Transcript for:
2022 GCSE Maths Higher Paper One Review

okay welcome to this video where we are having a look through the topics that have appeared on the advanced information that the exam board has released for the 2022 maths gcse we're going to be looking at higher paper one in this and as you can see on the screen i've done a list here or a checklist for all of the topics that are on the advanced information for that paper one now i have actually done this checklist for all of papers one two and three and if you'd like to have these or you'd like to download them and print them off you can get them in the description and i'll show you that in just a moment so these are free for you to download use them as you want give them to your friends but that's completely up to you now obviously if you've been following this channel for a while you'll know exactly how these videos work but just in case you're landing here for the first time i'm going to show you how it works and i'm going to show you obviously how you can get to this file so let's have a look at that now so when you're on one of my videos if you pause it to start with you can click onto the chapters on the bottom left and you'll see that i have bookmarked all of these topics that are on the list in order that i've drawn them on the list so that you are able to go through and pick the topics that you want as well as that if you go into the description you will see there that in the description there is a link where it says the hiring revision hiring foundation revision checklist where you can download that checklist that i was just showing you just below that will be all the links to all the different papers so you can access all the papers and have a look through the videos but just remember obviously i'm only going through one version of each topic so you will have to look further in depth into each of these topics as i'm only going to be covering one small snippet in a much larger topic so if you scroll down you'll also see that i will have all the links next to these timestamps this particular video i haven't actually put them on just yet but hopefully by the time that you're watching this all of the links will be on there as you can see there at 1 hour and 11 minutes you can see the exact values of trigonometry video is there so you can always click into that and have a look at the full video and that is obviously how this video works so hopefully that's useful and helpful if these videos are useful please don't forget to like the video don't forget to subscribe to the channel so that you get all the updates for all the future videos and don't forget to drop me a comment and let me know if there's anything else that you'd like making or just to let me know obviously how it's all going throughout the course of the year all right we've got a lot to get through so let's get started [Music] [Music] okay so this question says three-fifths of a number is 48 work out the number so this really tends to confuse people when it does come up because obviously when you see three-fifths of an of a number a lot of people are going to think okay straight away divide by the bottom times by the top but this is actually in reverse this is saying that three-fifths of a number is already 48 and if we think about this in terms of our bar again let's just think about if we had split this up into five portions this time because it is in terms of fifths there we go so if i split this up into five parts which i may not get perfect there we go good enough and it's what it's saying is that three of them has already been calculated as 48 and we want to know what the entire original number is so looking at this then okay well if only three of them were 48 well we wouldn't divide by 5 anymore but instead we would divide by 3 because we want to know how much is going to go into each of those three boxes just above so this time and it's obviously the complete reverse of before because this time we're going to divide by that numerator so instead we're going to divide it by 3 and that's going to tell us what number goes into each of the bars above so 48 divided by 3 is equal to 16 and again you can put that obviously the working out just to the side so if i put now 16 into each of these bars i would also put it into those additional two because we are looking at now all five of them so that number there if we add up all of those sixteens and again you can obviously add them all up or you can do sixteen multiplied by five so that's working out i would do to the side sixteen times because there are 5 boxes and that adds up to 80. there we go and that is my answer and again you could put the 88 next to your diagram if you prefer to use a diagram method but that is how we're going to go about approaching these questions so if you think about what we've actually done it's just the complete opposite of what we did before and that's why it's in reverse this time instead we are dividing by the numerator and multiplying by the denominator but you need to understand why and i think that this visual idea really does help you to understand why we would divide this time by the numerator and why previously we were dividing by the denominator so for the first one we've got work out three and four fifths add three sevenths and give your answer as a mixed number in its simplest form now when we look at this sort of question here uh one thing we need to remember no matter which type of calculation we're doing adding subtracting dividing or multiplying if we have mixed numbers involved we need to make them top heavy first now when it comes to adding and subtracting there is a little bit of a different approach that you can take that some of you may use but i'm going to use the same method for all of these types of questions and that is making any mixed numbers top heavy fractions first or okay or improper fractions first so the process for doing that is obviously taking our fraction here three and four fifths we want to turn that into an amount of fifths so to do that we want to figure out how many fifths are in three so we can do three times five which is fifteen add the extra four-fifths there and that makes nineteen-fifths so three and fifths three times five is fifteen at the four that we've got there as well as nineteen fifths okay so it's the big number times the bottom to make 15. add the top number on the fraction there 19 so we've got 19 fifths and we're going to add that to the 3 7. now when it comes to adding and subtracting we need to have a common denominator so we need that number on the bottom to be exactly the same now that means that we can multiply both fractions here as long as we do whatever do we do to the bottom we also do to the top we can make equivalent fractions with a common denominator so thinking about five and seven the lowest common multiple of those is 35 so in order to get them to be 35 the left fraction there i can times the top and bottom by seven and the right fraction we can times the top and bottom by five and that would give us 35 on the bottom of both so we've still got to work out 19 times seven there and obviously take your time doing so but 19 times 7 is 100 133 and at the moment that it's going to be over now 35 okay so you get some quite big numbers here but obviously don't be afraid just to decide to do some multiplication there i just is 7 times 10 and then 7 times nine and added them both together okay but you can show you're working out to the side on the right fraction there we're gonna times them both by five so three becomes fifteen and seven becomes thirty-five and now we're in a position where we can add these both together so we can just add together the numerators 133 plus 15 which gives us 148 and that is over 35 remembering you don't add together the denominators there we go we've got 148 30 fifths now obviously it says here to give your answer as a mixed number in its simplest form so it's up to you whether you try and simplify this or whether you obviously turn it into a mixed number first but we need to turn it back into a mixed number so to do that i need to know how many times does 35 go into 148 that's not the nicest so i'm just going to write down a few of the 35 times table so 35 add another 35 is 70. add another 35 is 105. add another 35 is 140 and then it's not going to go beyond 140 there because i get obviously our number there is only 148. so to go from here um we know that that 35 goes in four times so that's going to be a big four and what's left over from 140 248 is an additional eight thirty-fifths there we go so four and eight thirty-fifths left over okay so looking at that number there you just gotta decide as well does that little fraction at the end there simplify does eight over thirty-five simplify now in the case of this fraction here it actually doesn't there's nothing that goes into both 8 and 35 other than one and obviously that's not going to simplify it for us so that actually is fully simplified there so even though it says give it in its simplest form in this circumstance here that little fraction on the end there doesn't simplify okay but it's going to say that anyway because we didn't have to use 35 as our denominator we could have potentially used a bigger number like 70 or even larger if we wanted but because we used the lowest common multiple it's just ended up that doesn't actually simplify at the end there but do look out for that because that fraction quite often does need simplifying okay but there's our final answer 4 and 8 over 35. right okay so dividing with mixed numbers to start with it says so work this out one and one fifth divided by three quarters give your answer as a mixed number in its simplest form so again let's make these top heavy so one and one-fifth is six-fifths big times the bottom is five but the one is six so we're going to divide that by three-quarters okay now normal fraction rules apply here when we are dividing fractions we keep the first one as it is we're going to flip the second one over to the reciprocal of that second one and i'm going to multiply the fractions instead and multiplying fractions is probably one of the easiest little bit of the math you have to do because you know exactly what we'd like to happen happens we just multiply the top multiply the bottom so in fact rather than doing six fifths and doing a divide i'm just gonna times it by the reciprocal which is four over three literally just flipping it over and then from there we just multiply the top numbers so six times 4 is 24 and 5 times 3 is 15 and there we go we get our final answer and obviously it says to give your answer as a mixed number in its simplest form so i do want to have a look at creating a mixed number or simplifying it first okay it's up to you which one you go i'm gonna turn into a mixed number first so we get 15 goes into 24 once so we have one and the remainder there is nine from 15 to 24 so we have 9 over 15 left over now this is the first time this has come up look because that mixed number there okay this little fraction with it does actually simplify they both divide by three and i would need to simplify this now okay so that would end up being one and if i divide the top and bottom by three nine divided by three is three and fifteen divided by three is five so my final answer is one and three fifths okay so you just gotta spot and watch out for that okay but there is something else you could have done you could have taken a slightly different approach there you could have actually simplified at this point the top and bottom both divide by three and you get eight fifths when you do that and then you could turn it to a mixed number from there so five goes into eight once with a remainder of three and you get one and three fifths so it's completely up to you which method that you prefer to use whether you turn it into a mixed number first and simplify the fraction or whether you simplify the fraction and then turn it into a mixed number it's completely up to you okay but just obviously remembering that when you divide you keep the first one and you multiply by the reciprocal again a lot of people remember that like keep flip change okay you keep the first flip the second and change the sign into a times and obviously when we're multiplying fractions that's really nice and easy so it says write 0.47 with the recurring dot of the seven as a fraction in its simplest form now when it comes this we are going to use a bit of algebra to do so now the first thing we're always going to do is just write that x equals the decimal in the question naught 0.47 the recurring dot and it's important just to think about what's actually recurring here and it's just a 7. if i was to write it out as an actual decimal it'd be 0.4 and i could just keep writing sevens forever okay now when there's only one recurring decimal the thing that we can do here is if we can take this value of x which is 0.47 we can just multiply that by 10. and depending on how many recurring decimals there are we'll determine what we can what we actually want to multiply it by and you'll see the reason that we actually do this so if we multiply it by 10 we would get 4 seven and then it'd be another seven as it's recurring and i'm not going to write any more sevens i'm just going to write the same amount of decimal places as i have in my one above there two decimal places in 0.47 so i'll keep two decimal places in this version here now the reason this can help us is now we can take these away from each other and if we do take them away from each other those seven those recurring sevens are going to cancel out so i want to take them away and i can seem a little bit odd doing this because i'm doing the bigger number on the bottom take away the smaller number on the top so you can always feel free to realign it to the side which i can do over here i won't actually normally do this but 4.77 take away 0.47 okay so i tend to do it just looking at it up sort of upside down on the side but you can actually just realign it look the seven take away the seven makes zero the seven take away the four makes three and then four take away zero is four so it's 4.3 so when we take these away from each other 10x take away the 1x leaves us with 9x and then our decimals when we take them away we get 4.3 and i won't write the zero after i'm just going to write 4.3 now there are other methods of doing this this is just the way that i like to do it i always just like to remember one recurring decimal multiply it by 10 and then take them away from each other but you end up with this scenario here we have a decimal on the right hand side we've got 4.3 now i can't turn this into a fraction or i could but we shouldn't really have decimals in a fraction so before i turn it into a fraction i'm going to multiply both sides by 10. so once we multiply both sides by 10 we get 90x and that equals 43 and now we can turn that into a fraction we've got 90 in front of the x we can divide both sides by 90. and it's important to leave this in terms of algebra because some of these questions will say proof so i'm just going to keep it as x equals so x equals 43 divided by 90 or over 90. there we go now we've almost finished it to say give it its simplest form now you've got to think do these uh does this fraction actually simplify okay so there are different ways of actually doing recurring decimals and sometimes you might not get it in its simplest form now the way that i have done it does actually leave this fraction here in its simplest form okay 43 over 90 doesn't simplify it is worth just having a check though just uh you know let's just try and apply some little tricks are they both even do they both divide by two do they divide by three or five and keep just checking a few numbers that you know go into 90. and but as the way i've done this particular question it has actually finalized in its simplest form okay so that one doesn't actually need simplifying but just make sure you check in those final steps to see if it simplifies okay 525 is a product of prime factors so with these slightly harder numbers we're going to try and apply some number tricks here so let's have a look what does that divide by now things that i'm spotting it ends in a five so it definitely divides by five and ends in this 25 so it does also divide by 25 now i'm going to just stop to divide by five because i can do that quite easily over to the side and don't be afraid to do some working out for these so fives into five goes once five into two doesn't go so carry that over fives into 25 goes five times so 525 is a little split there it's five times 105 and again 105 ends in five so we can do the same again you might be able to do some of them in your head but i'm just going to show you obviously that you have do have these methods available five doesn't go into one five goes into ten twice and five goes into five once so 105 is five times twenty one now at a point here where hopefully we spot the twenty one is three times seven and there we go we've finished off our tree so we've got five five three and seven so three times five times five times seven and then putting these two fives as five squared so three times five squared times seven so this first one we've got 25 to the power of negative a half now when it comes to these powers we just need to remember what each piece does so if we make some notes on this the negative part of the power as we've seen before does the reciprocal it flips it over so it flips it over the number on the bottom underneath underneath that line there which i like to refer to as ground level the number underneath does the root so in the case for two that would be a square root and if it was a three it'd be a cube root and so on and the number on the top is just a normal power so these are the three things that we have to look at when we have these combinations going on and sometimes we might not have a number on the bottom sometimes we might not have a negative but if we have them all we have to think about all three of these pieces so it doesn't matter what order you do it in now i tend to always go for this number on the bottom first purely because it's easier to do the root of a number than it is to then do the than it is to do the normal power when the number is quite large if we imagine there was a normal power of two on the top it'd be a lot easier for us to square root this than it would be to do say 25 squared first so i always do the number on the bottom first that's just personal preference or you can do it in any order that you like so what i've got to do here first of all i'm going to deal with this number on the bottom so i'm going to go for this root first so if i do the square root of 25 we'll get the answer 5. so that's the root dealt with now i'm going to move on to the other piece so i've got the flip going on all the reciprocal so if i flip five over remembering five is five over one if we flip that over it becomes one over five now the normal power there is just a one and a normal power of one let's find a different color i've used quite a lot now a normal power of one doesn't actually change anything anything to the power of one is just itself so that one over five there would be my final answer let's have a look at another one though where that normal power on the top is something different to one okay here we go so write down the value of eight over 27 to the power of two-thirds so we've got eight over twenty-seven and our power there is two-thirds so again no negative in this one so we're not gonna have to flip it but we do need to deal with these two numbers so on the bottom i have a three which is a root and that is a three there so that's a cube root and on the top we have our normal power of two which is going to be to square the numbers right so again just like before you can do this in whatever order you like i'm going to stick with the root to start with so i need to do the cube root of both these numbers it's a bit of a hint in the numbers and they're both cube numbers so the better you know you square any cube numbers easier this becomes so the cube root of 8 is 2 and the cube root of 27 is 3. so we're at 2 3 at this point now the next thing i'm going to have to deal with is this square so i need to square both of them so 2 squared and 3 squared and 2 squared is 4 and 3 squared is 9. so my final answer there is 4 over 9 and i can leave them out so just like that it doesn't simplify so that is absolutely fine as a final answer so moving on to this one write the square root of 80 in the form k root 5 where k is an integer now the square root of 80 is a good one because there are a few square numbers that go into square root of 80. so let's have a look 1 4 4 goes in 9 16 25 36. and this is where you've got to be really really careful with sir so we know that four goes in but we do have to make sure and just check is there a bigger square number that actually goes into 80. it's not a very nice one to spot with the square root of 80 but 16 actually does go into 80. it goes in five times so the other way that i can write the square root of 80 is i can write the square root of 16 times the square root of 5 just like we did before and again the square root of 16 now actually becomes the whole number four so it's four times the square root of five and again just getting rid of that times sign we can say four root five there we go finishing off our question now i think it's quite important to talk about this at this point because if we only spotted that the 4 went in at this point we might have written the square root of 80 as square root of 4 times the square root of 20 and we may have gotten to the point here where we wrote 2 times the square root of 20 and then simplified down that to 2 root 20 okay but unfortunately at this point root 20 is not as simple as not small as it can actually go um so this would actually be wrong and we'd have to continue simplifying it i'm going to move on to another question here where we look at what to do in this scenario where there is a number in front of the of a third that can be simplified okay but four root five is our final answer there okay so we've got the square root of 40 plus the square root of 90. again i like to link this back to algebra if we think about something in algebra like 3x add 4y okay hopefully we know that we can't actually add those together we just leave it as it is but if we have something like 3x add 4x they can actually combine and they can make 7x okay and the reason they can add together is because they have this common letter here the x now we're going to take the same approach when it comes to thirds if we can have the same number here underneath the third underneath the square root okay then we can actually add them together now at the moment it's a 40 and a 90. so they don't add together but let's have a look let's see if we can do anything and simplify this down let's get rid of all that okay so the square root of 40. let's have a look at that to start with that is the square root of 4 times the square root of 10 and again remember just write them down 1 4 9 16 25 we'll see if we can write enough down to start with right so the square root of 40. square root of 4 times the square root of 10. well that looks like a 6. let's get rid of that square root of 10. there we go so that is 2 lots of the square root of 10 root 4 being 2 so 2 root 10. okay so root 90 can also have a root 10 we can add these together so let's have a look root 90 is the square root of 9 times the square root of 10. lovely so we've got that root 10 going on there lovely the square root of 9 is 3. so it's 3 lots of root 10 or 3 root 10. perfect so we've got 2 lots of root 10 and another 3 lots of root 10 we're getting combined together and 2 plus 3 is 5. so we have 5 lots of root 10 in total perfect so we could add those together because we had that 10 underneath the square root for both of them okay so writing some numbers in standard form we're going to make this a number between 1 and 10 times 10 to the power of however many jumps we have to do in whatever particular direction so this number here 5 600 i'm going to hop the decimal in between the five and the six so one two three so that becomes five point six times ten to the power of it's a big number there it's not a naught point number so ten to the power of three final answer the next one is a small number so we can have a negative power because we hope the decimal the other way this time so 1 2 3 would make it 3.4 and again this time it would be times 10 to the negative 3 because it's a naught point number that indicates that it's a small number when it comes to the other way around writing them as ordinary numbers i like to rewrite this little bit the start to start with so two and three and i like to just imagine why that decimal is between the two and the three so times ten to the power of five makes it a big number so we need to hop it five places one two three four five to the right and then filling in all those zeros underneath and if it is messy just remember to rewrite that but there's my answer 230 000. the next one 8.04 times 10 to the negative three so this is going to be a naught point number and again i'm just going to write these digits out at the start these bits here and imagine where the decimal is at the top eight point zero four so negative three means it's gonna be a small number so we're gonna jump it left three one one two three so it's gonna go there fill in all the zeros and tidy it up at the start so naught point naught naught eight zero four final answer okay so looking at this one four times ten to the five times three times ten to the minus two and there's an example of one more we probably won't want to write this out as ordinary numbers and work them out because it might get a little bit over complicated here let's look at applying that same little trick so let's do the four times the three which gives us 12 and that's going to be times 10 to the power of something i'm just gonna be a little bit careful here because although we can add these powers let's just have a look because one of them is a negative so the powers if i write this to the side we're going to do 5 the first power and we're going to add the next power which is negative 2 and 5 plus negative two is five take away two so my power there is going to be three so that's fine twelve times ten to the power of three and again we just need to put this in standard form because this twelve is not between one and ten so instead we'll make that one jump smaller we'll make it 1.2 and to balance that out we'll make the power one jump bigger so three goes up to four and there's our final answer 1.2 times 10 to the four same process again we've got three divided by six though so let's do this to the side three divided by six let's write that as a fraction three divided by six if we simplify that goes down to one half so to keep this as a number rather than a fraction i'm going to write that as 0.5 so over here we have 0.5 of 3 divided by 6 and then we've got that times 10 to the power of and again subtracting the powers on these so we've got 3 take away and negative 2 on the second one and 3 take away negative 2 turns into a plus so we have 3 plus 2 which gives us a power of 5 for this one so we have 0.5 times 10 to the power of 5. and again we can balance this out because it needs to be in standard form but this time to make 0.5 between 1 and 10 we make it one place value bigger to make it five and if we make the number bigger the opposite this time for the power we make the power smaller so five drops down to a four so five times ten to the four so a similar process but here we had to make the number bigger so we made the power smaller okay here we go so we've got some x squared pieces and some x pieces now although all the pieces are x and these x squared pieces we're going to treat separate so they're just the x pieces on their own so let's have a look at the x squared pieces we've got a 12x squared and we've got a minus 3x squared so 12 lots of x squared take away 3 lots of x squared leaves us with 9 lots of x squared and i'm just going to leave that piece separate to the x parts so look at the next one we've got 5x take away 2x and just like we did before five lots of x take away two lots of s leases with three lots of x and that's positive three lots of x so plus three x and there we go we'd leave it just like that we can't join up an x squared and the next piece they stay separate okay so expand and simplify this now i've got this minus 2 in the middle here so we just need to be careful but let's just treat it in exactly the same way we'll times the first bracket by 3 to start with so 3 times 2x gives us 6x and 3 times negative 4 gives us negative 12. now let's have a look at that second bracket so we have a negative 2 at the start here so we're just going to have to remember to times by negative 2 so negative 2 times x is negative 2x and negative 2 times positive 5 gives us negative 10 okay so the symbol's not just getting copied there it is getting affected by that negative 2 at the start be very very careful with this negative 10 here when you're doing these right so collecting it all together we've got 6x take away the 2x and that leaves us with four x's and we've got negative 12 and negative 10 here negative 12 take away 10 is negative 22 and there's that expanded and simplified okay so we've got another algebraic fraction here that has already been factorized on the bottom so actually let's have a think about how we'd approach this now we want to cancel off a common bracket or a common factor and the top and the bottom both divide by x plus two so actually if we imagine what this would look like if we were to sort of expand that bracket or just write it as a double bracket with x plus 2 on the top on the bottom it's x plus 2 and x plus 2 as a double bracket there we go right so if we cancel off one of the x plus 2's top and bottom we're just gonna be careful this time because we can cancel that one off and it will disappear but not to forget we are dividing by x plus two so when we cancel the one off the top it doesn't just disappear but if we divide it by x plus two it becomes one so writing our final answer we would have one on the top and x plus two on the bottom which again i'll just get rid of the bracket there so one over x plus two so just watch out with these sorts of questions here when the factor is all that there is on the top obviously that does become a one there it doesn't just disappear and let's have a look at one more before we ever go okay so in this question we've got a slightly harder quadratic on the top and a different sort of quadratic on the bottom so on the bottom there hopefully you recognize that it's a difference of two squares when there's no x piece in the middle there it's because the numbers are the same and they've cancelled each other out so if i start with the bottom i just think it's the easier one there the factors of 16 that are the same are four and four so our difference of two squares there is x plus four and x minus four there we go and that's that factorized on the bottom that gives us that a little helpful hint for the one on the top there which is slightly harder because our factors are four one and four or two and two now because we've got the little hint here that we you know we know we're going to have a four if it's going to simplify we know it's going to be the one and the four but i just need to have a look and figure out which way it goes around so we've got three x in one bracket and an x and another that'll allow us to get the three x squared again obviously check out my video on factorizing these quadratics or harder quadratics if you're unsure on these but in order to make 11 there we're gonna want the four to get multiplied by three to make twelve so we want plus four that'll make positive twelve and we want to take away one so i'll put the minus one over here in the other bracket and again now we've got it factorized it's nice and easy for us to cancel off these x plus fours and just write what we've got left on the top we've got three x minus one and on the bottom we've got x minus four and there we go that is our fraction there fully simplified okay so this says simplify this fully so we've got eight over x plus three add three over x plus eight now what we need is a common denominator when we're adding fractions just like we're adding normal fractions so i'm going to multiply the right fraction by the left denominator so i'm going to times the top and the bottom by x plus 3 which i'm going to write in a bracket there it's going to help when i actually multiply it so times the top and the bottom by x plus 3. and then i'm also going to times the right by the sorry the left by the right denominator so i'm going to times the left by x plus eight and also on the top times that by x plus eight just like with normal fractions here so if we times the top there by x plus eight we actually get a single bracket we get eight lots of x plus eight there we go over and we've got a double bracket on the bottom we've got x plus three that's also getting multiplied by x plus eight and we are adding to that three lots of x plus three and on the bottom we've got that double bracket x plus three x plus eight there we go x plus three x plus eight right so if we go back actually expanding these numerators and we can join it all together so eight times x and eight times eight that gives us eight x plus sixty four and on our other one we've got three times x which is three x and three times three which is nine there we go and we're going to add these fractions together so let's rewrite these denominators in we've got x plus three x plus eight and another one x plus three x plus eight there we go and we're adding these together right so if we add them together we've got the eight x up the three x which makes eleven x and we've got the sixty four at the 9 which makes 53 so 11x plus 53 all over that denominator which we can expand or we can leave it in brackets if you do want to expand it x times x is x squared times 8x and 3x so we've got 11x and 8 times 3 is 24 so plus 24. well you can actually leave the bottom there factorized as x plus 3x plus 8. but we'll expand it out and leave that as our final answer there so we have 11x plus 53 all over x squared plus 11x plus 24. okay so this question here we've got quite a lot going on we've got one fraction divided by another and it says simplify it fully now if we have a look what we've got to do um we know when we're dividing fractions we have to keep the first one and times it by the reciprocal so flipping the second one over so we can just write that out straight away if we want i'm just going to factorize any in the process so i'm just going to do that to the side well here before we actually write anything down so that factorizes by 3 and that would leave you with x plus 2. the bottom there doesn't factorize that'll stay as it is so we can write this out now so we've got three lots of x plus two all over x minus four we're going to times that by the flip version but let's see if we can factorize them first so the bottom here is quite a nice one factorizes by x and we get x minus four so that's going to now be on the top when we flip it over so we have x brackets x minus four and this one here we're gonna have to factorize this as well so we've got one and ten or two and five let's have a look now one of them is going to be a 2x because it's 2x squared and one's an x right so let's write these down we've got one and ten or two and five and we're trying to make nine so let's have i think we could double the two there we are that works so double the two so two in the other bracket plus two that would make four and plus five over here and four plus five would make nine so we can put that on the bottom now so we've got two x plus 5 and x plus 2. there we go now if you think about what we'd have to do here with times and fractions if we're times fractions which is times the top times the bottom so essentially everything there on the top is going to be on the top of our fraction and everything on the bottoms of those two fractions are going to be everything on the bottom so what we can do at this point we can actually cross cancel some of these if you have a look we've got an x plus two there and x plus two here because obviously if we if we actually did multiply these together one would be on the top one would be on the bottom so we can cross cancel them before doing so so they're gone we can also cancel this x minus four uh so that's gone as well obviously on the bottom there that's the only thing there so that will turn into a one and everything else has something with it so if we times together what we've got on the top now we've got a 3 and an x on the top of those two fractions so 3 times x is 3x and the bottom there we've got 1 multiplied by 2x plus 5 so that's just 2x plus 5. there we go so that simplifies all the way down to that okay so obviously just have a look when you're doing one of these dividing fractions or any of these algebraic fractions look to factorize anything you can and just follow normal fraction rules so here we factorized everything and then times it by the reciprocal flipping it over and just cross cancelled any common brackets there before multiplying them together just to save us having to write it out and then cancel them off okay so we've got these word questions here and you might recognize some of these questions as well in terms of the style that you might have seen these come up in little exams or little tests that you've done so we've got amy has some sweets bethany has twice as many sweets as amy and charlie has five more sweets than bethany in total they have 55 sweets how many sweets does charlie have and at first these sort of worded questions can seem a little bit confusing it might seem that you have to just apply a bit of logic or guess some numbers but there's a nice way that we can do this by forming and solving an equation now what we're going to do is we're going to allocate the letter and we may as well use the letter x as we've been using that for the rest but we could use any letter if we wanted we're going to allocate the letter x to one of these people so we have amy and i'm just going to write these to the side we have amy we have bethany and we have charlie now if we look at our statement here one of these people we haven't been told anything about so this is amy has some sweets it says bethany has twice as many as amy so we've been told something about bethany she's got twice as many as amy and we've been told that charlie has five more sweets than bethany they've been told something about charlie as well but the person that we haven't been told anything about there is amy okay amy's kinda like our little odd one out there we've not been told anything about her so if we look at amy and we're gonna just give her that letter we'll call her x okay we don't know how many sweets amy has but we know how many sweets everyone else has in relation to amy so look at that first line it says bethany has twice as many as amy so if bethenny has twice as many she has basically double what bethany has so we could call bethany 2x okay and 2x just means 2 times x whatever amy has 2x would double it so there we go there's our expression for bethany now if we look at charlie it says charlie has five more sweets than bethany so charlie has five more sweets exactly five more than what bethenny has now bethenny's was 2x so we're not gonna times it by five it's just five extra sweets so charlie is gonna be bethany's 2x plus an extra five okay obviously if i had five sweets less than bethany we'd have 2x minus five but it's five sweets more than bethenny so there we go we've got two x plus five right so we've got our three expressions for each person and now we can actually have a look at forming our equation just as a little side note obviously some of these descriptions could be given to you slightly differently we could have been told that charlie or somebody had five more than amy and then so and so had twice as many as that person and that would just mean doing it in a slightly different order for example i mean if it said charlie had five more than amy then charlie would instead be x plus five it might then say that bethenny had twice as many as charlie and i realized i'm saying quite a lot of different words here but then we would have to double the x plus five so we could have these in different orders but this is just one type of example and one type of question so obviously what we need to do is now think about how we're going to form our equation we've got the expressions for each person and now it says in total they have 55 suites or in other words if we add together all these expressions they are going to add up to 55. so if we add them all together we've got x 2x and 2x that's 5x and we've got the plus 5 so we've got 5x plus 5 and that is going to be equal to the 55 sweets so if we write our equation then so 5x plus 5 has to equal 55 and then we can just go ahead and solve that so take away the 5 from both sides we have 5x equals 50 and then we can divide by 5 and we get x equals 10 and then we can use that x equals 10 to find out how many sweets they all have and we can just use it based on the expressions that we've got over there we now know that amy is x so amy's got 10. bethany has double that so bethany's gonna have 20 and charlie has five more than bethany so charlie will have 25. and just double checking there they all add up to 55 and they do so in terms of this question how many sweets does charlie have charlie has the 25 that we've got just there right there we go so that's how to go about some of these questions and i think it's good for this question here although we could have just substituted 10 into charlie's expression end and 2 times 10 plus 5 i always think it's good just to find out how many all of them have or whatever the question's about and then just pick the one that you need for this particular question so as it's asked about charlie obviously we only need to give the 25 there as our answer okay so this inequality here we've got 6x plus 1 is greater than or equal to 2x plus 11. so obviously we've got an x on both sides here and i'm going to treat that just like when we have equations with x's on both sides i'm going to eliminate the smallest x from both sides to start with so the smallest value of x is that 2x there so i'm going to minus 2x from both sides which is going to get rid of it on the right and it's going to reduce this 6x on the left here so i'm just going to write my working out it's going to write minus 2x here from both sides and let's see what we end up with so 6x drops down to 4x we've still got the plus 1 and that is now greater than or equal to positive 11 so i won't put the plus in if obviously if that was a minus there i'd have to put that in but it's no need for me to put plus 11 there now we can go about taking away the one from both sides okay so just like on the previous question we'll get rid of the one we want to get to it so it says x is bigger than or equal to so we're going to have 4x now is bigger than or equal to 10 and then we need to divide both sides by four and obviously that's going to leave us with a decimal here because 10 doesn't actually divide by four perfectly so if you have a calculator you can type that in but if you don't let's have a look let's do this to the right here we're going to have x is bigger than or equal to and i'm just going to write it as a fraction 10 divided by 4. there we go so 10 over 4. now obviously if you've got a calculator you can just type that in but if you don't we just need to actually simplify that as a fraction so if we divide the top and bottom by 2 we get 5 over 2 and that's always quite nice 5 divided by 2 is 2.5 so we've got x is greater than or equal to 2.5 okay just remember obviously that's nice and easy to simplify 5 over 2 2 goes in twice with a remainder of 1 so it's 2 and a half or 2.5 so that's not the nicest to draw on this number lines it's going to be hard for me to get perfectly on 2.5 but i'm just going to do my best i'm going to put it in the middle of two and three and this time it's greater than so i'm going to point it to the right doesn't matter how long the arrow is there and obviously look we've got our extra ink on the inequality there so it can also be equal to that number so i'm just going to color that circle in there we go and that just shows that it's a greater than or equal to the extra ink on the arrow extra ink in the circle and there we go that's just obviously how we do that we've got x on both sides we'll just treat it in the same way hopefully you notice just by looking at it when it says x squared minus 2x equals 15 this question here doesn't equal zero or it doesn't equal zero yet so before we can actually factorize this and solve it we need to move that 15 to the other side that's easy enough we just have to subtract 15 from both sides and that will give us equals zero on the right there and if we do that and i'll slot the 15 right at the end we x squared minus 2x minus 15 is equal to zero there we go and it's worth noting at this point as well to be fair that 2x could have been put on the other side it could have been 15 plus 2x over here and we had to subtract the 2x as well now for the purpose of obviously this question we're just going to have a look at obviously a nice simple one but to be fair we could put as many pieces as we want on the right hand side it always needs rearranging to make it equal zero but we'll get rid of that obviously because the 2x was on the left and now we can just go about factorizing it and solving it as normal so for 15 we can have 1 and 15 or 3 and 5 and there are only options for this one and we're trying to make negative 2 there so we're going to have to use 3 and 5. so we want positive 3 take away 5 and that gives us negative 2 in the middle there so if we put that into a double bracket we get x and what do we say positive three and x negative five or x take away five there we go that's equal to zero always leave that equal zero in there don't forget it is an equation that we're solving so we don't want to get rid of that and then we can get our two solutions i'm just going to apply the little trick this time it's plus 3 in the bracket so that's going to flip to negative 3 so x equals negative 3 and for the second one there we have negative 5 in the bracket and again if we flip the sign for that we get x equals positive five and there we go there's our two solutions x is negative three and x is positive five and again just very briefly in terms of what the graph would look like there it would go through a negative and a positive solution i'm not going to do it to scale but it would look something like that that would allow it to go through a negative over here at minus three and a positive over here at five and that's all we're actually working out that's what we're solving these two solutions here i'm just thinking about what the graph would look like so i've got a couple of questions for you to have a go at obviously the main priority here is actually getting to these solutions x is negative 3x is 5. but if you want to do a little bit of an extension on top of that you could always think about drawing the quadratic curve as well and actually thinking about what it would look like and i think that's a nice little extension for this topic because okay so find the equation of the tangent to the circle with the equation x squared plus y squared equals five at the point one two now you don't have to have a picture for this it can sometimes just help just to visualize it to just draw a little sketch they're all centered at naught naught so we can just draw a little sketch it doesn't matter if it's any good or not one two will be in this quadrant here and then we can draw part of the tangent there and we can just label it up so one two obviously center naught naught there we go we can use that to try and help us visualize this so we are going to find the gradient of this radius here which we can do by drawing a little right angle triangle in so the rise goes from naught up to 2 and the run goes from 0 to 1. so we can just use that to find our gradient now so the change in y over the change in x is two over one there we go that's quite a nice one because that's just two so our perpendicular gradient doing our negative reciprocal of two remembering again that two is just 2 over 1 so it becomes 1 over 2 and swapping to a negative so there's our gradient and again from here it's pretty much the same steps as on the previous video there so let's just have a look we've got y equals minus a half x plus c and then our coordinate is one two and if we sub that in let's see what we get we get two equals minus a half times one which is just minus a half so i won't bother writing that in plus c add the half over we get two and a half equals c and that's fine to leave it as a mixed number and now we can just write our equation just replacing c with two and a half so y equals minus a half x plus two and a half and again you can write these as decimals if you want you could write minus 0.5 x plus 2.5 okay likewise you could even write two and a half as five over two you can leave it at the top of your fraction if you want to improper fraction absolutely fine for you to do as you can see that one there's a little bit nicer than the last one okay so just showing you that they're not all that nasty um still obviously a difficult topic but that's a nicer version of one of these questions okay so our first question we are starting with a quadratic now we know that a graph is a quadratic if it has a little squared in it so this particular graph has x squared and this is quite a nice one it's just x squared take away four now normally you're asked to complete a table and then also draw the graph now the my main priority for this video is going to be on actually getting these tables completed i like to hope that we can all draw these graphs once we've got the coordinates so i'm not going to stress too much about the actual drawing here in fact the actual graphs that i've got on the screen are quite small and mine aren't going to be perfect i'm just going to give you the general idea of how to go about drawing all these graphs here so for this one here we've got y is equal to x squared take away four now that obviously means just to find the y coordinate we square the x coordinate and then take away four so it's quite hopefully quite nice and simple we're gonna have a look at a trickier one after this we've just got to be careful with how you actually go about doing this so it says uh for all the values of x from negative three to three now if it doesn't give you that information you can find them along the x axis here if you have a look the x values stretch along from negative three along to three there so that's where we get them if it doesn't tell us okay or if we're not given a table but for all of these i am going to give you a table now obviously we are starting with this one on the left so we've got negative 3 there and we need to sub that into our equation so to find y i'm just going to write that working out we would do x squared so negative 3 squared and we've got to be very careful when we're writing negative 3 squared because particularly if this is a calculated question if you type that into the calculator you are going to get the answer negative 9 because what the calculator does is it squares the 3 and then puts the minus with it it doesn't actually do negative 3 squared so a really important little thing that you need to remember throughout this is that we're just going to put all our numbers in brackets like that okay so we're going to write minus 3 in brackets squared and then we're going to take away 4. of course if this is non-calculator we can work that out anyway because we can do 3 square negative 3 squared which is positive 9. so we've got 9 take away 4 and that gives us the answer 5 so we know that number there is going to be 5. we can then move on to the zero and again i'm not going to write all the working out down for every single one here but we would just do zero squared again we'll put the zero in brackets although it doesn't matter if the number's not negative and then take away four and zero squared take away four is minus four so we can just put minus four straight in then the rest and this next one here i'm going to do without writing the working out down we've got one squared which is one take away four which is negative three now it's not that easy to spot but obviously when it comes to straight line equations there's always quite a nice pattern in the table when it comes to quadratics there's not the nicest patterns okay and the others aren't going to follow a pattern but there is actually a bit of a pattern in a quadratic if you have a look we've got minus 4 in the middle and either side of that we've got negative 3. either side of that we've got zero and either side of them we've got five so there is a bit of a pattern although i certainly wouldn't rely on these because they don't stay there throughout all of the graphs and sometimes it's very difficult to spot some of these patterns but just thinking about the fact that there is a bit of a pattern we're looking at these so let's have a look then actually plotting this on the graph now we've got an x and a y coordinate there that little bit there x is negative three y is five so that's the coordinate minus three five now i'm going to plot these quite quickly minus three five is somewhere up here i'm not gonna be able to do them perfectly on the screen here then we've got negative two zero we've got negative one to negative three obviously getting your x and y coordinates the right way around here zero to negative four and then it repeats the pattern going up the other side there so we've got one negative three we've got two and zero and we've got three and five which is somewhere up there right so in terms of actually drawing the graph here we've plotted the points we now need to get a nice smooth curve that goes through all of the points so not with a ruler a nice smooth curve and it has to go through every single point so i doubt i'm going to get this perfect i'm going to give it my best shot i'm just going to do a nice smooth curve no little kinks in it i actually can't see them as i'm doing them there we go so i've done it as best as i can but you want to get yours nice and smooth okay so make sure you do it with a pencil rub it out if you have any um any kinks like i've got down the bottom there but then you can go back and have a go and actually rub it out and just make sure you get a nice smooth curve there right so that's what a quadratic graph looks like we're going to have a look at a slightly harder one before we move on there there are a couple of things to know about this quadratic graph and a couple of the points that are important we're not going to look at them specifically in this video but there are two places which are well technically three but there are two things that are quite important with a quadratic graph one of them is this point right here at the bottom where it sort of goes starts stops going down and starts going up that can be called a couple of different things it's got a specific coordinate there it's zero minus four it can be called the turning point there we go or it could be called the minimum point or the maximum point if it's a sloping another way so the minimum point or turning point there we go the turning or the minimum point that's a very key place on a quadratic graph and there's also two of the points here which is where it crosses the x-axis so we've got these two points there which is negative two and two let's just write that in there minus two and two and these have a special name as well they've got two names again they can be called the roots there we go or they can be called the solutions there we go roots or solutions so we're not going to focus on those points in this particular video but obviously just a couple of extra little bits to be thinking about there the roots and the solutions where it crosses through the x-axis and the turning or minimum point down the bottom obviously just adding a little bit of context to the language there the turning or the minimum point seems like quite an obvious place so just give that as a coordinate zero four but then the roots okay doesn't always necessarily have a bit of a logic there but i do like to think of this x axis as being ground level when you think about plants and things like that when they meet the ground is where the route starts so i do like to think of these sort of things as like the little roots there and that's sort of where the word comes from but you've also got the word solutions so there's no really any context there but it's just where it meets the x-axis there okay so what we're going to have a look at is some equations of lines where it isn't in the form y equals mx plus c so whenever we're looking at lines we need it to be in this form here now it can be written in other forms we can use them perfectly fine but in order to compare them and in order to find out the gradient it needs to be in this form here so looking at this question it says here are the equations the four straight lines two of the lines are perpendicular which two now if we look at this first one it's not in the form y equals mx plus c okay it has a two in front of the y so in order to get it into the form y equals mx plus c we're gonna have to divide everything by two okay so i'm just to write divide by 2. i'm going to divide both sides by 2. so 2y divided by 2 gives us 1y we've got 1x there 1 divided by 2 is a half so that 1x would become a half x and the one divided by two at the end would become a half so that's plus a half there you go half x plus a half so there's my first one let's have a look at the next one now this doesn't have anything in front of the y but it does need rearranging look the x there is on the left hand side of the equal sign so if we add or plus 2x to both sides we would get y equals and just to keep it so the x parts here although i don't have to i'm going to stick it in front of the 4 so y equals positive 2x so as it's in front of the 4 don't have to put the plus sign there we go we get 2x and then we do have to introduce the plus sign for the 4 there so plus 4 as it is positive 4. there we go 2x plus 4 and that's my second one let's look at the third one we're going to do for this one okay so we've got a 3 in front of the y so we need to divide everything by 3 so that it says y equals don't have to rearrange it at all but dividing everything by 3. so dividing that by 3 we get y equals 6 divided by 3 is 2. so 2x 6x divided by 3 is 2x and then add 5 and 5 divided by 3 let's leave that as a fraction 5 over 3. that just means 5 divided by three there rather than trying to actually work that out so five over three is fine as we are also looking at the gradients here to decide which two are perpendicular because that's what we're gonna have a look at which ones are perpendicular we don't really need to worry about the wine to set there okay but there we go let's have a look at this last one okay so uh nothing in front of the y but the 2x is in the way here so we need to minus 2x from both sides to get that out of the way and again just like before i'm going to put it in front of the three so y equals we're subtracting 2x so minus 2x there we go and then that is positive 3 again so plus 3. right so we need to have a look at which ones of these are perpendicular now as we've got and let's see if we can highlight this we've got a 2x and a 2x here so those two lines there are parallel not asking for which ones are parallel but they have the same gradient so we could say that those two are parallel so let's have a look at the other two so we've got these two here so the gradient of the one at the top so the gradient of a is a half and the gradient of d down the bottom there is negative two and if we think about this how do i if i if if half is my gradient my negative reciprocal would be negative two wouldn't it you'd flip it over it'd be negative two over one which is negative two so therefore those two lines must be perpendicular okay so it's just something to be thinking about in terms of different line equations what they can look like and the way we want to have them we want to have them as y equals we don't want any numbers in front of the y if we do we're going to divide it and we if there's anything next to the y we just want to move over to the other side okay but they're a and d there and my two lines are perpendicular but we're going to be having a look at some line equations next where they might not necessarily already be in the form y equals mx plus c and that just hints to you that you do have to change them first okay so this question says carol runs a race the graph shows her speed in meters per second t seconds after the start of the race calculate an estimate for the gradient at t equals four and explain what your answer represents now obviously this type of question we already know now what the answer represents if we find where t equals four which is just here there we go we obviously know that once we work out that gradient that is going to show her acceleration so let's just write that in straight away that shows her acceleration all right there we go so let's work out the acceleration at that point or the gradient at that point so t equals four again this is not the nicest for me to draw here but i'm going to do as best i can right there we go to be honest actually looking at that i'm not going to keep drawing this over and over again but i'm not actually very happy with the fact that there's quite a big gap here and there's not really much of a gap on the right hand side there so to be honest if i had this on paper right now with a ruler on the pencil i'd be rubbing now and doing it again but as we're just running through the process here and having a little practice i'm going to leave it as it is i'm just going to go ahead with it and see what we get but obviously if you are following along with me on the paper with this uh obviously trying to make yours a little bit more accurate than mine there now it looks like i've got quite a nice triangle that i can draw in i'm going to go across from here to here and again you can do this however you like you can make it whatever size you like you just need to read accurately but i've just gone along from three to six there and i'm going to go up and i look like i get some quite nice numbers here so i'm going to stick with this now the run there goes from three which is just here where the start of that orange line is along to six so that is a run of three and it goes at a height from seven up to ten if anything maybe it's just over ten but seven to ten there we go that gives me a height of three as well there we go so in this case i've got a rise over runoff or a change in y over change in x i've got three divided by three three over three and that gives me an answer of one there you go meter per second squared all right there we go so there is an answer for this one here now i'm just going to do that one more time i'm going to i'm going to do this quite quick i'm going to see if i get the same answer there we go because obviously i don't think it was super accurate so i'm going to have one more go so let's have another look okay let's just draw as best like that is even worse than the last one let's have a go one more time okay maybe one more there we go i'm a lot more happy with that one let's see if this works very difficult to do without a ruler there we go let's have a look so let's go across i'm going to make it a little bit smaller my triangle this time i'm going to go from here to here and then i'm going to go up there we go so my run there goes from four to six so that is a run of two and my rise there goes from 8 to 10.2 by the looks of it so that has a rise of 2.2 right there we go so in that for that working out there for the second time i've done it my rise is 2.2 the run is 2 so 2.2 divided by 2 gives me 1.1 meters per second squared and there we go so a slightly different answer there so you can see how obviously just changing the gradient slightly is going to slightly change your answer but these sorts of exam questions you are going to have an allowed amount of range there so it's going to be somewhere between 1 and 1.2 or something like that okay but obviously you just want to do this as best as you can make sure you show you're working out make sure you draw your tangent in nice and clear make sure you show where you've got your numbers make sure you show your rise over run write down change in y over change in x show those numbers okay and then even if your tangent isn't great you're not going to lose you're not going to lose more than one mark on a question like this but obviously just a little bit practice getting those tangents perfect okay so here's quite a different question it says the diagram shows a pattern made by two regular polygons work out how many sides tile a has and what's not very nice about this one is it doesn't show you um all of the sides here obviously that's why it's asking us how many sizes it just shows us part of tyler and then we've got the equilateral triangle there so have a go at this one see if you can work it out i'm going to go over the answer but see what you get see if you can figure out how to apply some of the methods we went over in that first um that first slide there and so you can get the answer so pause the video have a go we'll go in a sec okay so looking at this question then now first things first let's have a look at the angles around this point right here now we know in the equilateral triangle that's an angle of 60 degrees and if tile a is the same on both sides then what we can do is take that 60 away from 360 and split it in half so if we do that so 360 take away the 60 degrees in the triangle leaves us with 300 and if we divide that by 2 we've got 150 degrees in each of these tile a's here so 150 degrees and 150 degrees so what the question's now asking us is what polygon or how many sides does a polygon have to have to have that interior angle of 150 degrees when it's a regular polygon so two different ways we could do this now we could just guess i could just say okay let's just guess 14 sides and then if it is 14 sides i'll take away 2 times 180 divide by 14 and see if it matches and if it's too high i'll go i'll go lower and if it's too low i'll just go higher with my sides okay but we can take a little bit of an easier approach here now if you remember on that first one when we looked at the polygon if it had an interior angle of 150 we could have a look at what the exterior angle is and because it's on a straight line the exterior angle would be 30 degrees so if you think about this if you remember that all the exterior angles add up to 360. well if one of them's 30 obviously 150 and 30 make 180 then i could just say all right well if i do my 360 and i divide it by the exterior angle there which is 30 that gives me a total of 12. there we go and that tells me that there must be 12 angles around the polygon because obviously that 30 fits in 12 times so there we go working that out then we've got 12 sides as our final answer and again you can check that if i was to do um 12 take away 2 that would be 10 times 180 and then divided by 12 and you get the answer 150 so you can always check that out okay so this question here is not a right angle triangle it is a type of triangle though it's an isosceles and that is shown to you via these two lines here drawn on those two lengths obviously if we had another one on the bottom it would be an equilateral but you can already see it's not an equilateral as the side lengths are different but this is an isosceles triangle meaning both of those diagonal lengths there are both 17 matching this one just here now again with a triangle like this we're just going to use the same formula and that is that the area is equal to half base times height there we go which again you could write in a different way if you prefer you can write down base times height or bh divided by two so for this one here then we just need to identify the base and the height the base is 16 and as you can see the height is 15. so for this particular one we're going to need to do 16 times 15 and again ignoring that diagonal length so if we write that down we've got to do 15 times 16 now obviously that's probably not one that you're going to do in your head but you could obviously just write you're working out to the side now the answer for that comes out as 240 and we need to actually halve that so 240 if we divide that by 2 gives us an area of 120 and again we would write that with centimeters squared so it's 120 centimeters squared now particularly when you've got larger numbers like this you could save some time by using the first formula and that is to halve the base before you multiply it by the height so instead of doing 15 times 16 and dividing by 2 we could have just halved the 16 and an 8 times 15 instead and then and that as well would give us 120 straight away and we wouldn't need to halve the answer so up to you which method you use either one is fine but that is how you work out the area of a triangle you do the base times the height and divide it by two or you just do half times the base and then times that by the height okay so when we're looking at the area of a sector and particularly when we're doing it in terms of pi some of these questions can be easier sometimes they can be slightly harder but this one is sort of in the middle so it says here that the diagram shows a sector of a circle radius seven work out the area of the sector and give your answer in terms of pi so all we're going to do is do the working out as we would when we were using a calculator we're just not going to actually type it in so to start with we need to know our area formula so the area of a circle is pi r squared when we are looking at a sector we also multiply it by the fraction that we're looking at which i'll call theta over 360. so in this case the angle is 40 degrees and the radius is 7. so if we plug that into our formula there we have pi multiplied by 7 squared multiplied by 40 over 360. now we obviously want to simplify this a little bit so we get so we can get a bit of a simpler answer that we can give in terms of pi the first thing i probably simplify is the 7 squared so that's going to be pi times 49 which you can write as 49 pi and i'm going to multiply that by this fraction i want to simplify that as well so if i do that to the side 40 over 360 for starters you could divide them both by 10 which would give you 4 over 36 then of course both divide by 4 so that become 1 over 9. so we're going to times that by 1 over 9. now the easiest way to go about this is to just simplify those numbers so 49 times the ninth just means 49 divided by nine now that doesn't actually simplify it doesn't divide perfectly so that would just be the fraction 49 over 9 so i'd just say that this is 49 over 9 and put the pi symbol after it and that would be my answer there in terms of pi of course if they did divide you could simplify that fraction further giving it as a whole number but of course as it doesn't and we're writing it in terms of pi it's fine for us to leave it just as it is in a fraction let's work out the total surface area of the cuboid now when it comes to a cuboid and we're working out the surface area we need to remember we are going to work out the area of each individual surface and when it comes to some of these shapes you can't always see each of those surfaces now if we think about what we can see we can see that we have the face on the front and the same face that's on the front which i've labeled number one is also on the back so we're going to have two of those rectangles there for those two surfaces if we also think about what else we have we also have the shape here on the right which is another rectangle and that is also going to be on the left hand side of the cuboid as well even though we can't see that one and for the final shape that we have we have the rectangle that's shown on the top and that is face number three and that is the same as the face on the bottom so when it comes to a cuboid we actually have two of each face and when it comes to a cuboid we also need to know how to work out the area of a rectangle as all of these faces are just rectangles so when we're looking at each one we just need to identify the side lengths for each individual rectangle work out their areas and then we'll combine them all at the end so if we start with face number one if we have a look the width and length in either order is eight and ten so to work out that area we would just do eight times ten and that would give us an area of eighty and we will worry about the units at the very end now again that is the same as number one so number one is also eighty so we can label that for both of them we're going to then move on to another rectangle and as i've already labeled it number two we'll start with that green one so if we look at that one there the length of that rectangle down the bottom is five now the actual width or the height of this cuboid we can get from the other side and that is a height of eight so we have a height of eight centimeters so to work out that area there we would do five times eight so just make sure that you get those lengths from wherever you need them so if i write that down that is going to be 8 times 5 which is going to be 40 and that is another one of our areas again we can label that with our other side which is also 40. and on to our final one we've got the rectangle on the top on the bottom now again we're going to have to label both of these but if we look at the top we have the area and let's just get rid of that we've got five down the bottom and the other area the other length is 10 just down here so if we label that up near our rectangle this is 10 and this is 5 so to work out the area of number 3 we would do 5 times 10 and that is equal to 50 and again that is the same as the one down the bottom which is also 50. so again you might have noticed that in all of these multiplications that we've done to work out in each individual area each one has a different pair of numbers because all of the lengths on the cuboid were different numbers so for the top for the middle one there the front face we had eight times ten for the side we had eight times five and then for the top we had five times ten so it's up to you in terms of which process that you do when it comes to a cuboid what you can do is just work out the area of each individual face the three that we worked out you can add them all together and then double your answer or you can obviously work out all of the individual faces even though there are matching pairs and then add them all together so completely up to you but if we go ahead and add them together now we have 80 and 80. we then also have our green face that we worked out which is 40 and 40 and then our final two we have 50 and 50. and all we need to do is add all of those together so let's add them all up so we've got 80 plus 80 which is 160 plus 40 gets us to 200 plus the other 40 gets us to 240 and then we've got an extra 100 there which is 340 in total and now we want to give our units it was in centimeters and we're looking at an area so it's centimeters squared so our final answer for this question would be 340 centimeters squared now when it comes to a cuboid and we're working out the volume what a lot of people will do is they'll just multiply all the sides together that is absolutely fine that will work but it doesn't really help with the understanding of how to move through all the different shapes so what's important to note is that when we work out the volume of a shape we work out the area of what's called the cross section and then we multiply it by the depth or however far that goes through the shape and in regards to a cuboid the cross section can actually be any of the faces now all that the cross section is it's just one of the faces that goes or extends all the way through the shape now in regards to the cuboid here as it's the one facing us i'm going to look at this one as the cross section if you kind of imagine slicing it if it was you know something that you were able to slice down the middle that cross section would stay constant all the way through the shape here going in this direction which in this case will be the depth but as i said when it comes to a cuboid to be fair you could actually look at any of the faces as the cross section take the right hand side for example this one here we could actually slice it down that way and that face would also go through the shape in this direction if i can try and draw that in okay but as it stands i'm just going to have a look at the front face there and we're going to use that as the cross section now as that is a square sometimes most of the time that's going to be a rectangle but in this particular one here they're both 20. to work out the area of that cross section right there when it's a rectangle or a square we just do the length times the width so the first thing we work out is 20 times 20 which is 400 obviously that's an area there so our units would be centimeter squared but we're not going to worry about the units until the final step here when we work out the volume now the next pro part of that process is obviously to multiply that by the depth and in this case the depth how far it goes back this distance here is 40 centimeters so we're going to do to finish this off to get our volume is going to take the 400 the area of the cross section and multiply it by the depth which in this case is 40 and that gives us 16 000 and that'll be centimeters cubed okay not forgetting that when we have a volume our units there are going to be cubed or potentially when we're looking at liquids we might have something like liters or milliliters so that is how you work out the volume of a cuboid okay so this question says a rectangular frame is made from five straight pieces of metal work out the total length of the metal needed now in order to do this you can obviously see that the rectangle around the outside we have 12 on the bottom and five on this side so they're going to be matched up on the other side this will be 5 and this will be 12. we just need to find that diagonal length now obviously that means we're going to have to use pythagoras as we're finding a missing length in a right angled triangle and we've got the other two lengths and obviously pythagoras can appear in lots of different formats this is quite a nice one though so if we find this we can do and we're finding the longest length so we're going to square the sides and add them together so we've got 5 squared add 12 squared and that is 25 plus 144 which is a total of 169. now when you are doing pythagoras regardless of whether you're finding the longest side or the shortest side you want to square root your answer and if we square root 169 it's not one of the most commonly known ones to be done without a calculator but the square of 169 is 13. so that's going to come out as 13 and of course we're looking at meters with this particular question that's going to be 13 meters of course this was five and twelve so now we all we actually have to do is add them together up to you how you add them i'm going to add the two twelves to start with which is a total of twenty four then i'm going to add the two fives which is a total of ten just to make it a bit quicker to do the addition and then we're going to add the 13 into that and just add them all up so four and three is seven and then two three four so we get 47 meters in total and that'll be the total length needed to make this particular frame so there we go a little bit of pythagoras involved in a slightly different type of question okay so find the exact value of tan 30 times sine 60 and give your answer in its simplest form now if you had a question like this obviously we'd have to start drawing out our triangles and work out what tan 30 and sine 60 is quite a nice one because you only have to use the one triangle for 30 and 60 and if you can remember some of them you actually can just get away with doing half of the triangle which is what i'm going to do and we've got 2 1 and root 3 and again we just need to read these values so i'd be writing down soccer terror again just to make sure that you know which ones you're reading so there we go and let's just get these two values from the triangle so tan 30 is opposite over adjacent so we're looking at the 30 opposite the 30 is 1 and adjacent is root 3 so tan 30 is 1 over root 3. there we go there's your first one and the next one is sine 60 so moving on to the sine 60 looking at the 60 angle opposite over hypotenuse opposite is root 3 hypotenuse is 2 so it's root 3 over 2. there we go and it asks us to multiply these together so we've got to do is multiply them together and see what we get so on the top 1 times root 3 is root 3 and root 3 times 2 is 2 lots of root 3 so 2 root 3. there we go and that is finding the exact value there just say that give your answer in its simplest form and that's because of the top and bottom here divides by root 3 so we can divide the top by root 3 and we can divide the bottom root by root 3 and that'll simplify the fraction just like a normal fraction so root 3 divided by root 3 is 1 and 2 root 3 divided by root 3 is 2 so our final answer there would be one half okay there we go so it's quite nice and quick to find them once you know those triangles i think it's quite a nice visual way of actually just being able to find them quite quickly um but you have got to actually know how to draw those triangles and understand where these values actually come from okay so looking at this question it says b is the midpoint of ac that's this one here so it says that this is a midpoint now straight away if you have a look on that line look a to b is the vector a and if that's the midpoint of b to c then this here on the other side has to be exactly the same so we can label that a straight away so that's another array it says m is the midpoint of p b let's find that that's here m is the midpoint of p b and if it's the midpoint it's not a ratio this time but a midpoint means it's in the ratio one to one or one half and the half here and just also taking note of that that line there bp you can see doesn't have any vectors on every other line has a vector on the b to p or p to b whichever way we look at it has no vector on it so that line there is going to be one that we're going to have to find how to move up and down it says show that nmc is a straight line so if i draw that in let's have a look nmc quite hard here without a ruler but nmc there we go that's the line there we're trying to prove is a straight line now when it comes to these types of questions we've got to show it's a straight line we just need to have a look at this green line here so there's three vectors i can have a look at on this green line i can go right from the start from n all the way up to c that's a vector i can have a look at into c or i could do part of the line i could go from n just to m that wouldn't get me part way along the line so i've got n to n or i could go from m that line there with the cross on up to c there are three three possible vectors there that we could have a look up now when we're trying to prove something's a straight line i only have to find two of these and it tends to be that this one here is the easiest to find the full length of the line and i always look for that one first then i'll just pick and choose between one of the other two and you're going to see we're going to be able to prove something straight line because there's going to be something that is similar within the vectors there when we actually find them okay we'll go about actually just trying to work them out first and then we'll see at the end how it proves it's a straight line but end to see that's quite a nice easy one actually because i can go from n here i can go backwards up this 2b so i'm going to write this down the vector n to c so i can go minus 2b let's have a minus 2b because i'm going backwards through that 2b and then i can move all the way down the line here from a to c and that's two of those a's there so plus 2a there we go and there's the vector n to c i'm just going to rub these lines out let's get rid of those okay so that's my vector there and again i can rewrite that if i want to get rid of the symbols i can write 2a minus 2b okay so now we have the vector 2a minus b that is from n to c now that's one of our vectors done so let's tick that off one of the three next let's have a look at one of the other ones and it doesn't really matter which one we pick n to m or n m to c it doesn't really matter let's just give the second one as it's the second one that i wrote and we just need to think how we're going to get from n to m there okay so we have two options either we could go down here and then up there or we could go from n to a a to b and b down to m all the way around now that's three parts there so i'm just gonna i'm just gonna ignore that one i'm just gonna go down and up okay so to get from n to p that's okay that's just down this little vector b here but then i have to work out how to get from p to m now in order to get from p to m okay this part here i need to be able to move up the full length of the line so i need to know what the full vector there for that line is and that line is the vector p up to b so i need to work out p to b now so let's have a look at how we could go about getting p to be okay and then we'll think about adding all these up together in a sec so to get from p to b let's write this one down and write that down over here p to b i have to go up here from to a and then from a over here to b so to get from p to a let's have a look we have to go backwards through this b and backwards through another two b's so that's gonna be minus three b's so minus three b and then plus this a up here so plus a so minus three b plus a okay now that gets me from p to b now we don't want to know p to b do we want to know how to get from p to m which is going to be half of that okay so to get from p to m i'm going to have to halve that there we go because m is the midpoint we only want to go halfway up that line half of that vector so we just need to do what we did before we'll stick a half outside the bracket i'm not going to rearrange it here to to make it a minus 3b i'm just going to do it as we as we've done it here so minus 3b plus a i'm going to do half of that okay so if we expand that let's have a look a half times minus 3. remember you can always think of these as fractions over 1. so a half times minus 3 is minus 3 over 2 so minus 3 over 2 b plus and a half times 1a is a half a so plus 1 half a there we go and that's our vector p to m now we're almost done because if we can get from n to p which we can we just go down a b now we can get from p to m using this vector here minus three b plus half a so if we have a look at this one here let's label this up so we're now going from n to m the full the full bit there and in order to do that we do one b so b and then add to that this vector here that we've just worked out okay so we're going to bring that down here okay literally the one we've just worked out so b and i'm gonna put the the half a next to it let's have a look in fact no let's put the minus three so it's not gonna be plus because it's minus three and a half b so we could put add minus three and a half b i'm just gonna get rid of the plus and just put it in there just to avoid having extra symbols here so it's b minus three and a half b plus a half a there we go right so again we just need to simplify this down we're almost done we just need to simplify this and if we have i think let's have i think the b's there in terms of halves so one b is two halves b and two halves b take away three halves b is gonna leave us with minus a half b okay one take away one and a half is minus a half so if we work that out we've got minus a half b plus a half a that half a that we've got there and that's our final vector there to match that up now that means we've done end to end so we can tick that off okay now we've got two we just need to have a look and see if we can compare these now i mentioned in the last question it's very very important here that we factorize these vectors and we can see if there is anything common between them we'll have a look so if i come back to this end to see one the first one if i factorize this one here and rather than rearranging it i'll factorize this one i could take two out and if i take two out that would leave me with minus a i'm sorry minus b plus a inside the bracket okay and that there is going to be a really key vector so i'm going to highlight that two lots of minus b plus a if i do the same again for this one so if we factorize this one as well let's just see that was that one factorized if we factorize this one the only thing we can factorize out is a half so i could take a half out let's have a look so take out a half divide the first bit by half we get minus b a half divided by half is one so minus one b and then dividing a by a half we get plus a there we go and that's that vector there if we uh if we factorize it and again i'm going to highlight that one and let's just write them down over here next to these vectors that i was going to work out in the first place so for n to c we had we have two lots of minus p plus a and for enter m the other one we worked out we've got a half brackets minus b plus a there we go so if you have a look we've obviously got this similar bracket we've got this same vector in there minus b plus a now if we've got the same vector there minus b plus a it means that the lines are going in the same direction and if they're going in the same direction and they're connected here at m then they must be on the same straight line because if they go in the same direction they're connected together at that point m then they have to be going in the same straight line the two there in n to c means that's the full length of that so it's doing two lots of that vector the half means that from n to m is half of that vector which means if we had of worked out this line here m to c it would have been through the other three halves or the other one and a half of that vector okay so you'd get three halves of minus b plus a and you could work that one out as well but we only ever have to work two out now that is enough to show that it's a straight line but personally particularly if it ever says prove or anything like that i'd just like to write that they both share the same vector you can you can say that they share different multiples of the vector one's two lots of it one's half lots of it but you would i would normally put a little statement in just pointing an arrow at those brackets and saying that they share the same vector okay so obviously that is quite a tricky one um yeah it's not not some of the hardest of the vectors some of these new vectors seem to be very very tough but there's a lot going on in that question there and a lot to be thinking about so this one says work out twelve and a half percent of a hundred and sixty so again we'll work out ten percent so we get our ten percent here is going to be 16 when we divide it by 10. again we'll halve it for five percent so that gives us eight and again halving that for two and a half percent would give us a value of four so how do we get to twelve and a half percent well twelve and a half percent we could use the ten percent and then the additional two and a half percent and that would give us our twelve and a half percent so again for this one we just need the 16 and the four we just have to add those two together that gives us an answer of 20 and there we go final answer right so as you can see these ones aren't actually too bad in order to get a two and a half percent you just have to find your ten percent halve it for five percent and then have it again for two and a half percent and then just carefully think about your combinations that you're going to put together in order to get to that final answer so it says alice and james share 80 pounds in the ratio three to two work out how much they each receive now first thing says whenever i'm looking at a ratio i always bring it away from the words so rather than looking at these words here i'm going to have a and j as my alice and james i'm just going to bring that away from all the words there and it says that they share it in the ratio 3 to 2. and obviously not forgetting when it comes to ratios that first name that's mentioned is correlated or relevant to that first number that's mentioned there so when we obviously have a look at this we've also got james and james is going to be the two so we can pair those up and bring it all away from the words and it says that they share 80 pounds between them now when we're looking at the ratio here that is relating to both these numbers here so both these people are going to share 80 pounds now you can do this obviously just numerically or you can actually draw a little picture um but essentially when it comes to a ratio that we just need to think about how many parts that number is referring to so obviously we've got 80 pounds as our number and that is referring to the total amount of money that they both share and they both share look three and two and that makes five parts there we go so they've got five parts between them and that equals 80 pounds so when it comes to any sort of ratio question we want to know what the value of one part is so if we've got five parts that equal 80 pounds we can get the value of one part by dividing by five so we get the value here of one part there we go and one part therefore is going to equal 80 divided by five which is 16 pounds there we go so one part of this money of this amount that's being shared in this ratio is worth 16 pounds now obviously we've got alice that's getting three of those parts so alice is getting three so alice is going to get three times the 16 which is 48 pounds there we go and james is getting two of them and james's share then is two lots of 16 which equals 32 pounds there we go and you can always do a little quick check at the end there because those two numbers should add together to make the original 80 and 48 plus 32 adds together to make 80. now if you're not really the most fond of just doing it via the numbers you can actually take a different approach to some of these i won't do it very often because it does actually have sort of um just create a little bit more work for yourself but if you don't like doing it as the numbers you could have a think about what this actually looks like in terms of a picture i might talk about this a little bit but if alice is getting three if we imagine she's got three little piles of money there and james has got two there we go we know that 80 pounds got to be split between those five piles so we can do 80 divided by five it's a very similar process gives us 16 and you can just fill in each of those little piles of money with 16s there we go and then obviously just count them up there so alice is getting the three of them and those three are worth 48 and james is getting the two and that's worth 32. so you could rewrite it as a ratio 48 to 32. obviously it's about money so we should include the pound sign there but there we go now a lot of these questions i'm going to focus on to start with i'm just going to be looking at money but these questions can be to do with anything could be recipes and an ingredient ingredients in a recipe sorry or it could be something you know there's quite a lot of questions where it says they're sharing sweets or sharing that you know something out uh whatever that may be we're gonna take the same approach but i'm just gonna look at many questions just to start with for the for the starting for you let's have a look at our next question okay so this question looks very similar it says tom and amy share some money in the ratio five to three but this time it says tom gets 70 pounds more than amy work out how much they each receive so again let's bring this away from the words we've got t to a and it's in the ratio of five to three now what it says is tom gets 70 pound more than amy so tom who's this one here is getting a little bit more now if we have a look in terms of the parts look amy's getting three parts tom's getting five so tom is getting an additional two parts to amy so those extra two parts that tom's getting it's saying that those plus two parts they're a worth 70 pounds and from there we can take exactly the same approach that we did on the last one okay we can divide that by two to get our value of one part now again you could think about this in terms of a picture we could have tom's five parts and we could draw five little circles look and a means two sorry three and then we can just identify in terms of the picture how many more he's getting well he's getting these two extra here so those two parts here are worth 70 pounds and again you can just take a little bit of a logical approach there dividing that by two tells you the value of each part and then you can stick your 35 in each of these piles and start to work it out that way but i'm just going to do it via the numbers look we're just going to divide this by two and that tells us that one part which is what we're looking for every time equals 35 pounds and then from there you can take exactly the same approach we can do for tom he's got five lots of that so 5 times 35 there we are and 5 times 35 is 175 so tom's getting 175 pounds and amy who is getting three of those is gonna get three lots of 35. there we go and three times 35 is 105 pounds and again you can add those two together and to get their total this one actually doesn't ask for their total it says how much do they each receive but if it did if it said work out the total that they both get you could actually add those together look and we've got 280 as the total so this question doesn't ask for that but just think about what we could be asked so we've got tom gets 175 amy gets 105 and if they ask for it we know they both share 280 pounds between them so this is following on from what we've just done it's basically the same as what we've just done it's just a few more words and sort of all wrapped up in a problem that looks more complicated than it actually is it says tom and adam have a total of 240 stumps the ratio of the number of tom stamps to the number of atom stumps is three to seven then it says tom buys some stumps from adam and then they have a new ratio tom to adam is now three to five how many stamps does tom buy from adam now there's no extra stamps getting added in at any point we're always having 240 if extra stamps are being added in this problem does get a lot more complicated but this type of problem here look there's only 240 and they're just sort of interchanging some between them so if we have our first scenario look scenario number one we've got 240 stamps and we've got tom to adam in the ratio three to seven notice that 240 stamp share between them so the total amount of parts there is 10 so we've got 10 parts equaling 240 stamps so if we divide that by 10 we get one part equals 24 stamps so if we actually multiply both these numbers look tom and adam tom gets 3 times the 24 and adam has 7 times the 24. so if we work both of those out 3 times 24 is 72 and adam has 7 times the 24 which comes out as 168. there we go so that's how many stamps they start with but we've now got scenario number two or the second part of this scenario where tom and adam now have stamps in a different ratio so now it's in the ratio three to five so now we know it's in the ratio three to five we can go about splitting these 240 stamps up between them so we've got three to five and that is a total of eight parts so we've got eight parts that now equal 240 again so again we can divide that by eight and 240 divided by eight gives us a total of 30. so this time one part in this ratio is equal to 30 stumps so if we have a look i'm just going to do this in a different color we've now got tom who's getting the three parts gets three times 30 which is going to equal 90 and adam is gonna have five of them so adam gets five times 30 and he gets 150. right there we go so obviously we're looking at uh let's have a look at what the question said again how many stamps does tom buy from adam well tom started with 72 and now he has 90 so we can hopefully obviously work out the difference there if we want to show our working out we can do the 90 stamps that he now has take away the 72 that he started with and that leaves us with a total of 18 stamps that you must have bought there so 18 stumps would be our final answer and obviously we could have taken that from adam as well we could obviously look at the fact that adam who's here started with 168 and then dropped down to 150 which is a loss of 18 stamps as well so you could look at that in both different ways but there we go there's a difference of 18. okay so again there's nothing more complicated in this question it's just there's a little bit of different wording going on so it says mario dylan and kate share three thousand pounds the ratio that of the amount that mario gets the amount that dylan gets is in the ratio five to four and kate gets 1.5 times the amount that dylan gets work out the amount of money that dylan gets so we want to have a look how are we actually going to create this ratio because at the moment we've got mario we've got dylan and we've got kate but it only tells us the ratio and that's just highlight that dylan and maria or mario and dylan five to four so we've got five to four for mario and dylan but we don't know how much cake gets in terms of parts of the ratio but it does then say kate gets 1.5 times the amount that dylan gets so dylan gets four parts so if we times that by 1.5 that actually gives us a value of six so kate is getting six parts remembering multiplying by 1.5 if you don't have a calculator just adds an extra half on half of 4 is 2 add it on gives us 6. so now that we've multiplied that by 1.5 the rest of the question is nice and easy because all we're going to do now is split that three thousand pounds in the question in the ratio five to four to six so if i get rid of this little bit of working out this times 1.5 let's have a look in total that ratio adds up to 15 parts and now we know that those 15 parts equal the 3 000 pounds so we can find the value of one part again by dividing by 15 and 3 000 divided by 15 gives us a value of 200 there we go so they all get 200 so we can go about times in the more by 200 we're actually only concerned about dylan in this particular question and dylan gets these four parts here so we can just answer that straight away dylan's four parts each one of them is 200 quid so 4 times 200 gives us 800 pounds for dylan there we go and there is our final answer and again you could go about obviously finding out the amount that they all get we could do maria times 200 which is a thousand and kate's times 200 which is 1 200. and you can even add them all together then and show that there's 3 000 pounds there but that's how we'd approach a slightly different question here we've got to think about creating the ratio slightly ourselves so this question on the screens here it says only blue vans and white vans are made in a factory it says the ratio of the number of blue vans to the number of white bands is four to three and it says for the blue vans the number of small vans to the number of large bands is three to five work out the fraction of the number of vans made in the factory that are blue and large okay so we're looking at these ones are blue and large now as as this sort of a question develops and this sort of gets a little bit harder there's a specific approach that i'm going to take when i'm actually looking at these and that is i'm going to kind of structure this in the same way that you would structure a probability tree and looking at this as more more along the lines of work out the probability of picking a van that's blue and large out of all of these uh sort of vans in the factory and sort of a way that so i'm going to draw this then is i'm going to draw a bit of a probability tree so i'm going to start by saying uh blue to white okay so we've got the vans they're blue or they're white now over here we've got the ratio of blue and white we've got four to three now that's out of seven in total so four out of the seven there are blue and three out of the seven there are white so for the blue here we've got four out of seven and for the white here we've got three out of seven there we go and then it gives us the next piece of information now it only mentions anything to do with the blue vans here and as these sort of questions develop we're going to have a look at ones where it mentions something about both branches here but in this particular question look it just says for the blue vans we've got small and large so we're going to do another branch coming off the blue vans here and that's going to be our small ones and our large ones and again we've got a ratio here it's three to five so it's out of eight in total so for the small vans there that's the three and that's three out of eight and for the large ones there it's five and again that's five out of eight now when it comes to actually working out the fraction here that are blue and large we treat it in the same way that we do a probability tree we have a look okay we go up the blue branch and we get four out of seven for the blue and then we go up the large root or down the large root depending on how you've drawn it here and that's five out of eight okay and just double check that you've put those in the right place that's large five out of eight that's right so that's blue and large so all we do normally when we're going along a tree like this is we just multiply those fractions together and we're not going to do anything different in this question so what we're going to do is we're going to take that first fraction there for blue which is 4 7 and i'm going to multiply it by the large fraction there which is 5 8. there we go and multiplying fractions is nice and easy we just times the top and times the bottom so 4 times 5 is 20 and 7 times 8 on the bottom is 56 there we go and that's our final answer 20 out of 56 is our fraction of the vans in the factory here that are blue and large okay so hopefully that makes it quite nice and simple to understand but essentially we just pair together the two fractions that we're looking at and multiply them together just like we would on a probability tree we're going to deal with this one here well we've got y is inversely proportional to d squared and then we're going to deal with this one here well we've got d is directly proportional to x squared so treating it in the same way let's create a formula for this first one so y is equal to and it's inverse proportion so k over d squared and then again we can sub these numbers in we've got d is 10 when y is 4. so we've got 4 which pushes y is equal to k over d squared and again the working out to the side we've got 10 squared which equals 100 so k over 100 and again let's rearrange that so times both sides by 100 so we've got our value for k and we get k is equal to 400 and again we can write our formula for that so let's bring that down here let's write y is equal to 400 divided by d squared right there we go there's our first one done let's move on to our second statement here so d is directly proportional to x squared so there we go let's write our formula for that so d equals k x squared there we are and let's plug in those values so x is 2 when d is 24. so we have 24 equals k times if we can write that in times 2 squared and again let's do the working out on here 2 squared equals 4. so k times 4 or again let's rewrite that as 4k let's get rid of that and write 4k instead there we go okay so to rearrange that we're going to divide both sides by four so divide by four and we get k equals and we get a value of six there so we go k equals six and again we can rewrite our formula there but having k is six so if we write that down here again we have got d is equal to six x squared and there we go there's our two formulas written now again if we just read this question carefully it says find a formula for y in terms of x so we have this formula on the left here that says y is equal to 400 over d squared now that's a formula for y in terms of d so we need to replace this d here and we've got this formula over here that says d is equal to 6x squared so we're going to do is we're going to put that 6x squared in place of d and if we do that let's see what we get so we'll get y is equal to 400 over d squared now d is 6x squared so i'm going to put that in a bracket so 6x squared and that is going to be squared okay so we're nearly done right we just need to square what's on the bottom there so y is equal to we've got 400 on the top and if we square what's on the bottom six squared is 36 and x squared squared is x to the power of four remembering that you just times powers when they're in brackets like that there we go 400 over 36 x to the power of four now again it does say give it in its simplest form we're almost done but the top and bottom here both divide now they both divide by and you might not spot the you know the biggest number straight away but then both divided by four so on the top there 400 divided by 4 is 100 and on the bottom there 36 divided by 4 is 9 so 9x to the power of 4 and then that is in its simplest form 109 both don't divide by anything similar so y is equal to a hundred over nine x to the power of four and to finish this off really i should get rid of that little step of working out in the middle so i can write my final answer as y is equal to 100 divided by 9x to the power of 4 and there is our final answer so let's just highlight that one there okay so this is quite an interesting question because this is talking about density and obviously we're doing this non-calculator and normally density is a calculator-based question so hopefully we're going to have some nice numbers now as soon as you see that word density in a question you're going to want to write down your formula whether you use a formula or whether you use a formula triangle so i'll write down the formula triangle because that works for everybody then so that's going to be density is equal to mass divided by volume now this question says it wants us to work out the mass so straight away if i cover up the mass that means i'm going to do density and multiply it by my volume so this question gives us a cube and it says the rubber cube has side lengths of 4 centimeters so in regards to this image here we've got 4 centimeters across there four centimeters here and a four centimeter height so to work out the volume of a cube you would just multiply all those numbers together or in other words four times four times four obviously take that in stages 4 times 4 is 16 times that by 4 again and that is going to be 64. so 64 and the units are centimeters so it's 64 centimeters cubed now it gives us the density says the density is 1.5 so we know to do this we do the density 1.5 and multiply that by the volume and the volume is 64. so we just need to do 1.5 times 64 which isn't too bad when you are multiplying by 1.5 that just adds on an extra half obviously you can hop the decimal out you could do the working out to the side so if we do that we'll get 64 times 15 5 times 4 is 20 5 times 6 is 30 add the 2 is 32 put our placeholder in one times four and one times six is nice and easy if we add that together zero six and nine so you get 960 and then you just have to hop the decimal back in one jump as we took it out at the start so when you jump that in your 96.0 so that'll be 96 and in the units it said grams per centimeter cubed so that would be 96 grams as our final answer so there we go we'll get 96 grams so as you can see we can have density questions that are non-calculator but you're going to tend to get nicer decimals involved they're just going to be a little bit nicer for your calculations this question says in a bag there are red blue yellow and green counters a counter is taken at random from the bag and the table shows the probability of getting each colored counter so we've got red blue and yellow and we're missing green then it says later in the question look it says there are 45 green counters in the bag work out the total number of counters in the bag so if we have a look here we just need to figure out what's missing what's that probability of getting green to start with because it's slightly different it's not saying how many times they're going to pick the counters out or something along those lines it's actually given us an actual amount of counters allocated to that probability so at the moment if we add all of these up we've got 0.3 0.25 and 0.15 and they add up to 0.7 or 0.70 there we are so we're missing 0.3 which again is 30 percent there we go so 30 percent of the counters are green and it says down here there are 45 green counters so this is actually telling us that 30 is actually equal to 45 counters there we go so that 30 there is 45 counters and we need to find out the original amount of counters or the total amount of counters in the bag and in order to do that we need to turn that thirty percent back into a hundred percent so this is almost like a reverse percentage because we need to break this down in order to turn it back into a hundred we can't just multiply it by two or multiply it by three it's not going to get us to a hundred percent so if we go about breaking this down always look how can we get back to that 10 and to get 30 back to 10 we would have to divide that by 3 and that would tell us that 10 is equal to 15 counters when we divide that by 3. there we go so 10 is 15 counters and now we're able to turn that back into 100 and from 10 back to 100 it's nice and easy because we just need 10 of those so times it by 10 so that original 100 of the counters is a hundred and fifty there we go and there's our final answer so a slightly different bit of wording there actually allocates the amount of counters to that certain percentage so it's called a reverse percentage obviously because it's not given this earth the hundred percent it's given us part of it it's looking looking for us to go back or build it back to 100 okay so that's a different sort of approach a different sort of question now when it comes to independent events we're going to talk about it throughout the video but an independent of event is just two events where the second event is not affected by the first event when i look at why it's not affected in these types of different questions but it's just that's all that it means there just means that the one event that we're looking at is not going to be affected by the other so let's look at this question now in a lot of these questions you'll normally actually be given a tree but in another ones i'm going to have a look at i'm actually going to give the tree we're just going to draw them all ourselves because obviously if you're not given the tree it's important that you're able to do that anyway and if you are then that's you know you know easier for you so we're going to look at these slightly harder questions here so james has a bag of counters and in the bag there are four red and five blue and james takes a random account from the bag and notes its color he then puts the counter back in the bag and that's an important bit we're going to discuss there and takes at random a second counter and we're going to work out the probability that james takes two different coloured counters so that line there where it says he puts the counter back in the bag that there is what determines that this is an independent event because after we take the first counter if we put one back in the bag the second time we're just actually just taking a counter from exactly the same bag of counters so it doesn't actually matter whether we take a red or a blue let's have a look at constructing this via a tree so if we draw a little tree here we're going to draw two branches that represent red and blue okay and that is going to be representing our first pick of the counters there i'm going to label this part of the tree so that's the first time we take a counter now let's write the probabilities in so looking over there we've got four red and five blue so in total there we've got nine counters so you can do that working out to the side if you want four plus five equals nine so we're going to write the probability so i'm going to write them as fractions just keep it nice and easy obviously you can write it as decimals if you like but i think in the case of this question it's probably easier for us to write it as a fraction so for red it's four reds out of nine so the probability of getting a red is four out of nine and then we've got blue and that is five out of nine there we go i'm just thinking about these fractions obviously if you were given the tree and you were just and you you know you didn't have the information up there and it said that the red was four ninths you'd have to actually work out the blue there and obviously the numerators have to add up to whatever that denominator is if it's four out of nine that tells you there's nine counters four of them being red and therefore five of them have to be blue in order to make nine out of the four and the five so just thinking about that if you weren't given the uh you know the information in the question you'd have to actually do that yourself on the tree now obviously what we have we have a little scenario here and i always like to think of it as a bit of a journey so if we imagine a little person standing over here at the start not the best drawing there but we could either go on a little journey now we could take a red out of the bag in which case our little person was end up would end up here and then we'd have our second counter so at this point we can construct the second part of our tree and i'm gonna run out a little bit of space here and this is gonna represent the second pick now on this second pick here look at this part of the journey we've taken a red counter and now we're going to take another one and again that can be red or blue and because the counter's gone back in the bag again it's going to be four ninths and it's going to be five ninths again for blue and that is that one part of the journey but of course we might not have taken a red out so if we get rid of our little person here and let's imagine that actually instead we've taken out a bloom we've gone down this part of the journey here we've taken out a blue counter so here we are we've taken out our blue now again we're going to take a second counter okay and that again can be red or blue so we'll label that up red or blue and again the probabilities are just going to be exactly the same it doesn't matter whether i took a red or whether i took a blue it's still four ninths and five ninths as that counter has been put back in the bag so we've constructed our tree and let's have a look let's just get rid of all of this um we now just need to work out this probability and now the question says work out the probability that james takes two different coloured counters and there are two ways i can actually do that so to get different colors i could and i'm just going to squiggle along the lines and this is how i do it i go up the red and then i'd have to take a blue and the two probabilities that you go through on this journey that are really important so we went through four nights when we took the red and we finished on the five ninths so to work out the probability of that root happening all we do is we multiply those two fractions together and all we've got to remember is you times when you go across the tree so if we times them together we've got four ninths and we're going to times that by 5 9. and that's really nice and easy not to forget when you're multiplying fractions you multiply the top numbers so 4 times 5 is 20 and 9 times 9 is 81. so the probability of getting a red and then a blue is 20 out of 81. now not that that one can simplify but you do not need to simplify fractions when it comes to probability and the reason being is that that probability there 20 out of 81 tells us something quite important it tells us that there are 20 different ways of this happening out of a possible 81 different ways of picking all these counters that can seem a little bit hard to understand sometimes but if we labeled the red ones one to four and the blue ones let's say one to five there are there are 20 different ways of taking a red then a blue i could take red counter number one and then blue counter number one or i could take red counter number one and then blue counter number two etcetera etcetera obviously they're all still red and blue but if they would have different numbers on we think of them as you know all different counters there are 20 different ways of doing that now what we're going to have a look at now is the other way of getting two different counters because i could take instead of a red then a blue let's think i'm just going to do a different color here i could take a blue and then i could take a red and that would also give me different color counters and again that goes through these two different fractions so 5 9 and 4 9. and if we times them together you'll probably notice here that actually we're going to get the same answer because it's the same two fractions in reverse and four ninths times five ninths again on the top we get 20 and again on the bottom we get 81 but that there is taking a blue then a red okay and there are 20 ways of doing that as well so there are 20 ways of getting a red thunder blue and 20 ways of getting a blue then a red so the total probability there in order to get our final answer we just need to add these two together to combine them to get the total probability there so 20 plus 20 on the top gives us 40 out of the possible 81 ways remember and you don't add the the denominators together there when adding fractions and that would give us our final answer 40 out of 81. okay so a few different things to be thinking about there one was constructing the tree thinking about obviously our first pick and our second pick and then the next one was obviously mulching multiplying across the tree and any different combinations that we have at the end we just add them together and combine them together of course you might get a nice easy question here it might just say what's the probability of taking two reds in which case you only need to do the one journey have a look at the tree now that goes through the four ninths and the four ninths so you just times them together and that'll be your answer but if there are if there are multiple answers you just need to think about adding them together at the end these questions we've now got some decimals introduced so not necessarily harder in fact you might you know depending on which ones you prefer i think that i personally think they're just as easy as each other but obviously these are slightly different when we've got decimals involved so it says hannah is going to play one game of darts and one game of snooker and the probability she'll win at darts is 0.3 and the probability that she'll win at snooker is not point four work out the probability that hannah will win at least one game now win at least one game we'll talk about that in a sec and let's have a look at how we're going to go to go about this but first we need to construct our tree so looking at this we've got the first part of our tree could be our first game and the first game that it mentions is darts so let's just have darts as our first one now we've got in this scenario of winning or losing and it says the probability that she will win at darts is 0.3 now hopefully you already know that probabilities have to add up to 1. now 0.3 is 30 so obviously to get it up to 100 or to get it up to one is 70 or 0.7 so the property of losing this time is 0.7 now we've got the scenario where she's going to play her second game so she can win or lose if she wins then she goes on to play our next game here which is snooker let's label that just about fit that in snooker and again it's a win or a lose and the probability she wins this game is 0.4 now again that's 40 so 60 to get to 100 or 0.6 to get us up to one and that is the scenario there if she wins at darts she can then win or lose a snooker so obviously we could lose at darts in which case again she's going to want to go on to play snooker and again the probability that she wins is 0.4 and 0.6 it doesn't say that anything changes you know it might say in a question if she wins on the first game or if she wins at darts then she's got you know a higher chance of winning at snooker maybe but it doesn't mention anything like that in the question so both of those parts of the branches are just going to be replicated down below now it says here work out the properties she'll win at least one game now if we have a look at the outcomes at the end of each part of the tree here we've got this outcome up here which is a win and then a win we've got the one below that which is winning at darts but losing at snooker we've got the bottom here which is uh obviously going down our bottom branch there on darts which is a lose and then a win and then we've got a loss and then a loss down the bottom there now this language here says win at least one game now at least one game does mean that you uh could win both so we could have this route here well that's winning at least one game yes she wins both and that's definitely at least one and then we've also got the one below where she wins one and loses one and then the one below that where she loses one and wins one now that is the probability of winning at least one game the only one we don't want is that one down the bottom losing both so if we just go ahead and work out all the probabilities here for the top three then we can get all these probabilities these different probabilities of winning at least one game so for the first one there let's have a look and i'm going to have three branches here so i don't want to squiggle all over it but if we go the win-win we've got 0.3 multiplied by 0.4 and that is 0.12 0.12 we've got 0.3 times 0.6 for the one below and there we go 0.3 times 0.6 is 0.18 and then the one below that 0.7 times 0.4 which gives us 0.28 now there are our three different probabilities not 0.12 not 0.18 and 0.28 so just like before all we need to do is add them all together and that will give us our combined probability of all those three different outcomes so you can do them to the side you've got 0.12 you've got not point 18 and 0.28 and when you add those all together again you may or may not have a calculator here let's have a look what we get so 8 2 3 4 5 0.58 and there is our final answer which can write down here 0.58 as our total probability there we go and let's highlight that now obviously that's um one way of doing it there is another way of doing it again we could have actually just worked out the probability of losing both and then subtracted that from one now obviously that introduces another method here thinking about obviously just eliminating the one that we don't want but that is another method you could have done as well i'm thinking about how that works if we'd had done the probability of a loss and then a loss that would have been 0.7 times 0.6 which is 0.42 and you can do one take away 0.42 which would leave us with what's left over which is 0.58 okay so you can do that it's a little bit quicker but obviously if you just want to follow the same method and just work out all the roots of the ones you want then just follow this same approach here and just pick out the ones you want showing all you're working out okay so looking at a cumulative frequency graph when we have to draw one of these the first thing you want to have a look at in the table is if this word here says frequency or cumulative frequency so as it says frequency we're going to have to actually find the cumulative frequency which we can do just by adding up as we go down the rows so we start with 4 then we get an additional 18 that takes us to 22 then we've got an additional 24 that's going to take us to 46 and an extra 40 there so that's 86 adding on an extra 24 gets us to 110 and then we finish off with the extra 10 which finishes on 120 and that's good because that matches up to the top of our axis there when we have that 120. now when we're plotting one of these we're just going to plot against the n numbers so those n numbers there on the intervals and that means just plotting on the graph so we've got 10 and that goes up to four now this is obviously quite small on my screen but you just need to plot it as accurately as you can so i'll try and do it as best as i can but this should give you a good idea then we've got 22 as our cumulative frequency we're going to plot that against 20 so i'm going to just above 20 there for the actual value so just on 22 30 goes up to 46. again i'll try and do it as accurately as i can 40 goes to 86. there we go and 50 goes to 110 which is a bit easier to plot and then 60 goes to 120 which is right at the top so when you are joining these up make sure you look at the starting number which in this case is zero so we're going to go from zero with a nice smooth curve through all those points not using a ruler and there is our cumulative frequency graph drawn now for part b in this question it says to find an estimate of the interquartile range and it wants us to use the graph for that so when finding an estimate of the interquartile range we need to obviously find the quartiles so if we look at our total which is 120 and we divide that by four to find the quarter each of the quartiles is going to be 30 above the bottom and 30 down from the top so if we find 30 on our axes and again you can do this a lot more accurately when you have the paper in front of you but that is 30 just there i'm going to go across to my curve and i'm going to go down i'm going to read the value as accurately as i can so that for me comes out as 23 that is my lower quartile i'm going to do the same from the top so go down 30 from 120 that's going to be down at 90 and i'll go across again to my curve using a pencil and a ruler and then go down again as neatly as you can and i am going to land there on 43. okay so i've got my values i have my upper quartile which is commands 43 my lower quarter which has come out as 23. says he get the inter-quartile range we just do 43 take away 23 and that leaves me with an interquartile range of 20 and that would be my final answer and that's how i go about finding the interquartile range so it says here there are 10 boys and 20 girls in a class and the class has a test says the mean mark for all the class is 60 and the mean mark for the girls is 54 work out the mean for the boys so when it comes to a question like this obviously we know well hopefully we know that when we're normally working out a mean we get the total we divide by how many people there are or how many things there are and we get our mean okay so we do the total and we divide by the amount okay for whatever it is there we are on its most basic sense there with the mean usually the total divided by the amount well in this question here obviously it doesn't actually give us the total it tells us what the mean is it tells us the amount it tells us that there are 10 boys and 20 girls but it doesn't actually tell us the total so what we're going to do is we're going to go backwards to work out some of these totals okay so we're going to think about this in reverse all right and that is how you get the mean there the total divided by the amount okay so if we link some of these things up then it says that there are 10 boys it doesn't give us the mean for the boys so that's no good so far so that first piece of information that we are actually given in terms of the mean it says the mean mark for the whole class is 60. well if we were going to work out the mean for the whole class we'd get their total let's call that x we divide it by the amount and in this case the whole class well there's 10 boys and 20 girls so we would divide it by 30 and it would give us this mean here which it says is 60. so in other words we're looking for a number that divides by 30 that gives us the answer 60 and we can just do that in reverse we can just do 60 times 30. so actually all we're going to do in these questions is we're going to link the mean that we're given with the amount that is appropriate for that mean so i'm going to get rid of this little bit of algebra that i've drawn here but obviously you could think of it in terms of an equation but this is the sort of concept that we're going to run throughout this question and hopefully once you actually get your head around this and why we multiply instead of divide obviously we are doing a reverse this question obviously becomes a lot easier then okay so obviously thinking about this topic's actual name it's called a reverse mean normally when you work out the mean you do a divide and obviously then the opposite or the reverse of dividing is multiplying and in these questions we're going to be multiplying rather than dividing to go backwards let me get rid of that so let's have a look so obviously we've been given that the mean for the whole class is 60 and we've just established that there's 30 students or 30 people in the whole class so we're going to do 60 times by those 30 students and that's going to tell us their total so 6 times 3 is 18. add on the two notes that's one thousand eight hundred and that is the total for the whole class and it's usually worth just labeling these particularly if you haven't quite got your head around it let's just label that total for class there we go right so the next bit of information then we've also been told that the mean mark for the girls is 54. now in terms of the question look there are 20 girls so if there's 20 girls and their mean is 54 we'll take their mean 54 times it back by that 20 to find out what their total was so 54 times 20 there we go and that gives us 1080 and there we go there is the total for the girls and let's just label that again total four girls right so if just thinking logically about the information that we've got here we've got a class of boys and girls it says the total for the classes of 1 800. the total for the girls is 1080. so logically we can work out now the total for the boys we can subtract those away from each other so 1 800 take away 1080. let's write that down there we go take away 1080 leaves us with 720 there we go and that's the total for the boys let's label that up total for boys and now we can go about answering the question because now we know the total for the boys we can actually think about working out their mean and that's what the question wanted it said work out the mean for the boys so we've got the total for the boys of 720 and in total up here we're given in the question let's highlight in the same color there are 10 boys so to work out the mean just like normal we would just finish this off by saying the total for the boys is 720 i'm going to divide that by how many boys there are which is 10 and that gives us a mean of 72. there we go and there's our final answer and just think about what we did there just working backwards through the question multiplying the means by the amount to get the totals and then using a bit of logic there to find out therefore what the total for the boys must have been and then just working out the mean as normal okay that is the end of this video looking at paper one don't forget to have a look in the description if you want to have a look at papers two and three don't forget to download the checklist so that you've got something just to give you a bit more advice on some topics to have a look at again don't forget that this is not all the topics or just one variation of each topic within this video so you do need to do some further research have a look more into your revision books and what your teachers are giving you in terms of in your lessons and obviously make sure that you are looking at the full videos on each of each on every topic that you're not a hundred percent confident on so i hope that was useful and helpful if it was don't forget to like don't forget to comment don't forget to subscribe i'll see you for the next one [Music] [Music]