GCSE Maths June 2024 1H NON CALCULATOR Higher Tier Paper 5pm BST Edexcel | GCSEMathsPro

Added: 2026-05-09

[00:00:06]Welcome back everyone to the GCSE Math Pro YouTube channel. Today we are going through another ed Excel higher tier paper one. This is in preparation for your next week.

[00:00:18]If you want to download the paper or the mark scheme, I will have them linked in the description. It's live. And I also have an edited video of me talking through this paper. So if you want a little bit faster to paste or something that you want to refer back to, you can look at that in the description as well.

[00:00:36]So, we're going to get straight into it.

[00:00:38]Make sure you're not using a calculator for this because obviously you won't have one to exam. This is the June 2024 paper one. As I mentioned, it is for higher ed and it is edexl specifically.

[00:00:50]So, please let me know in the chat if you've got any questions. If at any point you're not sure about a question, I can go back over it. That's absolutely fine. just let me know in the chat if anything isn't.

[00:01:03]So going to jump straight into it.

[00:01:05]Question one. Here are the first four terms of an arithmetic sequence. 1 5 9 and 13. Find an expression in terms of n for the nth term of this sequence. See they've not given you much space. This is quite straightforward. So I just look at the difference. It should be the same every time. You can see here that we've got four. every time is a difference. So what you do with that is that is the number that goes in front of the n. Then just because I want to write a bit of working out, I would then therefore write down the four times tables. So 4 8 12 16 just kind of lining them up with the sequence there. And then I just have a look and say right if I'm currently at four, how do I get back to the actual number in the sequence in terms of what do I add or subtract? That one would be subtract three.

[00:01:57]Then I double check. Does that work for all of the others? Because it should do, right? 8 to 5. Yeah. Take away three.

[00:02:03]Same with 12 and 9. Same with 16 and 13.

[00:02:05]Perfect. So what that means is my nth term is 4n and then I take away three. Each time you multiply by four and then you minus three. That is literally it. And that is two marks to start.

[00:02:22]Question two. Obviously a lot of the time paper one we have got some kind of fraction. So part A is to work out 3 4s minus one and 2/3. So what I like to do especially as it's a mixed number here we've got as in a whole number and then a fraction it's mixed. I personally like to convert them into improper fractions.

[00:02:46]You don't really have to but I feel like it's just easier. I think it's more complicated if you don't. So I would just say for any kind of fractions operations, multiply, divide, add, subtract, where you've got mixed numbers, I would just convert them into fractions straight away and then carry on as normal. So converting them into um an improper fraction, we just take the whole number on the outside, multiply it by the denominator. So 3 * 5 is 15, and then you add the numerator. So that' be 19 as my new numerator. It's still out of five, that one. Same again for the second one. 1 * 3 is three. add two be five. Same denominator three. So now I've converted them. I'm actually then going to subtract.

[00:03:30]So the way that we want to add and subtract fractions is they have to have a common denominator. So what's the lowest common multiple of five and three? That would be 15. Which means I do need to change both of these fractions. So the first one I'd have to multiply top and the bottom by three. So 19 * 3. The way I did that was I did 20 * 3 which is 60. Then I took away three.

[00:03:50]But however you want to work it out mathematically in your head is completely up to you. Either way you should get 57 over 15 because we are changing the top and the bottom now. And then the second one to make it something over 15 I would have to multiply by five. So the top the numerator would be 25 over 15. Now that they have a common denominator we can actually subtract. So subtract just the numerators.

[00:04:18]57 - 25 take away 5 is 52 take away 20 is 32.

[00:04:24]Again, however you want to do the kind of arithmetic is up to you. I've got this now actually I think they would accept that because they haven't specified how to leave it. Normally I would convert it back into mixed number but they haven't actually specified have they in the question. So let me double check. Uh yeah, mark scheme says any kind of equivalent that would be fine if you did convert it into a mixed number. I'll just show you. How many times does 15 go into 32? Twice as a whole and then my remainder is another two out of 15. So either of these two answers is fine because they haven't specified how to leave your answer. Obviously if they do specify, you have to leave it in that format.

[00:05:06]So that is part A and that's two marks.

[00:05:12]There we go. Down part B. We've now got Kevin was asked to work out 2 and 1/3 * 58.

[00:05:22]Here is his working and his answer. So we need to check out they said that Kevin's answer is wrong which basically what was the mistake. So let's have a look at his working. You can also use this space here to work out yourself and then compare your answer and then you might find that easier to figure out where the difference is. So, we're multiplying. As I said, I would just convert them into mixed um in from mixed numbers into improper fractions anyway.

[00:05:45]So, have they converted it correctly? 2 * 3 6 + 1 is 7. Yep. 7 over 3 is fine. 5 over 8 wasn't a mixed number in the first place, so it stays the same. Now, multiplying across. Yeah, we just multiply straight. So, 7 * 5 is 35. Yes.

[00:06:01]And 3 * 8 is 24. So, this is fine. And this is fine. So, I'm then kind of looking at this. Okay, the conversion back into a mixed number here. So, how many times does 24 go into 35? One is correct. Yes, it goes in once. And then what's my remainder between 25? Uh 24 and 35. They put that the remainder was nine. Well, actually it's a bit of a trick question because in your head you I feel like nine and 11 are quite I don't know how to explain like not similar but when you're trying to subtract them if you subtract it the other way around like a four and a five the other way around it is nine and when you subtract it a different way it's 11.

[00:06:44]Do you know what I mean? So I can see why that would be quite easy to miss but actually it should be the one is fine, the 24 is fine. The problem is the numerator. So what is the mistake? Just put the most obvious answer you can. Um, and when I tend to do this one is I just tend to correct them. So, what is the mistake? I try to just say, well, it should be this instead. Should be one and 11 over 24.

[00:07:07]So, I find it kind of a little bit difficult to explain what the mistake was. Um, you might find it easy. That's fine. But I just say, well, what it actually should be is this. And that is fine. I will show you some other examples they said. Um, okay. So, they've said uh acceptable examples in the mark scheme. in his answer. 9 over 24 should have been 11 over 24.

[00:07:28]Effectively, that's the 9 should be 11. You can even just say that he has not got the numerator right in his final answer. You don't even have to say what it should be. Uh he simplified it into the mixed number incorrectly.

[00:07:44]He has not put the remainder as the numerator because technically they've got quite a few options. Um but that's fine if you just correct them.

[00:07:56]Okay, so that's another well that's actually one mark, isn't it? And therefore total three marks question.

[00:08:02]So question three, we've got the diagram shows a plan of a floor. Petra is going to cover the floor with paint. Petra has three tins of paint and there are 2.5 L of paint in each tin.

[00:08:16]Petra thinks one liter of paint will cover 10 m squared of floor. So for part A, assuming Petra is correct, does she have enough paint to cover the floor?

[00:08:27]Obviously, don't just say yes or no. You actually have to out. So we need to know area. This is obviously a compound shape, as in it's split um it's made up of two or more shapes. So you can split it. Now I think I would split it vertically. You can split it horizontally if you like across here.

[00:08:44]Fine. whichever you prefer. So now I've got two rectangles. So the area the first rectangle is kind of easier it this way is 8 * 6 and then because it's a compound shape I'd have to add that to the second one.

[00:09:02]Now the second one I have got 5 m but I'd have to work out the width. So that's why they've given me 10 because the whole width across the whole shape is 10. The width of this other rectangle on the left is six. So therefore, the width of the right side rectangle has to be four, right? Cuz it has to add up to 10. Now, if you did it, if you split it horizontally, you'd have to work that out vertically, right? So either way, you'd have to work it out.

[00:09:26]Either four meters or this one here, which is what's that? 3 m.

[00:09:31]If you split it that way, that's fine.

[00:09:33]You'd have to work it out. Then it's just 5* 4. So here we've got 48 and 20.

[00:09:39]If I add those, however, yeah, whatever way you've done it, you should get 68 m squared for the area. Now, let's look at the paint.

[00:09:52]So, Petra thinks one liter of paint will cover 10 m squared. So, how many liters of paint will I need?

[00:09:59]You could work it out from a different angle if you want, but I do it this way.

[00:10:03]So, I've got 68 m squared that I need to cover. Each liter we think is going to be 10 squared. So I'm going to divide that. That's quite a nice divide. 6.8 lers of paint is what I would need.

[00:10:17]Okay. Now I'm going to work out how many lers she actually has. Three tins at 2.5 L.

[00:10:26]I put multiply. You can also add that.

[00:10:28]You don't have to multiply decimal. Just put 2.5 plus 2.5 another 2.5.

[00:10:33]We should get 7.5.

[00:10:36]Okay. Okay, so Petra has 7.5 L. She thinks she will need 6.8 L. Does she have enough paint? Well, yeah, that's enough evidence there to say that yes, you have to say that it's or no.

[00:10:55]Look, that was four marks for part A.

[00:10:59]Now there is a part B which is worded question again. So actually one liter of paint will cover 11 m squared. So remember she thought that one liter would be 10 m squared. That's what we worked off of that assumption. Here we're saying actually in reality it would actually cover 11 m squared of floor. Does this affect your answer to part A? Obviously you must have a reason. Does it affect my answer? Well no because she already had enough paint and now we're saying that the paint actually stretches further like it will paint more floor. So we already had enough uh paint. So it doesn't matter that the paint that she has will then stretch further. That doesn't matter.

[00:11:39]No, I would say again something really basic. She already has enough paint is fine. You don't have to explain it in too much detail. You already have enough. Okay, it's fine.

[00:11:52]They do give you three lines. You don't have to go up with three lines, by the way. Just write the most obvious answer you can. This is because it's math. It doesn't have to be like some kind of deep analysis or anything. Okay, so that's another mark, five marks total.

[00:12:06]Question four, here is a ven diagram.

[00:12:09]We've got part A. Write down the numbers that are in the set P apostrophe. You have to recognize what this is. This is a set that is not P basically. So anywhere that's not the P bubble, this is the bubble, the circle if you want to call it, anywhere that's not there. and we have to write them down. So that would be 10 and 14. You don't have to write them in order. You can if you want. So whatever's in cube that's not in P. And then whatever's on the outside 11, 13, 15, and 17.

[00:12:42]All of those are not in Pic. That's your answer.

[00:12:47]There's no working out that you need.

[00:12:48]You just have to look at it like that.

[00:12:51]This part B. Now we need to work out an actual probability.

[00:12:58]So a number is chosen at random from the universal set. What that means is just the ven diagram as a whole, right? So we're picking a number out of any of these numbers here. Find the probability that this number is in the set P union Q. Again, you have to remember what that means. That means that's P or P. P or Q, even P or Q. Now, what part of my ven diagram is that referring to P or Q. When it's or it's basically both bubbles, E or Q.

[00:13:33]I like to think about it like this. When it says or, it's normally a bigger area of your vend diagram. When it says and, which is the like the N shape, it's normally a smaller area. So, P or Q. Okay. So, the probability of picking one. Well, how many chances do I have of picking in that area there? 1 2 3 4 5. There's five numbers there. That's five chances out of how many chances do I have in total in my whole ven diagram? 1 2 3 4 5 6 7 8 9.

[00:14:06]So, do you see how I've got 5 9ths of a chance that any of these numbers I pick happens to also be in the yellow bit that I've highlighted. That's literally it. They do give you space to work out.

[00:14:16]There's not really anything to work out.

[00:14:17]I think in this case, two marks effectively, one for each number, the three marks that whole question.

[00:14:28]Question five, Sophie drives a distance of 513 km on a motorway in France. She pays 0.81 for every 10 km she drives. Work out an estimate for the total amount Sophie pays. So, when it says estimate, it's obviously also non-cal. I'm just going to round them to normally one significant figure. It does depend on the context, but in this case that works. So I'm going to round the distance to just 500 life easier. And then the cost sign to I'm just going to round it to like 0.8. The one is a bit unnecessary, isn't it? So that's for every 10 km.

[00:15:08]Okay. So how many lots of 10 kilometers do I have in 500 km or divide?

[00:15:18]So 50 what that's saying is I've got 50 lots of 10 km and every 10 km it cost me 80 cents I think that is. So 50 time 0.8 however you want to work that out is up to you. I look at that I know we're talking about cents and euros or whatever. I look at that as just going, what's effectively 80%, isn't it? But you might prefer to use percentages to work that out. You might prefer to convert that to a fraction and work it out. However you want to do it is fine, as long as you obviously get end up getting um that it's a correct answer. So I just thought of it as 80%, well 10% of 50 is five. So 8* 5 is 40.

[00:15:58]That would be effectively 40, wouldn't it?

[00:16:01]That's how I worked it out. But again, however you want to work out, it's fine.

[00:16:03]Even if you round it slightly differently, as long as you do round it, um I think because they said estimate, don't literally what that's three marks.

[00:16:12]And then obviously part B is normally is your answer to part A an underestimate or an overestimate.

[00:16:18]So I rounded the distance down, which in that case means she likely would have cut me more because I've estimated for a shorter distance than what she actually drove. And I also rounded the price down. So either way, that's got to be an underestimate, isn't it? Because I rounded both down. That's your answer. You have a reason for your answer. Most obvious thing is because I rounded both of the numbers down.

[00:16:47]Rounded up like that. Rounded numbers down. Literally, that's fine. That's enough explanation.

[00:17:02]Scroll down. That's one mark of worded answer and four marks in total.

[00:17:13]Okay. Question six. Here is a straight line L drawn on a grid. Part A. Find an equation for L. So we need to remember that all straight lines have this format kind of to work with. Y= MX plus C has to be a straight line. That tells me I need two things. I need a gradient of a line and I need plus C, which is the Y intercept.

[00:17:39]So I normally work out the gradient first. Obviously, you can see the graph, so you can see what see it, but I I tend to like to work out the gradient first, just generate. So let's do that. Now, it's a bit easier because we can see it.

[00:17:49]You can choose two sets of coordinates.

[00:17:51]Obviously, don't choose like something like this because you're just guessing at what that number is. Try and choose kind of like all coordinates, I guess, is what I call it. So I'd probably choose this one here. Then maybe the next one along which is here. And then we're just saying change in y over change in x ar we could just count squares if you wanted in this case a little bit easier.

[00:18:13]So for the gradient it is so the formula change in y the triangle there just means change like difference over change in x. You can see that y we've gone up by three and x I've gone along by two.

[00:18:27]Do be careful of the scale but that does correlate. So my gradient here would be 3 over2.

[00:18:36]Now if you do the one where you subtract both of the coordinates that's also absolute. So however you want to work out the gradient you should get 3 over2 or in other words.

[00:18:45]Now that obviously goes in front of the x and then the plus c is really easy because we can see it just where it crosses the y axis which in this case scribbled all over it see three. So our answer is y is equal to 3 over 2x + 3.

[00:19:04]That's it.

[00:19:07]What you put here? Three marks.

[00:19:11]Obviously if you put 1.5x that's also fine right? any kind of equivalent.

[00:19:17]Now we've got part B. So m is a different straight line with equation y = 5x. So ignore part a. It's not relevant. Now it's a completely different straight line. They've told us the equation is y = 5x. Write down the equation of a straight line parallel to f. So any straight line that's parallel.

[00:19:38]So when we have parallel lines, they have to have the same gradient. So the letter the number in front of x the coefficient of x has to be the same.

[00:19:48]I'll put y equals the gradient for this one is five 5x but I have to then make it a different line. I can't just leave it like that. Any different y intercept doesn't matter. 1 + 2 + 5 - 3 - 2.5.

[00:20:06]Like it doesn't matter as long as you add or subtract something else at the end.

[00:20:11]I haven't specified any coordinates so I know that I can just make one. So that's my example. As I said, you could have any like one 25 - 73. Like do you know what I mean? You could do any as long as you have got the 5x and then you change the end. That's fine. And that's just one mark. That's why they said write down. If they ever say write down, normally it's a clue that the answer is kind of like there's not really anything you need to work out. Just have to write an answer. So four marks in total for the straight lines question.

[00:20:45]Question seven. A sim has some small jars, some medium jars, and some large jars. He has a total of 400.

[00:20:53]3/8 of the 400 jars are empty. For the empty jars, the number of small jars, the number of medium jars is 3 to 4. And the number of medium jars to the number of large jars is 1 to two. Work out the percentage of Kim's jars that are empty small jars.

[00:21:12]I'm just going to work through an order information.

[00:21:17]First of all, we've got 400 jars. Let's work out how many are empty. So, what is 38 and 400? And we don't have a calculator. We want to divide by the denominator and multiply by the numerator.

[00:21:31]You could do bus stop if you really wanted to, but I kind of went off the basis of I know that four divided sorry I know that 40 divid by 8 is five. So 400 divid 8 would be 50. That's how I worked it out but you want and then times by 3 is 100.

[00:21:48]There's 150 empty jars.

[00:21:54]Now looking at the next piece of information because I've used that now tick tick. I've got these two ratios but they overlap. They both have medium jars. So small to medium and medium to large. So I obviously need to have one ratio but for the three all together. So looking at where they overlap, we've got three to four and then if I put it below one to two, they overlap on medium.

[00:22:18]What's the lowest common multiple of four and one? That's just four. So I only need to change this second ratio four. So the 3 to four can stay the same and then the 1:2 would become four to 8.

[00:22:35]But now I can write it as a three-part ratio. 3 to 4 to 8. Okay, great. I need to share out this ratio, but I am only interested in small one. Right? So when we share out a ratio, you add up all the parts. So 3 + 4 is 7 + 8 15. They've obviously chosen the specifically you divide your total and because it is the empty jars the total by how many parts of a ratio you have that gives me 10. So what that's saying is there are 10 jars for every part ratio. I want to know the small jars. There's three parts. The small well small empty ones and such would be the 10 * 3 which is 30.

[00:23:21]So now I know how many of my jars are empty small ones. But that's not the final answer, right?

[00:23:31]Double check.

[00:23:34]When you've got a long question, don't just think, "Oh, numbers of work. That's probably the answer." Double check what actually was question. So the original question was, "Work out the percentage of Cassim's jars that are empty small jars." So percentage of all of the jar.

[00:23:47]So 30 of them are small empty jars out of originally he had 400. doing it out of what I said.

[00:23:58]Make it into a percentage we normally times by 100. Again, however you want to work that out. Um, I might just multiply by 100 to start with. That's 3,000 400. And then you can knock off and I've got zeros and then I've got 30 divided by four. So half it, half it again 7.5. But again, however you work it out mathematically, it's fine. You can do it differently. As long as you should get 7.5, that's the most important.

[00:24:31]You go about it a different way, you still get 7.5 there. Like 7.5% of his of all of his jars are small empty jars. That one's obviously a five marker because there was a lot to do.

[00:24:50]Question eight. Len has eight parcels.

[00:24:53]The mean weight of the eight parcels is 2.5 kilos. The mean weight of three of the parcels is 2 kilos. Work out the mean weight of the other five parcels.

[00:25:04]So this is what I call like reversing the mean. We kind of have to undo it. So if you think about what you would normally do, let's say if you knew the eight parcels, the weights, you would add them all up and then you would divide by eight and you've got 2.5. So the opposite of that would be to take the 2.5 and multiply it by eight again because that would give me the total of all eight. I obviously can't figure out what all of them are individually because I don't have that information.

[00:25:31]But I can figure out what the eight parcels would add up to as a total weight.

[00:25:36]That gives me 20. I don't know why I'm done. Pretty sure that gives me 20.

[00:25:44]Double check.

[00:25:46]Yeah. Yeah.

[00:25:47]So 20 kilos for eight parcels. Do the same with the other three. What's the total weight of those three sense? 2 kilos multiplied by three parcels obviously 6 kilos. So the other five parcels I could figure out the total weight of.

[00:26:06]Again I can't figure out what they are individually but I don't really need that much detail. So 20 minus 6 is obviously 14 kilos. You see how that's then the weight of the other five?

[00:26:18]And then how do I work out the average of five? Well, the total weight divided by five. Now, when I divide by five, what I tend to do is divide by 10 and then times it by two. I just find that easier. Up to you. So, divide by 10 would be 1.4 and then double it would be 2.4.

[00:26:35]I just find that easier when you especially when you got decimals.

[00:26:42]So, reverse the mean.

[00:26:46]Okay, that one is three marks.

[00:26:51]Question nine.

[00:26:53]In a sale, the normal price of a coat is reduced by R%.

[00:26:58]Given that the sale price is equal to 0.7 multiplied by the normal price, find the value of R.

[00:27:06]This is obviously a percentage decrease kind of like multipliers.

[00:27:11]When want to take a percentage off, obviously you can work out that percentage and then subtract it from the total. But there's a quicker way, isn't there? Is doing this.

[00:27:23]So this decimal here is your multiplier.

[00:27:26]That's effectively 100% minus the reduction and then put it as a decimal. But what have they subtracted here? That is obviously 0.3 which as a percentage 0.3 as a percentage is 30.

[00:27:44]They've reduced it by 30%, haven't they?

[00:27:46]But as a decimal, right, as a multiplier. That's only one mark. You don't need to do any working out. That's it.

[00:27:56]Question 10. We now have simultaneous equations. So 5x - 2 y = 23 and 2x - 3 y = 18. and they're both linear. So, you can either do substitution or elimination. I tend to use elimination.

[00:28:10]That's just my preference, but you can you can do it the other way. That's fine.

[00:28:15]Now, for this one, if I do elimination, I need to make either the x is the same amount or the y is the same amount.

[00:28:20]Unfortunately, I wouldn't say that one is easier than the other. Normally, I try to have a look and see which one might be easier, in which case maybe only have to change one of those equations. But in this case, I would have to change both either way. So, if I were to make the x's the same, I'd make them 10 x. And if I made the y's the same, I'd make them six. I think I do it that way just because the y's are smaller numbers. But it really doesn't matter that much. So, I made the y's the same. That was just my personal preference, as I said, because they are smaller numbers, which means I'm multiplying by smaller numbers and I'm keeping all my numbers as small as possible. That's the only reason. You can make the x's the same 10 x, no problem. You should get a sim. So, in order to make them both the same in terms of y, it' have to be six y, wouldn't it? So, I'd have to multiply the top one by three.

[00:29:09]Remember, the whole equation, not just that bit, and the bottom one by two. So, they're quite small numbers, that's why I've chosen it. So, the top one would be 15x - 6 y = 9. And the bottom one would be just times by two. 4x - 6 y and then 18 * 2 is 36.

[00:29:35]Then you go, how do I effectively cancel out the y's? Well, it's same sign subtract. They're both negative, aren't they? So, they're the same sign. Same sign means you subtract the equations.

[00:29:48]Same sign, subtract. If they are different signs, plus and minus, then you add. So 15x - 4x is 11x obviously - 6 y take away - 6 y that would make zero. The whole point is that they should cancel.

[00:30:04]69 - 36 is then 33.

[00:30:09]Now I've got an equation with just x. So solve that. Divide by 11 is 3. X is 3.

[00:30:14]Now you've done the hardest bit. Once you've got one of the letters, it's really easy to get the second one. So just choose one of your original equations. It doesn't matter whichever.

[00:30:22]I tend to choose whichever one looks easier, whichever one has smaller numbers. So I probably choose equation number two and sub in that value you've got. So 2 * x would be 2 * 3 - 3 y= 18. Now I can solve this for y.

[00:30:38]Obviously that makes six. What I'd probably do here actually because I've got a negative y is I would just add it across because if you add it or if you move it to the other side of the equation obviously it makes it positive and then I would take away the 18. So effectively swap those two is 3 y and divide by three. You can obviously solve it a different way but like to do it that way. So y you should get minus four. As long as you get4 you're so obviously x is 3, y is -4 that is four marks.

[00:31:23]Okay. Question number 11.

[00:31:26]Triangle A is translated by vector 6 - 4 to give triangle B.

[00:31:33]B is rotated 90° clockwise around the point two to give triangle C.

[00:31:40]And then we need to describe fully the single transformation that maps A to B.

[00:31:44]A to C even obviously make sure you read that correctly. That is very important.

[00:31:48]A to C. Okay. So we've got a vector. The top number is left or right. Because it's positive it means right by six. The bottom number is up or down. Because it's negative it mean by four. So, just pick a corner. I'm probably going to use this bottom one here and go one, two, three, four, five, six along and then down by four. One, two, four. So, that's that corner there. It hasn't rotated or anything. It's just the same shape. So, I'm going to draw it here.

[00:32:21]You can always draw, label, annotate the diagram on your So, this one is B. Now from here we then want to rotate it 90 degrees clockwise about the.12 which I would recommend obviously in an exam you should have tracing paper.

[00:32:37]I would use that. That's going to be the easiest way to do it. Obviously I don't have digital tracing paper. So I just figure it out by counting the squares but hopefully you can use tracing paper.

[00:32:48]So currently if I did it without tracing paper it is currently I'm going to go to this corner here.

[00:32:56]two, three, four down from that point.

[00:33:00]Four down and two right. Now, if I'm going 90° clockwise, what was down is now left, isn't it? So, it's four left.

[00:33:12]One, two, three, four. And then what was right becomes down. One, two.

[00:33:19]And remember that it has rotated as well. So, trying to that's why it's quite hard to do it without trying to imagine it. this on its side by 90°. I did this top corner here should be like this should be C.

[00:33:41]Hopefully you can just use your tricks of paper to do that much easier. Now, how do we get from A to C? Well, clearly it's another rotation, isn't it? But we just need to give some details about that because can't just write rotation.

[00:33:54]That's why they said describe fully. So what can we say about rotation? They've actually told us one rotation. So you can see you need to know the degrees, the direction, and then what's the center center of rotation. So I would say it's gone that way, isn't it? That's 90° clockwise again.

[00:34:18]But what is our center of rotation? What I would do is again, if you have tracing paper, just put it over here and just keep testing out different points. But I can kind of imagine that it's going to be here cuz what was up and right is then right and down. That's like a 90° rotation, isn't it? So that's how I figured it out. But kind of trial and error on your tricks and paper.

[00:34:44]Okay. Which bit are you not sure about?

[00:34:46]Do you want to just tell me like because obviously I'm speaking a bit ahead so I don't know at which point you're unsure.

[00:34:50]Just let me know in the comments. We can go back.

[00:34:54]Which part of this is it that we're not sure about that minus 41?

[00:35:00]I'll go back over it. It's quite hard to demonstrate because as I said I don't have the tracing paper but in your exam it'd be much better if you could use tracing paper. You can then just literally draw over it and test it out visually rather than count squares here.

[00:35:19]How do I know where to move the shape?

[00:35:21]Just from this piece of well, these two pieces of information here.

[00:35:26]That's I just basically went from a which was translated. Translated means you just move it up and down, left and right. Um, they told me that it was by this vector. And we should know that a vector like this, the way that it's laid out, means I need to go six to the right and then four down. So that was the first bit. So from A, I counted the squares six across to the right and then I counted four squares down and then I drew that shape. It hasn't rotated at that point, has it? So it's still like the direction is still the same. That became triangle B. And then triangle B, it told me rotated 90° clockwise about the point two to give triangle C. That's the bit that I would say use your tracing paper for.

[00:36:14]I just counted it because I don't have any tracing to kind of show you on on the screen if that makes sense.

[00:36:22]Yeah, it's it's just the instructions they gave. Basically, that was all. So, there we go. I've given enough information. You can see it's three marks. So, I guess a mark probably moving them around, but I've said rotation 90 degrees clockwise and I figured out where the center was. So, I've given the Okay, so that's three marks.

[00:36:45]This one I've kind of copied the second half of the page here so you can look at it at the same time. Obviously this piece here would be there on your actual page. So here are some graphs. We've obviously got nine graphs here.

[00:36:57]So you can see here write down the letter of the graph that could have the equation and then there's three options for the first one. This is where we just need to recognize graphs right this one x^2 - 4 that is a quadratic graph. So positive quadratic looks like a U shape.

[00:37:13]So first of all have a look at the ones that look like a U shape. We have got two quadratics here. But this one's an N shape which is a negative quadratic. So it's not that one. It would be h. And you can see that's the four that it's talking about. It's the y intercept, isn't it? There.

[00:37:31]So this one is h.

[00:37:35]Then the second one we've got y=x.

[00:37:38]So that's a cubic graph. Now this is again why you need to know the difference between a positive and a negative one because you probably know what a cubic graph is roughly you know how to identify which is positive and which is negative because here actually here we've got three cubic graphs got three here this one's a positive one this one's also positive and so we want the negative one which is f it's not this one it's not that one it's f but yeah there are three cubic graphs listed here and the Last one, y =5x.

[00:38:12]This is a reciprocal graph, but also a negative one again.

[00:38:18]Now, we've got two reciprocal graphs.

[00:38:20]We've got B and we've got J. You just have to know which is the negative one.

[00:38:25]Now, the positive one is B. So, therefore, obviously, the negative one is J. So, you might not only have to know roughly what graphs look like.

[00:38:36]Sometimes they want you to distinguish which is positive and which is negative.

[00:38:43]Scroll down. See what?

[00:38:50]See that? That's where it was. So I put H, F, and J.

[00:38:56]And that was three marks. That's just a kind of memorization really.

[00:39:03]Question 13. The table gives information about the amount of time that each of 150 people were in a shop. Part A on the grid draw a histogram for this information. So histograms we need to know the axes. One of them will be the I guess the intervals. So this column time in this case that's the x axis. The y ais is not the frequency. It's actually the frequency density which means we need to work out frequency density. So I like to just add I haven't really given space for working out. I like to add a column onto my table that's already there. Now work out frequency density like to use this formula triangle for histograms. Basically what you frequency goes at the top. The way I remember that is because it's the only one with a single letter. The other two have both got two letters. So I put them at the bottom of the triangle. We're looking for frequency density. So it's the frequency which is this column divided by the class width. So 20 / 10.

[00:40:04]You see how the the width there between 0 and 10 is 10. That would just be two.

[00:40:09]Then 70 divided by that width which is 20. Now do have to be really careful because often with histograms they tend to have different widths.

[00:40:21]Effectively that's.5 22 / 5. Remember I said dividing by 5, I divide by 10 and double it. That would be 4.4.

[00:40:33]30 / 15 2 and 8 / 10 is 0.8.

[00:40:40]I do that first just because can you see how they haven't actually labeled the axes at all. So I want to know how to scale the axis. Now the Xax is a bit easier because you can see it goes from 0 to 60 and we have got six big boxes I guess you want to call it but I'm just going to go 10 so on there to 60.

[00:41:05]Now I would label it as well. So as time and in the Y ais look at your frequency density. The biggest I have is 4.4.

[00:41:18]four. So let's say five and you can see that they are we have got five five big rocks. So that does also match up three five. You can see that I've scaled it appropriately. I'm not using just like half of the graph. I want to use it and then put obviously don't on your actual exam right don't write outside of these lines. I'm just doing it because I can't really I don't have space on this one.

[00:41:45]I've written too big. Frequency density.

[00:41:49]Obviously on your piece of paper be much bigger. So you be able to fit the words there. Do not write outside of this gray circle as you can see. Yeah. Do not write in this area on either side. Okay.

[00:42:02]So frequency density. Right now what I need to do is effectively like a bar chart in terms of rectangles. I guess they're bars. They're not um it's not a line graph. So the first one was between zero and 10 and it goes up to two. Get ruler. I would do it with a pencil just in case you do it wrong and you want to redo it there. Mine won't be as straight as yours will be will be able to do it with a pencil. Then we've got from 10 to 30 3.5 to as I said often times your bars will be different widths quite common.

[00:42:47]This one we go down.

[00:42:54]Next one I think was only five across.

[00:42:56]Yeah. 30 to 35 at 4.4. So we're going up again.

[00:43:01]Careful of the scale that you've chosen.

[00:43:09]extend it.

[00:43:14]Uh, we've got 35 to 50 and that's at two.

[00:43:23]Last one 60 which is 0.8 eight bad.

[00:43:35]So that would give you three marks as you can see. Now there is obviously a part B.

[00:43:43]Work out an estimate for the fraction of these 150 people who were in the shop between 20 and 40. Let's have a look at our histogram here.

[00:43:56]So 20 minutes was the starting point.

[00:44:01]See that?

[00:44:03]And then 40 minutes which is here.

[00:44:07]I try to get my highlight.

[00:44:11]We're talking about all of this time. Do you see how I'm kind of covering three bars? Not not fully, but I've crossed into three bars.

[00:44:30]You don't need to color yours, but just so you can visualize it.

[00:44:35]Go. I just need to work that out as frequency. Obviously, this middle bar, because it's the whole bar, this thin one. I know what that was. That was the 35, which was 22 people. I already know this one. It's basically the other two I'm working out. Now, have a look at the first one. The first bar, the whole bar goes from 10 to 30. It's a width of 20.

[00:44:56]20 is the midpoint, isn't it? It's like the halfway point of that bar. So, I can see that I've clearly got half of the bar. Now, the whole bar, yeah, you can just about see that the whole bar is 70 people. So, that's why they said estimate, right? We are just estimating. If all of those 70 people were distributed equally between 10 and was it 10 and 30 minutes, then I've got half of those people, haven't I?

[00:45:21]Effectively, from 20 minutes on, that's my estimation. So, I'm going to say half of 70, which is 35 people. That's my estimate. I know that this one's 22. So, I'll leave that there for now. And then over here on this bar, which it started at 35, it goes to 50.

[00:45:40]So, each of these kind of thinner squares columns is worth five. So in this case, I've got a third of that whole bar, haven't I? I've got a third of it. So again, imagine they were distributed equally along that bar. Was it 30 people?

[00:45:55]So a third of 30 just divided by it's just 10 people.

[00:46:01]My estimate for that one is it's 10. And now I just add it up. Got 35 + 22 + 10 gives me 67. But don't forget how they wanted you to answer it. It wasn't that's not just the answer, right? They want it as a fraction of all of the So literally just write it as a fraction of 50.

[00:46:24]I don't think you could simplify that, but even if you could, they haven't told us to, so I'm not going to. That's I think enough working too much.

[00:46:33]Yeah.

[00:46:34]67 out of 100. I'm assuming they would let you have any other equivalent fraction, but I just leaveest.

[00:46:43]So that whole question is five marks.

[00:46:47]Question 14, we've got expand and simplify. 3x - 1, 2x + 3, and x + 5. So triple brackets. The way that I expand triple brackets is I just pick the first two. You can pick any two, but just to pick the first two. Expand those first like double brackets.

[00:47:05]simplify it and then bring back third bracket at the end. So just going to expand first two got 6 x^2 + 9 x - 2x and then simplify that is well unsimplifus 3 and then bring back whichever bracket you're missing which in this case is the third one. Now I like to expand it. A lot of people like to expand it the other way. I like to expand it this way just because I'll show you.

[00:47:49]So 6 x^2 * x would be 6 x cubed. And then I take the 6x again and do second of that bracket. 6x^2* - 5 is - 30x^ 2.

[00:48:02]Then I know I've done the 6x^ 2. I've already multiplied that by both terms here. So I move on to the 7 x 7 x * x is just 7^ 2 and then 7 x again - 5 - 35x and lastly move on to 3 - 3 * x is 3x - 3 * 5 is positive there positive 15.

[00:48:34]Now, as I said, the reason why I do that is because then just a little bit easier to simplify. Do you see how I end up if I do it in that exact order, I end up with my like terms right next to each other. So, I'm just fine. See, that's just personal preference. It doesn't really matter. So 6 x cub - 30x^2 + 7 x^2 is - 23^ 2 - 35x - 3x is - 38 x and + 50. And that's my final one.

[00:49:08]Yeah, I don't know. I just find it makes sense to me in my head that way. If you want to expand it in any other order, that's also fine as long as we end up with the same answer at the end.

[00:49:19]That is three marks. Obviously, write your answer down here.

[00:49:25]Question 15. O A is a sector of a circle with center O and 6. The length of the ark A is 5 cm. Work out in terms of pi the area of the sector.

[00:49:40]So, I know the radius. I know the arc length. I might as well set up that equation, that formula. So the formula for arc length is 2 pi r* diameter. That's also fine. 2 pi r multiplied by the angle would be this bit here, right? To use theta for that over 360 and that equals the arc length. So if you substitute in, I know that the arc length is pi. I also actually know what radius is.

[00:50:15]Now that then kind of clearly tells me, oh yeah, the bit that I don't know is the angle. And then I could use this formula just to work out this equation.

[00:50:26]So let's just multip first of all what I like to do is got pi on both sides.

[00:50:30]Might as well cancel that out. Basically divide by pi. It all cancels. 2 * 6 is 12.

[00:50:37]That would go into the numerator.

[00:50:39]However you want to kind of simplify this is really up to you. I just Look at it like that. Go right. Okay, I can simplify that. What is 360 / 12? That would simplify quite nicely to 30. I'd be left with I can I can divide the top of that fraction by 12 the angle and then 300 30 then I can just multiply across.

[00:51:08]That's how I would do it. As long as you end up with 150 degrees, it doesn't matter how you've actually worked it out, right?

[00:51:15]So, this is now 150. And then I now have all the information I need to work out the area of the sector because that formula is p<unk> r 2 * 360. Now, I've got the angle, haven't I?

[00:51:33]And I already had the radius. So pi * 6^ 2 * 150 over 360. And then I'm just going to simplify that. They have said leave it in terms of pi. So you don't need to work out the pi bit. That's fine. Often when it's non calculator they don't make you work out pi. So that I'd obviously just knock off zeros. That fraction I can divide by three. That would give me five 12. Is that right? Yeah. 5 12. And then I would just again however you want to work it out. 36 / 12 is 3. So 3 * 5 is 15 and just leave it in terms of pi.

[00:52:14]Again, however you work it out doesn't matter. You might have done a different kind of arithmetic to me. That's fine.

[00:52:19]As long as you end up with 15 at the end, then you would have got it correct.

[00:52:24]And that one is question 16. There are only n orange sweets and one white sweet in the bag.

[00:52:42]Syra takes at random sweet from the bag and eats the sweet. She then takes at random another sweet from the bag and this sweet show that the probability that Sarah eats two orange sweets is n minus one over n + one. Now this one I just like to think of it as a probability tree just I feel like probability trees cuz they're quite visual. That's what helps me these ones.

[00:53:06]So let's just imagine you were going to set up your probability tree. We've got for the first suite two options, orange or white. Now, let's create probabilities.

[00:53:18]The chances of getting an orange suite is n orange suites out of all of the suites.

[00:53:25]Now, all of the sweets would be these two added together, wouldn't it? N + one. Now, technically, the white suite would be one, not an N plus one, but that's better question. Now, for the second suite, I'm only going to fill out this one here because we wanted Zara to take two orange sweets. I'm going to follow this path.

[00:53:45]This is then a little bit different because she's eaten the first sweet, which means she has not replaced the first sweet in what was it? Bag. Yeah, she hasn't put it back. So, there's no longer n let's say she has already picked an orange one. There's now one less, i.e. N minus one. And out of it's no longer n plus one in total. She's already eaten one of those sweets. There's just n sweets. Now take away one n. Now I'm not going to fill out the rest of the tree because that's irrelevant to the question. We're just looking at the options for orange and orange.

[00:54:24]And then when you're using a probability tree, remember that you'd multiply across, wouldn't you? So you've got n / n +1 multip.

[00:54:32]And if you were to multiply those together just in terms of algebraic fractions, the ns would cancel. So you'd end up with n -1 over n + one, which is exactly what they've got. That's how it makes that's just two as well. So if you were stuck on that, don't worry. Don't spend five minutes on it. No point any worse two marks. So just move on if you are stuck on any kind smaller questions. Try not to lose too much time on them. Okay.

[00:55:07]Now we've got some thirds. So question 17 a. Rationalize the denominator of 1 over<unk> 7. So when we've got just a singular third in the denominator, we just multiply by that third. But because it's a fraction, you have to do the same to the top as you do to the bottom. So I'm multiplying by root 7. That would give me roo<unk> 7 in the numerator, which is fine. And when they say rationalize they just mean the denominator can't but the numerator obviously can roo<unk> 7 * roo<unk> 7 in the denominator would be roo<unk> 49 square number so it's square roots to become just 7. So roo<unk> 7 over 7 rationalize that's why it's only one mark and part B we've got simplify fully roo<unk> 80 minus 5. Now with SS you can't add or subtract them if they're not the same number. So we're going to simplify first. Root 80 should break down. Root five is right.

[00:56:06]So what square number goes into 80. Now there's sometimes square numbers well sorry sometimes SS can have multiple square numbers that are factors. Same with fractions right when you're breaking down a fraction. If you use the smallest one then you might have to break down the third multiple. So, we're kind of trying to look for the highest factor that's the square number. Now, the clue here is because we know that in order to subtract, they both have to be root five. So, I'm thinking, well, what do I multiply by five to get to 80? That would be 60. A bit of a clue there.

[00:56:40]Then, we can square root the 16. That's why we want to square numbers because it will square root. So, we're left with 4<unk> 5 minus another roo<unk> 5. That doesn't get rid of the root fives, right? That's just saying how many root fives do I have? Like imagine 4x - x.

[00:56:56]It's still certain number of x's but now we have three effectively works the same as algebra.

[00:57:07]So 35 marks and three marks in total there.

[00:57:15]Question 18. Show that 0.15 recurring plus 0.2 2 27 recurring can be written in the form n / 66 where n is an integer. So this is the question where they basically say convert the recurring decimals and fractions but we just happen to have two to do both. Now there's actually two ways to do this.

[00:57:37]You can if you like add the decimals together and then just convert one decimal to fraction.

[00:57:45]Personally, I don't really like to deal with decimals. So, I would prefer to convert them both separately into fractions and then add fractions. That's my personal preference. As long as you get to the same answer at the end, it's so I'm going to convert them both into fractions. So, x is.15.

[00:58:06]And then for this one, we need another version with the exact same decimal.

[00:58:13]Because we've got two digits recurring, I'd have to multiply by 100. Actually, 100x and that would give me 15.

[00:58:25]You see how I've got exactly the same decimal. That's the whole point because then going to subtract 100 - 100x - x is 99x. But on this side, the whole point is that when I subtract them, the recurring part of the decimal cancels. 15 - 0 is just 15, isn't it?

[00:58:46]And then solve it to find x again. So x is 15 over 99. You can simplify that.

[00:58:51]I'd probably just leave it for now. Just hold fire on that. So if x is 0.15 recurring, it's also that fraction. Do the same for the next one. The next one is a bit more complicated though. Use y just so it's not saying we've already used a but you can use any letters that you like.

[00:59:10]So y is 0.227. Now this one's a little bit I guess a little bit harder just because you see how we've got this two which is not recurring. second two and the seven are recurring, but the first two is not. So, if you have that case where you effectively want to isolate the recurring part in the decimal and anything else out of the way, I'm going to use 10 y because that would move the two that is not recurring out of the decimal. You see how now it's a whole number and then I need a second version of that. So, like here we had x. I didn't have to change it for the first one with all of them.

[00:59:56]>> Sorry about that. That's my uh earthquake alarm. So, we didn't have to do it at the first one just because it was already that all of the decimals were already recurring. So, I didn't have to get any of them out of the way.

[01:00:11]This one I had to get out of the way.

[01:00:12]So, now I need a second version of that, which means I'd have to multiply twice again, which would be by multiplying by 100. but it's already 10, so it' be a th00and.

[01:00:23]So, it's a little bit tricky to get your head around, I guess. So, we'd have 227.27.

[01:00:29]But the whole point of this is that you see how I've got the exact same recurring decimal twice.

[01:00:35]And then same before, subtract 990 y.

[01:00:43]Then I can cancel out because obviously decimals will subtract and just cancel.

[01:00:47]But don't forget you are actually subtracting here. 227 minus 2 is 224.

[01:00:54]And then lastly divide.

[01:00:59]Okay. So I've converted both.

[01:01:03]I now need to add them. Right? Don't forget to add them. The reason why I didn't simplify them is because I'm just looking, okay, when I add fractions, they do have to have the same denominator. So I'm actually just going to multiply this one here, the first one, by 10 to make it 990. So I'd have 150 over 9990 just because I can't divide this one by 10, right? Otherwise, I would have made it smaller.

[01:01:26]Add 225 over 990.

[01:01:30]When I add them, I get 375 over 990.

[01:01:39]and then just double check because the format of the answer they have is obviously simplified version. So while I have got an answer, I do then need to by that.

[01:01:56]So you just have to kind of try and figure out. I know it's such a big number, but like what do I divide 920 by to get to 66 and then you buy that as well cuz I told you unfortunately you can use bus stop or you can work it out however you like.

[01:02:10]That even help. Yeah, it yeah it would go into 99. So then what's that 33 left over? Then 330 sure that's five times. It's half bigger. Yeah. So it goes in 15 times. So I need to then divide 375 by 15 goes into 37 twice. The remainder of seven goes into 75 times. Sure.

[01:02:39]25 to write the 25ly over 66. That would be final answer.

[01:02:49]So the 1,00 y does have the 27 because it's recurring, right? Effectively, if I were to write that out, if you had it in your calculator, if you were to make it uh written out, it would be like this, wouldn't it? It goes on 27 27 27 27 forever. So when I multiply by 100 or a thousand even in this case, there's still another 27. If I multiplied it by another 100 again, there'd be still another 27. Like the recurring decimal is always there.

[01:03:24]That's why it's recurring. So I wanted to cancel that out to me fractionally.

[01:03:34]The same as this one. When I multiplied it, it became 15.15. There was still a 15 there. 1515.

[01:03:46]Yeah. 25 over 66 should be your answer.

[01:03:50]Obviously giving you a lot of space to work it out and you get three marks.

[01:04:00]Okay. 19. This is the first one I thought.

[01:04:05]little bit difficult. They've given you a lot of space. You need the whole page when it's only three marks, but I think it's just to run an error type thing.

[01:04:12]So, we've got ABC and DAB, a similar isoclesles triangle. ABC is this kind of smaller vertical. And then DAB is this one, the bigger one, but it's on its side. They're both isosles triangles, and they are similar, which one's just like exact copy, bigger scale, the other one. And then they've told us which ones are the same. So which side we've got a and b uh ab sorry and ac are the same.

[01:04:39]And then on the bigger triangle we've got a d and bc as a whole. Right? Then they said that the ratio of BC to CD is 4 She want Okay, so I put 4 and 21 there. They're not actually like it's not 4 cm, right?

[01:05:25]It's not an actual measurement. It is just the scale, but just so you can have a look. So that means that the whole of BD as a scale, you could say would be 25.

[01:05:38]And because AD is the same, say that one's 25 just if you wanted to annotate it a little bit.

[01:05:46]So we've got find the ratio AB to A. So AB is this shorter side here to AD which is the side there. Now I think the easiest way was if I just highlighted that if I drew lines on it.

[01:06:12]We've got AB actually not sure is if you look at the bigger triangle that's the base of the tri base of it and then we've got to ad here which is one of these like one of the equal sides.

[01:06:41]And then we've got BC, which is the base of the smaller one, and AB is one of its two equal sides. So you see how I've got kind of that overlap there. So what I did, there's actually I think on the mark skin there's like three different ways to do this. So if you think about it differently to me that's also fine. I think the way that made sense in my head was we don't actually have proper numbers we just have scales is I was just looking at when we use similar triangles we think about scale factor don't we? So when you work out a scale factor we normally just divide. So what I was going to say was BC if I were to divide it by this side AB that would give me the same scale as if I were to divide the smaller or the base side of the bigger triangle which is also happens to be AB by its bigger side.

[01:07:46]If I actually knew what those numbers were and I divided it like that. You see how I've divided the small base side of each triangle by then the bigger side that there's two of them. It would give me the same scale because they are um similar triangles. So that's what I did.

[01:08:02]You you can set up an equation in multiple different ways but that's what I did. And then what I did with that was I said right well I do need numbers because I need natural ratio for my answer. I know that BC as a scale written as four and I know that A D that's why I mentioned it earlier as a scale as 25 and you see how now I'm left with AB and AB it's the same thing isn't it so I could kind of solve that to find a scale number for AB let's just solve this I would just multiply across right multiply Okay, I'd get 4 * 25 which is 100 and then would get ab squared. But that's fine because 100 is a square number. So I can square root to make just ab. So as a scale in comparison to all of these other sides AB is 10 as a scale. Now all I need to do then is I go well AB is 10 and I know A we've already said that 18 is 25 as a ratio 10 25 now you can leave it like that actually um it hasn't said simplify they haven't accepted that it does simplify two to five so they will also accept two to five or any kind of equivalent is fine.

[01:09:38]So, however you put it, you could just leave it like that. You can obviously but it is only three marks.

[01:09:51]All right.

[01:09:54]So, that's question 19. Question 20, we've got some indices, index um index laws, I guess, and fractionales.

[01:10:04]So 2^ x is equal to 2 ^ n over the cube root of 2 and then 2^ 2 given that x + y= 8, work out the value of n. Seems like a bit of a random question, but let's just work through it piece by piece. Don't worry about the x plus y for now. When you have something like this, I just want to kind of break it down to make it an equivalent on both sides as in just 2 to the^ of something on the right hand side as well. So currently 2^ x is equal to now the cube root I can write as a power as an indicy that is the power of third.

[01:10:48]Not only should we be able to recognize that power of a third means cube root, you should also be able to do the reverse. So you go, oh yeah, cube root can be written then as the power of a third. And then here I've got index laws, isn't it? When you're dividing two numbers, they basically have the same base number. You subtract them, don't you? So in other words, I can write that as 2^ nus3.

[01:11:16]So you see how now I've got an equivalent like 2^ x then 2^ n they're both just 2^ something. Therefore I can say that x is the same as whatever this power is.

[01:11:33]Then I can say get rid of the twos.

[01:11:35]Forget about the twos. X is n minus 3.

[01:11:39]Great. Okay. Pause on that. Do the same for the y one. That was actually easier.

[01:11:44]So same thing when it's a square root I can write it as a half power and then I've got index laws here when I've got the brackets in that would be as a fraction 5 over two you write it as 2 wanted. So again you can just look at it as y is 5 over two or 2.5.

[01:12:10]So now I can forget about the numbers uh two and just look at okay well if x + y is 8 let's sub those in shall we? We've got n minus a3 plus 2.5 is 8. You know I might leave it as a fraction. It's completely up to you again arithmetic very as to how you might like to do it.

[01:12:32]Now I just need to solve an equation basically and forget about powers. Solve the equation. So, we're just working with fractions again because I'm adding fractions here. I mean, yeah, I kind of need to have a common denominator, which would be six. So, that would be 2 over 6 for a third time by 3 15 / 6. Might as well do the eight as well cuz we're going to move it, aren't we? Um, any whole number is over one. So, if you want to ever make it into a fraction, currently the denominator is one. So, to make it over six would be 48 over 6. That's the same as 8. Okay. Now, when I add these together, I should get n + 13 / 6.

[01:13:17]48 / 6 and then I would solve the equation. So, take away 13 48 over six.

[01:13:31]35 over 6. Now, that doesn't simplify, but even if it did, they haven't told us to neess. simplify it. So that's fine.

[01:13:38]You can also convert it into a mixed number if you like. They would take any equivalent of that fraction. I guess if you decimal, right, but I would leave it as a fraction just as it is. There's no point doing extra work is there when you don't need to. And that was just three marks. So you might as well stop there because that was quite a lot of work.

[01:14:00]Question 21.

[01:14:04]Three left now.

[01:14:07]Question 21. A solid cuboid has a volume of 300 cm cubed. Cuboid has a total surface area 370 squared. The length of cuboid is 20 cm.

[01:14:19]And the width of the cuboid is greater than the height.

[01:14:23]Work out the height. That is again basically set up some solve it. So I could set up two equations here. We've got what to draw it sticky void.

[01:14:37]So we've got we say height, width, length. Now we know that the length is 20. The others we don't know. So volume of cuboid is just those three multiplied together, isn't it? Height, width, length.

[01:14:52]Now if you sub in what you actually know, I know that that equals 300.

[01:14:56]And I know the length is 20. So if I put that as a You could then simplify that if you wanted. Divide by 20, you divide by 10 person.

[01:15:06]Get the height and the width together is 15. That's just a simplified equation.

[01:15:12]Let's leave that there. Now, for the surface area, that's just all the areas added together. So, that would be the height and the width. There's two of those.

[01:15:22]Plus, we've got the width and the length. Oh, there's also two there.

[01:15:31]And we've got then two of the height and the length. Now again, if you sub in what we know, we know that that actually equals 370.

[01:15:40]And we know what the length is. So we've got two height widths.

[01:15:44]We times that by four uh times that by 20, we get 40. 40 width 40 height.

[01:15:52]Now, you could obviously simplify some of that as well if you wanted to.

[01:15:58]Divide across by two, I guess.

[01:16:04]And I guess you could sub this in actually, couldn't you? Sub that into there if you wanted that would be two times height width would be 30. And then, yeah, you could divide then by two. You can divide by 10, 20, whatever.

[01:16:21]However you want to simplify it. If I divide Yeah.

[01:16:28]Let's take away the 30 first, isn't it?

[01:16:31]That be 340ide by 10 34. 34 you'd have width to height.

[01:16:48]skip ahead here. Then you can divide by two again or however many steps you want. That would be 17, two width and two height. There we go.

[01:16:56]Okay.

[01:16:59]So the thing is obviously when you have an equation like this with two letters, you solve it. Need two sets of equations which is why I've got two. So what I think I would do is I'd rearrange one substitution. It's kind of like simultaneous equations.

[01:17:15]I could rearrange the linear. Normally, we rearrange the linear one and sub it into the harder one, don't we? Now, because I want to work out the height, I'm going to make the width the subject. When I substitute it in, it will get rid of the width. I don't really care about.

[01:17:32]So, making the width subject would be 17 - 2 h all over two. You could half 17.

[01:17:41]and then in here.

[01:17:45]So instead of the width, I'm going to write height by all of this.

[01:17:56]It's a bit annoying, but that's why I've got five mark question long. I would expand all of that out. So I'd get 17 2h 2. Now I probably multiply across by two because we don't really want fraction. That give me 30. See how now I've got a quadratic. So I know where I'm going now. I can see right. I'm going to move all of this to the left because I want the quadratic the well in this case the h squive.

[01:18:25]I move everything to the left just to now I can factoriize this 2 h into brackets. We are multiplying to make positive but when we're adding the negatives it's two negatives. factors of 30 that would make 17 when one of them is two should be you can solve a quadratic however you want you don't have to factoriize necessarily obviously so then if I solve this I get the height is 2.5 or six now I have got two options that is why they gave us the information about the width of the cuboid greater than the height. So that we could actually then decide.

[01:19:15]If you were to substitute these in to this equation here, effectively either the width has to be six 2.5 or the height has to be six because the height is the smaller one or it's like if you were to substitute them back in here to work out, okay, well therefore which one is the width? The width has to be the bigger one, right?

[01:19:37]So the height has to be the smaller one.

[01:19:39]You should get 2.5.

[01:19:46]Yeah, quite quite a long one. That just a lot of algebra.

[01:19:50]Now question 22. Sketch the graph of y= sinx or we've got this graph between 0 and 360. You should just have this memorized. It goes up to I think you label this actually one and minus one on the y ais. starts at 0 0 and then every 90 degrees it kind of changes direction. That's how I imagine it in my head. But goes from zero up here actually down to what's meant to be 180.

[01:20:20]Obviously do it with a pencil and across to 360. Right. Those are obviously meant to join up.

[01:20:42]That's better. It's meant to be this kind of Yeah, I can be louder. I don't know if the sound is really quiet. Let me know in the comments if it's really really quiet. I probably need to adjust the microphone to be fair. So, I've sketched it now. So that is two marks. And for part B, we've got solve the equation 2 sin x = -1 for that same part of the graph. What we need to do here is just solve it so that you've got sinx as your subject. So that means just divide by two and move the two across. So then I get sin x is equal to minus a half.

[01:21:24]Now for the non-cal exam you have to have memorized certain values. I know that one of them is not a negative but if you look at it's all it's all symmetrical isn't it? Sign of something is a half. What is this?

[01:21:44]We should know that that is 30.

[01:21:47]The reason why I say that is because then when you look at this graph, if you were to go across, I know it's not accurate, but if you were to go across at a half, that's where you get the 30 from right here. Again, I'm not draw accurately, so it looks more like 45, but it's not.

[01:22:04]That's 30. But because it's all symmetrical, if you were to go across at minus a half roughly here, where it crosses, which actually crosses twice, this kind of gap is also 30 and that gap is also 30 where you know where this gap is also 30 and then technically over here that would also be 30. So I just use this graph to kind of visualize. Okay, I know it's 30°, but the first value is from 180. So 180 + 30. This is kind of more A level stuff, I'll be honest, which is 210.

[01:22:47]That's my first solution there. And then it crosses again, but that's backwards from 360, isn't it? That would be 360 minus the 30.

[01:22:57]So 330. So those should be your two solutions.

[01:23:01]210 and 330.

[01:23:06]But again, I think this is kind of more leaning towards um Alevel maths. So if you weren't really sure where that's come from, I wouldn't worry because it is only two marks. Um again, don't spend 5 minutes trying to figure it out, 10 minutes trying to figure it out. It's not worth that much. It's only worth two minutes of your time. If you can't do it in two minutes, just keep moving on. Try and make sure you've at least answered the majority of the other questions in it. So four marks in total for that question.

[01:23:32]And the last question is 23. So C is a circle with center 0 0. L is a straight line. The circle C and the the circle C and the line L intersect at the points P and Q. The coordinates of P are 510. The X coordinate of Q is minus2.

[01:23:53]L has a positive gradient and crosses the Y ais at the point 0 K. We need to work out what K is. That's obviously a lot of information. I feel like there's often a circle question at the end. I like to sketch it backly. Uh you don't have to, but again, I'm quite visual in terms of this type of thing. So, I like to visualize it.

[01:24:14]They've said that the circle has a center O. I don't know what the radius is, but that's fine because this is just a sketch, right? So, let's just say that that's my circle. Now they've said that it intersects with a line at points P and Q. P is 510. So obviously that's on the circumference, isn't it? So let's just say five and then it would have to be higher up. So you know roughly I'm going to say that that is P more or less. It's in the right quadrant, isn't it? And then Q, they've said the X coordinate is minus 2. So here. Now this is the thing. Q could be here, but it also could be here. So, we don't quite know that yet, but just bear that in mind.

[01:24:58]That's Q currently. So, my line either goes through P and Q here kind of upwards or P and Q down here. Now, I'm assuming it's down here because it said that the line has a positive gradient, but obviously my drawing isn't very accurate, so I can't really work on that. So okay I want to know the coordinates of Q fully. Now you might be thinking oh I don't know how to do that.

[01:25:25]But what I would do is I'd say right the equation of a circle is x^2 + y^2= the radius isn't it? Now I have one full set of coordinates on that circle which is 510. So if I substitute those in, I could work out this second part like the radius part. So that would be 5^ 2 + 10^ 2 is 125, isn't it? So the equation of my circle is x^2 + y^2 = 125 or do root 125 that would be the radius.

[01:26:02]Then because q is also on the circle, isn't it? It's on the circumference. And I do have one set of coordinates as in one part of the coordinates which is minus2. I could then sub that in, couldn't I? So -2^2 + y^2 = 125 would then tell me what the ycoordinate of q is. I feel like it seems more complicated than that. It is the final question, but that bit is just as basic that. So that's four. Remember that is positive four when you're squaring a negative. Then if I solve that, I get 121 and I can square root. Now, when we square root, we do get a plus and a minus, don't we? So, y is either positive or negative 11. So, that's why I've got these two options, right? This is either positive 11. This is the 11 option. Now, I just need to go with whichever one would give me a positive gradient for L. That's how I would decide which one it is. So, when I work out the gradient, we do change in Y over change X. I'll work out the gradient of line L.

[01:27:06]I guess it's just trial and error. You can try them both. If you get the right one the first way around, then great.

[01:27:11]Because based on my rough sketch, I'm assuming it's this one here. That would be where y is minus 11. I'm going to use that one first. If I get a positive gradient, I know it's that one. If I get a negative gradient, I know it's the other one. Right? So remember y at the top.

[01:27:29]So 10 -1. I'm using the negative one because that's the one that's down here.

[01:27:34]5 minus it was also -2.

[01:27:37]You can put these the other way around.

[01:27:38]That's fine. But as long as you vertically match them up, right? 10 and 5 - 2 and - 11.

[01:27:45]That would give me 21 over 7, which would simplify to three. So yes, I've got a positive gradient. So it is this is Q here, the one that's a negative.

[01:27:56]That's just confirmed it for me.

[01:28:01]If you tried the other one, you would probably get the same number.

[01:28:05]Maybe even No, you wouldn't get the same number, but you'd get obviously um one that didn't make sense. Yeah, it's not. So, this one is not Q. We've now confirmed that that's Q. So, my line, if you you know, if you wanted to draw it on Oops. My pants decided to stop working.

[01:28:41]Okay, here I'll try and do it roughly like this would be our line. Obviously, it would carry on, right? So, last thing that we need to do.

[01:28:52]Yeah, it's still let me write. That's good. The last thing that I then need to do is I wanted to know the value of K. When you have so much working out, you kind of forget what the whole point of the question was. I want to know what the value of K is. K is the point where this line crosses the Y ais. So K is the plus C. So if I just work out the whole equation of that line, then K is the plus C, isn't it? Y intercept. So I know that the gradient is 3x plus C. And I've got well technically two sets of coordinates on that line. So you can substitute in either P or Q.

[01:29:25]I'll probably just go with P because that's the one they've given me. So it's, you know, more guaranteed to be correct and solve it for C. So that's 15 take away 15 is - 5.

[01:29:37]So I've called it C because normally Y= MX plus C, but that is um K, isn't it? Minus 5. And also I know again my drawing isn't accurate, but you can kind of see it was likely to go to be a negative number.

[01:29:55]Oh, let's try and move this up. Go.

[01:30:03]K is minus 5.

[01:30:06]You might have done that differently. I feel like there's probably different ways to do that, but that is the way it made sense to me. That's obviously five marks. It's a lot of work um there to give you five marks. Okay. And as you can see, that is then the total for the paper. So, that is everything. So just as a reminder that was the June 2024 paper one ed XL higher tip paper yesterday we went through the November 2024 paper. So you can go back and have a look at that. If you want to look at any of my explanations you can obviously go back on this uh live it will upload but I have got an actual edited version of this video. So one where I've actually gone through it and kind of succinctly edit it. You can go back and look at that. Um yeah any resources you want everything should be linked down below in the description. tomorrow. I'm hoping to do one more for this week likely. What would that be? Probably November 2023. We'll be working backwards now. Um again, paper one or higher tier ed. And then I'll do a couple of next week as well. Um but yeah, otherwise um thanks so much for joining me today guys and hopefully I will see you tomorrow.