For q24 you can just say xy-px-qy +pq=pq then factorise (x-q)(y-p)=pq so the greatest y can be as part of its bracket which is an integer factor is pq+p so x must be q+1 then same logic.
I was also doing this past paper yesterday and I xy-qy=px therefore y(x-q)=px so y=px/x-q. To make y largest as x-q has to be an integer x-q=1 so y=px/1=px so then we can say that y-x=px-x so y-x=x(p-1) and if x-q=1 as we established before x=q+1 therefore y-x=(q+1)(p-1). Happy I got this one correct if it was as you say a hard question and I am only in year 11
The 'easier way' for Q8 is to form rectangles by joining the interior points of two equilateral triangles on the left, and then doing the same on the right (or on any other two opposite sides of the shape). The resulting four 'quadrants' of this shape will be equal in area, and if you draw a similar rectangle with four quadrants in the middle of the hexagon, you can see that the shape is made up of 18 triangles of equal area, 6 of which are within the hexagon, therefore the area is 6/18 * 216 = 216/3 = 72
Thanks for the video, helped me massive to calm my curious and nervousness down of whether I did decently or not. Hopefully your answers (and mine) matches with the mark scheme. Keep up the good work, your are changing the lives of so many mathematicians.
For q20, if you draw in the whole circle, with a square inscribed in each quadrant, you get 5 squares of side length x (one per quadrant plus the one in the middle). Then we can see that (3x)^2 + x^2 = 20^2 (as diameter is 20), giving x^2 = 40. It's easier to explain with a diagram, but I thought it was a pretty way of doing the question.
thank you for the quick release RDrewwwwwww. I think Q19 is 3 solutions since two of the solutions are just x and y flipped so they count as the same pair
q8 notice the isosceles triangles have the same area as equilaterals so count 6/18 of total area shaded q24 y=px/(x-q)=p+pq/(x-q) maximised when x=q+1 as y is integer so y=p+pq
(6666666^2 - 3333333^2) 6^2 = 36 , 3^2 = 9 36 - 9 = 27 7 + 2 < 10 so you can add them without the sum changing (2 + 7) * 7 = 63 (7 being the number of digits in each)
A simpler way of thinking of the same thing is (9999999)(3333333) using difference of 2 squares, and then 9*3=27, 2+7=9, and this happens 7 times so 7(9)=63
This is completely unrelated but does anyone know what the answer format for the MAT is. I've searched online to try and see but there doesn't seem to be any info. Do you answer on lined paper? Or maybe on the question paper itself? I know there is a space for the multiple choice, but I can't seem to figure out where to put answers for the long answer questions. Also, are you allowed to ask for more paper to submit, or is there a set space where you can write your answers. Thanks in advance!
Q7 you have 66p and 85p - in particular the units digit of the sum must be 0. The only way to get a multiple of five that is a multiple of 66 is by time sing by a multiple of five - where 5 is the smallest. It follows that 85x2 works with this. (Just more robust for proving this is the smallest). Q18 you form two equations - you can just add them and then divide by four not all that weird stuff! Q20 there’s a right angled triangle - middle of the bottom edge of square to bottom right vértex of it to top vertex of quarter circle. Hypotenuse is 10, long side x+x/2, short side x/2 Pythag to solve! Q23 nicer if you get rid of the area sf of 10 then times by 10 to finish! Can use 1:2:root5 triangle similarity don’t need to faff with all that specific stuff.q24 you already saw my comment q25 you can set up simaltaneous pythag equations with finding the height - little simpler and I got a really nice cancellation when I did this
What do you think the grade boundaries will be like? I’m really hoping for a bronze but my heart will be crushed if I don’t get it because I knew the answers to questions like question 14 but I was too scared to lock my answers in because I was worried I’d get it wrong and loose marks. I feel like I got over 50 (with the 25 marks everyone gets) at least but idk if it will be enough…
Using symmetry in the trig graphs Essentially this idea: ua-cam.com/video/F3jEmmeBXK8/v-deo.html Except I never use the cast diagram I just draw the graphs (here I drew the cos graph in my head)
For q24 you can just say xy-px-qy +pq=pq then factorise (x-q)(y-p)=pq so the greatest y can be as part of its bracket which is an integer factor is pq+p so x must be q+1 then same logic.
Nice, I knew I was missing something there. I'll do this when I remake the video with some optimised solutions
@@mtsfun7183 yes
Interestingly the mark scheme uses my not so pretty solution rather that yours, very surprising
I was also doing this past paper yesterday and I xy-qy=px therefore y(x-q)=px so y=px/x-q. To make y largest as x-q has to be an integer x-q=1 so y=px/1=px so then we can say that y-x=px-x so y-x=x(p-1) and if x-q=1 as we established before x=q+1 therefore y-x=(q+1)(p-1). Happy I got this one correct if it was as you say a hard question and I am only in year 11
@@kubdnice work buddy
Was my first maths challenge and yikes this video does not make me feel confident but it’s informative and helpful, hope i do better for next year
Watching this video knowing full well I got 10/125
10 more than people who didn't bother
@@rtwodrew2 15 less technically; you start with 25 marks.
@@phantomfire5013 no becasue they dont get a mark?
The 'easier way' for Q8 is to form rectangles by joining the interior points of two equilateral triangles on the left, and then doing the same on the right (or on any other two opposite sides of the shape). The resulting four 'quadrants' of this shape will be equal in area, and if you draw a similar rectangle with four quadrants in the middle of the hexagon, you can see that the shape is made up of 18 triangles of equal area, 6 of which are within the hexagon, therefore the area is 6/18 * 216 = 216/3 = 72
Thanks for the video, helped me massive to calm my curious and nervousness down of whether I did decently or not. Hopefully your answers (and mine) matches with the mark scheme. Keep up the good work, your are changing the lives of so many mathematicians.
Seems tougher than previous years!
I am regretting not doing it last year it was way easier fr
For q20, if you draw in the whole circle, with a square inscribed in each quadrant, you get 5 squares of side length x (one per quadrant plus the one in the middle). Then we can see that (3x)^2 + x^2 = 20^2 (as diameter is 20), giving x^2 = 40. It's easier to explain with a diagram, but I thought it was a pretty way of doing the question.
Thank you, I sat this paper and was waiting for this video to release!
thank you for the quick release RDrewwwwwww. I think Q19 is 3 solutions since two of the solutions are just x and y flipped so they count as the same pair
Even if they were flipped that still counts as a separate pair.
Wolfram says 4 solutions: wolframalpha.com/input?i=y%5E2-x%3D2022%2C+x%5E2-y%3D2022
@@rtwodrew2 ah, thanks for the clarification, hopefully it is 4 since i put that down. love your videos, wish you were my teacher 😭
q8 notice the isosceles triangles have the same area as equilaterals so count 6/18 of total area shaded q24 y=px/(x-q)=p+pq/(x-q) maximised when x=q+1 as y is integer so y=p+pq
(6666666^2 - 3333333^2)
6^2 = 36 , 3^2 = 9
36 - 9 = 27
7 + 2 < 10 so you can add them without the sum changing
(2 + 7) * 7 = 63 (7 being the number of digits in each)
A simpler way of thinking of the same thing is (9999999)(3333333) using difference of 2 squares, and then 9*3=27, 2+7=9, and this happens 7 times so 7(9)=63
could you explain what you did here please? how does 3 and 6 correlate to 6666666^2-3333333^2?
I don't remember exactly what I did but I noticed 63 was the only answer which 7 could go into and therefore had to be right?
This is completely unrelated but does anyone know what the answer format for the MAT is. I've searched online to try and see but there doesn't seem to be any info. Do you answer on lined paper? Or maybe on the question paper itself? I know there is a space for the multiple choice, but I can't seem to figure out where to put answers for the long answer questions. Also, are you allowed to ask for more paper to submit, or is there a set space where you can write your answers. Thanks in advance!
Q7 you have 66p and 85p - in particular the units digit of the sum must be 0. The only way to get a multiple of five that is a multiple of 66 is by time sing by a multiple of five - where 5 is the smallest. It follows that 85x2 works with this. (Just more robust for proving this is the smallest). Q18 you form two equations - you can just add them and then divide by four not all that weird stuff! Q20 there’s a right angled triangle - middle of the bottom edge of square to bottom right vértex of it to top vertex of quarter circle. Hypotenuse is 10, long side x+x/2, short side x/2 Pythag to solve! Q23 nicer if you get rid of the area sf of 10 then times by 10 to finish! Can use 1:2:root5 triangle similarity don’t need to faff with all that specific stuff.q24 you already saw my comment q25 you can set up simaltaneous pythag equations with finding the height - little simpler and I got a really nice cancellation when I did this
Ur acc so smart
Yeah, seems like I got 32/100. Not sure how to feel about it.
What do you think the grade boundaries will be like? I’m really hoping for a bronze but my heart will be crushed if I don’t get it because I knew the answers to questions like question 14 but I was too scared to lock my answers in because I was worried I’d get it wrong and loose marks. I feel like I got over 50 (with the 25 marks everyone gets) at least but idk if it will be enough…
That’s enough, bronze is 48 👍
mate relax i got bronze without even doing any practice for it and i even came an hour late
I solved 16 by just sketching the function with some sample input lol
How did you get cos 135 without a calculator?
Using symmetry in the trig graphs
Essentially this idea: ua-cam.com/video/F3jEmmeBXK8/v-deo.html
Except I never use the cast diagram I just draw the graphs (here I drew the cos graph in my head)
@@rtwodrew2 Danke :)
got 100/125 😊 🎉 edit: nooo I got 95 this is big L
Broo how tho. That is not normal lol how r u that smart?
48 for a bronze. I got 46 😢
I got silver :)