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GCSE Maths June 2022 1H NON CALCULATOR Higher Tier Paper 6pm BST Edexcel | GCSEMathsProAdded:
Welcome back to the GCSE Master YouTube channel. In this live, today is the night before your paper one math exam.
We are going to be going through the EDXL June 2022.
Paper one obviously paper. This is higher tier.
that it's 2022, but obviously on the title should say that this is 2022 and I have got the link for this paper and the mark scheme in the description for this live. So if you want to open up or download them, feel free to do that.
So we are just going through paper one.
It's obviously a non-cal paper for ED XL. So make sure that you practice without a calculator as you obviously will not have tomorrow in your exam. And as always, if you have any questions, just let me know in the chat and I can go back through any of the questions that you are unsure of. There's a part of it that you weren't sure of, just ask me and I'll go back over it. That's absolutely fine. Um, but otherwise, let's carry on. So, we have question one. You solve 7x - 27 is less than 8.
So, an inequality, just a linear one.
Just imagine it's the same as if equals.
I want to get x on its own on one side and then move all the numbers to the other side. So the 27 is what I would move first. It works the same way as I said as any kind of equation. I'm just going to do the opposite. So add 27.
You'd have eight. You have to add this 27 which is 35. And then lastly, I need to move seven. So I need to divide by seven.
That would give me five. So x is less is your answer. That is literally it.
Just like any other equation, you just have to have the an inequality symbol there instead.
Question number two, we have to write 124 as a product of its prime factors.
So I like to break it down with a factor tree. And I like to just use prime factors. You can use any factors technically because at the end they will break down into prime factors. I just think it's easier to start off with prime factors. So for example, I like to start off with two if it's an even number, which it is. So I'm going to divide by two because that also makes the number a lot smaller. Breaks it down quickly, but I should get 62. It's still even, so I'm going to use two again.
Left with 31.
It's no longer even, so I can't use two.
But actually 31 in itself is a prime number. So I don't need to do anything beyond that. Obviously just write your answer as the product. So that means multiply 2 * 2 or 2 ^ 2 * 31. It's fine.
That is another two marks.
Question number three. A delivery company has a total of 160 cars and van.
Number of cars to the number of vans is 37.
Each car and each van uses electricity or diesel or petrol. We've got 1/8 of the cars use electricity. 25% of the cars use diesel and then the rest of the cars petrol. Work out the number of cars that use petrol. So when you've got this kind of word question, I just like to work through it more or less in order.
So starting off with the total of 160 cards and bands and then this ratio. I might as well split them up first of all share out the ratio. So the way that we share out a ratio is we have to add up the parts would give me 10. There are 10 parts in that ratio and then 160 is the total of both. So divide by how many parts there are. That's obviously quite straightforward. It's 16. What that tells me is there's 16 either cars or vans per one part of my ratio. Now the rest of the question is only talking about cars. So that's what I'm going to work out is the car side which is the three side basically isn't it? So three lots of 16 isn't it?
I often recommend for higher tier students to familiarize yourself a little bit with the 15 and 16 times tables just up to maybe five or six times each of them because I feel like it comes up quite a lot in the non-cal paper and I just think that then it's easier if you're already familiar aren't you? You don't have to actually try and work yourself.
Oh, thank you. That's a comment. Thanks for leaving that. I'm glad it's been helpful. Okay, so carrying on. So, we've got 48 cars here.
That's the total of the cars. So, the rest are obviously vans, which I'm not really interested in, am I? Because I've not really mentioned anything beyond that about the vans. So, now carrying on to the next bit of information. We've got 1/8 of the cars use electricity. So, 1/8 of 48, that's the cars, right? So, 48. Working out an eighth, it's just divide by the denominator, which is eight. So, six.
So, six cars are using electricity.
That's meant to be Oh, that was a T.
That's meant to be a T.
I've done that. Done this. I've done this. Right. You can always take it off as well. 25% of the cars use diesel. So, 25% of 48. The way that I would work that out myself is I would because I know that 25% is equal or the equivalent of one quarter, I would just divide by four. If you weren't sure about the conversion into a fraction, that's fine.
But you could work it out by doing 10%, so two lots of 10% and then 5%. However you want to work it out. It might be different to me. That's fine. Obviously, as long as we get the same numbers. So, what was that? That was diesel.
And then the rest of the cars use petrol, which is the one I'm actually interested in, isn't it? I've got a total of 48 cars. Six electricity, 12 diesel. The rest is what I want to know.
So subract six and the 12 which is effectively be 18 isn't it? Take away 18 we should get 30 in total which is cars and petrol.
All right it's just a worded question as I said work through it in order they give you the information I find that that's a little bit easier to organize yourself and that is actually a five mark question. So yeah quite nice.
All right, question four, part A. Write 1.63 * 10us3 as an ordinary number. So there's two ways to do it. One, you just move the decimal place. Or if you already know whenever you have a negative power, that's how many zeros you put in front.
So you might find that easier to think about. So either way, if I were to write it like this, when it's a negative power, it's getting smaller, right? So I'm going to move the decimal place to the left one, two, three times. And then you see how you'd fill in obviously a zero left of that and then two others.
So there's three zeros in total 0.00163.
That is obviously one mark. Part B, we've got right 438,000 in standard form. So the opposite. Now we have to put a number between I don't want to say 1 and 10 cuz it's not 10 but 1 and 9.9 either way that would be four wouldn't it 4.38 that would be within that range so the decimal places currently here I want to move it to between the four and the three how many times have I moved it one two three four five and because to get to my original number it would be much bigger four, wouldn't it? That's a positive power this time because my standard form number is smaller than the original.
Okay, so it's kind of the opposite of the first one.
Now, part C, I think, is where it gets a bit more interesting. Work out 4 * 10^ 3 * 6 * 10us 5. Give your answer in standard form. Obviously, this is if this was on a calculator paper, I would just put it in my calculator straight away and just see what it gives me. We can't do that here. So, the way that I split it up, they've obviously put brackets in, but if you think about multiplying, when you're multiplying multiple things, it doesn't matter which order you multiply them in. So, what I mean by that is I would pick out, let me get a different color, just the four and the six. Multiply those. That's quite straightforward, isn't it? And then I would pick out the tens and multiply those separately. And then at the end I'll bring it back together. So obviously 4 * 6 that will give you 24.
And then 10 ^ 3 * 10 - 5.
Do you see how that is effectively index laws?
So it's obviously still to the^ of 10. I would leave it as 10 to the^ something.
I wouldn't literally work it out. And when we've got a multiply here, we actually add the powers. So 3 + - 5 would be -2. And then you can bring it back together. We're still multiplying those together, aren't we? It's still tsing. I've got this. Okay. But that's not standard form. It kind of looks like it, but it's not because, as we said in part B, it has to be a number between 1 and like 9.999, whatever it is. 24 is obviously not in that range. So you'd have to then convert it into standard form either way. Now, it's up to you.
you might be able to figure out. Okay, I mean obviously it has to be 2.4, but you'd have to change the power. You might be able to just figure that out yourself. If you can, perfect. If you can't, then what I would do is just literally work this out. Same as standard form, it's telling you when you've got the power of -2, it's saying to move the decimal place, which is currently here, to make it smaller by two. So, actually, this as an actual number is 0.24. And then you go, okay, well, let me put that back into standard form. Clearly, we'd have to move the decimal place once to make it 2.4. But because my original answer here is smaller than the standard form number I've used, it would be a negative power.
That is meant to be a nice one. Not a very good one. It's meant to be. Okay.
Well, that's meant to be a one, right?
Negative one. Okay.
So, again, however you want to work this out, you might have done it differently to me. That's fine. As long as we get the same answer at the end, 2.4 * 101, then that is fine. Okay.
So, two marks for that, four marks for the whole question.
Question five, here is a regular hexagon and a regular pentagon. Work out the size of the angle marked X. So, with pentagons, hexagons, whatever it is, polygons, basically, we have this formula that you might have forgotten about. more so a keystage three thing but it does still come up in GCSE where we can work out the degrees or the angles whatever you want to call it in any polygon. Now obviously you know a triangle should do um obviously you should know a square quadrilateral that kind of thing but beyond that you might not have memorized them which is absolutely fine. So this is a formula n minus 2* 180 that will tell you the angles total angles in any polygon. So here a pentagon has five sides. N is the number of sides. So for the pentagon if you didn't know what the total number of angles or well total degrees I guess were then you could use this to work it out. So 3 * 180 that should give you 540°. So a total here is 540.
Then the key thing from there are the words regular. That means that all of the sides for each one, you know, they're not the same between them, but they're all the same.
So that means that all of the angles are the same, right? They are regular, which means if my total is 540°, I can work out one angle by dividing it by how many there are, which is five. So you might want to do bus stop there. You might be happy to. The way I divided this um off the top of my head was 500 divided 5 is 100 and then 40 divided 5 is 8. So luckily it just worked out quite nicely.
That's how I divided it. So 108. This one here is 108. And then I just did exactly the same for the hexagon. Obviously that's then six sides, isn't it? So you have to restart again. If you don't know the degrees in hexagon that's absolutely fine. We're just going to work out this way. So that would be four * 180. So what's that? 720.
And then because again it's regular, all of the sides are the same. So therefore all of the angles are the same. So divide that by six this time because there are six angles again. Then 72 / 6 I know is 12. So then add a zero. 20.
This one's 120. Hopefully you can see where I'm going with this. The last bit would be that we've got angles around a point. So angles around a point add up to 360. Take away the other two. So the 120 108. Add those together. So that would be what? 228. Subtract that. So you could do obviously column subtraction, but I'm working through it in my head. Um, take away 200 would be 160. Take away 20 140 takeway 8 132. I'm just going off the top of my head, but obviously you can write it down. Want to that's absolutely fine. Okay. 132 degrees should be your answer for X.
That was just three marks. I have a lot of work for three marks I feel like. But that is okay. Question number six. We have part A. Complete the table of values for y = x^2 - 3x 1. So a quadratic. We've obviously got some here filled in for us. Because it's not on calculator. we are going to have to work it out. So effectively just substitute in these values of x. Now if we can find where the line of symmetry is, it might be a bit easier but yet on that. So let's start with one. Be really careful when you're subbing in negatives because when you square a negative remember it becomes positive.
So that would be one squared is pos1.
Then we've got another double negative here.
- 3 * -1 is 3. So that be 4 + 1 is 5.
That is positive there. Same for 2.
Easier. It's positive. Anyway, so 2^ 2 is 4 - 6 would be - 2 + 1 is 1. Okay, there's our line of symmetry.
As a little tip here, this is a quadratic. You should be able to recognize that this is a quadratic, which means it's a U- shape and it is symmetrical. So, can you see how we've come we've come from five to one to minus one minus one again. Then it's going to have to go back up. So, it should be one here and five here because of that line of symmetry. That's a bit of a tip. So, you don't have to work out the last two. If you can't find the line of symmetry, then there obviously should be one, but if you're not sure what I mean by that, then obviously just go through and work them all out individually. So, substitute in three, substitute in four, you'll get the answers the same.
Just a little bit of a trick there. So two marks.
Now actually I do see that. So on the grid part B on the grid draw the graph of y= x^2 - 3x + 1 with the values of x from -1 to 4. So that's the same thing isn't it? So your table here you would just fill it out. - one 5 0 1 1 - one again up to one again to see where this is it's going back up again there. Now use a pencil when you do this just in case it goes really badly wrong to rub it out and try again. We need to join these up as smooth as possible. Smooth.
Uh I'm going to attempt it the best I can.
Uh digitally it is a little bit easier obviously in front of you on a piece of paper. So mine's Oh, that is mine is going to be really wobbly. Actually, I don't know if I had a way to do this.
Oh, let me try this because my actual drawings.
Oh, that's way better. Yeah, that's way better than me trying to make it a bit thinner just because I think we need to use it in part.
Yeah, there's part C. So, I'm going to do this again, but I'm just going to make the light thinner so we can see.
That's why I realize, you know, person.
When you draw it yourself, it won't be super super accurate, but try and make it as accurate as possible. Okay, that's pretty good. You can see also here where it was minus one minus one. Don't obviously draw a straight line across.
Like it still carries on technically lower and then goes back up again.
Okay, part C. Using your graph, find estimates for the solutions of the equation x^2 - 3x + 1 = 0. This is really easy. Easiest to Yeah. So just as a quick explanation what this means. Did you see how originally we had x^2 3x + 1 and here we have again x^2 - 3x + 1.
That has not changed. What has changed is originally we had y on the other side.
Now we have zero. So that means we're looking at when y is zero on this graph.
Now where is y z? That would be here.
This line, the x axis.
That is where we're looking at.
So you just have to see where it crosses. That's why they've said solutions as in plural because obviously it crosses twice. You see here, here, and here. So just read it off the best you can. This is why it's important that you try to get your part B curve to be as accurate as you can. And obviously I know it's not going to be fully accurate, but just roughly um as accurate as you can so that your answer to part C is it's within the range they offer. So what have I got here? Check the scale. That seems like every small square is 0.1. So I'd say I think I would say 0.4 for that first one. X is 0.4.
And then the other one here 2.6 six I would say now because obviously your curve my curve everyone's curve will be slightly different um because you're doing it with pencil you obviously have to go through the same coordinates but it's still going to be slight there will be a variation they have given a slight range of possible answers so their range was 0.3 to 0.5 for this one so if you've got anything between those or including those fine they will give you the mark. And then for the second one, 2.5 to 2.7. So it's not a massive range, right? But it is a slight range. So as long as you've got something between those two ranges or including the numbers there, then you get both marks, obviously one number. We don't need to write down what y is because a the actual equation they gave us only mentioned x, but b technically we know that y is zero. So it's implied.
That's another two marks, six marks for all of that question.
Question seven. Here are two cubes A and B. Cube A has a mass of 81 g. Cube B has a mass of 28 g.
Work out the density of cube A to the density of B. Give your answer in the form A to B where A and B are integers i.e whole numbers. You can also see that we've got cube A has a side length of three and B has a side length of four.
So this is screaming to me. We've got mass. They've mentioned density and technically we have enough information to work out volume. So I'm thinking density is mass over volume. You have to memorize that formula. They won't give it to you. As you can see in question, it's not given. I like to use it as a formula triangle. Um we are looking for the density. So I need to have the mass which I've got. Perfect. And that then tells me I need to work out the volume for A and B. That's where I'm going to start. The volume of A is a cube. So you're just multiplying 3 by 3 by 3, aren't you? 3 cubed.
So we should know that that is 27.
Then we'll do the same 7 27 cubed. The volume of B is 4 * 4 * 4. cubed which you know is 64.
Yeah, your answer is looking right. Um but we will we'll get to that. But yeah, it looks like you're on the right lines there. So we've got the volume for both.
We've got the mass for both even. So the density for a separately density is mass over volume. So for a 81 / 27 they've chosen these numbers deliberately. We are again paper they are or at least it's question seven right they're going to choose nice numbers at least for this bit I would say so 81 is a multiple of three hopefully for that and 27 is a multiple of three we've just seen that so that should help you work out that when you divide it is just three g per cime cub that's just the units same for the density of B we're going to take 128 over 64 they've chosen these numbers again deliberately because They're both I guess whatever you want to call it. They divide and they give you two. They divide quite nicely.
They're not there's not any decimals quant say. So we've got that the density of A is three. The density of B is two.
So they just want two. Basically that's all they wanted. Where they're whole numbers I've got whole. So that's fine.
Yes, you were correct. 322.
How many marks do I need to get in every paper to get a grade six? I think I have a video on I don't have it off the top of my head, but where I went through or I briefly mentioned the average grade boundaries like I've worked it out myself over the last few years for grade nine, grade eight, grade seven, six, five, four, so on. Um, one of my most recent video I'm trying to think which one it is because I can't remember what the numbers are off the top of my head. It's a lot lower than think. Yeah.
I don't think it'd be higher than 60.
Probably much lower than But yeah, it's not that high if you're on.
Um, yeah.
Yeah. I can't remember which video it was, but it's one of the videos where I'm like, it's not like a voice over like this. It's like an actual kind of sit down face the camera video. Um, I brought it.
It might have been the one that went up two days ago, like a seven minute video.
Uh, have a look at that. It might that one or if not, I think I've only got five or six sit down videos. You can have a little look through um because I do kind of put it on the screen like a little screenshot, but I honestly Yeah, have a little look, but it might be the one from two days ago. If not, then there's some from last year um where I definely kind of mentioned the fact that the grade boundaries are lower than you'd expect. So, yeah. Um there somewhere.
the sheet head. Yeah, it's definitely a lot lower than what you'd imagine.
Um, how to stop making random mistakes on questions.
Yeah, work slower. That's a really good uh really good tip. Try to not rush.
You've got about a minute per mark. So, if you're working at that pace, perfect.
If you're going faster than that, you don't necessarily need to go that far.
Um but also if you do a minute per mark you get 10 minutes at the that will give like let's say you did stick roughly on that you get about 10 minutes at the end to go back through the paper. Um I know at the end of once you finish paper but like use all the time they back double check all of your calculations and um but yeah just try to be really intentional. Also I'd say think when you get an answer think about it. Does it make sense in the context of the question? Sometimes you get an answer, let's say, like, you know, when you have Pythagoras's theorem, you might forget to square root at the end. And if you look at your answer and go, well, this answer is way like much bigger than the other sides of the triangle, then you can go, hang on a minute, something's not right here. Oh, wait. I forgot to square root it. That kind of thing. Just try and think about, you know, is my does my answer actually make sense to the context of the question. Sometimes it can give you a clue as to whether you think answer doesn't really make sense.
What might have gone wrong?
Y how many marks to pass foundation?
Yeah, foundation is much higher because obviously only go up to a five. It passes a four and probably about 70%.
Again, I can't remember off it is on that video I mentioned earlier. Um it does mention grade four and grade five as well. Um yeah, probably about 70% I would imagine. It's a lot higher than I to pass.
Check.
Yeah, again, how many marks you want to guarantee get an eight or a nine?
I have to find that that spreadsheet at least for Ed XL.
I do have the um head I mean at least 10 on each paper, I'd imagine. But don't take my word for that. I need to double check it.
Yeah, the end of the paper. Yeah, obviously that's the hardest stuff because then it is the grade five. You want to get to grade seven.
I think it's possible. Obviously, there's, you know, tomorrow is tomorrow, right? Whatever happens tomorrow will happen, but you've then got another three or four weeks for paper two and paper three. It depends if you're going to commit to maths or if you also have a lot of other subjects that you need to put time into. That's the issue. three or four weeks for maths if that was your main focus then yeah sure I'd imagine that you could do that um but if you have to also focus on English and science it depends on priorities doesn't it depends on how you allocate okay right I'm going to carry on just so that we get through the paper and I'll see if I can find that spreadsheet later so question eight the table shows the amount of snow in centimeters that fell each day for 30 days so we've got frequency table work out an estimate for the mean amount of snow day. This is just mean from a frequency, right? So if you know how to do this, it's the same for every question. Quite nice. So what we have to do is because we have ranges here, you have to have another column. I just write it up here, which is the midpoint. The midpoint of these ranges.
Now here it's probably quite obvious what the midpoint is. But just for other questions whenever you're trying to find the midpoint if you're not sure like if it's not obvious to you what it is you add both of them together and you divide by two that will always give you the point. Also works for like coordinates and things like that. So we can see here see that's five 25 35 and 45.
And then you want another column which is the frequency multiplied by the midpoint. So these two multiply together. So 8 * 5 is 40. 10 * 10 * 15 is 150. I was going to say 7 * 25 is 175. I think 2 * 35 is 70 and 3 * 45 is 135. Yes.
Once you've done that, so that's kind of like that's step one, that's step two.
Then you have to add this column together for step three.
So add all of these up. Let's I've not drawn them very much in line, but the five and five would give us 10. Then we've got 10 here, 17, 22, 26, 27. Yep.
Seven here. Zero. The two there. One, seven. Oh, don't count the seven again.
This is why you got to line up properly.
Don't count the seven again. 100 200 and 300, 500. Okay, however you want to add it up. It should be 570, though.
There we go. Now, your final step is working out the mean where, you know, when you add them all together, divide by how many there are. We've just added them all together. That's what we've just done. Divide by how many there are.
Now, if you don't know that, you can add up the frequency column, but they have told us it should be 30. Divide by 30, which means I can knock off the zeros.
57 / 3. You could do bus stop, however you want to divide it. Oh, it started raining. Hopefully, that's not too loud for you guys. It goes in what? Once remainder of two then 27 that would be n 19. There we go. 19. Whoever said something in the chat that was 19, you are correct. It is 19. But yeah, you just follow those steps every time.
That's why I quite like it. Let me have a look. You got 19. Yes, you are right.
You also got 19. Perfect.
Yeah.
What do you do if you finished every pass paper one? Well, perfect. If you have, I mean, if you have, that's the most ideal scenario to be in. Um, oh, you class width. Yeah, class width is to do with histograms. Uh, it's a completely different topic. It's still a frequency table, but it is a completely different topic. Um, yeah. Oh, yeah. I also have a predicted paper. I don't know if you've seen that. It uploaded, I think, yesterday. If you just go on to my channel as an actual video, not as a live. I have got my own predict paper which you can have a look at. Uh, or some Yeah, just Google someone else's sort of. Um, okay.
Right. Question nine. A cube is placed on top of a cuboid as shown in the diagram to form a solid. The cube has edges of length 4 cm. The cuboid has dimensions 7 cm by 6 cm x 5 cm. We can see all of that on the diagram. Work out the total surface area. Okay. So surface area is just all of the areas of all the faces but on the outside, right? But this kind of bit here where it overlaps is not going to be counted. We just want the outside.
So I'd probably you can start wherever you want, but I'm going to start with the box. Well, cube if you want to call it that box on the top. So the surface area of that to find out one side is four * 4, isn't it? Length time width. They're both four because it's cube. So four * 4. And then normally a cube has six sides, but as we said, we're going to ignore this one here. We don't want that one because that's kind of inside. So there's five other sides on the outside of that. 4 * 4 is 16. You see how we've got that 16. Again, I was saying at the start, I think it's um a good tip to be familiar with the 15 and 16 times tables because I feel like they come up a lot in non-capage papers and it's just a bit easier, a bit quicker if you've already got it on top of your head, isn't it? So 5 * 16. What you can also do with a five is just multiply by 10 and then half it. Do that 4 * 4 of them. 80 squared. Okay. Now the bottom one. So that was the top one, right? The bottom one. There's a few more because it's a cube. We've got different dimensions. So the front and the back would be 7 * 5. The front and the back there's two of them. Add that to let's say the left right the left side and the right side. 6 * 5. There's two of those five.
Add that to then the bottom which is 6 * 7. And then this top bit is also 6 * 7.
But as we said we keep saying we don't want this bit here. So that bit we know is 16, don't we, from the previous one.
So we want to add, I guess, well, yeah, you could add two of those. And then you can take away the 4* 4 in the middle, the cutout bit.
So we've got what? 35. So that's 70.
Plus here we've got 30. So 60 plus here we've got 42. So what's that?
54 and then take away 16.
130 214 98. So I'm just saying numbers, but that's just me working out. Obviously you can work it out using whatever method you like. If you want to do it on top of head, fine. You want to use column addition, column column subtraction, that's fine. But you should get 198 for the bottom one. Now we're just adding those together basically, aren't we? Total surface area would be those two added together which should get 278.
You might have done it in a different order to me. That's fine. Again, as long as you get 278 at the end, then it's fine. That's three mark. Okay. Let me have a look.
Okay.
Yeah, you got 278.
I haven't seen her predicted paper actually. Go have a look at it. Is it really good? I'll have a look.
Um, okay, cool. Next one. Question 10. So the table shows some information about the profit made each day at a cricket club on 100 days. Again, we've got another frequency table, but in a different context. Now, we've got to complete the cumulative frequency table.
It's only one mark because it's really straightforward. All we're going to do is just add them up. So, for example, the first row is exactly the same. You can see it's 0 to 50, 0 to 50, exactly the same. But then the second row is now 0 to 100. So, that includes the first two rows up here, which is then five if you add it. You see what I mean? Every time I'm going to add another one. So add 25 here which is 50. Add 30 which is 80. Add five. 85. Add 15. 100. And your final row should add up. If they have told you in the question what the total is it should add up to that. If it doesn't there's something wrong. So that's one mark forward.
Part B. On the grid draw a cumulative frequency graph for this information. So obviously I need to have page. you will have it on the same page. But see when you open your exam page, it'll be two people's sheets. You'll be able to see it on print screen. I can and copy it over.
So we can see it here. Okay.
Now for cumulative frequency, we do not use the midpoint where we have to try not get used between all of these versions of frequency tables. We're not going to use the midpoint. We are just using these numbers on the right hand side of your ranges. Okay.
So the first one will be at 50 along and 10 up. So accurately 100 and then 25. Be careful of the scale. This is here is not 25 is 30. So 25 would be in the middle of that about there. Obviously on the line ideally um 150 would be 50 which is here then 80 should be constantly going up. If you're doing aative frequency graph the point of the word cumulative is adds and adds and adds it carries on going up. Um 250 85. So again be careful because this is 90. So halfway.
Always double check the scales of all of your graphs because they tend to change.
They might not be what you expect them to be. And then obviously the last one should be 100.
Okay.
Okay. So I'm going to draw I'm going to attempt at least to draw a nice smooth curve.
It's not a line graph in terms of a straight line.
Something like that, you know, roughly.
Again, with a pencil.
Okay. Part C. Use your graph to find an estimate for the number of days on which the profit was less than £125. So I'm going to get Well, you would get your ruler, right? I'm going to get my straight line. 125 and my x axis is here.
Then across. Yours might be slightly different answer to mine obviously because we will have drawn our graphs a little bit differently. We should have the same coordinates again as we mentioned before but no one can draw it the exact same.
So what have I got? That's I would call that 36. Now they have given a slight range always because it is slightly variable. So as long as you get anywhere between 35 and 39 then that is fine. So I'm kind of on one end. Yeah. Just look at the chat. Yeah. I think it's a bit quiet, isn't it?
I don't know if that's just my settings on the microphone.
It might be something I have to look into for when I do paper two and paper three as well. I'm not really sure why it's so quiet for you guys. I'm trying to speak a bit louder, I guess, or I'll get a bit closer to the microphone.
Yeah, that's something I'll definitely look into and I'll try and fix for next time. It's really quiet.
Um, part D. Use your graph to find an estimate for the interquartile range. So intercortile range is when we work from the y- ais that is the upper quartile minus the lower quartile.
The upper quartile is the equivalent of 3/4 or 75% however you want to think of it of the total which was 100 wasn't it?
So 75% of 100 is just 75. So again be careful this is 70 here. Okay. So 75 would be like halfway through this square and then down.
What would I read that off as? That's 175. Each square on this x axis I think it's five, isn't it? 55 60 65 70. Yeah, 75 80 I would say that's what I've got.
185. And then we want the lower quartile which is 25%. So 25% of your total relative frequency. So 25 in this case straight across.
I think that was actually one of our numbers. I'm just looking at it now, isn't it? That so that should be 100. I guess everyone should get 100 for that, right? Because that was actually one of the coordinates.
So 100.
And then the interal range is where we subtract them.
So 85 is what I have got.
Now what's their range? They've said you could get anywhere between 85 and 93. So that tells me that mine would while mine is correct, I was quite close um to the edge of that range. So hopefully you'd get something a bit higher than 85. Yeah. Okay. Let me have a look at the chat again.
How did I do C and get 36?
So C is the purple line.
Right? Do you see I've gone here which is 125 as they said 125 pounds. I've just drawn a line up at 125 until I hit my graph and then I went across and I read off the number here which for me was 36. As I said, it might be slightly different to you. As long as you've got somewhere between 35 and 39, then you should be okay.
Any other questions on the whole of question 10.
Yeah. Okay. Move on. I'm just having a look into if I can change the microphone thing whilst I'm doing this.
I think I can change it in settings, but I don't know if that would actually take effect whilst I'm currently using the microphone, if you know what I mean. So, um, look, I can do anything right now.
I could add a filter on it. We could try that.
Let me try and add a filter really quickly to this. Makes it louder and then I'll fix it for the Next slide.
Okay, I'm adding a little bit there. So, you tell me if that's louder. If it's still not loud enough, I can add a bit more. You just guys let me know in the chat. Um, okay.
Got any more questions before I carry on?
What grade was the question?
I'd say about a grade six. That's just my kind of estimate. That's not necessarily factual, but that's my estimate. Um, okay.
So, when you use the midpoint for a graph, that is for the frequency polygons.
is a very specific topic, but again it comes from a frequency table. They're all stemming from frequency tables. That is where um you use the midpoint. Okay. All right. So, Colmarmac has some sweets in a bag. The sweets are lime flavored or strawberry flavored or orange flavored. In the bag, the number of lime sweets to the number of strawberry sweets to the number of orange sweets is in this ratio 9 to4 tox.
I like to just label it like this.
That's just my personal preference. So I can quickly refer back to it. Is going to take a random sweet from the bag. The probability that he takes a lime flavored sweet is 37. Work out the value of x.
Okay. So what I did was I because they've given us x. I'm thinking right I need to set up an equation. That's what I would like to do. So probability of getting probability line I know is 37th but I'm going to put something here with this ratio. So because probability you can use a fraction can't you see there's a fraction here I can convert my ratio into a fraction as well and then I can make them equal to each other. So the probability of picking lime would be if I were to convert this ratio into a fraction but for the lime section would be nine. See lime is the nine one out of and then you when you're going to create any kind of fraction from any kind of ratio you add them all together to put in the in the denominator don't you? So 9 plus the four plus the x and then that would equal to 37th.
Do you see how I can create an arrow? I can create my own equation just by doing that. Now all I need to do is solve for the equation because all I want to do is work out x. So we've got obviously add that together.
That would be 13 + x, wouldn't it? Which at the same time is equal to 3 over 7. I would probably cross multiply. Okay, what I mean by that is multiply by the denominators. You can do it in two different steps separately if you want.
I like to just do it in one step. So 9 * 7, which should get 63. And then on the other side, you're kind of expanding a single set of brackets. That'll be 3 * 13, which is 39. 3 * x, which is 3x. So do you see where I'm going with this?
Now, it's just basically a very straightforward equation. I'm going to move 39 by taking it away. you should get 24 is equal to 3x and then divide by 3 8.
That's what I would have done for that is just creating an equation and solving it.
Okay, let's look.
You got 134. Yeah. See, that's the thing because if you think about it, right, you know what I was saying about how can you pick up on if you've made maybe some small mistakes and if your answer makes sense? Well, if you've got nine lime sweets, four strawberry sweets, then orange, even if you don't know what it's going to be until the end, you can kind of guess is going to be in the ballpark of this, isn't it? Maybe maximum like 15. But if you get 134, do you see how you might think, well, surely that doesn't make sense because of the context? Like if I've got four and nine, then why would I have 134 of another? Do you see what I mean? Like you can kind of have a look at your answer and go, does that make sense? In this case, I would say, hang on, something does seem a bit off there. Like it seems too big.
Uh why is it equal to 37th? Because they told me here that it was equal to 37.
That's where I got that information from. What do I mean by cross multiply?
So I just did it in one step, but I can go over it kind of in more detail if you'd like.
9 over 13. I'll just rewrite it here.
And then if I were to do it in more steps, right? I obviously in order to solve it, I don't want a I don't want fractions because that's quite like kind of annoying. But also I don't want x in my denominator. So also very annoying.
So let's say I wanted to get rid of uh the fractions. The way I get when I say get rid of fractions, I mean you multiply by the denominator. Now obviously you have to do that on both sides. It's technically like multiply by 13 plus x on that side and you see what I mean the same on this side multiply by oh that's meant to be a plus look plus and then you see how if I multiplied the left hand side by 13 plus x it would cancel out the denominator right that's what I wanted to do I don't want it as a denominator but it still has to go somewhere it's not just going to be disappear is now here and Then I can multiply again by 7 to get rid of that denominator which would obviously multiply by 9 63 and end up with this.
Do you see what I mean? I just did it all in one step. That's when I said cross multiply. I'm multiplying across on both sides. I just did it in one step. If you want to do it in two or three steps, that's absolutely fine. And then I obviously just do you see what I mean? I expanded those brackets.
How do you remember to put it into an equation? Uh, I would say two things.
One, for me personally, I quite like algebra. So, when I have questions, I kind of naturally tend towards, oh, can I make an equation here, maybe? But the other thing is the fact that they've given you X. Whenever they've given you X or they've said work out X, the only way you can work out X is if you have an equation. You know what I mean? Like, it's telling you, oh, you could probably use algebra here. That would be like the most ideal way to do it.
That's the kind of hints that I would take from that question.
Yeah.
Yeah. So, how is it how is 42% equal to the ratio? because I guess it would be 9 to 4 to 8 and 9 is 42%.
If you were to work it out, you don't have to obviously work out percentages in this question at all because it would take too much time. But yeah, just so you get get what I'm saying, nine out of well, what would that add up to? 30 9 out of 21 is 42% I'm assuming. Never actually worked out as a percentage to be fair.
Yeah. 42.85.
Okay. Um, someone asked me to go through it again. I don't I went through kind of half of it again. I don't know. Do you want me to go through like where I've got this bit from? Because I guess that's the key bit. You just let me know.
First 15 questions. Yeah. Normally, how many questions are there? kind of maybe 21 to about 24 25 on average. It does change obviously, sometimes it's fewer.
Um, okay. As long as you've understood that's um so 15 out of let's say 22 questions, but then a lot of the heavier marks towards the end.
I don't know. It might be close. Again, I can't remember exactly how many what the average was for a grade seven. It's obviously lower than you might think, but Yeah, you're still cutting off quite a few questions at the end, aren't you?
Um, okay. Question 12, express 0.117 recurring as a fraction. So, this is just again if you've memorized the method, then you can just apply it. It's quite nice. So, I always say let x you can use any letter you want equal the one 0.117 recurring. Do you see how just here the clear thing is that this first one is not recurring?
then the second two digits that are. So we want to isolate the recurring part of the decimal. So I've kind of got two steps here because as I've just said one of those numbers is not recurring. So I want to get it out of the decimal. So I would do 10 x because that will move my decimal place along by one. And now do you see how I've isolated the recurring part in the decimal? Then you need two versions of this. So we've already got one here. I need another version of this with the exact same recurring decimal in the same order. So that would be kind of moving the decimal place twice more, wouldn't it? Which is multiplying by 100, but we've already started with 10, haven't we? So that's actually a,00 x would be 117.17 recurring. So you see how I've got exactly the same recurring decimal.
That's what I want. And then you subtract them. So that would be 990x over here. here. Obviously, the whole point is that when I subtract them, the recurring decimals cancel. But don't forget that you do need to do 117 minus the one that's still there. That' be 116.
Then we're going to solve the equation.
So divide across by 990. Now that's a fraction. They said leave it as a fraction. So you can leave it like that. They haven't told you to simplify it. So you can, but yes, I've seen someone said in the chat, yeah, you got 58 over 495. Exactly. it would simplify if you divided by two. You don't have to because they haven't said you have to. So, it's completely up to you. Um, you know, just make sure you read the question. If they tell you to simplify it, then obviously do. If they give you an answer that you have to match, then you might simplify to get to that matching answer. But in this case, technically they'll they'll accept 116 over 990 or they'll also accept 58 over 495. And of any equivalent is fine.
Yeah, this one's quite a nice one if you know exactly what to do.
Um, is there another difficult ratio question? Uh, because we're going through this paper. I don't have any others, but I'm pretty sure I have a video as in not a live, but an actual video where I've done the hardest ratio questions. I think you could always have a look on my channel and have a look at those because yeah, I go through all the edex the hardest what I deem as the hardest ratio questions. That might be helpful to you. Um, that's three marks for question 12. Question 13.
So a right angle triangle is formed by the diameters of three semic-ircular regions A, B and C. As you can see on the diagram, we need to show that the area of region A is equal to the area of region B plus the area of region C. Now this one, this is where you might be like, okay, I'm going with this because it's a prove kind of question, isn't it?
But yes, this is a higher tier. So, because we've got a right angle triangle and we've clearly got diameters that are the length of all the sides, I'm kind of leaning towards Pythagoras just from the context, if that makes sense. So, I think there's different ways to do this.
If you have a look on the mark screen, there's different ways to lay out. You can also use numbers. That's fine. Um, I'm going to use I think letters.
I would use A, B, and C, but obviously Pythagoras's theorem itself is A, B, and C. So, it's a bit confusing. So, I think what I saw in the mark scheme that was suggested is that you could use X, Y, and Z. But you whatever letters you want. You can also use as it said actual numbers. It's fine. I'm going to do it this way. That's just the one that I thought made sense. So, we obviously know that Pythagoras's theorem is true. That's not We're not trying to show that. That's just a true fact.
So, we know that x^2 + y^2 should obviously equal z^2, but also they've talked about areas, haven't they? So, we might as well also use areas. So, from a second, I'll just pull it to the side. The area of semicircle a circle is p<unk> r², isn't it? Pi r 2, but it's a semicircle, which is half. So over two or multip I've labeled as the side length, right?
So the diameter is zed. So therefore the radius is half of zed. So you got to be a little bit careful. Half of zed is the radius, isn't it? Squared. But because it's a semicircle, I also want half of that. Now that looks a bit confusing.
That's meant to be a zed there. And that's a two. The difference there. So hold on to that for a second. And then you can do the same for B and C. So the area of B would be p<unk> * the radius which is half of x squ and then over two because it's only a semicircle and the same for c y / 2 is the radius squared over two.
Okay, we've got those.
Now what we can do is because there we're basically saying that when you add them together it should equal to area A.
If I add those and then we'll simplify it. Hopefully we'll end up with this is what I'm 2 plus the B1 which was X to 9.
Oh, hang.
I'll put X first so it matches this. And then where's my Y?
Yeah, it's a really tricky one. I feel like for question 13 y / two and that should equal the bigger one, right? The z one.
I make that clear that that's a zed, not a two over two.
If I were to simplify this equation, we can easily cancel out the pies, right?
If you were to divide across by pi, it would cancel out, wouldn't it? Get rid of those. That's kind of irrelevant. If I were to multiply across by two the whole equation, it would cancel out those denominators, wouldn't it? Just cancel those out. Get rid of those.
Now, if we actually square these, we get x^2 over 4, cuz remember, you'd have to square that two as well, plus y^2 over 4 = z^2.
And then, do you see how again, if I multiply across by four, it would cancel them all out?
You see how I've ended up with the same thing?
And that's that basically proves it because you've started here and then we've ended in the same place.
Yeah. Have a look at the chat.
Yeah, this one is a really really tricky one I I honestly think. And also, you know what the good thing about this one though is? is like very niche. It's very specific. It's unlikely you're going to get exactly this type of question again.
You know what I mean? So, there are some questions where you go, you know what?
That doesn't make sense to me. So, I'm going to try and get however many marks is how many marks is it? Three mark.
It's only three marks. You might think, well, I'm going to just try and get three marks on a different topic that is more predictable. You know, one that you can revise and if it comes up, you'll actually be able to apply it rather than this one that's such a random question really, isn't it? like a proof question.
I wouldn't necessarily worry. Um, hang on. You've asked a question about the recurring decimal.
Uh, why when it's sorry, a th00andx is the recurring of seven rather than in front of the decimal because you only have recurring in terms of decimals.
There's never recurring in terms of whole numbers like units, tens, hundreds. There's never recurring parts there. It's always just decimals. Dude, I didn't draw it on the the 170.
It's out of the decimal.
Yeah, how is this question 13? I agree.
As I said, there are quite a few ways to do it in the mark scheme. You might have a look at the mark scheme and actually decide that you prefer a different that's also absolutely fine. What kind of makes sense in my head would, you know, could be different to what makes sense in your head. That's absolutely fine. They've given what, two, three, they've given like three or four different ways of doing it. So, it doesn't have to be my way. Um, that's the one that when I looked at mark scheme, I thought, okay, that kind of makes sense to me. Um, you might prefer a completely different method. That's fine. But that's why I think it's quite good to look at the mark schemes because you can pick up on how they've suggested to answer it.
Um, I have linked the mark scheme in the description of this live as well, so you can have a look.
What is this topic called? I don't even know what you call the topic because proof I guess proof overall because it's saying show something but you have got a bit Pythagoras in there. You've got the area of semicircles. It's not really one topic. That's why I think it's quite hard as well. It's not one topic that gets repeated over and over. It's a combination of multiple, isn't it? Are you supposed to be specific about the wordings of circle theorems? Yes.
Uh yeah.
Is capture recapture on here? I don't think it's on this paper. though I have got a video on it though. You can have a look at my video separately to this if you like.
I'm going to move on just because I think there's no point going in detail over this question. As I said, it's unlikely to come up exactly the same again. This was also 2022, so it's kind of far away. So, I wouldn't worry if you feel like that doesn't make sense. You can have a look at the marking yourself if you like, but it's not a topic that repeats itself exactly the same every time. So, I I really wouldn't stress about it. Uh, question 14. And this one is a much better one to know how to do.
This one does repeat like itself. So here is a speed time graph. Part A, work out an estimate of the gradient of the graph at t is two. So where t is two on the x-axis here. When it says work out the gradient at a specific point and it's obviously a curved graph, this is the one where you have to basically do a tangent. Draw a tangent. Now before I draw a tangent, what I would recommend is you write down the exact coordinates of well in this case t being two. So we've got two and then what is that exactly? Because obviously your tangent is meant to go through that point. So that would be the most accurate point along your tangent, wouldn't it? Because everyone's tangent is slightly different. So I would use the set of coordinates that they've actually given you. So then at least half of your coordinates are accurate.
So 2.8 I would say for that. Now draw your tangent. So it's meant to be a straight line back. A straight line that kind of comes along. It just skims along. It touches ideally at two and then carries on. That's kind of the idea. Um obviously use your ruler and pencil so you can rub it out if you need to do it again. I'm just going to kind of give it a few goes. See that to me I think is do you mean that's too slanted?
So move it again. You can obviously move your ruler around a few times until you're happy.
Not quite.
I'm just going to keep moving it until getting something that I like.
No, still not quite. It needs to be a bit No, I've gone past it now. See?
Gone past it. Not quite as there is much harder digitally, I think. But I think that's a bit further. See, I could play around with this all day until get something that's accurate, but I'll try and I don't think that's very good, but we'll go with that just so I'm not here for five minutes messing around with it.
Maybe there. So, I've already got my set of coordinates, right, that was on the point, which is the most accurate. And then what I'm going to do is because we have to find the gradient of the tangent, right? Straight line. So, you need two sets of coordinates. That's why I chose one set of coordinates that was exactly on the point I was supposed to technically have my tangent. My tangent I don't think is very accurate to be honest now that I'm looking at it. But now I'm going to choose another set of co uh coordinates. You could choose obviously the one where it goes through zero. Um which one is a nice set? Maybe here I've got four 4.8. That's what I've got. As I said, I don't think this is a very good tangent to be honest with you.
So we'll see if my answer is even in the range of what they've given as a suggestion. But we want to find the gradient of the straight line. So remember gradient of a straight line is change in y over change in x. It doesn't matter which way round you put the coordinates as long as you obviously put them vertically correct. So as in this is y, this is y. If I put 4.8 first, take away the other 2.8. Then I'm going to put the matching x coordinate vertically there. Do you see how that matches? And then the two goes with this one. just so that you get the right positive or negative.
If I subtract that, that gives me two.
Oh, I've got two over two, which is one.
I've got one. Now, let me actually have a look at what the answer should be because don't know if that's even close to being accurate. Oh. Oh, yeah. Okay, that's fine. So, my answer was one, but obviously because everyone's tangent will look slightly different, they have given quite a good range. If you get anything between 0.7 and 1.1, I was within that range, so I've got the three marks there. If you're in that range, you also get the three marks.
Don't rub out your tangent cuz they're going to want to see it as part of the marks. Just leave it where it is. But that's it for that one. And part B, what does the area under the graph represent?
So the second kind of half of that topic is finding the area under the graph. But remember what that means. If it's a speed time graph and you're doing area, you're effectively multiplying those together. When I multiply speed and time, it gives me distance. And if you think about uh formula triangle, speed equals distance / time. So distance equals speed time time. So the area under a curved graph where you've got speed and time is just distance. That's it. Distance one mark, right?
So that is four marks in total. This is a question that's more likely to repeat itself, right? So it's helpful to know.
Okay, I need to draw a tangent, find the gradient, so on. Okay, 15. Now we are on to vectors. So A, B, and C are three points such that A is 3 A + 4 B and A C is 15 A. It's 20 B. Prove that A, B, and C lie on a straight line. Two marks.
Okay. Proving that they lie on a straight line is really straightforward.
Again, if you know how to do it, if you remember, like just memorize it, right?
To prove that two vectors lie on a straight line, they have to first of all be proven to be parallel. And what does that mean? That means that they have to be a multiple of each other. So, is one just a bigger version or a smaller version of the other? What you're going to do is compare the coefficients. What I mean by that is the numbers in front.
So for a and then separately b the scale between them should be the same. So for example if I do 15 / 3 that gives me five. So for the a coefficients it's telling me that it's five times bigger right ac now that should be the same for b. So if I do 20 / 4 oh yeah it also gives me five. So what that's saying is I just need to take a and multiply it by five. That will give me AC.
So I would write it like this. You can write it different ways, but this is how I'd lay it out. I'd say right AC is equal to five lots five times AB.
It's a multiple of AB. Therefore, you don't have to write a lot. It's only two marks, right? That means it's parallel.
When vectors are multiples of each other, they're they're parallel. That's just the definition of it. They are parallel.
If they are multiples, they are parallel. Now, a step further because it's asked for a straight line, not just parallel.
AB and AC. Yeah, okay, they're parallel.
But how do I prove that they actually actually are connected? They create a straight line. Well, this is the easiest bit. They both have A in common, don't they? The point A, they both have them in common. So, if you've got two parallel lines, they both go through the same point, then they must overlap fully, mustn't they? As in create a straight line. So, the second bit is you just have to say that um something like they share point A or whatever point it happens to be in the question. In this case, it's A in common or they both have point A, whatever you want to call it. Therefore, that means not only are they just parallel because they could be parallel but just floating around separately, they go through the same point. So therefore, they must actually create Oh, I just wrote it straight a straight line.
That's all you need to know if you're ever proving that they're parallel. They say prove these two vectors are parallel. find out what the vectors are and then show that they have the same like multiple scale between both coefficients and then if they actually ask for you to prove that it's a straight line beyond that then you just have to show that okay not only are they parallel they also have the same point which normally you just look at it and go oh yeah both have point A so clearly they cross through the same the same points how do you know when to write methods they will tell you in the question yes exactly said it says write a reason for like give a reason for each stage if you're working or it says write down each circle theorem that you use.
Normally it's only one mark so you probably only need to write one. You might have used two so write down two.
Obviously don't write down a random one.
Write down the one that you did use but it will tell you.
So that's the first I think part A there's all right. It's part B that's a little bit harder to get your head around.
So D, E and F are three points on a straight line such that so already we're saying that they're on a straight line, right? Just bear that in mind. This case they are already on a straight line. D to E is 3 E + 6F. E to F is 10.5 E minus 21F. Now remember with vectors the positive and the negative tells you a direction as well.
So find the ratio of the length DF to D.
I in this case like to just sketch it because I feel like otherwise it's really hard to visualize. Now if we read this correctly, we've got D to E.
D to E is obviously the smaller one.
That's the most obvious thing I think to start with. So let's say that's D.
That's E. This direction we've got 3 E plus 6F. Now how do I interpret E to F?
It's obviously bigger. You can clearly see that. And they're on the same straight line, so they are multiples of each other. Exactly. But it's a it's clearly a bigger version, but the other thing is the negatives.
DE was positive, wasn't it? So I've drawn it in doesn't really matter which direction, but I've drawn it in that direction. E to F is then negative. So it's on the same straight line, but it's going the wrong way as in the opposite direction, and it's obviously much bigger. So, do you see how that's how I've interpreted it and said, "Right, well, it's not D E F in that order. It's D to F D to E and then back to F."
So, that's why I think that one's a little bit complicated. Now, the whole of that from E to F is this one, not just D to F, right? Or little points. This is obviously not to scale, but just a rough idea, just so you can visualize it. Okay. Now, they've asked for DF to D.
Now I know what DE is, but I don't have DF, do I? What I've actually got is EF.
I've got the whole thing, haven't I?
That's what I've got here.
So the first thing I would probably do is just work out as a vector DF.
So going D to F, I would probably have to, if you imagine walking along a line, I would go this direction, go to E, to be a six. And then I would go the other way.
Effectively, we're just adding them. You don't have to think about it the way I've just said, but we're effectively just adding them. These Let's collect the like terms. So what would that give me? - 7.5 E minus 15 F. So that is the green bit.
Now I want to know the ratio between the top two. And as we said they're multiples of each other. So just have a look compare the coefficients. Don't worry about the negative now because a ratio doesn't care about direction does it? It's just size scale. So ignore the negative. But 7.5 / 3 those are the coefficients of e. What does that give me? That gives me 2.5. It should obviously be the same for the other one.
You don't technically need to check because I know it's a straight line. So, I know they're multiples, but 15 divided by 6 would be the exact same answer.
They're both the scale is 7 uh 2.5, sorry. Now, how do I make that into a ratio? Well, the 2.5 is obviously the bigger side, isn't it? I mean, I can see that clearly. So, DF D if if DE is just one, then the other one would be 2.5. Now, they accept that or do they does it need to be it doesn't say it needs to be a whole number, does it?
doesn't say that they need to be integers.
This question, yeah, they said they would allow this or obviously if you want it as whole numbers as integers, it' be 5 to2. So any kind of equivalent as long as it's in a ratio and obviously it's you've got the correct things on the right side of it. The bigger one should be on the left and that's fine.
I'll put two because that looks a bit nicer, doesn't it? Uh that one I think is just a bit harder just because you have to figure out the way it's laid out. But beyond that, the actual math isn't that that difficult. Uh so that's three marks, five marks in total for question 15.
Question 16. A first aid test has two parts, a theory test and a practical test. The probability of passing the theory test is 0.75. The probability of passing only one of the two parts is 0.36.
The two events are independent. We need to work out the probability of passing the practical test. So this is just Oh, you said one more time.
Just go back.
Sorry, I think the chat's a little bit behind where I'm at. Did the diagram make sense to you? Because that's kind of a key bit. Or was it is it the diagram that didn't make sense or was it the bit after that that didn't make sense? So I can narrow down the lag.
Yeah. Oh, for part A.
Yeah, the summary for part A is just that if you're wanting to prove that it's on a straight line, first of all, prove that they are parallel. What that means is you have to prove that they are multiples of each other. So, just figure out what scale there is between a coefficients. If it's the same scale for the B coefficients, then yes, they are parallel. And if they have the same letter, do you see how they both got a in this particular case? It might be a different letter for a different question. But if they both got the same letter, then they're obviously going through the same point. So imagine two parallel lines, but they actually go through the same point. Well, they obviously have to create a straight line. Yeah.
But I have got um again like if there's any of this uh paper that you want to kind of go through in more detail more questions like it a lot of these I have got videos on. So like for example I've got videos on vectors where I've gone through vector questions just like these ones. Um so that might be helpful if you have a look at those as well just so that we can obviously get through the paper and then I can obviously answer some questions at the end as well. But have a look through my channel. You'll see that I've got vectors questions where I go through it in more detail.
That might be more helpful to you as well. Um, just so that we can get through the paper as well. All right.
So, this one I am thinking about it as a probability tree. That's just how I like to look at it. So, first of all, you have to do the theory.
Now, we know the theory pass is 0.75. So, I might as well fill that out. So, therefore, if I fail the theory, what's the chance of that? Well, it has to add up to one. So that would be the other 0.25.
That bit's fine. Then either way, we have to do a practical test.
Can there we go? So pass the practical.
We don't know currently, do we? We don't know the fail. We don't know here. We don't know here. So this is where when there are bits that are unknown to you, this is where again I like to add in algebra. That's just how I like to look at it. Let's say that the probability of passing the practical is x. The probability of failing the practical, I'm not just going to put y. Try not to put two letters if you can avoid it because often times there's only enough information to create one equation. And if you've only got one equation, you can only solve for one letter. So we've just done here. If the pass was 0.75, then I just did one minus 0.75 to get the fail.
So for the fail, it would be 1 min - x, wouldn't it? You see how I can write them both in terms of x. And because they're independent, it means whatever happens in the theory test, it doesn't impact the practical test. So they're the same exact probabilities.
That's how I would lay it out. Now, the key part here is this sentence. The probability of passing only one of the two parts is 0.36.
So imagine you were asked to work that out. What is the probability of passing only one of two parts? Okay, I could pass the theory but then I'd have to fail the practical. Or the other option is I could fail the theory and pass the practical.
Now imagine you're working through a probability tree. When you work through them, we always multiply between the branches. So for the blue option that would be 0.75 ultipli 1 - x and then for the green option that would be 0.25 25 * x and then when you have probability trees when you've done multiple options so here I've got two possible options haven't I when you've got multiple options that means that we have to add them because it's either the blue option or the green option isn't it one or the other and then that is what is equal to 0.36 you see how I've created an equation and that's why I only wanted to use x and not a second letter cuz I only had enough information to create one equation.
So from here, let's expand it. So we get 0.75 minus 0.75x plus the other 0.25x. It's just a bit annoying cuz we've got decimals, but that's okay. You can obviously multiply the whole thing by 10 if you wanted. Uh I might do that in a minute. Let's see.
So this would add together to give me -0.5.
Hopefully the rain isn't too loud.
x.
Now to solve it, I want x to be positive. So I'd probably add the x to the right hand side and then I would take away. So move the 0.36 across, take it away, which gives me 0.39. You, as I said, you could multiply by 10 if you don't want to deal with decimals. That's fine. Or well, I guess by 100.
Going to let me move it. Yes. Now this might be the tricky bit. Technically, yes. divide by 0.5. Hopefully, you know that dividing by 0.5 is the same as dividing by a half, which dividing by a half is the same as tsing by two. So, I'm just going to double it. So, yes, you are right.
Uh, if you get 78, but did you leave it as 78 or did you say 78%.
Because if you just put 78, I don't think they'd accept it. Obviously, probabilities have to be less than one.
If you put 0.78, fine. If you put 78%, fine. If you put 78 over 100, that would be fine. Uh, let me have a look.
Or equivalent. Yeah, if you just put 78, I don't think they'd accept it because it has to be a probability.
So, any kind of equivalent of this 78% like that, that's fine. Um, 78 over 100 would be fine. You could obviously simplify that to you might have made it into fractions. That's also fine. Um, any kind of equivalent is okay.
Could we use fractions? Yes, exactly.
You can use fractions up here. Like instead of using 75, well 0.75, you could use 3/4 and one quarter. That's also okay. Um it might be a bit tricky because 0.36.
You're happy to convert that into a fraction then. Great.
That's okay. So yeah, that's four marks for that one. That's um I would do it.
Just create an equation. Whenever there's unknowns, I try and create an equation, I think. Okay, question 17.
We've got y is directly proportional to the square root of t. We've obviously got our matching pair. Then we've also got t is inversely proportional to the cube of x. Again, matching pair. Find a formula for y in terms of x. So for proportion, most of the times it's three steps. Here it's just those three steps but twice because there's two different relationships. So step one is to write the correct formula. So if for direct proportion it' be y equ= k multiplied by this bit which is the square root of t. k is multiplied by that.
You can do the other one at the same time. Oh hang on. You can do the other one at the same time if you want. So the formula for the second bit is t is whenever it says is that's where you put the equals inversely proportional. That is where we do the fraction k over x or in this case x cubed.
So it's quite hard to remember that sometimes I think you put k in the denominator but you don't. K is in the numerator. The x or whatever it is in this case x cubed is in the denominator.
So those are my two formulas. Step two is then you have to sub in without coloring it in. Nope. Sub in the matching pair so that you can work out what k is. Now, actually, what I probably should do because I've got two relationships here, obviously K is not the same for both. So, mathematically, I should use a different letter. So, I'll put C, but just think that that's the same thing. The sub in Y is 15. Step two, and I don't know what K is, so that's what I'm trying to find. The root of 9 is three. So, you can see I could solve this quite easily. Divide by three. So, for the first formula equation, K is five. It's obviously not five for the second one. That's why I changed the letter just to distinguish.
I like to therefore write out the same formula but with the number just so that I've got just y and t in that particular case. Same over here. Sub in your matching pair. So t is 8 and two. Okay. So we've got 2 cubed which is 8. If I multiply it across I get 64. So you see how this one is 64.
They're not equal. That's why I put a different letter only for that reason.
Oh no it's not one.
t is 64 over x cub. So I've got both of my relationships there.
That's done. That's step one and step two. Step three is actually answer the question. I didn't even have to worry about what the question was until this point. Now, so the question was find a formula for y in terms of x.
That means if you translate that that they want a formula where it's y is the subject and then you've obviously got numbers but everything else in terms of letters is just x. So what they're saying is they don't want t. So what I would do is quite simple just substitute in all of this where it says t. So y equals 5. Instead of root t, I'm going to put all of this.
You see how now I've got y as the subject and then I've only got x as a letter. I don't have t anymore. I obviously don't have k or c or anything.
So that's that is a formula for y in terms of x. But they also do want it in its simplest form. So at this point I just need to simplify.
So square rooting when you're square rooting a fraction you can square root the numerator and the denominator separately. So 64 if you square root is obviously eight. But how you're thinking, well, how do I square root this?
You could I think you can leave it like that. Do you know? I think you can leave it like that. I was thinking about fractional powers, but I don't think necessary to be honest because it's not really that much more simplified. Um, I would leave it like that. I wouldn't even bother square rooting because it doesn't really square root, does it?
Just at least square root. Oh, yeah.
Square root the 64. And then we can multiply the five and the eight, can't we? When you have a whole number, so the five is a whole number, isn't it? And you're multiplying by a fraction, it only multiplies the nuator, not the otherwise it would just cancel out.
So that's what they mean by simplify. I wouldn't Yeah, I wouldn't bother simplifying the root x cubed. Don't worry about that. There are some things where you think this is a non-calculus.
It's a bit of a step too far, isn't it?
They're not going to ask me to do something that. So yeah, that's it. That was uh technically the question and that is four mark.
Okay, question 18. We've got work out the value of 5 and 4 9th the power of a half * 4 2/3 over 2 ^ of3 you must show all you're working you want to show all your working okay so what I would do here because we've got mixed numbers is I would just make them into improper fractions just to start with because apparently the numbers are a bit random so to convert into an improper fraction I multiply the five as in the whole number by the denominator. So 5 * 9 is 45 add 49 over 9. That is a slightly better set of numbers for me. And then this one 4 * 3 is 12 add 2 is 14 and three. That's a bit rounded still but guess we don't need a bracket there.
And then the two I would convert that into a fraction as well. So the two the power of negative a thirdg3 sorry the negative is the reciprocal isn't it? So flip the fraction upside down two as a whole number any whole number you can write it over one. So obviously if you flip it to make the reciprocal it is 1/2 and then obviously you then need to cube that.
Okay this top thing here we've just said a negative power is where you flip it upside down. So I might as well flip it upside down. Then I've got the power of a half. Just leave your 14 over here for Oh, sorry. One.
Leave the 14 over here for now. I thought 14 over three. This one we just said cub it. So one cubed is 2 eight.
Okay. The power of a half we should know means to square root. So if I square root I get three. That's why I said I like better because I could see that I was going to square root. 3 over 7 * 14 over 3. Hope you going. We've got rid of all our powers.
When I multiply these together, obviously I'm multiplying across, aren't I? Like numerator, numerator, denominator, denominator.
Oh, won't be down there. Yes. And the threes would just cancel, wouldn't they?
If I There's no point multiplying 14 by three and then seven by three if I'm then going to try and simplify the fraction and divide them both by three.
So those would just cancel. You would end up 14 over 7 / 108 which you could divide it straight away but that 14 over 7 will simplify beyond won't it as in 14 divid 7 is two I'm doing 2 / 1/8 we're almost there that's the same as keep flip change isn't it like if I write it like this keep flip change keep the first one flip the second one change it to a multiply 8 over one is the same as eight.
Either way, you should get 16. However you've kind of worked through it, you should get 16. Yes.
Have a look at the chat. Um, oh yeah, the three dots that I did on the last question just means there. It's just a very quick way. This thing like it makes it horrible. It's meant to make it like a triangle. this thing here like a little triangle but in three dots that just means therefore because I can't be bothered to write therefore so I just I just use the three dots a lot in math. Yeah. Yeah. You've just said it. Uh perfect.
16. You also got 16. Okay. Full marks for that one. Question 19. We've now got algebraic fractions. So solve 1 / 2x - 1 + 3x - 1 = 1. Give your answer in the form P plus or minus roo<unk> Q over2 where P and Q are integers.
When I look at this, I can already see A, it's going to be a quadratic.
B, I can't factoriize because you can't factoriize and get thirds, right? So, I'm thinking it's going to be a quadratic. When I solve it, I can't use um factorizing. I have to use Well, you could complete the square, but I probably wouldn't. I would use quadratic formula which isn't on your formula sheet and you just simplify it.
So sometimes you look at questions and you can kind of get a rough idea of where you that's what my evaluation of this question is so far. So we're coming back to cross multiplying again that we mentioned previously.
As I said you can do it in multiple steps if you find that easier. I like to just do it all in one step kind of know what pieces are going to go where. When I solve an equation, I don't want to have a fractions and b even less x in a denominator. So I have to get rid of the fractions. The way we get rid of fractions is we multiply by those denominators to cancel them out. So I have to multiply first of all everything. So this is a term, this is a term, and this is a term, right? All three by 2x - one. So all three of those multiply by 2x - 1. And then that would cancel out the first denominator. Then the second denominator is different. So I'd have to multiply all three of those again by x - one to cancel out that one.
Now personally I would do that all in one step. You can do it in multiple steps if you find that's easier. That's absolutely fine. So I know that when I multiply by x - one it's going to be multiplied by one. So that's just x - one and the one over here - one. And obviously it cancels it on this one.
Right? That's just not there anymore.
Then when I multiply by 2x - 1, I have to multiply. Well, first of all, it cancels out here, right? It's not there anymore. Then I have to multiply it by 3. So you see how I get this bracket.
And then I have to multiply it at the end as well.
So I know it's going to end up like this. As I said, if you want to do it in multiple steps, that's fine.
Now I've at least got everything on one kind of line. I'm going to expand everything and simplify. So do all of your expanding and simplifying, collecting the like terms, all of that.
We've got double brackets here. 2x^2 - x - 2x. Be careful with negatives and all of that. So easy to mistakes. We've got 7 x - 4 on the left, 2x^2 - 3x - uh one even on the right.
And then because I can see that I've got a quadratic, i.e. the highest power of x x^2. So therefore overall it's a quadratic. The way we solve quadratics is they have to have all of the terms on one side of the equation. Now I'm going to move everything to the right because I want my x squ to be positive and it already is on the right hand side. So I'm going to move the terms from the left to the right.
So subtract 7 x. So 2x 2 is still there.
- 3x - 7 x is - 10 x and then add four would be + 3 and that should equal to zero I made a mistake I feel like oh yeah yeah I was going to say I think I made a mistake that's what I was just saying it's so easy to make mistake minus one * minus one is obviously positive one right so plus one plus one just be really careful so when I add five add four I get five there we Okay, so we've got our quadratic. Now, normally, especially in non-calculus, I'd be like, "Okay, great. Quadratic factoriize two brackets, whatever."
There's no You can try. Obviously, there's no point trying because I can see the format of my answer is like this. So, there's no way I' like if I were to factoriize, I get either whole numbers or maybe a nice fraction or something like that. This is not a nice fraction. It's obviously not a whole number either. So, I'm not going to do that. I am going to use the quadratic formula. So you if you don't know it that's fine. It's obviously on your formula sheet. Uh I'll write it over here.
Minus b plus or minus the square root b ^ 2 - 4 a c over 2 a. Obviously a b and c are in the order as long as you've written it in correct order. Those are a b and c.
So subbing it in. So to work out x I want minus b. Now be careful because it's negative B in the formula. I've already got a negative B which means it's double negative. So that would become a positive plus or minus technically - 10^ 2 which I know would become positive take away 4 * 2 * 5 over 2 * 2. And then from here I'm just going to simplify it until I get to something that looks like basically where I get a denominator of two.
That's what I'm going to do. So the 10 can stay there for a second. Squaring -10 is positive 100. Take away 4 * 2 is 8 * 5 is 40 over 4. And then just keep working through it.
So that would give me 10 plus or minus<unk> 60 over 4. Now you might think, yeah, that looks quite nice and simplified. Um, but as we just said, the denominator here is two. not four. So it tells me I need to do a bit more. And what that is is I need to simplify the third. Remember how you simplify thirds?
You break it into two thirds, two factors. One of them has to be a square number though. So square number factor of 60.
I think it's just four. I'm trying to think there's a bigger one. I think four is the biggest one.
No, four and 60. Four and 15. And then you can square root the square number.
So 2 15. There we go. the 10 plus or minus 2<unk> 15 over 4 then do you see out of ignore the third but out of these three terms they're all divisible by two aren't they I can simplify them all by two down here that would give me five plus or minus technically one but you don't need to write one roo<unk> 15 over two that would be your final answer that's the format that they wanted question uh 5 + over two. Yes, you're um how do you deal with a fraction square?
Is that a completely separate question or do you mean because I mentioned earlier that you could complete the square because if it's related to this question I can go through you mean or do you mean just like separate two I move on.
Yeah. You mean in this question uh what calculations do you need to know? You need to know uh density.
That's probably one of the key ones on because do you get is it force pressure get that one density you need to know uh prisms to know they will give you anything that's not a prism. So if it's a sphere they'll give you that. If it's um cone they'll give you that cuz it's not prison. But prisms I would recommend. Uh okay. So if we were to complete the square here because that's the other option right? There are three ways.
Well, technically there's four ways to solve quadratics. One, ideally factoriize put into brackets. Uh, two, quadratic formula, which you often do with obviously um calculator. In this case, we could use it. We just had to simplify it rather than getting an actual decimal answer. Three is completing the square. Four is graphically, but obviously graphically you would only ever do if they actually told you to do it graphic. Don't just try and draw a graph if they haven't.
So, you would kind of ignore that one.
So completing the square, if I was to complete the square, it would be at this point here, right? So I would go leave the zero on this side so you can see. I would factoriize out two, I would only factoriize it out of the first two parts, right? That's just me. There's no there's no need to take it out of the third part. You can, but I don't see the point. Then I would complete the square.
So from there, we get square brackets x minus a half of this. So unfortunately, yeah, I guess that's what you mean by the fraction. We're getting fractions here. You just have to be really good at fractions, I guess, is the answer to your question. You subtract this squared. So that would be a positive, but I'm subtracting it. 25 over 4 in the brackets. I still have five here. Expand it all.
So 2 x - 5 / 2^ 2. And then I'd have to multiply the two by this as well. That would be minus 25 over two. That would simplify to that plus five. Obviously, I'd want to I guess add those together.
So, um, you could make it into a decimal because it's it's only 0.5, isn't it?
It's not that hard. Or if you leave it as a fraction, then you would have to do five. That would be 10 over two, wouldn't it? Like if you make them have a common denominator, that would be 10 over two. If I add those together, what's that? Minus 15 over two. That's right.
That's not what is meant to be five.
15 over two. That's completed the square version. And obviously that's much harder. That's why I didn't really bother doing it here. Then to solve it, I would probably add that across.
Kind of unpick it, I guess. You see what I mean? Like un undo that equation, times it by two. Uh, sorry, divide by two. So actually that' be 15 over four, wouldn't it? Cancel out, but it's not.
I guess this is where our square root would come in, isn't it?
I would square root both sides. But this is where you have to remember when you square root anything you get plus or minus. You should have two solutions for any quadratic unless they happen to be the same number. In which case that's fine. And then I would add the 5 over two. 5 over two plus this.
Obviously plus or minus roo<unk> 15. I can actually square root the four can't I? So that'll become two. And then you see how you just end up with the same thing. Yeah.
Have a look at chat. Yeah, you could do it that way. That's absolutely fine. I just think that's more work. I don't see How do you do the third thing? Um roo<unk>4 roo<unk> 15. So the third bit is where I'm simplifying roo<unk> 60.
So you see how 4 * 15 would make 60.
They're factors of 60. But four specifically is a square number which means it then square roots. 15 obviously does not square root. It stays. That's just simplifying ss. That bit there I kind of pulled it to one side. Um do I also teach IIT and G? I don't know what those are. I'm going to say no.
I'm not sure.
Uh if you've just joined the live stream, you are welcome to obviously at the well you can go backwards but this will upload once I finish anyway you can go back. Uh if not then you're in time for the last two questions basically.
Yeah if your exam is calculator I would assume that you are OCR.
I might be wrong, but that's my assumption because AQA and ED XL tomorrow is paper one, which is noncal.
I think OCR, I don't know if there are any others, but I know that OCR tomorrow is a calculator paper. So, this wouldn't be very helpful. It would be better to look up OCR or whatever your example is.
Um, oh yes, let me move it over.
These are just two options, right? These are two options of working out.
So it it's whichever you prefer.
Your options are you either use quadratic formula here or you complete the square. But as you can see I feel like completing the square was just more working.
Is this for IGCSE? It's not. But I would imagine it's not.
If you're doing ED XL IGCSE it's not massively different because it's still ed XL. The only thing is that IGCSE you have two papers that are they're slightly longer and you also have calculator for both but it is kind of similar. It's not that far off. I would recommend you look up IGCSE specifically though just that you're really Yeah. Just that you're used to that one.
Okay.
OCR. What What is your um example?
Because you should know that. I'm assuming it's OCR that like letters.
Okay.
Carry on. Last two questions. I think the last one is actually all right. Uh the center of circle um can't even read that. Let's try again.
The center of a circle is the point with coordinates -1 3. The point A with coordinate 68 lies on the circle. Find an equation of the tangent to the circle at A. Give your answer in the form Ax plus B Y + C equals Z where A are integers. This is I think quite a nice question. For some reason, I feel like tangent is always one of the last questions, but this one is a very nice one. You just have to follow exactly the method that hopefully you can memorize.
All right.
So when we work out um what you can do I kind of recommend again I'm quite a visual person so I like to look at it.
You don't have to if you're not a very visual person then I'm saying right now I like to sketch it out just really really quickly. Don't waste your timeffecting it right. Um I like to justly sketch it. So my center is minus3. So let's say roughly here again it's not scal let's say here. And then it goes through six eight. So I don't know something like that. Right?
It doesn't have to be exactly to scale.
So my center is about here about there. And then let's say cross.
Let's just say it's right. This is a 68.
So my tangent is there straight line something like that you know more or less.
So the way that we work out the equation of a tangent is it's obviously a straight line. The whole point of a straight line is y= mx plus c. So I need the gradient. I need c. I always start with the gradient. Now to find the gradient, we need change in y over changing x as we said earlier. But we don't have two sets of coordinates on that line. So you have to start from the radius. That's the radius there, the one I've just drawn. because I do have two sets of coordinates for that. That's n. So the gradient of the radius change in y over change in x 8 - 3 / 6 - -1 which will give me 5 over 7. You can see I've obviously not drawn it super scale, but you can see that it is positive gradient. That makes sense. Right? Now this is where we have to remember your circle theorem that the radius and tangent are always perpendicular 90° so negative reciprocal. I can convert this gradient into the negative reciprocal.
Oh, I can also see from my sketch that it should be a negative gradient. So if you're your gradient's already positive, make it negative. If it's negative, make it positive. And then the reciprocal, we keep saying about reciprocals, is where you flip it upside down to 7 over 5. So, I've got the gradient for my tangent.
I'll put t for tangent, r for radius.
So, now I've got y = -7x + c. The only thing then I'm missing is the plus c, which is the y intercept.
Now, I can see that it's somewhere up here positive. I obviously can't I can't look at my diagram and actually guess because that's obviously not going to be accurate, right?
So, I'm going to actually work it out.
Now the way we always work out the plus C is you just substitute in a set of coordinates on the line which is 68.
I've got those. So remember the eight is for y. Don't sub in six there. It can be quite easy to put the six first.
The six is for x. Again a lot of fractions basically that we have to work around here.
So when you're multiplying a whole number with a fraction, it only multiplies the numerator. So that would be 7 * 6 42 that went out there over 5 + c. Now I'm going to add it to the other side. Now when I want to add it to eight, I'm going to need to have a common denominator again.
So 8 currently any whole number I keep saying this is over one. So to make it a common denominator, I would have to multiply this one by five. That would be 40 over 5, wouldn't it? Plus 42 over 5.
is C. That is 82 over 5. That's how I would work it out. I guess you could use decimals if you wanted. I don't really want to, so that's why I've used fractions, but I'm just going to leave it like that now. So, let's rewrite it.
We've got y = -7 over 5 x + 82 over 5.
Now, that is the tangent, but remember to double check what format they want it in.
So, we want ax plus b y + c equal z.
First of all, everything on one side, but also they're all integers. They're all whole numbers. So, when you get that, to make them all whole numbers, do you see how they both have the denominator five?
So, in order to get rid of a fraction, in order to make them whole numbers, I just multiply across the whole equation by five. So, that would also make it 5 y - 7 x + 82.
X. So they're all whole numbers now. And the last thing was we wanted everything on one side. So I'd probably make it so that X is also like if I moved five Y it would be negative. So I think it'd be tidier like neater if I moved everything to the left. I think they wanted X first. So 7 X 5 Yus 82. It's fine. It's negative. But X and Y are positive. Equals Z. That's just formatting at the end. That's all that is.
Yeah, it's a pretty straightforward tangent question. Just the last bit you had to format it. That's for Okay, the chat.
Yeah, Cambridge is an international example.
Yeah, be that I guess. Um, but this won't be Yeah, if you're doing Cambridge that'll be different, but you probably look that up on YouTube. People going through those as well specific.
This one is from 2022. It's not my predicted paper. However, I have got a predicted paper that's not a live. It is just a video which I uploaded yesterday. So if you have a look at my channel should be the second you can have a look at my like which would be based on edex though. So uh all right checking if there's any questions.
How come the eight is 8 over one?
Because eight as a whole any whole number which eight is a whole number. If you want to make it into a fraction which I needed to do because I wanted to add it to another fraction it's over one. So any whole number you can just put it over one because if you did eight divided by one it would still be eight, wouldn't it? It be the same thing. And then I had to in order to get a common denominator so I could add them. I had to times top and the bottom by five. That's where that came from.
Um how come we don't multiply the 42 and the one just leave?
Yeah. So the 42 going to how come you don't multiply the 42 one and just leave it because it the lowest common multiple between one and five is five. So if it's already like if one of your fractions already has the denominator which is the lowest common multiple, you don't have to change that fraction. If they were both different, then yeah, I'd have to multiply both. But because one was already, it already had the denominator five and my common denominator was going to be five. I didn't have to change the 42. That would have been easier. This isn't IGCSE. No, this is normal GCSE. If you're doing ED XL IGCSE, this might be kind of similar. Not that far different, but it's technically not IGCSE.
Times it will always be positive.
Which bit are you referring to when I'm It's always positive the 42 over 5 became positive cuz I add it like the other side of the what you're referring to.
you're talking about and then go on to there's just one question left. It is I think the hardest question.
Yeah. Okay.
That was four marks. Final question.
This is I think one of the hardest question.
I mean again it's one of those things where I feel like the maths necessarily isn't Yeah, it's a little bit hard but I think knowing what to do is the hardest bit. Figuring out what to do is the hardest bit.
So we've got the diagram shows three circles each of radius 4. The centers of the circles are a b and c such that a b c is a straight line and a to b equals b to c which is 4 cm. So like you could they're all in the same level. I mean and the distance here it's the same. It is four. Work out the total area of the two shaded regions.
obviously give your answer in terms of pi. They're not going to make you work out what pi to exam unless it's an estimate.
So what I did here I think you got things like this. I'm always looking for triangles. What I would suggest always look for triangles.
So if I draw a line from A to B, that is the radius of circle A. It's also the radius of circle B, isn't it? See that?
And then if I draw a line from B to this corner here. Oh, it's not really, you know, anymore where they intersect. That is the radius of circle B still. So the purple lines are both four. They're both a radius.
And then if I do the same thing, get a different color though.
If I had A to B, know that's four. We know that that's the radius of A as well as B. And then if I do this line, this is also the radius of A. So you see how then they're all four, which is an equilateral triangle. Now, how is that helpful to me? Because equilateral triangles have angle 60°. So at least I've now got an angle to work with, not just sides.
So let's think about the order I would do this in. Okay, I'm going to leave that there.
We look on obviously the other side is the same, right? And below is the same.
There's like four triangles that are identical. Okay, I'm just going to draw here.
Let's work with this as if it were a sector. Now, think of it as a sector.
That section there, that's also going to be 60° because, as we said, if I were to draw a line here, it would be the same equilateral triangle. There's a sector there. I know it's four and four. So, let's work out the area of the sector.
We should know that formula.
Yeah, I don't think they give that sheet. Yeah. Area of a sector, someone was asking what formulas. Area of a sector and then arc length. You should know those. So area of a sector is the area of the whole circle, which is p<unk> r^ 2 multiplied by the angle of the sector over 360. Now that's why I wanted to know the angle. So in this case, we know that the radius is 4 and we know the angle is 60.
So just go through and kind of rearrange that. Simplify it.
That' be 16 pi. This would simplify if I knock off the zeros to one six.
That's not six at all.
Six.
So 16 over 6 pi. You could simp further if you want. It's not. What's that? 8 over3.
If I highlight it, we can What's that? One sector, right? This bit here.
Now, there are four of those, aren't there?
See how there would be four? There'd be one down here, one on the left, or two on the left, one up and down. There would be four of those. So, we could times it by 4. That would be 4 * 8 is 32 over three.
That's all four sectors.
Now I then have the bit that's then missing is like this little slither here. I guess we call that a segment, don't we?
That's why I'm going to come back to the triangle.
That segment there is the same as this segment here, isn't it? They're all the same. So if I can work out the whole sector which I've got one sector is 8 over 3 pi and then I subtract this triangle then it leaves me with a se the segment even isn't it segment so area of a triangle now this one I think you would have to use the trigonometric area of a triangle because we don't have a vertical height you probably could work it out use pyagor But I can't really be bothered to do that. So I'm going to use the half a sin c that's on your formula sheet. Now that's obviously the angle and then the two sides between it. So half * 4* the other four time s of the angle 60. Now remember that for this for the non-calculated paper you're supposed to have memorized certain values. All right? So certain values in this case sin 60 is roo<unk>3 /2. So you do half * 4 is 2 * 4 is 8.
What have we got here? Root two. And then obviously if you multiply that you get 4<unk>3.
43.
Oh, thank you to someone who has just sent something in the chat that was really that's lovely.
All right. So, this is my triangle.
Let's color code that separately. So, we'll go with green.
Tidy there.
That's what that is. Okay.
So, segment area of sounds like sector but I'm not saying sector I'm saying segment the orange bit would be the whole blue bit 8 over 3 pi minus the whole green bit wouldn't it that's not fraction there now I can't simplify that I'm l just going to leave it like that. It looks kind of messy and annoying, but you know what I mean. I've got SS. I've got I not going to work that out.
So, that's the orange bit. I like that.
We can keep track of what is what.
That's the orange bit.
Okay. Finally, where are we actually going with this? If I want to find the shaded region, I could work out the whole try and ignore all of my diagrams.
the whole of circle B which was that center circle and then I could take away I've got four segment uh sectors sorry and I've got four segments haven't I so I could take them away yes um someone who's asked about 5 over2 I'll go back to that once I finish this question all right so it's just all in at least I'll carry on with this and I'll go back to it um okay so the Area of circle B as a whole p<unk> r² isn't it radius is four that is 16 pi that's the whole of circle b then I want to take away the bit that isn't shaded subtract as I've just said I've got four sectors which I worked out here didn't I that's four lots 3 pi and take away four segments.
Let's work that. Well, if I times that by four, I get 32 over 3<unk> - 16<unk>3 take away 32 over 3.
Obviously, that's messy, right? So I don't want to leave it like that. But that's where I'm getting to I'm just going to have to go away so that I'm actually following some logic here.
Okay, let's see. So these have got both pies. They both got a common denominator. So I can that would be 64 over three, wouldn't it?
And then just be careful here because you're doing double negative.
You see how that would be a double negative? So minus minus 16 roo<unk>3. Okay, these are both pies.
So they're like terms. If you think about algebra, they're like terms. So I'm actually going to work those out.
This is again, we're coming back to this over and over. If you ever want to convert a whole number, an integer into a fraction, you just put it over one because that is still the same number, isn't it? Now the lowest common multiple because when I subtract fractions I need to have a common denominator. The lowest common multiple between 1 and three is three. So I don't need to change this fraction. But I do need to change the first one by multiplying the top and the bottom three. That would be 48 again. 16 times tables. I keep saying that comes up.
And now do you see how I can take them away? 48 - 64 actually be 16 over 3 I plus the other roo<unk>3 there. Now I would probably just write it the other way around because just for the sake of the fact that I don't like to have any start but it doesn't really right it's the same thing. This is my final answer.
That's the most simplified I can get.
Okay, that is yeah there's a lot to keep track of there. I think that's why it's hard. There's a lot to keep track of there. That is a obviously five mark question. So that would be the shaded region here being roo<unk>3 - 16 over three.
So the last bit is here.
I know I've not really drawn it in very logical order to be honest with you.
That's fair enough.
I did the area of circle B, which is just pi squ.
And then I subtracted four of these blue segments because if you imagine it here, there's a segment below the blue one. There's a segment to the left and then there's one like the orange and the green add together to make a segment. So that's four. Sorry, add together to make sectors. So that's four of them. But there's this little bit here, little slither, this little slither, and this little slither. It's four of those, which are segments, which is the orange.
So I had four of the blue and four of the orange, which I then subtracted from the whole circle, and that should leave me with the shaded region. And then from there it was basically just simplifying like fractions effectively.
The last bit is just fractions.
It answers your question about the last bit but you can see here that's fine. Uh and that's the last bit uh you asking about the flip.
So here when you have two lines that are perpendicular.
So because of circle theorems, we should know that a radius and a tangent are always perpendicular.
That's the radius and this here was the tangent, wasn't it? They're perpendicular. Whenever two lines, straight lines are perpendicular.
You can convert the gradient between each other. You can flip between one and the other by doing what I would call not just obviously is called this negative iron recipro.
So there's two parts to that. The negative means you swap symbol. So it was already positive. So I made it negative. If it was already negative, I would make it positive again. You know what I mean? So you see how obviously if you have two perpendicular lines, they're crossing through each other. So one is going up and one is going down.
So one's a negative gradient, one's positive gradient. And the second part is the reciprocal. So if you have a fraction already, that's why I kept it in a fraction. I don't like use decimal questions. Easier to have a fraction.
You flip it upside down. The reciprocal means flip your fraction upside down.
It's just a fancy way to say that.
That's why we not only made it negative but also then flip just because of perpendicular.
That's that bit.
Yeah, that was all of that paper. That was as a reminder the edexl June 2022 one H. So higher tier somewhere should have remember where I saved the um spreadsheet earlier saved it some a quick look obviously that is the end of the paper so leave the live yeah but I will have a look for it just in case if not then have to look for it as um as like yeah like a post. You can see a screenshot but it is in my other videos as well where the grade boundary.
Yeah, I have the old version. I don't update it.
Hot loes.
video.
It's great.
train.
Oh, there we go. Found it.
very but I'm going to print screen it'll read it just this will actually upload later this week so you see it there we go that's what I was looking for have to find the actual original spreadsheets so it looks a little bit nicer but you can read that this is for edex right so AQA doing AQA just ignore what I'm saying this at XL average grade boundaries, right? I've worked out the average grade boundaries.
Okay. So, this is not to say that they won't be a higher grade boundary this year, a lower grade boundary this year.
This is the average um I think yeah, here I've said from 2017 to 2025. So, it's a bit skewed because of co I guess, but just so you've got a rough idea, you're looking to get a seven. Your total amount of marks you would need to get is roughly 136. So roughly 45 per paper which is only 56%. Now to be safe because as I said it might be a bit higher, it might be a bit lower to be safe maybe round it up just to cover your B. So, if you were saying, "Right, I need to get a seven." Maybe try and aim for like at least 50 on each paper, maybe 55, just so that you're really safe because I don't want to say this and then you then this year the grade was higher and you'll be like, "Oh, yeah, but you said the grade were going to be lower." I don't know what they're going to be obviously. I have no idea.
But take these numbers and round them up a bit just to cover yourself, right? So, you can see there for four, five, six, seven, eight, nine for higher tier and for foundation. If you're looking for a four, 46% uh sorry, 46 marks per paper, that's 57%. So again, round that up to like 50 55 just to be safe, just to cover yourself. So if anyone was looking for the average EDXL grade boundaries from 2017 to 25, those are what they are. Um I don't think that includes November last year. Yeah, that won't. Yeah, I don't think it includes November because I'm not sure if they would have released. Yeah, they might have done.
Um, but I updated this like a month ago.
So, it's with the most up to date information that I had. So, just going to keep repeating it. If you're using these, just to give you kind of an idea of what you need to aim for, round them up just to be safe because there are obviously some years where it wasn't 45, for example. It might have been much higher than that. So, just be really careful. Just round them up and then you've got a good idea. um of something to aim. But I think you'll find particularly for higher tier that the grade boundaries are a lot lower than what you might think. So whoever was asking how many marks can I lose um per paper to get a grade nine, I said as a guess probably 10. So I think that's a safe guess because here you can see it's 14. But obviously as I said it might have been the grade boundary that was higher one year. If you say maybe 10 per paper maximum, that would probably a good estimate.
Safe, maybe eight or 10 per paper, that's a safe bet. You're looking for a grade. If you're looking to just pass, but you're doing higher tier, you effectively only need to get 20%. Which is insane. But yeah, so maybe round that up. Maybe say, "Okay, instead of 14 marks per paper, I'm going to aim for 20 marks per and then you'll probably be quite safe." So yeah, you can screenshot that. Uh, as I said, I know it's quite blurry, but this will be uploading. Uh, I have a video uploading tomorrow afternoon. This is when you look through the video, you can screenshot it from there. It should upload a little this.
Um, and otherwise, I'll try and see if I can actually find the original spreadsheet that I did use because obviously the spreadsheet will look a lot lot clearer than this, but I've saved it. So, right now, that's that. But yeah, so that is everything um for today.
You can obviously go back and look through. I've done other live videos yesterday, last week. If you want to those um if you just want to rest the rest of the evening and just pay yourself mentally for tomorrow, that's also absolutely fine. If you want to look at a predicted paper, I have my own predicted paper that I uploaded yesterday. Um that you can have a look through as well. Uh if there's any specific topics that you want to look through, again, check if they're already in my backlog catalog of videos. I might have done specific videos on certain things that came off this paper as well that you can look through. Completely up to you. But best of luck to everyone who is going to do their paper, whatever exam paper you're doing tomorrow for maths. Just I think the most important thing is to stay as calm as possible.
Try not to get in there and panic because that is worse, isn't it? Because then you might not answer the questions as you actually capable. Try and stay calm and I know it's a stressful environment. Try and just focus on what you're doing. you've got keep an eye on the time. We don't want to run out of time. So, keep yourself moving. If you're stuck on a question, leave it.
Okay? Just leave it. Move on. Try and get marks elsewhere, right? And then come back later. Um, the other thing, yeah, my other tip is to make sure you write down something for every question. I don't care if you think it's stupid. If you're like, ah, I'm going to think really stupid, write this. They're not judging you. They don't know who you are, right? So, it doesn't matter what you write. Just as long as you're not contradicting yourself, as long as you're not saying one thing and then you're working out something that's completely the opposite, then it's fine. Just write down anything that you think might be relevant. If you're not sure on how to answer the question, don't leave it blank. Write down anything you think is relevant. It's a formula. If anything to do with the topics you can figure out, even if you're not sure, oh, I don't think it's meant to go this way. It doesn't matter. I'd rather you just wrote down something because you never know where those method marks are going to come from. you could get a higher grade from not necessarily answering the questions fully, but just from half answering all of the questions. You know what I mean? So, yeah, try to stay calm, keep an eye on the time, make sure you write down as much as you can for every question is what I would say. Those are kind of my top three. Um, yes, this video, this live will be uploaded afterwards. You can go back and watch it. Don't know how long it takes to do that. I don't know if it's automatic, but it will upload itself. Yeah. And you can also, as I said, look through my previous lives. Like yesterday we went through June 2023. Last week we went through a few others as well which were obviously closer to 2024 time. You can look at more recent ones as well. Yeah, as I said, best of luck to everyone tomorrow. Thank you so much for coming to the live. Hopefully helpful. And I will be continuing lives. Obviously I'll give you a bit of a break after paper one. I think half term we'll continue with paper two and paper three.
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