Pre Maths Paper 1 Help

Man City 10

Could anyone help and take their time out to tell me how to do Q.2 (c)
Thank you, much appreciated

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Bluestrike

Man City 10 wrote: »

Could anyone help and take their time out to tell me how to do Q.2 (c)
Thank you, much appreciated

I had that for Christmas and i got full marks so here, it won't be very good quality as its my phone but hopefully u can see it

Bluestrike

Bluestrike wrote: »

I had that for Christmas and i got full marks so here, it won't be very good quality as its my phone but hopefully u can see it

Sorry, my phone won't upload it for some reason
just put in stuff u know for sure (stuff that says only). To get A etc you have to add up the centre and the parts where that circle and another collide and take that away from amount in A

Eg: 36 in A, 7 in A and B, x in A and C, 10 in A and B and C

you would do 36-(7+x+10)
36-(10+x)
36-10-x
24-x

Then put that figure in A and do same for rest
to find x u put the total on one side and add up all the figures in the different sections. You should get the total on the left and i higher sum minus x on other. Simply use algebra to find answer then

Man City 10

Bluestrike wrote: »

Sorry, my phone won't upload it for some reason
just put in stuff u know for sure (stuff that says only). To get A etc you have to add up the centre and the parts where that circle and another collide and take that away from amount in A

Eg: 36 in A, 7 in A and B, x in A and C, 10 in A and B and C

you would do 36-(7+x+10)
36-(10+x)
36-10-x
24-x

Then put that figure in A and do same for rest
to find x u put the total on one side and add up all the figures in the different sections. You should get the total on the left and i higher sum minus x on other. Simply use algebra to find answer then

any chance you can tell me the answer because how did you get 36 if there is only 31 in a

Bluestrike

Man City 10 wrote: »

any chance you can tell me the answer because how did you get 36 if there is only 31 in a

I was just using that as an example but the answers are
(i) A is 12
A union C (AUC) is 10
A union B (AUB) is 4
B union C (BUC) is x
A union B union C (AUBUC) is 5
B is 12-x (21-(4+5+x) = 21-9-x = 12-x)
C is 7-x (22-(10+5+x) = 22-15-x = 7-x)

(ii) The maximum it can be is 7 as C is 7-x so if x was 8 it would be 7-8 and you cant have a minus number of people

(iii) I'm not fully sure with this one but i think it is 7 again because the C thing with 7-x?????

(iv) 48 = 12+4+5+10+x+12-x+7-x
48 = 50-x
x+48 = 50
x = 50-48
x = 2

I think thats it, if you have any other querries i would be happy to help :cool:

Man City 10

Bluestrike wrote: »

I was just using that as an example but the answers are
(i) A is 12
A union C (AUC) is 10
A union B (AUB) is 4
B union C (BUC) is x
A union B union C (AUBUC) is 5
B is 12-x (21-(4+5+x) = 21-9-x = 12-x)
C is 7-x (22-(10+5+x) = 22-15-x = 7-x)

(ii) The maximum it can be is 7 as C is 7-x so if x was 8 it would be 7-8 and you cant have a minus number of people

(iii) I'm not fully sure with this one but i think it is 7 again because the C thing with 7-x?????

(iv) 48 = 12+4+5+10+x+12-x+7-x
48 = 50-x
x+48 = 50
x = 50-48
x = 2

I think thats it, if you have any other querries i would be happy to help :cool:

Thanks

bosco9871

Can anyone show me how to do this Quadratic equation please.

Solve the equation x²-2x-8=0
Hence solve the equation (y-3)² -2 (y-3) -8 =0

Thank you.

dh0011

solns to quadratic are 4 and -2
soln to second equation is got by letting y-3 equal to the solutions in the first eqn giving 7 and 1.