## CA2 Science 2009 Section A

### October 17, 2012

1 (3) 11 (2) 21 (4)
2 (4) 12 (1) 22 (2)
3 (4) 13 (3) 23 (2)
4 (3) 14 (4) 24 (2)
5 (1) 15 (4) 25 (3)
6 (3) 16 (2) 26 (3)
7 (2) 17 (4) 27 (4)
8 (2) 18 (2) 28 (1)
9 (1) 19 (4) 29 (4)
10 (2) 20 (3) 30 (2)

## SA2 2011 Maths

### October 17, 2012

Paper 2
For Q1 – 5, apply the following:
Correct method and correct answer — M1A1
Correct answer with no working — A2
Correct method and wrong answer —M1A0
Incorrect method but correct answer (fluke) — M0A0

1. \$3 ÷ 12 = \$0.25 (M1) or 60 ÷ 12 = 5 (M1)
\$0.25 x 60 = \$15.00 (A1) \$3 x 5 = \$15 (A1)

2. (A2) or Area of triangle = x 1 x 4 = 2 units2
Area of square = 16 units2
Fraction = = (M1 / A1)
3. 120% x 5 = 6
160 + 6 = 166 (M1)
(bar to be drawn accurately for 2011) (A1)
4. Method 1:
∠OPT = 90° – 65° = 25°
∠RPQ = 90° – 37° = 53°
∠TPR = 90° – (25° + 53°) = 12° (M1 / A1)

Method 2:
∠PTR = 180° – 65° = 115°
∠PRT = 90° – 37° = 53°
∠TPR = 180° – 115° – 53° = 12° (M1/A1)

5. 2 cm x 2 cm = 4 cm2
5 + 5 + 10 = 20
20 x 4 cm2 = 80 cm2 (M1 / A1)

For Q6 – 18, apply the following:
Deduct a maximum of ½ m for missing universal units.
Deduct a maximum of ½ m per question for wrong number sentence for 4 & 5-mark questions only.
Misread error — follow through the working; if answer is correct, deduct M1 from the maximum method (M) marks awarded for the question
6.
26 x 2 = 52 (M1) or 26+26+40 = 92 (M2,A1)
52 + 40 = 92 stamps (M1 / A1)

7. 3.17 + 1.06 = 4.23l
6 units + 3 units = 9 units
4.23 ÷ 9 = 0.47l (M1)
0.47 x 2 = 0.94l (M1/A1)

8. Method 1: 120 cm x 80 cm x 90 cm = 864 000 cm3
x 864 000 cm3 = 288 000 cm3 (M1)
288 000 cm3 ÷ 12 = 24 000 cm3 (M1/A1)
= 24l
Method 2: x 90 cm = 30cm
Volume of paint used = 120 cm x 80 cm x 30 cm (M1)
= 288000 cm3
= 288l
Volume of paint used for 1 wall = 288 ÷ 12 (M1)
= 24l (A1)
9.

∠ PQT = 180° – 20° – 20° = 140°
∠ TQR = 180° – 140° = 40° OR ∠ TQR = 2 x 20° = 40° (Int Opp Angle) (M1)
∠ QTR = 180° – 40° – 40° = 100° (M1) ∠ RTS = 40° + 20° = 60° (Int Opp Angle) (M1)
∠ RTS = 180° – 20° – 100° = 60° (M1) ∠ x = 60° (A1)
∠ x = 60° (A1)

10. 3 units  66 marbles
1 unit  66 ÷ 3 = 22 marbles
5 units  22 x 5 = 110 marbles (M1)
100% – 45% = 55%
55%  110 marbles
1%  110 ÷ 55 = 2 marbles (M1)
100%  2 x 100 = 200 marbles (A1)

11. 13 x 130 = 1690
11 x 120 = 1320 (M1)
1690 – 1320 = 370 (M1) or
370 – 20 = 350 (M1) 370 + 20 = 390 (M1)
350 ÷ 2 = 175 390 ÷ 2 = 195 (A1)
175 + 20 = 195 (A1)

12. 8 units  age difference = 57 – 25 = 32 (M1)
1 unit  32 ÷ 8 = 4 (M1)
9 units  4 x 9 = 36
57 – 36 = 21 years ago, Nancy was 9 times as old as her son (M1)
2010 – 21 = 1989 (A1)
OR
Son’s age is 25 – 4 = 21 years ago. (1 unit) (M1)
2010 – 21 = 1989 (A1)

13. (a) 40g (A1)
(b) Method 1:
Total mass + 3 cans  200+120+190 = 510
Total mass (2X + 2Y + 2Z)  510 – 120 (M1)
= 390
Ave of 3 objects  390 ÷ 6 (M1)
= 65 g (A1)

Method 2:
Mass of X + Y = 200 – 40 = 160
Mass of X + Z = 120 – 40 = 80
Mass of Y + Z = 190 – 40 = 150
Total mass  160 + 80 + 150 = 390 (M1)

Ave of 3 objects  390 ÷ 6 (M1)
= 65 g (A1)

14. (a) ∠ ABE = 35° (Base angles of an isosceles triangle)
∠ AEB = 180° – 35° – 35°
= 110°
∠ BED = 360° – 118° – 110° (M1)
= 132°
Since BE is parallel to CD
∠ x = (180° – 132°)
= 48° (A1)
(b) ∠DCE = 48°(Base angles of an isosceles triangle)
∠CED = 180° – 48° – 48° = 84°
∠BEC = ∠BED – ∠CED
= 132° – 84° = 48°
∠y = (180° – 48°) ÷ 2 = 66° (M1/ A1)

15. Method 1: of the remainder  25%
of the remainder  25% ÷ x = 31.25% (M1)
100% – 31.25% = 68.75% (M1)
68.75%  385 sweets (M1)
1%  5.6 sweets
100%  5.6 x 100 = 560 sweets (A1)

15. Method 2:
 4 units
 16 units (total at first) (M1)
16 – 5 = 11
11 units = 385 (M1)
1 unit = 385 ÷ 11 = 35
16 units = 35 x 16 = 560 (M1, A1)

16. Method 1:

15 cm x 15 cm = 225 cm2 (M1)
2362.5 cm3 ÷ 225 cm2 = 10.5 cm (M1)
– =
4 units  10.5 cm
1 unit  10.5 cm ÷ 4 = 2.625 cm (M1)
6 units  2.625 cm x 6 = 15.75 cm (M1, A1)

Method 2:

4 units  2362.5 (M1)
1 unit  590.625 (M1)
6 units  590.625 x 6 = 3543.75 (M1)

3543.75 ÷ 225 = 15.75 cm (M1, A1)

Method 3:
4 units  2362.5 (M1)
1 unit  (2362.5 ÷ 4)
= 590.625
15 units  590.625 x 15 (M1)
= 8859.375
8859.375 ÷ 225 = 39.375 (M1)
39.375 x = 15.75 cm (M1,A1)

17.

Short
Long

Method 1
Length of longer piece = (105 + 9) ÷2 = 57 cm (M1)
Length of shorter piece = 57 -9 = 48 cm

3 units  9 cm (M1)
1 unit  9÷3 = 3 cm
2 units  3 x 2 = 6 cm (M1)
Length that was cut away = 48 cm – 6 cm = 42 cm (M1 /A1)
Or
5 units  3 x 5 = 15 cm (M1)
Length that was cut away = 57 – 15 = 42 cm (M1 /A1)

Method 2
3 units  9 cm (M1)
1 unit  9÷3 = 3 cm
2 units  3 x 2 = 6 cm (M1)
5 units  3 x 5 = 15 cm (M1)

105 – 6 – 15 = 84
84 ÷ 2 = 42 cm (M1,A1)

18. Length + Breadth of rectangle = 78 cm ÷ 2 = 39 (M1)
Length (cm) Breadth (cm) Area (cm2) Check
25 14 25 x 14 =350 No
26 13 26 x 13 = 338 No
27 12 27 x 12 = 324 Yes

Perimeter of the figure = length of rectangle + 2 x breadth of rectangle + 2(side of 1st triangle + side of 2nd triangle + side of 3rd triangle + side of 4th triangle)
= 27 + 12 + 12 + (2 x 27) (M1) OR = 78 + 27
= 105 cm (A1) = 105 cm
—–THE END —–