St Gabriels Prelim 2009 Em P1 Solutions

St. Gabriel’s Secondary School Mathematics Department Secondary 4E5N4N Elementary Mathematics 2009 ‘O’ Preliminary Examinations Marking Scheme (P1: Prepared by Koh Keng Wee – Q1 to Q15 & Teo Chen Nee – Q16 to Q23)

Q1

Loss = 12.5% × $480 = $60

Q6(a)

Sellg Price = $480 – $60 = $420 [M1, A1]

Q2

Ben: 12 days = 1 work. 1 day =

1 work 12

Q6(b)

Q7

Q3(b)

12  3 days 4

Jane’s age = (x + 6) years old

Mean Age 

3

74 cm = 7.4 × 10 m

1

[B1] [B1]

1

(2 x3 y 2 )3  4 x 2 y 2 3

1

[M1, A1]

7

 2x y

[M1, A1]

[B1]

1  x  2( x  6) 3

1  (3 x  12)  ( x  4) years old 3

Q4

–5

1 1 4   work 12 4 12

Both: 1 work =

Q3(a)

3

 8 x9 y 2  4 x 2 y 2

1 Ali: 4 days = 1 work. 1 day = work 4 Both: 1 day =

74 cm3 = 74 × (0.01)3 = 0.000074 m3

 32  V    845  50 

[B1]

Q9(a)

3

 16  V    845  25 

Q8

4

7   x 1    5112  100  5112 x 1.07 4  $3899.92

Actual length

[M1, A1]

= 2500 × 9 = 22500 = 225 m

3

[M1, A1]

Q9(b)

Scale of map

[B1]

= 1 : 2500 = 1 cm : 2500 cm

3

4 V     845 5 V  432.64 cm

= 1 cm : 25 m Scale of Area on map

[M1, A1]

= 1 cm2 : 252 m2

3

= 1 cm2 : 625 m2 Area of path on map = 450 ÷ 625 = 0.72 cm2

Q5(a)

63 = 3 x 3 x 7 or 32 x 7

Q5(b)

14 = 2 x 7

[B1] Q10(a)

AC  82  152  289  17 cm

[B1]

Q10(b)

cos ACD  

15 17

[B1]

Q10(c)

 15  BAC  tan 1    61.9 8

63n = 3 × 3 × 7n When n = 2, 63 × 2 = (3 × 3) × (7 × 2) = 9 × 14 n=2

[B1]

[B1]

Q11(a)

22

Q14(a)

Area of ACR AR 3   Area of BCR RB 4 3 Area of ACR   Area of BCR [M1, A1] 4 3 Area of ACR   98  73.5 cm 2 4

Q14(a)

Area of CPQ  PC   5      Area of CBR  BC   7  25 Area of CPQ   Area of CBR [M1, A1] 49 25 Area of ACR   98  50 cm 2 49

Q15(a)

7  2  5       1 9  3  6 5 10   9 3 35 8  or 3 9 9

2

3  2 1  2 1

7  3  4  3  22 16  7  9  7  32 32  16  16  16  4 2

[B2]

57  32  25  32  52 93  57  36  57  6 2 142  93  49  93  7 2 Q11(b)

1  211 2  22 1 4  231 8  2 41

[B1]

16  25 1  nth term = 2n 1

Q12(a)

 AB  52  (3)2

[B1]

 25  9  34 or 5.83 units

Q12(b)

Q13(a)

   5   20  AB CD     k    3   6x  5 1 5  20k  k   [M1, A1] 20 4 1 12 3  (6 x)  x    2 4 6

3 xy  2 y  12 x  8  y (3 x  2)  4(3 x  2)

[M1, A1]

 ( y  4)(3 x  2) Q13(b)

32 x  27 x 1  0 32 x  33( x 1) 2 x  3x  3 x  3

[M1, A1]

Q15(b)

2

32 110  1 2 83 1 1   18 2 4  9

2

 2     5 [M1, A1]

[M1, A1]

2009 E Maths 4E5N Prelim Paper 1 Marking Scheme (Q16 – 23) Charges ($) 25

20

15

10

5

0

25°

20

40 60 Number of Photographs

80

100

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y y

x

x

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x

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