Cambridge Additional Mathematics

(singke) #1
456 Answers

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EXERCISE 1G
1a 7 b 14 c 14 d 7 e 5 f 9
2ab+c b c+d c b
da+b+c e a+c+d f d
3a i 2 a+4 ii 4 a+4 iii 3 a¡ 5 iv 5 a¡ 1
bia=6 iia=^325
Since a 2 N, there cannot be 31 elements inU, but it is
possible to have 29 elements.
5a 15 b 4 6a 18 b 6 7a 7 b 23
EXERCISE 1H
1a

bi 9 ii 3 iii 3

2a

bi 4 ii 2

3 13 players 420 people
5a b i 16
ii 33
iii 14
iv 7

6a 29 b 6 c 1 d 11
7a 3 b 5 c 5 d 21
8a 3 b 4 c 9
REVIEW SET 1A
1aS=f 3 , 4 , 5 , 6 , 7 g b 5 c 31
2ayes byes c no dyes
3aX^0 =forange, yellow, green, blueg
b X^0 =f¡ 5 ,¡ 3 ,¡ 2 , 0 , 1 , 2 , 5 g
c X^0 =fx 2 Q:x>¡ 8 g
d X^0 =fx 2 (¡1,¡3)[[1,4]g
4afx 2 R:¡ 26 x< 3 g, neither
b fx 2 R:x< 3 g, open
5a b

6a b

c

7a ifs,p,r,i,n,g,b,o,k,w,a,t,e,u,cg
iifr,b,kg iii fg,i,n,o,p,sg
bifthe letters in ‘springbok’ or ‘waterbuck’g
iifthe letters common to both ‘springbok’ and
‘waterbuck’g
iii fthe letters in ‘springbok’ but not ‘waterbuck’g
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(3) (2) (6)
(3)

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(4) (5) (3)
(2)

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(3)

(2) (17)

(16) (5)

(5)
(14)

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100 100
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100 100 IB HL OPT
Sets Relations Groups
Y:\HAESE\CAM4037\CamAdd_AN\456CamAdd_AN.cdr Tuesday, 8 April 2014 8:17:32 AM BRIAN

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