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MAT Prep Thread - 2nd November 2016

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Original post by m0.4444
The short cut was 5 choose 3 or you could have calculated 5 choose 2


No I got combinations right, I got the c values wrong.
Original post by Matte0
I also got something like that. Therefore using the previous part I believe n had to be divisible by 6?


yeah thats right. I think i put n had to satisfy the previous part and be even. Not sure whether thats full enough.
Original post by AmanGottu
No I got combinations right, I got the c values wrong.


How many marks do you think the c value bit was worth? I didnt have enough time to do the c value bit.
Does anyone know stats on how MAT score varies based on maths with joint schools? the feedback shows results only for maths but maybe maths and comp sci is higher/lower usually?
Original post by pressurize
Does anyone know stats on how MAT score varies based on maths with joint schools? the feedback shows results only for maths but maybe maths and comp sci is higher/lower usually?


They are probably the same. If they were markedly different they would publish the boundaries for each i think. Thats something to ask them.
What did you guys get for the mc that asked: when the origin would be in the circle?
Original post by dfbenjamin
0 isn't possible and I just realised why. I put the same as you thinking for some reason that 5^2 ended in 0 lol

ah well


5*2 ends in 0 though
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Original post by redsquared
Thats exactly how i think i did, hopefully we're on low end of the scale for getting an interview, and i put the answer for J was that if there is only one distinct prime factor then there are some final digits which are impossible, not sure if its right though.


For this question was it asking which one is false?
Also what was option A?
Original post by AmanGottu
What did you guys get for the mc that asked: when the origin would be in the circle?


For origin to be in circle, distance from centre to origin < radius. After some algebra I think it was c > 0
Original post by AmanGottu
What did you guys get for the mc that asked: when the origin would be in the circle?


i got that c>0
Original post by m0.4444
i got that c>0


Do you remember which choice it was. I think it was choice (a)
Original post by m0.4444
i got that c>0


Do u remember the exact question? It was something like

x^2 + ax + y^2 + by =c?? Idk
Original post by louisforrest
5*2 ends in 0 though


Yeah but P(5*2) = 2 not 1
For the last part of Qn 2, it sums (A^m)(B^n) some number of times to get 214x + 92 m, n >= 1. Surely since AB is 6x+5, then any combination of those two funcs is a multiple of 6. 214 isn't multiple of 6... Is this right way?
Original post by AmanGottu
Do u remember the exact question? It was something like

x^2 + ax + y^2 + by =c?? Idk


Yeah that was the equation.
Original post by collmr
For the last part of Qn 2, it sums (A^m)(B^n) some number of times to get 214x + 92 m, n >= 1. Surely since AB is 6x+5, then any combination of those two funcs is a multiple of 6. 214 isn't multiple of 6... Is this right way?


m = 1, n = 0 gives you just A, so that's not a multiple of 6
Original post by collmr
For the last part of Qn 2, it sums (A^m)(B^n) some number of times to get 214x + 92 m, n >= 1. Surely since AB is 6x+5, then any combination of those two funcs is a multiple of 6. 214 isn't multiple of 6... Is this right way?


Ah that was probably it, unless you could have one of the functions to the power of zero?
Original post by m0.4444
Yeah that was the equation.


I got the answer that had like 4c or something?? Is that wrong?
Original post by dfbenjamin
m = 1, n = 0 gives you just A, so that's not a multiple of 6


Idk, that's why I ask i can't remember whether it was m, n >= 1 or m,n >= 0
Original post by m0.4444
Yeah that was the equation.


For that I got (x+a/2)^2 + (y+b/2)^ = c + (a+b)/4

This is a circle with radius (c + (a+b)/4)^(1/2), centre (-a/2, -b/2)

So the origin is ((a/2)^2 + (b/2)^2)^(1/2) away from the centre

so c + a/4 + b/4 > a^2/4 + b^2/4
so 4c > a^2 + b^2 + a + b

what have I done wrong here?

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