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Maths, Addmaths and Statistics: Post your doubts here!

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Can someone explain what's going on here? This question went through my head!
Its from 4024_w02_qp_1
upload_2014-5-6_15-57-43.png
 
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Universal Set = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18}
A = {2,4,6,8,10,12,14,16,18}
B = {5,10,15}

i) U for Union, N for Number of Elements
So 'A U B' = {2,4,5,6,8,10,12,14,15,16,18}
n(A U B) = 11

ii) Inverted ∩ for Intersection
A ∩ B' ∩ C' => (A ∩ B') ∩ C'

Solve Part by Part:
1)
A = {2,4,6,8,10,12,14,16,18}
B' = {2,3,4,6,7,8,9,11,12,13,14,16,17,18}
A ∩ B' = {2,4,6,8,12,14,16,18}

2)
{2,4,6,8,12,14,16,18} ∩ C' = {2,6,14,18}
This {2,4,6,8,12,14,16,18} is same as Set A
C' is the set containing elements which are not in set C.

C' ∩ A = Common Elements of C' and A.
So C will not contain {2,6,14,18} as they are in C'.
C is a subset of A.
C will only contain A - C' which is {4,8,12,16}

iii) Multiples of 4

See this
 
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View attachment 41541
View attachment 41542
this one too facing problem in part b and c(ii):cry:
Let the Radius of Bigger Circle be R

So

(R^2*pi)-(15^2*pi)=1206

Solution:

Solve second Bracket, (R*R*pi)-(706.95)=1206
Replace Pie (R*R*3.142)-706.95=1206
Add 706.95 to Both sides
R*R*3.142=1912.95

Now Divide both sides by 3.142,
R*R=608.83
R=Square root 608.83
R=24.68

Check by Replacing in Equation or Marking Scheme

Thanks
 
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Let the Radius of Bigger Circle be R

So

(R^2*pi)-(15^2*pi)=1206

Solution:

Solve second Bracket, (R*R*pi)-(706.95)=1206
Replace Pie (R*R*3.142)-706.95=1206
Add 706.95 to Both sides
R*R*3.142=1912.95

Now Divide both sides by 3.142,
R*R=608.83
R=Square root 608.83
R=24.68

Check by Replacing in Equation or Marking Scheme

Thanks
ok thanks :)
 
Messages
316
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Universal Set = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18}
A = {2,4,6,8,10,12,14,16,18}
B = {5,10,15}

i) U for Union, N for Number of Elements
So 'A U B' = {2,4,5,6,8,10,12,14,15,16,18}
n(A U B) = 11

ii) Inverted ∩ for Intersection
A ∩ B' ∩ C' => (A ∩ B') ∩ C'

Solve Part by Part:
1)
A = {2,4,6,8,10,12,14,16,18}
B' = {2,3,4,6,7,8,9,11,12,13,14,16,17,18}
A ∩ B' = {2,4,6,8,12,14,16,18}

2)
{2,4,6,8,12,14,16,18} ∩ C' = {2,6,14,18}
This {2,4,6,8,12,14,16,18} is same as Set A
C' is the set containing elements which are not in set C.

C' ∩ A = Common Elements of C' and A.
So C will not contain {2,6,14,18} as they are in C'.
C is a subset of A.
C will only contain A - C' which is {4,8,12,16}

iii) Multiples of 4

See this
i got it thanks for the link
do you have any link like this for transformation as well?
 
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