GCSE Maths Paper 1 - Top 10 Topics Revise - Predicted May June 2026 - Higher AQA Edexcel

Added: 2026-05-09

[00:00:01]Welcome to Mr. Tommpkins edtech. Math paper 1 is days away and if you're sitting the AQA or Ed XLOO exam boards, that means you have a non-calculated paper to do this coming Thursday.

[00:00:13]In this video, I'll identify the top 10 most likely topics you will encounter on a non-calculated paper and run through some of the typical exam questions to help you prepare for that paper one.

[00:00:24]There is a worksheet that goes with this video and I strongly recommend that you try the questions first yourself before watching. You can then mark it and uplevel your answers and identify any weaknesses or misconceptions that you still might have. Also, in the description underneath the video, I've put links to further videos and worksheets on each of these specific topics. That way, you can delve deeper into any topics that still need further work.

[00:00:51]If you find this video useful, please give it a thumbs up. That really helps me out. Once paper one is out the way, I'll be posting more videos covering paper two and three. So, make sure you're subscribed with a bell notification turned on so you don't miss out on those. Any further questions that you might have, just write them in the comments below the video and I'll try to answer them for you.

[00:01:12]Now the non-calculate paper is the only opportunity to test certain non-cal skills which [music] means these topics come up nearly every paper. So the topics we'll be looking at today are standard form, calculating in terms of pi, inequalities, expanding and factorizing, solving quadratics, indices, commumulative frequency, recurring decimals, SS, exact trigonometric values, and algebraic fractions. Now, if you're solid on all of these topics, you should be looking at a very respectable grade for this paper. Now, if you haven't already downloaded the worksheet yet and given the questions a try, I'd suggest you pause this video now and do that first. You'll find the link to the worksheet in the description below the video. Right, got the worksheet? Good.

[00:02:06]Let's move on to the first topic.

[00:02:10]So, first up, let's take a look at standard form. Now when you write a number in standard form you need to write in the form a * 10 ^ n where this a here it's called the mantisa if you want to know the real name for it. Uh and the mantisa value needs to be a number between 1 and 10.

[00:02:30]Can't actually be equal to 10. Uh it's got to be like a one digit one digit unit value. It can be decimal of course.

[00:02:37]Uh and then the next part over here the 10 to the n is called the exponent. The value of n needs to be an integer basically. So n is an integer. That's a symbol for an integer if you didn't if you wanted to know. So n is an integer.

[00:02:51]Has to be like a positive or negative whole number. And a needs to be a number between 1 and 10. So write 0. 0.97 in standard form. Okay. Very small numbers going to have a negative uh power of 10.

[00:03:05]And the way I usually do, I just count the number of zeros in front of the first digit like that. So that's got like four zeros in front of the nine. So I just use that. And then I write down the digit part with the decimal point after the first digit. 9.7 and then times 10 to the 1 2 3 4. So it's going to be 10^ the minus 4 like that. Okay.

[00:03:25]So the number of zeros at the beginning is the number you're going to get up there. And the mantisa part here is just going to be those numbers with a decimal point in between them. Right? Part B.

[00:03:37]Write down 3 * 10 5 / 4 * 10 3. When you've got a quotient like this where you got numbers in standard form on top and the bottom, then divide like with like. We're going to divide the mantis with the mantissas and we're going to divide the exponents with the exponents.

[00:03:54]Okay. So 3 / 4 that's 3/4 or 0.75 as a decimal, isn't it? Okay. Okay, so that's 0.75 and 10^ 5 / 10 3 numbers with the same base, you subtract the indices, don't you? So 5 take away 3 is 2. So I get 0.75 * 10 to the 2. Now this question is asking me to write this as an ordinary number. So let's evaluate it. So 10^ 2 is 100, isn't it? So 0.75 * 100, what's that? Then 0.75* 100 that is 75. So that's my final answer. Okay.

[00:04:32]Now again, I've got two numbers here in standard form that I'm multiplying together. And again, I'm going to do the same thing. We're going to multiply like with like. So we're going to multiply the number part with the number part.

[00:04:42]I'm going to multiply them uh the exponent part with the exponent part. So 2 * 8, that's 16.

[00:04:50]And 10 the 6 * 10 4. Two numbers with the same base add the indices. That's going to give me 10 to the 10. Now, I've got a similar problem here in that 16 is not a suitable mantis, is it? It's not a number between 1 and 10. But if we divide it by 10, if we take 10 out of it, we're going to get 1.6. It's 1.6 * 10 is 16, isn't it? What we going to do with that extra 10? We're going to roll it into the exponent part. So, I get one more 10 in my block of 10 at the end there. So, 16 * 10 becomes 1.6. 6 * 10 11.

[00:05:30]Part B. Well, this is very similar to the one we've just done. So, divide 2 by 8. What's that? 28 1/4. That's 0.25.

[00:05:40]Uh, dividing 10 6 by 10 4. That is going to give me 10 2. Now, this time it wants me to write it as an ordinary number. Uh, so 10^ the 2, that's 100, isn't it? Okay.

[00:05:57]100 * 0.25, that's just going to be 25. Okay.

[00:06:03]25.

[00:06:05]Question three. Work out 2 * 10 14 / 8 * 10 9. Another very similar sim similar question. 2 / 8 uh that is 0.25.

[00:06:19]10^ 14 / 10^ the 9. That's given us 10^ the 5. Uh, and again, that isn't a suitable mantis mantis. So, I'm going to borrow 10 from my 10^ the 5, make it 10^ the 4 and multiply my 0.25 up by it.

[00:06:34]That's going to be 2.5 * 10 4. And then part B says 6,200.07 that is equal to 6.2 * 10 C and 7 * 10 D. Okay. Well, I suppose we could say that 6,200.07 is 6,200 plus 0.07 then, isn't it? Okay. So, writing those two numbers in standard form, 6,200, uh, take the digit part, six and the two, going to write it as a mantisa, 6.2.

[00:07:10]Uh, and that's going to be multiplied by an appropriate power of 10, isn't it?

[00:07:14]Now, if the decimal point started here and we want to move it there, we're going to move it three jumps, aren't we?

[00:07:20]So, it's 10 to the three. Another way of thinking about it is the thousands has got three zeros. It's 10^ the 3. Uh 0.07. Remember earlier we were counting the number of zeros in front of the first digit. So, I've got two there. So, I think this is going to be 7 * 10us2 because I got two digits. Uh, and so C is the the power of my 10 on the 6.2. So that's three. And D must be my power of 10 on the other one. It's minus two.

[00:07:51]Moving on to calculating in terms of pi.

[00:07:54]Don't forget your two most important formulas for this one for circumference and area. Cherry pie delicious apple pies are 2 C= P<unk> D A= P<unk> R².

[00:08:06]They're the main formulas you're going to use. Here is a quarter circle of radius 6 cm. Work out the area of the quarter circle. Give your answers in terms of pi. Okay. Well, the area of a quarter circle is going to be 14 of the area of a circle, isn't it? So, it's going to be a quart of pi r 2. Here the radius is 6. So, I'm going to get 14 of<unk> * 6^ 2. 6^2 is 36. Divide it by 4. Fours into 36 go nine times. So that ends up being 9 pi. And that's my answer. We're always going to give it in terms of pi on the calculator non-calculated paper uh because they don't want you multiplying through by 3.14 whatever. So yeah, you just leave your answers as a multiple of pi. So 9 pi is our final answer. Question five. This semicircle is a radius of 5 cm. Work out the perimeter of the semicircle. Give your answer in terms of pi. Okay, so perimeter is the distance all the way around the shape. And for a semicircle, the perimeter is made up of two parts, isn't it? You've got like the arc here, which is half a circle, half the circumference of a circle. And you've got this straight line part here, which is made up of two radi. Okay. So the perimeter part then the yellow part is going to be half of the circumference of the circle which so we can use either uh pi d but we can also use 2 pi r which might be better in this case. So it's p<unk> * 2<unk> r. Uh and then we're going to add on two lots of the radius which is the green bit. Okay. So the green bits is 2 r. It's radius and radius. And the yellow bit is half of 2 pi r the circumference of the circle.

[00:09:53]Okay. So let's just tidy that up a little bit. half times two. Well, they just cancel each other out. But we know r is five, so that's going to be 5 pi.

[00:10:04]5 pi from the first part plus 2 r. Well, two lots of five, that's 10. Okay, so 5 pi + 10. That's my final answer.

[00:10:15]Question six. Four identical circles just fit inside a square as shown. Work out the area of the shaded section. give your answer in terms of pi. Okay, so this is a bit like a compound uh area problem. You've got a square shape and you've got four circular holes punched out of it. Okay, so the work to work out the area the combined area is going to be the area of the rectangle minus four lots of the area of a circle then isn't it? Okay, I just tend to do little not notation like that so I don't confuse myself. So let's just work out those two areas. So the area of the the square part first. Well square formula is just L squares, isn't it? Length time length.

[00:11:02]And the length there is 12. So it's going to be 12 squar that's giving me an area of 144.

[00:11:08]Uh and then the area of the circle that's just going to be pi r 2 isn't it?

[00:11:14]So what's r here? So you can see it's 12 cm all the way across. So that's the same as two circles. So that means one circle it must be half of that must be six. And the radius is going to be half of that. It's going to be three, isn't it? So the area of the of one circle is going to be uh<unk> * 3^ 2. 3^ 2 is 9.

[00:11:37]That's going to be 9 pi. Okay. So now we've worked out the various bits. We can put it into our kind of master area formula down here. So area of the of the square we worked out to be 144 and the area of the circle is 9 pi. So I'm going to get 144 - 4 lots of 9 pi. 4 9 are 36.

[00:11:58]So that's 144 - 36 pi. Okay. Again I can't really evaluate that any further other than working out what pi is and and and multiplying it through but not doing that. So that's my final answer.

[00:12:12]Inequalities next. Inequalities solve exactly the same way as equation. So if you know what to do if there was an equal sign in between these things, you do exactly the same thing if there's an inequality symbol. So we just need to try and get uh all the x's on one side and get everything over to the other side and evaluate it. Okay. So first up, I've got a bracket on the left with a number on the outside. Now, normally I just double check if I can't just divide that number outside the bracket over on the other side, but I can't. I can divide 4x um two but not 4x - 1 doesn't work. So in this case it's going to be easier just to multiply out these brackets. So uh two lots of 7 x is 14x and 2 lots of 3 is 6. So that gives me 14 x + 6 and that is less than 4x - 1.

[00:13:00]Okay. Now uh you can then either think about this as subtracting 4x from both sides. That's one way of thinking about it. The other way is change sides, change signs. [snorts] Another way of thinking about it is moving things over the equal sign. One side if they're positive, they move over, they become negative. I kind of think of it that way. I think it's slightly quicker to do it that way. So 14x minus that 4x is going to give me 10 x on the left.

[00:13:25]Similarly, if I take six away from both sides or uh if I take this plus 6 and move over the inequality similar, it's going to become minus 6 on the other day on the on the other side. uh combining that with the minus1 is going to give me -7. So I've got 10 x is less than -7.

[00:13:42]Dividing through by both sides by 10, I'm going to get x is less than0.7.

[00:13:49]Okay. Question eight. Solve -3x is greater than 6. Now beware here. There's a trap. Okay. uh because I've got a minus sign in front of the x and inequalities behave really weirdly when with negative numbers. Now there are two ways to approach this. Either you just avoid dividing through by negatives and you can do that uh by if if your unknown is negative. Just shift it over to the other side of the equation make it positive. So switching these two numbers around. So adding 3x to both sides or change sides, change signs with the 3x, I'm going to get plus 3x on this side.

[00:14:28]And doing the same thing with the six.

[00:14:29]Either taking away from both sides or just change sides, change signs. I'm going to get min - 6x is greater than 3x. Now to get x, I can divide through by positive uh 3. That's no problem at all. I can just go ahead and divide, and that gives me x is less than -2. Okay, so that's one approach. Just simply avoid doing it. Uh the other way of doing it is just remembering if you divide through by a negative number, you have to switch the inequality around.

[00:14:59]Okay, so I could just divide through here by -3 on both sides. Okay. Uh so that would give me -3 / -3 would give me positive 1 or just x. Uh 6 /3 would give me minus2. But I have to remember if I do that, I have to switch the inequality symbol around like that. Okay. So either avoid doing it or remember to switch your inequality symbol around if you divide through by a negative. Okay, question nine. Work out all the integers that satisfy the inequality. A is less than 2 n, which is less than equal to 16. See, I've got 2 n in the middle. Be much nicer if I just had an n in the middle. And I can do that by just dividing all parts of the inequality through by two. So 8 / 2 is 4. 2 n / 2 is n. 16 / 2 is 8. So that tells me that n lies between four and 8. If you imagine it on the number line, four 5 6 7 8, then you'd have one of these things going from four up to eight, wouldn't you show it? And because a is part of the part of the range, we fill it in. And because four isn't because n has to be strictly greater than four, it's it's left open. Now we want the integer values that satisfy those. So which integer values are between those two limits? Well, four isn't because four can't be. It has to be greater than four. But five is, six is, seven is.

[00:16:24]Eight is as well because n can be less than or equal to 8. So it's those four there. So my answers are going to be 5 6 7 and 8. Okay. Question 10. N is an integer. List all the values of n such that minus1 is less than equal to n + 3 which is less than five. Okay. Now again at the moment I've got n plus three in the middle in the middle. Now I can simplify that. Get n by subtracting three from all parts. So take three away from there. Take three away from there.

[00:16:53]Take three away from there. These ones in the middle will cancel just leaving you with n in the middle. Uh on the left hand side you're going to get min -1 minus 3 which is -4 and I've got 5 takeway 3 that is 2. Okay. So again, just imagine that on a number line. So I've got -4, -3, -2, -1, 0, 1, 2. Then this time the left hand limit is part of my inequality. So I can fill that one in. My right hand limit has to be strictly less than two.

[00:17:26]So that one is left open. So again, what are the integer values that lie between those two limits? my four, my minus3, my minus2, my minus one, zero, and one.

[00:17:37]These are all fine. So, I'm going to include those. I'm not going to include two because it needs to be strictly less than two. Okay, so it's those question 11 - 17 is less than or equal to 4x + 3, which is less than 11. Uh, slightly more complicated in the middle. Got two two steps to make it uh just x. First, take three away from every part. That's going to give me -20 is less than equal to 4x which is less uh or sorry strictly less than which is less than 8. Okay. And then dividing everything through by four I'm going to get -5 is less than equal to x which is less than equal to 2. Now this question doesn't want us to fuff around with uh finding integer values. So that's just my final answer. - 5 is less than equal to x which is less than 2.

[00:18:28]Uh, part B says, work out the product of all the integer solutions to -17, which is less than or equal to 4x + 3, which is less than 11. Oh, they do want us to fuff around with uh with finding integers after all. So, again, I'm just going to imagine our well, we've already simplified it down to -5 is less than equal to x, which is less than 2. So, imagine that as a number line again. -5 -4 -3 - 2 - one 0 1 2 Putting a a filled circle at five because it's greater than or equal to five and an empty circle at two. Okay, so what are the values that match? -5 -4 -3 -3 - 2 -1 0 1. Okay, you might be thinking, I don't want to spend time doing all that numberline business, but I do think it gives you a little bit of clarity uh and oversight and makes helps you avoid making silly mistakes. You know, you've got plenty of time to do this paper. Uh you don't have to rush.

[00:19:39]Um it's easy to make little mistakes with these sort of things. So, just to visualize it like that just helps you prevent you making those silly mistakes.

[00:19:46]Oh, now I just reread the question and it says work out the product of all the integer solutions. Uh, the product is multiplying them all together, isn't it?

[00:19:56]So, if I'm going to multiply all these together, -5 * -4 * -3 * -2 * -1 * 0 * 1, that's going to give me zero, isn't it? Because anything that's got a zero in it basically is going to give you zero. So, because it's got a zero there, I know I don't have to worry about the rest of it. It's going to be zero. Okay.

[00:20:15]Uh but yeah, just just top tip, always read through the question again before you move on. Just check you've answered it correctly because I almost moved on without actually answering that question and that would be a silly place to throw a mark away. Yeah. So reread a question before you move on. Top tip. Next up, expand and factoriize. Like putting brackets in, taking brackets out, that sort of thing. So first up, I've got factoriize x^2 + 7 x + 10. Uh this is a quadratic. Quadratics always factor into factoriize into two brackets like this.

[00:20:44]Uh now to work out the contents of the brackets I normally start with the bit on the end and I think about the factor pairs of that number. So what are the factor pairs of 10? Got one and 10. I got two and five. Not a lot of choices there. And then once I got the factor pairs I then think about well which factor pair is going to deliver me that seven in the middle. It's this pair here as it's the two and the five. 2 + 5 is seven. So that must be the pair I need.

[00:21:09]So, I'm going to get x + 2 x + 5. Does it matter which way around they go? No, it doesn't. Uh, next one. Factoriize x^2 + 9 x + 14. Going to do it the same way.

[00:21:21]I know it factorizes into two brackets.

[00:21:23]I know it's got something to do with the factor pairs of this 14 here. What are the factor pairs of 14? It's 1 and 14.

[00:21:29]Two and seven. Which one of those pairs is going to deliver me nine in the middle? It's those two, isn't it? It's the two and the seven. So I'm going to get x + 2 x + 7 as my um two brackets.

[00:21:42]Question 14. Factoriize 4x^2 minus y^ 2.

[00:21:46]Now this one is commonly known as the difference of two squares. This is a square. This is a square. And this mathematically is the difference. So it's the difference of two squares.

[00:21:55]Okay. Now difference of two squares always factorizes into two brackets. Um the identity that I think about is a^ 2 minus b^ 2 is equal to or is identical to I should say a + b a minus b. Okay.

[00:22:10]Now notice that a is the square root of a squ. B is the square root of b^ 2. So my two things in the brackets are going to be the square roots of those two terms I've got at the top there. So what's the square root of 4x^2? Well roo<unk>4 is 2. Root x^2 is x. So I'm going to get 2x and 2x at the front of my brackets and at the back of the bracket. Then what's the square root of y^ 2? It's y. So + y - y. Okay. So I'm going to get 2x + y 2 x - y. That's my solution.

[00:22:46]Question 15. A. Expand and simplify 2x + 1 x - 2. Going the other way this time removing the brackets. Uh there are different ways of doing this. First, outside, inside, last. uh two eyebrows, a nose, and a mouth. Or you can do it as a grid. Uh if you're not very good at doing this, I might suggest you do it as a grid. Um because it helps you work out what all the values need to be, and you won't miss one out. So 2x * x is 2x^ 2, x * 1 is x, - 2 * 2x is -4x, and 1 * -2 is -2. Okay? So putting it in a grid like that ensures that you won't miss one of the terms off which is the most common silly mistake that students make when multiplying out double brackets. Uh notice that you get one term which is in x^2. So then this is going to be equal to 2x^2.

[00:23:39]You're going to get two terms in x along the middle here. I've got x and - 4x.

[00:23:44]Need to add these together. So x + - 4x that's - 3x. So I'm going to get 2x^2 - 3x and then on the end I'm going to get this number down here minus 2. Okay, minus 2. So 2x^2 - 3x - 2. Part B says factoriize fully 3x^2 - 48 y^ 2. Okay.

[00:24:05]Now again if I just bring up my difference of two squares identity. We said that we need to square root the two terms, don't we? The two squares. This is a square. This is a square. This is a minus sign in between or a difference of them. So, it is a difference of two squares. It's going to factoriize into two brackets. But before I do that, I've noticed that uh three and 48 have got a common factor. So, I'm going to factoriize those out first. And it should make then finding um the the difference of two squares a little bit more simple. So if I if I factoriize the three out first, I'm going to get three lots of x^2 minus 3 into 48 go 16 times and don't know. So I'm going to get x^2 - 16 y^ 2. Okay. And then using my identity, I just need to square root those two um squares that I'm finding the difference of there. So the square root of x^2 is x and the square<unk> of 16 y^2 is 4 y. So I'm going to get two brackets x + 4 y x - 4 y like that. So that's my final answer. 3x + 4 y x - 4 y. Question 16. Factoriize fully.

[00:25:28]144 - 4x^2.

[00:25:31]Now again going to factoriize out four first. Make it a whole lot easier then to uh do the difference of square part.

[00:25:39]So 4s into 144. What's that then? um 40 80 120. So 30 and 24 / 4 is 6. So 36. So it's going to be 36 - x^2. Well, luckily 36 is a square number. So square rooting 36 is going to give me 6. Uh so 6 + x 6 - x is the solution to that one. So four lots of 6 + x 6 - x.

[00:26:13]Okay. Question 17. Write two lots of 7 x + 4 - 4 lots of x + 6 + 1 in the form a + b x + c. Okay. So here I need to kind of multiply everything out. Get rid of the brackets and then put brackets back in again. Make your mind up. Right.

[00:26:28]Let's do it then. So uh two lots of 7 x and two lots of four. That's 14x + 8. um multiplying out the second bracket. Just take care cuz that's - 4x. So -4x * x is -4x and -4 * 6 that's -4. And then I got plus one on the end. Okay. So collecting like terms, what have I got? I've got a term in four in x there and in x there.

[00:26:57]So 14x - 4x. That gives me 10 x and I've got an eight, a minus 24 and a plus one.

[00:27:05]So 8 + 1 is 29. 29 from 24 would be 15.

[00:27:10]So it's going to be - 15, isn't it? So 10 x - 15. Is that in the form a lots of bx + c? Not quite. But I can see that 10 x and 15 both have a common factor of five. So taking the five out, that's going to leave me with five lots of 2x - 3. That is in the correct form. So that is in my final answer. Five lots of 2x - 3. Question 18. Expand and simplify 2x + 1 3x -4. So multiplying out brackets again. Uh I'm going to do it as a grid method. Don't have to do it as a grid method. But if you drop a term, that's on you. Okay? If you do it in the grid method, you won't drop a term. So 3x * 2x is 6x^2. 3x * 1 is 3x. Uh 2x * - 4 that's [snorts] - 8 x and -4 * 1 that's -4. Okay, so again going to get one term in x squ there that's 6x^2 that goes at the front. Uh I've got two terms in x I've got 3x and - 8x. So adding those together gives me - 5x and then I've got that minus4 on the end -4. Okay. uh factoriize 6x^2 - 2x - 4. Again, I know it's going to factoriize into two brackets like this. It's a little bit more complicated though because I've got that 6x at the front. Okay. Now, 6x could be 6x and x. It could be 2x and 3x. Could be either of those combinations. So, when we're when we're thinking about our factor pairs and thinking about numbers, we need to be thinking about those different possible combinations. So, it could be 6x and x at the front, or it could be 3x and 2x.

[00:29:00]Okay. Uh, other than that, again, I'm just going to do it like I did before.

[00:29:04]I'm going to take the number on the end.

[00:29:06]I'm going to going to think about what the factor pairs are, what the factor pairs of four, 1 and 4, 2 and 2.

[00:29:14]Okay. So, how am I going to get minus 23 out of those numbers? Okay. Well, I need to It's going to quite a big number, isn't it? So, I'm going to have to kind of multiply the bigger number with the bigger number. So, if I Let me look at the 3x and the 2x first. If I put the two and the two here, that's never going to work, is it? I'm not going to get anywhere close. Okay. If I put the four there and the one there, then 3x * 4, that's 12x. Again, I'm not going to get close to - 23. So, it's not this way round. Let's try the other one. 6x. 6x * 4. That's close, isn't it? 6x * 4 is 24x already. Okay. So if I put a minus sign in front of it, that would be -4x.

[00:29:55]And then if I multiply 1 by x, that's going to be a positive x. So -4x + x is - 23x as required. So it's that way round. And uh it's 6x + 1 x - 4 as uh of my final answer. Okay. So if you got a lead term here, so a lead coefficient here, think about what the possible combinations are at the front. um find your factor pairs and just try them in all the brackets until you find a nice one that fits. That is my advice.

[00:30:25]Simplest way of doing it in my opinion.

[00:30:28]Okay. Well, now expanding factoriize is going to carry on being helpful for this next section. Solving quadratics. When x^2 = 16, the only value that x can be is four. Is this true or false? Tick a box. It's false, isn't it? Because when you uh square root something, you're going to get a positive and negative solution, aren't you? So square<unk> of x^2 equ= the square<unk> of 16. So x is going to be + or minus 4. Uh so there are two solutions.

[00:31:05]Okay, I think that's good enough.

[00:31:06]Question 20. Solve x^2 - x - 12 is equal to 0. So we're going to start off like we did earlier by looking to see if we can factoriize first. So, does it factoriize into two brackets like this?

[00:31:19]Let's take that number on the end, 12.

[00:31:21]Let's think about the factor pairs of 12. 1 and 12, 2 and 6, 3 and 4. Now, which combination of those is going to give me a minus one in the middle. Okay.

[00:31:31]Well, these two are only one apart. So, if I make one negative and one positive, then I'm going to end up with negative one in the middle, aren't I? Have to make the bigger one the negative one.

[00:31:43]So, I end up with a negative number overall. So x + 3 x -4 is going to be equal to zero. Now if you got two brackets multiplied together and give you zero then one of those brackets must be zero to start with because that's the only option isn't it? Two things multiply zero one of the things must have been zero to start with. So if this first bracket here was equal to zero then if x + 3 is zero then x must be minus 3 or if this bracket is equal to zero. If x - 4 is equal to 0 then x must be four. So 4 takeway 4 is zero. So my second solution is x = 4. So just take note in the factoriize form and the final answer the signs flip over don't they? So x = -3 x = 4. Question 21.

[00:32:31]Circle the two roots of 2x + 3 5x - 2 is equal to 0. Again let's look at this one here. If 2x + 3 is equal to 0.

[00:32:41]If 2x + 3 is equal to 0, then 2x must be equal to minus3. So x must be -3 / 2. So that's one of them. I need to find the other one. If I've got 5x - 2 = 0, that means 5x = 2x = 2s. Then so it's that one there. So it's those two. Uh 22.

[00:33:03]Solve the equation 5x^2 + 14x - 24. Now I have a lead coefficient of x squ here but because it's prime if it factorizes it's only one way it's going to go isn't it that's 5x and x you know five is only got two factors one and five uh so that makes things easy let's have a look at the other number minus 24 so let's think about the factor pairs of 24 that unfortunately has lots of factors 1 and 24 2 and 12 3 and 8 4 and six it's going to be one of those pairs now slight complication because one of has to be multiplied by that 5x and I need to get 14 in the middle. So, it's a matter of trial and error going through putting the numbers in and seeing which one works. Okay. Now, if I put the eight here and the three here, that doesn't work, does it? So, 58 43. That's not going to get me down to 14. I'm too far away. The other way around, uh, three lots of 15. Three lots of five is 15 and eight. That isn't going to get me to 14x either. I think it's the last possible pair. uh the four and the six must be those two. So, how can I use those to get 14? Well, if I put the six here and the four here and I made this one positive, then 6 * 5 is 30. And if I made this one negative, that doesn't work. That gets me to 26. What about the other way around? So if I put the four here and the six here. So that's going to give me five times 5 * 4 is 20 - 6 * 1 is -6. That will work. Okay. So it's that combination. Took me a time to find that one. But that's kind of how you have to do it. You have to just try different pairs and different positions until you find the pair that works.

[00:34:48]Okay. Quite tricky that one. on that one. Uh 5x - 6 x + 4. Question 23. The area of the rectangle is 8 cm squared.

[00:34:59]Work out the value of x. Okay, so area is length time width. So the area of this one is going to be the length, which is 3x + 1 * the width, which is 2x + 5. Okay, so we got a pair of double brackets. If I'm going to multiply those out. So if I've got 3x and 1 and 2 x and 5. 3x and 2x is 6 x^2. 2x and 1 is 2x.

[00:35:26]3x and 5 is 15x. 5 and 1 is 5. Okay. So I've got a term in 6x^2.

[00:35:33]It's going to be 6x^2.

[00:35:35]I've got two terms in x. I've got 15x and 2x. That makes 17x.

[00:35:43]and I've got a five.

[00:35:45]Okay, so plus five and that is equal to eight I'm told.

[00:35:51]Okay, so just taking eight off both sides, I'm going to form an equation equal to zero then, aren't I? So that's going to give me 16x^2 + 17x - 3 is equal to zero.

[00:36:09]H, this is going to be a tricky one to factoriize.

[00:36:13]So, I've got a lead coefficient of 6x uh 6x squared on the front. So, again, I've got two possible pairs. It could be it could be 6x and x or it could be 3x and 2x. Don't know which one it is just yet.

[00:36:27]I'm just going to try both. I'll cross out the one that doesn't work. Okay. So, but the number on the end is a prime.

[00:36:34]So, that makes things easier. It only can only be one and three. So, I'm just going to try one and three in different combinations in those two brackets until I can work out a way of getting 17. I think I've seen it already. If I put the three in this one here, so I'll get 6x * 3 give me 18x and I put like a minus one there, that 18x take away that one be 17x. So, it kind of works that way around. Okay. So, I won't need that one.

[00:37:01]Uh, so it's 6x -1 x + 3 is equal to zero. So then take one bracket at a time make it equal to zero. See what the value of x is. So if 6x -1 is equal to 0 then x is going to be 6x is going to be 1 x is going to be uh 16 and if x + 3 is equal to 0 then x is going to be minus3.

[00:37:25]Okay. Now that that those two numbers are definitely solutions to that equation that we had but not both of them but both of them are not solutions to the problem. And the reason for that is this is a problem steeped in shape and space and geometry. And imagine x is minus3. Then this side over here is going to be two lots of x. That's -6 + one. That means this side would be minus one in length. This one over here would be 3 * - 1 - 9 + 1. That'd be minus 8.

[00:38:01]Now, have you ever seen a rectangle which has got sidesgative8 and negative 1? Not in this dimension. So, whilst minus3 is a solution to the equation, it's not a solution to the problem. So, we need to cross that one out and say no, not a valid solution. Okay. So our only valid solution is the one that is 16. So the solution is x = 16. Okay. So if you are if you have got a quadratic and it's a problem like that, just check your answer is going to be sensible in the in the shape and space problem. Okay?

[00:38:40]Because not every solution to an equation is a solution to the problem.

[00:38:43]Next up, we're looking at indices. So you need to know all your laws of indices here, don't you? So if you got two numbers with the same base multiplied together, we add the indices.

[00:38:55]If you got two numbers with the same base that would divide in, we subtract the indices.

[00:39:01]If you've got a number and then you raise it to another power like this, an index and an index, you multiply those indices together. Okay, they're the three main rules. Plus, on top of that, you need to know what a negative index means. That a minus n is equal to 1 / a to the n. And you need to know what a fractional index is. So a to the power of 1 / n that is the nth root of a.

[00:39:24]Okay. So that is the kind of assumed knowledge for this topic. So you need to know all that. That's a bit chunky.

[00:39:32]Just try that again. Okay. Rainbow pen.

[00:39:35]Uh so yeah that's that's the info we need for this topic. Let's let's go through the questions. Question 24. Work out 3 12 / 3 7. I've got numbers with the same base. It's that second rule, isn't there? I'm going to have to then subtract the indices. So 3 12 / 3 7 that is going to be 3 12 - 7. That's going to be 3 ^ 5. Now it's asking us for an answer as a whole number here. So I'm going to have to evaluate 3 ^ 5. It's quite a lot of work, isn't it? So 3 to the 3 to the^ 2 is 9 uh to the^ 3 if you're going to cube it going to get 27.

[00:40:13]times it by another three it' be three to the four that's 81 and times that by 3 that's 243 isn't it? So that's my answer 243.

[00:40:24]Question 25. Work out the value of 3 12 / 3 5 / 3^ 2 * 3. Oo this is not very nice.

[00:40:35]H I think I might write this first bit as a a quotient. So that's 32 / 3s and then I want to divide it by 3^ 2 * 3.

[00:40:50]Let's just tidy it up a little bit. So 3 12 / 3 5 that's going to be 3 12 takeway 5 that's 7 / and then I've got 3^ 2 * 3 that's 3 cubed. Just combining those two together and then I've got 3 to 7 / 3 cubed. So same bases subtract the indices that's 3 to the 4 isn't it? Now we need the value here. So let's evaluate it. 3 to the^ 4 3 * 3 is 9 * 3 is 27 * 3 is 81. So the value of that is going to be 81.

[00:41:28]Question 26. Simplify 8 * 2 6 * 2 ^ 4 give your answer as a power of two. Uh well that first number there eight that's a power of two isn't it? 8 is 2 cubed. So I'm just going to replace that as 2 cubed to start with. And then I'm going to notice that I've got same base 2 2 and two. So I'm going to add the indices 3 6 and 4. Add those together.

[00:41:50]So 3 + 6 + 4 that sums to 13. So that's going to give me 2 ^ 13. Okay. 2 ^ 13.

[00:41:58]Question 27. Work out<unk>3 * <unk>12 * 5 -2.

[00:42:05]give your answer as a decimal.

[00:42:08]Well, this looks almost like a surge question rather than a index question.

[00:42:13]Root three is roo<unk>3. Root 12 though I can break down. Root 12 is 4 * 3, isn't it? Root 4 * 3. And the roo<unk> 4 is 2.

[00:42:23]Um 5 to theus 2 is 1 over 5^ 2, isn't it?

[00:42:30]So then I've got roo<unk>3 times well roo<unk>4 is two so that's going to be 2<unk>3 and that's divided by 5^ 2ar but roo<unk>3 * roo<unk>3 is 3. So the roo<unk>3 * 2<unk>3 is going to be roo<unk>3 * roo<unk>3 is 3 * 2 is 6. So it's going to be 6 over 25. Okay.

[00:42:52]And that's my final No, it isn't because that's not a decimal. If I want to get that into a decimal, uh, if I times top bottom by four, I'm going to get 24 over 100. And then that as a decimal is going to be 0.24.

[00:43:05]Then, isn't it? Question 28. Work out the value of 57 to the minus2. Give your answer as a mixed number. Okay. Negative index means one over the same number to a positive index. So, what I can do here is I can just flip the fraction over and say that's the same thing as 7 over 5 to the^ of two. Okay. Uh that's the easiest way of doing it. Uh because 57th is 7 fths to the minus one. Okay. And then multiply minus one with -2. That gives you two. So if you got a ne negative index of a fraction like that, you can just flip the fraction over and change to a positive index. So 7^ squar is 49.

[00:43:41]5^ 2 is 25.

[00:43:44]Mixed number. So 25s into 49 go once remainder 24 don't they? So that's equal to 1 and 24 25ths. Okay. So 1 and 24 25ths. 29. Work out the value of 2 ^ 14 / 2 9 raised to the^ of 2. Okay. So I've got an index and an index here. Going to deal with that first. So 2 9 raised^ 2.

[00:44:09]That is 2 to the 9 * 2. It's going to be 2 to the 18. So I've got 2 to the 18 which is being divided into 2 to the 14.

[00:44:17]Got two numbers with the same base.

[00:44:18]Subtract the indices. 14 subtract 18 that is 2us4.

[00:44:25]Now we need to write this as a fraction in its simplest form. So something that's got to the power of negative a number is going to be reciprocal of that number with a positive power. So I'm going to get 1 / 2 ^ 4. And 2 ^ 4 is 2 * 2 which is 4 * 2 which is 8 * 2 which is 16. So as a fraction it's going to be 1 over 16.

[00:44:50]Part B says work out the value of 25 ^ 3 over 2. So I've got a kind of a fractional index here. Uh I've got a numerator and I've got a denominator.

[00:45:02]Now you can ch treat these as two separate processes. So I'm going to do the the denominator first. Denominator of two that means the square root of 25.

[00:45:13]So what I'm going to get is the square<unk> of 25 cubed because I haven't dealt with the three part yet the numerator. So I just leave that there. I work out what the square root of 25 is. That's five. And then I can apply the cube. So 5 cubed that's 125 then isn't it? Uh similarly work out the value of 9 the -2. Now again I can treat this as two separate things. Uh the negative symbol and the square root symbol. Um, so a negative index means one over the positive index. So that's the same thing as 1 over 9 to the^ of a half. Uh, and then power of a half.

[00:45:53]That's a square root, isn't it? So that's the same thing as 1 over the square<unk> of 9. And the square<unk> of 9 is 3. So that's going to be 1/3.

[00:46:01]That's my answer. 1/3. Question 31. Work out the value of 64 to the power of 2/3.

[00:46:08]Again, I'm going to treat these as two separate operations. Going to find the cube root and then I'm going to square my answer. So, what is the cube root of 64 then squared? That's going to be well the cube root of 64 I know is four because 4 * 4 is 16 * 4 is 64. So, that's going to give me four. And then 4 squared that's 16 then, isn't it? Moving on to community frequency. Here is some information about the test marks of 120 students. Now, the first table is just a frequency table. It's drawn sideways, but it's a frequency table. And the next table is a communicate frequency table.

[00:46:43]Again, drawn sideways, but it's still a communive frequency table. Just takes up less space if you do it that way. Uh, now first frequency is always the same as the first frequency. The next communive frequency is going to be all the values up until that point, isn't it? It's going to be the the 20 and the 40 and the 20 added together. So that gives you the 48. Then the next value is going to be those three numbers added together. Well, I already know the 20 and the 28 give me 48. So add in another 40 on that's uh 88. So that's going to be 88. And then next communicate frequency is going to be those. So adding the 20 on to 88. That's 108. And then the last communicate frequency is all of them added together. That is 120.

[00:47:29]Now your last cumulative frequency the 120 should match the question up here.

[00:47:34]If it doesn't, you've made a mistake.

[00:47:35]Okay. Uh right now, once I've got my table of values, I could then plot those on the graph. Now, you always get like a bonus point uh that you can plot here.

[00:47:46]This lowest value here, we plot against zero. So, I'm going to have a point at 0 0. So, going to plot a point here at 0 0. And my next point is going to be the upper class bound versus uh the community frequency. So I'm going to do 10 versus 20. So 10 along 20 up is there. My next one upper class band against community frequency. So 30 and 88.

[00:48:16]Now this scale five boxes spans 20 units. So each box is little box is going to be four, isn't it? So 88 is going to be two little boxes above 80.

[00:48:25]It's going to be there. 108. Again, that's going to be two little boxes above 100 and 120. That's up there. That wasn't too hard to plot. Okay. Now, join it with a nice smooth typical S-shaped commul frequency curve. Okay. Single line. Make sure you skewer each of your points. Make sure you don't feather.

[00:48:51]Okay.

[00:48:54]Went a bit low on that last bit. Let's just do that bit again.

[00:49:00]Okay, topping out with your last point there. Now, it doesn't have to be Michelangelo, but it does need to be a single line non-feathered and skewering each of your points one after the other.

[00:49:09]If you miss the point, then you will miss a mark. Okay. Right. Have we got any questions on this? Yes. Uh, students who scored 15 marks or fewer take another test. Use your graph to estimate the number of students who take another test. Well, 15 marks is here. Look, that's 15 marks. So what we're going to do is we're going to use our ruler.

[00:49:27]We're going to read up to the graph and then across and then that is going to be our cumulative frequency. So these people here reading across.

[00:49:38]I do recommend you drawing these things on a graph. It's like your method. It's always good to show method. So each one of those little squares was four, wasn't it? So it's four less than 40. That's going to be 36. So, the students who scored 15 marks or fewer, how many were there? There were 36. Okay. Oh, I just noticed I've forgotten to plot a point there. I should have plotted the the point uh 20 at 48. So, oops. I should have had a point here.

[00:50:10]Okay. Uh so, me bad. When you do it, make sure you plot all the points.

[00:50:20]Right, next one up. I've got here are some information about the miles per gallon of 60 cars. Uh, so we've got miles per gallon and frequency. We're going to have to first find the commute frequencies ourselves here. So this is going to be X again here and this is going to be the cumulative frequency. So this one we're going to have less than equal to 50, less than equal to 60, less than equal to 70, less than equal to 80. Uh so you'll notice I'm using the right hand limits here. Okay.

[00:50:54]Uh commutive frequency just add them up there. Our first frequency is our first commutive frequency. Our next communicate frequency is those two added together. 6 and 16 makes 22. Then add on the 28. That's going to take me up to 50. And then add on my 10. That takes me to 60. Just double check I get the same number as I should have. Yes, I do.

[00:51:14]Okay. Now I can go on and plot the point. Now remember, I can use a bonus point down here. So I'm going to plot 40 against zero on my graph on this one. So at 40, let me see both the things together so I can plot them. So I'm going to have a zero point at 40.

[00:51:33]Okay. And then 50 goes with six. So again, I've got five divisions representing 10 units on my scale. So six is going to be three little squares up, isn't it? It's going to be there. Uh 22 for 60, that is going to be one square above 20. Uh 50, that's a nice easy one to plot. And 60, that's a nice easy one to plot. Again, nice smooth kind of distinctive S-shaped chameleia frequency curve topping out at that top number. It should kind of be flat at that top number there because that is the to not shouldn't be angled up. It should be kind of flat at your your upper value and also kind of flat at your lower value down there. That's that's what gives it its kind of S shape. Okay. Do we need to use this? Yes, we do. Uh use your graph to work out the interquartile range. Now interquartile range is interquartile range is the upper quartile Q3 minus the lower quartile Q1.

[00:52:34]Now the upper quartile lies three quarters of the way through our data set and the lower quartile is one quarter the way through. Now I've got 60 uh people on this table. So a quarter of 60 would be 15. So my lower quartile I'm going to find by reading across from 15 and down. Okay. So 15 is going to be halfway along that square, isn't it?

[00:52:58]Going through the middle of that square.

[00:53:00]So reading across at 15 to my graph and then changing directions and reading down, it's going to give me Q1.

[00:53:14]Okay, so that's Q1 there.

[00:53:17]And then Q2, uh, it's going to be 3/4 of the way through. So three lots of 15 that's 45.

[00:53:26]So reading across from 15 won't draw there. Okay there. Done it now. Then reading down again using this method to mark your axes like uh mark your graph like this is a good way of showing method. So the inter quartile range then is going to be the value we got at Q1 not 15 but the value we look up at 15. I make that 56 for that's the Q1 value. So that's going to be 56. And then my Q2 value, what's that? That's uh 68.

[00:54:20]68.

[00:54:21]So my inter quarter range is going to be 68 minus 56. I make that 12. Now again, yours may vary slightly on how you've drawn your community frequency curve. So you might not get exactly 12. Don't worry, the examiners will mark your graph, your numbers correctly if you've done it correctly. Okay? But you drawing a graph is there's always a slight wiggle to it. So you might get like a a value either side of my 12. Okay.

[00:54:46]Recurring decimals express 0.15 recurring. Sorry, I'm it's a really warm day today. I've got the door open and my neighbors are taken at the hedge with something very loud. Do excuse me, but I can't close the door cuz it's swelling.

[00:55:02]So express 0.15 with a dot on top as a fraction in its simplest form. So I'm going to let that be equal to x. x is equal to 0.1. I'm just going to draw some of these fives in so I get a feel for the number. Okay. Now if I multiply that number by 10, I'm going to get 10 x and that is equal to 1.555 and so on forever. Okay. Now if I then subtract one from the other. Okay, so 10 x - x that's 9x. Uh and if I subtract the the numbery parts, look, all these fives can cancel each other out to there. And then I've got 1.5 - 0.1 left, haven't I? So 1.5 minus 0.1, that is just 1.4. So I've got 9x is equal to 1.4. So x must be equal to 1.4 over 9.

[00:55:59]Now, we don't tend to write fractions with a decimal point in there. So, if I multiply top and bottom through by 10 just to kind of get rid of that decimal point, shift along one, that's going to give me 14 over 90. Does that simplify?

[00:56:16]Well, they're both divisible by two, so I guess that's the same thing as 7 over 45. Go any further?

[00:56:27]No, don't think so. So 7 has only got factors 1 and 7 and 7 does not go into 45. So that is my most simple answer.

[00:56:36]7 over 45. Similar question here.

[00:56:40]Convert 0.28 reoccurring to a fraction.

[00:56:43]Give your answer in it simplest form.

[00:56:44]Again going to start off with x and that's equal to 0.288.

[00:56:49]I'm going to show some of the eights and so on. And now if I multiply through by 10 I'm going to get 10 x is equal to 2.888. 8 88 88 88 88 and so on. Uh if I then subtract one from the other then you can see that all these cancel by 8 and it leaves me with a 2.8 takeway 0.2 and 2.8 takeway 0.2 is uh 2.6 and that must be equal to 9x. Dividing both sides through by 9 I'm going to get x is equal to 2.6 26 over 9, [snorts] which is the same thing as 26 over 90, which is the same thing as 13 over 45.

[00:57:32]And that is as far as I can get. 13 over 45. Moving on to SS. Now, you need to know how to combine, to simplify, and to rationalize SS. If you do all those three things, you should be good for these uh for these questions. Question 36. work out the value of roo<unk>3^ 2 * <unk>2 2. Well, when you square a square root, you just get your original number back, don't you? So roo<unk>3 2ar is just 3 and <unk>2 is just 2 and 3 * 2 is 6. That's not so hard, is it? So six.

[00:58:06]Question 37. Work out<unk> 5^ 2 +<unk> 6 2 -<unk> 7 2. That's the same question, isn't it? So that's going to be 5 + 6 - 7. Is this a higher question? Look at that. That's so easy. So 11 takeway 7 that's four.

[00:58:24]Okay. Question 38. Show that 14 over<unk> 7 can be written as a over roo<unk> b. So this is an example where we have to rationalize the denominator.

[00:58:35]But we've got a third as the denominator of our fraction. In maths we kind of consider this to be a bit fugly to be honest. um hard to imagine fractions which have got thirds as the denominators. You know, if you had a seventh, you could imagine a pizza divided into seven lovely pieces, but can you imagine a pizza divided into root seven pieces uh pieces? I would struggle. So, yeah, it's um doesn't make a lot of sense. So, we don't like them as mathematicians to have thirds as our denominators. So, we see them, we get rid of them, get rid okay. So, and we do that by multiplying top and bottom by that denominator. And that has the effect of rationalizing the denominator because roo<unk> 7 * roo<unk> 7 is 7. Uh 14<unk> 7 goes on the top. 14<unk> 7 uh so it's 14 roo<unk> 7 over 7. Now you'll notice the seven goes into the 14. So you can cancel that out. So that divides into that leaving you with two. Uh so my final answer then is going to be 2<unk>7 which is in the form a roo<unk> b as required. Okay where a is 2 b is 7.

[00:59:41]Question 39. Work out<unk> 18 - 28 28 over<unk> 50. Now we have to do two things here. First off we need to simplify that roo<unk> 18 because we can write it in a more simple way and we have to rationalize the denominator of the second part of the fraction. Now if you like you can do that separately and recombine. Um that's one way of doing it if it's difficult to manage everything.

[01:00:04]So we can work on this in two parts. If we work on the root 18 part first we can say well root 18 that is the same thing as the root of um when you're simplifying these you're looking for factors of your number that are also square numbers. So think about the square numbers 1 4 9 16 25. Can you think of any of those numbers being a factor of 18? Well 9 is isn't it? 9 * 2 is 18. And the reason why you're looking for square factors is that you can then square root your squares. Square root of 9 is three. So that's the same thing as 3 <unk>2. So you can break it into two different um parts. You can square root the nine part, get three, and it just leaves you as 3<unk>2. So we know 18 is 3 <unk>2. We simplified that part. Now let's see if we can simplify the 28 over<unk> 50. Now doing the same thing as with last question, multiplying top and bottom through by<unk> 50.

[01:00:58]So that's going to rationalize your denominator. Root 50 * roo<unk> 50 is 50. Uh and that's going to give me 28<unk> 50 on the top. Now root 50 might also simplify. I can certainly divide 28 and 5 by two. That's going to give me 14 over 25, isn't it? 14 and 25. I've divided both by two. Uh but roo<unk> 50 can also simplify as well because 50 has got a square factor of 25. So I can rewrite that as 14<unk> 25 * 2 over 25 and 25 rooted is 5. So that is going to be 14 * 5 <unk>2 over 25. Now rather than multiplying the five with the 14, I've seen it can cancel into the denominator. uh give me 1 and five. So that's the same thing as 14 <unk>2 over 5. Okay. So now I've got both of my um my um surge simplified down. So this part here roo<unk> 18 I can write as 3 <unk>2 and this part 20 over 28 over<unk> 50 I can write as 14 <unk>2 over 5. Okay. So now now I can recombine them. I can then say then um<unk> 18 which I know is 3 over uh 3 <unk>2 and I'm going to be minusing 14 <unk>2 over 5. Okay. So factoriize I could factoriize out the the root and then just deal with the the fractional part and then put it back or I could just work on the two fractions and work out what that is and write it at the front.

[01:02:53]What should we do? Let's let's let's do it properly. Let's factoriize out the roo<unk>2 and think about what is 3 - 14 over 5. Let's just do that bit. So what's 3 - 14 fths? Well 3 is 15 fths.

[01:03:08]So 15 fs - 14 fs that's 1/5 isn't it?

[01:03:14]Okay. So I'm going to get then that is equal to 1/5 of <unk>2 or <unk>2 over 5.

[01:03:23]Okay. So that's what it simplifies down to <unk>2 over 5. Question 40. Root 150 minus<unk> 6 over <unk>2 *<unk>3 simplifies to an integer. Does it? Let's find out. Well, again looking for square factors. cuz I'm seeing 150 and I'm thinking 25 is a nice square factor of 150. 25 * 6 is 150. Okay. Uh and I notice I've got a root six in the other part there as well. So that's kind of handy. So what we've got then so roo<unk> 150 is 20 25* 6. So that becomes five 5<unk>6 on the top and I've got that minus<unk> 6 there and then I've got <unk>2 *<unk>3. Well<unk>2 * roo<unk>3 is the same thing as<unk> 2 * 3. Uh 5<unk> 6 minus<unk> 6. If I've got 5<unk> 6 and I take a roo<unk> 6 away, I'm left with 4<unk> 6. So 4<unk> 6 and<unk>2 * 3 is the same thing as roo<unk> 6. So I've got 4<unk> 6 /<unk> 6. So I can cancel that with that and lo and behold it leaves us with an integer which is 4.

[01:04:44]Okay. Question 41. Work out the value of 5 over<unk>3 minus the square<unk> of 6 and 3/4. Give your answer in the form k<unk>3.

[01:04:56]Okay, I think I'm just going to do the same thing as I did earlier. I'm going to write work on those two separately first and I'm going to bring them back together again, reunite them after I've simplified them down a bit. So, let's start with 5 <unk>3. Uh, rationalizing the denominator, multiplying everything through by um<unk>3 over<unk>3.

[01:05:18]That's going to make the denominator three on the bottom and I'm going to get 5<unk>3 on the top. Okay, so that's going to be my new yellow bit. Uh, now the green bit, six root of 6 and 3/4. 6 and 3/4. A mixed number. That doesn't look right, does it? So, let's turn that into a topheavy fraction first. 6 * 4 is 24 + 3 is 27. So, that is 27 over.

[01:05:47]So, 6 and 3/4 square root. It is the same thing as uh 27 over 4. Okay.

[01:06:00]So when you're square rooting a fraction, you can square root both parts separately. That's the same thing as roo<unk> 27 over roo<unk>4. Root 4 is two. So I've got um<unk> 27 over two.

[01:06:15]Now 27 has got a square number as a factor which is 9. 27 is 9 * 3. So I can say that that's the same thing as the square<unk> of 9. 3 * <unk>3 / 2. Okay.

[01:06:31]So that's my green bit.

[01:06:34]That's about as simple as going to get that one. So I've got my yellow bit.

[01:06:38]Let's recombine them then. So starting back with what I started with. So 5 over<unk>3 minus the roo<unk> of 6 and 3/4 that's equal to the yellow bit then 5<unk>3 over 3 minus the green bit which is 3 <unk>3 over 2.

[01:06:59]Got fractions with different denominators here. If you want to subtract them you need the same denominator don't you? So got threes and twos here. So common denominator would be six. So just multiplying um top and bottom through by two for the first fraction. So time 2 * 2 is going to give me a denominator of six. And if I times three by and times three on the second one, I'm going to get six. So that's going to be equal to 10 <unk>3 over 6 - 9<unk>3 / 6. So 10<unk>3 / 6 - 9<unk>3 / 6.

[01:07:38]They've got the same denominator. So I'm put them over one denominator then. So that's 10 <unk>3 - 9<unk>3 all over 6.

[01:07:46]Well 10 roo<unk>3 - 9<unk>3. If I've got 10 roo<unk>3's and I take nine of them away, I'm just left with 1 roo<unk>3, aren't I? So that leaves me with roo<unk>3 over 6. Okay. Is that in the right form? Well, I've got it as roo<unk>3 over 6. I could just write it as 16 of um three. It doesn't say that K has to be an integer. So I can just write it as a sixth of roo<unk>3 just to get into the the form that they want it to be in. Okay, moving on to exact trigonometric values. These are the ones where you need to know the sign, cosine, and tangent of various angles. You need to know the angle of 0, of 30, of 45, of 60 and 90. And you need to know that for s, cosine, and tangent. Okay. Now, the easiest way to remember it is in a kind of a simplified version of these um these values. But if you do it this way, you might have to simplify them a little bit more after you write them down.

[01:08:51]Okay? because it's an easy way to remember but you got to do a little bit more work to find the final um search sometimes. Okay. So what I do is I just take the numbers roo<unk> 0 roo<unk> 1 roo<unk>2 roo<unk>3 roo<unk>4. Can you remember that? 1 0 1 2 3 4 and put a root sign over it. And then put all those over two over two over two over two over two. They're they're your sign values. Okay. Now your cosine values kind of give myself a bit more space.

[01:09:26]Cosine values are going to be the same values but the other way around. So rather than starting with roo<unk> 0 I'm going to start with roo<unk>4. So roo<unk>4 roo<unk>3 <unk>2 roo<unk> 1<unk> 0 over two over two over two over two over two. There my co values. And then finally tan. tan you take the sign the numerator of the sign and you divide it by the numerator of the cos. Okay. So roo<unk> 0 over<unk> 4<unk> 1 over<unk>3 <unk>2 over <unk>2<unk>3 over<unk> 1<unk> 4 over<unk> 0. Now some of these values then simplify further. Okay. Uh I'm not going to do all of them. I'm just going to pick out the ones when I need them and simplify them as I go through and I'll show you because obviously look root one is just one. Um, and I've got roo<unk>2 over roo<unk>2 somewhere that's just going to be one. So a lot of these things just simplify down but that's how I remember it. Super simple way of remembering it. Uh, and then I just kind of simplify as I pull them out the table as I need them. Okay. Right.

[01:10:36]So question 42 circle the value of cos 30. So straight from my table, I can see that the value of cos 30 is that one, isn't it? Roo<unk>3 over 2. Okay, so I've kind of scored all over my question paper here, but it's that one there.

[01:10:53]Root 3 over two. Uh, question 43. Show the value of cos 30 * cos 60 plus sin 30 is an integer. Well, cos 30 is the one I just had. That was <unk>3 over 2. I'm going to times it by the tan of 60. Tan 60 on my table says roo<unk>3 over roo<unk> 1, but I know that roo<unk> 1 is just one. So that's the same thing as roo<unk>3 over 1 or just roo<unk>3. So this one simplifies down to roo<unk>3.

[01:11:20]So that is roo<unk>3. And then plus the s of 30. So in my table I've written roo<unk> 1 / two. Well root one is just one. So that's the same thing as one over two. It's just a half. Okay? So that gives me a half. Okay? Okay, so this is what I mean about simplifying these values as you pull them out and use them. Okay, so I've got roo<unk>3 over2 * roo<unk>3 + half. Well, if I've got roo<unk>3 / 2 * roo<unk>3 over 1, that is roo<unk>3 * roo<unk>3, which is 3 over 2 * 1, which is 2 plus a half. So what's 3 over 2 + a half? Well, that's 4 over two, isn't it? Or just two. So it's an integer. It is an integer. It's two.

[01:12:00]Question 44. Show that the value of 5 sin 30 * cos 30 * 8 tan 30 is an integer. Okay, so I remember from the last question sin 30 was a half and cos 30 was <unk>3 over 2. What's the tan of 30? Let's go back up and look at the table. Tan of 30 is this one here. Look, roo<unk> 1 over<unk>3. Well, the root of one is just one. Um, so that's going to be time 8 * roo<unk> 1, which is just 1 over<unk>3 down there. Okay, so I need to show that this is an integer. Now I've got things on the numerator. I've got five, a 1, a roo<unk>3, an 8, and a 1. And I've got things on the denominator. I've got a two, a two, and a minus three. Now I can cancel anything in the denominator with anything in the numerator. That's how math works, isn't it? So I'm going to div cancel cancel that min that roo<unk>3 into that one because roo<unk>3 / roo<unk>3 is one. Uh I can cancel that two into the 8 giving me four and one and I can cancel that 2 into the 4 giving me 2 and 1. Okay. So now I've simplified all my denominators down to 1. 1 * 1 * 1 is 1. And I've simplified a lot of my numerators as well. So now I've got 5 * 1 * 1 * 2 * 1.

[01:13:18]I make that 10. 10 over 1 which is just 10. The answer is 10. Question 45. Angle X is acute. Cos X = sin 60 * tan 30.

[01:13:30]Okay. So remembering earlier sin 60 is <unk>3 / 2. Tan 30 was 1 / <unk>3.

[01:13:43]Uh so multiply those together and that's going to give me cos x. So cos x is <unk>3 / 2 * 1 over<unk>3. Now again I could see that cancels with that. That leaves me with a half. Cos x= a half.

[01:13:58]Work out the sign of angle x. Well coming back to my table. Which which angle has a co of a half? Well uh it's that one there, isn't it? It's um roo<unk> one over two. That one. Because root one is just one. One over two. So if cos 60 is a half that means that um x must be 60.

[01:14:23]x is 60 because it's the cosine of 60 that has a value of a half. Okay, x is 60.

[01:14:31]Uh work out the value of cos 30 * the s of 45 time the s of the tan of 60 all squared. Okay. So I can remember cos 30 is roo<unk>3 / 2. Um sin 45 was what was the sign of 45? That was <unk>2 / 2.

[01:14:56]So times <unk>2 / 2. And the tan of 60 that was roo<unk>3 over roo<unk> 1 which is just roo<unk>3.

[01:15:11]So times roo<unk>3 and then we're squaring that. Okay. So again I've got some things on the numerator. I got roo<unk>3. I've got roo<unk>2 and I've got roo<unk>3 and I got some things on the denominator. Two and two. So combining the numerators roo<unk>3 * roo<unk>3 is 3 * <unk>2 that's 3 <unk>2. So my numerator becomes 3 <unk>2. My denominator is 2 * 2 which is four. And I need to square all that.

[01:15:40]Okay. Now, when you're squaring things multiplied or divided together, that square gets applied to each part. So, it gets applied to each three of those parts separately. So, I'm going to be squaring three, that's going to give me 9. I'm going to be squaring <unk>2.

[01:15:55]That's going to give me two. And I'm going to be squaring four. That's going to give me 16. Okay? So, I'm going to get 9 * 2, which is 18 / 16. Uh, that's the same thing as 9 over 8.

[01:16:09]Okay? Or one and an eighth, but I'm happy with it being 9 over 8. Final section, algebraic fractions. Show that 2x + 1 over 3 + 5 x - 2 over2 simplifies to 19x - 4 / 6. Now, the thing to remember about algebraic fractions is they behave exactly the same way as fractions. Okay? So, what you do with fractions, if it was a number, that's what you do with fractions if they're algebraic. you do the same thing. So here, if I've got two fractions with the different denominators, I have to make the denominators the same. So if I if I've got three and two, the common denominator is six, isn't it? So how do I change them into six? Well, three goes into six two times. So multiplying top and bottom through by two, basically, aren't you? So 2 * 2 * 3 is six on the bottom. Two lots of 2x + 1. Um, that's going to be two lots of 2x + one like that. Okay. And then 5x - 2. Well, 2 goes into six three times. So, I'm going to multiply this by three and this by three to get six. So, on the top I'm going to get three lots of 5x - 2. Okay.

[01:17:19]Uh, so putting that over one denominator and multiplying out the brackets. So, I've got two lots of 2x and two lots of one. That's going to give me 4x + 2. And I've got three lots of 5x and three lots of minus 2. So, that's giving me 15x - 6. Okay, collecting like terms, I got a 4x and a 15x. And I've got a 2 and a - 6. So, that is 4x + 15x is 19x. And 2 and - 6 is -4. So I'm getting 19x - 4 all over 6. Oh, look. That's the same as that one. I did it. Question 48. Simplify fully 6 / a - 11 over 4 a. Okay. So what's the common denominator out of a and 4 a? Well, 4 a is four times bigger than a. So if I multiply top and bottom through by 4, I'll get 4 a on both of them. And then I can just do the subtraction then, can't I? Okay. So that's the same thing as 4 * 6. That's 24 over 4 a - 11 over 4 a. And now I've got the same denominator. I can just simply subtract the numerators. 24 take away 11. That's 13. So I'm just going to get 13 over 4 a. That's my final answer.

[01:18:38]13 over 4 a. Question 49. Simplify 4x^2 - 1 over 4x^2 + 12 x + 5. Now, with these ones, you tend to look to factoriize top and bottom and then see if you can cancel any of the factors out and simplify it that way. On the top, I've got 4x^2 minus one. Now, that is a difference of two squares. Again, I have got a square and a square. One is a square. One square is one. And I'm subtracting them from each other. Okay.

[01:19:05]So, the top half is a difference of two squares. So using our um our identity a so using our identity a^ 2 minus b ^ 2 is ident= to a + b a minus b. All we need to do then really is is um square root those two squares and write them in that form instead. So that is going to be equal to square<unk> of 4x^2 is 2x square<unk> of 1 is 1. So I'm going to get 2x + 1 2x -1 on the top. On the bottom I've got 4x^2 + 12x + 5. Now if it factorizes it's going to factoriize into two brackets but annoyingly it's got a lead coefficient of four. So it could be it could be um 4x and x or it could be 2x and 2x. Now because of the nature of the question I'm suspecting it's the second one. So, uh, we can cancel some of the 2x plus somethings with the 2x plus somethings underneath.

[01:20:08]So, I'm I'm beginning to think it's probably not that one. It's going to be the one at the bottom one. Uh, the number ends in five. Uh, only one factor pair of five. That's 1 and 5. That makes things easier. So, if I did have a one and a five here, does it work? So, 2x * 5 is 10. 1 * 2x is 2. 10 and 2 is 12.

[01:20:26]So, that does work, doesn't it? So, it does factoriize like that. So, I'm going to get um 2x + 1 and 2x + 5.

[01:20:36]Now, notice, look, I've got a 2x+ one on the top and a 2x + one on the bottom.

[01:20:40]And when you divide things at the same, they just cancel each other out. Uh, only works if they're factors. Here, they're factors. Uh, and that leaves me with 2x -1 over 2x + 5. Okay, that's as simple as this one's going to get. We're done. Question 50. Simplify fully. y^2 - 3 y * y^2 + 10 y + 21 over y^2 - 9. So again, I've got on that fraction on the right hand side, I've got a quadratic on the top and I've got a difference of two squares on the bottom. So I'm just going to leave the first part as is at the moment. No, I'm not. I'm going to factoriize out a y. So I think I'm going to factoriize a y out of the first part.

[01:21:26]make it y * y - 3. And then the second bit of the fraction, I'm going to factoriize into two brackets on the top by factorizing the quadratic. And I'm going to factoriize the bottom as well using a difference of two squares.

[01:21:41]Difference of two squares. Easy. We do that first. Y + 3 yus 3. Super simple.

[01:21:45]Uh then the other one, I've got 21 on the end. Factor pairs of 21. Uh 1 and 21. 3 and seven. I'm suspecting three and seven because it adds up to 10. So that is y + 3 y + 7. Okay. Oh, look.

[01:22:03]Brackets that cancel.

[01:22:05]Those two go. Oh, but also brackets that cancel.

[01:22:09]There you go. And that leaves you with y and y + 7 in the numerator. So that is all equal to y + 7. Now, I think that is as simple as it needs to be. Um, you could multiply out and get y ^2 + 7 y. I'm not sure that's any simpler than you had already, though. Um, different versions of the same thing. I'd be quite happy to leave it like that. Question 51. Again, I've got a difference of two squares on the top. I've got a quadratic underneath.

[01:22:40]And I've got uh two factors top and bottom in the second fraction. This time, not look though. I've got a divide. Uh, but I can keep change flip on this, isn't it? KCF. Uh, keep the first fraction the same. Change the divide to a times and flip the second one over. Okay. So, the first fraction, difference of two squares on the top.

[01:22:59]That's going to factoriize into two brackets. I've got a quadratic underneath that hopefully factorizes into two brackets. I'm going to change the divide into a times and I'm going to flip those two factors and the last fraction over like that. Okay. And then again, hopefully I've got some bits that cancel a bit further down the line.

[01:23:21]Right, so I've got 9x^2 - 1. Okay, so the square<unk> of 9x^2 is 3x. So I'm going to get 3x and 3x. Uh square root of 1 is 1. So + one minus one. That was easy. Uh 3x^2 + 2x - 1. If it factoriize, it's going to have 3x and x there, isn't it? Okay, got one at the end. So it's going to have to be one and one. How do the signs go? If I make this one positive, that's going to give me plus 3x and minus x will give me the plus 2x in the middle. Okay, that all seems to work. Now, what we got that matches that we can cancel? Well, I've got uh a 3x -1 on the top there and a 3x -1 on the bottom there. I've got a 3x + one on the top here and a 3x + one on the bottom here. And that just leaves me with x - 2 x + one on the top and bottom. So x -2 remains over x + one that remains also. And that's my finished final form. You can find more exam question compilations over here. For more past paper walkthroughs, click down here. If you want to visit my Amazon shop with my recommendations for calculators, revision guides, and other math related stuff, click down here. Good luck in your revision and in your exams [music] and see you again next time.