Cambridge International Mathematics

ANSWERS 1

CHAPTER 33(CD question answers)

31 a i equilateral ii 3 : 2 m iiih¼ 4 : 37 m

b¼ 18 : 1 m^3 c h¼ 5 : 76 m

32 a$750 bi 2 : 5 x+ 5(200¡x) = 905 ii x=38

ci$6300 ii $105

33 aVC=6+

p

62 ¡ 3 : 62 =10: 8 cm

bi¼ 905 cm^3 ii ¼ 147 cm^3 iii¼ 83 :8%

ci¼ 37 : 7 cm ii 7 times iii 15 : 2 o

34 a 30 o b OZ=^3

cos 30o

¼ 3 : 464 cm

cOW¼ 4 : 899 cm dV¼ 25 : 5 cm^3 e ¼ 54 : 7 o

35 a i^14 ii 127

bi 12 and 9 , 11 and 10 , 10 and 11 , 9 and 12 ii 361

ci 11 ii 10 iii 8 : 9 di 26 : 2 cm^2 ii 0 : 0291

36 a i^12 x(x+1)=5 ) x^2 +x=10, etc.

ii x¼ 2 : 7 or¡ 3 : 7 ,PR¼ 2 : 7 cmfPR> 0 g

biAB^2 =3y^2 +6y+4

ii 3 y^2 +6y+4=49 simplifies to y^2 +2y¡15 = 0

iii (y+ 5)(y¡3)

iv y=¡ 5 or 3 , but we reject¡ 5

) AC=3cm, CB=5cm

37 a i Half size

ii BE¼ 9 : 42 cm iii ¼ 9 : 42 cm

biDA¼ 5 : 77 cm iiAE¼ 24 : 2 cm

iii BE¼ 31 : 4 cm iv ¼ 31 : 4 cm

cBE¼length of semi-circular arc BA

38 aTa x= $3000£0% + $17 000£25%

= $4250

b$4800

ciT= (20 000¡x)£ 0 : 3 ii x= 3500

39 a i a OC¼ 17 : 7 cm b¼ 9 : 59 o c ¼ 18 : 8 cm

ii a ¼ 113 cm b 60 o

bia 1:2 b1:4 c 1:8

ii 18 ¼cm^2 iii^78 Vcm^3

40 a i ii^1532

iii^3364

b 12831

cimode=7, median=5 ii mean=4: 94 iii 0 : 72

41 a i Alexis$2000, Biatriz$1200, Carlos$400 ii $9450

b$960

42 a i 24 cm ii¼ 135 o iii ¼ 30 : 6 cm iv ¼ 37 : 7 cm

bi¼ 653 cm^2 ii ¼ 30 :8%

43 a i $ 10 : 20 ii^100

x

pounds

bi

100

x+1

+1=

100

x

and on multiplying both sides by

x(x+1)reduces to x^2 +x¡100 = 0.

ii x¼ 9 : 51 fthe other solution is< 0 g

iii $ 10 : 51

44 a 18 cm b ¼ 6 : 65 g c¼50 000

di 0 : 07 cm^2 iiequilateral

iii A=

p

3 x^2

4

,AB¼ 4 mm

45 a ² RbSX=PbQX fequal alternate anglesg

² SbXR=QbXP fvertically oppositeg

) ¢s are equiangular and so are similar.

bPX=5: 25 cm, QX=3: 5 cm c ¼ 8 : 9 cm^2

d¼ 29 o

46 a i μ¼ 38 : 7 o ii l=

¡ 77 : 364

360

¢

£ 2 ¼(12)¼ 16 : 2 km

bTime saved=3: 778 ::::min¼ 4 min

47 a V¼ 93 : 4 cm^3 bs=^3

q

V

2 : 6

ciA=^3

p 3

2 s

(^2) ii¼ 2 :598 076s (^2) ¼ 2 : 60 s 2
48 a i 9 : 52 =x^2 +8^2 ¡ 2 x(8) cos 60o fcosine ruleg
and this simplifies to 4 x^2 ¡ 32 x¡105 = 0
ii x=10^12 or¡ 212 butxcannot be< 0
) x=10^12
So, YZ=10: 5 cm
b¼ 46 : 8 o
49 aAbOC= 2 sin¡^1 (0:9)¼ 128 : 3 o b 4 : 48 m^2
cOAC¼ 1 : 57 m^2 , ABC¼ 2 : 91 m^2
dAbOC¼ 128 : 3 o ) ObAC¼ 25 : 85 o ) AbFE¼ 64 : 15 o
So, sin 64: 15 o¼
d
5
) d¼ 4 : 5
e¼ 26 : 0 m^2 , ¼ 28 : 9 m^2
50 a i 8 m/s ii 28 : 8 km/h b ¼ 7 : 97 m/s c ¼ 55 : 8 o
di¼ 118 o ii¼ 298 o
51 a

¡¡ 8

6

¢

bD(¡ 6 ,0) c AB=10,BC=5,AC=5

p

5

dAC^2 = 125 = 100 + 25 =AB^2 +BC^2

) AbBC is a right angle

fconverse of Pythagoras’ theoremg

So, the parallelogram is a rectangle.

e 50 units^2 fi¡^34 ii (0, 134 )

52 aa+b bi¡PT!=

³ 1

(^34)

́

ii 114 km

ci 12 tacks ii ¡PQ!= 12(a+b) d 053 : 1 o

53 a¼49 100m^2 (¼ 4 : 91 ha) b¼ 26 : 0 o

54 a 3 girls

3cm

30°3cm

B

C

DEA

9cm

good wind

kite flies

kite does not fly

not a good wind

kite flies

kite does not fly

Er_

Qr_

Ti_

Ei_

qA_y_

Qq_Ty_

IB MYP_3 ANS

cyan magenta yellow black

(^05255075950525507595)
100 100
(^05255075950525507595)
100 100
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