WRP Solutions General Mathematics 2012

8/13/2019 WRP Solutions General Mathematics 2012

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WESTERN REGION

2012Preliminary Final

EXAMINATION

General Mathematics

SOLUTIONS


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Preliminary Final Examination General Mathematics 2012

Multiple Choice Answer Sheet

Name ______________________________

Completely fill the response oval representing the most correct answer.

1. A B C D

2. A B C D

3. A B C D

4. A B C D

5. A B C D

6. A B C D

7. A B C D

8. A B C D

9. A B C D

10. A B C D

11. A B C D

12. A B C D

13. A B C D

14. A B C D

15. A B C D

16. A B C D

17. A B C D

18. A B C D

19. A B C D

20. A B C D


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Question 21 Final Preliminary Examination General Mathematics

2012

Part Solution Marks Comment

(a) y = 2x

6 or 2x

y

6 = 0 2 1 for correctgradient

1 for correcty-intercept.

(b) 2x15 = 3x4

4(2x15) = 3x

8x60 = 3x

5x

60 = 05x = 60

x = 12

2

1 for correctexpansion

1 for solution

(c) i) 12-6

= -2

1

ii) 3(-2 + 12)

= 3 10

= 30

1

iii) 32

(-2)2

12

=9 412

= 3

1

(d) i) $25 from graph 1

ii) before 2100 (9:00pm) 1

iii) 5.5 hours would cost $20. Two lots of under 3

hours costs $12. The exercise would save $8.

1

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2012


(a) i) $63 920 1

ii) $3572 + 32.5% of (63 92037 000) = $12 321 1

iii) 1.5% of $63920 = $958.80 1

iv) $550 26 = $14300.14 300-12 321-958.80= $1020.20 refund.

1

(b) i) = P(1 + r)n

= 5000(1.055)5

= $6534.80

1

ii) = P(1 + r)n

= 5000(1.0135)20

= $6538.00

6538-6534.80= $3.20 better off with new bank.

2 1 for compoundinterest result.

1 for difference.

(c)i) 45.6

35 + 68.4

5 + 91.2

3 = $2211.60

2 1 for correctvalues forovertime/effectivehours worked.

1 for result.

ii) 24% of $2211.60 = $530.78 (tax)

2211.6 530.78 = $1680.82 net pay

1

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2012


(a) cosA = 6.511.3

A = 5453

A = 55

1

(b) 2 = 62

+ 92

2= 117

x= 10.8cm

2 1 for lengthof 9 cm

1 for correctanswer.

(c) (i) 1

(ii) tan 56=79x

x= 79

tan 56

x= 53.286

x= 53.3m

2 1 for correctratio

1 for correctanswer

79m

56o

x


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(d)

B BBBB

G BBGB

B BGBG

G BGG

B GBB

BG GBG

GB GGB

GG GGG

2 1 for any

method usedto find thesample space

1 for correctlist ofoutcomes.

(e) i) P(spanner away from car) = P(spanner under car)

= 1

0.85

= 0.15

1

ii) 60 0.85 = 51 1

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2012


(a) i) 130 cm 1

ii) 134.25cm134.75cm 1

iii) 0.25 134.5 = 0.186% 1

(b) i) 100 11 = 9.09 km/L 1

ii) 65 9.09 = 590.85 (591 km to nearest km) 1

(c) i) V= r2h

= 22

2.5

= 31.415. . .

31 m3

1

ii) 31000 300 = 103..

3

so 103 hours 20 mins are needed to fill the tank.

1 Accepteither.

iii)

SA = 2r(r+h)

= 2 2(2 + 2.5)

= 18

= 56.55

57 m2

2 1 for correctsubstitution

1 for correctanswer

iv) 57 240 = $ 13 680

Also accept $13 571.68 using 18

1

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(a) 620

are tagged, or 30%

30% = 1510% = 5100%= 50

2 1 mark for correctpercentage/decimalor alternate method1 for solution

(b) i)3 7 8 94 2 3 6 7 8 95 0 2 2 2 3 3 5 76 1 3 5

1

ii) range = 65 37 = 28

median = 51Q1 = 44.5Q3= 54

interquartilerange = 54 44.5 = 9.5

2 1 for each response

iii) 37, 44.5, 51, 54, 65

37 44.5 51 54 65

2 1 for summary,

1 for plot

Plot should haveappropriate scale

iv) x = 50.1 sample standard deviation = 7.9 2 1 for each response

(c) median as it is less affected by outliers. 1/10

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WRP Solutions General Mathematics 2012