Transcript for:
Essential Topics for IGCSE Maths Success

okay so if you're looking for a video which is a crash course in igcc0607 maths and this is the video for you i'm going to go through the 10 main topics that appear every year and if you have these in the bag and you really understand these topics then you'll do really really well in the igcc067 course and you'll be well prepared for 60 even 70 off questions so my very first topic is of course going to be gdc skills so we have a typical gdc question here so f of x is equal to x cubed minus five x plus three we are given the scale here and we need to sketch it this is the classic gdc question so i am now going to open up my gdc i'm going to type that function in so in the graph function here we just press x cubed now the easiest way of doing this is clicking on this button here giving you the index and then you can put in the 3 there we go minus 5x and then we are going to go to plus three so we type that in and you'll notice ah it doesn't quite look like what we need because if we go back to our powerpoint you'll see the scale is given from -3 to 3 and from -10 to 15 so we need to go back to my gdc and we go to menu window window settings and then we put in the correct settings so i'm going to put -3 here for here i'm going to put 3 following the instructions of the question and then we have minus 10 and positive 15. so you must change the window to fit the question that you are given so we have our typical cubic shape here notice it crosses the x-axis three times it crosses the y it y axis once so we're going to copy that across and make sure we have a good sketch so let me get my pen ready so it goes up it crosses in the positive quadrant here and then we go upwards like so so there's a two mark question i imagine the mark scheme would be looking for it crosses it three times where two are in the positive and one in the negative and it crosses here once as well and now we need to find the coordinates of the local minimum point so notice this is our minimum point here so we are going to go straight back to our software we are going to go to menu we go to analyze graph and we are going to go to minimum and when we do this we know the minimums around here somewhere we click before it we click after it and then we can read off the value for our minimum okay notice that you are going to need both coordinates here and we want to do both two three significant figures so we have 1.29 and minus 1.30 so we write in 1.29 and minus 1.30 always play on the safe side make sure you write this to three significant figures now question c has been quite tricky for a number of students over the years where they've tried to fit in or crowbar in this rotational symmetry into a gdc question i've seen this now a couple of times so be aware of this notice it's a three mark question so we need three pieces of information i need to describe fully the symmetry of diagram the first thing we should recognize here is there is a rotational symmetry imagine this is our center of our rotation it would spin around and actually look the same as we spin it around and it'll look the same twice so we have to write down rotational symmetry so for one mark order two because if you visualize this in your mind you can rotate it around once when it's the same and then back to the beginning so to give you twice and we also need to know where the rotation point is well this is going to be our y-intercept now we can actually without using the gdc actually read this off because the y-intercept is always the number in our function so this is actually going to be three so we just put in zero three as our center again three marks they require three pieces of information to make sure you do not leave anything out okay and now we go on to part d this is very very typical for a gdc question where they give you another function in this case a straight line and then you need to solve the equation the amount of times i've seen people try to equate these two things together and then solve it you simply can't do that with our gcse methods possibly with ib methods so we need to use the gdc way of doing this so let me show you very quickly how we do this so if i go to our gdc like so we press the tab button this is a very useful shortcut if you press the tab button you can then type in the second function straight away and let's remind myself yep it's 2x minus one so you go to x minus one press enter and you'll get a straight line that goes across like so now the question wants us to solve this notice you've got three marks and that makes sense because you have one two and three intersection points so we're going to work out these one at a time so we go to menu analyze graph and then we are going to go to intersection and we do it in the same way that we did it with minimum maximum so we click four we click after you can just about see we scan this in and we get minus 2.90 so we're going to type that in our first part minus 2.90 notice we're looking to solve so we only want the x values okay that's important to realize okay and then we're going to do the same process for the other two points so we go menu analyze intersection we click before we click after and this time we get 0.603 again be accurate with that three significant figures so zero point six zero three let's go back again almost therapeutic doing this menu intersection and we click before we click after and we get two 2.29 so go back to our powerpoint and write in 2.29 again one mark for each of these points now this is the part that is really that discriminator question between a and a star so we need to be careful here use your answers in part one to solve f of x is greater than g of x so if i was to do a quick look at my gdc what we're looking for is when the function the blue function is bigger than the red function okay and we've already worked out the intersection points and hopefully you agree here the blue function is bigger than the red function between this point and this point and after this point okay so we're going to write that now in mathematical notation so the way we do this is using inequality so using the points that we just did we have minus 2.90 is less than x just be careful with the inequalities notice we've got strict inequality here not less than or equal to so we know between these two values we just worked out that x the function f of x will be bigger than the red function that we had but also and this is what people forget x is also bigger than 2.29 so you're using these values and then you're saying to yourself okay when is the blue function bigger than the red function okay i need to write this as an inequality okay and this gives us the two marks one for the first one and one for the second one okay and my number two topic is going to be statistics and it can appear on paper four or it can appear on paper two as well and really be prepared for those cumulative frequency star questions so let's get on to question four so the marks x of 300 students in a chemistry test are shown in the table below and we need to calculate an estimate of the mean mark now notice we have group data here therefore we do not know how many people got exactly one mark or five marks or seven marks and so on so what we need to do is work out an estimate and the way we do this is working out the midpoint of these different categories so we're looking for half way between 0 and 10. that is going to be 5. halfway between 10 and 20 is 15 and so on so we go through this process like so however you do have to be careful so the halfway between 40 and 60 that's going to be 50. so watch out for that sneaky trick where they change the class widths between 60 and 80 that's going to be 70 and then finally between 80 and 100 is 90. now we could do this manually but one thing that we do have in our possession is a graphical display calculator so i'm going to show you how this works i can work this out really really easily okay so you see i've got my graphical display calculator here so i'm using the ti inspire so we go to on we're going to go to new and we're going to create a new spreadsheet a very useful function what i'm going to do is i'm going to call this m for midpoints and i'm going to type in all the midpoints and now here for the frequencies i'm going to type in all the frequencies okay and once i've got this i go to menu i go to statistics stack calculations one variable statistics notice we only have one variable in this situation and now my x list is going to be m frequency list is going to be f for frequency makes sense i click ok and you'll see the very first value you get here is our mean this x bar which is 39.817 so i'm going to go over and write that into my powerpoint and this is going to be our answer 39.8 remember always round to three significant figures if you are not sure question b we're going to complete the cumulative frequency table here now remember this phrase cumulative frequency this means a running total this is a really important topic to understand if you want to do well on this igcse06707 course so the way we work out cumulative frequency we start with our value 41 and then we keep adding these values on so we do 41 plus 32 that's equal to 73. then i go back i take 73 and then add on 44. so 73 before if we do that we get 117 i go back and now i want to add on the 50 so 117 that previous value plus 50 and then we get 167 and we do this same process going back adding this value popping it in like so until we get all the values and one thing you can do to check in the exam is when you do the very last one here so 280 and then we add on that 20 we should get to that last number here of 300. to make sure you're really really careful about actually putting these values in what we're going to do now is actually put these values onto our cumulative frequency graph where we're going to plot the 10 against the 41 the 20 against the 73 and so on so this one is 20 and 73 so we go 20 along and 73 up try and not get dizzy when looking at these so 60 70 just before here 73 just keep doing this in a nice logical manner now be careful with this one here notice we've got 60 and 232 so don't do 50 make sure you do 60 here okay so i've got all the points here so when we connect this up this must be a smooth line that goes through all the points and make sure it's increasing at all times so i'm going to start at 0 0 work my way along again try my best to get a smooth curve as possible that's not too bad considering i'm working on a computer so part d we're going to use this curve to now work out an estimate for the median mark so notice we have 300 pieces of data here so to work out our median officially what you should do here is do 300 plus one divided by two it's important for the plus one but notice it on the graph it's not going to make much difference here so we get 150.5 and we're going to look for the 150.5 value so i'm going to get a different pen color here and i go to 150 i'm going to go all the way across and then read down so we just go dot dot dot make sure you use your ruler at home very important to do if you're liking this video as well feel free to give myself a like does go a long way to help the channel okay and if we read the scale here going along in two so 32 34 36 so i'm going to pop in 36 and those in the mark scheme they are to go between 35 and 38. now we go to do our interquartile range so to work out the interquartile range we need to work out two things first we need to work at the lower quartile and we need to work out the upper quartile so one quarter of the data and three quarters of the data so this formula is very similar we do 300 plus one divided by 4. very similar to this but we didn't just halve it so notice we're going to get the 75.25 value again 75 will also work here as well so i'll use a different color here for interquartile range so i'm going to do some magenta lovely color so again 75 which is here they go across downwards so i'm getting that value there to be 20. now let's choose a different color for the upper quartile on the graph so this one here we're going to take this value just the lower quartile and times by three if we do that we'll add these two together as well we're going to get 225.7 value so let's choose a lovely color here let's go with cyan why not so 225 so it's roughly here a little bit above this line here so again we have to be really careful to go across try making this process essentially as accurate as possible okay downwards like so we're going to read that value off again hopefully i've been accurate enough for the mark scheme so i get here 58 pop that there and the way we from those figures there actually work out the interquartile range is we're going to take one away from the others so 58 which is our answer here minus 20 giving us the answer off and that fits in within the boundaries i'll show you the answers here so it fits in within this boundary of 35 to 39 okay so 35 of the students passed the test and we're going to use that curve before to find an estimate of the minimum mark needed to pass so if 35 of the students pass the test we actually need to find where it is actually equal to 65 so we know that cutoff between either passing the test or failing the test so we need to find sixty five percent of the three hundred so sixty five percent three hundred again you can use your calculator to work this out by doing zero point six five times three hundred but if you do that process you will get 195. so i'm going to do a very similar process to what we did before but this time we're going to go to 195. so i go back to my graph like so and i'm going to use yellow this time that's a nice color to use we go to 195 here we draw a line all the way across similar to how we did the quartiles and the median and then we go down at this point and this will help us work out the cutoff point between the students passing or failing the test so you'll see the value that we get here is equal to 48 and that's the value we're looking for as our estimate for how many marks are required to pass the test so our answer here is going to be 48 marks and what this value means is this is kind of cut off point where anyone above 48 that you're into that 35 passes the test okay so you can check quickly check your answers here again appreciate any likes and if you want to subscribe that also goes a long way to support the channel as well and this is the last page of the answers for question four okay and my number three which is coordinate geometry often appearing on a paper two occasionally appears on the paper four and what i'm gonna do with you is really go through a detailed question here going through perpendicular bisectors going through gradients and how to apply that also in a geometric context so a is the point 3 2 and b is 0.95 so we are given these on our diagram i tend to like to write this in just so it reminds me what they are and we need to find the length of a b so a pretty generous three marks so let me draw this line in so we've got the line a b i'm really not known for my line drawing and we want to find the length of this so the way we do this is the standard procedure so we are going to make a right angle triangle out of what we have here okay as soon as you see a line segment kind of question you need to be thinking right angle triangles and then pythagoras that's what comes with it so let's work out then what the lengths of my triangle are so notice the x coordinate here is three the x coordinate here is nine what's the difference between three and 9 that's going to be equal to 6. we do the same thing for the y so what's the difference between 2 and 5 that's going to be 3. and at this point we can use pythagoras so a b squared is the same as 3 squared plus 6 squared again most of these things you can use a calculator to work out 3 squared is 9 6 squared is 36 36 plus 9 is 45 remember this is a b squared so to get our final answer for a b well what's the opposite of squaring well square rooting so we need to square root the 45 at the end and this is the paper four so we're just going to use a calculator to work this out here's a handy one i've got so i go ctrl square root 45 press the enter button and i generate my number here 6.7082 remember we always round to three significant figures so i'm going to go back and what i would do in the exam is write down what's on my calculator and then round it afterwards so gonna round to three significant figures first second third because we've got an eight here we're going to round the zero up to one so we get six point seven i don't think there are any units involved here we aren't told any units so that's fine next question we need to find the equation of the line a b give your answer in the form y equals mx plus c now the good thing about doing the method that i've done here is that we kind of worked out some of the information that we need remember the gradient is equal to rise over run uh depends on your teacher they will teach you different things or the change in y over the change in x or that formula um y2 minus y1 over x2 minus x1 the thing is here we've actually got the information we require okay so the rise is going to be 3 and the run is going to be 6. so the gradient is going to be equal to 3 over 6 which if we simplify our fraction to a half so by using this method we've kind of got some bonus information for this question here at this point i can then write my line in the form of y equals mx plus c but now i know that m the gradient is equal to a half so i just write down y equals a half plus c but i need to work out this constant and the way we can do that is just taking any of the coordinates we can take this one or we can take this one and we substitute it in so when x is equal to 3 y is equal to 2 and we're going to put that in here so where i see a y i'm going to put a 2 where i see an x i'm going to put a 3 and let me continue this slightly over here so we've got a half times 3 is 3 over 2 plus c now we're going to minus 3 over 2 from both sides which is 1.5 you can use a calculator for any of this and two minus three over two that can be one over two and again use your calculator to confirm okay and then we have to write our answer in so i'm gonna write y equals a half x plus a half notice the marks can actually use decimals but again fractions is absolutely fine that's a nice six marks to get in the middle of an exam yeah it should make you feel comfortable here to get those six marks and then we'll go on to the harder questions okay so now we're given an extra point c is the point eight two so let's put that roughly on our diagram so we've got an idea of where it is um eight two well that's going to be along here somewhere because this is two so let's pop it here let's pop it there h2 to make it more or less you know in the right place and what do we need to do with it find the equation of the line perpendicular to a b which passes through c so let me draw this visually so what we're looking for is a line like this that goes through c and is perpendicular to a b okay so it's slightly different than the perpendicular bisector we're just looking for a perpendicular line that goes through c right so the first thing we need to do is work out the gradient of the green line well let's have a think about that there we are so the gradient of a b was a half so the gradient of the green line well there is a formula here but the way i tend to teach this is to work out the gradient of a perpendicular we flip the fraction so we get two over one and then we switch the sign so we have a plus here so we're going to switch that to a minus again we don't usually write this minus two over one we just write this as minus two so the gradient of the green line the steepness of the green line is equal to minus 2. now this point we do this very similar to how we did the previous question we write it in general form y equals mx plus c y equals minus 2x plus c and at this point we substitute this coordinates in because we know it's on the line so when x is equal to eight y is equal to two let's pop that in so two is equal to minus two times eight plus c always running out of room here there we are we're close so i think at this point you add 16 on both sides then we get c is 18 and so we put this back into the general equation so perpendicular line will have the equation y equals minus 2x plus 18. just be careful it just wants a perpendicular line doesn't want a perpendicular bisector i.e you do not need a midpoint in this case okay and we are going to continue so find the coordinates of the points where the line in part c intersects okay so we see this green line i drew kind of randomly and now we actually want to find this point here do we have a name for this point we're just going to find that random point now notice this is the intersection of two lines now we've got the equation of both lines we've got this one so let me write this down so we've got y equals a half x plus a half and then we've got the previous one so y equals minus 2x plus 18. now if they intersect um that means that these two equations are equal to each other so we set the right hand sides equal to each other what that means in practice is a half x plus a half is equal to minus 2x t at this point we solve this like a normal equation so what i would do first is bring the x's to one side so i'd add 2x on both sides that's going to give me 5 over 2x okay because this is 4 over 2. okay if you're not sure about that you can convert it to a decimal so that makes your makes you feel better plus a half is equal to 18. there's a reason i've kept it as a half as we'll see shortly so if we minus a half from both sides we're going to do it nice and steadily ignore my alarm so minus a half from both sides this cancels with this like i should have said this cuts through this that's why it disappears then we get 5 over 2 x well 18 minus a half is 17 and a half which is the same as 35 over 2. the reason i've kept it all like this i can now cancel this from both sides and notice i got a really nice equation out of this and i'm going to get the coordinate of x equals seven i'm not finished at this point so we've got the seven coordinates but now we need to work out what the y coordinate is you can substitute now 7 into either of these equations so i'm going to substitute it into this one but it doesn't matter which one so when x is equal to 7 y is equal to -2 lots of seven plus eighteen we work through algebra and then we get the answer of four again please do feel free to stop the video if i've gone through a little bit too fast that's absolutely fine that's why we have the videos it makes your life much easier okay uh part e d is the reflection of c in the line a b okay in the line a okay so right so d is going to be up here somewhere so continue this line and this would be d it will be on the same line because it's perpendicular and what we need to work out is the coordinates of that yeah coordinates of d okay the first thing will come to mind is writing that intersection coordinates that we just worked out so seven four so this was seven four and then we think to ourself well this distance needs to be the same as this distance so there's a gap of one between eight and seven seems to be a gap of one between seven and this coordinate therefore this would be equal to six and the same principle applies here so there should be a gap of two between two and four so we need a gap of two between these coordinates so we need to add on an extra two and that gives us our coordinate of six six for two marks um be interesting to see what they say for working here um okay they don't give any marks working here they just say you get one mark for each coordinate okay and now we need to know the special name of quadrilateral a c b d make this so a b d okay let me use a different color so we can see exactly what shape they want so a to c this now the better you make your drawing the easier it is to see what shape this is i think i've done a pretty decent job of this i think we can see that this is a kite yeah because this distance is the same as this distance that's one of the um definitions of it being a kite so we just write in the correct mathematical word which is like so okay and on to question f so we need to find the area of this quadrilateral a c b d and i did have a think about this and i don't like the way the mark scheme does it so what i'm going to do is draw this out again i'm going to draw out the rough kite that we have try and do it know to decent scale so we've got a okay and the general formula for working out a kite is to take these two sides i'm putting in times them together and divide by two now we already know a b we already know this is root 45 so all we need to do is find dc which we can do by using right angle triangles so the same way we did the previous questions so d was 6 6 coordinates c was 8 2 down here so we can work out the length of the line segment using pythagoras in the same way we did before so we've got the height difference between 6 and 2 is 4. the difference between 6 and 8 is 2 so we can use pythagoras to work this out so we get x squared is equal to 2 squared plus 4 squared x squared is equal to 20 so x is equal to root 20 so then to work out the area of this all we do the area is going to be root 45 times root 20 divided by 2. if we put that into a calculator so i'll just do that now so root 45 20 divided by 2 that gives you the answer of 15 which is the exact answer the precise answer so we get the answer of 15 there are no units in this case no centimeters squared or anything like that um i think the masking wanted to use like ac and cb and do it in a similar kind of way but i think this makes much more logical sense but again feel free to leave a comment in the comments uh what you think whether you think there's an easier way than doing it what i how i did here i'll just show you the master scheme though it's always interesting to see how they do it as well and where you pick up the marks again pause your video you can go through yourself you're doing it with me and then the last question here which i didn't follow any of these methods as far as i'm aware actually i think i did this method here instead um i think that makes more logical sense than this one before we move on of course the exams report so most people did well in the first three parts of the question which makes sense it's pretty standard 90 perpendicular lines etc but yeah as we went through into d for example realizing that you can just um equate it i think that was part d or part e and also um you need to find the midpoints that wasn't the case you need to find the intersection point instead again clearly drawing diagrams notice i thought it was a little bit overwhelming so what i did is i drew what i needed for the question and then put in the correct information that helped me shape my thinking for working out the correct solution okay and my number four is probability now this generally appears in a tree diagram question and this is the question i'm going to go through with you and it's the most likely to appear in your exam so we have a probability question trisha's got two bags in the first bag we have six white balls four red balls in the second bag four blue balls three white and two red she takes a ball out of random out of the first bag then takes a ball at random out of the second bag so there is no replacement um it is with replacement ie there's no change in probability between bags one and two so it stays as is and we need to complete the tree diagram for two marks so if we take our first bag the number of white balls there are six and we have ten balls in this total so this will be six out of ten and then for the probability of red there will be four red balls and ten as well now we have a completely different bag so for the second bag there are four blue balls so it'll be four and then how many balls are there in this bag well let's be careful four plus three plus two is equal to nine so we get four of nine likewise with white we have three out of nine red we have two out of and notice it'll be the same regardless of whether we chose white or red beforehand it will still be the same probabilities so four out of nine we have three out of nine and finally here two out of nine nice for two marks certainly that's marks you should be picking up in this paper for exam now they ask you some probability questions linked to this so first of all we want to work out the probability the tricia's two balls are both white so let's go back so we want probability of white white tend to write it and the way we do this is we take the probability of the first one being white 6 over 10 and then we multiply along the branches so we multiply this by 3 over 9 and remember to multiply fractions we times the top so 6 times 3 is 18 and then 10 times 9 is equal to 90 and we should simplify at the end if we can i'll leave this there for a future calculation generally speaking igcc cambridge they don't mind if you leave it like this but we should try and simplify it and if we divide top and bottom 9 we get then 2 out of 10. bottom should be 10 and not 9 because we're dividing by 9 top to show you what we did there here here and then we get to our answer over five but i'll just show you the mark scheme here quickly so notice if you leave it as 18 over 90 on this exam or oe or equivalent that's absolutely fine you get your one method mark for six tenths time ninths that's wrong it should be uh nine here not ten so a mistake in the mark scheme but that's fine we know what we're doing 18 out of 90. so write that so i'm writing the working 6 over 3 over 9 that gives you 18 over 19. notice that's all i need to write now for this one we need to be slightly careful we need one white and one red so that can be either way round this is important so we want the probability of red white plus the probability of red probability of white red there we are let's not get those mixed up so let's go back so the probability of uh white red so let's do this one first well that's going to be six tenths times two-ninths notice i'm not going to simplify this point so 12 over 90 and then we work out the probability of red white so white no wets just reds so we get four tenths times three over nine there we go and we also get 12 overnight and so we want to use these two answers here i mean i would usually write this working on the other page but i'm working between two slides so our answer is 12 over 90 plus 12 over 90 is equal to 4 over 90. now i know with the mark scheme i can leave it like this kind of they different to what they teach you in year 7 8 and 9 you'll be simplified whenever you get the opportunity tree diagrams is one of the few exceptions to that so we get 24 over 90. uh biggest mistake made on this question is you don't take into account both ways around so you do white red and red white let me show you the mark scheme here okay so 24 over nine fine now part three we want to make this as easy as possible for us so we want to work out the probability that they're different colors so notice you can either have two balls that are different colors or the same color okay it's one or the other and all probabilities have to add up to one so a quicker way of doing this question rather than working out everything individually is doing one minus probability of the same color because whatever we're left with after we work out the probability of same color is that it must be different colors the reason i do it this way is that we've already worked out some of these things so we've already worked out the probability of both white which is 18 over 90. so all we now need to do is work out the probability of red red and blue blue i think the other color was blue blue exactly and then we can just work it out from there so what i'm going to do is just work out the same color in fact it's even easier than that because we only have blue in the second ball not the first ball so all we need to do then is actually work out the probability of red red which if we times these two fractions together we get 8 out of 90 and that's the only way we can get two of the same color probability of white white or red red okay so i'm going to bring that over this answer here minus 8 over 90. now remember 1 on its own is 90 out of 9. this is why i did not simplify to make the attracting fractions much easier and then if i do this calculation well 90 minus 18 well that's going to be equal to 72 72 minus 8 well that's going to be equal to 4 over 90 yeah which as we saw in the previous mark scheme not need to be simplified so i'm going to leave it as 64 over 90. and there's our answer there so one thing i want to mention with this question is don't always think you have to do things in always the same way look for opportunities to save yourself time and this is a good opportunity think ah of different colors well if i work out probability of things being the same color and do one minus that i will get the answer much much more okay and you'll see the actual exams report actually mentions this so the probability of different colors was a little more challenging but well answered by many candidates because they took this longer method working out wet wet again white red red white red blue blue red blue white white blue and added them all together whereas if you use the more efficient method like i just mentioned here you'll find that's much much more so particularly an exam which is two hours 15 minutes long you want to look for ways to make your calculations more efficient and this is one possible way i notice this question is generally well answered you need to be getting all the marks here to make sure you're heading for those top grades and my number five is percentage calculations now this generally appears near the start of a paper so it goes through things like compound interest simple interest and working between them as well as some reverse percentage change right question one so acid buys a one-year-old car he pays 19 975 dollars which is 15 less than its price when it was new calculate the price when it was new now one thing that i do notice in these kind of questions it's something i highlight to my pupils as well is this question is in the past tense so it says when it was new see that's what we're looking for so as soon as you're working backwards in time you'd be thinking reverse percentage problem that's the first thing that comes to mind now we're paying 15 less than the price when it was new so essentially this price that we have here is essentially 85 of the original price when it was new so we can say that 85 percent that's 100 percent minus 15 is equal to the price that acid bought it for which is nineteen thousand nine hundred and seventy five dollars what we want to find is a hundred percent we want to find the original price when it was new so we need to find some way of going from 85 percent to 100 now we can go directly that is a method that we can use what i tend to recommend to students is finding some kind of intermediary step and what i would recommend to do is work out one percent it's so-called unitary method and then from one percent we can really easily find then what a hundred percent is so the way we can go from 85 to one by multiplying or dividing well we can divide by 85 and then once we've divided by 85 because 85 divided by 85 is 1 then we can multiply by 100 to give myself the original price doing it in two steps but you can see my very very clear working here whatever we do to the left-hand side of an equation we of course do to the right-hand side so we're going to divide this side by 85 and times by 100 in the same way so this is where i get out my gdc and then actually work out what this is going to be so let me now open up my gdc so i've changed the format slightly here so i've made it kind of bigger so you can really see what i'm doing so i've got the calculator on the left hand side here so you can see what buttons i'm pressing so you've got one nine nine seven five we're going to divide that by 85 this gives me 235 so i'm going to go back to my powerpoint and i'm going to write that in 235 and at this point you can use your calculators times by 100 that's fine but remember times by 100 we're just going to add two zeros on so we get our final answer of 23 500. germany on paper fours they give you the units here already so i don't need to worry about putting dollars in at the end so my final answer will be twenty three thousand five dollars nice and clear working so even if i got this wrong for whatever reason on my gdc i would still get my method mark for the working i've shown here all good all right let's move on okay so in part b um asif's got two different options to actually pay for the car the first option is you can pay 10 of the 19 975 which he's paying for the car and then pay 345 a month for five years option two you can actually borrow the 19975 and pay this back at the end of five years at an interest rate of two point five percent we'll look at that in part two so as you can pay using either option using option one how much will acid pay in total for the car so we're going to work out option one well the first thing to notice here is that paying 10 of that 19975 well ten percent is essentially the same as saying one over ten so what we can do to work out the total amount here is we take the 19975 we multiply that by 1 over 10 that's essentially the deposit and then we're going to add on the extra that he pays over the five years well he pays 345 dollars a month so we're going to get 345 times 12 12 months in the year and then he's doing that for a total of five years so we're gonna times that by five notes being very efficient with the working here i haven't written any more than i need to so at this point i'm going to take this calculation and i'm going to put this straight into my gdc to generate an answer okay so i'm going to type this straight into my gdc you can see what i'm doing over here so i'm going to put in 19975 multiplied by control fraction one over ten okay and then i'm going to add on i'm gonna put this in brackets so 345 dollars times that by 12 12 months in a year then times that by five and if we put that all together we get a fraction but that's fine we can just press ctrl enter to give us the decimal that we're looking for which is two two six nine seven point five we're going to pop that back into my calculation so we get two six seven five zero now usually we always say round to three significant figures that's really really important however when it comes to money we need to put the exact value in here so notice we're in dollars and cents here so we do need to put two two six nine seven point five zero that's really really important you don't want to lose an accuracy mark at this point you'll get two method marks for the calculation are shown and then one for the answer here okay on part two we look at by how much is option two cheaper than option one so now we need to work out what option two is so i'm gonna write very clearly here we're going to do option two now here we use the compound interest formula so the way we do this is we take 19975 so 19 975 dollars we multiply it by the multiplier now if we're increasing by two point five percent a year the way we work out the multiplier is we do a hundred percent plus two point five percent that will give you a hundred and two point five percent and we want to convert that to a decimal multiplier the way we do that is divide by 100 and we'll get 1.05 so we put this into our compound interest formula so we put 1.025 and then we work out how many years we're going to be paying this interest for which is five and by doing this calculation directly this will give you the total amount of money that acid has paid for this uh borrowing from the bank so we pop that into our gdc and let's get an answer okay and we're going to type this in so 19975 multiplied by and i tend to put this in brackets brackets 1.025 close brackets the power button is over here on the left hand side we pop five we press enter and we get our answer like so two two five nine nine point nine so we pop that in okay at this point we just then take one away from the other so in order to work out how much it's cheaper and good this is cheaper than this so that's very very good we get two two six nine seven five we take away 2 2 5 9 0 and then if we work this out we'll get the following answer so let's just work this out now and i'd round this now to the nearest dollar and cents so i get 97 and 62 cents like so so just show you the mark scheme here so you can see 97.62 is the precise answer for your 4 marks you can see the method here i went straight into 1.025 writing rather than writing as a fraction which is absolutely fine so you can check your answers here one thing to show you before we move on to question two is the examiner's report you should be aware that this thing exists so generally this is a well-answered question and the first part was a reverse percentage question that was generally answered well just make sure you don't add on 15 but you divide by 0.85 or the method i showed you and already mentioned this point here which is that you do not want to round any money answers to three significant figures what you want to do here instead is you want to actually write the exact value here in dollars and number six is not to be underestimated transformation so make sure you've got tracing paper make sure you ask your teacher to have tracing paper as this also comes up very very often as well okay and question two so a very typical transformation question this is something you really should revise if you want to excel at the paper for so let's go through this so describe fully first of all the single transformation the maps triangle a onto triangle b this two mark question always use the amount of marks to guide how much detail that you need to put the first thing we need to notice here is that going from a to b we're just moving the shape essentially which we call a translation so you need to know these keywords so i'm going to write down translation which essentially means moving the shape and as soon as you talk about translation you need to be thinking of vectors so we're going to do translation by the vector and then we're going to write how it's actually moved we put this in bracket notation like so the top number tells you how much it goes left or right so let's work out what's happening here so to go from a to b i'm going to take a point here and match it up to a point over here so we're going to go one two three four so we're gonna go four to the right so we put a plus four here and if you're in the exam and not entirely sure you can always check another point so if we take this point and this point for example then we can check one two three four to make sure we've done it correctly and notice that we're not going up or down with our triangle so the number here is going to be zero so we get a translation by vector four zero one mark for translation one mark four four zero our next part is we're going to take triangle a and we're going to reflect that in the line y equals no subtitles y equals minus x so what we're going to do is we're going to take the original triangle here the triangle a always read this carefully if they want to do triangle b or triangle e etcetera and we're going to reflect that in the line y equals minus x the first thing we need to remember is what does the line y equals minus x look like so y equals x is this line going all the way up so the line y equals minus x is the 45 degree line going in the opposite direction so let me draw this in i'm going to try this first time i've been using the pad so let's see how straight this is not too bad probably needs a little bit of work if you're not sure where this comes from well let's have a look here when y is equal to minus 1 then x has to equal to positive 1 so that's true so you can check coordinates here to make sure you've got the correct line y equals minus x and what we're going to do here is do a reflection now i'd highly recommend that you use tracing paper to do this this is perfectly allowed on igcse without tracing paper this is also fine as well so let me choose a different pen color let's go with magenta why not so we take a point on our triangle we take the corresponding point that's perpendicular so at a right angle to that point and then we transfer it across to the other side so we've gone one to the left and one down so we need to go one to the left and one down so this purple point here we reflect in this line reflects over here again use tracing paper it will make your life much much easier let's take another point let's do this in blue so if i take this point here and i make a perpendicular line so a line 90 degrees to that blue point notice there's a right angle here i go two to the left and two down so we go two to the left and two down and then we get to this point here notice this should be the same length as the line over there and let's do it for completeness let's see if we can get a different color let's go with i don't want magenta i've already used magenta i want to use yellow there we are so we take this point here we do the same analysis this is a bit trickier because we're halfway in a box so we're going to go from here to here so you can see this is a right angle we are going essentially one and a half along one and a half down so we need to do the same thing and then we'll come to here again tracing paper makes your life easier once you've done the three points either using tracing paper or using method i've done here then you can just draw in the triangle like so not the best triangle in the world we're going to label the image c so make sure you label this for your examiner c so they know the triangle that you've actually drawn that's probably the worst triangle i've ever drawn but you can see the idea here for part c we're going to take go back to this original triangle a and we're going to rotate this 90 degrees clockwise remember clockwise goes in this direction following the clock about the center one minus one notice we're not using the origin we're not using the center point here we're using a different point so we're going to be using this point here one minus one i'm going to put in black again use tracing paper so put tracing paper over the shape draw around it mark in the center of rotation and then spin it 90 degrees however we don't have access to such clever things here so we're going to do this manually to notice that in order to work out the rotation we know it's roughly going to be over here because we're turning this direction the way we would find this out um to keep it fairly straightforward is that it's going to make a right angle with where the new point is going to be and it's going to be the same distance as well so this point here is going to be rotated to here when i've got this reference point then i can fill in the rest so i know that it's going to look like this again you can visualize that in your mind that this is coming around to here but in order to know where it's going to be which is the most important thing you need to find a reference point so you use this point to hear but again use tracing paper it will make your life much easier we are going to label this d so we've done this question this question and this question and the reason we label it is exactly for this type of question down here d describe fully the single transformation the map triangle c to d so essentially we want to go from this triangle to this triangle here now the first thing to notice here is there's a reflection going on so we need to write down then reflection and it's a two mark question i'm using that as a guide here so we need to write another bit of information and that is the reflection line so we're going to have a reflection in the line remember when i first talked uh gcse and i gcse they're very fussy about the prepositions you used but now they're fine just say reflection in the line and i think we can see here that the line is going to be this vertical line going all the way down and we can check this at this point here is two away from the reflection line and this is also two so we can use that as a check now let's think about how we identify this line well if we take all the coordinates on this line so this is 1 5 this is 1 4 if i go down the bottom this is 1 minus 5 etcetera what do all these coordinates have in common and they all have an x equals one in common that gives me the name of this line then of x is to one one mark for reflection one mark for x equal now for the last part i'm going to get rid of the pen because we need to be able to see what is going on there we are black magic and what we're going to look at here is the single transformation the maps a to e so going from this to this and you first look at this and think well it's not quite an enlargement because it's not in the same proportion it's not a translation either because it doesn't look the same like a to b so then we have to think it's like an enlargement and it's kind of like a translation what would be kind of in between and that would be a stretch now make sure that you have learnt about stretches this is something that often gets forgotten by igcse teachers make sure you know how to do stretches as well now this is a three mark question in comparison to these two mark questions so we need three bits of information so first of all we have a stretch uh the first next thing we need to work out is our stretch factor okay so i'm gonna put stretch by factor and now let's have a look at this so let's take uh corresponding sides well this side is one and this side here in comparison is two so like an enlargement we compare the sides and go oh well this is one and this is two so we must have a stretch then of factor two and the last part which is probably the most trickiest part here is actually working out what the stretch where the stretch is actually occurring from and we call this the invariant line or the invariant axis yeah i'm going to call this invariant line because you don't know whether it's an axis or not and then we need to work out where it's being stretched from so let me go through that nice and carefully this is probably the hardest part of the question the easiest way of doing this is a kind of trial and error basis so for example i take the x-axis which is going to be our correct answer but let me show you why that's the case we look at corresponding points so this point here and this point here of our shape the distance from here to here is one whereas if i take the distance from the my supposed invariant axis and this point here that is going to be two now let me take another point let me take the top point of our triangle this magenta if i go from this point to this point this is going to be two but if i go from this point it's all the way up to the top this is four so you can see there is a consistent stretch going on from this particular line and that's what we're looking for likewise if i do this for completeness from here to here is one whereas from here to here is going to be two so for every point in my triangle there is a consistent stretch upwards and if you find this consistency then you know you found the correct line so actually here we have an invariant axis not an invariant line so we're going to call this by invariant line the x-axis i'm going to write in purple because why not so we get one mark for stretch one mark for factor two notice it's not an enlargement factor it's a stretch factor so i'm just writing factor you usually don't lose them up for this but just be careful and then our invariant line so where the stretch is happening from is the x-axis think of this as if you're putting your foot on the actual line and then you're stretching the shape upwards that will give you a good idea of what is going on so let me show you the mark scheme so we've got a translation for zero these the coordinates of the triangle are reflected and rotated reflection x equals one and then stretch we'll mark factor two or stretch back to two and the x-axis is our invariant line okay i put invariant line but as long as you've got invariant there that is important um quickly go through the exam review after this exam was done last year as well done by most candidates um yeah you do need to put translation that's really important you can't just put move for example or shift not good enough um most b c and d were generally done correctly and many students got the stretch of the transformation with scale factor two but finding that invariant line is that discriminator question between an a and a star so do make sure you know it and my number seven is a tricky topic for the igcc standard this is functions so working with composite and inverse functions i'm going to go into a detailed question that goes through the very typical kind of questions you are going to see okay and on to some functions for question five a very typical paper for question so we're going to start with some substitution here so work out f of -2 so to work this out wherever we see an x in our function we're going to replace it with minus two so i'm going to copy across the two i'm gonna put a minus two where the x was and then leave everything else alone and then we just have to do some calculation here so 2 times -2 that's equal to minus 4 minus 4 minus 1 that's equal to 5. and for the second question it's a very similar kind of process here so we're going to work out g of minus 2 first of all so i look at my g function here wherever i see an x i'm going to put a minus 2 so we get 3 minus minus two notice we've got two minuses there and if we put those together we get minus and the minus is the plus so three plus two is a five now this question is slightly different because we take that answer now and now put it into our h function so we're going from right to left our final step is to work out h 5 and we put a 5 where the x is in our h function and we get 5 squared which is equal to 5 which we pop over here the next one is a typical follow-up question to this where you need to take the whole function and make it equal to something so notice our function here is 2x minus 1. so we're going to write in 2x minus 1 where the f of x is equal to 7 and from here this is a straightforward solving equations so we want to get the x on its own what's the opposite of minus 1 well i'm going to add 1 on both sides this cancels here leaving us with 2x is equal to 7 plus 1 which is 8 and then what's the opposite of timesing by two almost gave you the answer that dividing by two and if we do that to both sides we then get our answer of x is equal or we'll drop in over here so you really need if you want to get that grade c on this paper you need to make sure you're getting those five first five marks correct 100 of the time question c is a typical composite function question i'm going to write here the word composite and we always work from right to left so what do i mean by that well we start with the g of x function so g of x is equal to 3 minus x so i'm going to replace that here with this x and in order to work this out so i'm going to put the entire function here so let me put this under magenta so wherever i see an x in my f function i'm going to replace this x with the 3 minus x now what does that look like well let's put that in so we get 2 stays as normal open brackets 3 minus x so i've replaced the x here with a 3 minus x minus 1 and remember this is not the finished uh answer here you need to simplify we're gonna expand expand out the bracket using the hook so two times three is equal to six two times minus x is equal to minus two 2x don't forget our minus 1 over here and then if we bring it all together 6 minus 1 is 5 and get our final answer of 5 minus 2x which you need to have simplified to arc then we go to question d which is a kind of slightly strange question here it's gonna kind of untypical so we need to solve this kind of function equation so again we just substitute in the functions that we have here into the equation so f of x is equal to two x minus one we need to multiply that by g of x which is three minus x then we're going to add on two lots of h of x so two lots of squared and this is equal to zero so essentially this is a quadratic question so wrapped up in function notation at this point we need to then expand the bracket using foil first outside inside last so 2x times minus x is minus 2x squared so then we can do the first term here 2x times 3 is 6x minus 1 times 3 is -3 and then minus 1 times minus x positive x don't forget our 2x squared over here and that's equal to 0. then you can do this in lots of different orders but as long as you make sure you multiply everything by everything now notice here this is something quite convenient we have minus 2x squared here we have a plus 2x squared here these two things cancel that's super we have 6x plus x well that's going to equal 7x minus 3 can't be simplified and now we've got actually a really straightforward equation from that it looked very complicated at the start so if we now simplify this while we do it in the same process process we did before so let me just pop this over here we've got seven x minus three is equal to zero what's the opposite of minus three well plus three on both sides this cancels we're left with seven x is equal to three and now at this point what's the opposite of times in by seven divide by seven on both sides giving us our final fractional answer of x is equal to three 7 which i'm going to write in like and the decimal would also be acceptable as well and now we've got the classic inverse question here so we're going to find the inverse of the g function here so we do this in three easy steps number one we set this function equal to y so y is equal to our g function three minus x step two is we swap over the x's and y's so we do x is equal to 3 minus y and our third more complicated step is to make y the subject okay so i'm going to write that down because that's really important for inverse functions make y the subject if you follow three steps you can do any inverse question algebra is complicated but you can do it so the way we simplify from here i'm going to add y on both sides thinking why is that is that going to help well we'll see in a moment so we get x plus y is equal to 3 because these cancel and then finally because we want y on its own we do what's the opposite of adding x well we're going to minus x from sides and if we do that process we then get our inverse function which is y is equal to three minus x which is you know kind of cool really because you know we have the initial function three minus x and its inverse is the same function so it's equal to three minus x you can show this actually graphically it's quite interesting to see and number eight also likes to come up regularly as well this is the sine cosine rule bearings question this can be a good question with lots of marks and it's relatively straightforward as long as you see the patterns within the question which i'm going to go through now very very typical sine cosine rule bearing question so we've given some various angles and some various sides and we need to calculate various lengths and angles and areas so our first thing we need to do is calculate the length of a b so that's this length here notice we have a right angle triangle we have an angle on the side that really looks to me like socatoa or basic trigonometry so if i draw a very quick sketch like so this is 150 this is my right angle like so and i've got an angle here of 55 degrees as you know with sokatoa we label the sides and angles this angle here well it's opposite this side this is the hypotenuse the longest side and of course the other side is going to be our adjacent like so now we want to find a b the adjacent so we want to find the adjacent we know the hypotenuse so we're going to use cosine so i tend to teach it using triangles it's not the only way so cos times h and a so these are the triangles or normal soccatoa nh now we are looking for the adjacent so we cover up the adjacent so we do cos of the angle times the hypotenuse so we do the calculation of cos 55 times 150 like so and now we work that out on the calculator so i'll do it on the gdc on my desktop so we need to make sure it's in degrees that's very important so let's start a new document so we might save that because we've already done that add calculator you can even type in cos 55 but i'm just going to use the standard function here so cos 55 what times y and let's just go back and check 150 type in 150 we get 86.0 and a quick check of the units it's meters yet so generally we do it to 3sf so we get 86.0 okay all good now calculate the bearing of d from a so we want the varying of d from a so we start at a and then work our way around to d so we need to find this angle here the d from a so we start at a work clockwise around until we hit d so we've already got two of these angles already this one and this one so all we need to do and let me use the eraser let's find this angle here in the triangle uh i always forget where the eraser is on this okay it doesn't matter but we need to find this angle here specifically okay and we've got well one two three sides okay so we've got three sides we want to find an angle so we need to use the cosine rule but for an angle so the cos of c is equal to a squared plus b squared minus c squared all over 2 a b where c is the side opposite the angle so it'll be 235 in this case now you're probably wondering how did i get this formula how do i know this formula well first i learned it off by heart but if we go to our formula sheet right at the start let me scroll up quickly notice actually in the formula sheet here we have it only in terms of the sides so a squared so this formula i just given i'd recommend learning off by heart so you don't need to put yourself under pressure and rearrange this formula to get cos a in this case the subject notice with this formula because a equals b squared plus c squared minus a squared over 2bc so it's a is going to be the angle you focus on whereas here i'm using c both are absolutely fine you can use so if we work this out so we've got 120 squared plus 150 squared minus that minus size would be sticky minus 235 squared all divided by 2 times 120 times 150 and once we've worked this decimal out so we'll get a decimal of some kind so let's go back to my gdc so we want to do it as a fraction so we go control and divide 120 squared plus 150 squared minus and it was 235 that's it 235 squared oh the answers nice squared calculator is pretty clever at working out business you've got good math sorry as you go along so times 120 times 150. but we'll do a common sense check at this point so we get that and if you press control and enter we'll get a decimal and you need to ensure that this decimal is between minus one and one otherwise you've done something wrong however we want to that was the cause of the angle so we need to find the inverse cost of that answer so we go to trig pause inverse and i want to work out the answer so i go control here and then press enter i get my angle which is 120.599 so i go back and so my angle c is nine 120.599 that's not the end of the question so this angle here is 120 so we still need to add on 1755 to work out our final answer kind of running out of space so our answer is equal to 120 dot dot plus those 12 angles so 70 and 55 make sure this is a three figure bearing as well to go back to my calculator again try and never round too early it's one thing i say to my students a lot so i go control answer plus 70 plus 55 and then get the total bearing which is 2 3 sf 246. so i'll come back to here write in bearing as 2 4 6 3 like so okay excellent all right one more question to go and then i'll stop the video for the moment to calculate the area of the field abcd essentially we've got two triangles here um i think the easiest way of doing this is to work out one triangle at a time so let's call this t1 and t2 because we've got all the information that we need to work these things out let's work out t1 first of all so we know our angle here that's what we worked at earlier 120.599 and we're going to use the other formula on the formula sheet with regards to triangles that is half bc sine a and we're essentially going to use that twice so let's scroll down what i just did so let's work out t1 so t1 going to be equal to well remember use the formula we need the angle in the middle and the two sides next to it so we do half times 120 times 150 times the sine of the angle we worked out in the previous question you'll notice i'm putting as many decimal places as i can down to reduce the idea of any rounding error so let's type that into the calculator so i'm going to use 0.5 times 120 150 times the sine of 120.599 so and so i get this answer here as many decimal places i should in the future make sure the float though the amount of decimal places used is a bit longer otherwise my answer will be slightly inaccurate so i'll go back and go 7746 0.76 so okay so that was t1 for t2 we do very similar process so we start with the half and now 150 but we also know this side as well because we worked this out earlier as 86.0 um let's see if we've got any more decimal places on that so i scroll up yeah so it's 86.0365 so it's worth writing that in okay 86.0 the thing is better when you've got a calculator next to you three six five let me just write that in got it there so we need to use the formula again so a half times 86 times 150 times the sign of the angle which is a nice precise answer in this case got 55 excellent and if we do that on our gdc let's go back so it doesn't matter what order you're multiplying of course so let's write down the decimal before i forget times 0.5 times 150 times the sine angle which i think was 50 55. so it's worth always checking these things go along we get an answer so 5285.77 okay and just to make the video a little bit shorter we just add together so 7746.76 we add on five to eight five point seven seven and we get i will do this to 3sf as always so 13 000. so i'll write that in 13 000 meters squared remember we're doing it to 3 f and then the 3 doesn't round the zero up at all so answer obviously it takes a bit longer on the video working between the gdc and the document but you can see the clear method that i used in order to work it out and my number nine is proportionality or variation and different teachers use different words for this this can appear on paper two and paper four make sure you're happy with some twists on this as well and you'll see that in the very last part of this question so why is inversely proportional to the square root of x i'm underlining the key information here and then we're given some initial conditions which we'll use later and we want to find y in terms of x so we're going to write down our proportionality statement as soon as i see the word proportional i think proportionality statements and let's go through this carefully so y is proportional to or inversely proportional so we're going to write 1 over because it's inverse and then the square root of x so we're going to write the square root of x this thing gives me the equation so we transfer this into an equation and the way we do that is putting a k for where 1 is so we get k over root x check out my video on proportionality that will go through this in much more detail at this point we then substitute these numbers in so when x is equal to 25 y is equal to four this reminds me the x goes here and the y goes here so we go four equals k over root five just take our time here so 4 is equal to k over well the square root of 25 is equal to 5. you're not sure on the exam check on your calculator it's the way to go at this point well we want to get k on its own so what's the opposite of dividing by five we want two times by five both sides cancel and so we're left then with the answer of k e twenty now that's not our final answer our final answer is when we put this k is 20 back in this original equation at the start so our actual answer here and i'll just remove some of the working so we can see it nice and clearly is equal to well we know k is 20 so we put 20 back in here and then over the square root of what are two marks now what's typical here is that it gets you to substitute y to find x or substitute x to find y and you can see both questions are here in b and c so find y when x is 0.25 so when x is 0.25 y is equal to 20 divided by the square root of 0.25 at this point okay that looks more like a 6 and a 5. let me try and write that a bit more clearly as a5 just put this into your calculator okay you can use your gdc or otherwise 20 divided by the square root of 0.25 and that's equal to 40. paper 4 use your calculator for straightforward calculations this question works the other way around so when y is equal to 5 you put 5 equals 20 over root x i've got a slightly strange equation here but we treat it in the same way we want to remove the denominator so i'm going to times root x on both sides cancel then i get 5 root x is equal to 20. i'm going to copy this across over here so i have a bit more space and we want to get the x on its own so we divide by 5 on both sides that gives us then vertex is e4 and at this point you can do this mentally or then think to yourself what's the opposite of square rooting well that's squaring both sides that then gives us the answer of x squared which is 16. pop in there again at any of these points you can use your calculator to help you work this out good the last question here is a little bit sneaky so this time they've introduced a new variable z z is proportional to y plus two then they give you the initial conditions in terms of x and z so i'll show you how we go about this kind of a a stars kind of question if i write my proportionality statement so z is proportional to there's no inverse here so it's going to be proportional to just i plus 2 which i'm going to put in brackets that means that if we transfer this to an equation this way we have to be careful then z is equal to some constant k open brackets y plus two okay brackets are really really important here okay this is why i put it in brackets here at this point you think ah well we don't have an x to work out what k is we actually do because if we go back to question a here we have y equals t over root x we already worked this out so at this point i can replace the y with 20 over root x so i get z equals k 20 over root x is two the substituting answer in from part a now at this point i can then actually work out my new constant k using the initial conditions so when x is equal to 4 it is equal to 84 and we just work through the algebra so 84 is equal to 8 open brackets 20 over root 4 2 84 okay open brackets well 20 divided by root 4 20 divided by 2 is equal to 10 use your calculator at any point here transfer this upwards here we get 84 to 12 k plus 2 is 12 then k and you can do this on your calculator or mental methods gives you seven so a new constant k is seven and so we want to copy across this equation made here but replace the k here with seven so we get is equal to seven open brackets x plus two for the sneaky proportion variation questions so that's definitely a a star part it's quite unseen it's not very very standard by any means it answers on the board so you can check through what i've done here and see where you get your method marks and the examiner's report a and b were very standard and was done correctly um some people didn't do this last part correctly which is slightly embarrassing so make sure you've got the square root of x equals the four you have to square both sides but this part was pretty well done considering it's not really a standard kind of question so well done if you've got question 10d correct and definitely not to be forgotten number 10 is volume surface area and similarity questions and often be combined together or they can be separate as well so make sure you know how to use your formula sheet here and i'm gonna go for a pretty tricky question here so you really have to concentrate hard in the very last part of the video okay let's continue so this is a pretty tough question this is question nine so we're going to look at some similarity first of all and we've got two similar triangles e a b and e c d what the word similar means in this case is that they're in the same proportion okay imagine one is an enlargement of the other and we need to show that ec is equal to 120 centimeters and so what that is is the entire length of this line equal to 120 centimeters what we're looking to show first of all notice we've got two corresponding sides here so we've actually got inside here d and then we've got this side 20. so i teach it slightly differently maybe to other teachers and the mark scheme is also different the way i do it is by working out the scale factor which i generalize as big divided by small get 20 with a 15. okay which if you wanted to you could simplify down of course if you divide by five top and bottom you get four over three okay please avoid decimals decimals will make your life a little bit more difficult so that's our length scale factor here so we've got four over three what we need to do now is use that to then work out this missing length so imagine i don't know this length so if i wipe this out imagine i call this blue length that i highlighted earlier x yeah so this length at ec then the entire length of the line is equal to x is 40. this length is x then to go from a to e we just need to add 40 to x okay now notice these are the corresponding lines of our small triangle and big triangle so what we're asking ourselves is okay if i take x a smaller triangle i times it by four thirds our scale factor i should then get to my long length which is x40 it actually gives us a reasonably straightforward equation to solve so if this times together we write this in good algebra as four thirds x this is actually a pretty straightforward equation we minus x from both sides so four thirds x minus x this is a 1x so we get one third x remember one is the same as three thirds one third x well these two things cancel so 40 and what's the opposite of dividing by three will be times by three most you can see the answer here at this point this cancels here and we're left with x equals 120 which is good because that's what the answer should be because that's what it says there i've noticed many teachers don't teach it that way but i think it's probably a more intuitive way than doing ratios of sides i don't know it's up to you how you want to interpret that question okay we'll see the mark scheme at the end once we know that's 120 we need to work out the following things so we need to work out ed and db so let's highlight what ed looks like it's this length here and if we think about it we've actually got a right angled triangle so i'm looking for e d e d okay so let's just take that triangle and so let's draw it separately essentially what i have here triangle where i don't know this i know this is 120 from what we just did and we know this is 15. so essentially we can use pythagoras on this question just a straightforward pythagoras question so let me draw it onto this slide 120 this 15 is x and so pythagoras says the x squared is equal to 120 squared plus 15 squared so x is equal to the square root of those two things since i'm writing the minimum of working enough to guarantee me all the marks here and this is where it's gdc so you're going to normal scratch pad you're familiar with and go control 120 squared so you use the square button plus 15 squared parallel to here of 120.93 again we're going to round to 3sf but we need to write down decimals first and two three significant figures which is generally expected it's 120 centimeters okay all good let's try and go back to the black pen there we are black pen and now we need to find db so let's have a look at db and where that is so db is this length here okay uh i think the most straightforward way because we've already got this length here we already know d e we can use pythagoras to work out the entire length of the line having fun with my colors here and use pythagoras to work out the entire length of the line uh 20 and this is 160 and take one away from the other to work out the short length so let me go on to the next slide i'll show you what i mean just 20 with 160 we've already worked out this part it was 121. so we just used pythagoras for the entire length if we called the entire length a sum y and y is equal to the square root of 160 squared t squared okay and then to work out this length call that s is going to be i minus 120.93 from before and i will just skip to the answer at this point you do that calculation then get 40.3 significance okay so the way i do part three again slightly different to the mark scheme i think they recommend use ratio similar sides but i just use pythagoras twice and then take one away from the other look for the most straightforward ways of being able to answer the question okay and now the next part i do like it's hard certainly i can see this is the aa star material but it really does test your understanding of similarity and your knowledge of volume of shapes the diagram shows an open waste paper bin made from metal now this is actually important words that we'll talk about in the second part of the question from waste paper so this is hollow so it speaks only this circle exists this is the circle top is 20 centimeters raised to the bottom is 15 perpendicular height of the bin is 40. and using part a notice these lengths 40 20 15 these all correspond to this diagram here yes and notice that we're going to use that later on so using answers in part a calculate the volume of the waste paper bin now this shape is called a frustum if you're pronouncing that right waiting to work out anyway um and the way you work out a frustum is to think if i extend these lines i'm going to kind of go over the question here but just to show the point essentially you're going to make a cone yeah a very big cone so if you take the entire cone okay that's the entire shape including this bit so the way to work out a frustum is to work out the in this case volume the big cone then minus the area or the volume sorry of the smaller code now this is where you need to use your formula sheet so let me write this up here the volume of a cone equal to one third pi r squared h this is on your formula sheet don't worry you do not need to remember this absolutely fine so to work out the volume of this we do with the whole shape we take the volume of the bigger cone which is one third times that our radius is 20 now you're probably thinking at this point ah i don't really know the height i don't know the perpendicular height of this but if we think about it we actually do because we go back to the question we looked at before the perpendicular height remember so the entire length of the line is the perpendicular height which in this case is 160 we use the 420 plus 40. 160. so this is where you're using your passes from part a so actually we do have the entire perpendicular height it's 160. we minus muller so we get one third times pi times 15 squared times this perpendicular height again we've also worked that out as well that was actually given to us right at the start is 120. it's this length here but we type in 120. okay we do this and this is where we're going to type this all into our calculator so carefully just need to be very careful as we do it so i'm going to put this in brackets so we've got control one third times [Music] times 20. very important yeah we're using the answers from part a if we put that all into our calculator we get the answer of 38 746.3 always we're going to write down a few decimal places and then round to 3sf so the 3s7 is going to be 38 000. which is the answer right here okay and very last part which is the really tricky part but we just have to think about this quite carefully now we need the area of metal needed to make the bin okay so we need to think about exactly how we are going to work that out the area of metal needed to make the bin so just to remind you okay we've got the circle at the bottom that does exist we've got then the curved surface area which we'll need to work out as well and then remember the top is empty so we don't actually have a circle here so give yourself a minute or so have a think about it how would you go about working that out okay well it's gonna be a very similar way to how we did it before so we're gonna find the curved surface area of the larger cone and we're going to take away the surface area of the smaller cone in the same way did it before the thing is however the formula for the curved surface area of a cone is equal to pi r l okay where l is that diagonal length so if i draw a cone for example pi r and l is this length here and the first size we don't actually have that length yeah we don't actually have the length needed in order to work it out but then we think ah actually we've already worked this out we worked this out in the previous question here already worked out that green length okay which was part two 120.933 we've already worked out this purple length here as well which was using pythagoras we actually do have these lengths so let's put this all together so we can actually work this out let's put it all together let's start with the curved surface area of the frustum i always forget there's an r in there maybe you can correct me in the comments so we're going to do pi r l of the bigger one so that's going to be pi times 20 l was this calculation here so i'm going to round now normally i would say not to round this but just to show you exactly what's going on here we actually do need to work out the length of the entire line so we want eb so we want this one here i don't actually have a value for that so i'm going to write it as this so that's the curved surface area of larger one and we're going to minus the surface area of the smaller one so that was pi times 15 and then the diagonal length so let's go back i think it was the ed was ed so it's this one here i'm trying to avoid rounding errors it's not like everything there should be a type uh plus plus this calculation works out the curved surface area of the first strum so this part here okay and what we need to do now is add in this extra circle at the bottom okay so this is the bottom of the waste paper bin so we need to add in a pi times 15 squared the area of the circle so go back plus squared and at this point i'm going to put this all into my calculator this is all about being very accurate here making sure that everything written correctly so go brackets let's see how this goes my times 20 times and open another bracket so i'm going to put a square root in you'd have to do this all in one step but it's good to get used to being able to do this in one step 20 squared we're going to close that okay i'm going to close that bracket again make sure everything so it should be brackets wise good and then minus we'll do a similar calculation now times open brackets square roots 120 and then we want this part now so we've got i impart let's where we hope everything works out to answer which is good that should be exactly as it is 5139 okay in fact i think my answer is more accurate from theirs because they rounded too early so we've got a thousand as you can see the answer was acceptable 5139.3 dot dot dot that gives us the answer of 5139 as you can see falls within the range of so it doesn't matter sometimes to a certain degree it doesn't matter in terms of the rounding because it lies within the range that they expect okay so you get four marks for that notice i haven't really written too much working there but it's very condensed yeah this is the curved surface area of the large frustum this is the curved surface area of the small first room we take one away from the other and then we don't forget of course to add in the base circle as well and i'll put the mark scheme up on the board so you can have a look it's been done slightly differently some of these questions particularly part three and part one i've done it a little bit differently maybe you find it easier maybe not okay but you can see you don't need to write lots of working but it needs to be clear it needs to be is what you are doing to the examiner before i finish this video i'm just going to look at the examiner's report so the similar triangle situation that we encountered here is actually one of the most challenging questions on the paper in some ways i find a bit surprising there are other harder topics but in a topic like those are my pupils i've taught them in year 10 but then they haven't had a chance to really go through it so it's something that gets easily forgotten and so ai was actually quite badly answered even though it's the first part part two was pythagoras which is good problem with getting a part i incorrect well you still get that value to work with but it probably has an effect on your performance in part three so you don't see the similar idea and a lot of people forgot about a to answer out b as well okay so they forgot they could actually transfer their knowledge from a to b and with the international mathematics course is a little bit on the non-linear side unlike some of the other igcse courses you do need to actually look through the previous question the investigation paper paper six is quite like that as well where you have to use previous calculations to use in subsequent calculations um however most people found the volume of frustration that's sort of standard uh working out of the shape so that's very very good and as we can see that favorite word discriminator the surface area question is quite hard because many peoples did not realize to actually use the sloping heights from before and as i pointed out the area of the circular base was often emitted it wasn't there so this is a good question if you want to get those a a star grades you need to be getting these questions right yeah it's an error if you're looking for those higher grades really to focus on so congratulations if you made it to the end of all two hours of video hopefully this has got you really well prepared for the igcc-060 course please do check out my playlist just over here so i have my paper 2 and paper 4 playlist which goes through a lot of the things that we've gone through in this video as well as my paper 6 playlist which goes through the investigation modelling paper all right best of luck with your exam