GCSE Maths November 2023 1H NON CALCULATOR Higher Tier Paper 5pm BST Edexcel | GCSEMathsPro

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[00:00:03]Welcome back to the GCSE Math Pro YouTube channel. We are going through today the Ed Excel November 2023 non-cal paper one higher tier paper. I have left a link for it in the description of this live if you want to download it. You can also download the marking as well. We do not need a calculator for this paper. So do make sure you're doing it without just so that you know that next week in your exam you're not surprised with the fact that actually I don't have a calculator and I can't rely on it. So just make sure you're doing this as you would the actual exam.

[00:00:38]And if you have any questions or anything that you're not sure about if you want me to repeat anything to go over it again that's absolutely fine.

[00:00:43]Please let me just let me know in this below and I will go back over anything.

[00:00:52]So question number one, we have work out 6.3 * 2.4. So decimal calculation.

[00:01:00]What I'm going to do here out of preference is make them both whole numbers. You don't I guess have to do that, but I just I like to do that. So for both of them, I'm just going to multiply by 10.

[00:01:16]Multiply by 10. So I'm going to get 63 * 24 instead. That looks in my head at least a little bit easier multiplying.

[00:01:24]I'm just going to use column multiplication.

[00:01:27]So 63 at the top, 24 beneath.

[00:01:32]And we just have to remember to basically use this method. How do we use this method? So starting with the four, 4 * 3 is 12. So I can put the two, but I will have to carry the one because it's obviously bigger than 10. And then 4 * 6 would be 24, but then add the extra one.

[00:01:50]So that would be 25. So 252 in total.

[00:01:54]Now I've dealt with the four. I'm now going to look at the two. But the two is actually a 20, isn't it? So I would have to put a zero here just to account for the fact that it's not literally two. It is bigger than the four in that sense.

[00:02:06]So 2 * 3 is 6. Don't have to carry anything over. 2 * 6 is 12 again. So, I mean, carry the one, but because that's the final one, I'm just going to write it down straight. So, I've got 1,260.

[00:02:19]And then at the end, I just need to add these. So, I get two 11. So, carry the one that makes five and obviously the last one.

[00:02:30]So, the answer to this question here is 1,512.

[00:02:37]However, we obviously made both numbers bigger. In reality, we're looking for something effectively close to 6 * 2.

[00:02:45]So, I have to scale it back down again.

[00:02:48]Now, either you could look at it and go, "Yeah, it's roughly 6 * 2, which is 12.

[00:02:53]So, what's the closest like wherever I can put the decimal place? What would that give me that's closest to 12?" You could look at it like that. Or you go, "Right, well, I did multiply each by 10.

[00:03:03]I moved the decimal place twice initially. So I have to then do the opposite which is moving the decimal place twice backwards or dividing by 100. So either way you should get 15.12 and that is three marks.

[00:03:23]Just don't forget you do have to scale it back at the end if you have adjustments. Okay, that is three marks for question one. Question two. A I write down the value of 5 to the^ of 0.

[00:03:35]This comes up quite a lot. It doesn't really matter what the number is. The fact that it's the power of zero means the answer is one. Okay, anything to the power of 0 is one. Any number, any letter to the power of zero, it's one. So, bit of a trick question there because I feel like you want to put zero, but it's not.

[00:03:57]Question I I we've got write down the value of 5 to the^ of minus two. So a negative power just remember what that means. It means take reciprocal.

[00:04:08]Reciprocal is flipping the fraction upside down. So currently because five is a whole number, it's 5 over one. So if I were to flip it, I'd get 1 over 5.

[00:04:15]And then I'm just going to square that.

[00:04:17]So obviously 1 square is still 1.

[00:04:21]So that 5^ 2 is 25. So 1 over 25.

[00:04:29]Then for part B, we've got write 2 ^ 5 * 2^ 4 over 2 ^ 3 in the form 2^ n where n is an integer. So basically simplify it.

[00:04:40]So using my index laws, I'm going to start with the numerator. When I'm multiplying and I have the same base number, that means I can add the powers.

[00:04:47]That would give me 2^ 9. Now I feel like they've done this deliberately because obviously 9 is divisible by three. So you might think, yeah, I'll just divide it. But the second index law is not divide. It is in the case of we can see here I need to divide as a whole. But in terms of the index law that means I have to subtract powers.

[00:05:09]Just be really careful there. You should get two to the^ six. Don't actually divide powers.

[00:05:15]So four marks for all of that question.

[00:05:23]Question three. A write 56 as a product of its prime factors. So I like to just use a prime factor tree. Now I normally divide solely by prime factors. You might just break it down in a general way and then at the end find out the prime factors. That's okay. Normally I start with two if it's an even number just to half it. So we' go what? 75 + 3 78. If it's still an even number I just keep halfing until it's null.

[00:05:51]We've got 39. Right now I need to change tactics. So then I start looking at three or five, seven, that thing.

[00:05:58]And because 13 itself is also a prime factor, I know I've got to the end.

[00:06:04]I've got 2^ 2 or you can write 2 * 2 if you want * 3 * 13. That is the product of prime factors for two marks.

[00:06:21]Part B, we've got Find the highest common factor of 156 and 130. So they've done that because we already have the values for 156. I'm just going to lay them out like this. 2 2 3 and 13. I would probably then do the same for 130.

[00:06:38]That's just again my preference. There are multiple different ways to do question by five I think as well actually. Yeah. And then that's it. 13. I've got two, five, and 13.

[00:06:54]And then I just like to lay out as ven diagrams. You don't have to do this as I said, but I just kind of scribble it out. It doesn't have to look particularly neat. And then the highest common factor is the bit in common, isn't it? So, they've both got a pair of twos. I'll write that. And then they've also both got a 13.

[00:07:12]You don't have to fill it out beyond that because we're not looking for the lowest score multiple, but you can obviously the rest will just go on the right. The highest common factor would be the two and the 13 multiplied which is 26.

[00:07:27]That's how I'd work that out. As long as you get 26 then we can assume you have used a method. It's also correct. If it's different to mine that's okay. So four marks in total for that question.

[00:07:39]Question four. The mean length of five sticks is 4.2 cm. Now measured the length of one of the sticks as 7 cm.

[00:07:50]Work out the mean length of the other four sticks. So we are reversing the mean here.

[00:07:56]I've got five sticks. An average for each of them is 4.2. So I could work out the total of all of the lengths. If I just multiply again, you think about when you you work out the mean, you add them all up and divide by in this case five. I'm going to then times by five to reverse that.

[00:08:16]So I would probably do 5 * 4 which is 20. Then 5 5 * 2 is one. So it should be 21.

[00:08:30]You can obviously use column multiplication if you that now one of the sticks is actually 7 cm. So I'm going to take that away from the total length of all five sticks and just that would leave me with the other four total length of the other four. So that would be 14. So this is the total length of the other four sticks. Now how do I work out the mean? We would add them all up which is 14 in this case and divide by how many there are which is four.

[00:08:58]Now dividing by four you can obviously do bus stop. What I would do is just half it and half it again. So half of 14 is seven. off it again it's 3.5 so you should get 3.5 question number four we have got part B and a worded question here the no made a mistake the stick is not 7 cm long it is actually 17 cm long so we worked this out based on 7 cm what would happen if it was actually 17 how does that affect our answer. Well, imagine what we've done. We would have done 21 take away 17 and then divided by four. So, how does that affect my answer? My answer would actually be a lot lower, wouldn't it? My the mean would be a lot lower. So, any kind of version of that, your answer would be less. It would be a lot lower.

[00:09:54]It will be a lower mean something. So, that is fine.

[00:10:03]And that is one mark.

[00:10:05]So, four marks in total.

[00:10:12]Question five. The point P lies on the line AB. Use ruler and compasses to construct an angle of 90° at P. You must show all your construction lines.

[00:10:23]Obviously, don't rub anything out. So, I think what I would do here, I think I grab my ruler.

[00:10:32]Here's my ruler.

[00:10:39]I would measure between A and P. Obviously, this is digital, so it's not accurate, right, to what you would see on your piece of paper in front of you if you had it printed out in A4. It would be a different amount. I pretty sure when it's actually measured, uh, here I've got just a bit more than 3.5. Yours should obviously be slightly different if you had it printed out, but either way, you would measure it, right?

[00:11:04]And then I would bring it over and have that same amount on the right hand side.

[00:11:10]So that was just over 3.5 for me.

[00:11:17]About here.

[00:11:24]Obviously, you will line yours up a lot better than mine.

[00:11:40]Let's just go with that. So, that's there.

[00:11:43]That you should have an equal amount, an equal distance on both sides. What I would do from there is get your compass.

[00:11:55]compass and I would put the point on A. Let's start there. Stretch it so that it's a little bit further.

[00:12:08]Obviously, not at the point P, but just a bit further, but not beyond your other mark. And then you can draw an arc.

[00:12:16]Let me Yeah, draw an arc. Oh, it lets me do it one way. Let's see if I can swing it the whole way. There we go. And then keep it the exact same width and we'll bring it to the point that you've made. That should be the equal distance away.

[00:12:36]Draw it exactly the same. Another arc.

[00:12:42]Obviously, it's digital, so it's a bit funny.

[00:12:47]Once you've done these two, you can get your ruler and draw a straight line vertical.

[00:12:58]You see where you can join up the two points? Now, obviously, if you've done it more accurately than mine, yeah, it's not too bad. They should match up. Mine isn't as accurate because it is digital.

[00:13:07]If you draw a straight line joining up these bits where they cross, it should be exactly a straight line 90° at the point P. That's the idea really along this. Oh yeah, it does. Amazing.

[00:13:24]There we go. So, something like that.

[00:13:26]And I just wouldn't rub anything out, right? Just leave your construction lines there.

[00:13:35]Okay. So, that one is a two mark question.

[00:13:41]Question number six. The diagram shows an isoclesles triangle A D and the straight line ABC.

[00:13:48]We've got here information that BA equals BD, which you can see they've drawn these little lines here that show that those two are the same side, which tells me that this triangle ABD is actually isoclesles more specifically.

[00:14:02]And then we've been told that the ratio of X to Y is 2:1 and we have to work out the value of W. So using the information about isoclesles triangle the base angles are the same. Now I know it's rotated but effectively the other base angle would be this one here wouldn't it? Both of these are x.

[00:14:25]Okay both of these are x. So the way that I looked at it was if we had a ratio of all three angles, we've got x to x to y. In terms of numbers, that would be two, two, and one because obviously x are the same. Now, let's share out the ratio because I know that the angles or the degrees in a triangle add up to 180. So if I share out a ratio, we always add up all the parts, which is five. So I would need to do 180 / 5 When I divide by five, I actually divide by 10 and then double it. I find that that's easier. You can obviously use bus stop as well. You should get 36.

[00:15:06]So that is one part of my ratio. Now y is also worth one part of my ratio, isn't it? That tells me that y is 36.

[00:15:18]You can obviously work out what x is, but you don't need to in this case because now we're going to look at w. So this is 36. W and Y are on a straight line and angles on a straight line also add up to 180, don't they? So all I need to do is just subtract it.

[00:15:45]Now you can take it away, take away 30, take away six or you can do column subtraction. Again, all of these arithmetic um the arithmetic parts are all up to you. You should get 144.

[00:16:00]So W is 144 and that is four marks for that question six.

[00:16:10]Question number seven, we've got Mano has three shelves of books. There are X books on shelf A. There are 3X + 1 books shelf B. And there are 2 X - 5 books on shelf C. There is a total of 44 books on the three shelves. All the books have the same mass. The books on shelf B have a total mass of 7,500 g. Work out the total mass of the books on shelf A. So quite a lot going on here. We just need to take it step by step. Let's have a look at this first bit here.

[00:16:44]I can set up an equation there because I've been given algebra. I'm kind of leaning towards making an equation. So, if there's a total of 44 books, I can just add all of the shelves together.

[00:16:55]x 3x + 1 and 2x - 5. That should equal to 44. So, before I even worry about the mass, don't worry about that yet. I could solve this.

[00:17:08]We get 6x - 4 is 44.

[00:17:12]Just a normal equation. Now, add four, we get 48.

[00:17:17]divide by six we get x is eight. So for every x we've got eight books. So the first one is eight and so on. You can work those out. Now let's move on.

[00:17:27]They've said that the books on shelf B have a total mass of 7,500 g. So let's actually see how many books that is. So shelf B is 3 * 8 + 1.

[00:17:42]That's 25 books.

[00:17:45]And then because they all have the same mass, if I divide it, so total of 7,500 divided by 25 books.

[00:17:56]If you divide 75 by 25, you get three and there's no remainder. So we should get 300.

[00:18:04]So that tells me that each book is 300 g.

[00:18:11]Now the final bit is we've been asked to find the total mass of the books on shelf A. So A, there were eight books, weren't they? Which is X, which is eight.

[00:18:22]So therefore, for shelf A, for eight books, they all weigh 300 g.

[00:18:29]Multiply it, I get 2,400 g.

[00:18:36]And that is a five mark question.

[00:18:45]Question number eight. The normal price of a mattress is reduced by 40% in a sale. The price of the mattress in the sale is £660.

[00:18:55]Work out the normal price of the mattress. So reverse percentages here.

[00:19:01]The normal price was reduced by 40%. So therefore, what percentage is left?

[00:19:08]If you think about it, it's always starting at 100%, isn't it? If I reduce it by 40%.

[00:19:14]Then my current sale price is worth 60%, isn't it, of the original? So 660 is the equivalent of 60% of the original price. So then how do I go back to 100%. Well, there's kind of different ways to do it. I think what I would do is clearly I could divide both sides by six and it would give me 10%. That's one way of doing it.

[00:19:40]If I divide both by six, I would get 110 would be the equivalent of 10%, wouldn't it? And then how do I go from there to 100%?

[00:19:56]I would just have to times by 10. That's a fairly straightforward way to do it when you don't have a calculator.

[00:20:04]So, you should get £1,100 if you've done it differently, but you do get £1,100, and that should also be fine. You might also want to divide it by three to get 20%. And then times it by five to get 100%. That's also okay. Normally, if it was on a calculator, I would just divide by the decimal. You could, I guess, do that if you wanted, but I'd rather not have to deal with that um if I don't have a calculator. So, that's why I like to do it this way. But that's only two marks.

[00:20:35]Question nine. So to cook rice, the number of cups of rice, which is x, to the number of cups of water, which is y is 4 to 5. Use this information to draw a graph to show the relationship between the number of cups of rice and the number of cups of water needed to cook rice. So it's effectively going to be a straight line, isn't it?

[00:21:00]For every four cups of rice, we need another five cups of water. That's basically a linear relationship, isn't it?

[00:21:11]So, what I'd probably do is just say, well, use those as coordinates. So, the number of cups of rice, I could go to four. It should then give me five cups of water, shouldn't it? And then just draw a straight line from zero. If you've got zero cups of rice, you water.

[00:21:31]So, I know 0 0 is going to be there as well. A straight line from 0 0 and make it as accurate as you can.

[00:21:43]There. That's what I would do for that one.

[00:21:49]Now, we've got part B. Let me scroll down a bit.

[00:21:55]Part B. I find the gradient of the line drawn in part A. Well, if you think about gradient of a line, it's a straight line, isn't it? We know that it's the change in y over the change in x. So, you can just use that go, well, yeah, the change in y is five. Let's say if we use those same numbers again, the change in y is five and the change in x is four. You can just leave it like that.

[00:22:20]5 over 4 if you want. That does translate to 1.25. 25 as a decimal, but they'll accept either any kind of equivalent of this is fine.

[00:22:33]And then part I explain what this gradient represents.

[00:22:37]So what this represents is for every cup of rice we have to add or we need 1.25 cups of water. So you have to relate it to the actual context of the gradient of the graph. Can't just be like what does gradient mean? Generally we need to relate it.

[00:22:56]So for every cup of rice that every kind of one that we go along the x axis we need 1.25 of water which was the y ais.

[00:23:16]That's how I've interpreted it.

[00:23:22]And that is four mark. So total four mark.

[00:23:28]Question number 10. The circumference of a circle is 10 m. Work out the area of the circle. Give your answer in terms of pi. So if the circumference of the circle is 10 m. Then what formula can we use for that? Obviously the circumference formula, but do we know what that is?

[00:23:50]that we should well you could use 2 pi r.

[00:23:55]You can obviously also use pi * diameter if you want. I'm just going to use 2 pi r because I know for area I'll need the radius. So I just find it easier, but that's completely up to you.

[00:24:12]So if 2 pi r is equal to 10, literally that number there, the only thing that's missing from that formula that from that equation is the radius. So I could just use that to work out the radius. Let's divide.

[00:24:28]If I divide first by two, because I can actually work that out, I get five. And obviously then I want to divide by pi.

[00:24:34]Now because they've told me to leave it in terms of pi, I'm not going to, you know, try and calculate anything. Just leave it as a fraction like that.

[00:24:43]Then now I know what the radius is in terms of pi. I can work out the area.

[00:24:49]The area of a circle is p<unk> r squ.

[00:24:52]So I * 5<unk> squared.

[00:25:00]Now we just have to kind of simplify that. So if I do square it, I would get 25 over<unk> 2. And these are multiplied here, aren't they?

[00:25:15]So then if you were to simplify that fraction, you would have pi, which is over one. So that would only affect the numerator. Effectively, what I'm saying is the pi in that sense would cancel. So you would have just 25 over pi singularly on its own, not squared anymore as your area. So that's just effectively simplifying fractions really.

[00:25:39]And that's three marks.

[00:25:45]Question 11. Alice recorded the number of cars going into a village on each of 80 days. The incomplete table and the incomplete box plot give information about her results. So, first of all, we need to use the information in the table to complete the box plot. So, we've got the lowest number, 300.

[00:26:06]Now, double check the scale. 300 is obviously going to be in the middle here or a line.

[00:26:12]That's kind of like our little tail bit, isn't it? And then we've got the median, which is at 900. Another line here.

[00:26:22]And then finally we've got oh here we've actually got the range. So the range is the smallest or the least amount the least number to the biggest number. So if my least number is 300 and then my range is a th00and my biggest number I'd have to work out if you add that would be 1,300. It's obviously got to be bigger than 1,200 anyway. That does make sense. I'll draw that on.

[00:26:51]That's the other tail bit.

[00:26:53]So, we've completed the box plot. We then have to use the information from the box plot to complete the table, which you can see is the lower quartile and the upper quartile. Those are the other two lines that were already there.

[00:27:04]So, the lower quartile is here. Now, let's be really careful about this scale.

[00:27:09]This is 700.

[00:27:12]So, each square must be 20, mustn't it?

[00:27:14]So, that would be 780, I would say.

[00:27:19]And then the upper quartile is a bit easier. You can see it's just on 1,200.

[00:27:25]So that's three marks to complete both of those.

[00:27:29]We do have a part B. So on some of these 80 days, Alice saw fewer than 1,200 cars going into the village. Work out an estimate for the number of days Alice saw fewer than 1,200 cars going into the village. So there's 80 in total. Have a look at your box plot I would suggest for this because 1,200 is quite a significant number.

[00:27:55]So 1,200 or fewer would be this section of my box plot, wouldn't it?

[00:28:04]See?

[00:28:06]So from the beginning from the least amount up to the upper quartile. Now the upper quartile, what we need to know about that is that it represents 75% or in other words, you might think of it as three quarters of the total.

[00:28:24]So we're saying it's that 75% there, isn't it? So what is 75% of 80 is effectively all I need to work out. So you can work it out however you want or 3/4 if you like. We could divide by four to find one quarter, times by three to find three. That's the same as 75%, isn't it? However you want to work it out, it's up to you. That' be 20 * 3 which is 60.

[00:28:48]Just write down some kind of working out so that they can give you a mark there.

[00:28:54]So that question as a whole if I scroll up is five marks. Two marks for that bit and then five marks in total.

[00:29:04]Question 12. The straight line L has equation 2 y = 3x - 7. Find an equation of straight line perpendicular to L that passes through 6. Now that to me is kind of an ideal question really just the most straightforward perpendicular line question you can get.

[00:29:24]So first of all what we will have to do is you can see that the original line L is not currently in the format of Y= MX plus C because we've got 2 Y. So what I would do is just divide by two first of all so I can look at it properly. That would give me 32x - 7 /2. You can leave them as decimals, but I would recommend using fractions.

[00:29:46]This is straight line L.

[00:29:49]Now, if they're perpendicular lines, that means that they have a negative reciprocal gradient.

[00:29:58]So, that's all we need to do.

[00:30:02]We need to do the negative reciprocal of the gradient. Now the gradient of line L is the number the coefficient of X the number in front of X which is 3 over2.

[00:30:12]So the negative reciprocal of 3 over2 is what we need to do.

[00:30:16]If my gradient of line L is positive 32 the gradient of I'll just put P for perpendicular right is negative because we're swapping the symbols it's currently positive it becomes negative and then the reciprocal is you flip them upside down. So that would be 2/3.

[00:30:33]That's why I suggested don't use decimals because then you're like, well, how am I supposed to flip a decimal upside down? I think it's much easier to use fractions.

[00:30:55]Now that we've got that, it's quite easy from here. So we want to rewrite our format of straight lines which is y = mx + c. So m is the gradient. Our new gradient is -2/3.

[00:31:11]And then we just need to work out the plus c which is the y intercept. And we do that by using well effectively the only other information we haven't used yet is this set of coordinates. The coordinates that your new line is going to cross through.

[00:31:25]So substitute those in. Make sure you do something in the correct way around. It can be quite easy to put six first and then - 5 second. But it's not - 5 is y and 6 is x.

[00:31:38]And now I just solve that for c. So we are dealing with fractions again.

[00:31:45]So what I would do whenever you're multiplying a whole number by a fraction it it only multiplies into the numerator. Reason why is because any whole number over is over one, right? Any whole number as a fraction is over one. So when you're multiplying fractions denominator, if you times it by one, it's still the same thing, isn't it? That's why whenever I multiply a whole number by a fraction, I'm only dealing with the numerator. So that would become - 12 over 3, which does literally simplify, doesn't it? To 4 - 4. That's a lot easier. Don't have to then convert my 5 into a fraction.

[00:32:20]That's nice. So add four. We still have minus one then for C. So put it all together. y = -23 x - 1.

[00:32:36]That is literally it for three marks.

[00:32:38]Just make sure you do obviously rearrange it at the start. You can't just take three as your as your gradient. It's not. It has to be terms of y and then you can look at the coefficient of x as gradient.

[00:32:54]Okay. Question 13. Solid A and solid B are similar. The ratio of the height of solid A to the height of solid B is 2 to 5. And the volume of solid A is 12 cm cubed.

[00:33:09]Work out the volume of solid B.

[00:33:13]Okay. So whenever you've got this type of thing, we've got ratios, heights, sometimes a radius, we've got some kind of volume, sometimes it's surface area, that kind of question. I always like to just really simplify it and think of it as onedimensional measurements ratios two dimensional and threedimensional in this case we wanted one dimension two threedimensional don't we so we've got a to b as a ratio 2 to 5 that is the height which is a onedimensional measurement how do I then go to volume which is threedimensional don't worry about any formulas we're not worried about what type of shape it is the most simplified way to go from 1D one dimension to 3D is cubing it as in cime cubed right so just cube both of these to get the ratio of the volume from a to b so 2 cubed would be 8 and 5 cubed would be 125 now I can actually literally compare that to the actual numbers the actual volume is 12 isn't it and and I'm looking to find out the actual volume for B. So I know the scale between the volumes for A to B. I now just need to literally translate that into cime cubed. So you go what how do I go from 8 to 12? How do I scale that up? If we were to divide simp or 1.5, this is basically one and a half times as big as the actual scale, isn't it? in this case as the ratio. So I just need to times by 1.5 on both or 3 over2, however you want to look at it. That's the scale, isn't it? So 125 times Yeah, you might prefer 3 over2 or 1.5, whatever it is. I would just kind of half it, which is what? 60 62.5. And then add that on. It's like adding%, isn't it? So over here, if I were to add that So get 57 187.5 that would be 1.5 times 125. So the actual volume of B in cm cubed is 187.5.

[00:35:37]That's it. That's all I do with that.

[00:35:40]So three marks for that one.

[00:35:45]Question 14. Work out the value of 27.

[00:35:49]2/3 plus a half to the^ minus 3. So just fractional indices negative indices here. Working out the fractional indices first of all the denominator of the fraction is the type of root the cube root of 27 and then the numerator and the fraction you just apply as a square.

[00:36:11]So here I would get cube root of 27. You should know they won't give you any numbers that are decimal answers, right?

[00:36:16]They will give you obviously if they want you to cube root a number and it's a non-c paper then it will be a cube number that they've given you initially.

[00:36:24]So that should be three and then when I square three I get nine. That's that first term there. Then our negative power negative power means take the reciprocal. We keep mentioning that in this paper. That's where you flip the fraction upside down. So it's currently a half. I flip it upside down. I get two over one. Which in other words is just two, isn't it? So I don't need to write it as two over one. Just write it as two. And then your power, the three is just a normal power now. So cube it. 2 cubed is 8. So add it up to work out the value. You should get 17.

[00:37:04]And that's another three marks.

[00:37:10]Question 15. An object falls from rest.

[00:37:13]Here is the distance time graph for the distance d m fallen by object t seconds after it starts to fall. So work out an estimate for the gradient of the graph at t three. You must obviously show how you get your answer. Whenever it says this gradient of the graph and it's like a curve, it's not obviously a straight line. They want us to draw a tangent. So a straight line. So where t is three is here. Actually, before I draw anything, what I like to do is just write down the coordinates there because that is the most accurate coordinates you're going to get because the examiners have drawn that for you, haven't they? So, just have a look where t is three.

[00:37:53]What is it on the y axis? About halfway here, isn't it?

[00:37:57]Half. That's meant to be halfway. That's 25. So, each square is worth five. So, 30, 35, 40, and then halfway. So, what's that?

[00:38:07]42.5.

[00:38:11]The reason why I write down these ones is because we're going to work out a gradient in a minute and we're going to need two sets of coordinates. So, you might as well get the exact coordinates from the point they've asked you to get because that's the most accurate set of coordinates you will have because when we draw our tangent, everyone's tangent will be slightly different.

[00:38:29]So, it's easier to limit kind of the risk of your tangent being too far out of the range if you can use a set of coordinates that they've actually given us. So, now let's draw a tangent. So that means I need to draw a straight line. It will come along. It should just about touch my curve here at three and then carry on. It's kind of the idea. So let me t and draw as long as you want. I tend to draw it longer just so you've got more options of coordinates. Obviously, draw it with a pencil so you can redo it. You can obviously play around with your ruler. I'm just retesting it. That's not bad. I might ro maybe I should have rotated it a little bit, but they do give you a slight range for your answer.

[00:39:31]Rotate it just a more or less something like that. You can draw it the whole way through the graph if you like. That's also fine. So, we're going to find the gradient of the tangent. Now, when we find the gradient of a straight line, we mentioned just earlier change in y over change in x, which is why I needed two sets of coordinates that are on the tangent.

[00:40:01]So, I chose a set of coordinates that were literally on the point that my tangent should technically touch because that's the most accurate. And then another set of coordinates from my tangent which won't be as accurate obviously because my tangent is not ex it's not going to be exactly the same as your guys, is it? So, have a look along your tangent. See if you can find a set of coordinates look quite exact. So, I would say here for me personally, four that crosses through on my tangent.

[00:40:28]it along there, which is what? 55 60 65.

[00:40:34]That's what I've got. Yours will likely be slightly different. As long as our answer is within a certain range at the end, then it's all right. But we want to make sure that it is. So hopefully this will be in the range. Change in y 65 - 42.5 over change in x 4 - 3. I get what's that? 22.5 over one. So luckily I don't have to actually divide the decimal. Makes my life easier.

[00:41:11]Now what they've said for their example they've given a range of as long as you get somewhere between 22 and 28 then you get the marks. Now mine was within that range. So yes I get the marks but I was quite close to one side.

[00:41:25]So effectively that's telling me that my tangent wasn't very good. So obviously hopefully you would get something more so within the middle of that. But as long as you are within that range then that's absolutely fine.

[00:41:36]So leave your tangent there. That's how they said you must show how you get your answer. So leave the tangent there so they can see that you've drawn it.

[00:41:45]Move it. Yeah. That should be three marks there.

[00:41:56]Question 16. At the start of year n, the population of a species is PN.

[00:42:04]At the start of the following year, the population of the species is given by this formula. So P N + one, what that means is the population of next year.

[00:42:14]That's all that's saying. N it's one, it's one year. So the population for next year is equal to k multiplied by the population of this year. So if you were to translate that, that's all that's saying is every year we're taking our current year's population and we're multiplying it by a factor scale factor.

[00:42:31]K is also a positive constant saying the population of the species at the start of year 1 is 8 million and the population of the species at the start of year two is 6 million. We've actually got some numbers now. work out the population of the species at the start of year three. So in order to work out year three, I need to know what K is.

[00:42:50]And that's why they've given me two pieces of information because I can sub those in using my formula. The next year, which would be year two, would be 6 million.

[00:43:00]Just put six, but you can put six want.

[00:43:02]There's no put zeros in this case though. And then that would be equal to K.

[00:43:08]I say K multiplied by the previous year, which is 8.

[00:43:14]So you see how I've set up this equation. I've just literally substituted it in. And then if I solve that to find out what K is 6, but what does that simplify to? 34 or in other words, 0.75.

[00:43:31]Now, now that I know what the actual formula is, we're saying that every year we're multiplying by 0.75 or 3/4 or in other words, 75%.

[00:43:41]I can then take year two and multiply that again to get year three. Multiply as I said remember whole number just by the numerator. So that would be 18 over four which if I half it in half again I get 4.5. And then that's in terms of million isn't it? That's why the zeros because they've said million already. 4.5 million. So every year we we are basically losing 25% aren't we?

[00:44:10]Currently that's what that's saying.

[00:44:14]And then for part B at the start of year five the value of K is increased by 0.3 to a new constant value. Louise thinks that from the start of year five the population of the species would increase year on year. Is Louise correct?

[00:44:32]Now, as we've just said, K was 3 over4 or 3/4, which is the equivalent of 0.75 as a decimal. If I add 0.3 to that, I get 1.05.

[00:44:50]Now, think about this as a multiplier.

[00:44:53]This is almost compound interest, but it's not interest. It's in terms of population, but it is compounding, isn't it? If I were to multiply a number by 1.05, 5 would the answer get bigger or smaller? Well, because it's bigger than one or in other words, it's bigger than 100%. That's 105%, isn't it? That tells me that yes, my answer would always get bigger. Now, in the previous part, in part A, my answer was getting smaller because my multiplier was less than one every time it was getting smaller. But when your number is bigger than one, your multiplier, your answer will always get bigger. So she thinks that it's going to increase year on year. Is she correct? Yes, Louise is correct because K would be 1.05.

[00:45:42]So the population would increase something like that. That's something really basic. That's the reason why it would increase because it's bigger than one.

[00:46:02]And that's just one mark. For part B, four marks in total.

[00:46:09]Question 17. We've got part A.

[00:46:11]Factoriize 6x^2 - 5x - 4. So factoriize into double brackets because it's a quadratic.

[00:46:19]Now, you can well, yeah, you could do it as trial and error. You could also use splitting the middle term method if you like. because it's got a big number at the front. Actually, I might do that just to show you. You can use trial and error. It's fine. So, splitting the little term a b c. These are the numbers. We do a * c. That'll be -4, isn't it? Now, we need two numbers that multiply to make -4 and add together to make5.

[00:46:50]So, that it's still a bit of trial and error, but just maybe a bit less. So, I'm thinking maybe eight and three.

[00:46:57]- 8 + 3 would give me -5 and - 8 * 3 would give me - 24, wouldn't it? There we go. Split the middle term. So, rewrite it. Doesn't matter which uh letters or numbers you um well letters, not even letters, numbers in this case.

[00:47:14]It doesn't matter which of these two numbers you put first and which of the two you put second is what I'm trying to say.

[00:47:20]Cuz either way, oh x either way, they would add up to make minus five. That's the whole point. And then look at it halfway. I'm going to factoriize into single brackets the first one. So I can factoriize out two and then an x actually 2x and then I'm left with 3x - 4.

[00:47:40]Then the second one I can only factoriize out one, can't I? And then the bracket would be 3x - 4. The whole point is that those two brackets should be identical. And therefore that is one bracket 3x - 4. and the other bracket is what's left on the outside. So 2x + one.

[00:48:00]So however you want to factoriize that as long as you end up with these two brackets then that's fine. And then we have got a part B. Hence or otherwise solve 6x^2 - 5x - 4 is less than 0. So they say or otherwise you can either using that information or completely differently you can solve it. I would obviously use this information. You might as well. So when we are solving quadratics first of all I look at it as I guess ignore the inequality for a sec.

[00:48:31]Imagine it's just equal to zero. If you were to solve that I would get x is = pos 4/3 and I would get x is equal to a half.

[00:48:41]That's the first thing the actual values but I obviously then need to put it back into an inequality. Now, this is where you either just memorize it because there's only two possible answers here.

[00:48:53]When it's facing zero, it will be a certain format of answer. When it faces the quadratic the other way, it'll be a different format of answer, right? So, you can just memorize it. When it faces zero, that means that it's a range. So, as in x is between those two numbers, obviously put the smaller number on the left, minus a half, and then the bigger number on the right.

[00:49:16]You can just decide to memorize that.

[00:49:19]You can also sketch a graph and visualize it as well, but I won't go through that. It is in my video of quadratic inequalities if you want to look at it in any more detail. But that should be your answer.

[00:49:30]Another two marks, four marks in total.

[00:49:35]Question 18. Spinner A and spinner B are each spun once. The probability that spinner A lands on red is water. The probability that both spinner A and spinner B land on red is 124th.

[00:49:52]Work out the probability that one spinner lands on red and the other spinner does not land on red.

[00:50:01]So this one is effectively a probability tree. I like to use trees to visualize.

[00:50:05]So we've got spinner A.

[00:50:08]We've got options of red and basically not red.

[00:50:12]So landing on red is one quarter. Not red would therefore be the other three quarters. Right?

[00:50:19]Then we would spin spinner B. We've got another choice for red. And for not red.

[00:50:30]And then here if you did not red you can still get red again.

[00:50:34]Not red. Now, I don't have this information currently because this line here is telling me A and B.

[00:50:45]If I get my highlighter, that's saying A and B.

[00:50:50]The probability of both of those is 1 over 24. So, imagine that you can just call this X if you want.

[00:50:58]Well, the way that we work out probability trees is we multiply between the branches, isn't it? So, if I were to work that out, I would have done * x whatever that probability was and it gives me 1 over 24. So I can just rearrange that kind of like undo that calculation to work out what x is. If I divi e either you divide by a quarter or an easier way to look at it is you just multiply by four isn't it the same thing would be 44 which is 16.

[00:51:27]So I now know the probability of red on spinner B is 16.

[00:51:38]Therefore the probability of not red is 56. And this is the same because whatever happens on A does not affect what happens on B.

[00:51:47]So I've done that. Now I can actually answer the question now that I've got my completed tree. So, we wanted to have a look at probability of one spinner lands on red and the other does not.

[00:51:59]Let's highlight my options.

[00:52:02]A could be red, but then B would have to be not red. But I've also got A could be not red and red. It hasn't really specified.

[00:52:14]So, we've just said when you work through the branches of a probability tree, you multiply. So for the green option, we've got a quarter * 56, which is just fractions. So 5 over 24.

[00:52:27]And then for the yellow option, we've got 3/4 * 16, which is 3 over 24.

[00:52:37]This is the yellow option and then the green option. Now because we've got two possible answers as in like two possible ways that we can satisfy this um probability here only one or the other can happen. So now we're saying or either it's green option or yellow option. So when that happens you have to add if you end up with more than one option from your probability tree you have to add them which is helpful because they are still the same denominator. I can just add them to make 8 over 24. Now you can leave it like that because probability they never really ask you to simplify it or you can obviously simplify fraction to 1/3 and that is four marks.

[00:53:32]Question 19. Here is the graph of y= sinx or - 180 to positive 180. Part a.

[00:53:40]Use the graph to find estimates for the solutions of sinx is equal to 0.3 for that same value.

[00:53:49]So what we do here is it's actually fairly straightforward.

[00:53:55]Let me highlight her again.

[00:53:58]The sinx and the sin x are the same, aren't they? The thing that has changed, where it used to say y, it now says 0.3.

[00:54:07]So, what that's telling me is I need to look on my graph where y is 0.3. That's all that's saying. I'm going to get my ruler, draw a line at 0.3. Try to be as accurate as you can.

[00:54:20]That would be here. Positive 0.3 there.

[00:54:27]and then check where does it cross through with the curve.

[00:54:33]So obviously it crosses through twice a there and then bit I guess it's a bit further. So I've drawn it there, but it's a little bit further along, isn't it, than that line? And now we try and interpret it. So every five we've got kind of one of these mini squares is 30, isn't it? So that's 30 here.

[00:55:10]And then this one backwards be 150. And then so if that's 30, then every lot of these three squares, every square, sorry, is six. So three squares would be I would say it's 18 or maybe a bit less cuz I think it was a little bit on the left hand side of it. And then this would be 18 backwards, wouldn't it? It's because it's symmetrical. So 180 minus 18.

[00:55:33]So you've got 18° and then 180 take away 80 would be 162.

[00:55:39]Yeah, that's all I would do for that one.

[00:55:46]Two marks. Now they have given a slight range of values because you might read it slightly differently. They've said anything between for the first one 15 to 18 degrees is fine. And then for the second one anything from 162 to 165.

[00:56:05]As I said it was technically a little bit on the left hand side of that line and a little bit on the right hand side wasn't it? So 165. So as long as you've got anything between these two ranges, then you can get that answer right.

[00:56:19]And then we have a part B.

[00:56:22]Write down a value for x such that we've got sin x + 20 is equal to zero for that same range. So you can kind of use the graph if you want to visualize it. We're saying it's equal to zero. So that would just be the x axis there, wouldn't it?

[00:56:38]You can see it crosses at minus 180 and then positive 180. But we have got this well we have got this transformation that if you think about it in terms of graph transformations that's what that is. So try and remember I guess what that does when you've got plus or minus number in the brackets with the x it's affecting the x axis left or right but it's always the wrong way round isn't it? So where it says plus 20 it actually means move imagine that whole curve 20° to the left taking away 20. So where it currently crosses through at - 180 and 180 positive it would cross through at - 200 minus 20 and then 160.

[00:57:23]So they're saying write down a value that's between this range.

[00:57:34]The three values that we originally saw that crossed at zero and now these new three values. This one is no longer within that range. So ignore that. You can choose out of these two. They've said a value. So one of those two it doesn't matter which one. So I put minus - 20 but you can put positive 160. That would also be a correct answer.

[00:57:55]Think about graph transformations there.

[00:57:58]That's one mark. So three marks in total for question 19.

[00:58:07]Question 20. Here is triangle ABC. Find the size of angle ABC.

[00:58:13]That is this one here.

[00:58:17]So this is effectively I mean we're basically the end of the paper. We've got a scaline triangle got three sides. We're looking for an angle.

[00:58:25]It's obviously going to be sign and cosine rule generally. But which one?

[00:58:29]When you've got three sides and one angle, it's always cosine rule.

[00:58:34]Now cosine rule is on your formula sheet. So you would be able to refer to that or the formula.

[00:58:40]The only thing is I don't think they rearrange it for the angle. So you might have to rearrange it yourself as a formula to get the angle as the subject or you might just memorize what it is.

[00:58:52]I'll just show you what it is.

[00:58:58]A is a squ B^ 2 + C^ 2 - A 2 over 2 BC. That's the rearranged version. As I said, you might have memorized it. You might not.

[00:59:34]Now, because it's based in the angle A, and here we're looking for angle B. What I would recommend, and you can do, is just get your pencil, and you can scribble out the way that they've labeled it just because I think it's easier to change the triangle than it is to change the whole of the formula, right? So, I would just say, let's call that A. This one then can become B. C can stay the same. So, this is side A, this is side B, and this is side C.

[01:00:00]That's what I'd recommend. Now we can just sub them all in. So co of the angle a is 5^ 2 + 10^ 2 - 5<unk> 7^ 2 over 2 * 5 * 10.

[01:00:19]Now we obviously don't have a calculator. So we will have to simplify it ourselves. So we get 25 + 100 minus and then 5 2 is 25 and obviously<unk> 7 is just seven, isn't it?

[01:00:36]Over 2 * 5, which is 10 * 10, which is 100.

[01:00:41]And just keep working through. We'll deal with the cause in a minute. So, we get 125 and then 7 * 25 gives us 175. So, 125 - 175 is -50 over 100. that simplifies to minus a half which is deliberate choice obviously from them. So remember that that is cause of the angle. I actually want to know what the angle is. Obviously we can't do co inverse in our heads but what you can do is obviously there are some values that you should know. Not necessarily negative. But if we sketch the graph of cause one, isn't it 90 180 and come up at 270. You can just sketch this literally roughly. There's our cause graph. We are looking at minus a half. This is minus one, isn't it?

[01:01:42]That's like halfway. Think about what we did in the previous question. What you know is that these are symmetrical.

[01:01:52]So obviously I've not drawn it very nicely.

[01:01:57]This here would be 60.

[01:02:00]The distance there right to the distance here is the same, isn't it? Because we know that. So we know that this is 120.

[01:02:13]because they're symmetrical and we know that this is a half. That's that's the one that you can memorize.

[01:02:20]Cause the half is 60. So you use that knowledge to then sketch out the graph or if you do it in your head that's also fine. And you go, oh yeah, therefore cos 60 is the same as cos 120 in terms of a half and then it would be minus a half.

[01:02:39]It's the same symmetrical distance from the 90 point. This is kind of leaning towards more like a level stuff.

[01:02:45]Basically, this is like the top. Even if you can get this far just to the minus a half, that' be the majority of the marks. Obviously, this last bit to translate it into an actual uh angle is is really tough. This is like the top end of GCSE. I mean, barely. This is mostly I would say. So, it's the equivalent of 120 degrees.

[01:03:09]You can also see that it's clearly not an acute angle. You should know it's going to be it's going to be bigger than 90, isn't it?

[01:03:16]There's four marks, but you can at least get to, as I said, minus a half and get probably two or three out of four. So, don't necessarily worry about that last step. If that made no sense at all, I wouldn't like I wouldn't try and figure it out in the next week. Worry about it.

[01:03:28]Leave that last mark and get it somewhere else.

[01:03:35]Okay, question 21. So on the grid draw the graph of x^2 + y^2 = 169. So that is we should recognize a circle graph and we should know that circle graphs are x^2 + y^2= the radius squared. So square root 169 which is a square number as I said they won't make you square root or cube root or whatever root anything that is not already a square or a cube number if it's not calculator. So I know the square root of 169 is 30.

[01:04:07]So then we can draw a circle. We also know from that information there that the circle center is zero, don't we? And go to 13.

[01:04:16]Make sure it's what I would do. Uh obviously I'm using a digital circle thing here is in your actual exam.

[01:04:26]I would just obviously use your compass.

[01:04:33]Use your compass that you can obviously there's no way you can draw a circle just like freehand. So yeah, use a compass. Try and drop it. It might Oh yeah, that's not bad. Okay.

[01:04:48]Little bit. Yeah, that's a little bit less than 13. Okay, they're all a little bit less.

[01:04:58]there. Halfway just because we're going to be using this in a minute. So, I'm trying to make it accurate. Okay, that's not too That's not too bad. Right. So, there's part A, right? That's two marks just literally for identifying the circle radius 13 and drawing it.

[01:05:18]Now part B we've got use your graph to find estimates for the solutions of the simultaneous equations x^2 + y^2= 169 which is what we've got and 2 y = 3x.

[01:05:30]Now that's the new bit. What that is first of all is a straight line. The second of all is a straight line that we are going to draw on the graph. That's why it says using the graph. There's not really space to work out any.

[01:05:45]So, I need to draw that straight line.

[01:05:47]Now, it's up to you. I'd probably just rearrange it to make y the subject. You know, we have y= mx plus c. That' be y = 32x. So, I can see that my gradient is 3 over2 or 1.5. My plus c, my y intercept is zero. So, when you're drawing or plotting a straight line graph, you just need to plot two sets of coordinates. The one here that's the easiest is normally the plus c, which is zero. I know it crosses through 0 0 and then just choose any kind of other value x. So when x is let's say six y would be 3 over 2 * 6 which iside by two and then times by 3 9.

[01:06:38]So 6 9 six across 9. Just make sure obviously your the coordinates you choose are still on the graph out there.

[01:06:46]Now get your ruler, draw a straight line, but make sure you extend it the whole way through the graph, not just between these two coordinates.

[01:07:00]Extend this way.

[01:07:06]Make sure it goes through six and nine.

[01:07:08]Yeah, I think that's the most accurate I've got. Okay, so that is that graph.

[01:07:12]Now it's just simultaneous equations but from a graph color. So we're looking for all of the solutions. So it crosses down here left that isn't it?

[01:07:34]Just read it off to I guess one decimal place. What would that be?

[01:07:39]Careful of the scale. So, we've got -7 is halfway.

[01:07:45]That's probably - 7.2 - 7.3.

[01:07:48]Maybe - 7.3 - 7.4.

[01:07:52]I say - 7.3.

[01:07:55]That one.

[01:07:58]That's for X.

[01:08:01]And for Y cross lower than okay is a little bit lower than where I've dri I've driven where I've drawn the line. So this would be not it's not quite as far as minus 11 it. So maybe - 10.8. 8 - 10.8. That's the first set, right? But we've also got a set over here.

[01:08:43]So for X, it's a little bit left of that line there almost, isn't it? Um I guess that should be symmetrical really. So, if I've said - 7.3, it should be positive 7.3. This looks a bit more like seven, but it should be a bit more to the right than where I've drawn it.

[01:09:08]You could say 7.3 or 7.2 or even seven.

[01:09:13]Fine.

[01:09:15]But I know that it should technically symmetrical.

[01:09:20]And then vertically vertically the same should be symmetrical, shouldn't it? Circle center O. And also my line goes through the center. That's why it should be symmetrical.

[01:09:36]Cross.

[01:09:39]That looks also like 10 to be fair.

[01:09:42]Positive though.

[01:09:49]There we go. So, you should have four solutions for that. Four solutions. Two for X, two for Y. As I say often, they do have a slight range of values because your circle and the line might not be drawn exactly the same as everyone's as long as it should be close. So, for X, you have to get somewhere between plus or minus 7.

[01:10:16]27.4 is the max. So - 7.3 positive 7. It's fine. And then for y again plus or minus we should get between said 10.6 to 11.

[01:10:29]So my 10.8 was probably spot on there.

[01:10:31]It's right in the middle.

[01:10:34]So as long as you got things between these ranges, then you will get all the marks. And that is three marks.

[01:10:43]Okay, we're on to our final question now. Question 22.

[01:10:48]The second term of a geometric sequence is 3 + 2<unk>2.

[01:10:54]The third term of the sequence is 13 <unk>2. Find the value of the common ratio of the sequence. Give your answer in the form a plus roo<unk> b where a and b are integers.

[01:11:08]So geometric sequence we need to know would you call it a formula? I guess you call it a formula. It's a r n minus one.

[01:11:19]So a is a starting number. R is the common ratio i.e. what we're multiplying by every time and they've told us to find that. So that's helpful. And then n is the number in the sequence. So like the second term is n is two. The third term n is three and so on.

[01:11:35]So for the first one, the second term would be a r because it's just the second term, isn't it? 2 - 1 is just one. And then that would be b + 2<unk>2. And then for the second for the third term, sorry, it would be a r 2, which is 13 <unk>2. What I would do here is because I'm trying to find the common ratio. I would just divide them.

[01:12:04]Just write it the way around really because we would divide it this way around, wouldn't we?

[01:12:08]You rewrite smaller one because if I divide these then I would be left with just r which is the common ratio.

[01:12:19]You see I can just divide both sides of the equations. So when I divide the left, you see how a r would cancel and a would cancel. I'd be left with just r and on this side I am left with thirds.

[01:12:36]You think oh yeah okay what do I do then? Because yeah I have found what r is but it's not in the format they want.

[01:12:43]So now I'm almost forgetting about the fact that it's a geometric sequence and I'm going to focus on the fact that it's thirds and I can rationalize the denominator of this third.

[01:12:56]So when we rationalize denominators, when we've got two terms, you have to multiply by that denominator, but you have to change symbol. So it would be 3 - 2<unk>2. And then we're effectively expanding double brackets. So 13 * 3 is 39.

[01:13:13]13 * - 2<unk>2 is - 26 <unk>2.

[01:13:18]9 <unk>2 * 3 27 <unk>2 9 <unk>2 9<unk>2 * 2<unk>2 would be 18 on the outside and then <unk>2 * <unk>2 would cancel to get two as a whole number that's our numerator denominator I'm just going to skip because I know that it's obviously 3 * 3 is nine I know that the middle bit will cancel that's the whole point of expanding when you've got 1 + 1us and then I would have 2<unk>2 * -2<unk>2 which is -4 on the outside and then the thirds would cancel two.

[01:13:58]Now we just simplify it. So we've got 39 take away 18 * 2 which is 36. 39 take away 36 is just 3 - 26 <unk>2 + 27<unk>2 is just <unk>2 and then 9 takeway 4 * 2 which is 8. 9 take away 8 is one. You know when you divide by one it's effectively just the numerator isn't it no longer a third. Now if you have a look at the question that's the format that they wanted. So you just have to know if you've got fractions with thirds in them you might as well rationalize and that is four marks at the end. So that is it for that paper. As a reminder, that was the November 2023 Ed Excel higher tier paper one non-calculated paper.

[01:14:48]I will be doing some more lives just like this next week leading up to your exam. Thinking of doing Tuesday and Wednesday because I'm pretty sure you have an exam. Yeah, you have English literature on Monday. So, I don't think Sunday or Monday make much of a much sense. So, Tuesday and Wednesday I'm thinking I might go a poll up to see if you want to do more than that, but that's one I need to do. Um, keep an eye out because over the weekend and Tuesday I'll be uploading my predicted paper which I've written myself. That is for Ed XL students specifically for higher tier as well. Um, but otherwise thank you for joining us on this live.

[01:15:24]Hopefully you found it helpful and I will see you next week.