Edexcel GCSE Maths Higher November 2025 Paper 3 Walkthrough Explained

Hey guys, welcome back again. This time we're going to be doing the, again, 2025 November paper 3 this time, and this will be the last one for November 2025, in fact the last one for 2025. So let's jump straight into it. The first question, a nice simple one where you just enter in numbers into your calculator. I would recommend making sure you have the right calculator, so a calculator that looks maybe like this, or the new black one's quite good, so you can actually put fractions in there, because otherwise this is a little bit tricky, and it says write down all the figures, and I get 1.426874635, and then nice and easy just rounding to three significant figures. To round to 3SF, all you're doing is you're counting the non-zero numbers, so one, two, three, we're rounding it to that number. If I look at the digit next to it, it's a six, so we're going to round up to 1.43, and then the value of the reciprocal, this is a classic question they love to do, reciprocal just means one over, so we're going to do one over 1.25 in order to get our value, and again you can just use a calculator, or 1.25 is 5 over 4, so you flip it, it's four fifths. Now you don't have to give it as a fraction or a decimal, but I'll probably just write both just in case. Okay, so this is quite an interesting one, there are actually two ways to do this, and the first thing we could do is we could just keep, so it says two numbers in a ratio of 5 to 3, the difference between the numbers, so if we subtract the two numbers we should get 18. There are a few ways to do this, one way to do it is if I look at the difference between those numbers, that's two, it's two parts, and I want to make that number 18 instead, so what do I times by? I times by 9 in order to get 18. So if I times my ratio by 9, then I should get the values, so 45 to 27. Now if you subtract those two numbers, again you can use a calculator to check this, you do get 18, so those are my two numbers. So again quite a few different ways to do that, but that's the way I was thinking of. It then says three numbers are in the ratio 4, 7, 2. The largest number is 210, find the sum of these three numbers. So if I start off with this, if they're saying the largest number is 210, then that means that the number in the middle, the one with the biggest number already, must be 210. So I'm going to write 210 there. Now what do I times 7 by to get 210? But what I could do is I could do 210 divided by 7, and I get 30. So in other words, if I times these numbers by 30, that's what the other numbers should equal. 4 times 30 is 120, 2 times 30 is 60, and then the sum of those three numbers, so we're just adding them all up, gives us our answer. So it's a very interesting and weird ratio question, but hopefully that makes more sense as to how I would approach it in the exam. So it's 390 for three marks. Okay another question that usually comes up, complete the table of values. So what I would do is instead on my calculator, instead of the x, I'm going to put a bracket. So I'm going to put bracket squared minus two bracket minus one, and then all I'm doing on my calculator is I'm going to start from the left. I'm going to sub in minus two into those brackets, then I get my answer, rub it out, then I'm going to sub in zero, sub it in, rub it out, and then so on. So if I start with minus two, we get minus two squared minus two times minus two, which is also two squared, minus one, I get seven. Then again I'm going through on the calculator and changing the minus two to a zero, and doing the exact same thing again. So it gives me minus one, then I'm going through deleting the zero, changing it to a one. So again each time I'm just changing that red number for whatever the top number is in the table. And then whatever answer I get, I put that in the square, the box directly beneath it. And keep going, so three, I could probably do this faster without a calculator to be honest, and two. Now part B is something that some students get a bit wrong, then we need to draw the graph. So each of these pairs in the table make up a coordinate. So for example, the first coordinate here is minus two, seven. Then here it's minus one and two, and then so on and so forth. So it just keeps going. And we're going to plot those coordinates on the grid. So minus two and seven, minus two is here, seven is right here. So let's go, there you go. Then minus one and two, minus one, two, that goes down there. Then I think it's zero minus one, perfect. Then we have one and minus two, then we have two and minus one, then three, two, and four, seven. So this is one of the only graphs that I've actually seen them ask you to do, where the actual turning point, like the lowest point, is actually given to you, which is quite interesting. Normally they, you know, there'll be two points next to each other, and you have to know to go below those two points. And then you're going to connect them with a smooth curve. Now again, this might be a bit tricky for me, but again, I'm not joining the dots, it's a smooth curve, not taking my pen off the page, like that. That's actually not too bad. And then for part c, they tend to ask you to find estimates for a solution. Now, what does that mean? Well, if you notice, we have x squared minus two x minus one equals y. In this equation, it's x squared minus two x minus one equals three. So basically, they want us to work out when this whole graph equals three. Now bear in mind, it does say using your graph, so you can't solve this as a quadratic, they give you zero marks. Instead, I'm going to go to three on here, and I'm going to draw a line. And I'm going to stretch it all the way across both sides, like this, like this. And what I'm going to do is I'm going to look for where the line and the curve intersect. So one of them is right here, that's one point, and that's the other point over there. And I'm going to read off their x values. I already know their y values are minus three. Let's read off their x values. So if I read that off, that is minus 1.3. How do I know? Because between minus one and two, minus two, that's a gap of one, and there's 10 squared, so it goes up in 0.1. So minus 1.1, minus 1.2, minus 1.3. I'm going to put that down here as my first solution. I'm going to do the same thing on this side. That'd be 3.2. And again, in terms of for the answer, they do give you a range, which is nice. Yeah, that'll be it for that question. This question is quite interesting. All I really need to do is put down a calculator for two marks, but I need to change it to standard form form afterwards. So 3.2 times 10 to the 5. If I enter that in my calculator, I get 21 over 80, I get this. I've shown in quite a few videos how I do standard form. What I do is I look for the first non-zero number, which is 2, and I put a decimal point. Then I'm going to write every other number that is not zero, so 625. Now it's going to be times 10. Now I ask myself, this number, is it bigger or smaller than this number? Well, this number is bigger, which means I want to make my number smaller, which means it's going to be to the power of minus. And all I do is I count how many times the decimal point is away. So if I want to make this zero point, I need to move the decimal point one space, so it's the power of minus one. And then all I do is I get rid of that, and there you go. That would be your answer. And then error intervals, the way I do this, super, super easy. I'm going to write the 0.83. The upper bound, I just add a 5 onto the end of the number. That is it. Now, if it was something like 255, if I wanted to add a 5 onto the end, I'd have to do a So hopefully that makes sense. I'm not making the number 10 times bigger. And then what I do for the bottom part is I just minus one from the penultimate, the second last digit. So in this case, it'd be 0.825. That's my shortcut, and it works every single time. It's super, super easy, especially considering that this is actually a kind of a high grade topic, and our students make quite a few mistakes with it. If you look at the video for 2025 for June, which should have been uploaded by now, there's actually one where it says it's rounded to like the nearest 5 or the nearest 10 or something like that. And I'll show you how I do that as well. But this is just what I do. It's so quick. It's so easy for two marks. Eric buys a car for £19,950. It loses 5% each year. And he tries to work out what the value would be after four years. And he does this. He does 4 times 5%. It's 20%. He takes 20% of 19,000, blah, blah, blah. He subtracts it. His method's wrong. Why? Well, the simple thing is, ironically, is this is not a simple interest question. He's doing it as if this was simple interest. It's not. It's compound. So at the end of the first year, it decreases by, let's say, 5%. Then you take 5% of the new value and take it away. Then you take 5% of that value and take it away. And then you take 5% of that value and take it away. He's not doing that. He's trying to take it all off at once, which is wrong. So how do you work this? Well, you can either show the equation he should have used, which would be 19,950 times 0.95 to the power of 4. And you can say you should have used this because it's compound. Technically, it's not compound interest. It's compound depreciation. But honestly, it's the same thing. It doesn't make a difference. But you need to imply that that's the equation he needs to use or that it should be compound, not simple, something along those lines. And then it asks you, what's the actual difference? Okay, so between 15,960 and the real value of the car. So if I actually work out what this is, so again, choosing my calculator, 19,950 times 0.95 to the The actual difference is that these two things minus in each other. So if you notice, it's actually, you know, it's worth more money than he thinks, which is, you know, pretty good thing for him. And that's £289.40. Which, honestly, for four marksman, damn, like that's not much working out. But that genuinely is all you have to do. And it's nice and easy. Not too bad. Drawing inequalities is actually surprisingly, not that well done. So what I do is I look for the number, which is one, I'm going to put a circle, then I'm going to ask myself, is this circle coloured in? If it is equal to that number, then I colour it in. So if this was just less than one, I'd leave it like this. Because it says less than equal to, I'm going to shade it in. And now, what do we want? We want m is less than it. If I look for less than on this, on this number line, that's to the left. So I'm going to draw a line going to the left, and you have to put an arrow. Now, I'm going to explain why you have to use the arrow, you have to use the arrow, because it's all the values less than one, that goes on to negative infinity, it doesn't stop at minus six, it keeps going. And you have to make that clear. So in this diagram, we have two quarter circles on either side, and then a rectangle in the middle, work out the perimeter of the shape. Cool. So perimeter is just you just add up all the outside lengths, that's fairly okay. So just so you know, for this one, this would also be eight and eight for these bottom two sides. So what we can do straight away is we can add up the links that we have, which is 20. For the top, we have a curved piece, then eight plus 20 plus eight. So we can start with that. And can I move on? Yeah, sure, fine. It's 56. Now, these curved pieces are a little bit tricky, but can you see how they'd be the exact same size. So what we're going to do is we're going to do two times, whatever this curve length is. Now, this curve length, you can call it an arc length if you want. But if we keep it simple, this is a quarter of a circle, right? Well, how do you work out the outside length of a circle, you do two pi r. Now, because it's a quarter of one, we're going to divide that by four. So it's a half pi r. Now, if you notice, we're actually times here by two, and that'd be eight, the radius is just eight, because that's what it says. So that's just eight pi, which is about 24. But let's do it properly. We get 25.13. And it says to three SF, so I'm going to add those two numbers together, and I'm going to round it. So I get 81.1327. So 85.1, sorry, 81.1, I can't read, gives me my final answer. So hopefully not too bad. So we have these two parallel lines, and we have a length and all that fun stuff. And we're trying to work out the angle QRS. Let's call that x, why not? Now, this is a right angle triangle. And so if I wanted to work out an angle, I need two sides, and at the moment, I only have one. So if you have a look at this left hand side, they give us two sides. So I could work out this opposite side over here. And if I do that, then I have two sides, and I can use trigonometry. So the question is, how do I work out why? Well, if this is a right angle triangle, and we have two sides, and we want the third side, we're going to use Pythagoras, which is a squared plus b squared equals c squared. The only thing that matters is that the longest side the hypotenuse has to be c squared. So I'm going to say that this is y squared plus 15 squared equals 39 squared. So we get 36. That's handy. Now, if we start looking at this right hand triangle, I'm going to start labeling it. So the opposite side is hypotenuse, which we don't have. This side here would be the opposite side because it's opposite the angle, the leftover one is the adjacent. So for right out Soh Cah Toa, I'm going to cross off whatever side I just don't care about. In this case, I don't care about hypotenuse. I'm gonna cross up the two with hypotenuse. So I'm left with Toa. So that means tan x equals opposite, which is 36 over adjacent, which is 50, which means our angle is inverse tan 36 over 50. In fact, I'll write that out. So inverse tan 36 over 50. To get inverse tan, you just press shift and tan, which gives me 35.8 to one decimal place. Not too shabby, hopefully for four marks. Not bad. Okay, so we get given a bunch of marks for students in class x, and we need to find the range and interquartile range. Thankfully, they've been nice to us. They have just given us the values in order. So the range is going to be the biggest number 69 minus the smallest which is 30. 69 minus 30 is 39. Now the interquartile range is the upper quartile minus the lower quartile. And the way I do upper quartile is I just take three quarters of the number of students. Now it doesn't tell me what the number is. It's a bit annoying, but fine. So there's 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, double it, and then minus one, so it's 23. So it's three quarters of 23, which is of course going to be a decimal number, but we're just going to round it up. So three quarters times 23 gives me 17.25. We're going to look for the 18th value. So the 18th value is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, so 56, minus the same thing except it's going to be one quarter of n and a quarter of 23 is 5.75. So we're going to look for the sixth value. That's 1, 2, 3, 4, 5, 6 minus 42, which is 14. Yep, and that's it for three marks. Not too bad. A little bit annoying, but it's okay. And then here's some information about the students in class y. And Tanya says that this information shows that x did better than y. Is she correct? Now when we say people do better, we talk about two things. One is we talk about the average, and the average for x is higher. However, we also talk about the interquartile range. So is she correct? I would say yes, because x had a higher average. And bear in mind, when we talk about the average, in this case, we're talking about the mean, this is the median, the median doesn't get affected by extreme values. And so yeah, half of them got above 48, half of them got below, whereas for y half of them got above 45. And while it says that the information in the table shows that the marks of x vary more than y, is he correct? The answer is no. The reason is the interquartile range is what we use to measure spread, how much something varies, not the range. So I'm saying no, the interquartile range is smaller for x than y. And that's it. Okay, a classic question where we have five graphs, we have to match it up. And in this case, we have five equations. So that means every single one of the graphs has an equation. And sometimes it might be four equations, but then like eight graphs or something like that. But let's have a look. Or sometimes they give you more equations than graphs, which is always fun. So I'm gonna start with the easiest one, which is probably this one, which is a straight line with a negative gradient. The only one here that looks like a straight line is this one. And if I move it to the other side, it is a negative gradient. So that means that must be e. So what I'm gonna do is I'm gonna cross off this one. And then the rest of it really is just based on how well you know the shapes of graphs. So x equals three, well, x equals three all along this line. So it's gonna be a straight line. So it's gonna be d. And again, I would start with the ones that I know for sure. That is a quadratic, and it's the only quadratic. So it must be the only quadratic here. So that means that must be a and so on. Again, in terms of working it out, it'd be very difficult to work out in the exam. For example, the one over x, maybe you could work out by by plotting it yourself. And then that means b must be that one over there, which which does make sense. There you go. Not too bad, hopefully. Okay, so here we have shape p can be transformed to shape q by an enlargement scale factor s and the center a b. So p onto q. So what I would do first is I'll just work out the scale factor by matching up the sides. So this is a length of 123456. And that's the length of three. Now that means it's gone from six to three. So that means it must be a half. But and this is the really, really key thing is that's not it. Because if you have a look at what else has happened, it's also flipped upside down, right? Like if you look, this whole shape's been rotated or flipped or whatever you want to call it. If that happens, then it's also a negative. So we need to write scale factors, if it changes shape, and flips is negative. Otherwise, what you'd get is you'd get this exact same shape shrunken down. So be careful with that. So it should be the same orientation if it was positive. In this case, it's not. Now, the center, all you want to do is you want to put a ruler through each of the corresponding points. So for example, I know that this point here corresponds to this point over here. So if I draw a line straight through it, and then I'm going to do the same for all of the other points. So those two corners there correspond to each other. Cool. Now, I'm actually done now. But if you really want to be rigorous, you can also do the last two points as well. And what you'll find it. Come on, it's perfect for a second. Oh, come on. Alright, so either way, they all go through the same point, that point that they go through is the center of enlargement. So it's gonna be 2, 1. So 2, 1, which means A is 2, and B is 1. So here we have algebraic proof, when you prove that something is a multiple, you do is you expand it and then show that you can factorize out 20. So if I expand the first thing, we get 25m squared, plus 20m, because five times two twice, plus four, then we're going to get subtract 25m squared, then we're going to get minus 20m, then we're going to get plus four again. So have a look at this, what cancels what 25m squared minus 25m squared, then four minus four. So we're left with 20m minus minus 20, which is 40m, which is the same as having 20 times 2m. So this is always a multiple of 20. Assuming that this is an integer, which in this case, it is. Ali then says 80 is a factor of this for any integer where m is bigger than zero, which makes sense for seven, zero is not gonna work. And so what you're trying to do is trying to find a value where this doesn't work. Now, if you think about it, if you have a look at this, we've shown that it's always a multiple of 20. So that 20 is always a factor, we haven't shown 80. So if I have a look at this, if m equals one, then I get two times one, which is two times 20, which is 40. For 80 is not a factor of 40. So when m equals one, it equals 40. So 80, not a factor. So what that means is he is wrong. Right? You're just trying to find a value where it doesn't work. Here we have a circle theorem question, or at least I'm assuming it is because we have circles. And it says you must give a reason for each stage of your working out, I'm going to verbally say the reason, but you need to write them down. Because this is a four mark question. So what could I work out? Well, the first thing I could work out is I could work out the size of these two angles. Why is that these two sides are the same length, because the circle theorem says that two tangents from the same point are always the same length. So that would be your first reason. And that gives you the answer of ACD and CAD. And that'd be 180 minus 40 over two, which is 70 degrees. So we've worked out both of those being 70. And again, the reason I'd write down is that the lengths of tangents from the same point are always equal. Okay, what could we do now? Well, we're trying to work out the size of angle ABC, which is this one over here. Now, from here, it's a little bit tricky. But what I would do is what's the other circle theorem that we use when it comes to equations of tangents? Well, we use the fact that at the centre, the radius and the edge of this and the tangents would always be 90 degrees, right? So that means this angle here, this whole angle here, OCD, should be 90, which means these inner little bits here are both 20 degrees. So again, and then you'd say, you know, sorry, radii and tangents meet at 90 degrees, which means this angle here is just 180 minus 20 minus 20, which is 140 degrees. Okay, now what's my very last step? Well, that means CBA, or let's call it ABC, is half of that. Why is it half of that? Well, the whole point of me getting that, and bear in mind, a lot of the questions, if you've done a lot of practice on circle theorems, you'll actually recognise that this question comes up a lot, and they actually give you the two red lines that I've drawn. But in this case, they haven't. So just add them, right? It's because angles at the circumference are half the angle at the centre. And if you notice, both of these angles are made from A and C on the circumference, which means we can use that relationship. Angles at the centre are twice that of the circumference, or angles at the centre. And that would be it. Make sure you give at least two reasons, a four mark question. So you get two marks for working out, two marks for reasons, you need to give two valid reasons. Okay, and bounds are really, really interesting. Three marks, they can sometimes be a lot more. You get marks just for writing out what the upper and lower bound of the values are. And again, we've done this earlier, I'm going to add a five on to the end. So 715, then minus one from the penultimate number, I'm going to do the same thing for y. Right, so 24.35, subtract one from the second last number 24.25. Now we're going to actually look at what w is, w, it says work out the lower bound, trying to make the number as small as possible. If you're trying to make the number as small as possible, that means we want y to be as big as possible, because we're taking it away. Like imagine you have 10 suites, right? If I take away five, or I take away seven suites, in which option do you have less suites? When I take away seven, right? The more we take away, the smaller the number is. Likewise, if you look at x, we want x to be as small as humanly possible, because we're timesing by it, right? If I times by a smaller number, the number gets smaller. So what we're going to do is we're going to take the lower bound of x, which is 8.705. And then we're going to take the upper bound of y 24.35, shove it into our calculator, and we are done. Honestly, I think bounds questions are quite nice, especially when you get good at them. And they're worth so many marks, they can be worth up to five marks, which is kind of insane. So definitely make sure you're practicing that, especially with these little tricks that I've shown you. Okay, so quite an interesting question here, there are only blue and red counters in a box, there are five times as many blue counters as red counters. So I'm going to do is I'm gonna say red equals x, blue equals 5x, which means the total is 6x. Again, that's the first thing I do anytime we have conditional probability. Anytime we have probability that has algebra, I love doing this because it's a grade nine topic, it's tricky. If you write out this, it makes a lot more sense. Now if we read it, Ashley is taking a counter from the box, recording the colour, so it's red or blue, then putting it back in, which means the number of counters never changes. He does it three times. And it says find the probability takes the counter of the same colour three times. Well, there's only two possibilities, right? Red, red, red, blue, blue, blue. So let's do red, red, red first, that'd be x over 6x, times x over 6x, times x over 6x. Why? Well, simple, because it never changes, right? The number of counters never changes. So there's a one in six chance of him taking out a red, then he puts it back, there's a one in six chance of him getting the same red again. There you go. Blue, blue, blue is going to be the same thing, just the opposite. So instead of x is 5x. And if you notice, all of the x's cancel out. I still write them though, because I like to be really precise when working out. And yeah, we can just cancel this all out. And all we're going to do after that is we're going to add up the values. So what I'm going to do is a sixth, that should just be one over 256, isn't it? I'm sure that's the case. 216. And that'd be 125 over 216. So we add it together, because again, it's all right, there's two possibilities. Add it together, we get 126 over 216. I could simplify that, but I don't want to, so I'm not going to. So yeah, quite a nice question. And again, I like to use the x's. Why do I like to use the x's? Because if he didn't put the counter back in the box, we'd have to subtract one from the top and bottom. So if you said it was one over six for a red, and that means the next red would be zero over five. No, that's not right. Always use the letters, makes it more precise. It then says the number of blue counters is doubled and red counters is doubled. Ashley takes three counts in the exact same way. Ashley thinks the probability should be the same. Is he correct? Yes. Now the question is why is he correct, right? It's because the ratio. So if you look here, if we go back here, it says there's five times as many blue as red, right? So blue to red is five to one. If I double both of those numbers, is that an equivalent ratio? Yes. Okay, fine. Maybe you don't believe me. Let's go back up here. If I double both of these, I get 2x over 12x, which is still one over six. And same with the blue, I get 10x over 12x, which is still five over six. It doesn't change anything. So all you have to do is say yes, the ratio or proportion, whatever you want to write, stays the same. That's it. Nice. In terms of showing that an equation has a solution between two values, all you want to do is sub in one and two, and you're going to find what they equal, right? So let's start this one. That's just 2 plus 1 is 3, minus 8 is minus 5. 2 cubed is 8, times 2 is 16, minus 8 is 8, so that'd be 10. Change in sign implies, so then you just write change in sign implies existence of a root. Now what does that mean? Well, what it means is, if I've subbed in one and I get a negative, and I've subbed in two and I get a positive, it means that between one and two, it should equal zero. That's what the equation is asking for, right? The solution is when this whole thing equals zero. Now for this bit, showing that it can be written in the form, a lot of students don't like this because it's x on both sides, a bit scary. Let me help you. First of all, look at this as a cube root. What does that imply? It means that they've moved the cube root, they've kept the cube on the same side. So I'm going to move the x and the 8 to the other side. And if you have a look, I've got the same thing up here, right? Makes sense. Now what do I do? Well, I need to divide by 2. Well, I've got a 2 here, let's divide by 2. So now I've got the exact same thing, then cube root. Again, you do, sadly, you only get one mark for that, but oh well. This is such a beautiful question. Three marks is crazy, right? Because guess what? You can get it all in about five seconds, genuinely. What you want to do is you're going to do this. You're going to type in 1.5 and then the equal sign on your calculator. That means the ans button is now equal to your last answer, 3 over 2. Now I'm going to write this exact same equation in my calculator, but instead of xn, I'm going to press the ans button. So I get 8 minus ans. This is exactly what I'm typing into my calculator. When I say ans again, it's not like shift a and shift b. No, no, no. There's an answer button on your calculator. If I press this once, I'm going to get x1. If I press it twice, I'm going to get x2, and so on. It wants me to do this three times, so I pressed it once, twice, three times, and I get 1.483 for three marks. Now you do get method marks for writing the previous two numbers down, and in some questions they insist, so I would recommend doing that as well. For this question, we have a ratio equaling another ratio, and then we also get given that x plus y equals one. The first thing I notice, whenever we have two ratios equaling each other, we can do a really nice trick, which is we can write it like this. We can write y over x squared equals six over one, and you can do this with any two ratios that equal to each other, because what they're saying is, is the proportion should be the same, right? Like if I say one to two equals two to four, then what I'm saying is one over two equals two over four, which is true. So this works anytime you have two ratios equaling each other. It's very, very nice. Now this thing over here, I'm going to come back to it in a bit. This just gives me y over x squared equals six, which is interesting, and then I can times both sides by x squared. So okay, I have a little equation here, and what's the goal of the question? It's to find values, this is possible values, for x and y. So if I wanted to solve this, well y equals six x squared, and x plus y equals one, this is a quadratic simultaneous equation again. So there's quite a few of these in the 2025 papers. So what I'm going to do is rearrange this for x or y. I think I'm just going to do y, one minus x, and then substitute it straight in, and that's quite easy to solve. So we have one minus x equals six x squared, and move everything over to the right hand side, which gives me six x squared plus x minus one equals zero. Technically zero equals all of this, but that's completely fine, and then we can just either factorise or quadratic formula. So let's check if we can factorise first. So times the negative six, adding to one, well I think that would be three and minus two, and again I'm going to use my trick for factorising, which I quite like. I'll do a video explaining it, where we're going to write six x in both of the brackets, and then because that's going to be six times too big, right, because six times six would give me 36 x squared, I'm going to divide it by six, zero. Then we're just going to put the two numbers that we've worked out into here, and all we need to do now is simplify this. So for example this first bracket I can divide by three, so I'm going to divide the first bracket by three, and the second bracket by two, and if you notice that means we're going to divide the bottom by three and by two. Six divided by three is two, two divided by two is one, so now we get two x plus one, and we get three x minus one equals zero. So my two values for x are either x equals negative a half, or x equals positive a third. Now in the question it doesn't mention that x can't be negative, so both of those are valuable answers. It also says possible values plural of x and y. So now how can I work out y? Well I think the easiest thing to do is sub it into y equals one minus x, so we're going to have y equals one minus minus a half, so that's y equals one plus a half, which is just three over two, and we're going to have y equals one minus a third, so that'd be, by the way it's a calculator paper so you don't, you can just do it in a calculator if you need to, but those are my pairs of values. Pretty interesting, that's not not too bad of a question, hopefully, and yeah I like it. For five marks, not too bad, hopefully. Imran asks a hundred people if they speak French, German, or Spanish, and he draws this Venn diagram. Cool, wonderful. Now it does say that the diagram is not fully correct. Explain why the diagram cannot be fully correct. Now honestly, if you look at the question, they haven't given us any information about who speaks what, so I can't check if the number of French people is the case, German speaking people, and Spanish speaking people is correct. I can't check that, so the only other thing I can do is make sure this whole thing adds up to 100. So if we write in our calculator 17 for the number on the outside, plus 42, plus 5, plus 1, plus 10, plus 2, plus 8, plus 13, I actually get 98, so it can't be correct because the numbers do not add to 100. Why do they have to add up to 100? Because here it says 100 people, so there should be 100 people in this Venn diagram, but there's not, there's only 98, so something must have gone wrong. And then we have Jody asking which are the social network sites, P, Q, and R. That's very creative. The Venn diagram gives information, and they've already filled it in for us, which is really nice. Given that a person selected at random from the 200 people uses site Q, find the property of this person also uses site P. Well that blew my mind for a second. So given that they use site Q, that means our denominator, the bit on the bottom of the fraction, should just be all of the people in site Q, because they're saying that's already true. So on the bottom we're going to have 7, plus 8, plus 56, plus 18, and I'm not going to wipe that out because again it's a calculator paper, I'm not going to make my life harder. Then it says find the property that they also use site P. So how many in this circle here, in fact let's trace over it, how many people in this circle here also use site P? Well that's going to be these two groups of people, because these 8 people use P and Q, and these 56 people use P, Q, and R. So we're going to do that, and let's wipe that out. So 8 plus 56, almost at times, over 7 plus 8 plus 56 plus 18 gives me an answer of 64 over 89. Perfect for two marks. Then we have two more questions left. Well this looks fun. So here we have two equilateral triangles. What they mean by that is the overall big triangle, which again I'm going to attempt to trace, this should auto-correct I'm hoping. Yeah whatever man, if you've watched any of my videos you know at this point. Actually, not bad. That's an equilateral triangle, and they tell us that the top bit as well is also an equilateral triangle. So I'm just going to cover that with just one red line there, okay. And we need to work out the ratio of the areas. Cool. Now for me the area of a triangle I'm going to say is half a b sine c. And I'm going to say that because I think it's just the easier way to do it, instead of having to find the perpendicular height. I think that makes more sense. Now the good news is working out the area of this top thing is actually pretty easy, because what's important about an equilateral triangle? Well all of the angles are 60 degrees, so if I knew these two sides, which by the way are the same, then I could work it out. But remember all three sides should be the same for an equilateral triangle. It tells me here that g b d e is a square of side x, so all of these sides are x. So what do I know? Well I know the area of a b g is a half times x times x times sine 60, which you can work out. Now I know that there's an x squared in here which might put you off, but you should keep in mind that I'm willing to bet the x's are going to cancel out, so I don't think that's going to be too much of a hassle. Sine 60 is root 3 over 2, half that would be root 3 over 4. Now again if I look at my answer, actually yeah there's no square roots yet, but let's see what happens. Now ACF is going to be a bit more interesting, because the area of ACF, well again keep in mind it's still 60 degrees, right? That doesn't change at all. But the problem is this kind of bottom length here, and we're going to have two of them, so we need to add this length, this length, add x for our sides. Now the good news is we can actually work out what the length fe or dc is, because this here makes a right angle triangle, and we know that this side is x. So what I can do is I can use trigonometry, trigonometry where this is the hypotenuse, that is the opposite side, and that's the adjacent side. So again we're going to write out SOH CAH TOA as we've done a couple of times, either this paper or in the last papers, I don't actually remember. We're going to get rid of h, because that's the only one we don't care about. We're looking for the adjacent, so there. So we have tan 60 equals x over the adjacent side, so that adjacent side is just x over tan 60. Okay, it's going to be a bit of an ugly number. Tan 60 is root 3, so this would be 1 over root 3 x, or if you write 1 over root 3 in your calculator you actually get root 3 over 3 x, which I think might be a bit nicer. Sure. And again, remember we have two of these, so when we do half ab sin c for the blue triangle, a and b are the same thing, but it's going to be x plus two lots of this, because it's one for this side f to e, one for this side d to c. So we're going to have x plus 2 root 3 over 3 x squared, and then again sin 60. Yeah, this is going to be fun. All right, sin 60 we already worked out is root 3 over 2, so it's half that, so it's root 3 over 4. And then I'm just going to expand this. It should be fine, because everything's just x anyway, so what we could do is we actually factorise out the x squared, and then we're just left with... In fact, I'm going to do that instead. So this rearranging is going to be a little bit interesting. It's something that comes up a lot more in A-level maths, but if I factorise out x from here, because it's squared, we're going to be factorising out x squared. Another way to visualise this is to write it out twice. So I'm going to do this just to, and again, what I've just done is the shortcut. Here's the four-way. If I factorise out x from here, I get x 1 plus 2 root 3 over 2, and then I'm going to factorise an x from the next one to be x 1 plus 2 root 3 over 2, which makes it eventually x squared 1 plus 2 root 3 over 2 squared. So what I just did there was a bit of a shortcut, which is quite nice, and all we have to do is expand this. Or should we put it in the ratio now? I want to put it in the ratio now, because I can see the front's actually going to cancel. So the ratio of ABG to ACF is root 3 over 4 x squared to root 3 over 4 x squared, and then all of that squared. I've said squared so many times it's kind of losing meaning. If I divide both sides by root 3, times both sides by 4, divide by x squared, I get 1 to all of this expanded. So let's expand this out now. So we're going to get, in fact, I'll write it out twice just to hopefully make it a bit more clear to everyone. We get 1 times 1, which is 1. We get 1 times 2 root 3 over 2, which is 2 root 3 over 2. We get the same thing again because it's 2 root 3 over 2 times 1. Then we get 2 root 3 over 2 times 2 root 3 over 2. So the 2 times 2 makes a 4. I've just realised we didn't actually need to do that because the 2's cancel. Oops. So that's 4 times 3 over 4, so those two cancel. So we get, in the end, 4 plus 2 root 3. 4 plus 2 root 3. If I look at the form, they're all integers. That's good. So A is 1, B is 4, C is 2, and D is 3. That's fairly interesting. Perfect. Not too bad. Just realised something. Somehow I changed my 3 into a 2 over here. So I need to actually go through this and change it ever so slightly. Luckily, it's not too bad. Like, what happened is not the end of the world. We can actually edit this quite nicely. If I edit this now, we still get 1. That would be 2 root 3 over 3 for both of these, and that will stay as 4 times 3, but on the bottom, instead, it should be over 9. So now I need to quickly edit this because that's a bit embarrassing, and we can actually get our final answer. So yeah, apologies for that. That should be 1 plus, and then 2 plus 2 would be 4 root 3 over 3, and that would be 12 over 9, which I can simplify to 4 over 3. And so when we write out our working out, we are going to leave the left-hand side as 1, but you're going to notice something, right? We get 1 plus 4 root 3 over 3 plus 4 over 3. Now, it says everything has to be an integer. So now what I'm going to do is I'm going to times both sides by 3, and I get 3. That becomes a 3. That becomes a 4. 4 plus 3 is 7, plus 4 root 3, which is the actual final answer. Oops, I can't believe I missed my own handwriting. This is a very interesting question for part A because there are two ways to do it, right? What we could do is we could find the inverse function of this, which isn't really that difficult, right? It's fairly easy to work that out. You can just make this y equals 3x minus 6, rearrange for x, so you get y plus 6 over 3, and then your inverse function is just this. But I want to show you something else which will help your understanding of what you're trying to do, especially if you're trying to aim for those top grades and you want to do A-level maths. There's a much faster way to do it, and functions do come back in A-level maths. If you think about what the inverse function means when we sub in for x, they're trying to say if I sub in 9 in place of f of x, what's my answer? So what we're going to do is we're going to do it both ways. If I sub in 9 into this equation, what do I get? I get 9 plus 6 over 3. 9 plus 6 is 15, over 3 is 5. But if you notice, if I sub in 9 into this, plus 6 to both sides, I get the exact same thing. That's because an inverse function just says if I put this into my answer, what value, what input gave me that output? That's what they're trying to say. Anyways, the next bit, the composite function for the final question of the exam. Find fg of x, giving it as a single fraction in its simplest form. So just to make it a bit more spicy. So fg of x, what does that mean? It means you're taking the function g of x and you're putting it in place of x into this equation. So this is going to be pretty heinous, I'm not going to lie. This is going to be a bit of a war crime, but it's, you know, it's fine, right? NXL wants us to do this, so we're going to do it. So 3, open massive brackets, 2x squared over x squared plus 1, then minus 6. Okie doke. Now if I expand this, we get 3 times 2x squared, which is 6x squared over x squared plus 1. Now personally, I'd be happy leaving it like this, but single fraction. So we have to combine these. And there's something actually some a level students struggle with is when you have a fraction minusing a number or a number minus a fraction, you can combine them, because any number can be turned into a fraction by just putting over one. Now we have to make the denominator is the same. So we're going to times this one by x squared plus one, and this one by x squared plus one, if you'd like, you can times the first one by one as well, you know, you do you man, it's fine, whatever works for you. Over x squared plus one, take away. And if I times this out, I get six times x squared is 6x squared. Six times one is six over this. And now we can actually subtract the two fractions. But we need to be really careful here. I'm going to show you how you're supposed to write this. It'd be 6x squared minus bracket 6x squared plus six. So many people would write 6x squared minus 6x squared is zero, then they just write six on top, it's actually a negative six on top, not a positive six. So be really careful when you're doing that. Again, I see this all the time in from year 10 to year 13. Unfortunately, that's why I'd combine it with brackets first in an exam, you're under pressure, you're going to make stupid mistakes, like I did in the last question. And there you go. That's it for the final answer. Overall, I'd say that's actually a pretty tricky paper. Out of paper one, two and three, I think the paper three is probably the hardest. But yeah, hopefully you found that interesting or useful. Again, there's links to all my tuition stuff in the description in the pinned comment. Hope you have a wonderful day. See you in the next one.