Transcript for:
Revision Guide for GCSE Maths Higher Tier

Okay, welcome to this video where we're going to be having a look at some essential revision for GCSE maths paper one. Now, in this video, we are going to be focusing exclusively on the higher only content. However, if you are sitting the higher tier paper, there is another video worth watching which will be linked in the video under the foundation section and that's because that video focuses on a lot of the crossover content. So that being the content that appears in higher or foundation. Now in this video we'll be covering a lot of the higher tier content. However, just to make sure you know how these videos work. This is the previous video looking at the crossover content. You can see things like nonlinear graphs are included in this video. Now with all these videos you can click on the description at the bottom. From there, you can download the booklet, which is the top link in the video, and you'll have all of the paper videos that come out after this one. As well as that, you can download things like the revision guide and the checklist. But if you scroll down, you can also find video lessons to all of the topics mentioned in the video. If you hover over the chapters feature in the video and click onto the chapters, you can also go through here and it's labeled with all the question numbers that correspond to the booklet. So, don't forget to download the booklet for this so that you can follow along with the questions and you can do them on paper as well. Now, hopefully you've got your revision under control and you are making great progress towards paper one. But if you're struggling to know where to start and need to maximize your maths in minimum time, then you should sign up for my rapid revision program upgrade as you can use it to build your very own personalized revision plan. Which means all you need to do is complete my revision quiz and you'll know exactly which lessons or topics you need to work on to maximize your maths in minimum time. This is my paid program, but you can sign up for a free trial on my website now. And if you feel like you need top tier support direction from me, then sign up for my live weekly interactive lessons which include revision specific lessons focusing on the topics that I expect to appear in exam papers one, two, and three. [Music] Okay, so looking at our first question on reverse percentages says James buys a television 20% VAT is added to the price of the television. He then has to pay a total of £600. What is the price of the television with no VAT added? So here we have been given a price after 20% has been added on. Now if 20% has been added on to the original price and we think of the original price as 100% then we've been given 120% of the original price and that is equal to £600. Now similarly if we had been given a question that had been reduced by 20% instead of going up to 120 if it had been reduced we would go down to 80%. You can have one or the other with these questions but ultimately we want to get back to 100%. Now as we are doing this without a calculator typically I'll always try and go down to 10%. Simply because the jump back to 100 is really nice and easy to just times by 10. So, we just need to think to ourselves, how do we get from 120% down to 10%. Well, in this scenario, that would be dividing by 12. So, all we're going to do is do the same on the right hand side here, and we'll get to our answer. Now, without a calculator, we need to do 600 / 12. So, we may want to do some bus stop to the side. 600 / 12. 12 goes into 60 five times. And then we have our zero. So 10% is equal to £50. When we times that by 10, we get our final answer, which will be £500. So you just need to be really careful in these questions when it is asking you essentially to go back into the past and find out what a price was before a certain amount was added on or taken away. In this case, of course, it was added on. Now I've got some fraction calculations. So we've got some mixed numbers here. We have a subtraction at the top. So before we can do a subtraction with mixed numbers, we need to make them improper fractions. To do that, big times the bottom add the top. So 4 * 5 is 20, add the 1 is 21. So we have 21 over 5 take away 2 * 3 is 6 + the 2 is 8. So take away 8/3. In these questions, we need a common denominator. So we'll times the left fraction by three, top and bottom and the right fraction by five again top and bottom. That's going to make two new fractions. We'll get 63 over 15. So 63 over 15 take away 40 also over 15. We can then just subtract the numerators which will leave us with 23 over 15. And then reading the question, it says give your answer as a mixed number. So 15 goes into 23 once and there is a remainder of eight. So 8 over 15 left over. Now you could have made a different common denominator. I went for the lowest common multiple. But if you did and you got a different denominator here, just looking at the language, it didn't say it had to be given in its simplest form. So you might have got a slightly different answer here. Maybe you did a denominator over 30, but I went for the lowest common multiple. Pretty much the same process. When we're multiplying, we still have to make them improper. We're still going to have to make it into a mixed number at the end. But when we're multiplying, it's the nicest of the bunch because we just multiply the tops and multiply the bottoms. So, make them improper first. 3 * 2 is 6 + 1 is 7. So, 7 over two multiplied by 1 * 5 is five. Add the three 8 over 5. Multiplying these together, we get 7 * 8 is 56. 2 * 5 is 10. And then we want to convert this back into a mixed number. 10 goes into 50 five times and there's a remainder of six. So 6 over 10 left. Now it does say in this question to give it in its simplest form. So we do need to simplify this little fraction. It does divide top and bottom by two. And when we do that, we get five. and 6 / 2 is 3. 10 / 2 is 5. So 5 and 35ths would be our final answer. Onto some recurring decimals. It says here, prove algebraically that 0.73 with a recurring dot above the three can be written as 11 over 15. Now there are different methods to this. When there's one recurring decimal, I take the approach of multiplying by 10. So if x is equal to 0.73 I go with 10 x which will be equal to 7.3 and I balance it out with the same amount. So 33 from here I can subtract these straight away and I get 9x will be equal to I've got to be a little bit careful because I have got to do 7.3 take away 0.7. The recurring threes are going to disappear. So, it's only really that part there that I need to worry about. Now, it's a little strange as well because the bigger number is on the bottom here. So, if I just write it to the side, 7.3 take away 0.7 and I just need to do some borrowing. So, borrowing from the seven makes that 13. 13 take away 7 is 6 and 6 take away 0 is also six. So, there we go. We get 9x is equal to 6.6. From there, you can either turn it straight into a fraction, which tends to be what I do, turn it into 6.6 over 9, although fortunately, we do get this situation where there is a decimal within the fraction, which is obviously not allowed. So, we need to turn this into a fraction where it's just whole numbers. And that's easy for us to do. We just times the top and bottom by 10. And that gives us 66 over 90. And then I just need to simplify it and hope that I get 11 over 15. Now I can use a little bit of a approach here which is just going to save me a bit of time. If I know that the numerator has to get to 11, I just need to show what I need to divide by to get there. So 66 / 6 would get me to 11. I just need to check that this denominator is also going to go down to 15 when I divide it by six. So if I just show that working out to the side, 90 divided by 6, 6 goes into 9 once, remainder 3, and then 5 times into 30. So I do get that 11 over 15 as my final answer. I've also shown it very clearly. It does say prove in this question. So you do need to show you're working out really clearly. Now you might have learned a slightly different method as well. You might have learned tsing this by 100 as well when there are when there's only one recurring decimal at the end and different methods but that's absolutely fine. If you have a different method please do apply it but that's the method that I tend to tend to use when I'm using these. But there we go. There is our recurring decimal question. So 84 is a product of prime factors. So with we are going to draw a prime factor tree for 84 that would be 2 and 42 and circling the number if it's prime. 42 will be 2 and 21. 2's prime as well. And then of course 3 * 7 for 21. So they are the prime numbers that multiply together to make 84. So my final answer would be 2 * 2 * 3 * 7. And if there is a repeated number, we can write that to a power if you choose. That's 2 ^ 2 * 3 * 7. You can write those numbers in any order. And it doesn't necessarily matter to write the power, unless of course it says to write your answer in index form. But there we go. Either one of those answers would be fine. The next part asks us to find the lowest common multiple of 60 and 84. Now, there is a method of course where you can do this using a ven diagram as well, but for 60 and 84, I'm going to hope that it goes up or it doesn't take too long to get to my final answer. Because 60 ends in a zero, it gives me a bit of a hint that I know I'm either going to be having to times 84 by 10 or by five. Otherwise, it's not going to end in a zero. So, instead of going up in the 84 times table, I'm going to first times it by five and then I'm also going to times it by 10. So, I'm just using a little bit of a sort of a easier way to get some numbers here. Of course, you can just count up in 84s, absolutely fine. But for this one here, I'm going to try and apply this method and hope that it works. Now, 84 * 10 is 840. So, tsing by five would be half of that, 420. Now, as long as 60 becomes one of these numbers, obviously, I won't have to go any further. If I do, I'd want to keep going up in these sort of multiples of five in terms of their times tables. So, times by 15, time 20. Otherwise, they're not going to end in a zero, and 60 is never going to match. So 60, you might know it goes into 420, but if you don't, it's not going to take us too long to get there. And 60 is a relatively nice one to go up in 360. 420. And there we go. We've got our match. 420 would be our lowest common multiple of 60 and 84. So you can apply some of these methods which just make it a little bit nicer to get to that answer rather than writing out all of the 84 times table. So for negative and fractional indices, we need to know what rules are applied to those different elements within the power. A negative in a power does a reciprocal. If there is a fraction involved, the denominator will do a root. And if there's a fraction involved, the numerator is just the normal power. what would be there if there was no fractions and no negatives? That number would just be a normal power. So for 81, I first want to use that reciprocal element. It's the easiest one to do. So I just have to write 1 over 81. And that's the reciprocal done. It's just that I also need to do this root. So in the case of a two, that is a square root. So we're just going to write the square root over both these. You can put a two with that root if you want. The square root of 1 is 1. The square root of 81 is 9. And there we go. There's our final answer. 1 over 9. Now, the next one doesn't have a negative involved. So, we don't need to do a reciprocal. But in this particular question, we have a three as our root, which just means I need to do the cube root of these numbers. So for starters I will do the cube root of 64 which is four and the cube root of 125 which is five. I then need to have a look at this number on the numerator. And notice that I did the root first because it's easier to do the roots when the numbers smaller. If we squared both of them, we might not even know what the cube root of those numbers are. So for squared, this one here, I'm just going to put that squared symbol with both of my numbers and then square them both. So 4 squared is 16 and 5 squared is 25. And there we go. That's using negative and fractional indices. Now with thirds, there's lots of different types of questions that you could have with SS. This one here is expanding a single bracket. So for this one, we have root five on the outside. We then have roo<unk>8 plus roo<unk>8. It says here show that it can be written in the form a roo<unk> 10 where a is an integer which of course just means a whole number. We want to find the value of a. Now not forgetting when you multiply thirds it's just like times tables. So 5 * 8 is 40. So we have roo<unk> 40. For the next multiplication when we expand this bracket 5 * 18 is 90. So we have<unk> 90. And these are being added together. Now the only way that they're going to combine to make a single third like the a roo<unk> 10 is if the number under the root when we simplify them is the same. And it has given us a massive hint that that number is going to be 10. So for root 40 that could be written as the square<unk> of 4* the<unk> of 10. For roo<unk> 90 that's the<unk> of 9 * the<unk> of 10. And again now we can simplify these. So the square<unk> of 4 is 2. So that is 2<unk> 10 plus<unk> of 9 is 3. So + 3<unk> 10. And adding those together we get 5<unk> 10. So the value of a is the number in front of the root 10. So our answer is five. And there we go. That is some multiplication with ss. For standard form conversions we have large and small numbers. remembering that number has to be between 1 and 10. So we want our decimal here between the three and the two. So we'll have 3.246 up to the last whole number time 10 the power of and we'll count the jumps 1 2 3 4 5 6 7. So 10 ^ of 7. When we have the small number below, we want our decimal still to make this between 1 and 10. So 5.62. But when we write our power here 1 2 3 this time as it's a 0 point number it's going to be a negative power but with three jumps still and that's it just indicates there that the decimal essentially has to move the other way when we turn this back into an ordinary number. There we go. That's converting into standard form. We also have to convert from standard form into ordinary numbers. When we do this, I like to write out all the digits within that number first. Put the decimal just above so that when I do my jumps, I can hop over the top. This is tsing by 10 the^ of three. So, tsing by 10 once, twice, three times. The decimal go at the end here. We don't need to put 0. We can just write,452. For the next one, we have a negative power. So, we are going to get a npoint number. But I'll write the digits. Put the decimal just above. And this time I need to go one, two, three jumps to the left. My decimal will land here. And underneath each loop, tidy that up with a zero. And just stick that zero at the start. Rewrite the number nice and neat. So 0.00496. And there we go. That's converting into ordinary numbers. Now when we have to do calculations we can have additions, subtractions, multiplications and divisions but we can apply little tricks to them. Most of them with addition unfortunately there are no little tricks. We just have to convert these to ordinary numbers. Add them together and convert it back to standard form. So the first one is four and a two. We got to times it by 10 1 2 three times. So that's 4,200. Then the next one we have a five and a three. And we've got to times that by 10 twice. So 1 2 and putting our zero in. It's asking us to add these together. Now I already have them aligned in the right place value. But if you didn't, you just rewrite those again and line them up obviously from that final zero. We can now add these together. 0 3 7 and four. And we can convert this back into standard form. So to convert it back 1 2 3 jumps. So there we go. That would be 4.73 * 10 ^ of 3 as it is a large number. There we go. That's our final answer for the first one. For the next one, we have quite a tricky question as it's wanting us to do a division. Now when multiplying and dividing all we actually have to do is look at these starting numbers. So 2.52 and in this case divide it by four. We have to do that without a calculator. So I may want to do some bus stop division. Let's have 2.52. Four on the outside. Four doesn't go into two. So carrying the two it goes six times into 25 remainder 1. And then three times into 12. So when we do this division we get 0.6. 63 * 10 and when we are dividing we are going to subtract powers. So if we look at the powers here we have a five and a -3 and if we do 5 take away -3 not forgetting when we take away a negative that is adding so we do get a power of 8. So it's 0.63 * 10 to the power of 8. It may look like it's in standard form, but of course it's not between 1 and 10. The decimal does have to move here. So that would make it 6.3. And when we make the number bigger, we make the power one smaller. So 10 ^ of 7. That's our final answer. But just bear in mind as well, if we were multiplying, it's quite common that you do get a number that is bigger than our little threshold there between 1 and 10. So if you have to make the number smaller, you have to make the power bigger. So just watch out. There are two different scenarios there. Just thinking about essentially laws of indices and what we do with powers when multiplying and dividing. But there we go. There's some standard form calculations. When expanding brackets, of course, there's lots of different types of brackets, but we have two here where we're expanding two sets of single brackets, and we also have a set of three brackets. So when we're expanding single brackets, we just need to multiply what's in the bracket by the number on the outside. So here we get 5 p plus 5 * 3 is 15. Now the next bit you can approach in two different ways. You could times what's in the bracket by two and then remember you have to take them both away. But when there's a negative in the middle, I prefer just to times by -2. So -2 * 1 is -2 and -2 * -2 p we have a negative * a negative there which is going to make positive so we get + 4 p. From here we just need to do the simplified part. So adding together what we have we have a 5 p and + 4 p. Now in total that is 9 p. As well as that we have + 15 minus 2. 15 takeway 2 is 13. So it's 9 p + 13. And there we go. That's our final answer for that one. For the three brackets, we need to pick two of them to expand first. You can't multiply three things at the same time, particularly not things with algebra. So pick two. It doesn't matter which two. I do prefer to pick the one that looks the hardest. Now, you might argue the two on the right look the hardest because they have the coefficients of x bigger than one, but actually I think this one with a negative for me is the hardest. So, I'm going to pick these two to start with. So, expanding our double bracket, we have x * 2x and x * 3. That gives us 2x^ 2 + 3x onto the bottom. - 3 * 2x is - 6x and -3 * posit3 which is - 9. Now before we try multiplying that by the next bracket, we just want to tidy this up. So we get 2x^ 2 + 3x takeway 6x is - 3x and then - 9 at the end. I can then put this back into a bracket and we're going to multiply it by the last bracket that we had in the question. So just bringing that one down. 4x + 5. Now this is just my personal preference and you don't have to do this but I do prefer to write that final bracket actually at the start in front of the previous bracket. So I prefer to put that here. Now again that is just personal preference. You don't have to do this. I just think it looks a bit nicer when you are now expanding. So, I've got to do 4x * all of these three pieces. Just take your time with it. 4x * 2x^ 2 is 8 x cubed. 4x * -3x is -12 x^2. 4x * -9 is -36 x. Then onto the 5. 5 * 2x^ 2 is + 10 x^ 2 5 * -3x is -5x and 5 * -9 is - 45. Now, if you do have that final bracket at the start, it does mean that your x squares and x pieces are sort of separated, which doesn't happen if you have it the other way, but just being really careful now with the symbols. We're going to join up these x squ pieces and we're going to join up these x pieces and we're going to write our final simplified answer. So looking at what we have, we have 8 x cubed. The -2x^2 and the positive 10 x^2 are going to join up. In total, we have - 2x^2. We have - 36x and a -5x. In total, that is - 51 x. And then we have the - 45 at the end. And that is now simplified. You can see it's simplified because all of the different powers of x now there are no like terms. We just have them all sort of collected together and simplified. So there we go. That was expanding two single brackets and also expanding a triple bracket or three sets of brackets. Okay. So on to some factorizing and simplifying. Now this first one is an algebraic fraction. And what we want to do on this one is cancel off any common factors. And what we have, if we were to write this out in full, we have 8 x - 4. And on the bottom, we actually have a double bracket x - 4 and another x - 4. Now, in this scenario, all we actually need to do is get rid of this common factor of x - 4 on the top and the bottom. And if we write what we have left, we have 8 on the top and on the bottom we have x - 4, which you can put into a bracket if you want, but as it's the only thing there, it doesn't necessarily need to be in a bracket anymore. And there we go. That'll be our final answer. The next one is a little bit more complicated because it's asking us to factoriize this fully. Now, in the case of this one, it is a difference of two squares, but it's also been multiplied by something. Now, if you have a look, if you look at these two pieces, we've got a 50 and a 2 y^ 2. They both divide by two. So, you can factoriize 2 out, which would leave you with 25 - y^ 2. Now, that there is not the common way that you see a difference of two squares, but 25 - y^ 2 is actually a difference of two squared. Now, sometimes you see it more regularly like this. So maybe x^2 - 100 and that factorizes to x + 10 and x - 10. Now this is the same thing but the numbers first and the letter is second. So just this part here factorizes to 5 + y and 5 - y. But we also have the two and the two would go at the start as well. So we've had to factoriize out the two and then factoriize again what's in the brackets. But there we go. That is a an interesting difference of two squares question. When we are doing calculations with algebraic fractions, we obviously have adding and subtracting and then also multiplying and dividing. When we are adding and subtracting, we need a common denominator just like we would do when we are looking at normal fractions. So here our left denominator is x + 2 which means I need to multiply this side by x + 2. And I tend to write that in brackets just to remember that I'm actually going to have to form some brackets when I do that. The right fraction I'm going to times by the other denominator. So multiplying by x - 4. And again the same to the top. So just writing that in just to remind myself I need to multiply everything by the what I'm multiplying the denominator by. So the numerator on my left fraction is x brackets x - 4 and that's going to be over now a double bracket. So x + 2 and x - 4 and the right fraction I've got to do 2x multiplied by x + 2. And again that's going to be over a double bracket x + 2 and x - 4. So from there I want to expand these numerators and that will be x^2 - 4x and on the right I'm going to have 2x^2 + 4x and we're adding all of this together. Now at this point you can keep writing two denominators if you want but essentially we're just going to have to add all the numerators together. So here I'm just going to turn it into one big fraction which is x + 2 and x - 4 on the bottom. So I need to collect like terms now on the top to simplify this. We have an x^2 and a 2x^2. So in total we have 3x^2 and we have a - 4x and a + 4x. Now they just cancel each other out. So they're gone. So I am only left with 3x^2 on the bottom. I don't need to expand these brackets. I can leave them factorized. I can leave it as x + 2 and x - 4. And there we go. That would be my final answer. I would only want to look at this point to see if there was a common factor on the top and the bottom, but the top you cannot factoriize into a bracket. So, we're not going to be able to form any common factors. That would just be my final answer there as 3x^2 over x + 2 x - 4. Now we have a more complicated question here where we are multiplying algebraic fractions. This one here we have a lot of factorizing to do before we can even start multiplying. So if we start factorizing anything we can we have a difference of two squares on the left fraction. That one there would be 2x + 3 and 2x - 3. If you're not recognizing difference of two squares it might just be a nice topic to revise. But this is the kind of topic here where you need to just be able to recognize those difference of two squares on the bottom that divides by three. So we could have three lots of and we would have 2x + 3 inside the bracket. We are tsing that by 2x on the top and the denominator there we can only take x out of the bracket. So x out of the bracket would leave us with x - 3 inside the bracket. We now want to look and see if we can cancel off any common factors. Now, one that's really nice and easy to spot is this 2x + 3 on the top and the bottom of our left fraction. So, they can go on the right fraction. You can see that there's an x on the bottom and 2x on the top. So, we can actually divide this fraction top and bottom by x, which gets rid of the x here and gets rid of the x here. Now, if we times these together, now we've got rid of some things. It should look a little simpler. So we have 2 * the left numerator. We'll write that as brackets. 2 lots of 2x - 3 and 3 * that bottom right bracket. So 3 lots of x - 3. It does say in this question to write it in the form ax + b over cx plus d. It doesn't want it to be left factorized. So we need to multiply these brackets out. So 2 * 2x would give us 4 x and 2 * 3 gives us 6. So 4x - 6. And on the bottom 3 * x is 3x and 3 * 3 is 9. And then again with that negative sign it says a, b, c, and d are integers. One thing that might throw people a little bit here is the fact that they put these plus symbols in. And of course we have negative symbols in ours, but that's absolutely fine. just means that B is -6 and D is -9. That's absolutely fine for that to happen. But there we go. That was multiplication. The only thing you need to watch out for is if it's a division to make sure you do the reciprocal of that second fraction and then multiply them. But otherwise, exactly the same process. Okay, so for this question, we have some forming and solving equations. Now, it says here the diagram shows a right angle triangle. All measurements are in centimeters and the area is 27.5. Work out the length of the shortest side of the triangle. Now we can kind of visually see the shortest side of the triangle. We have this one over here which is x -2. Now whatever x is x takeway 2 is going to be smaller than x + 4. And the other side that's left is the hypotenuse. So that there are the two lengths that you would use when working out the area. you would do base time height and divide it by two. So in the case of these, we have a double bracket that we can do base time height. We're not going to actually be able to divide it by two, but we can write that we're dividing it by two. And that's going to give us this area here of 27.5. And that's where we're going to start to create our equation. So if I start by multiplying these brackets, we have x + 4 and x - 2. And if we multiply those out, we get x^2 - 2x + 4x and then - 8. And if we simplify these middle two here, we get x^2 + 2x - 8. So that's our expression. Not yet for the area because we need to divide it by two. So, if we want to write our expression for the area, we'll have x^2 + 2x - 8 all / 2. Or you could write times by a half and that's going to equal 27.5. So, from here, we need to get rid of this divide by two. So, to do that, we would multiply both sides of this equation by 2. And that would leave us with x^ 2 + 2x - 8 is equal to 7.5 * 2 is 15. So doubling that is not too bad. You could add two of them together, but 40 add 15 is 55. So we have a quadratic equation that we need to solve. In order to solve a quadratic equation, it has to equal 0. So in order for this to equal zero, so we can factoriize and solve it, we have to take away 55 from both sides. That leaves us with the quadratic x^2 + 2x - 63 is now equal to zero. So in order to solve this, we need to find the two factor or the factor pair that multiplies to make 63 and adds to make two. So if we write down some numbers that times to make 63 we have 7 * 9 which is the first that comes into my head and actually they have a distance of two between them. So when we factoriize this it will be 7 and 9. So we've got that straight away. If you don't get it straight away obviously write down more factor pairs. So to make positive2 we would want + 9 and minus 7. When solving quadratics hopefully you know we just flip the symbol. So x will equal -9 from this first bracket. The number that can be substituted into that bracket to make it equal 0.9 add 9 is 0. And for the next one, x is equal to pos 7. Again, the number that goes into that bracket to make it equal 0. 7 takeway 7 is zero. So they are our two solutions. However, this has context in this question. It's about the length of a triangle. Now you can't have -9 because if you look at the shortest side there which is x - 2 well -9 takeway 2 would make1. So we can't have this solution. It has to be x is equal to 7. If we substitute it into the side lengths the top one here would be 7 + 4 which equals 11. And this length here would be 7 - 2 which equals 5. So, our answer for the shortest side of the triangle, and it does say it's all in centimeters. Our answer is 5 cm. We've not really got much space anywhere here, but I'm going to put 5 cm over here. That is our final answer. Making sure that that's nice and clear. There we go. So, 5 cm as our final answer. When looking at inequalities, we could have to solve them or we could have to draw them on number lines. This one here, we have an inequality with unknowns on both sides. quite an interesting one because we have just the 14N on the left of that inequality. When there's unknowns on both sides, identify the unknown which has the smaller coefficient. In this case, the 11N. So, we're going to minus 11N from both sides of our inequality. And that leaves us with 3 n is greater than 6. 14 just leaves us with the 3 n here. This does divide nicely. We're going to divide by 3. So we get n is greater than two. And again, we're just dividing both sides by three. Not forgetting when you do that division, if it doesn't divide perfectly. For example, if it was a seven on the right, and it was 7 dividing by three that we had to do, we just leave it as a fraction 7 over 3. But this one did divide, we get a nice whole number. So n is greater than two. For the next one, we have to draw this set of values on this number line. But we have this plus three in the middle. So before we put our circles onto the number line, we need to subtract three from both of these numbers to get rid of this plus three. When we do that, our left number will be -5 and our right number will be positive one. So now we can actually draw this. We can put the circles above those two numbers five and one. We can join those up and then having a look at the inequality to see if any of them have the additional line on. This one does. We have the less than or equal to there. So we will color this circle in as it has the extra line on the inequality. We have the extra coloring in in the circle. Okay. So looking at some algebraic proof now. It says here prove that the square of an odd number is always one more than than a multiple of four. So with algebraic proof, we have two forms that we're going to use. We have 2 n. You don't have to use the letter n, but I tend to. 2 n is an even number. And to make that number odd, we would have 2 n + one. You can use 2 n minus one as well, but 2 n + one is much easier because we keep everything positive. So 2 n + one. So it says, prove that the square of an odd number is always one more than a multiple of four. Well, our odd number is 2 n + 1. So an odd number squared would be 2 n + 1 in brackets squared. Now that is a double bracket. So if we expand this double bracket, hopefully we will get to something which we can make look like one more than a multiple of four, but we'll wait and see when we get there. The first multiplication gets us 4 n^ 2. We then get plus 2 n underneath plus another 2 n and then plus one. This is already starting to look quite nice because we do have a four involved. Something's getting times by four and we do have a plus one at the end. So it is starting to look like it's going to work. If we combine these, we have 4 n ^ 2 + 4n and then plus one. Now, in terms of writing or showing or proving that something's a multiple of four, it means we need to factoriize four out. However, we want that plus one to stay at the end because we want it to be one more than a multiple of four. So, all we're going to do is look at these pieces here that can be divided by four. And we'll just factoriize that bit. So, four multiplied by n^ 2 + n. And then we have the + 1. Now, that there shows that whatever number we choose for n, you put it into that bracket. Whatever we get out, we're going to multiply it by four, which gives us a multiple of four. And then we're going to add one right at the end. So, it's always going to be a multiple of four, but with one added on right in the last step. And there we go. There's our proof for that question. You can always write some words along with it just to say it's always multiplied by four and then one is added on at the end if you want, but you don't need to. Okay. So looking at some graphical simultaneous equations where we have the equation of a circle says here the diagram shows the graph of x^2 + y^2 is 30.25. So for 30.25 which is the radius squared you can kind of see the square root of 30.25 where the circle is crossing through the axis. So you can kind of see that number there. I can't say it for for sure but that's where the number is going through. It says use the graph to find estimates for the solutions of the simultaneous equations. Then we have our circle equation and we have a linear equation y - 2x = 1. Now we want that in the form y = mx + c. So once it's in that form y = mx + c, it's much easier for us to draw. So if we add 2x to the other side, just rearranging it slightly, we get y is equal to 2x + 1. Now from there we can actually draw this straight line. You can either draw a table and you can do some values and you can sub it in. So x and y. We're going to want to choose a few values here. You only have to necessarily do all of them but -1 0 1 2 and three. Just having a look at this region on the x- axis here. We're going to need quite a few more than that. But this is one way that you could approach it. So tsing the x value by two and then adding one. Just following this equation down here. Starting with the positive numbers. 3 * 2 is 6 + 1 is 7. 2 * 2 is 4 + 1 is 5. And you can now spot the pattern. It's going down in twos. So be 3 1 - 1. You can plot these points on 0 and one just here. One and three just here. Two and five just there. three and seven just here. And actually, it's probably good that I only selected those ones because it would now go off the graph had I have gone any further. The other one I had was minus one and minus one. Now, I could just get a ruler now and extend the line, but I want it to be as accurate as possible. So, what I'm going to look at is seeing if I can spot the pattern between the points. And if you look, it goes across one square, up two squares. Now, if I follow that pattern going down, down two across one is here. Down two across one here, down two across one here. I then want to join this up using a ruler and making a nice straight line. Now, once I've drawn that, it wants me to use the graph to find estimates for the solutions of the simultaneous equations. So, I actually want to find the x and y values at the point where they cross over. That's quite hard to read. So, we are only finding estimates. So, I want to find these numbers here. So, on this particular one, not the easiest to see. It's on two though and about 5.2 by the looks of it. Each square goes up by two. So, or 2. So, x is 2 and y is 5.2. And then for the one down here, which is just as difficult to see and it's in the negatives, we have X on minus 2.8 by the looks of it. And Y is just above the five, maybe -4.8. Y is equal to minus 4.8. Now, obviously when you're doing this, you can get your ruler and you can draw or just hover your ruler above the point just up to the X and Y axis so that you can see it a little bit clearer. But there we go. I've managed to read them. I'm pretty happy with those answers there. Just remember it does say we're finding an estimate. So, we're just doing as best as we can. The important part there is as well that we've drawn the line on the graph. We've shown that we where we are reading the values from. But there we go. there is using some graph graphical simultaneous equations. Now with roots and turning points if we are asked to read from a quadratic graph here for part a it says write down the coordinates of the turning point. Now the turning point is where it stops going down and starts going up. That point where it turns around and that's just here. Now that coordinate looking above from that point is one and to the left is -3. So that coordinate there is 1 and minus3. The next part says, write down estimates for the roots of where this function is equal to zero. It's not actually given us the equation of the graph. It's just called a function. So, whatever the equation is, when it equals 0, what are the roots? Well, that's the points where it crosses through the x-axis. So, this one here looks like 2.8. And this one here looks like -0.7. Obviously, you can get as close as you can to it to see that accurately, but there's our two roots, also called the solution sometimes. So, there we go. That would be part B. And then part C says, use the graph to find an estimate for when the for the function of 1.5. Now, you'll notice the function of X, the X has changed to 1.5. So, that just means what's the answer basically when X is 1.5. So if we go to 1.5 which is just here 1.5 trace down to the graph the only other thing you can read at this point is the y value. So if we go along from this y value there we go I get minus 2.8. There we go. It's the only thing you can read is an x or a y value. So in that one there it asked us to go along from x where x is 1.5 and read the graph. and the graph there we have -3 well just above so minus 2.8 eight. Okay. So, the only other thing you could be asked is you could be asked in this scenario as well potentially to find where the function is equal to something like two. Now, if I do this in a different color because it's different to what we're being asked here, but if we were asked where it was equal to two, instead of finding it on the x- axis, we would draw essentially drawing like a new x-axis just slightly higher up. And we'd be looking at the points here and looking and reading those y values instead. You just got to be a little bit careful just to make sure it definitely is asking you for when it equals zero which is essentially at ground level and not as a different line slightly higher or lower. But there we go. That is looking at roots and turning points. Now with parallel lines, if we're asked to show that lines are parallel, then we just want to compare that gradient. When lines are parallel, just means that they're going in the same direction and therefore have the same gradient. Now the first line equation that we've called L1 has a gradient of three. But this second line line equation is not in the form y = mx + c. So we want to get it in this form. At the moment it's not. So we need to move the pieces around. So it is in that form. I'm going to start by moving the 9x. So I'm going to add 9 x to the other side. It's going to be a little bit closer. Then we have 3 y + 5 is equal to 9x. I'll also move the five. So - 5 to the other side. We get 3 y = 9x - 5. And now I need to divide by 3. And we get y is equal to 3xus. I'm not going to do 5 / 3. I'm just going to leave it as a fraction. 5 / 3. The actual number at the end isn't overly important. The important part here was that we have got the same gradient. So I would write some words along with this that just says they both have the same gradient of three. And there we go. That would show that these lines are parallel. Okay. So looking at a quadratic simultaneous equation. So quite a lot of steps when we have to solve one of these, but ultimately we are going to use a substitution method. So here I need my second equation to say y equals or x equals. And we can rearrange that in whichever way we see fit. But we want to look for the easiest way possible. So here the easiest way is going to be for it to say y equals because I can make it say y equals by adding 3x to the other side of our equation. So once we do that we get y is equal to 13 + 3x. Now normally when we are writing these expressions that 3x piece tends to be at the start. So, we can write that the other way round if you prefer. You certainly don't have to, but I'm just going to swap those around just so it says 3x + 13. Now, I know what y is equal to as an expression. I can substitute that expression, the 3x + 13 in place of the y up in that first equation. Now if I do that and do nothing else, so just writing it in the place of the y, I'd have x^2 plus now the 3x + 3 is our new expression for y. So I'm going to put that in a bracket because it is being squared. So that would be 3x + 13 close brackets squared is equal to 25. I now have an equation that only has x in. And as long as it only has one letter, whether that's squared or x's, but as long as I only have one letter, I can definitely solve that. I can either solve it using a linear equation method if there's no x squares, or I can solve it by factorizing, how we solve quadratics. So, this is going to form a quadratic equation, which ultimately is going to need to equal zero. But I need to expand and simplify and tidy everything up. So to start with, I want to expand this double bracket. So I would write that out in full. Don't try and be quick and expand it without writing it out nice and clearly. So I get another 3x + 13. I'm now going to expand that as a double bracket. So we'll use our double bracket expanding method, whichever one you prefer. I'm going to do my two multiplications here, which gives me 9 x^ 2 plus 3 * 13 is 39. So 39 x and then underneath 13 * 3x is another 39x. And 13 * 13 is one of those square numbers that you either need to know or be happy to work out without a calculator. So that's 169. If I then simplify this quadratic. So we have 9 x^2 plus 39 + 39 is 78. So 78x plus 169. So that is the quadratic. And but that's just the part in the double bracket. I still need to bring this x^2 and this equals 25 back down to this quadratic. So if I write those in, we have x^2 plus that double bracket that's been expanded and that's all equal to 25. So I still need a bit of simplifying cuz I now have two of those x squared pieces and I need it to equal zero. So we've got a lot of steps here just to get to the point of solving it. So let's start by adding these together. So that gives us 10 x^2 + 78x + 169 is equal to 25. Now we're almost at the point where we can solve it. We just need it to equal zero. So I need to take away 25 from both sides. And that leaves me with 10 x^2 + 78x + 144. that now is equal to zero. Now, at this point, I want to see if I can simplify this quadratic in any way. So, to do that, I'm going to look at the three numbers in the quadratic. We have a 10, a 78, and 144. And I just want to think, is there anything that they all divide by? I can divide these because it's an equation that equals 0. Whatever you do to the left, you got to do to the right. But it doesn't matter what I divide zero by, it's still going to equal zero. So that's the only reason I can do it. And in this instance, they all divide by 2. So my new quadratic would be 5x^2 + 39x + 72. And that equals 0. Now that quadratic I now need to factoriize and solve to get my solutions for x. And of course we've already done a lot of steps even up to this point. So I'm going to write that quadratic up here. 5 x^2 + 39x + 72. I'm going to try and factoriize it. I'm not going to put the equal zero in for the moment just to not clog up so much space. So for 72, we could have the factor pairs 1 and 72. We could have 2 * 36. We need to figure out if three goes into 72. And you do need to test all of these because it can be the worst factor pair out of the options. So if you don't know if 3 goes into 72, let's just check. 3 goes into seven twice, remainder 1. And four times into 12. So 3 and 24 is another option. Does four go into 72? Well, it's a few less than 80. But again, let's just do some division. 4 into 72. Four goes into seven once. Remainder three and it goes into 32 eight times. So we have 4 * 18. Five doesn't go into 72. 6 does. Six goes in 12 times. So 6 * 12. Seven doesn't go in. 8 goes in. So 8 goes in nine times. And then eight and nine are next to each other. So we have definitely found every factor pair. So it's going to be one of these options that goes in the brackets. We have a five at the start. So one of our brackets has to start with a 5x and x in the other. And we're trying to make 39 in the middle. And remember, the number that goes in that right bracket has to be multiplied by five. So when I'm looking through my list, I'm just thinking to myself, well, which one of those numbers could be times by five, add to the other, and still get 39. Now, take your time to work that out. I've just spotted though as I'm going down, it has to be 3 and 24. If three goes in this right bracket, that will turn into 15. And 15 added on to 24, which goes in this bracket, is going to equal 39. Thankfully, it's all plus symbols in our quadratic. So, it's plus and plus in our two brackets, which leaves us with the two solutions. On the right, x is equal to -3, flipping the symbol. And on the left, x is equal to -24, but there's a five in front of the x, so it's -24 over 5. So, without a calculator, we have got some pretty horrible numbers that we have to deal with here. But, we're going to follow it through. We know our equation is y is equal to 3x + 13. Not so nice for the x is equal to -24 over 5. But if we start with the x is equal to -3. So when x= -3, y = 3 lots of -3 add 13. 3 lots of -3 is -9. Add 13 is equal to 4. So when x is -3, y is equal to 4. The other solution is going to be a little trickier for us to do as it's a negative fraction, which isn't the nicest, but we get three lots of, and this is when x is equal to -24 over 5. We get y is equal to three lots of -4 over 5 + 13. So we need to work this out. We're not using a calculator. When you times that fraction by 3, it just timeses the numerator. So 24 * 3 will equal 72. So we get y is = -72 over 5 + 13. And to simplify this, we may want to write that as a mixed number instead of a 72 over 5. 5 goes into 70 14 times. So that is min -4 and two fths + 13. It's a little easier to see now. So it's going to be adding 13 to the -14 which would give us -1 and we still have the two fths. So that is equal to -1 and 2 fths which you could leave as a mixed number or you could write that as an improper fraction - 7 /2. Either one of those would be fine, but they are our solutions there. There they look a little bit messy as they are. So on these dotted line, we'd want to write that nice and clearly. But we have when x is -3, y is equal to 4. And for our second solution, we have when x is equal to -24 over 5, y is equal to 1 and 2s. And there we go. That's our solution for that quadratic simultaneous equation. That's obviously a very tricky topic and there are different variations of this question. So it would be one topic I would suggest making sure you go into the description having a look at the full lesson for this particular topic. Okay, onto some angles in polygons. Now this particular polygon is an interesting one as it is not a regular hexagon, but it does say it shows a hexagon and it has one line of symmetry. Now straight away you can kind of see that that line of symmetry is going straight down the middle. And there we go. It's the same on either side. It says here angle B C D is two lots of angle C D E. So what let's find BCD. So BCD is this angle here and that is two lots of C D E. Now wherever we're given a statement like this that means it's in a ratio. It means it's in the ratio 2:1. The one on the left BCD is two lots of the one on the right. You could pick any variation of that ratio, but 2:1 is the one with the smallest numbers in or the simplest form of that ratio. That can also be written in terms of algebra. So you could say it's 2x to 1x. You can write it like that if you want. Just depends whether you prefer to use ratios or prefer to use algebra. Now I prefer to use algebra in this question because it means that I can put 2x in this angle and x in this angle. Now I know all the angles on the other side are all the same as well. So this is 117. This is also 2x and this is also 1 x. There are other variations that you could look at here. You could look at this the shape on the right after we've split it up. We have some right angles here and here. So you could look at the shape on the right there instead and sort of treat that as a different shape and work from that one. And that's fine as well. I'm just going to go for the whole shape here and just see what we get. Now it says it's a hexagon. So regardless of the fact that it is an irregular hexagon, we want to do our working out to find the in the sum of the interior angles. So our formula is n minus 2 * 180 where n is the number of sides. A hexagon has six sides. So 6 takeway 2, which is 4 * 180. And we need to work that out. So 4 * 180, well 2 * 180 is 360. So double that is 720°. So we know the sum of the angles is 270° within a hexagon. We just need to find what's left. Now the only angles we know are these 117° at the top. And if we add both of those together, that gives us 234°. So if I start by taking that away from my 720. So 720 take away 234. I'm going to have to do some borrowing for this. So, borrow from the two. Four from 10 is six. Have to borrow again. And then 3 from 11 is 8. 2 from 6 is 4. So, there are 486° left in the hexagon. And the four angles that we have, we've got written with algebra. We have 2x, 2x, 1 x, and 1x. As we're looking at all four of them, you could write that as a ratio. 2 1. It's completely the same process though because ultimately we want to add together all of these angles that we have. So we have the 2x + 2x is 4x + 1 x is 5x and plus another is 6x. So we have six x's that equal this total 486. Now once we've done that and we have solved this little equation to find the value of x, we will be able to find the size of angle a fee which is in this particular 2x here by just tsing our answer by two. So I'm going to start by dividing by six to find the value of 1x. To do that I'm probably going to have to do some bus stop division. So let's do that to the side. 6 into 486. So that goes into 48 eight times and then it goes into six once. So my value of x is 81. For that value of 2x that means we just need to do 2 * 81 which gives us a value of 162 and that would be our final answer. So the answer for this one is 162° and that would be the size of the angle A F E which is obviously of course the same as angle B C D. There we go. And there's some angles in polygons. Okay. So looking at some angles in parallel lines says here A C is a triangle. It then tells us that AE D and ABC are straight lines. And it tells us that EB is parallel to DC which we can see by the arrows on the diagram. Now it says here to work out the size of angle EAB. EAB is this angle here at the top of the triangle. And it says you must give a reason for each stage of your working. And those reasons there are going to make up the large majority of the marks on questions like this. So for starters before we try and find that angle, all I'm going to do is I'm going to look at this diagram thinking about angles on straight lines which have been mentioned and angles in parallel lines which have been mentioned as well. and of course angles within a triangle. Now if you look at this diagram, there is a small triangle sitting on the top. None of the angles we have within that triangle. There is also a large triangle all the way around the outside. We do have one of the angles in that triangle in the bottom left. We know that with angles in a triangle, we need at least two, unless of course it's something like an isoclesles triangle where we can match up and find the base. However, it doesn't tell us anything to do with that. So in this question, I'm just going to have a look and see if there's anything that I can find. Well, we have these angles here on a straight line, and there's two angles next to one another. So straight away, I can find this angle here next to the 148. Now, when I label that, I am just going to label it next to the diagram. I'm going to write my working out next to it. So 180 take away 148 leaves us with 32°. So I know this angle here is 32. And of course I need to give a reason. So I would write angles on a straight line equal 180°. Now once we've done that and we have found one of the angles, we need to have a look and f and see if we can find any other angle at all. So looking at the angles on straight lines, there's nothing left to be found. There's still nothing within the triangles. So we have to think about the parallel lines. There are two types of angles primarily that we're going to have a look at. Now here, because there is no additional lines sort of shooting out of the shape, we can't really find any of those alternate angles that make that nice Z shape. So, we're really going to be focusing on looking at corresponding angles, which are on the same side of those parallel lines, which make this sort of shape. And we're also going to be focusing on co-terior angles which are on the insides but on the same side of parallel lines. So we have those two that we're going to focus on. So thinking about co-terior angles we could use that just here to find this angle. That would be one option and there's a lot of options we can take here. So that's one option we could do. We could also have a look on the other side and we could use co-terior angles to find this angle. That's another option. The final option and it doesn't matter which one of these we take. We could use corresponding angles. We could find this angle just here which is the same as the 63. Now it doesn't really matter which one we go for. Any of them would be fine as this is the last one that I happen to draw. And this is purely by chance not because this is my favorite. It's just this would be 63. Now next to that I would want to label it as well. You can write this you know nice written proof down the bottom. You can say angle a eB equals 63 and write your reason next to it or you can just label it near the diagram. So I'm going to say here that a and I'll write it as you could down below. We'll say a e b equals 63° because corresponding angles are equal. And there we go. There's another reason and another angle that we found. Now we just need to have a look and think based on that angle that I've actually found, can I now find that top angle in the triangle? And the answer is yes. Now, because we have this triangle here, we've got the two angles down the bottom, and we know angles in a triangle add up to 180. So, we can find that final angle. Now, this is the only problem with labeling the diagram is that sometimes you are going to just run out of space to actually label that. So, it is good to be able to write the proof down below as well. So, I'm going to say E A B is equal to 180 subtract the total of say 63 + 32 We need to work that out. So that is 180 take away 63 + 32 is 95 and that is equal to 85°. Next to that I need to finish off by putting my final reason which is that angles in a triangle add up to 180°. I'm just going to put angles in a triangle equal 180°. Okay, just for simplicity of our example here. So there we go. that angle is 85. And of course, there were other ways that you could do that. If you did go and find the bottom right angle using co-terior angles down where that letter C is, you'd find that that one's 32 does match the co-terior angle that we'd already found anyway. And then you could have looked at the big triangle around the outside. Or you could have found the angle just underneath the 63 that we've just drawn in as the co-terior angle and then found 63 using angles on a straight line. So lots of different ways that you could have approached this but there we go there's one method along with all of our reasons at each stage of our working. So looking at some circles where we are using terms of pi and a little bit of a problem style of question here where we have a circle or circles with inside within the circles within a square. So it's a strange diagram says we have a square with side lengths of 20. Says it also shows a semicircle and an arc of a circle. AB is the diameter of the semicircle and AC is an arc of a circle with center B. Show that the area of the shaded region over the area of the square is pi over 8. Now straight away there's something really easy that we can find in this one and that is this part down the bottom, the area of the square. So straight away we can put that into a fraction because to get the area of a square we would just do the length multiplied by the width and they're both the same. So 20 * 20 or 2 * 2 is 4 and there's two zeros so 400. So that's the area of the square. So let's put that down the bottom. 400. We now need to find the area of the shaded region. Now, to find the area of a shaded region, typically you'll either find the area of the whole shape, take away anything that's not shaded, or find the area of the shaded shape and take away whatever's not shaded. Just depends on how the diagram is drawn. Now, in this case, we can actually work out the area of the shaded shape if we kind of imagine that semicircle sitting on top of it. this whole quarter circle there. And I'm saying it's a quarter circle because it's a square and this is a right angle. That whole quarter circle, we can actually work out the area for that. So to work out the area of this shape, we'd use p<unk> r squ, but we're going to have to divide our answer by 4 as it's a quarter of a circle. So we would do pi * the radius squared. The radius for this one is 20. So p<unk> * 20^ 2. And we're going to divide our answer by 4. 20 is 400. So that is 400 pi / 4 and that is 100 pi in total. So that semicircle there which we've just worked out has an area of 100 pi. The sorry the quarter circle the semicircle is what we're about to work out. So here the semicircle being really careful because 20 is the diameter which means half of the diameter would be 10. So the radius is 10. It's also a semicircle. So we're going to be dividing our answer by 2. We're going to use the same approach though. P<unk> R^ 2. So P<unk> * 10^ 2. And then we're going to divide our answer by 2. So P<unk> * 10^ 2 is 100 pi. Dividing that by 2 leaves us with 50 pi. So the area of the semicircle we've got now as 50 pi. So for the shaded region we would want to do and this is the area of the shaded region we're looking at now 100 pi the shaded part take away 50 pi the part which is unshaded and then total that leaves us with 50 pi. So the area of the shaded region is 50 pi. We can now put that on top of our fraction. So 50 pi being the area of our shaded region. And now we have a fraction. Now that fraction is going to need simplifying. So to simplify this fraction, we need to think what we can divide the top and the bottom by. Now you might spot that they both divide by 50. And if you do, that is going to simplify in one step. But don't feel like you need to. You could divide by something else. There we go. There's me jumping steps already. So we're going to divide by 50, which leaves us with one pi on the top. And 400 / 50 is 8. And there we go. That's what it wanted us to show that the area of the shaded region over the area of the square was equal to pi over 8. There we go. There is a little bit of a circles problem and keeping things in terms of pi. Okay. So, on to some surface area. Says here the diagram shows a cube with edges of length x and a sphere of radius 3 cm. We've shown that in the diagram and we're also given the surface area of a sphere. It says the surface area of the cube is equal to the surface area of the sphere. Show that x is equal to the square root of k pi where k is an integer. So to start with we want to find the surface area of both of these shapes. Now the cube we can only find in terms of x. So to find the surface area of this cube, we want to first find the area of one of the surfaces, not forgetting that each of the lengths are x. As it's a cube, they're all the same length. So the area of the front square would be x * x, which would be x^ 2. On a cube, there are six faces. So we would do 6 * x^2, which is equal to 6 x^2. So that is the surface area in terms of x of the cube. We can now look at finding the surface area of the sphere. Now the sphere has a radius of r which is 3. Now here we are given the formula for the surface area of the sphere. So we can do 4 *<unk> * 3^ 2. We want to multiply the numbers together. But let's just simplify the 3^ 2. So 4 *<unk> * 9 4 * 9 is 36. So that is equal to 36<unk>i. And that is the surface area of the sphere in terms of pi. So it now tells us that both of them are equal to one another. So we can set them equal to one another. We can say that 6x^2 is equal to 36 pi. And we want it to say x equals. So to make it x equals, we're going to want to divide by 6, which would get rid of the six in front of the x^ 2. We'll have x^2 is equal to 6<unk>i when we divide by six. We now want to square root both sides. So if I root this side, we have x. And if I root this side, we can just put the square root over the 6 pi. And there we go. We've got it in the form that it asks for. we have x is equal to the square root of 6 pi. So there we go. That was all just really about forming an expression for the surface area of the cube, the surface area of the sphere in terms of pi. And then because it said that they were equal to one another, we could just set them equal to one another and solve it or or in other words rearrange it to make x the subject. There we go. There is some surface area using a bit of algebra as well. Now in this question we are looking at the volume of a prism. This particular question tells us the volume of the prism and wants us to work out the height. Says the prism has a volume of 750 cm cubed. Work out the height of the prism. So the height of the prism is going to be this length here. But to get the volume of a prism we want to get the area of the crosssection which is this face here this triangular face and multiply it by the length. So in order to get the volume we would have multiplied the cross-section by 25. So if we do 750 divided by 25 that will tell us the area of the cross-section and then we can think about what the height would be. Now 25 goes into 75 three times. So it goes into 750 30 times. So the area of that cross-section would be 30 cm squared. That's the area of our cross-section. We now want to think about the area of a triangle. Well, the area of a triangle is base * height / 2. Or you could write half base time height. So that base time height divided by two has to be equal to 30. Now if we put the numbers in, we have five as our base. So five times the height has to equal 30 when we divide it by two. So 5 * something has to not equal 30, but it has to equal 60. We can find that by multiplying this two or just thinking about it logically. But we know that 5 * the height has to equal 60. Now from there we can get the answer because we know we can either divide by five or you could just think to yourself well what number * 5 equals 60. Now that answer is 12. So the height is equal to 12 cm. And if you want to check it you can always put it back in on the diagram and then work out the volume. So 12 * 5 is 60 / 2 is 30. And we already know that when we times that by 25 we get 750. So there we go. That was a sort of reverse volume of a prism where we are working out the area of the cross-section. Just making sure you know that the volume of any prism is the area of the cross-section multiplied by the length that it goes through the shape. Okay. So in this question we have some exact trigonometry. Now there are lots of values that you need to know for exact trigonometry. This particular one here is tan 45. Tan 45 is a relatively nice one. The answer to that is just one. But there are lots of values that you need to know. You need to know sin 30, sin 45, and sin 60. You also need to know the same for cos and for tan. So obviously I will link the video in the description how to learn those. But you do need to learn them even if you just write them onto a revision card and just make sure that you know them before you go into your non-cal exam. So there we go. That's one of the exact values. 1045 is equal to 1. Another good two to remember. One of them's on the screen. cos 60 is equal to 0.5 and sin 30 is also equal to 0.5. So there are three really key ones just to make sure you definitely know but ultimately if you want to know the others as well definitely want to revise says here here is a right angle triangle work out the value of x. So if we're going to use SOCOA for this, we want to think about the sides and what they're called. Opposite the angle is not given to us and we're not looking for it. So we don't need to label the opposite. But we do have the hypotenuse and the adjacent. Now whatever method you use, if you use formula triangles or whether you write down the ratios, it's completely up to you. I'm going to go with a formula triangle for this one. So with the A and the H, we're going to be using cos. We kind of already know that because it does tell us that cos 60 is equal to 0.5. I just need to know if I'm going to multiply or divide by cos 60. So in this one we are looking for the h. So we're going to have to do a divide. The a is four. The co is the cos 60 part. I'm going to write cos 60 here. But it has just told us that that is 0.5. So instead I'm going to write it as 4 / 0.5. And when you divide by a half it doubles the answer. So for this it would be 8 cm. You can apply a bit of logic. Anyway, we know that the hypotenuse is the longest side. So, it's definitely not going to have four because that would make no sense in terms of our diagram. The hypotenuse cannot be longer than the other two sides. So, the answer for this one was 8 cm. Looking up some Pythagoras theorem, and this is an interesting one because there's no diagram. Says here, triangle ABC has a perimeter of 20 cm. A to B is 7, B to C is 4. by calculation deduce whe the triangle ABC is a right angle triangle. Now to start with if it is a right angle triangle let's just imagine what it would look like. So let's just draw a little sketch of a right angle triangle and let's find what that missing side would be. If the perimeter is 20 well at the moment we have 7 and 4 which is 11 cm. Now if 20 is the total perimeter, if we take away 11, we are left with 9 cm as that final missing side. 9 is the longest of those sides. So 9 cm would have to be the hypotenuse. 7 cm would be either side, but as I've drawn that one slightly longer, I'll label that as the seven. And then 4 cm is our final side. I'm going to get rid of this right angle because we don't know if it's a right angle triangle. But if it is, we could use Pythagoras to check to see if we get the answer 9 as the hypotenuse. Now, Pythagoras theorem is a^2 + b^2= c^². We can use it to find the longest side or one of the shorter sides, but I'm going to go with the longest side. So, a and b are our shorter sides. So, 42 + 7. We're going to see if that equals c ^ 2. Now, 42 is 16. 7 2 is 49. When we add those together, we get well 49 + 6 is 55 + the 10 is 65. Now the square root of 65 is not equal to 9. Likewise, we could to the side do 9 squared which is equal to 81. So we were looking for the answer 81. In this case, we've not got the answer 81. We've got the answer 65. And therefore, this is not a right angle triangle. So my answer would be no. There we go. It is not a right angle triangle. There we go. And that would be an absolutely fine bit of working out there where we've shown all of our working to show that we've used Pythagoras and to show that it is not a right angle triangle. Okay. So looking at some vector proof. Now this is undeniably one of the hardest topics on the GCSE. Most people tend to agree. You might disagree, but it is a very difficult topic. So, for that purpose, I will link a few videos down in the description for this one. Just looking at some different styles of questions. This is a relatively complex one and it gives us lots of information. It says here X is the midpoint of the line AC after it's told us it's a parallelogram. So, if we imagine the line A to C, we will have to draw additional lines on here. The midpoint if we visualize that would be just here. And then we'll call that X. says OCD is a straight line. We can't see point D, but it says O to C. C to D is K to one. So, we need to draw a little point over here. And there we go. We'll imagine that that point is now the extended part of the line. We don't know if C to D is longer than O to C, but that's okay. We're not drawing it to scale. Now, we know that that is in the ratio K to one. So, it's going to be a certain amount of C's. We just don't know how many but we will find that out. So it tells us next that X to D is the vector 3 C minus half A. So if I draw X to D in this line here with my arrow indicating the direction we have 3 C minus 12 A. It says find the value of K. So to find the value of K here, we're going to need to think about how much or how many C's or what proportion of C this one is here. And I say C because O to C is the vector C. So we want to know how many C's it's going to take to get to that particular point. And we've said C so many times. The letter C sort of seems a little bit irrelevant now. But we want to have a think about how we're going to do this. Now, in pretty much every vector question, you need to find the vector for any lines that you've not been given, particularly ones that you've drawn on. So, here we've drawn on the vector A to C. So, finding A to C is going to get us a mark on a question like this, as it's one of the vectors we don't know. Now, if we think about this, it's told us how to get from X to D. It's told us that that is 3 C minus a half a. So if we can figure out how to get from X to C, which is going to be something we can find, then we can make a comparison between those two vectors and see what we're missing that's going to complete that vector to turn it into 3 C minus a half a. So this whole problem is how do we get the vector X to C? Once we've done that, we'll be able to see what's missing. But we're going to start by having a look at the full vector. So to get from A to C. So to get from A to C based on the vectors we've been given, we'd have to go down to O and then across over there to C. So if we label this the vector A to C. So from A to C indicating the direction there, that would equal minus A as we're going backwards through the A plus C. Now that if we get rid of this this highlighter now that is the full vector now from A down to C. Now we only wanted to know half of that and it tells us X is the midpoint. So we can use the fraction 1/2. If it's split into a slightly different ratio like 2 to 3 or something like that, we have to use slightly different fractions. But this one's a midpoint. So from X to C, which is moving in the same direction as A to C. So it's fine for us to use the same vector. We would want to use half of this minus a + c. If we expand that bracket out, we have minus a half a plus 12 c. Now, just to make it match the vector we've been given where c is positioned first, we're going to swap those pieces around. So, we have positive a half c minus 12 a. And in the process of saying that, it's reminded me that we need to make sure that the symbols that are originally with the A and the C stay the same. So you see that I've not changed the symbols. I've just swapped them around. The minus is still with the minus half a and the half C is still positive. So that's the vector X to C. Now if we look at these two vectors to get from x to c we have the vector minus a half a and we have the half c. Now the minus a half a is also in our other vector that was the vector x to d. So let's just have a look at what's different to get from x to d we have 3 c minus a half a. So the minus a half a part is the same in both. The difference is we have this half C and this 3 C. So we want to know what the difference is between the two. Now to get from 1/2 to three that's 2.5 or 2 and 1/2 which means that the amount of C's to get to there must be 2 1/2 C. Now we could write that in. We could write down 2.5 C. I'm going to keep it as a fraction, which I always prefer with vectors. 2 and a2 C. I tend not to write anything as mixed numbers, and you'll kind of see why. But if we convert this to an improper fraction, 2 * 2 is 4 + the 1 is 5. So 5 over 2 C. And that's the vector C to D. We've got that just by comparing the two and seeing what was missing. So we are almost done now because we know that that little part is 5 /2 and it's told us that the ratio of O to C and C to D is K to 1. Well, let's just have a look. So currently our ratio if we look at what we have, we have O to C, C to D is in the ratio. Well, it's 1 C from O to C. Let's just write 1 C. And from C to D we've just worked out is 5 over 2 C. Now we can get rid of the C's here and just write that as a numerical ratio. So it's 1 to 5 /2. But the question says the ratio is in the in the ratio K to 1. So in order to get down to a K to 1 ratio, well on the right we want that to say one. On the left we want to actually work out what K is. So to get from 5 over2 to one, we have to divide by 5 over2. It's always the case when we're looking at a ratio which is something to one or one to something. So we need to do the same on the other side. we need to divide by 5 /2. So we have a little bit of a strange division to do here because we have to do 1 / 5 over2. But actually that's really nice to do because when dividing fractions you just multiply by the reciprocal. So actually that just becomes 1 * do the reciprocal 2 over 5 and that's equal to 2 over 5. So we know that that is going to be 2 over 5 just down here. And there we go. We've got our ratio in the form K to 1 and K is equal to 2 over 5 and that's our final answer for this question. Now the thing with vector proof there are so many different types of questions and different variations. This is a sort of difficult question but also a little bit in the middle ground. There are some questions that are really very challenging on vectors. This is already a very challenging version, but there are some slightly harder versions and I will link them in the description so you can have a look at those as well. And onto some sharing in a ratio. This is an interesting version of a question because it says here a shop sells packs of black pens, packs of red pens, and packs of green pens. There are two pens in each pack of black pens, five pens in each pack of red pens, and six pens in each pack of green pens. It says on Monday, packs of black pens sold to packs of red pens sold to packs of green pens sold is 7 to 3 to 4. A total of 212 pens were sold. Work out the number of green pens sold. Now, the difficulty in this question here is look, we've been given all of this information about how many pens are in each pack. And then later on it tells us the total amount of pens that were sold, not the amount of packs that were sold. But the ratio that's been given to us is telling us the amount of packs that have been sold. Now, we want that ratio to be the amount of pens because we all we really want to do in this question is share 212 in the ratio and not 7 to 3 to 4, but the ratio of the pens that were sold, not the packs of pens that were sold. So, in order to turn this into packs, well, here the ratio is in packs. So, we'll label that. We want the ratio in pens. So, if we want to turn this into the amount of pens, well, seven packs of, and let's just label up the colors here. That was black to start with, then red, and then green. So, the information down here tells us two pens in each pack of black pens. So, we'll times the seven by two, and that's 14 pens. In those seven parts, there are five of the red pens in each pack. So, we need to times the 3x five, which is 15 pens. And then the final one, there are six pens in each of the green packs. So, tsing that by six would give us 24. Now, that's a really strange process to do because normally when we are multiplying or manipulating a ratio, we times all the numbers by the same thing. But here, we're tsing all the numbers by different things. And that's quite a strange process. But we're doing that to turn it into a ratio for the amount of pens in each of those packs that were sold. So now we have a ratio in terms of the pens. From here, all we actually need to do is share that in that ratio. So if we add together how many parts we have 14, 15 and 24. Take your time to work that out. So 14 + 15 is 29. Add the 20 is 49. Add the four is 53. There are 53 parts in this ratio. And that means we have to do 212 / 53. But of course, think we're not using a calculator here. So it's got to be a relatively nice number. And we'd assume it's also going to be a whole number. So, if we do a few 53s, if you're not sure, 53 106 159 add 50 is 209 add the 3 is 212. So, there we go. It fits in four times. So, let's write that down. 212 / 53 is equal to four. So, each part there in our pens ratio represents four pens. Now, we want to know how many green pens there are. So, let's just write this 53 parts to the side because I can manipulate now this ratio. And this is the part now where we will times everything by the same number. Now, of course, it only wants the green pens. So, I do only actually have to times the 24. So, I'm going to times that by four. And 24 * 4. I probably just want to work that out to the side. I don't want to do all of this and then get this final part wrong. 4 * 4 is 16. 4 * 2 is 8 + the 1 is 9. So that becomes 96. You could go and times the rest by four, but it did only want the green pens. So I'm not going to times them all by four, but if you needed to know them all, we could times them all by four. There we go. Our final answer is 96. And that's the final part of this sharing in a ratio question. Now, when combining ratio ratios, we need to be really careful because when we have two ratios in a question, we are looking for something that overlaps. And by overlaps I mean the same thing in each ratio. So in this one here we have flats in our first ratio and we have flats in our second ratio. Now if I write this out I can write it out as what we would normally call a three-part ratio. It mentions houses first, then it mentions flats and then it mentions bungalows. There are multiple different ways to solve this question, but combining the ratios is going to help with lots of different styles of questions. We'll go with combining them and making a three-part ratio. So here the number for flats in one of the ratios is a four and one of the ratios is an eight. So I can't just write this into one ratio straight away because I haven't got a fixed number to put in the position of the flats. So I need that number to be the same which means I need to find the lowest common multiple of four and 8. That's a relatively nice one because four turns into eight. To do that I just need to multiply this top ratio by two. So that would become 14 and the four would become eight. The eight now matches and I can just put all my numbers in my three-part ratio. So houses is 14, flats is 8, and bungalows is five. That hasn't changed that second ratio. So there we go. It now tells us there are 50 bungalows in the village, while the bungalows is the number on the right here. And it tells us that that becomes 50. So, this is one of my favorite types of ratios when it just tells you what one of the numbers is turning into. Rather than making us add them all together and share it out, it's just told us straight away to get from 5 to 50. We can probably spot that that is just tsing by 10, which means we need to times them all by 10. So, flats will be 80. 8 * 10 and houses will be 140. Again, just tsing all of these by 10. It says, how many houses are in the village? Well, we've just answered that. We have 140 houses that are in the village. And there we go. That's one of my favorite types of ratios there where it does tell you what one of the numbers is turning into. So just watching out there when you have to combine them, making sure that that overlap number is the same in both. Our next question is looking at a ratio problem. Very similar to our previous ones, except this one here is looking at where there's a difference between them. So it says Robin, Mary, and Isabelle collect stamps. The amount of stamps they own are in the ratio. Robin stamps to Mary's stamps to Isabelle stamps is 4 to 7 to 15. But it says here Isabelle has 24 more stamps than Mary. We'll deal with that first. But then it says Isabelle has more stamps than Robin. How many more? Well, we can deal with that once we know how many stamps they have. But here, Isabelle has 24 more stamps than Mary. So if I label this so that the letters are very clear, Robin, Mary, and Isabelle, their current numbers are 4, 7, and 15 for their ratio. And it tells us that Isabelle has more than Mary. So from here to here, from that 7 to 15, that is 24 stamps. Well, we know it's 24 stamps because that's told us or was given to us in the question. But how many ratio parts does that relate to? How many parts of that ratio? Well, to get from 7 to 15, that is an additional eight parts. So, we know that eight parts is equal to 24. If you wanted to work that out, you could write 15 takeway 7, which is equal to 8. So, eight parts is 24 stamps. Now, with all ratios, we want to know the value of one part, not eight parts. We want to know what one part is. So, one part of the ratio to get that, we wouldn't really write parts as it's singular. So one part if we divide that by 8 we do the same to the other side 24 / 8 is three stamps. So each part of this ratio is worth three stamps. Now we know the ratio which is 4 to 7 to 15. We've just worked out that each part is worth three. So let's times them all by three like we did in one of our previous questions. So 3 * 4 is 12. 7 * 3 is 21. 15 * 3 is 45. Says Isabelle has more stamps than Robin. How many more? Well, we have Isabelle at the end. I've kind of written kind of looks like 15. So let's get rid of that. So Isabelle is at the end. Isabelle has 15 stamps and Robin has four. So nice and easy to work out this final step now that we have the numbers. Anyway, so to work out how many more she has, well, 15 take away 4 gives us 11 stamps. So, our final answer, Isabelle has 11 more stamps than Robin. Okay, so this question here, one that always tends to throw people, it's about ratios to fractions. Says that there are four types of cards in a game. Each card has a black or white circle or a black or white triangle. So, there are four types of cards. The number of cards with a black shape to the number of cards with a white shape is 3 to 5. The number of cards with a circle to the number of cards with a triangle is 2 to 7. Express the total number of cards with a black shape as a fraction of the total number of cards with a triangle. Okay, so this question is a relatively interesting one. So here it says in this final line express the total number of cards with a black shape as a fraction of the total number of cards with a triangle. Now looking at our ratios we can look at the total number of cards with a black shape as a fraction because it tells us that here black to white is 3 to five. So the fraction of cards that are black shapes and we can label this down here. So the black shapes, the three represents the black shapes. So it's three out of in total eight. So that's the fraction of cards that have a black shape on. We can then do exactly the same with the circles and triangles. And it wants to know out of the total number of cards with a triangle. Well, this ratio here adds up to nine and the seven represents the triangles. So let's just draw a triangle. Well, that triangle therefore is seven out of nine. Now, it's a little bit strange because we are dealing with fractions and it wants us to express the total as a fraction. So, the total number of cards with a black shape is three out of eight. And if we express that as a fraction out of the total number of cards with the triangle, which is seven out of nine, we kind of break a rule here. We have fractions within fractions. If we ever have a scenario like this, that fraction line means a divide. So actually what we have is 3 over8 / 7 over9. And actually it doesn't need to be written as a fraction. So that one there we are now actually just dividing fractions. We know when dividing fractions we are going to multiply by the reciprocal. And the reciprocal is 9 / 7. So that's nice and easy for us to do. We now just need to multiply the tops. 3 * 9 is 27 and the bottom 8 * 7 is 56. So actually that whole question there was just about dividing fractions. But we had to first convert the part of the ratio that we needed into a fraction and the second ratio the part that we needed converting it into a fraction just following the instructions. It said then express the total as a fraction of the total. So we got the fractional amount for the black cards on the top out of the fractional amount for the triangles on the bottom and that was then converted into a divide calculation rather than writing just as a fraction. So there we go 27 over 56 and that is ratios as fractions. When looking at direct and inverse proportion we need to know our formulas that we're going to use or our equations of proportion. It says here h is inversely proportional to p and p is directly proportional to roo<unk> t. It then says given that h is equal to 10, t is equal to 144 when p is 6. Find a volume or a formula for h in terms of t. So find a formula for h in terms of t. So we're going to start by writing our inverse proportion formula which is h is equal to k over p. We have a direct proportion formula for P. So P is equal to K *<unk>T. That's our inverse proportion formula and our direct proportion formula. We can now put these numbers that we're given into those formulas. So for the first one, we have H is 10. So 10 is equal to K / P, which is six. We can solve that or we can rearrange it. So we know what K K is equal to. We can times both sides by six and we get 60 is equal to k. We can then put that into our formula. So up here we have h is equal to 60 / p. For our second one, we can deal with that as well. We know that p is equal to 6. So 6 is equal to k * the square<unk> of 144. Square<unk> of 144 is 12. So 6 is equal to 12k. And we can divide both sides by 12. Divide by 12 and divide by 12 and we get 0.5 or 6 / 12 or you could write it as 1/2 is equal to k 6 / 12. So p is equal to 0.5<unk> t. Now, this is where the difficult part comes in because we now have a formula for H in terms of P and we have a formula for P in terms of T. But it wants us to find a formula for H in terms of T. So, we have the formula for H, but the P here we're going to need to replace with what we've just found. P is equal to P is equal to 0.5T. So, spotting that is the difficult part because putting it actually into the formula is relatively nice. we have h is equal to 60 / p and p is equal to this 0.5 roo<unk> t. So now we've done that, we do actually have a formula for h in terms of t. It says h is equal to and then we've got something with a t in which is means in terms of t. So to actually find this formula, we do need to make sure that there are no decimals within the fraction there. So we need to get rid of that decimal. There's two ways that we could do that. We could times the top and bottom by 10 which would remove it. Or we can look for the smallest number that we could times by. Now to get 0.5 to become a whole number, the smallest thing we could times by is two. And if we times by two, we get h is equal to 120 over<unk> t. And that's our nicest way of getting rid of that decimal there. Otherwise, if we did times by 10, you'd have 600 over 5. And then you'd need to simplify that down as well. Dividing the top and bottom by five. It doesn't actually say it needs to be in its simplest form, but we've got it in its simplest form there. H is equal to 120 over root t. So you need to know both your direct and inverse proportion formulas. And be prepared to have to put them together sometimes as well. But of course, you might just have to do one of these or maybe even just one where you're just finding the formula. And onto some density, mass, and volume. This is quite a tricky one to up here particularly when we're looking without a calculator. So we need to know our formula for density, mass and volume. Now density is equal to mass over volume. You can either write the formula down or you may use a formula triangle. So density which is equal to mass divided by volume. Now when we have a mixture question and this one here tells us straight away that A and B are getting mixed to make liquid C. I do tend to draw all the information into a two-way table. So for each of these mixtures, we have A, B, and C. So we're going to want three columns, A, B, and C. And for each one, we have a density, a mass, and a volume. It says liquid A has a density of 70 kg per meter cubed. We want to check all the units are the same. Liquid A has a mass of 1,400 kg, but we don't know the volume. Liquid B has a density of 280 and it has a volume of 30, but we don't know the mass. Now, it wants us to work out the density of liquid C. Now, to work out the density of anything, we need to know the mass and we need to know the volume. So, we need to find these missing pieces that are currently missing the mass from B and missing the volume from A. So, we can use our formula to do that. number just to cover up the letter or rearrange a formula, whichever method you prefer. To find the volume, we would do mass divided by density. So, for this one here, we need to do 1,400 divided by 70. You can make that easier for yourself without a calculator as well. We can cancel off some zeros when dividing. So, it's 140 / 7. 7 goes into 14 twice and there's an extra zero. So, 20. So, that is a volume of 20. That's in meter cubed. When we look at volumes, we can combine those two volumes. So if they're mixed together, that gives us a total volume of 50. We can then look at the mass of B. And once we've got that, we can get our overall mass. Now to find mass, it's density times volume. So for this one here, our density is 280. And we want to times that by our volume of 30. We want to take our time to work that out. You've got quite a lot to work out there. But again, we can do 28 * 3 and add the zeros on. So just to the side, maybe I'll just do 28 multiply by 3. 3 * 8 is 24. 3 * 2 is 6 plus the 2 is 8. So that becomes 84. So 2 28 * 3 is 84. And we've got the two extra zeros. So in total 8,400 and that is in kilogram. 8,400 kg. We can now add those masses together. In total, that gives us 9,800 kg. And to work out density, we need a mass and a volume, which we now have, we just actually need to do the calculation 9,800 / 50. So, if we're going to do 9,800 / 50, could write that as a fraction or as a divide. But again, when we're dividing, we want to get rid of one of these zeros top and bottom. just to make that division a little easier. So instead we're going to do 980 divided by 5. We can do bus stop division for that. So 5 into 980. 5 goes into 9 once remainder four. It goes into 45 nine times which would leave a remainder of three. And it goes into 30 six times. So that leaves us with 196. That is our density. So we get 196. If we look at the units in the question, that's in g kilog per meter cubed. And there we go. That'll be our final answer. 196 kg per me cubed. Okay. So, on to some speed, distance, and time. So, it says a car travels for 18 minutes at an average speed of 72 km hour. How far will the car travel in these 18 minutes? Now, you obviously know your formula hopefully for speed, distance, and time. However, when we are looking at speed, distance, and time on a non-cal exam, we're going to take a slightly different approach to using our formula. Speed is distance over time. Now, when we have a speed that's given to us, we can put that speed into a table. Now, that speed, we're going to need to write down the distance, and we're going to need to write down the time. Now a speed of 72 km hour means that in 72 km it will take us 1 hour. Now we are dealing with minutes in this question. So it is easier to write that as 60 minutes. So if we want to find out how to get to 18 minutes on the right hand side of the table, we're going to need to think about how to break this down. Now if we needed to go to a smaller number like 30, well that would be nice and easy. we could just divide by two. However, there's not anything obvious that we can divide by to get down to 18. So, ultimately, we want to get to 18 minutes, but we need to think about is there something in the middle that we could divide down to and then get to 18. So, doing this in two steps. Now, thinking about what we can divide by, there's lots of numbers that go into 60. You can divide by 6, 10, 12, 5. lots of numbers that go into 60, but we want to divide down to something which is going to turn into 18. Now, thinking about what a number that we can divide down to that does turn into 18. Well, six would be quite a nice one. So, could we go down to six? And if it works, then we'll stick with it. To get down to six, we would have to divide by 10. 60 / 10 is six. And then we would times by three. Whatever we do on the right, we just have to do exactly the same on the left. So, we need to divide by 10 and then times by 3. 72 / 10 is going to give us a decimal, which isn't the nicest if we're not using a calculator, but it gets us down to 7.2. And that's not going to be awful to times by 3. I would probably do this to the side. I'd do 72 * 3 and then we'll put the decimal back in. So 3 * 2 is 6. 3 * 7 is 21. And then I need to put the decimal back in one place. So 21.6. 6. So there we go. Our distance would be 21.6. So how far will the car travel? Well, it will travel 21.6 km. And I've got kilometers there from the question as it was in kilome/ hour. So there we go. A much easier method than having to use the formula because for this question here to find distance, we would do speed multiplied by time. And you can take this approach, but it's not quite as nice. The speed is 72, but the time is in minutes. So that amount of minutes has to be written in hours. And to get it into hours, we have to write it as a fraction. 18 / 60. And it might simplify down to a nice fraction. If we divide the top and bottom by six, we get 3 over 10, which isn't awful, but it means I have to do 72 * 3 over 10, which isn't the nicest. You could convert it to a decimal potentially and multiply it by a decimal. There are lots of ways that you could do this and you might prefer that. However, actually looking at the method we've used, although I've drawn a table, which seems like more work, although we are just drawing a table to keep it organized, you could actually do this without the table. You could just write an equal sign between the two. But keeping it all organized in a table and just visually seeing how do we get down to 18, it's a really nice method that you can apply to different topics. So there we go. There is a way of approaching this question of speed, distance, time without a calculator. Okay, so this next question is about probability from a table. Although you've probably noticed already there is no table in the question, but it says there are only red counters, blue counters, and yellow counters in a bag. And the ratio of red to the number of blue is 3 to 17. Shawn takes at random a counter from the bag. And the probability that the counter is yellow is 0.2. Look at the probability that Shan takes a red counter. Now, if we were to draw a table for this, and normally a probability table, you'll have the colors and the probabilities. So, we have red, we have blue, and we have yellow. And it gives us one of those probabilities. It tells us that yellow is 0.2. It also tells us then that red to blue is in the ratio 3 to 17. And that's given to us just here. So in terms of what we've got at the moment, that's all the information that we've been given. But we know with probabilities from a table, they've got to add up to one. So if I do 1 take away 0.2, that leaves us with 0.8. So these probabilities here have to add up to 0.8. Now it might may or may not help you to do this sort of next numerical calculation, but remember these decimals here are just percentages. So that's 80% that's left, and this 0.2 is 20%. So we have 80% that needs to be shared in the ratio 3 to 17. And once you write it as a percentage, that's actually not too bad because it's 80 that needs to be divided by 20 because there are 20 parts there in total. So if I want to share this out, we've got red to blue. It's currently in the ratio 3 to 17, which is 20 parts. So 80 divided by 20 is four parts or four per part. So, we just need to multiply these by 4. 3 * 4 is 12. 17 * 4, we want to take a little bit more time working out. 17 * 2 is 34. So, it's 68. Or you could do 82 taken away from 80 as well. So, 12 to 68 is our ratio of red to blue. Obviously, in our table, it's not a percentage. It's written as a decimal. So, let's write it as a decimal. So, 0.12 for red and 0.68 68 for blue. The question did want to know work out the probability that Shan takes a red counter. So red we have in our table 0.12. We could write that as a decimal as a percentage or even as a fraction if you wanted to. I tend to just leave the answer in the form that I've got it written. So I'd just write it as 0.12. Better than converting it to a percentage and potentially making a mistake. Just write down the number that's given to you in the question. So we're using decimals. We'll keep it as decimals. 0.12 is our answer. Okay, so for this question we have some functions. It says here the functions f and g are such that the function of f is 3x -1 and the function of g is x^2 + 4. And to start with, it wants us to find the inverse function of f. Now when finding the inverse function, I use a particular approach. I write the function as y equals. So y = 3x -1. And to find the inverse function, we just want to rearrange it. So x is the subject. To do that, I want to add one to both sides. And that would give me y + 1 is equal to 3x. From there, I've only got one more step, which is dividing by 3. So, a nice rearranging formula question. We have y + 1 / 3 is equal to x. Now, typically inverse functions questions on your dotted line will say f -1 of x is equal to and then dot dot dot dot dot. and you just got to write the function in there. Now the function that we've worked out is just here. However, we swapped the function of x with that letter y. So in the final step, you just have to swap the y back with the letter x. So just be x + one instead of y + one / 3. And there is our inverse function. Now part B of this question is trickier. It says given that fgx is equal to 2 gfx show that 15x^2 - 12x - 1 is equal to 0. Now by saying show that and then giving us that equation there all it wants us to do is find what fgx is find what two lots of gfx is and then just set them equal to each other. And in this one they've rearranged it so that there's zero on the right hand side. So if we start by finding fgx. Now fgx is when you substitute g into f. So subbing g into f. The way I do it is I put a bracket around the g function and around the x and I just put it in place. So that'll be three lots of x^2 + 4 and then minus one at the end. And we need to expand and simplify that. So we may as well do that first. So we have 3x^2 + 12 and then minus 1 and that'll equal 3x^2 + 11. So there is my function f gx. Now for the next one I need to do gfx and I need to multiply it by two. So let's do that one to the side. So gfx I'll do first before I do 2 gfx. So this particular one we are now putting f into g. That's a little bit more tricky because in g the x is being squared. So we need to put this function in and it's going to be squared. So we have there 3x -1 the f function and that's being squared and then we are adding four. So we need to expand the double bracket. You can do that to the side. However, it is just two brackets. So 3x * 3x is 9 x^ 2. 3x * the -1 would be - 3x -1 * 3x is another 3x - 3x and then -1 * -1 is + 1 and we have the + 4 at the end. Now I've done that without writing it down but of course to the side just write down 3x -1 3x -1 just expand and simplify it. Now if I simplify what we've got there, we have 9 x^2 the two -3x's we have - 6x in total and plus one and plus4 makes plus 5. So there we go. That is my function gfx. Now we want 2 g ofx. So 2 gfx just means multiplying it by two. So if we times what we have there by two, let's just write times two here. We get 18 x^2 - 12x + 10. Now at this point we're almost done. You can kind of see that we're going to get those pieces that it said to show that cuz in one of them we have an 18 x^2, the other we have 3x^2. Well, the difference between them is 15. So if we set these equal to each other, it doesn't matter which way we put them, but let's put them in the way the questions put them. So we have 3x^2 + 11, which is the fgx is equal to the 2 gfx, so 18x^2 - 12x + 10. Now, although they've got the zero on the right, it doesn't actually matter what side we get the zero. It only really matters when we have an inequality involved. when it's an equal sign, it doesn't matter either way. So, I want to minus 3x^2 and - 11. And I'll put those underneath the appropriate pieces here. So, - 3x^ 2 and - 11 from the 10. And that would give us 0 equals we get the 15x^2 we were looking for - 12x isn't changing. And 10 takeway 11 is -1. So, we get all the matching pieces there. All I really need to do at this point is just write it in the way they've they've written it. So 15x^2 - 12x - 1 = 0. And there we go. I have shown that 15x^2 - 12x - 1 is equal to 0 by doing those composite functions. So we had an inverse function in part a, relatively nice one. And we had these composite functions in part B there and quite tricky ones because we had both variations of the composite function and we had that 2 g ofx which is also a little bit rarer when it does appear. So 2 gfx just finding gfx and tsing by two and there we go there's some functions. Okay so our next question is looking at cumulative frequency. Now drawing a cumulative frequency graph I think is really nice. So all we do need to do is remember to use these end points and plot it against the cumulative frequency. Just be really careful. Make sure you have been given the cumulative frequency because if you are only given the frequency, you need to add them up as you go down and get that cumulative total. However, this has given us the cumulative frequency. So on 40, we need to go up to five, which is just here. On 60, we need to go up to 25. On 80, we need to go up to 35. On 100, we're going up to 38, which is very close to our top here because it only goes up to 40. And 120 goes up to 40. Now, for a cumulative frequency graph, the other thing we need to remember is our starting point. And it's this very first number in the table. So, our starting point is on 20, not to be confused with a starting point on the origin, 0. This doesn't start at zero, it starts at 20. So, we just want a nice smooth curve that connects these together. Not using a ruler, just a nice smooth curve. And there we go. There's our cumulative frequency graph drawn. It says, "Use your graph to find an estimate for the interquartile range." Now, if it's out of 40, we want to think about where the median is. Now, if we asked to find the median, that would be at 20. But the quartiles are at each quarter. So, we're looking halfway between the start and 20 and halfway between 20 and 40. Now, that's visually quite nice to see. You can see that the quarter will be at 10 and at 30. So, what we're going to do is go along to the graph. You would use a ruler and a pencil for this, indicating and marking out your lines. And we're going to read that number there. Now, just be careful. It is 50 in the middle. So, each little box goes up in twos. So, that would be 46. And again, there's some variation in answers that you're allowed. So, I'm going to go along from 30 and then go down for this one. That's three. Looks like three boxes after the 50 or two boxes. I think it was three boxes, but I've gone slightly off. So, you got to be really careful when you're doing these ones. Okay, it's three boxes. So, that would be 66. So, they are the values of the lower and upper quartile. For the interquartile range, that's the distance between them. We'll take the bigger quartile 66 subtract the smaller quartile and write down that range. So that range is 20. So there we go. A couple of things that you can find from a cumulative frequency graph. You could have to find the median. We' have just gone along from 20. Wouldn't have had to worry about finding two things. And you can also find things like how many people were longer than 80 minutes. So you could go up from 80 along and see how many extras are above that. I think by the looks of it, if you go up from 80, you land on 35 people on the cumulative frequency. So that' be an extra five people took longer than 80 minutes. So little things like that you can also be looking for on a cumulative frequency graph. But there we go. There was interquartile range. For this question, we have a reverse mean. So with a reverse mean, the sort of trick to this question is in the title. When we are given or when we are looking for a mean, we get the total divide by how many there are and we get our mean. With a reverse mean, typically we are given the mean but not the total and we have to do a multiplication instead. So here it says there are 15 children at a birthday party and the mean age of the 15 children is seven. Now before we even carry on that means that at the total age of all the children which we don't know we could give that a letter if we wanted. As we divide that by 15 we get a mean of seven. So something divided by 15 is seven. Now if you know what that number is in your head, great. You don't need to do the multiplication part. But we want to multiply by 15 and multiply by 15. And that's something that we're looking for, which we should have given a letter really. You don't even have to use algebra here, but we can do. If we times 7 by 15, we then get well 10 * 7 is 70. 5 * 7 is 35. So that would be 105. So that there is the total age of all 15 children. Okay. So that's the total age. It then tells us some more information though. It says nine of the 15 children are boys and the mean age of the boys is 5 years. Look at the mean age of the girls. So nine of the 15 children are boys. So this tells us that whatever age the boys are, let's use a letter this time rather than a question mark. So x divided by the nine boys gives us a mean of five. Well, in this scenario, just like before, we'd now just times by 9. So 5 * 9, much easier. We get x is equal to 45. And that's the age for the boys. So let's write that down. That is the boys age or the total boys age. Let's put the total total boys age. It says work out the mean for the girls. So the mean age of the girls now we're just doing a normal mean. So when we want to know the mean we need to know the total. We can find the total because we know that all of their ages were 105 and the mean age for the boys is 45. So for the girls well we can do 105 take away 45 which is equal to 60 and that is the total girls age. And to find their mean, we just need to divide the total by how many girls there are. Doesn't explicitly tell us how many girls there are, but it says nine of the 15 children are boys. So if nine of them are boys, that must mean that there are six girls. So therefore, we would do 60 divided by the six girls and that gives us a mean of 10. So there we go. The mean age of the girls is 10. And there we go. That is a reverse mean. And onto our final question looking at capture recapture. There's a lot of words in this question, but the actual maths is quite short to do. There are lots of different methods for this question, but I'm going to show you one method thinking about proportion. Says here, Sylvia wants to find an estimate for the number of birds in a reserve. On Monday, she catches 90 of the birds. She puts a ring on each of the birds and returns them to the reserve. On Tuesday, she catches 120 of the birds and she finds that 20 of the birds have rings on them. work out an estimate for the total number of birds in the reserve. Now, this is all just about writing down the proportion as a fraction for each of these captures. So, on Monday, she catches 90 of the birds and puts a ring on them. Now, that is 90 out of the total population, which is of course what we're looking for. We don't know that answer. So, that's 90 out of something. The second time she does the capture and we have the rings on the birds. There are 20 birds that have the rings on them. So that's 20 out of and she caught 120 birds. So that's 20 out of 120. Now when we're looking at this question, all we actually want to do is put an equal sign between them and assume that these are proportional. Now if they are proportional we just need to think about how to create that fraction 90 over something as an equivalent fraction to 20 over 120. Now there are a few different things you can do. If you can spot without a calculator how to get from 20 to 90. Brilliant. But you might be able to simplify this fraction on the right. And actually it does simplify. You can divide the top and bottom by two. You can divide by other numbers as well. But if we divide by two we get 10 over 60. And that's a really nice number to think about how it turns into 90. So if I write it again over to here, let's have a think. We'd have 90 on the top. How do we get from 10 to 90? Well, that's just tsing by 9. We do the same to the bottom. As we're creating an equivalent fraction, we're just going to times by 9. And we need to work that out. So you might want to do 60 * 9. So 6 * 9 is 54. So 6 * 90 or 9 * 60 would be 540. So there we go. So that's 90 out of 540. Now that X on the bottom, that 540 on the bottom is what we are estimating to be the total number of birds in this reserve. So our final answer is 540. We are making some assumptions in this question. And it does say in this part here, Sylvia assumes that none of the rings have fallen off between Monday and Tuesday. And that's one of the assumptions we make in this question. There are more assumptions that we make than that, but we assume that the rings haven't fall off. We assume that none of the birds have died. We assume no additional birds were added into this reserve. So, they're all things that we assume because all of those things could affect our answer. But it says here in part B, if Sylvia's assumption is wrong, explain what effect this would have on your answer to part A. Well, if she is wrong, and we might want to write that if she is wrong. Okay, if she is wrong, and we're specifically referring to the answer here as it says, what effect would it have on your answer? We would say the answer could be higher or lower. And that's absolutely fine for us to just say the answer could be higher or lower. The important part there is for us to say it could be either. If the rings have fallen off, for example, that would be different if more birds had been added to the reserve or if some of the birds had flown away or mysteriously died in this particular scenario. That could all affect our answer in different ways. We don't know which one it is, but in this case here where none of the rings have fallen off, well, if they have fallen off or if they haven't, we could have some different answers. Now, in this scenario, it is actually an interesting one because it does say this particular assumption says the assumption here is that none of the rings have fallen off. Now if she is wrong and some of the rings have fallen off well actually it does affect our answer in one particular way. So we need to consider that with this question. Now let's imagine in a scenario where the rings have fallen off and let's just imagine that actually 30 of the birds were supposed to have rings on. Now that would change our answer because if it was 30 out of 120 instead and we were trying to get to 90 well in that scenario we would only multiply by three and in that scenario the total would be 360. So in this one assumption which is very unique question actually our answer couldn't be higher or lower. If she's wrong and some of them have fallen off then our answer could only be lower. So, in this scenario here, I'm going to change that answer ever so slightly just because it's different to sort of the usual question. If she is wrong, the answer could be lower. And I'm going to stick with that lower one for this one because we've just proved an example there to show that it could be lower. Okay, so there we go. That was our final question. Over [Music] it. Over it. Over it.