The Ultimate Year 12 Maths Baseline Assessment – Full Student Test

If you're looking at doing A-Level Maths then this baseline assessment is going to be extremely useful for you so just give me literally two minutes to explain what this is. So what I've done here is I put together a baseline assessment for where you, just to give you an idea of how strong you are going into year 12. So it's all based on GCSE stuff, there's no A-Level content in here at all and again you can use a calculator. When I go through it I'm not going to use a calculator but yeah it assumes no prior A-Level knowledge, it's for the UK exam boards. Now basically what you're going to do is go through this and yada yada yada. At the end I have put a mark scheme so you can just quickly check all your marks obviously and afterwards, after all of that I've given you a little diagnostic profile thing. So essentially you can give yourself an idea of how many marks you've got on each section and then use that to help tailor any revision you want to do before you go into year 12. So again all of this is just GCSE knowledge anyway but then you can kind of just see how strong you are. Obviously you don't have to do the whole thing right obviously but it could be a good idea to see if you want to go through it and there you go again. So it gives you kind of an idea of what your overall score is, what you would want to do. So if you do get between 51 and 60 you can actually start doing the A-Level content which links into the video I made a week or two ago, I can't remember now and so on. So yeah all of that is for you, you can do the whole thing. I've got the link in the description so you can start that now if you really want to but I'm going to go through it and see if I can start doing A-Level maths and I guess you could use this also as like a guide on how to do it. Again I'm not going to use a calculator because I can't be bothered. So the first one says 5 over 6 minus 3 over 8, so 5 over 6 minus 3 over 8. You need to make the denominators the same. You can make them both 24 by timesing this one by 4 and timesing this one by 3 so that gives me 20 out of 24 minus 9 out of 24 and then we can just subtract the numerators. So for the first one it'd be just 11 out of 24. For this one when we divide is keep flip change, so we're going to do two thirds, change the divide to a times and then it'd be 9 over 5 instead of 5 over 9. 2 times 9 is 18 out of 15. It does say simplify, I can divide that both by 3 which gives me 6 out of 5. So those two I'm hoping are correct. Our next one indices comes up so much when we do A level maths. It's really really important. There's multiple chapters that kind of rely on it as well. So for this one when we have a multiply we add the powers. So that's going to be 5 plus minus 2 which is 3 so that gives me a cubed over a and don't forget when we divide we subtract powers. Now if there's no number it means there's a 1 so 3 minus 1 is going to be 2. So that's just a squared. With this one when you have a kind of an indice inside of a bracket, then what you do is you times the powers. So that's going to be 6 times a third, which of course is 2. And 27, that's gonna be the cube root of 27, which is 3. Now for this one solving the equation again, I've done this on some shorts and some TikToks as well. But the idea here is you want to make the bases the same. So can I write 16 as 2? Well, yes, it's going to be 2 to the power of what? Well, it's 2, 4, 8, 16 is 2 to the power of 4. So what we have is we have 2 to the power of x plus 1 is equal to 2 to the power of 4. The reason why we do that is we can now equate the powers because the bases are the same. So now we have x plus 1 equals 4, which means x is equal to 3. Not too bad. Working with thirds, simplifying, okay, that should be fine. Simplifying a third just means that you're breaking up into two numbers. Ideally, one of them should be a square number. So you can actually square root it. So I'm thinking square root of 36 times square root 2. Square root of 36 is just 6. That's going to be 6 square root 2. Cool. Expanding the brackets, again, rationalising is extremely important. So I'm going to do it fully, you can take a shortcut here, but I'm going to do it fully. So we have 2 times 2, which is 4. 2 times minus root 3, which is minus 2 root 3. Then we have root 3 times 2, which is plus 2 root 3. And then root 3 times minus root 3. So it's gonna be a negative number, it's gonna be negative square root 9. So here we can see that these two will cancel out. And then we have 4 minus square root 9. Square root of 9 is just 3. So 4 minus 3 is equal to 1. And rationalise the denominator of 5 over root 3. So when you rationalise the denominator, you're getting rid of the third and the bottom. So we're going to just times the top and bottom by square root 3. If it had something like this in there, then of course, we'd have to do the same thing, but change the sign in the middle. So it gives me 5 square root 3. Square root 3 times square root 3 is square root 9, which is just 3. Cool. Expand or factorise completely. So the expanding should be fine. So 2x times x is 2x squared. Then we're going to get 2x times 5, which is 10x. Then minus 3 times x is minus 3x. So 10x minus 3x is 7x. And then minus 3 times 5 is minus 15. When you factorise, you're looking for two numbers that times the number on the end, add to the number in the middle. So look for two numbers that times to minus 14 and add to minus 5. Because it times is to a negative, we know that one number is negative, one number is positive. And because it adds to a negative number, the negative number must be the bigger one, right? So for example, 7 and 2 works. So we're going to use the bigger number being negative. So minus 7 plus 2 does give me minus 5. So we get x minus 7 and x plus 2 as our bracket. Now, when there's a number in front of the x squared, it changes slightly. So what we're going to do is we're going to take the number in front of x squared and times it by the number on the end to give me negative 12. So we're looking for two numbers that times the negative 12 that adds to give me 1. So I'm thinking that's going to be a number that differs by 1, so 4 and 3. And again, in this case, because it adds to a positive number, we're going to make the bigger number positive. So it's going to be 4 and 3. Now, how do you write it when there's a 2x squared in front? I'm going to show you a trick that I showed to my year 11 top set, and that we use a lot in sixth form. The reason why we use this in sixth form is because factorising a quadratic doesn't even give you a mark. It's just something they expect you to be able to do automatically. So we need to do this as quickly as possible. We're going to put a 2x in both of the brackets. Then we're going to put the numbers, we have plus 4, and then we have minus 3. That should be a minus 3. Now, as you can see, when I expand this, I'm going to get 2x times 2x. I'm going to get 4x squared, not 2x squared. So everything's going to be 2 times 2 big. So what I'm going to do is I'm going to divide everything by 2. Now, the question is, which bracket can I actually divide by 2? Well, this has a 3 in there, so I can't divide it by 2. But this, I can divide 2x by 2, and I can divide 4 by 2. So we're just going to divide the first bracket by 2, which gives me x plus 2 and 2x minus 3. And of course, we've already divided by 2. So we don't need to divide that by 2. And there you go, that's how you factorise nice and quickly. I go to the next one, okay, this is gonna take a little bit longer, although there is a trick that I can show you. I don't know if it's worth it in this case. But just to write this out properly, we have 2x minus 3 over 5, minus x plus 1 over 2 equals 1. Now, the normal trick that you would use in GCSE is you'd make the denominators the same, combine the fractions, then get rid of the fraction, right? So you times by 10 afterwards. Here, I'm going to skip ahead. If I times everything by 5 and by 2, I will get rid of the fraction on this left hand side. So in total, I'm just going to times by 10 straight away. Now, when I do this, I have to be really, really careful. Because on this side, in fact, let's break it down into 5 and 2 again, just to show you why. And this is the same as timesing by 10, by the way. If I times this first fraction by 5, the denominator goes away. But then I'm timesing it by 2 afterwards. So I'm going to get 2, 2x minus 3, right? Because the 5 cancels with the 5, but the 2 stays there. This one, the 2 and the 2 will cancel, but I'm still left with 5. So we're gonna have 5x plus 1. I think that's a decent way to show it. And then obviously timesing 1 by 5, then by 2. So 1 times 5 is 5 times 2 is 10. Let's continue up here. We're just going to expand and simplify. We're going to have 4x minus 6 minus 5x minus 5 equals 10. Let's just make sure I didn't make a mistake with the signs. So 4x minus 6. Cool. Gathering the like terms, we get minus x minus 11 equals 10. I want to add 11 to both sides. So minus x equals minus 21. So that means x is equal to, so that should be positive 21 equals to minus 21. I must have done that last step in my head. Cool. Simultaneous equations. There's a few ways to do this. I think substitution is the best way to do it. Now elimination is the way you're most used to from GCSE. However, and that works in this case, by the way, elimination works completely fine. However, in A level, we're going to do simultaneous equations with things that aren't just linear equations. And so we're going to have to substitute at some point. So what I'm going to do is I'm going to show you the way that we do it. So x is equal to 1 plus y. We're going to sub that in to this top equation. So we get 2y plus 1 plus 3y equals 12. This way works all the time, even when we have something that's slightly weirder. So now we're going to expand. We get 2y plus 2 plus 3y equals 12. So we get 5y plus 2 equals 12. So that means 5y equals 10, which means y is equal to 2. And now we know that x is just equal to 1 plus y. So x is just equal to 1 plus 2, which is 3. Pretty quick, right? So yeah, not bad at all. Solving the inequality. So again, this is the same as solving a linear equation. Essentially, all you're going to do is expand both sides. The 6x minus 3 is less than equal to 5x plus 7. You're going to move the x's to one side in the same way you do with a normal equation. So I'm going to get rid of the smaller ones on the minus 5x on both sides. And we get x minus 3 is less than equal to 7. Add 3 to both sides, and you're done. You can treat this like an equal sign. It only matters if you times divide by negative 1, by negative number, actually. So the equation here has five quantities. Make A the subject. Again, rearranging formulas. It's so, I would say it's like 30% of the year, 13 marks. Like they'll say, show that this can be written as this, or differentiate this and show that it can be written like this. It comes up so much. So rearranging to make A the subject. Well, let's write out the equation first. When you make something the subject, it means you want it to be by itself. So we want A to be by itself. So I'm just going to put an arrow at A. So what's stopping the A from being by itself? Well, the u squared, the 2 and the s. Get rid of what's not attached first. So we're going to minus u squared from both sides equals 2as. And then we want to get rid of the 2 and the s. The 2 and the s are timesing the A. So we want to divide both sides by 2s. So we have v squared minus u squared over 2s equals A for the answer. And then since find A when v equals 13, u equals 5 and s equals 12. I'm going to regret not using a calculator for this, but let's go. So we have 13 squared minus 5 squared over 2 times 12, which is 24. So that's 169 minus 25 over 24. So that's 144 over 24. Does that actually go in? Well, we can just keep halving it, right? So we have 72 over 12. And that does. It goes in six times. Cool. There you go. A is 6. Perfect. Told you I didn't need a calculator. Don't doubt me. Solving the equation. Again, it's just factorising, right? So times to negative 6, add to negative 5. The only possibility is minus 6 and 1. So we're going to have 2x minus 6, 2x plus 1, all over 2 equals 0. Which bracket can actually be divided by 2? It's the first one. So we get x minus 3, 2x plus 1 equals 0. So your solutions now, you have to make each one of these, not negative, we make it equal to 0. So here we have x minus 3 equals 0, which means x should equal to 3. Here we have 2x plus 1 equals 0. So 2x equals minus 1. So x equals minus a half. There is a way to do this on the A-level calculator, but you should bear in mind that there are some questions on A-level papers that say something like, what is it? Solutions based on a calculator are not accepted. So you have to actually show yourself factorising it. Here we have completing the square. As I've said before, it's nice and easy. The formula is this. So x plus b over 2 squared minus b over 2 squared plus c. So here we're going to literally have x minus 4, because b in this case is negative 8, squared minus minus 4 squared plus 3. And then we just expand and simplify. So just so you know, this time here will always be negative. It's designed that way. So minus 4 squared is of course positive 16. But then we have a negative here. Perfect. Simplify fully. Okay, algebraic fractions. Nice. Just done a video on this one as well. So hopefully you saw that. So the way you do all of the algebraic fractions questions is the first thing you do is factorise. So here we're going to have x plus. That means, of course, on the top, we have difference of two squares. So again, you're going to square the 9 to give you 3. And you're going to have x plus 3 and x minus 3. And on the bottom, again, two numbers are times the negative 6 but add to plus 1. So it's going to be 3 and 2. Because again, if it adds to 1, that means the numbers have to have a difference of 1. And because it adds to a positive number, it means that the negative number, sorry, the smaller number is negative. So we have x plus 3 and x minus 2, which means we can cancel out the x plus 3 from the top and bottom to get our final answer of x minus 3 over x minus 2. Functions again, another really, really big topic. It honestly comes up more in year 13 than year 12. But the skills are still relevant. So when you have function notation like this, this x here is going to be your input. So for example, in the first part, it's basically telling me to sub in minus 2 in place of that x. So the way I do it, and I know you can use a calculator for this, because year 13, sorry, year 12 and year 13 is always calculator. But it's very importantly entering incorrectly. So this is what you do. So instead of x's, I'm going to put big brackets. And it's very important that you put the brackets because if you don't, when you have negatives, your calculator will do the wrong thing. And then all you do is you put the minus 2 or whatever the number is into there. So what does that give me? Well, that's minus 2 squared is 4, 4 times 2 is 8, 8 minus 3 is 5. Now for b, this one's slightly more interesting, because there's f g of 1. Now there's a couple of ways to do this. In my opinion, I keep things simple. All this means is you're going to find out what g of 1 is, get that answer and sub it into f. So let's first work out g of 1 using the exact same technique we did back there. So instead of x, I'm going to put a bracket. And then I'll write everything else as normal. And then I'm going to put a 1 in there because it tells me it's a 1. That gives me a 5. Now, because the answer to g of 1 is 5, that is what I'm going to sub into f. So now I'm going to do f of 5, which again, using the exact same techniques we've already been using, we get 5 squared is 25, 25 times 2 is 50, 50 minus 3 is 47, like that. And for part c, the solving part is again, something that's very useful in year 12, in particular, but also comes back in year 13, is solve f of x equals 15. So what they're telling you is this entire function here is equal to 15. So we get 2x squared minus 3 equals 15. So I'm going to add 3 to both sides. Remember, you want to do the square rooting last, ideally, that gives me 18. Then we'll divide both sides by 2, because again, you want to square root last. And then we're going to square root. Now, something that is relevant in GCSE maths, especially in recent years, but is very important in A level maths is that this is plus or minus 3. There are two solutions. And you have to be really careful with that. The solutions are x equals 3, x equals minus 3. Why? Because when you square root any number, any positive number, there are two values. So for example, minus 3, all squared is 9, as is 3 squared. So you have to be really careful with that. Cool, let's move on. Such big quadratics functions of graphs. This is where we're getting into more coordinate geometry is a little bit more challenging. But for the first part, finding the gradient of a line, I know it seems simple. But you'd be surprised what I see as a maths teacher, the gradient is the change in y of a change in x. Now I do like to tell some students if they're struggling to draw a little diagram. So we have point A is minus 1, 4. So let's say that's over here. And then B is 3, 12. And we're trying to work out the gradient of the line there. Now, the reason why I do this is because when you're doing change in y of a change in x, you need to make sure you you're subtracting the correct numbers. Technically, it should just be this coordinate minus this coordinate. Although it doesn't actually really matter. But some people confuse it, they do 12 minus 4, and then minus 1 minus 3 or something like that, which is weird. But again, it's always the second minus the first. So 12 minus 4 over 3 minus minus 1, that is 8 over 4, which is, of course, 2. Then the equation of that line. Well, now we know the gradient. When you want an equation of a line, you always need two things. You need a coordinate and you need a gradient. We have the gradient and they've given us coordinates. So at the moment, we have y is equal to 2x plus c. I'm going to pick one of the coordinates. Let's pick the positive ones to keep my life easy. So y is 12 and x is 3 plus c. So 6. So 12 equals 6 plus c. So that means c is equal to 6, which means the equation of the line is y is equal to 2x plus 6 is our final answer. And then part c, this is a little bit trickier, is the equation of the line that is perpendicular to L and passes through B. So if I change the colour here quickly, that would be something going like this, where this is perpendicular. The key thing to remember is that when a line is perpendicular, its gradient is the negative reciprocal. So what do I mean by that? It means you put 1 over, so this is going to be 1 over 2, and then you just make it negative. That's it. And now we're using the exact same technique we did to work out part b, because we have a gradient, it's minus a half x, and we have a coordinate and the coordinate has to be b. It can't be a because the line does not pass through a. So when they say use a coordinate, it's a coordinate that the line actually passes through. So here we have that again, y is 12. And x is a tree of C. So that's gonna be minus three over two and 12. So that's gonna be 12 plus three over two for C, which if I double this as 24 is 27 over two. So y is equal to minus a half x is 27 over two. It kind of makes sense because again, this is quite steep one. Perfect. Now we're doing sequences. Okay, so the first thing we need to do is check what type of sequence this is. Very important. So if I look here, the difference is five and seven, then that's nine, and that's 11. So I can see that the difference is increasing, which means it's not linear. Now, if I look at the difference of the difference, it's constant, which means this is a quadratic sequence, which means it has the form a n squared plus b n plus c. And the only thing that this is the method I personally use, the only thing we need to remember is that the common difference if we half it is what a is. So in this case, a is equal to two divided by two, which is one. So we have n squared plus b n plus c. And then I like to use simultaneous equations because very, very quick and easy. So here, if I sub in n equals one, remember what n is, n is just the position you are in in the sequence. So when n equals one, I should get two. When n equals two, I should get seven. When n equals three, I should get 14. So that means if I sub in n equals one into this equation, the whole thing should equal two. So one squared plus b times one plus c equals two. So if I rearrange this quickly, I'm going to minus one from both sides, because that's one squared, which is one, I get b plus c equals one. I do the same thing with n equals two, I'm going to get two really, really easy equations that I can do simultaneously. So that's going to be two squared plus b times two plus c again equals seven. And remember, two squared is four, so I'm going to minus four from both sides. So I get two b plus c equals three. So if I write those two equations on top of each other now, I have two b plus c is equal to three. And I have that b plus c is equal to one. And now I can just subtract the two equations, two b minus b is b, c minus c is zero equals two. So I know that b is equal to two. If I sub that straight back into here, I get two plus c is equal to one. So c is equal to minus one. There you go. So the equation for the nth term, I've not given enough space for myself, but that's okay, is n squared plus 2n minus one. And then in terms of finding the 10th term, you're just subbing in the number 10 into here. So that's going to be 10 squared is 100 plus 20. So that's 120 minus one is 119. Cool. Describing transformations is something really, really important. They do like to use a lot of words in A level maths, unfortunately. But it's really, really good that you know these transformations, again, year 12 and year 13, because you're going to do this with different functions as well, which is why I haven't given you an equation, because you need to be able to do this with all different types of functions, trig functions, exponentials, logarithms, etc. Even stuff you haven't come across yet. So the first one, if I look at this, inside the bracket affects x, outside the bracket affects y. So this minus three does something to the x coordinate, this plus two does something to the y coordinate, and you can do these independently. So the only other thing you need to remember is that if it's inside the bracket, it's lying to you, you do the opposite. So what's the opposite of taking away three from x, adding three to x, so you're going to plus three to the x coordinates. If you wrote this as shifting to the right by three, that is also 100% acceptable. And if it's outside the bracket, it does exactly what it says, that's going to be adding two to the y coordinate. Again, if you said something like shift up by two, that is also completely acceptable. Cool. So how does that apply to something like part b, minus f of x? Well, it's outside the bracket, so it affects y. And what's actually happening? Well, that's going to be a minus one f of x. And remember, when we don't write an operation, it means times. So this is going to be timesing y coordinate by minus one. If you'd like, this means it's going to be a reflection in the x axis. Because if you think about it, if we have a graph like this, when I times the y points by minus one, what happens is anything on top goes to the bottom, and anything on the bottom goes to the top. So you literally have the exact same graph, but upside down. So it's a reflection in the x axis. And lastly, but not least, we have y equals f of 2x. Again, it's inside the bracket, so it affects x, and it does the opposite. So 2x means timesing by two, which means we're going to divide the x coordinate by two. Now, how would you call this? You would say that it's a, like a, you can say an enlargement by a scale factor of one half, so you're shrinking it. So you could say it's squished by a factor of two, something along those lines. So honestly, I'll probably leave it as that. But either way, you can write something else as well. So with the exact trig values, I know that people might think, why the hell do we need to learn them if we have a calculator in year 12 and 13. But again, it's super, it's just a lot more convenient. It's like saying, what's the point of knowing your times tables if you have a calculator? It's like, do you want to calculate every single time you need to work something out? And this is really useful for sketching the graphs and things like that. So I do have a video on how I personally choose to like, not necessarily remember it, but at least work the things out in case you do forget, which over time will obviously lead to you remembering, right? Here, we have trigonometry, some basic trig, we use it all the time, things like bearings and mechanics, etc. So you need to be very, very quick and efficient with it and accurate, of course. So here, you label up the sides. So this is our opposite side. This is our hypotenuse because it's opposite the right angle. And of course, that means the leftover one is the adjacent. Now we have this angle and we want the opposite side and we have the hypotenuse. So if you think about SOHCAHTOA, S-O-H-C-A-H-T-O-A. The only side we don't care about is the A, which means we don't care about this one or this one. So this is the only one that's left. So we have sine 38 equals opposite, so H over 12. The thing I'd recommend strongly, if you really want to get a head start in A-level mechanics, is basically being able to do this in your head. So being able to rearrange this equation to get H is equal to 12 sine 38 in your head is so, so, so important. Yeah, it'll make your life so much easier. And it'll make you like mechanics more. And that's like a third of your grade. So it's quite important. And it's just 3SF, so 7.39, according to the calculator. Now we have some probability. Again, we have statistics in A-level maths. So probability is quite important. There's a whole chapter on it. And then chapters on probability distributions as well. We need the basics. So it says here we have red counters, blue counters, green. We want the probability that there are two counts of the same colour. What I do is I write how all of the combinations that would give me two counts of the same colour, which means it's either gonna be two red, or two blue, or two green. So we have red, red, blue, blue, green, green. Now I'm gonna work out the probability of getting each individual one. Well, there's five red counters. And in total, we have five plus three is eight, plus two is 10. So to get a red count, that's a five out of 10 chance. And then because we've taken one red counter out, there's only four red counters left. And of course, there's only nine counters in total left. So it's gonna give me 20 over 90. Like that, with the blue ones, again, we have three blue counters out of 10. And again, because we've taken one out, that's gonna be two out of nine, to get six out of 90. And lastly, GG, it's going to be two out of 10, times one out of nine, the same logic. And then because we have all three of these possible combinations, we need to add them up. Because remember, the question isn't what's the probability of finding two reds, two blues, two greens. Instead, it's what's the probability of getting two of the same colour? Well, that's two reds, or two blues, or two greens. So we have to add those probabilities up because there's three different ways we can get two counts of the same colour, which gives me 28 out of 90. You do not need to simplify these fractions. There you go. So that brings us to the end of the paper. I'm going to quickly show you how to use the rest of this. So when we look at the rest of it, obviously, we have just the mark scheme. These are the work solutions that I've given you. So you can have that to double check. And at the end is something that's really quite useful, which is you're going to put your total number of marks in each of these sections. And that will help you determine what your priority is. Again, if you look at this, I've given you a lot more to do. So there's three available marks, for example. So let's say hypothetically, you got two, that would be you change that to a percentage, which is obviously going to be 67%. Because two over three times 100. If you do that, it allows you to kind of prioritise the different steps that you need. All of this is useful in year 12. So a lot of people have been asking me, what do I study before year 12? This will help you decide exactly what you should study. I have got a video on the five topics that I would make sure I've mastered. But this is very personalised, which is, of course, very, very useful. And I mean, just look at that, right? Like you can look at your total score, I've given you an interpretation where maybe you can start doing the A level work. Whereas anything below 50, 50 or below, you do need to actually revise the content. Whereas again, 51 to 60, you can just, you can actually learn A level. And again, just a little personalised action plan just to help you plan out exactly what you need to do. If you have a tutor or a teacher, you can email and they can add anything they need to. Perfect, right? If you are going to book one of these year 12 meetings, the one-on-one meetings that I've put on there, I'll put a link to that in the description as well. Having this completed will allow me to give you a much kind of better idea of what you want to do as well. So it's extremely useful. Again, the link to this is obviously in the description. The mark scheme is in the description. And of course, you have the video solutions for when you need them.