# HSC 2023 Maths Standard 2

Are you looking for a quick review of your answers in the HSC Maths Standards 2 exam? You’re in the right place!

Let’s compare notes and see how well you did in this year’s exams. In this blog, we’ll be taking a closer look into the questions and solutions from the exam. This can help you in verifying your answers and improving your performance to excel in next year’s Maths Standard exam.

Section I

1. An amount of $2500 is invested at a simple rate of 3% per annum.

How much interest is earned in the first two years?

A. $75

B. $150

C. $2575

D. $2652

**Q1: This question asked you to find the interest earned in the first two years for $2500 invested at a simple interest rate of 3% p.a.z**

B

*I = PrtI = 25 000 x 0.03 x 2I = $150*

2. In a normal distribution, what is the approximate percentage of scores with a z-score less than 1?

A. 50%

B. 68%

C. 84%

D. 97.5%

**Q2: This question asked you for the approximate percentage of scores with a z-score of less than 1 in a normal distribution.**

C

3. The number of bees leaving a hive was observed and recorded over 14 days at different times of the day.

Which Pearson’s correlation coefficient best describes the observatioons?

A. -0.8

B. -0.2

C. 0.2

D. 0.8

**Q3: This question requires you to find Pearson’s correlation coefficient from a scatter plot.**

D – no need to do calculations – almost a straight line (so strong correlation) and a positive gradient

4. A delivery truck was valued at $65 000 when new. The value of the truck depreciates at a rate of 22 cents per kilometre travelled.

What is the value of the truck after it has travelled a total distance of 132 600km?

A. $35 828

B. $29 172

C. $14 872

D. $14 300

**Q4: This question asked you to find the value of the truck after it had traveled 132 600km given that the truck was valued at $65 000 when new, and depreciated at a rate of 22 cents/km.**

A

0.22 132 600 = 29 172

65 000-29 172 = 35 828

5. Four petrol pumps are shown, each with the amount of petrol purchased and its cost. Which one represents the best value?

**Q5. This question provided four petrol pumps with prices for a fixed amount of litres, and asked you to find which option represented the best value.**

B

$46.7/24L = 1.94583

$48.50/25L= 1.94

$52.3/26L = 2.01154

$54.8/27 = 2.02963

6. An item was purchased for a price of $880, including 10% GST. What is the amount of GST included in the price?

A. $8.00

B. $8.80

C. $80.00

D. $88.00

**Q6. This question asked you to find the amount of GST included in the purchase price of an item, where the item’s purchase price of $880 was stated to include 10% GST.**

C

$880 = 110% of the price

880/110 = $8

$8 * 10 = $80

7. City A is at latitude 34°S and longitude 151°E. City B is 72° north of City A and 25° west of City A. What are the latitude and longitude of City B?

A. 16°N, 126°E

B. 16°N, 176°E

C. 38°N, 126°E

D. 38°N, 176°E

**Q7. This question asked you to find the latitude and longitude of City B given that City B is **72°** north of City A and 25° west of City A (City A had a latitude of **34°S** and a longitude of 151°**E**).**

C

North and south → 34 – 72 = -38 → 38 N

East and West → 151 – 25 = 126 E

**Q8. This question requires you to find the probability of getting a score of 7 or more in a game involving throwing a die and spinning a spinner.**

D

10 out of 24 chances = 5/12

1 | 2 | 3 | 4 | |

1 | 2 | 3 | 4 | 5 |

2 | 3 | 4 | 5 | 6 |

3 | 4 | 5 | 6 | 7 |

4 | 5 | 6 | 7 | 8 |

5 | 6 | 7 | 8 | 9 |

6 | 7 | 8 | 9 | 10 |

9. The length and width of a rectangle are measured to be 8 cm and 5 cm respectively, to the nearest centimetre. What are the lower and upper bounds for the area of the rectangle?

A. 28 cm^{2} and 54 cm^{2}

B. 36 cm^{2} and 42 cm^{2}

C. 38.25 cm^{2} and 41.25 cm^{2}

D. 33.75 cm^{2} and 46.75 cm^{2}

**Q9. This question asked you to find the lower and upper bounds for the area of a rectangle with a length of 8cm and a width of 5cm (both to the nearest centimetre).**

D

Lower bound = 4.5 * 7.5cm = 33.75

Upper bound = 5.5 * 8.5 = 46.75

10. An amount of $25 000 is invested for six years. Interest is earned at a rate of 8% per annum, compounding quarterly. Which expression gives the value of the investment after 6 years, in dollars?

A. 24 000 x 1.02^{24}

B. 25 000 x 1.02^{6}

C. 25 000 x 1.08^{24}

D. 25 000 x 1.08^{6}

**Q10. This question asked you to find an expression to give the value of an investment after 6 years. You are given that $25,000 was invested for 6 years at an interest rate of 8% p.a., compounding quarterly.**

A

8%/4 compounding periods per year = 2% per quarter

2%/100 = 0.02

R= 1+0.02 = 1.02

4 compounding periods per year * 6 years = 24 compounding periods over all

Therefore, 25 000* 1.02^24

11. A bag contains 150 jelly beans. Some of them are red and the rest are blue. The ratio of red to blue jelly beans is 2:3. Sophie eats 10 of each colour. What is the new ratio of red to blue jelly beans?

A. 2:3

B. 4:9

C. 5:8

D. 11:17

**Q11. This question asked you to find the new ratio of red to blue jelly beans in a bag of jelly beans given that the bag initially had 150 jelly beans (in a ratio of 2:3 for red:blue), and that 10 jelly beans of each colour were eaten.**

C

Initially, 150 jelly beans → 2 parts red to 3 parts blue → 5 parts in total

150/5=30

30*2 = 60 red

30*3 = 90 blue

After eating 10 of each colour, 50 red and 80 blue

50:80 = 5:8

12. A cylindrical pipe with a radius of 12.5 cm is filled with water to a depth, d cm, as shown. The surface of the water has a width of 20cm.

What is the depth of water in the pipe?

A. 2.5 cm

B. 5.0 cm

C. 7.5 cm

D. 12.5 cm

**Q12. This question asked you to find the depth of water in a pipe.**

B

Pythagoras theorem

12.5^2 = 10^2 + x^2

156.25 = 100 + x^2

156.25 – 100 = x^2

Square root of 56.25 = x

x= 7.5 cm

Then, x + d = 12.5

d= 12.5 – 7.5

d= 5 cm

13. An item is discounted by 30% and then a further discount of 20% is applied to the reduced price. What is the percentage discount?

A. 25%

B. 44%

C. 50%

D. 56%

**Q13. This question asked you to find the total percentage discount for an item discounted by 30%, with a further 20% discount being applied to the reduced price.**

B

100% → 100 * 0.7 = 70% of original → 0.7 * 0.8 = 0.56

New price is 56% of the original price meaning that the discount is 100-56 = 44%

14. A network with source A and B is shown. The capacities of two paths are labelled. The cut shown on the diagram on the diagram has a capacity of 30.

Which of the following statements is correct?

A. The maximum flow is 30.

B. The maximum flow is 35.

C, The maximum flow is 30 or less.

D. The maximum flow is 30 or more.

**Q14. This question asked you to select one correct statement regarding the maximum flow of a network.**

C

If the cut has a capacity of 30 that means that at a maximum, 30 could flow through that section. The capacity can not increase from there as the cover spans across the network. From the cut, there could be a reduction in flow due to the distribution of 30 across the 4 paths. Therefore, maximum flow is 30 or less.

15. Ashan’s mathematics class needs to complete six tests, each worth 100 marks. After completing the first five tests, Ashan calculated that he would need a mark of 90 in the final test in order to have a mean mark of 80 for the six tests. What was Ashan’s mean mark after completing the first five tests?

A. 78

B. 74

C. 70

D. 65

**Q15. This question asked you Ashan’s mean mark after completing five tests. For information, you were told that Ashan needs to complete six tests (each worth 100 marks), and that after completing the first five tests, Ashan would need a mark of 90 in the final test in order to average 80 for the six tests.**

A

Mean mark of 80 for 6 tests = overall combined mark of 80*6 = 480

Last test is a score of 90 → 480 – 90 = 390

First five tests add up to 390, so the average is → 390/5 = 78

16. (2 Points)

The graph shows Peta’s heart rate, in beats per minute, during the first 60 minutes of a marathon.

A. What was Peta’s heart rate 20 minutes after she started her marathon?

B. Peta started the marathon at 10am. At what time would her heart rate first reach 140 beats/minute?

**Q16a) This question provided a graph of Peta’s heart rate in beats per minute during the first 60 minutes of a marathon. Part a) asked you to find Peta’s heart rate 20 minutes after starting her marathon.**

120 beats/minute

**Q16b) Part b) asked you to find the time at which Peta’s heart rate would first reach 140 beats/minute, given that she started the marathon at 10am.**

140 beats/minute at 30 minutes in → started at 10am so will reach this HR at 10:30am

17. (3 marks)

The table shows some of the flight distances (rounded to the nearest 10 km) between various Australian cities.

a. Use the information in the table to complete the network diagram where the edges are labelled with distances.

b. Mahsa wants to travel from Hobart to Darwin. She wants to change planes only once.

Using the network diagram, calculate how many kilometres she will travel by plane.

**Q17a) This question asked you to draw a network diagram for the information in the table provided. The information in the table provided flight distances between Australian capital cities.**

**Q17b) This question asked you to use your network diagram to calculate how many kilometres Masha will need to travel by plane if she wants to travel from Hobart to Darwin, but only change planes once.**

Only wants to change planes once → Hobart to Sydney to Darwin

1040 + 3150 = 4190kms

18. (2 marks)

The histogram shows a summary of scores on a test.

Provide TWO features of the histogram that indicate that the data comes from a normal distribution.

**Q18. This question provided a histogram for a summary of scores on a test. You were asked to name two features of the histogram that indicate the data came from a normal distribution.**

Approximate mode, mean and median are in the centre of the histogram

From the score of 49.5 which is the approximate centre, the data lies both 35 marks below and 35 marks above the centre, meaning the spread is symmetrical.

19. (4 marks)

A network of running tracks connects the points A, B, C, D, E, F, G, H, as shown. The number on each edge presents the time, in minutes, that a typical runner should take to run along each track.

A. Which path could a typical runner take to run from point A to point D in the shortest time?

B. A spanning tree of the network above is shown.

Is it a minimum spanning tree? Give a reason for your answer.

**Q19a) This question provided you with a network diagram for a series of running tracks that connected multiple points. The edges represented the time in minutes required for a typical runner to run that track. Part a) asked you to find a path a typical runner could take to run from point A to D in the shortest time. **

A→ F→ G→D takes 15 minutes

**Q19b) This question provided you with a spanning tree and asked if it was a minimum spanning tree, asking you to give reasons for your answer.**

No it is not a minimum spanning tree. C can be connected to the tree via D through a shorter path.

20. (3 marks)

On another planet, a ball is launched vertically into the air from the ground. The height above the ground , h metres, can be modelled using the function h = 6r^{2} + 24r, where t is measured in seconds. The graph of the function is shown.

A. based on the graph, what is the maximum height reached by the ball?

B. Based on the graph, at what TWO times is the ball are ³ / ₄ of its maximum height?

**Q20a) This question gave you the graph of the trajectory of a ball that was launched vertically in the air. The ball’s trajectory was said to be modelled by the equation 6t ^{2} + 24t. Part a) asked you to find the maximum height reached by the ball.**

24m above the ground. Read straight from the graph

**Q20b) Part b) asked you to use the graph to find two times at which the ball was ³ / ₄** **of its maximum height.**

1 second and 3 seconds

¾ of the maximum height → 24m * ¾ = 18m

From the graph, the ball is at 18m at 1s and 3s.

21. (5 amrks)

Electricity provider A charges 25 cents per kilowatt (kWh) for electricity, plus a fixed monthly charge of $40.

A. Complete the table showing Provider A’s monthly charges for different levels of electricity usage.

Provider B charges 35 cents per kWh, with no fixed monthly charge. The graph shows how Provider B’s charges vary with the amount of electricity used in a month.

B. On the grid on the image above, graph Provider A’s charges from the table in part (a).

C. Use the two graphs to determine the number of kilowatt hours per month for which Provider A and Provider B charge the same amount.

D. A customer uses an average of 800 kWh per month. Which provider, A or B, would be the cheaper option and by how much?

**Q21a: The question provides that Electricity Provider A charges 25 cents/kWh for electricity , plus a monthly charge of $40. You are asked to complete the table showing Provider A’s monthly charge for 400 kWh of electricity usage.**

Monthly charge =

**Q21b: Part b) asks you to graph provider A’s charges from the table in part ****a)**Using the two points already given in table A. (0,40) & (1000, 290)

**Q21c: Part c) asks you to use the graphs to determine the number of kWh/month for which Provider A and B charge the same amount.**

The two lines intersect at 400kWh. The providers will charge the same rate for 400kWh.

**Q21d: Part d) provides that a customer uses an average of 800 kWh per month. You are then asked which provider would be cheaper for the customer and by how much.**

After 400kWh, Provider B’s rates are higher than Provider A’s. At 800kWh per month, Provider A offers a better deal.

Provider A → 40 + 0.25 * 800 = $240

Provider B → 0.35 * 800 = $280

Provider A is cheaper than provider B by $40 for this customer.

22. ( 3 marks)

The breaking distance of a car in metres, is directly proportional to the square of its speed in km/h, and can be represented by the equation.

braking distance = k x (speed)^{2}

where k is the constant of variation.

The braking distance for a car travelling at 50 km/h is 20 m.

A. Find the value of k

B. What is the braking distance when the speed of the car is 90 km/h?

**Q22a) This question provides that the braking distance of a car is provided by ***braking distance = k x (speed) ^{2}*

**, where**

*k*is the constant of variation. You are told the braking distance for a car travelling at 50km is 20m.**Part a) asks you to find the value of k.**

50km/h = k * 20m^{2 }

50 = k * 400

50/400 = k

k = 0.125

**Part b) asks you to find the braking distance when the car’s speed is 90 km/h.**

Q22b) Braking distance = 0.125 * 90^{2}

Braking distance = 0.125 * 8100

Braking distance = 1012.5m

23. (2 marks)

One hundred tickets are sold in a raffle which offers two prizes. Hazel buys five of the tickets. A ticket is drawn at random for the first prize. A second ticket is drawn from the remaining tickets for the other prize. What is the probability that Hazel wins both prizes?

**Q23: You are told that 100 tickets are sold in a raffle which offers two prizes. You are also told that Hazel buys five of the tickets. Given that one ticket is drawn at random for the first prize, followed by a second ticket being drawn from the remaining tickets for the second prize, you were asked to find the probability that Hazel wins both prizes.**

Probability of winning the first prize = 5/100

Probability of winning the second prize = 4/99

Combined probability = 5/100 * 4/99 = 0.002

24. (5 marks)

The diagram shows the cross-section of a wall across a creek.

A. Use two applications of the trapezoidal rule to estimate the area of the cross-section of the wall.

B. The wall has a uniform thickness of 0.80 m. The weight of 1m^{3} of concrete is 3.52 tonnes. How many tonnes of concrete are in the wall? Give your answer to two significant figures.

**Q24a: You are provided with a diagram of the cross-section of a wall. Part a) asks you to use two applications of the trapezoidal rule to estimate the cross-sectional area.**

**Q24b: Part b) provides that the wall has a uniform thickness of 0.80m, and that the weight of 1**m^{3}** concrete is 3.52 tonnes. You are asked how many tonnes of concrete are in the wall to two significant figures.**

Volume = 0.8 x 18 = 14.4 m^{3}

Tonnes of concrete = 14.4 x 3.52 = 50.688 tonnes

25. (5 marks)

A table of future value interest factors for an annuity of $1 is shown.

A. Micky wants to save $450 000 over the next 10 years.

If the interest rate is 6% per annum compounding annually, how much should Micky contribute each year? Give your answer to the nearest dollar.

B. Instead, Micky decides to contribute $8535 every three months for 10 years to an annuity paying 6% per annum, compounding quarterly. How much will Micky have at the end of 10 years?

**Q25a: This question provides you with the future value table. Micky wants to save $450,000 over the next 10 years.**

**Part a) asks you to find how much Mickey should contribute to his savings each year to reach his goal. You are given that the interest rate is **6%** p.a. You also need to round to the nearest dollar.**

**Q25b: Part b) provides that Micky contributes $8 535 every 3 months for 10 years to an annuity paying 6% per year, compounding quarterly. You are asked to find how much Micky will have at the end of 10 years.**

26. (5 marks)

Kim is building a path around a garden at the back of a house, as shown. The path is 0.5 m wide.

A. Find the area of the path.

B. Kim is mixing some concrete for the path. The concrete mix is mad crushed rock, sand and cement in the ratio of 4:2:1 by weight.

Kim needs 2.1 tonnes of concrete mix in the correct ratio.

Calculate how many 15 kg bags of cement Kim needs to buy.

**Q26: This question provides you with a diagram of a path that Kim is building around a garden. The path is 0.5 m wide. The dimensions of the rectangle (path and garden included) is 8m x 3m.**

**Part a) asks you to find the area of the path.**

Area of the path → 2.5 * 0.5 = 1.25

→ 8 * 0.5 = 4

Therefore, 1.25 * 2 + 4 = total area of path

Total area = 6.5 m^{2}

**Q26b: Part b) provides that Kim is mixing concrete for the path, with the concrete being made up of crushed rock, sand and cement in a ratio of 4:2:1 by weight. You are told that Kim needs 2.1 tonnes of concrete mix in the correct ratio. You are then asked how many 15kg bags of cement Kim needs to buy.**

2.1 tonnes made up of 4:2:1 (7 parts total) → each part is 0.3 tonnes

0.3 *1000 = 300 kg of cement

300kg/15 = 20 bags of cement required

27. (4 marks)

The diagram shows the location of three places X, Y and C.

Y is on a bearing of 120° and 15 km from X.

C is 40 km from X and lies due west of Y.

P lies on the line joining C and Y and is due south of X.

A. Find the distance from X to P.

B. What is the bearing of C from X, to the nearest degree?

**Q27: This question provides you with a diagram showing the location of X, Y and C. You are told that Y is on a bearing of **120°** and 15 km from X; that C is 40km from X and lies due west of Y; and that P lies on the line joining C and Y and is due south of X.**

**Part a) asks you to find the distance from X to P.**

Using cos (60) = a/15

a = 15 cos (60)

a = 7.5km

**Q27b: Part b) asks you to find the bearing of C from X, to the nearest degree.**

Using sin (b) = 7.5/40

29. (3 marks)

A plumber leases equipment which is valued at $60 000.

The salvage value of the equipment at any time can be calculated using either of the two methods of depreciation shown in the table.

Under which method of depreciation would the salvage value of the equipment be lower at the end of 3 years? Justify your answer with appropriate mathematical calculations.

**Q28: You are told that a plumber leases equipment valued at $60,000, and that the salvage value of the equipment can be calculated using straight-line depreciation (depreciating by $3500 p.a.) or declining-balance depreciation (depreciating by 12% p.a.). You are asked which method of depreciation would yield the lowest salvage value at the end of 3 years, and to justify your answer with appropriate calculations.**

straight-line method → 60000 – (3500 * 3) = 49500

Declining-balance method → 12% per annum depreciation → 1 – 0.12 = 0.88

→ 0.88^{3} x 60000 = 40888.32

The salvage value is lower in the declining-balance method

29. (4 marks)

The table shows monthly repayments for each $1000 borrowed.

A. A couple borrows $520 000 to buy a house at 8% per annum over 25 years.

How much does the couple repay in total for this loan?

B. Chris borrows some money at 7% per annum. Chris will repay the loan over 15 years, paying $3596 per month.

**Q29a: You are provided with a monthly repayment table showing monthly repayments for each $1000 borrowed. Part a) asks you to find how much a couple repays in total for a loan of $250 000 at 8% p.a. Over 25 years.**

8% over 25 years → $7.72 per month per $1000

$520 000/$1000 = 520

Monthly payments will be 520 times the rate compared to $1000 loan

$7.72 520 = $4014.4 per month

The payments are monthly for 25 years → 25 x 12 = 300 payments

300 payments x $4014.4 per payment = $1 204 320

Total repaid is $1 204 320

**Q29b: This part asks you to find how much money Chris borrows given that he borrows an unknown amount of money at 7% p.a., will repay the loan over 15 years, and pays $3596 per month.**

7% over 15 years → $8.99 per month per $1000

$3596/$8.99 = 400 times the payment amount

Principal is 400 times $1000 → 400 x $1000 = $400 000

Chris borrowed $400 000

30. (3 marks)

A receipt from a supermarket shows a total of $124.87. The GST shown on the receipt is $3.86.

GST, at a rate of 10%, is only charged on some items.

What was the value of the items which did NOT have GST charged?

**Q30: This question provides that a receipt from a supermarket shows a total of $124.87 but that the GST shown was $3.86, and GST at a rate of 10% is only charged on some items. You are asked to find the value of items that did not have GST charged.**

$3.86 GST → x 10% = $3.86

x = $3.86/10%

x = $38.60

GST was charged on items totalling $38.60

$124.87 – GST = $124.87 – $3.86 = $121.01

$121.01 – items with GST charged = $121.01 – $38.60 = $82.41

The value of the items which did not have GST charged was $82.41

31. (4 marks)

A function centre employs staff so that all necessary tasks can be completed between the end of one function and the beginning of the next function.

The network diagram shows the time taken in hours for the tasks that need to be completed.

A. Find the TWO critical paths.

B. The function centre wants to decrease the length of each critical path by 3 hours. They can do this by hiring more staff to do ONE of the tasks so it takes less time to complete.

For which task should the centre hire more staff, and how long should that task take to ensure all tasks can be completed in 14 hours?

**Q31a) This question provides you with a network diagram of tasks that need to be completed between functions/events for a function centre. Part a) asks you to find the critical path.**

Critical paths = H→I→K (17)

H→I→G→C (17)

**Q31b) This part provides that the function centre wants to decrease the length of each critical path by 3 hours, and that this can be done by hiring more staff to do ONE of the tasks so it takes less time to complete. You are asked which task the centre should hire more staff for, and how long should that task take to ensure all tasks can be completed within 14 hours.**

The two critical paths share tasks H and I

H can only be reduced by 2, bringing the critical path to 15 not 14

Have to reduce I by 3 to bring both critical paths to 14 hours.

Task I should be reduced to 4 hours.

32. (4 marks)

Ali has a credit card which has no interest-free period. Interest is charged at 13.5% per annum, compounding daily, on the amount owing.

During the month, Ali made only one purchase of $450 using the credit card. The full amount owing was repaid 21 days later.

A. Calculate the amount of interest charged on the purchase, assuming that interest is charged for the 21 days.

B. What percentage of the full amount repaid is the interest? Give the answer to two decimal places.

**Q32: You are told that Ali has a credit card with no interest-free period, with interest charged at 13.5% p.a. Compounding daily on the amount owing. You are also told that during the month, Ali made one purchase of $450 using the credit card, with the full amount owing being repaid 21 days later.**

**Part a) asks you to calculate the amount of interest charged on the purchase, assuming interest was charged for all 21 days.**

R = 13.5/(365*100) = 0.000369863 applied per day

$450 x 0.000369863^{21} = $3.5081631

$3.51 of interest charged in total for 21 days

**Part b) asks you to find the percentage of the full amount repaid that is interest, giving your answer to two decimal places.**

Full amount repaid = $450 + $3.51 = $453.51

$3.51/$453.51 = 0.0077396

0.0077396 * 100 = 0.77% of the amount repaid was interest

33. (4 marks)

The diagram shows a shape APQBCD. The shape consists of a rectangle ABCD with an arc PQ on side AB and with side lengths BC = 3.6 m and CD = 8.0 m.

The arc PQ is an arc of a circle with centre O and radius 2.1 m and ZPOQ = 110°.

What is the perimeter of the shape APQBCD? Give your answer correct to one decimal place.

**Q33: You are provided with compound shape APQBCD, and are asked to find the shape’s perimeter correct to one decimal place.**

Perimeter = 8 + 8 + 3.6 + 3.6 + b – a

34. (6 marks)

A university uses gas to heat its buildings. Over a period of 10 weekdays during winter, the gas used each day was measured in megawatts (MW) and the average outside temperature each day was recorded in degrees Celsius (°C). Using x as the average daily outside temperature and y as the total daily gas usage, the equation of the least-squares regression line was found. The equation of the regression line predicts that when the temperature is 0°C, the daily gas usage is 236 MW. The ten temperatures measured were: 0°, 0°, 0°, 2°, 5°, 7°, 8°,9°,9°, 10°. The total gas usage for the ten weekdays was 1840 MW.

B. What is the equation of the regression line?

C. In the context of the dataset, identify ONE problem with using the regression line to predict gas usage when the average outside temperature is 23°C.

**Q34a)** **You are provided with a series of data for the average daily temperature across a series of days **** and the total daily gas usage across those days (****).**

**Part a) asks you to plot (**x,y**) and the y-intercept of the least-squares regression line on the grid shown.**

Average of x variables → 50/10 = 5

Average of y variable → 1840/10 = 184

(5,184)

**Q34b) ****Part b) asks you to find the equation of the least squares regression line**

Two points on the line → (5,184) & (0, 236)

Gradient = (236 – 184) / (0 – 5) = – 10.4

y = -10.4 x + c

236 = -10.4 * 0 + c

236 = c

Equation of the line → y = -10.4 x + 236

**Q34c) Part c) asks you to identify one problem with using the regression line to predict gas usage when the average outside temperature is 23°**C.

The data points for temperature fall between 0 and 10 degrees. By predicting for 23 degrees we are extrapolating and the predicted daily gas usage may be unreliable.

35. (3 marks)

The diagram shows triangle ABC

Calculate the area of the triangle, to the nearest square metre.

36. (4 marks)

The following formula can be used to calculate an estimate for blood alcohol content (BAC) for males.

N is the number of standard drinks consumed

M is the person’s weight in kilograms

H is the number of hours of drinking

Cameron weighs 75 kg. His BAC was zero when he began drinking alcohol. At 9:00 pm, after consuming 3 standard drinks, his BAC was 0.02.

Using the formula, estimate at what time Cameron began drinking alcohol, to the nearest minute.

**Q36: This question was a BAC question and asked you to estimate what time Cameron began drinking alcohol to the nearest minute. You were provided with the BAC formula for males; that Cameron weighs 75kg; that his BAC was zero when he began drinking; and that having consumed 3 standard drinks by 9:00pm, his BAC was 0.02.**

0.02 = ((10 x 3) – 7.5H)/6.8 x 75

0.02 = 30 – 7.5H/510

10.2 = 30 – 7.5H

30 – 10.2 = 7.5H

19.8/7.5 = H

H = 2.64 hours

Cameron began drinking 2.64 hours ago

0.64 x 60 = 38.4 minutes

2 hours and 38 minutes ago → current time 9pm

Cameron began drinking at 6:22pm

37. (3 marks)

The table shows personal income tax rates for different taxable incomes for a particular country.

A person with a taxable income of $90 000 pays 25.8% of that income in tax (excluding any levies).

What is the value of X in the table?

**Q37: This question gave you a tax table. You needed to work backwards to find the value of X in the table. To assist, you were given that a person with a taxable income of $90,000 paus 25.8% of income in tax (excluding any levies).**

Tax on $90 000 = 0.258 x 90 000 = $23 220

$23 220 = $18 292 + (X x (90 000 – 78 000))

$23 220 = $18 292 + 12000X

12000X = $4 928

X = $4928/12000

X = $0.41 = 41 cents

38. (4 marks)

A random variable is normally distributed with a mean of 0 and a standard deviation of 1. The table gives the probability that this random variable lies below z for some positive values of z.

The probability values given in the table are represented by the shaded area in the following diagram.

Z score = (11.93 – 10.40) / 1.15 = 1.33

Probability of Z score of 1.33 = 0.9066 → 0.9066 x 400 koalas = 362.64 koalas

The graph shows equal to OR less, we want to find greater than

400 – 362.64 = 37.36

37 koalas are expected to weigh more than 11.93 kgs

Ready to excel in next year’s HSC Maths Standard Exam? Enrol to Dymocks Tutoring today and get the support you need to master all the key concepts and ace every question on the exam.