2007 NSW HSC General Mathematics Exam Q’ns 23 – 28 Worked Solutions
Question 23
Question 24 (continued)
(a) (i) $5000
(a) (ii) The outlier is 50.
Without the outlier:
(ii) Rose.
$11 000 – $9 000 = $2 000
3 + 5 + 5 + 6 + 8 + 8 + 9 +10 +10
9
64
=
9
Mean =
(iii) 15 years – 14 years = 1 year.
(b) (i)
V =
=
≈
≈
€
πr 2 h
2
π × ( 4.78) × 2
143.56073
144 m3.
≈ 7.1
€
Median is 8 both with and without the outlier.
€
So by removing the outlier, the median drops
€ from 11.4 to 7.1, but the median stays the same.
€ 000 ÷ 7 500 = 12 hours.
(ii) 90
€
€
(iii) 20 mm = 0.02 metres.
€
t
3
t
1.2 =
3
(b) D =
V = 400 × 0.02
= 8 m3
= 8 000 Litres.
€
= 3.6 seconds
(c) (i) €40% = 18 crocodiles
10% = 18 ÷ 4
= 4.5 crocodiles
100% = 4.5 × 10
= 45 crocodiles.
€
(ii)
€
(c)
€ (i)
US$75 = A$100
US$150 = A$200
So Sandy has A$800 to convert to euros.
A$100 = 60 euros
A$800 = 480 euros
40% = 24 crocodiles
€10% = 24 ÷ 4
= 6 crocodiles
100% = 6 × 10
= 60 crocodiles.
€
(ii) The gradient will increase (ie get
steeper).
Question€24
(a) (i)
3 + 5 + 5 + 6 + 8 + 8 + 9 +10 +10 + 50
10
114
=
10
Mean =
€
= 11.4
t = 1.2 × 3
(d) (i) Females are skewed towards older ages,
while males are skewed towards
younger ages.
(ii) 64
(iii) 5 × 35 = 175
(iv) Using a grouped frequency table to
calculate the mean will only give an
approximation. Caitlyn’s method will
€
instead give an exact answer for the
mean.
€
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Page 1 of 5
2007 NSW HSC General Mathematics Exam Q’ns 23 – 28 Worked Solutions
Question 25
Question 25 (continued)
(a) Many possible answers, such as:
A bag contains 3 red and 1 white
marble. If a marble is selected at
random, the probability that it will be
red is three quarters.
(c) (iii) P (same rating)
⎛ 5 4 ⎞ ⎛ 3 2 ⎞ ⎛ 2 1 ⎞
= ⎜ × ⎟ + ⎜ × ⎟ + ⎜ × ⎟
⎝10 9 ⎠ ⎝10 9 ⎠ ⎝10 9 ⎠
(b)
∠MJK = 90° – 75°
= 15°
€
sin∠JMK
sin∠MJK
=
20
18
sin∠JMK
sin15°
=
20
18
€
sin15°
sin∠JMK =
× 20
18
€
⎞
⎛ sin15°
∠JMK = sin −1⎜
× 20⎟
⎠
⎝ 18
€
≈ 17°
€
€
€
€
€
€
(d) (i) z-score of Test 1
63 − 60
=
€
6.2
≈ 0.48
z-score of Test 2
62 − 58
€
=
€
6.0
≈ 0.67
∠LMK = ∠JML − ∠JMK
€
= 75° – 17°
€
= 58°.
€
So the angle of elevation from M to K is 58°
(to the nearest degree).
€
€
(c) (i)
(ii)
2
20 6
+ +
90 90 90
28
=
90
14
=
45
=
So he scored better in Test 2.
€
€(ii) 64 is one standard deviation above the
mean (since 58 + 6.0 = 64).
50% scored below 58 and 34% scored
between 58 and 64. So we would expect
84% to score above 64.
2 1
= (or 20% or 0.2)
10 5
84% of 150 = 126 students.
Question 26
€
(a) (i)
2 h 48 min = 2.8 hours
15 ÷ 2.8 ≈ 5.357
≈ 5 km/h
€
(ii) Let q be the distance from Town A to
€
Town B.
€
q2 = 15 2 +10 2 − 2 × 15 × 10 × cos87°
= 325 − ( 300 × cos87°)
q = 325 − ( 300 × cos87°)
€
≈ 18
€So the distance from Town A to Town B
is approximately 18 km.
€
€
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Page 2 of 5
2007 NSW HSC General Mathematics Exam Q’ns 23 – 28 Worked Solutions
Question 26 (continued)
(a) (iii) First calculate ∠QAB using cos rule.
2
2
15 +18 −10
2 × 15 × 18
€
449
=
540
€
€
⎛ 449 ⎞
∠QAB ≈ cos−1⎜
⎟
⎝ 540 ⎠
€
≈ 34°
2
cos∠QAB ≈
€
€∠QBA = 180° − ∠AQB − ∠QAB
€
(angle sum of a triangle)
≈ 180° – 87° – 34°
€
= 59°
€
Let the North pointer at Town A be N1
€ and the North pointer at Town B be N2.
€
€
Question 26 (continued)
(b) (iii) Tax payable:
$4 500 + 0.3 × ($41145 − $28 000)
= $4 500 + 0.3 × $13145
= $4 500 + $3 943.50
= $8 443.50
€
€
(c) Value of investment
€
⎧ (1.06) 22 −1 ⎫
€
⎬
= 100 ⎨
⎩ 0.06 ⎭
⎧ 3.6035374 −1 ⎫
⎬
≈ 100 ⎨
⎩
⎭
0.06
€
∠N1 AB = 149° − ∠QAB
≈ 149° − 34°
= 115°
€ BA = 180° − ∠N AB
∠N
2
1
€ €
(cointerior angles)
≈ 180° −115°
= 65°
€
∠N 2 BQ = ∠N 2 BA + ∠QBA
€ €
≈ 65° + 59°
= 124°
€
€
Bearing
of Town Q from Town B
€
=
360° – ∠N 2 BQ
€ €
≈ 360° – 124°
= 236°
€
(b) (i)
€
⎧ 2.6035374 ⎫
⎬
= 100 ⎨
⎩ 0.06 ⎭
€
⎧ 2.6035374 ⎫
⎬
= 100 ⎨
⎩ 0.06 ⎭
€
€
€ €
€
≈ 100 × 43.392290
≈ $4339.23
Total investment:
= 22 × $100
= $2200
Interest earned:
= $4339.23 – $2200
€
= $2139.23
Question 27
(a) (i) Any two positive values where one is
six more than the other, such as 4 m
wide by 10 m long.
$5000€- $3635 = $1365
(ii) A = l ( l − 6)
$1365 ÷ 3 = $455
$455
× 100% = 9.1%
$5000
€
€
(iii) If the length is less than 6 metres than
the width would be negative.
€
€
(ii) Yearly income:
52 × $800 = $41 600
(iv) Length = 15 m (from graph)
Width
= 15 m – 6 m
=9m
(v) See graph next page.
Taxable income:
$41 600 − $455 = $41 145
€
€
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Page 3 of 5
2007 NSW HSC General Mathematics Exam Q’ns 23 – 28 Worked Solutions
Question 27 (continued)
(a) (v)
(a) (vi) Company A would charge $50 000.
Company B would charge:
135 × 360
= $48 600
So Company B would charge less.
€
(b) (i) c = ( 4 × 6) + ( 365 × 5d )
(ii)
250 = ( 4 × 6) + ( 365 × 5d )
250 = 24 +1825d
226 = 1825d
226
€ d =
1825
€
d ≈ $0.124
€
€
Question 28
(a) (i) There are 10 cases where the difference
is 1. (1,2 2,1 2,3 3,2 3,4 4,3 4,5 5,4
5,6 6,5)
10
So probability =
36
5
=
18
(ii) Financial€expectation
=
(iii) Total cost:
€
(4 × 6) + (365 × 10d)
€
≈ 24 + ( 365 × 10 × 0.124 )
= $476
So the cost does not double. (This is
€
the initial purchase price
€because
€
remains
the same).
(6 × $3.50) − (10 × $5) + (20 × $2.80)
€
36
$21 − $50 + $56
36
$27
=
36
=
€
€
= $0.75
(iii) He expects a loss of 25 cents.
€
(iv) 5000 is two standard deviations above
the mean. So the mean is:
5000 − 2 × 170
= 4660 hours.
(although this is only on average, if he
plays the game many times. If he only
plays once, then the most likely outcome
would be winning $2.80, which is a gain
of $1.80)
€
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Page 4 of 5
2007 NSW HSC General Mathematics Exam Q’ns 23 – 28 Worked Solutions
Question 28 (continued)
(b)
H 2 = 22 + 22
H2 = 8
€
€
€
A = 2.598H 2
= 2.598 × 8
= 20.784 cm2.
3.6
3.6
(c) (i) € A ≈
(5 + 4 × 4.6 + 3.7) + (3.7 + 4 × 2.8 + 0)
3
3
= (1.2 × 27.1) + (1.2 × 14.9)
€
€
= 32.52 +17.88
= 50.4 cm2.
€
(ii) Let the curved surface area be a cm2.
€
7480.8 = (200 × 5) + (200 × 14.4 ) + (50.4 × 2) + a
7480.8 = 1000 + 2880 +100.8 + a
7480.8 = 3980.8 + a
€
a = 7480.8 – 3980.8
€
So the area of the curved surface is 3500 cm2.
€
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Page 5 of 5