# 2 UNIT

Based on the 2014 syllabus and the BOS specimen paper.

Working time 2 ½ hours

DIRECTIONS TO CANDIDATES

Section 1: 25 marks

 Attempt Questions 1  25

 Allow about 35 minutes for this section

Section II: 75 marks

 Attempt questions 26  30

 Allow about 1 hour and 55 minutes for this section

 A formula sheet is provided at the end of the paper.

 Board approved calculators may be used.

 Answer each question of section II on a separate page.

 In questions 26 to 30 show relevant mathematical reasoning and/or calculations

 You may ask for extra paper if you need it.

## Practice Paper 5 Answers Section I 25 marks

### Attempt Questions 1-25 Allow about 35 minutes for this section.

\$580 per month for 2 years = 580 x 24 = \$13920
Interest = \$13920m- \$12000 = \$1920
I = P r n 1920 = 12000 x r x 2
r = 1920/(12000 x 2) = 0.08 = 8%

Volume of box = 20 x 10 x 9 = 1800cm3
Volume of pyramid = (20 x 10 x 9) / 3 = 600cm3
Volume of polystyrene = 1800cm3  600cm3 = 1200cm3

m0 = 1 m1 = 3 m2 = 9
1 + 3 + 9 = 13

Note that this is a CUMULATIVE frequency polygon so the number of students who worked 4 hours is the number who worked 4 hours or less (6) less the number who worked 3 hours or less (4).
6  4 = 2

The number of combinations of girls is 8C2 = 8 x 7 ÷ 2
The number of combinations of boys is 12C2 = 12 x 7 ÷ 2
Number of combinations of girls and boys is 8C2 x 12C2 = D
Order is not important in combinations hence we divide the number of ordered ways that the combination can be chosen by number of ways each combination can be arranged, in this case 2.
Mary and Betty is the same combination as Betty and Mary.

Angle AFB = 180  67  73 = 40o
The side opposite the 67o angle is shorter than the side opposite the 73o angle.
Hence we are after the length FB
From the sine rule: a/sinA = b/sinB = c/sinC
AB / sin ∠ AFB = FB / sin∠ FAB
10/sin 40o = FB / sin 67o
By cross multiplying
FB = (10/sin 67o) / sin 40o

By drawing the image (in your mind or on paper) you should see the following. A trapezium (4 sided figure with 1 pair of || sides).

95% of bottles are within 2 standard deviations of the mean i.e. 520 ± 20 mls
So 2.5% are above 540mls and 2.5% are below 500mls (the stated contents)

By finding the intersection of 18 periods and 4% we get 25.65.
\$100 per month = \$100 x 12 per year.
The annuity is compounded annually so there are 18 periods.
Hence the future value is: \$25.65 x 18 x 12 x 100

There are no numbers in the survey so the only answer that can be interpreted from the list above is the most common rating i.e. mode.

Nathans income = \$20 x n = 20n
Nathans costs are cost of market stall + \$15 x number of cakes = 50 + 15n
Nathan will break even when income = costs
20n = 50 + 15n
Rearranging gives 20n  15n = 50

Chiara can choose 2 black socks OR 2 brown socks

Probability of 2 black socks:
There are 20 socks in the drawer, 12 of which are black so the probability that the first sock is black is 12/20
There are now 19 socks in the drawer, 11 of which are black so the probability that the second sock will be black is 11/19
The probability that Chiara chooses 2 black socks is 12/20 x 11/19

Probability of 2 brown socks:
There are 20 socks in the drawer, 8 of which are brown so the probability that the first sock is brown is 8/20
There are now 19 socks in the drawer, 7 of which are brown so the probability that the second sock will be brown is 7/19
The probability that Chiara chooses 2 brown socks is 8/20 x 7/19

The probability of one of the outcomes is the sum of the probability of each outcome.
Probability of 2 black or 2 brown = 12/20 x 11/19 + 8/20 x 7/19

Since Sydney is GMT + 10 and Calgary is GMT  7 there is a time difference of 10 - - 7 hours between Sydney and Calgary i.e. 17 hours.
Since +10 > -7 then Sydney is ahead of Calgary. (i.e. Sydney is later)
17 hours ahead of 6pm Tuesday is 6 + 11 hours ahead.
6 hours until midnight and 11 hours until 11am Wed.

d αs2 so d = ks2
30 = 202 k = 400 so k = 30/400 = 3/40
d= 402 k = (40 x 40) (3/40) = 120m

Initial amount of electricity used in a year = 1.2 x 3 x 52 = 187.2 kWh
Initial cost of electricity used in a year = 1.2 x 3 x 52 = 187.2 x \$0.30 = \$ 56.16
Cost is cut in half so amount saved = ½ x \$56.16 = \$28.08 = \$28 (nearest \$1)

B x 6.8M = 10N  7.5H
6.8MB = 10N  7.5H
10N = 6.8MB + 7.5H
N = 0.68MB + 0.75H

1 Australian dollar buys 75 US cents so A = 0.75U
But the number of Australian dollars converted to US dollars is A - 10 because the bank charges a \$10 fee.
i.e A - 10 = 0.75U or by rearranging; A = 0.75U + 10

(x3 + x) (x2  1) = x3(x2 -1) + x(x2 -1) = x5  x3 + x3  x = x5 - x

After 19% of 455 kL is recycled then consumption is reduced to 81% of the previous level.
81% of 455kL = 368.55 kL = 369kL (nearest kL)

Mean of 15/20 for 5 tests is 5 x 15 = 75 marks total
Total marks for first 4 tests = 12 + 14 + 16 + 16 = 58
Marks needed for fifth test = 75  58 = 17

The error is ± half the limit of reading of the measuring instrument.
Since the ruler is graduated in centimetres the error is ± 0.5cm

This works on the compound interest formula: A = P(1 + r)n
Since interest is charged daily we have to work out the number of days
May = 5days (27, 28, 29, 30, 31)
June = 14 days
Total = 14 + 5 = 19 days
Interest + 20% p.a. = = 20/100 p.a. = 20/(100 x 365) per day = 20/36500 Amount to pay = 850(1 + 20/36500)19 Bearing = angle clockwise from north = 360  60 = 300oT

There are numbers involved so the data is quantitative.
The divisions can be broken down into smaller and smaller units so it is continuous.

Area of base = 6 x 6 = 36 units2
Area of each side = ½ x 6 x 4 = 12 units2
Total area = 36 + (4 x 12) units2 = 36 + 48 = 84 units2

## Question 26 (15 marks)

(a) (i) \$5400 = 90% of asking price
? = 100% of asking price
(100/90) x 5400 = \$6000
> (ii) Registration transfer = \$30
Stamp duty = 3% of \$5400 = \$162
Amount to pay = \$162 + \$30 = \$192

(iii) Total = \$1540 + \$462 + \$616 + \$77 = \$2695
Discount 5% (loyalty bonus) = \$134.75
Price before GST and stamp duty = \$2695 - \$134.75 = \$2560.25

(iv) \$2560.25 + 10% GST = \$2816.28
\$2816.28 + 5% stamp duty = \$2957.09

(v) Average driving speed = 6800/200 = 34 km/h
2 hours at 34 km/h = 68km between rest stops.

(vi) 6800km = 68 x 100 km = 68 x 7.2 = 489.6 litres = 490L (nearest L)

(vii) \$719/490 = \$1.4673 = \$1.47 (nearest cent)

(viii) (54/36) x 10 = 15m/s
15 x 1.2 = 18m

(ix) d 7alpha; s2 d = ks2 28 = k 362 k = 28/362 = 28/1296
d2 = (28/1296) x 542 = 63m
(x) BACmale = (10N  7H) / 6.8M BACmorgan = (10 x 3.2  7 x 2) / 6.8 x 75 = (32  14)/510 = 0.0353 g/100ml
(xi) 0.0353/0.015 = 2.353 hours = 2hours 21 minutes(nearest minute)
26. (b) (i) Length of sides = 16 ÷ 4 = 4m
Area = 42 = 16m2

(ii) 2 π r = 16
r = 16/2π = 8/π = 2.54648m

(iii) A = &pi r2 = π (2.54648)2 = 20.37m2

(iv) 20.37/16 = 1.27 times bigger

## Question 27 (15 marks)

(a) (i) Gradient = rise/run = (80  70)/(23  13) = 10/10 = 1 cm/month

(ii) Mean age = (12 + 15 + 18 + 21 + 24) / 5 = 18 months

(iii) Using your calculator, standard deviation = 4.24264 = 4.24 (2dp)
(iv) (69 + 72 + 75 + 78 + 80)/5 = 74.8

(v) Using your calculator, standard deviation = 3.96988866 = 3.97 (2dp)

(vi) 1 = r x (3.97/ 4.24)
r = 4.24/3.97 = 1.07 (2 dp)

(vii) Height = 74.8  (1 x 18) = 56.8cm

(viii) Frieds Formula: Dosage = (18 x 20) / 150 = 2.4mls

(ix) Youngs Formula Dosage = (1.5 x 20) / (1.5 + 12) = 30/13.5 = 2.2mls

(x) Clarks Formula Dosage = (8.1 x 20) / 70 = 2.3mls

(xi) 5 grams/litre = 5000mg/litre = (20/1000) x 5000mg per 20ml = 100mg

(xii) (2.3/20) x 100 = 11.5mg

(xiii) 200/2.3 = 86.9565 = 86doses with 95.65% of a dose remaining.
> (b)(i) The number that tested negative and did not have cancer can be found in 2 different ways:
Method 1:
Total number that tested negative (745)  The number that tested negative but had cancer (50) = 745  50 = 695
OR
Method 2:
Total number that did not have cancer (764)  The number that tested positive but did not have cancer (69) = 764  69 = 695

(ii) The patients that were incorrectly diagnosed fall into 2 groups:
Group 1: Those who had cancer but tested negative (50 patients)
Group 2: Those who did not have cancer but tested positive (69 patients)
Total number of patients incorrectly diagnosed = 50 + 69 = 119 patients.
119/1000 = 11.9%

## Question 28 (15 marks)

(a) = = 40m3

(b) (i) 82.8% full means (100  82.8)% i.e. 17.2% to be added to fill to capacity
17.2% of 97190ML = (17.2/100) x 97190 = 16717ML (nearest ML)

(ii) 97190 ML = 97190 x 1000 x 1000L
Since 1000L = 1m3 then 97190ML = 97190 x 1000 = 97 190 000 m3

(iii) Area = 8.5km2 = 8 500 000 m2
Depth = Volume/surface area = 97 190 000 / 8 500 000 = 11.43m

(iv) Volume added (in m3) is catchment area (in m2) x amount of rain (in m)
= 130 x 1000 x 1000 (catchment area) x 0.003 (amount of rain) = 390000m3
Since 1m3 = 1000L = 0.001ML then 390000m3 = 390000 x 0.001ML = 390ML

(c) (i) Cost of home supply (91 days at \$0.70 per day = \$63.70
Cost of electricity used = \$223.86 - \$63.70 = \$160.16
Number of kilowatt hours = \$160.16 ? \$0.22 = 728kWh

(ii) \$223.86 less 12% = 88% of \$223.86 = \$196.9968 = \$197.00 (nearest cent)
\$197.00 + 10% GST = 110% of \$197 = \$216.70

(d)
Multiply all terms by LCM i.e. by 6:
2m + 4 + 42 = 3m  15
4 + 42 + 15 = 3m  2m
61 = m m = 61

(e) (i) 720MB = 720 x 1024kB = 720 x 1024 x1024B = 720 x 1024 x 1024 x 8 b
= 6039797760 bits = 6039000000 bits (4 sig figs)

(ii) 256 kb/s = 256 x 1024 b/s = 262144b/s
6039797760 / 262144 = 23040 seconds
OR
6039000000 / 262144 23037 seconds (nearest second)

(f) Length of track = ½ x 2 x πr (left end) + 2r + 2r (two straight sides) + ½ x 2 x πr (right end) = 2 πr + 4r = 400m
r(2? + 4) = 400m
r = 400 / (2? + 4) = 38.89845m = 38m 89.845cm = 38m90cm (nearest cm) or 3890cm

## Question 29 (15 marks)

(a) (i) C ? 1/N
C = k / N k = constant = CN
100 = k / 80
k = 8000
C = 8000/60 = \$133.33

(ii) C = 8000/N

(iii) 50 = 8000/N
N = 8000/50 = 160 people

(b) 12/30 of the marbles in the second draw had been marked
This indicates that Izabella marked 12/30 (i.e. 2/5) of the total number of marbles in the first draw.
So 2/5 of the marbles = 40 marbles
5/5 of the marbles = ? marbles
5/2 x 40 = 100 marbles

(d) (i) Difference in latitude = 43  19 = 24o
Distance = (24/360) x 2 x π x 6400 = 2680.83 = 2680 km (nearest 10 km)

(ii) 2680/670 = 4 hours

(iii) 10.00am + 4 hours = 2.00pm Hobart time.
But Hobart time = Townsville time
So lands in Townsville at 2.00pm Townsville time.

(e)
(i) 7.2% p.a. = 0.6% per month i.e. 0.006
3 years = 36months
The intersection of 0.006 and 36 gives 32.290749
\$12000 / 32.290749 = 371.623
Monthly repayments = \$371.62 (nearest cent)

(ii) \$371.62 x 36 = \$13378.32

(iii) \$13378.32 - \$12000 = \$1378.32

(f)
(i) Number of Combinations of girls 12C4 = 495 i.e. (12 x 11 x 10 x 9) / (4 x 3 x 2 x 1)

(ii) If Bill is chosen then there are 5 boys left.
The probability that Joseph will be chosen is 1/5

(iii) There are 495 combinations of girls.
There are 5 combinations of boys where Bill is one of the members.
Total number of combinations = 495 x 5 = 2475

## Question 30 (15 marks)

(a) (i) Bearing 248o - Bearing 131o = 117o

(ii) Use the Cosine Rule: BC2 = BO2 + CO2  2 x BO x CO x cos ∠BOC
BC2 = 182 + 162  2 x 18 x 16 x cos 117o
BC2 = 841.5 BC = 29.0m
(iii) Area = ½ ab sinC = ½ x 18 x 16 = sin117o = 128.3m2

(b) (i) & (ii) ((iii) C = 40 + 5N

(iv) The break-even point i.e. the number of belts where income received is equal to the cost of making them.
(c)
(i) Median of all marks is 6 :There are 9 numbers above and 9 numbers below.
Taking the median of the numbers above the median: 6, 7, 7, 7, 8, 8, 9, 10, 10,
The upper quartile, Q3 is 8

(ii) Similarly the lower quartile, Q1 is 4
The interquartile range is Q3 - Q1 = 8  4 = 4

(iii)

(iv) QL  1.5 x IQR = 4  1.5 x 4 = - 2
0 is not less than  2 so 0 is NOT an outlier.