120 that's the 20 increase we need to start with all of the pairs and then we increase 20 from there so hopefully that makes sense now if you know that then i mean this problem gets really easy from here this is a 1.2 and then we know that type a trees produced 144 pairs so we can plug that in and then solve for the pairs in type b trees just divide by 1.2 on both sides doing that we're going to get that number of pairs in type b trees are 144 divided by 1.2 that's 120. and that's a blue so that is choice b so you can think about this in your head honestly you don't need to go through the whole equation thing and and look i'm writing this part just so you understand where these variables are coming from you when you go through and do this you're not going to be writing that down you're you're going to be you already know because you created those variables so you know exactly what they are and so unless it really helps you you don't have to write that down you can get to this equation really quick and honestly you might even be able to do that in your head you're like oh well type a produces 20 more than type b that means that if i multiply type b by 1.2 by 120 that'll get me to type a so if i have to multiply this guy to get to type a after if i have to multiply it by 1.2 then if i know type a i would need to divide that by 1.2 and that'll give me the number of pairs that type b trees produced and that's exactly what we did here we divided the number of pairs in type 8 trees by 1.2 so that would be my thought process of doing this in my head and that's how i would pretty much do it okay so hopefully that all makes sense so moving on to problem 43 now this problem says in the x y plane the line determined by the points 2 comma k and k comma 32 passes through the origin which of the following could be the value of k okay so for this problem here we're given a few points we're given two comma k we're given k comma 32 we're trying to find the value of k though and just with this information i don't see how we're going to be able to do that we need some other kind of information because just the fact that it's a line it's got these two points on the line yeah you might be like oh we have two points we can find the slope but what would the slope do for us here if we found the slope with these two we would get what a 32 minus k over k minus 2 that doesn't help find the value of k we don't know what this slope is equal to so we've got to have some other bit of information here that we just haven't gotten out of this word problem yet and we do if you look here it says that the line passes through the origin as well and that means that we have another point and now we can do that slope thing that we were thinking about before but of course you know you weren't thinking about the slope thing i was thinking about the slope thing and just assuming that maybe you would think of it eventually so why should you be thinking of the slope why is that something that i would have thought of well because for this problem if we're given just multiple points in general on a line the slope is just something that should be crossing through your mind oh that's something that we can find and if we have three points like this on the same line well think about it we can find the slope between these two and that's going to be the exact same as the slope between these two it's got to be they're on the same line the slope can't be different you can't have like a slope of 2 on one part of the line and then a slope of 5 on another part of the line that's not how that works the slope has to be the same so we can set up an equation with the slope of you know one of the slopes that we calculate that's got to be equal to another slope and maybe that's how we can solve for k so that was my whole thought process with this that is my whole thought process with this so let's start off finding these slopes here i'm going to start off by finding the slope between these two points remember just in case this wasn't clear the slope is y2 minus y1 over x2 minus x1 that's how you find the slope between two points on a line and if we're using these two points and i'm going to call this one the second point so it goes first here i'll have 32 and the x coordinate would be k and then i have minus minus these are both zeros so that actually makes my life really easy i know that the slope is 32 over k that's why i chose the origin point i didn't want to choose both of these because that's going to be like really gross like i even showed you what it would be before it's not fun so that's one of the slopes let's calculate another again it's the same formula let's instead this time use the two comma k and the zero comma zero i'll make this guy go first again so it's going to be 2 minus 0 sorry not 2 minus 0 k minus 0 because k is the y coordinate so we'll have k minus 0 all over a 2 minus because x coordinate here x coordinate here that's going to give me a k over 2. and now remember what we were talking about these two slopes they are of the same line lines aren't going to have different slopes like that if we're talking about one line then the slope is k over 2 and it's also 32 over k that means that k over 2 is equal to 32 over k and so what we can do to solve for that just cross multiply we get the k squared is 64. right we have a k times k that's where i got the k squared from and 32 times 2 gives me the 64. from here i square root both sides and i get the k is equal to a plus or minus 8. so there we go that gives me the value of k now from here i do want to note the answer doesn't seem to be here we don't have a plus or minus 8. but read the question again it says which of the following could be the value of k it's not saying what are all the possible values of k it's just asking for one of the possible values and we have one illustrated right there we could have a positive 8 be the value of k and so c works so moving on to problem 44 now this problem is involving a system of inequalities it says if the system of inequalities y is greater than or equal to two x plus one and y is greater than one half x minus one is graphed in the x y plane above which quadrant contains no solutions to the system you guys hear my dog barking yeah see i'm almost done with this video i'm i'm on my last leg here where we have six more problems to do and my dog has to bark all right so hopefully he'll stop if he doesn't i'm just gonna have to cut the video but let's just start up let's start setting these inequalities here okay we have first off y is greater than or equal to two x plus one and so if i just graph that here 2x plus 1 it's gonna look about like i mean give it a positive y-intercept there because it's got a positive y-intercept of one it's you know relatively steep it's got a slope of two and we know that we have to shade for these inequalities right we that's part of it we are talking about all of the y's that are greater than or equal to 2x plus one so not just all the y's that are equal to 2x plus 1. all the y's that are equal to two x plus one is right here but let's say that we're talking about like x equals two we're not just talking about this y value here we're talking about all the y values above it too and that goes the same for any x value here so that's why we get a shading hopefully that makes a little bit of sense we're going to be shading all the y's that are greater than this line and now you actually might see an issue here look at what quadrant doesn't have any shading quadrant four and when we are asking for the solution to a system of inequalities we are asking for where the shadings intersect and the shadings can intersect if there's no shading in quadrant four to begin with for one of these inequalities so that means that quadrant four is going to be our answer here but we know that there's definitely no solutions in quadrant four so we already know that that's the answer however if you didn't see that yet let's go ahead and graph this next one now this is something i don't really think you need to care about for the sat to be honest but uh if y is greater than it's not a greater than or equal to and that goes the same for less than if it's a less than and not a less than or equal to you draw that with a dotted line i don't know if you remember that so i'm going to draw this one with a dotted line because that means that we're not talking about where y is equal to one half x minus one so the dotted line doesn't count here this solid line it does count because we are talking about y's that are equal to two x plus one and we're also talking about y's that are greater than two x plus one but here we're only talking about y's that are greater than one half x minus one so it's a little bit different so again we're going to give it a nega it's got a y intercept down here now it's negative and it's gonna be at more of a shallow slope because it's one half so it's going to only rise one for every two that we go over so it's going to look something like this right just i mean i'm not drawing these exact but now we can shade and you can clearly see now this actually worked really well you can see exactly where the intersection's happening you see some intersection in quadrant one right here you see some intersection in quadrant two you see some intersection at the bottom or at the top rather of quadrant three but there's no intersection in quadrant four in quadrant four we only have one little bit of shading we don't have the shading from the solid line and so that further says that oh yeah quadrant four is definitely the answer here all right moving on to the last problem for this set of five and then we're on to our last set of five for this entire video which is which is awesome all right i'm excited too all right so this problem says in planning maintenance for a city's infrastructure a civil engineer estimates that starting from the present the population of the city will decrease by 10 percent every 20 years so first off i mean you see this decreased by 10 every 20 years you might look at your answers to we're going to be making an exponential function here that's exactly what we're going to be doing and so you might already be thinking oh well i know what the form for an exponential function is we'll talk about that later okay but so we should probably start underlining this type of information if the present population of the city is 5000 so that's their starting population i want to get nice underlines for all of this there we go so if the present population of the city is 50 000 which of the following expressions represents the engineer's estimate of the population of the city t years from now so we are trying to find p we're trying to find the population of the city after t years and to do that we need to use the formula that we have for exponential functions it's p equals p zero times one plus r to the nt power and you might be like oh my god i have no idea what any of that means don't worry i will define it for you so p is going to be the population at time t at some time t so you plug in two years okay well p is going to be the population at two years so that's what p is p zero is your initial population and actually a little fun fact for you i don't know if you know this i didn't know this when i was in high school the reason why we write it is p sub zero it's an abbreviation it's like a shorthand for writing p at t equals zero it's the population at your starting time so you know if you start writing this in a more compact way you can say p sub t equals zero to say oh that's your population at t equals zero and that got shortened to p sub 0. so that's where that all comes from i don't know i think that's kind of cool maybe you don't even care but i mean i need to entertain myself guys i've been doing this for like hours so r is going to be your rate and n is your number of compounds this is something we need to go over a lot because everybody gets confused at that point but i think something i need to touch on before we even start talking about the other stuff just first off what i'm naming and it's the number of compounds in this case per year but it's whatever units of time you're talking about so like here we're talking about years so n is the number of compounds per year if we were talking about like hours instead then n would be the number of compounds per hour it depends on the units of time okay so that's just something to especially review before you go into the sat because i know most people they just get confused with that point and so t is going to be time but that's it so now let's just start naming the information that we've been given here first off we were given that our initial population is 50 000. we were given this 10 percent every 20 years okay the rate here we are losing 10 so that's going to be negative 10 or negative 0.1 okay it's got to be negative because we're decreasing 10 percent i should probably underline the decrease part 2. and i'll talk about the n in a second like i said so if we just start making this function here p is going to be 50 000 times that's the p0 part we have 1 plus r r is negative 0.1 people get lost with this because they will say oh well we're decreasing by 10 so there needs to be a 0.1 that goes here but that actually decreases by 90 percent because that would make the r negative 0.9 and think about it after every certain amount of time you'd be multiplying by 0.1 that would be taking 10 percent of this 50 000. and that is a decrease of 90 so hopefully you know you can think about it that way but at least we have this part figured out now so we know that p is 50 000 times a 0.9 and so that already narrows down our choices quite a bit oh yeah okay i did that right so now we need to see if there's going to be a 20t or a t over 20 in the exponent this is where everybody starts getting confused if we're talking about the number of compounds per year it's the number of times per year that we're multiplying by 0.9 it's the number of times per year that we're decreasing by 10 percent and we're not doing that a certain amount of time every year it's actually one time every 20 years so if we did 20 times t that would be 20 times every year we're doing once every 20 years so it's not 20 times t it's a 1 over 20 times t but yeah that is me just trying to explain every single aspect about exponential functions and that brings us to our answer of choice d so that does it for i guess that's our ninth set of five problems for this video so take your final break seriously do it okay because i know you're like oh let me just jump into this next five take a break and then you'll be able to just cruise through these next five okay so assuming that you've taken your break let's hop into this last set of five problems for this video we start with problem 46 and this problem gives us a parabola and it says which of the following is an equivalent form of the equation of the graph shown in the x y plane above from which the coordinates of vertex a can be identified as constants in the equation that that paragraph was one sentence okay so anyways what are we trying to do here we're trying to find what equation will allow us to see the vertex this problem you don't need to actually complete the square at all because remember when we want to find the vertex of a quadratic or of our parabola we complete the square and you don't need to do that here because it's just saying which of these equations will show you the vertex immediately and that's choice d because that's the only one here that's in vertex form remember vertex form is y equals a times x minus h squared plus k and your vertex ends up being that h comma k and i really want to go over this whole thing one more time just to make sure that you don't mess up with this because on the sat you might have to identify the vertex so it might not be this easy where you're just like oh well which one's in vertex form there could be sat problems where you need to identify the vertex there also could be sat problems where you need to complete the square and if you need to complete the square if you want to review that because i don't think we did that actually in all 50 of these problems if you want to review completing the square i have a full youtube video on completing the square i'll put that thumbnail right up here somewhere but if you need to identify the vertex let's talk about that so everybody messes up with this x minus h squared for the same reason that people mess up we talked about this before in problem 38. it's the same reason that people mess up with the equation of a circle because again you have the x minus h squared and in addition to that you also have a y minus k squared in both of those you do the opposite you do the exact same thing for the h in problem 30 in problem 46 where we have vertex 4. so for example what is the vertex here for the x minus 1 squared minus 16. well we can definitely see the k that's negative 16. but what's the h well yeah you might see a negative one here but h is one we do the opposite why do we do the opposite well i can show you oh and and actually just so you know then the vertex would be negative one or sorry it'd be one comma negative 16. but yeah let me show you why we do the opposite again just to make sure especially if you ended up skipping that circle problem now you can see it first off a is one you can see that there's no number out here in front of the x factor there and that you know a it doesn't matter what number out is out there it's not going to change your vertex because your vertex is based off of the h and the k not the a so what number would do we need to put here what number do we need to plug in for h in order for us to have an x minus 1. the minus sign is already there it's not part of the h and so if we want to have an x minus 1 all we need to do is plug in an h of positive 1. and then for k well we want a negative 16 we just plug in negative 16 for k so that's where we get the 1 comma negative 16 from because we're not talking about this minus sign it's not part of h and just to make sure you really got this let's say we wanted an x plus 1 squared in here what would h have to be well it would have to be negative 1. we do the opposite and you can even see that if i just pull this apart here if h was negative 1 then we would get an x plus 1 squared so hopefully that whole thing makes sense to you and that's some some good review of vertex 4. so on to problem 47 this problem says why it can husk at least 12 dozen years of corn per hour and at most 18 dozen years of corn per hour based on this information what is a possible amount of time in hours that it could take wyatt to husk 72 dozen years of corn so we are asking for what here we're asking for the time in hours that it would take wyatt to husk 72 dozen years of corn so let's let t equal time just to make sure we're all on the same page here and the dozens of years of corn that he has husked we'll call that c it's for corn and now well i mean he needs to husk 72 ears of corn i love that word husk i don't i actually don't even know what that means and i grew up on a farm so that's pretty bad but he needs to he he can husk 12 dozen ears of corn per hour at the lowest so at the lowest the amount of corn that he husks is 12 times t because after one hour he would have husked 12 years of porn so you'd get 12 times 1 and that's 12. after 2 hours it would be 24 year 24 dozen years of corn and so on and so forth so that's the minimum now for the maximum amount of corn that wyatt could husk we would then have c equals 18 times t and if he needs to husk 72 years of corn then we just plug in the 72 for c all right so now divide 12 on both sides and you can see that the minimum amount of corn that wyatt would be husking in i i can't i can't do that with a straight face anymore this would be six so this is the at his minimum pace it would take him six hours but at his fastest pace it would take well 72 divided by 8 is 4. you don't need to do that in your head there's a calculator for this so that's at its fastest pace and so what's the answer for this problem we have the t equals 4 and t equals 6. well it's asking for what's one possible amount of time in hours that it would take him and so there's actually a couple different answers that you could put here so you could put anything between four and six because this is the quickest you could do it this is the longest he would do it so actually if you put anything in between four and six that will work so uh for instance i'd put six into the little scantron thing and i would be right if i put four in there i would be right as well if i put five in there i would be right so that's how that one works all right moving on to problem 48 here here we have in a circle with center o central angle aob has a measure of 5 pi over 4 radians so this is a central angle that means that it connects with the center of the circle i'm going to assume that that's about where the center is probably going to be a little bit down there yeah maybe right there we go okay so we have a little central angle here aob and now we know the angle measure is five pi over four the area of the sector formed by central angle aob is what fraction of the area of the circle so we're asking okay we're saying the area that's formed by this central angle so this is the area we're talking about i'm going to switch my highlighter color because this is green section so this area here is what fraction of the area of the entire circle so to know this you need to know a very important thing about circles okay there's a lot of different proportionalities when talking about circles and i'm going to name i'm going to name everything that you need to know so this is the proportion that you need to know about circles i know it seems like a lot like there's a lot going on here but once i explain it to you you're going to be it's not going to be like oh wow that's so new i would have never thought of that it's going to be like oh yeah that's that makes sense and if you know how to solve this problem already it would have been like oh yeah that's how i would have solved it so let's talk about this here this equation is talking about the fractional area that's equal to the fractional angle and that's equal to the fractional arc length and that might that totally does not clear anything up for you so i'll actually draw you a picture and show you exactly what that means in this context so with the circle i've just drawn out here i drew a right angle that's my central angle here for this circle and in this case all of these fractions are one-fourth because i have traced out one-fourth of the entire circle for instance the sector area if i say that this area right here is four then what's the area of the entire circle well the area of the entire circle would be 16. it'd be four times that because again i've traced out one-fourth of the entire circle and so if you take the area here and you divide it by the area of the entire circle which i said was 16 you'll get exactly that one-fourth now what about the central angle here well for the central angle divided by 360 part that's talking about how much angle you've traced out out of the angle of the entire circle here we've taken up one-fourth of the angle of an entire circle and i'll show you that too this is 90 degrees the angle of an entire circle is 360. so again that's one over four it's the same ratio and lastly for arc length and circumference if i tell you that the arc right here measures let's say six what's the circumference of this entire circle well there's four of these arcs on that circle so it'd be six times four which is 24. and so let's divide that the arc length here divided by the entire circumference which we just talked about was 24. that again is 1 over 4. and so now that we've talked about this proportion extensively we've talked about all the different moving parts we can actually use that to help solve our problem here and we're not going to use the arc length or circumference part because it just doesn't mention it in this problem but some problems will here though we do talk about area we do talk about angle that's what we're going to be using so again this problem gives us that the angle measure here is 5 pi over 4 it says that the area of this sector right here formed by that central angle is what fraction of the area of the circle so we're asking for the fractional area that's exactly what i've just circled here the sector area the area of the little sector divided by the area of the entire circle that's exactly what this is in math terms and we know that that's going to be the exact same as the fractional angle the amount of angle that we've covered out of the entire circle it's the same ratio so well let's do this here the central angle is 5 pi over 4 let's divide that by the angle of an entire circle which is 2 pi that's going to give us the fractional area now to divide these two fractions just think about it as 2 pi over 1 then we have two fractions all you got to do bring this 2 pi over 1 up to the numerator and flip it we've talked about that a couple times in this video and so you're going to get 5 pi over 4 times a 1 over 2 pi so the pi is cancel you get a 5 over 4 times 2 which is 8. and so we've traced out 5 8 of the entire circle so moving on to problem 49 this problem says that a dairy farmer uses a storage silo that is in the shape of a right circular cylinder above if the volume of the silo is 72 pi cubic yards what's the diameter of the base of the cylinder in yards so first off big thing here we're asking for the diameter that is extremely important and so we're not looking for the radius because the radius is what we use for the volume speaking of which if you want to figure out the volume for a cylinder it's in the formula sheet that the sat gives you okay right right at the front it should be i think something like that but the formula will be pi r squared h and you can remember that some of you probably do it's just the area of the circle which is the base of the cylinder times the height of the cylinder because you have to bring it into three dimensions so it's not only the area here but it's times that h so that's how i remember it all right so if we know that we know the volume is 72 pi that's going to be equal to pi times the radius is what here well that's we don't know that's what we're trying to find the height though we do know that's 8. and we don't need to really worry about the units here this is a math problem it's not physics so what i'm going to do is i'm going to divide each side by 8 pi i want you to understand why i'm doing that because it might be a little weird here since i have the pi over here and the 8 over here so i want to solve for this r squared so that means i need to get everything that's not an r squared over to the other side it's being multiplied by pi and 8. so if it's being multiplied by both of those things i need to divide both of those things off the eights will cancel and the pies will cancel so that's how i figured to do that so then we get the r squared is 72 2 pi divided by 8 pi the pi's are going to cancel here you can use your calculator if you want but i already can see that this is 9 and the reason why i can see that is because i just think of 80 divided by 8 that's 10 so 72 divided by 8 well that's 8 less than 80. so i'm going to have 1 less 8. so instead of having 10 8 it's 9 8. so that's how i got 9. so if r squared is equal to 9 that means the radius is a plus or minus 3 because you take the square root of both sides and if you have that the radius is plus or minus 3 of course we know that the radius is not going to be negative that's not a real world thing so from there we can see that the radius is 3 and it looks like that's our answer right no no it is not because remember we are talking about the diameter here that's what we need so remember that the diameter is twice the radius so we need to multiply that radius by 2 and that will give us that the diameter is 6 and that's the correct answer here all right so that is problem 49 now we're going to move on to the last problem for this video i don't know who's more excited me or you probably probably me because you still have an sat to take i i now have to edit 50 problems and then i got to put an intro and an outro in i got to put like an ad for my sat course and like i'm probably going to do something else too so that's going to take some time but you have an sat to take and i think that's worse so we are on problem 50 now and this problem says in the x y plane if a point with coordinates a comma b lies in the solution set of the system of inequalities above what's the maximum possible value of b okay let's take the slow here we're trying to find the maximum possible value of b what is b that's probably the first thing that you would ask there well b is the y-coordinate of what well it's a point that's in the solution set of the system of inequalities above so we're asking what is the largest y value in this solution set so how do we find that well honestly just thinking about it at like face value just just trying to get a good picture of this in your head you might be like my dogs are barking like i'm just trying to finish this video oh my god all right so so as i was saying before my dog so rudely interrupted me we're looking for that maximum y value in the solution set so how do we find something like that you might be looking for some different ways and it might not come to you but if we have something that we can graph here let's just graph it then and maybe that will give some light on what to do we're kind of just searching for different things that we can do here and actually if you graph this as a calculator problem you could probably just find the answer that way you probably could i know the graph is a little bit messy with inequalities but you could probably do it but honestly you don't even need a calculator to do this problem and i'll show you why you don't need an exact sketch of the graph i'll just give one right here that'll give you enough information to solve the problem let's think about this first inequality here y is less than or equal to negative 15 x plus 3 000. so we've got some positive y-intercept it's 3 000 in this case which is insane but we have a negative slope so let's draw that so yeah something like that right on some scale it will look like that and how do we shade this well it's all the y's that are less than or equal to negative 15x plus 3000. so we're all these y's right here all the y's that are beneath this line all right great so now that we have that let's look at the other one that's 5x so 5x first off the y intercept there would be 0 and it's going to be something with some positive slope going through the origin and how is the shading on this one again it's less than or equal to so we're going below this line so it's everything underneath here there we go nice and shaded there we go so since this is the y-axis this is the x-axis here we're looking for the highest value in this solution set remember the solution set is where these two shadings are intersecting that's happening right here that's going to be where our highest y value is in this whole shading here all these other y values are lower and you see what that is that's the intersection of these two lines it's the intersection of the lines y equals negative 15x plus 3000. and y equals 5x remember that can happen these are less than or equal to so part of the graph are these lines so if we want to find that maximum y value let's find where these lines intersect let's find exactly where this point is and then just find what the y value is the y coordinate is of that point so we can do that because both of these guys are equal to y so they're equal to each other we set them equal and now let's solve for x i'll add 15x on both sides that'll give me a 20x is equal to a 3000 and then i divide by 20 on both sides when i do that i get that let's see zeros cancel here i have 300 divided by 2. that's 150. so is 150 my answer well don't be so quick there remember what we wanted we wanted the y-coordinate so we need to plug into i think it'd be easiest to plug into the 5x we need to multiply that guy by 5. so it's 5 times x is 150 so y is actually 750. that is the y-coordinate of that point and that is the answer for this final problem for the video so yeah guys that is all 50 problems and at this point i am way too tired to do an outro for this video so i'm gonna leave that to my future self i'll cut to that at some point here uh so i'll cut so that is going to do it for this video and remember if you're looking to review more sct math problems specifically some harder ones i have a whole nother video where i do 25 harder sat math problems and if you're looking for more help with the sat i go through every single topic that you need to know in my full sat math course which i've linked in the description down below lastly make sure that you're subscribed to this youtube channel look you and i we're escaping poverty together you're going to get to go to the college of your dreams and i am eventually going to be able to move out of my parents house all you have to do to make that happen is hit the sub button and share this video with anyone you think it could help alright guys i hope this video helped and i'll see you soon

Channel: Ludus

Views: 141,892

Rating: 4.9725504 out of 5

Keywords: sat math, sat math review, sat geometry review, sat algebra, sat geometry, sat math prep, sat, sat prep, sat math crash course, sat help, algebra, sat questions and answers, sat math study guide, sat math online course, sat hard math questions, sat functions, sat math ratios, probability, ratios, sat algebra and functions, sat math overview, sat math tips, sat math practice, sat math help, sat math test problems, sat test, sat test prep, sat math test, sat practice, ludus

Id: vnF2zqwI8oY

Length: 203min 55sec (12235 seconds)

Published: Sat Mar 06 2021