Consider the function `f:[−3,2] -> R, \ \ f(x) = 1/2(x^3 + 3x^2 - 4)`.
- Find the coordinates of the stationary points of the function. (2 marks)
The rule for `f` can also be expressed as `f(x) = 1/2(x - 1)(x + 2)^2`.
Aussie Maths & Science Teachers: Save your time with SmarterEd
Consider the function `f:[−3,2] -> R, \ \ f(x) = 1/2(x^3 + 3x^2 - 4)`.
The rule for `f` can also be expressed as `f(x) = 1/2(x - 1)(x + 2)^2`.
a. `text(Stationary points when)\ \ f´(x)=0,`
`1/2(3x^2 + 6x)` | `= 0` |
`3x(x + 2)` | `= 0` |
`:. x = 0, − 2`
`:.\ text(Coordinates of stationary points:)`
`(0, − 2), (− 2,0)`
b. |
c. | `text(Avg value)` | `= 1/(2 – 0) int_0^2 f(x) dx` |
`= 1/2 int_0^2 1/2(x^3 + 3x^2 – 4)dx` | ||
`= 1/4[1/4x^4 + x^3 – 4x]_0^2` | ||
`= 1/4[(16/4 + 2^3 – 4(2))-0]` | ||
`= 1` |
Evaluate `int_1^4 (1/sqrtx)\ dx`. (2 marks)
`2`
`int_1^4 x^(−1/2)\ dx` | `= 2[x^(1/2)]_1^4` |
`= 2[4^(1/2) – 1^(1/2)]` | |
`= 2(2 – 1)` | |
`= 2` |
Let `f′(x) = 1 - 3/x`, where `x != 0`.
Given that `f(e) = −2`, find `f(x)`. (3 marks)
`x – 3log_e(x) + 1 – e`
`f(x)` | `= int 1 – 3x^(−1) dx` |
`f(x)` | `= x – 3log_e |\ x\ | + c` |
`text(Substitute)\ \ f(e) = − 2,`
`−2` | `= e – 3log_e(e) + c` |
`c` | `= 1- e` |
`:. f(x) = x – 3log_e|\ x\ | + 1 – e`
The company prepares for this expenditure by establishing three different investments.
Determine the total value of this investment at the end of eight years. (2 marks)
Determine the total value of this investment at the end of eight years.
Write your answer correct to the nearest dollar. (1 mark)
Deposits of $200 are made to this account on the last day of each month after interest has been paid.
Determine the total value of this investment at the end of eight years.
Write your answer correct to the nearest dollar. (1 mark)
a. | `I` | `= (PrT)/100` |
`= (7000 xx 6.25 xx 8)/100` | ||
`= $3500` |
`:.\ text(Total value of investment)`
`= 7000 + 3500`
`= $10\ 500`
b. `text(Compounding periods) = 8 xx 4 = 32`
`text(Interest rate)` | `= (text(6%))/4` |
`= 1.5text(% per quarter)` |
`:.\ text(Total value of investment)`
`= PR^n`
`= 10\ 000(1.015)^32`
`= 16\ 103.24…`
`= $16\ 103\ \ text{(nearest $)}`
c. `text(By TVM Solver,)`
`N` | `= 8 xx 12 = 96` |
`I(text(%))` | `= 6.5` |
`PV` | `= 500` |
`PMT` | `= 200` |
`FV` | `= ?` |
`text(P/Y)` | `= text(C/Y) = 12` |
`=> FV = −25\ 935.30…`
`:.\ text(Total value of investment is $25 935.)`
It is estimated that inflation will average 2% per annum over the next eight years.
If a new machine costs $60 000 now, calculate the cost of a similar new machine in eight years time, adjusted for inflation. Assume no other cost change.
Write your answer correct to the nearest dollar. (1 mark)
`$70\ 300\ \ text{(nearest $)}`
`text(Find value)\ (A)\ text(in 8 years.)`
`A` | `= PR^n` |
`= 60\ 000(1.02)^8` | |
`= 70\ 299.562…` | |
`= $70\ 300\ \ text{(nearest $)}` |
A company purchased a machine for $60 000.
For taxation purposes the machine is depreciated over time.
Two methods of depreciation are considered.
The machine is depreciated at a flat rate of 10% of the purchase price each year.
The value, `V`, of the machine after `n` years is given by the formula `V = 60\ 000 xx (0.85)^n`
a.i. | `text(Annual depreciation)` | `= 10text(%) xx 60\ 000` |
`= $6000` |
a.ii. `text(After 3 years,)`
`text(Value)` | `= 60\ 000 – (3 xx 6000)` |
`= $42\ 000` |
a.iii. `text(Find)\ n\ text(when value = $12 000)`
`12\ 000` | `= 60\ 000 – 6000 xx n` |
`6000n` | `= 48\ 000` |
`:.n` | `=(48\ 000)/6000` |
`= 8\ text(years)` |
b.i. | `1 – r` | `= 0.85` |
`r` | `= 0.15` |
`:.\ text(Annual depreciation is 15%.)`
b.ii. `text(After 3 years,)`
`text(Value)` | `= 60\ 000 xx (0.85)^3` |
`= $36\ 847.50` |
b.iii. `text(Find)\ n\ text(when)\ \ V = $12\ 000`
`12\ 000` | `= 60\ 000 xx (0.85)^n` |
`(0.85)^n` | `= 0.2` |
`:. n` | `= 9.90…\ \ text(years)` |
`:.\ text(Machine value falls below $12 000)`
`text(after 10 years.)`
c. `text(Sketching both graphs,)`
`text(From the graph, at the end of the 7th year the)`
`text(value using flat rate drops below reducing)`
`text(balance for the 1st time.)`
Khan paid $900 for a fax machine.
This price includes 10% GST (goods and services tax).
He considers two methods of depreciation.
Flat rate depreciation
Under flat rate depreciation the fax machine will be valued at $300 after five years.
Unit cost depreciation
Suppose Khan sends 250 faxes a year. The $900 fax machine is depreciated by 46 cents for each fax it sends.
a. `text(Let $)P = text(price ex-GST)`
`:. P + 10text(%)P` | `= 900` |
`1.1P` | `= 900` |
`P` | `= 900/1.1` |
`= 818.181…` | |
`= $818.18\ \ text(nearest cent)` |
b.i. `text(Annual depreciation)`
`= ((900 – 300))/5`
`= $120`
b.ii. `text(Value after 5 years)`
`= 900 – (250 xx 0.46 xx 5)`
`= $325`
Khan wants to buy some office furniture that is valued at $7000.
The balance is to be paid in 24 equal monthly instalments. No interest is charged.
Another store offers the same $7000 office furniture for $500 deposit and 36 monthly instalments of $220.
Write your answer as a percentage correct to one decimal place. (2 marks)
A third store has the office furniture marked at $7000 but will give 15% discount if payment is made in cash at the time of sale.
a.i. | `text(Deposit)` | `= 25text(%) xx 7000` |
`= $1750` |
a.ii. `text(Installment amount)`
`= ((7000 – 1750))/24`
`= $218.75`
b.i. | `text(Total paid)` | `= 500 + 36 xx 220` |
`= $8420` |
b.ii. `text(Total interest paid)`
`= 8420 – 7000`
`= $1420`
`I` | `= (PrT)/100` |
`1420` | `= (6500 xx r xx 3)/100` |
`:. r` | `= (1420 xx 100)/(6500 xx 3)` |
`= 7.282…` | |
`= 7.3text{% (1 d.p.)}` |
c. | `text(Cash price)` | `= 7000 – 15text(%) xx 7000` |
`= 7000 – 1050` | ||
`= $5950` |
Gas is generally cheaper than petrol.
A car must run on petrol for some of the driving time.
Let `x` be the number of hours driving using gas
`y` be the number of hours driving using petrol
Inequalities 1 to 5 below represent the constraints on driving a car over a 24-hour period.
Explanations are given for Inequalities 3 and 4.
Inequality 1: `x ≥ 0`
Inequality 2: `y ≥ 0`
Inequality 3: `y ≤ 1/2x` | The number of hours driving using petrol must not exceed half the number of hours driving using gas. |
Inequality 4: `y ≥ 1/3x` | The number of hours driving using petrol must be at least one third the number of hours driving using gas. |
Inequality 5: `x + y ≤ 24`
The lines `x + y = 24` and `y = 1/2x` are drawn on the graph below.
On a particular day, the Goldsmiths plan to drive for 15 hours. They will use gas for 10 of these hours.
On another day, the Goldsmiths plan to drive for 24 hours.
Their car carries enough fuel to drive for 20 hours using gas and 7 hours using petrol.
Maximum = ___________ hours
Minimum = ___________ hours
a. `text(Inequality 5 means that the total hours driving)`
`text(with gas PLUS the total hours driving with petrol)`
`text(must be less than or equal to 24 hours.)`
b.i. & ii.
c. `text(If they drive for 10 hours on gas, 5 hours)`
`text(is driven on petrol.)`
`=>\ text{(10, 5) is in the feasible region.}`
`:.\ text(They comply with all constraints.)`
d. `text(Maximum = 18 hours)`
`text{(6 hours of petrol available)}`
`text(Minimum = 17 hours)`
`text{(7 hours of petrol is the highest available)}`
The Goldsmiths car can use either petrol or gas.
The following equation models the fuel usage of petrol, `P`, in litres per 100 km (L/100 km) when the car is travelling at an average speed of `s` km/h.
`P = 12 - 0.02s`
The line `P = 12 - 0.02s` is drawn on the graph below for average speeds up to 110 km/h.
Write your answer correct to one decimal place. (1 mark)
The following equation models the fuel usage of gas, `G`, in litres per 100 km (L/100 km) when the car is travelling at an average speed of `s` km/h.
`G = 15 - 0.06s`
The Goldsmiths' car travels at an average speed of 85 km/h. It is using gas.
Gas costs 80 cents per litre.
Write your answer in dollars and cents. (2 marks)
a. `text(When)\ S = 60\ text(km/hr)`
`P` | `= 12 – 0.02 xx 60` |
`= 10.8\ text(litres)` |
b. `text(When)\ s = 110,`
`G = 15 – 0.06 xx 100 = 8.4`
`:.\ text{(0, 15) and (110, 8.4) are on the line}`
c. `text(Intersection of graphs occur when)`
`12 – 0.02s` | `= 15 – 0.06s` |
`0.04s` | `= 3` |
`s` | `= 75` |
`:.\ text(Gas usage is less than fuel for)`
`text(average speeds over 75 km/hr.)`
d. `text(When)\ x = 85,`
`text(Gas usage)` | `= 15 – 0.06 xx 85` |
`= 9.9\ text(L/100 km.)` |
`:.\ text(C)text(ost of gas for 100km journey)`
`= 9.9 xx 0.80`
`= $7.92`
The Goldsmith family are going on a driving holiday in Western Australia.
On the first day, they leave home at 8 am and drive to Watheroo then Geraldton.
The distance–time graph below shows their journey to Geraldton.
At 9.30 am the Goldsmiths arrive at Watheroo.
They stop for a period of time.
After leaving Watheroo, the Goldsmiths continue their journey and arrive in Geraldton at 12 pm.
The Goldsmiths leave Geraldton at 1 pm and drive to Hamelin. They travel at a constant speed of 80 km/h for three hours. They do not make any stops.
a.i. `text(Let)\ \ y = g(x)`
`text(Inverse: swap)\ \ x harr y,\ \ text(Domain)\ (g^-1) = text(Range)\ (g)`
`x = 2 log_e (y + 4) + 1`
`:. g^-1 (x) = e^((x – 1)/2) – 4,\ \ x in R`
ii. |
iii. `text(Intercepts of a function and its inverse occur)`
`text(on the line)\ \ y=x.`
`text(Solve:)\ \ g(x) = g^-1 (x)\ \ text(for)\ \ x`
`:. x dot = – 3.914 or x = 5.503\ \ text{(3 d.p.)}`
iv. | `text(Area)` | `= int_(-3.91432…)^(5.50327…) (g(x) – g^-1 (x))\ dx` |
`= 52.63\ text{u² (2 d.p.)}` |
b.i. `text(Vertical Asymptote:)`
`x = – 1`
`:. a = 1`
ii. `text(Solve)\ \ f(0) = 1\ \ text(for)\ \ c,`
`c = 1`
iii. `f(x)= k log_e (x + 1) + 1`
`text(S)text(ince)\ \ f(p)=10,`
`k log_e (p + 1) + 1` | `= 10` |
`k log_e (p + 1)` | `= 9` |
`:. k` | `= 9/(log_e (p + 1))\ text(… as required)` |
iv. | `f prime (x)` | `= k/(x + 1)` |
`f prime (p)` | `= k/(p + 1)` | |
`= (9/(log_e(p + 1))) xx 1/(p + 1)\ \ \ text{(using part (iii))}` | ||
`= 9/((p + 1)log_e(p + 1))\ text(… as required)` |
v. `text(Two points on tangent line:)`
`(p, 10),\ \ (– 1, 0)`
`f prime (p)` | `= (10 – 0)/(p – (– 1))` |
`=10/(p+1)` | |
`text(Solve:)\ \ 9/((p + 1)log_e(p + 1))=10/(p+1)\ \ \ text(for)\ p,`
`:.p= e^(9/10) – 1`
Michelle intends to keep a car purchased for $17 000 for 15 years. At the end of this time its value will be $3500.
a. | `text(Depreciation)` | `= (17\ 000 – 3500)/15` |
`= $13\ 500` |
`:.\ text(Annual depreciation)`
`= (13\ 500)/15`
`= $900`
b. `:.\ text(Flat rate of depreciation )`
`= 900/(17\ 000) xx 100text(%)`
`= 5.29…`
`= 5.3text{% (1 d.p.)}`
Michelle decided to invest some of her money at a higher interest rate. She deposited $3000 in an account paying 8.2% per annum, compounding half yearly.
Write your answer correct to the nearest cent. (1 mark)
a. `text(Compounding periods)\ (n) = 2 xx 2 = 4`
`text(Interest per half year) = 8.2/2 = 4.1text(%)`
`:. A` | `= PR^n` |
`= 3000(1.041)^4` | |
`= 3523.093…` | |
`= $3523.09\ \ text{(nearest cent)}` |
b. `text(After 4 years)\ (n = 8),`
`A` | `= 3000(1.041)^8` |
`= 4137.396…` |
`:.\ text(Interest)` | `= 4137.396-3000` |
`= 1137.396…` | |
`= $1137.40\ \ text{(nearest cent)}` |
Jevin has a banThe bank statement below shows the transactions on Michelle’s account for the month of July.
Interest for this account is calculated on the minimum monthly balance at a rate of 3% per annum.
a. | `text(Deposit)` | `= 6870.67-6250.67` |
`= $620` |
b. `text(Minimum Balance) = $6120.86`
`:.\ text(Interest)` | `=(PrT)/100` |
`= 6120.86 xx 3/100 xx 1/12` | |
`= 15.302…` | |
`= $15.30` |
An event involves running for 10 km and cycling for 30 km.
Let `x` be the time taken (in minutes) to run 10 km
`y` be the time taken (in minutes) to cycle 30 km
Event organisers set constraints on the time taken, in minutes, to run and cycle during the event.
Inequalities 1 to 6 below represent all time constraints on the event.
Inequality 1: `x ≥ 0` | Inequality 4: `y <= 150` |
Inequality 2: `y ≥ 0` | Inequality 5: `y <= 1.5x` |
Inequality 3: `x ≤ 120` | Inequality 6: `y >= 0.8x` |
The lines `y = 150` and `y = 0.8x` are drawn on the graph below.
One competitor, Jenny, took 100 minutes to complete the run.
Tiffany qualified for a prize.
a. `text(Inequality 3 means that the run must take)`
`text(120 minutes or less for any competitor.)`
b.i. & ii.
c. `text(From the graph, the possible cycling)`
`text(time range is between:)`
`text(80 – 150 minutes)`
d.i. `text(Constraint to win a prize is)`
`x + y <= 90`
`text(Maximum cycling time occurs)`
`text(when)\ y = 1.5x`
`:. x + 1.5x` | `<= 90` |
`2.5x` | `<= 90` |
`x` | `<= 36` |
`:. y_(text(max))` | `= 1.5 xx 36` |
`= 54\ text(minutes)` |
d.ii. `text(Maximum run time occurs)`
`text(when)\ \ y = 0.8x`
`:. x + 0.8x` | `<= 90` |
`1.8x` | `<= 90` |
`x` | `<= 50` |
`:. x_(text(max)) = 50\ text(minutes)`
Tiffany decides to enter a charity event involving running and cycling.
There is a $35 fee to enter.
The event costs the organisers $50 625 plus $12.50 per competitor.
The number of competitors who entered the event was 8670.
a. `R = 35x`
b. `C = 50\ 625 + 12.50x`
c.i. `text(Break even when)\ R = C`
`35x` | `= 50\ 625 + 12.5x` |
`22.5x` | `= 50\ 625` |
`x` | `= (50\ 625)/22.5` |
`= 2250` |
`:. 2250\ text(competitors required to break even.)`
c.ii. `text(When)\ \ x = 8670,`
`text(Profit)` | `= R – C` |
`= 35 xx 8670 – (50\ 625 + 12.5 xx 8670)` | |
`= 303\ 450 – 159\ 000` | |
`= $144\ 450` |
The golf club management purchased new lawn mowers for $22 000.
a. `text(When)\ n = 4,`
`text(Depreciation)` | `= 22\ 000 xx 0.12 xx 4` |
`= $10\ 560` |
`:.\ text(Depreciated Value)`
`= 22\ 000 – 10\ 560`
`= $11\ 440`
b. `r = 16text(%)`
`text(Value)` | `= 22\ 000(1 – r)^n` |
`= 22\ 000(0.84)^4` | |
`= 10\ 953.169…` | |
`= $10\ 953.17` |
c. `text(Reducing balance gives a greater depreciated)`
`text(amount after 4 years.)`
`text(Greater depreciation amount)`
`= 22\ 000 – 10\ 953.17`
`= $11\ 046.83`
The golf club’s social committee has $3400 invested in an account which pays interest at the rate of 4.4% per annum compounding quarterly.
Write your answer in dollars correct to the nearest cent. (1 mark)
Write your answer in dollars correct to the nearest cent. (2 marks)
a. `text(Interest rate per quarter)`
`= 4.4/4`
`= 1.1text(% …as required.)`
b. `text(Compounding periods = 12)`
`A` | `= PR^n` |
`= 3400(1.011)^12` | |
`= 3876.973…` | |
`= $3876.97` |
c. `text(Compounding periods) = 6 xx 4 = 24`
`A` | `= 3400(1.011)^24` |
`= 4420.858…` |
`text(Interest earned over 6 years)`
`= 4420.86- 3400`
`= $1020.86\ \ text{(nearest cent)}`
Rebecca will need to borrow $250 to buy a golf bag.
Calculate the interest Rebecca will pay in the first month.
Write your answer correct to the nearest cent. (1 mark)
Calculate the annual flat interest rate charged. Write your answer as a percentage correct to one decimal place. (1 mark)
a. `text(Interest in the 1st month)`
`= 1.5text(%) xx 250`
`= $3.75`
b. `text(Annual interest) = 12 xx 6 = $72`
`:.\ text(Annual flat interest rate)`
`= 72/250 xx 100text(%)`
`= 28.8text(%)`
Another company, Cheapstar Airlines, uses the two equations below to calculate the total cost of a flight.
The passenger fare, in dollars, for a given distance, in km, is calculated using the equation
fare = `20` + `0.47` × distance.
The charge, in dollars, for a particular excess luggage weight, in kg, is calculated using the equation
charge = `m` × (excess luggage weight)².
Suzie will fly 450 km with 15 kg of excess luggage on Cheapstar Airlines.
She will pay $299 for this flight.
Determine the value of `m`. (2 marks)
`m = 0.3`
`text{Total cost = fare + charge (luggage)}`
`text(fare)` | `= 20 + 0.47 xx 450` |
`= $231.50` |
`:.\ text(Amount left for luggage)`
`= 299 – 231.50`
`= $67.50`
`67.50` | `= m xx 15^2` |
`:. m` | `= (67.50)/(15^2)` |
`= 0.3` |
Luggage over 20 kg in weight is called excess luggage.
Fair Go Airlines charges for transporting excess luggage.
The charges for some excess luggage weights are shown in Table 2.
Fill in the missing (excess luggage weight)² value in Table 3 and plot this point with a cross (×) on the graph below. (1 mark)
charge = `k` × (excess luggage weight)²
Find `k`. (1 mark)
Write your answer in dollars correct to the nearest cent. (1 mark)
Fair Go Airlines offers air travel between destinations in regional Victoria.
Table 1 shows the fares for some distances travelled.
The fares for the distances travelled in Table 1 are graphed below.
Draw this information on the graph above. (1 mark)
Fair Go Airlines is planning to change its fares.
A new fare will include a service fee of $40, plus 50 cents per kilometre travelled.
An equation used to determine this new fare is given by
fare = `40 + 0.5` × distance.
How much will this passenger save on the fare calculated using the equation above compared to the fare shown in Table 1? (1 mark)
What is this distance? (2 marks)
fare = `a` + `b` × maximum distance.
Determine `a` and `b`. (2 marks)
a. `text(250 km)`
b. |
c. | `text(New fare)` | `= 40 + 0.5 xx 300` |
`= $190` |
`text(Fare from the table = $220`
`:.\ text(Passenger will save $30.)`
d. `text(In table 1, a fare of $220 applies for travel)`
`text(between 250 – 400 km.)`
`:. 220` | `= 40 + 0.5d` |
`0.5d` | `= 180` |
`d` | `= 360\ text(km)` |
e. `text(Equations in required form are:)`
`100` | `= a + b xx 100\ \ …(1)` |
`160` | `= a + b xx 250\ \ …(2)` |
`text(Subtract)\ (2) – (1)`
`60` | `= 150b` |
`:. b` | `= 60/150 = 2/5` |
`text(Substitute)\ b = 2/5\ text{into (1)}`
`100` | `= a + 2/5 xx 100` |
`:. a` | `= 60` |
Simple Saver is a simple interest investment in which interest is paid annually.
Growth Plus is a compound interest investment in which interest is paid annually.
Initially, $8000 is invested with both Simple Saver and Growth Plus.
The graph below shows the total value (principal and all interest earned) of each of these investments over a 15 year period.
The increase in the value of each investment over time is due to interest
Give a reason to justify your answer. (1 mark)
Find the amount of interest paid annually. (1 mark)
Write your answer as a percentage correct to one decimal place. (1 mark)
a. `text(Simple Saver has the highest annual)`
`text(interest rate because after 1 year,)`
`text(the value of investment is higher.)`
b. `text(Total interest earned)`
`= 21\ 800 – 8000`
`= $13\ 800`
`:.\ text(Interest paid annually)`
`= (13\ 800)/15`
`= $920`
c.i. `text(Using)\ A = PR^n,`
`24\ 000 = 8000 (1 + r/100)^15`
c.ii. | `(1 + r/100)^15` | `= 3` |
`1 + r/100` | `= 1.0759…` | |
`:. r` | `= 0.0759…` | |
`= 7.6text{% (1 d.p.)}` |
$360 000 is invested in a perpetuity at an interest rate of 5.2% per annum.
a. `text(Monthly repayment)`
`= 1/12 xx (360\ 000 xx 5.2/100)`
`= 1/12 xx 18\ 720`
`= $1560`
b. `$360\ 000`
The cash price of a large refrigerator is $2000.
She does not pay a deposit and will pay $55 per month for four years.
The following year it will rise by a further 2.0%.
Calculate the cash price of the refrigerator after these two price rises. (1 mark)
a.i. `text(Total amount paid)`
`= 55 xx 4 xx 12`
`= $2640`
a.ii. | `text(Total interest)` | `= 2640-2000` |
`= $640` |
a.iii. | `I` | `= (PrT)/100` |
`640` | `= (2000 xx r xx 4)/100` | |
`:. r` | `= (640 xx 100)/(2000 xx 4)` | |
`= 8text(%)` |
b. `text(After 1 year,)`
`P = 2000(1.025) = $2050`
`text(After 2 years,)`
`P = 2050(1.02) = $2091`
Let `x` be the number of Softsleep pillows that are sold each week and `y` be the number of Resteasy pillows that are sold each week.
A constraint on the number of pillows that can be sold each week is given by
Inequality 1: `x + y ≤ 150`
Each week, Anne sells at least 30 Softsleep pillows and at least `k` Resteasy pillows.
These constraints may be written as
Inequality 2: `x ≥ 30`
Inequality 3: `y ≥ k`
The graphs of `x + y = 150` and `y = k` are shown below.
What is the maximum possible weekly revenue that Anne can obtain? (2 marks)
Anne decides to sell a third type of pillow, the Snorestop.
She sells two Snorestop pillows for each Softsleep pillow sold. She cannot sell more than 150 pillows in total each week.
Inequality 4: `3x + y ≤ 150`
where `x` is the number of Softsleep pillows that are sold each week
and `y` is the number of Resteasy pillows that are sold each week. (1 mark)
Softsleep pillows sell for $65 each.
Resteasy pillows sell for $50 each.
Snorestop pillows sell for $55 each.
a. `text(Inequality 1 means that the combined number of Softsleep)`
`text(and Resteasy pillows must be less than 150.)`
b. `k = 45`
c.i. & ii. |
d. `text(Checking revenue at boundary)`
`text(At)\ (30,120),`
`R = 65 xx 30 + 50 xx 120 = $7950`
`text(At)\ (105,45),`
`R = 65 xx 105 + 50 xx 45 = $9075`
`:. text(Maximum weekly revenue) = $9075`
e. `text(Let)\ z = text(number of SnoreStop pillows)`
`:. x + y + z <= 150,\ text(and)`
`z = 2x\ \ text{(given)}`
`:. x + y + 2x` | `<= 150` |
`3x + y` | `<= 150\ \ …text(as required)` |
f. | `R` | `= 65x + 50y + 55(2x)` |
`= 65x + 50y + 110x` | ||
`= 175x + 50y` |
g. |
`text(New intersection occurs at)\ (35,45)`
`:.\ text(Maximum weekly revenue)`
`= 175 xx 35 + 50 xx 45`
`= $8375`
Anne sells Resteasy pillows.
Last week she sold 35 Softsleep and `m` Resteasy pillows.
The selling price per pillow is shown in Table 1 below.
The total revenue from pillow sales last week was $4275.
Find `m`, the number of Resteasy pillows sold. (1 mark)
`40`
`text(Total Revenue)` | `= 65 xx 35 + 50 xx m` |
`4275` | `= 2275 + 50m` |
`50m` | `= 2000` |
`m` | `= 40` |
Anne sells Softsleep pillows for $65 each.
`C = 500 + 40x`
Find the cost of making 30 Softsleep pillows. (1 mark)
The revenue, `R`, from the sale of `x` Softsleep pillows is graphed below.
In the plan below, the entry gate of an adventure park is located at point `G`.
A canoeing activity is located at point `C`.
The straight path `GC` is 40 metres long.
The bearing of `C` from `G` is 060°.
`CW` is a 90 metre straight path between the canoeing activity and a water slide located at point `W`.
`GW` is a straight path between the entry gate and the water slide.
The angle `GCW` is 120°.
Write your answer in square metres, correct to one decimal place. (1 mark)
Straight paths `CK` and `WK` lead to the kiosk located at point `K`.
These two paths are of equal length.
The angle `KCW` is 10°.
a. | `x^@ + 60^@` | `= 180^@` |
`:. x^@` | `= 120^@` |
b. `text(Bearing of)\ G\ text(from)\ C`
`= 360 – 120`
`= 240^@`
c. |
`text(Let)\ d = text(distance north of)\ C\ text(from)\ G`
`cos60^@` | `= d/40` |
`:. d` | `= 40 xx cos60` |
`= 20\ text(m)` |
d.i. | `A` | `= 1/2ab sinC` |
`= 1/2 xx 40 xx 90 xx sin120^@` | ||
`= 1558.84…` | ||
`= 1558.8\ text{m² (1 d.p.)}` |
d.ii. `text(Using the cosine rule,)`
`GW` | `= sqrt(40^2 + 90^2 + 2 xx 40 xx 90 xx cos120^@)` |
`= sqrt(13\ 300)` | |
`= 115.32…` | |
`= 115.3\ text{(1 d.p.) …as required.}` |
e.i. `DeltaCKW\ text(is isosceles)`
`:. angleCKW` | `= 180 – (2 xx 10)` |
`= 160^@` |
e.ii. `text(Using the sine rule,)`
`(CK)/(sin10^@)` | `= 90/(sin160^@)` |
`:. CK` | `= (90 xx sin10^@)/(sin160^@)` |
`= 45.69…` | |
`= 45.7\ text{m (1 d.p.)}` |
Tania takes out a reducing balance loan of $265 000 to pay for her house.
Her monthly repayments will be $1980.
Interest on the loan will be calculated and paid monthly at the rate of 7.62% per annum.
Immediately after Tania made her twelfth payment, the interest rate on her loan increased to 8.2% per annum, compounding monthly.
Tania decided to increase her monthly repayment so that the loan would be repaid in a further nineteen years.
Write your answer to the nearest cent. (1 mark)
a.i. `text(By TVM Solver:)`
`N` | `= ?` |
`I(text(%))` | `= 7.62` |
`PV` | `= 265\ 000` |
`PMT` | `= −1980` |
`FV` | `= 0` |
`text(P/Y)` | `= text(C/Y) = 12` |
`=> N = 299.573…`
`:.\ text(After 300 months, the loan will be repaid.)`
a.ii. `text(After 12 months, by TVM Solver:)`
`N` | `= 12` |
`I(text(%))` | `= 7.62` |
`PV` | `= 265\ 000` |
`PMT` | `= −1980` |
`FV` | `= ?` |
`text(P/Y)` | `= text(C/Y) = 12` |
`=> FV = −261\ 305.74…`
`:.\ text(Amount paid off after 1 year)`
`= 265\ 000 – 261\ 305.747…`
`= 3694.252…`
`= $3694.25\ \ text{(nearest cent)}`
b. `text(By TVM Solver:)`
`N` | `= 19 xx 12 = 228` |
`Itext(%)` | `= 8.2` |
`PV` | `= 261\ 305.75` |
`PMT` | `= ?` |
`FV` | `= 0` |
`text(P/Y)` | `= text(C/Y) = 12` |
`=> PMT = −2265.043`
`:.\ text(New monthly repayment is $2265.04)`
Tania purchased a house for $300 000.
She will have to pay stamp duty based on this purchase price.
Stamp duty rates are listed in the table below.
Write your answer to the nearest thousand dollars. (1 mark)
Tania bought her house at the start of 2011.
a. | `text(Stamp duty)` | `= 2870 + 6text(%) xx (300\ 000 – 130\ 000)` |
`= $13\ 070` |
b. `text(Using)\ \ \ A = PR^n`
`text(Value)` | `= 300\ 000(1.0317)^5` |
`= 350\ 661.7…` | |
`= $351\ 000\ \ text{(nearest $1000)}` |
c. `text(Find)\ n\ text(when)\ \ \ A > $450\ 000`
`300\ 000 xx 1.0317^n` | `= 450\ 000` |
`n` | `~~12.99…` |
`:.\ text(After 13 years, in 2024, the house value)`
`text(will be over)\ $450\ 000.`
Tom and Patty both decided to invest some money from their savings.
Each chose a different investment strategy.
Tom's investment strategy
• Deposit $5600 into an account with an interest rate of 7.2% per annum, compounding monthly.
• Immediately after interest is paid into his investment account on the last day of each month, deposit a further $200 into the account.
Patty's investment strategy
• Invest $8000 at the start of the year at an interest rate of 7.2% per annum, compounding annually.
At the end of twelve months, Patty has more money in her investment account than Tom.
Write your answer to the nearest cent. (2 marks)
Write your answer correct to one decimal place. (1 mark)
a. `text(Tom’s investment after 1 month)`
`= 5600 xx (1 + 7.2/(12 xx 100)) + 200`
`= $5833.60`
b. `text(After 1 year,)`
`text(Value of investment) = 8000 xx (1 + 7.2/100)`
c. `text(Tom’s investment after 12 months,)`
`text(by TVM Solver,)`
`N` | `= 12` |
`I(text(%))` | `= 7.2` |
`PV` | `= 5600` |
`PMT` | `= 200` |
`text(P/Y)` | `= 12` |
`text(C/Y)` | `= 12` |
`=> FV = −8497.58…`
`text(Patty’s investment after 12 months)`
`= 8000 xx (1 + 7.2/100)`
`= $8576`
`:.\ text(Extra value of Patty’s investment)`
`= 8576 – 8497.580…`
`= 78.419…`
`= $78.42\ \ text{(nearest cent)}`
d. | `text(Using)\ \ \ I` | `= (PrT)/100` |
`1000` | `= (8000 xx r xx 1)/100` | |
`r` | `= (1000 xx 100)/8000` | |
`= 12.5text(%)` |
Tony plans to take his family on a holiday.
The total cost of $3630 includes a 10% Goods and Services Tax (GST).
During the holiday, the family plans to visit some theme parks.
The prices of family tickets for three theme parks are shown in the table below.
If Tony purchases the Movie Journey family ticket online, the cost is discounted to $202.40
a. `text(Let $)P\ text(be the cost ex-GST)`
`P + 10text(%)P` | `= 3630` |
`1.1P` | `= 3630` |
`P` | `= 3630/1.1` |
`= $3300` | |
`:.\ text(GST)` | `= 10text(%) xx $3300` |
`= $330` |
b. `text(C)text(ost to visit all 3 parks)`
`= 82 + 220 + 160`
`= $462`
c. | `text(Savings)` | `= 220 – 202.40` |
`= 17.60` |
`:.\ text(Discount)` | `= (17.60)/220` |
`= 0.08` | |
`= 8text(%)` |
An area of a club needs to be refurbished.
$40 000 is borrowed at an interest rate of 7.8% per annum.
Interest on the unpaid balance is charged to the loan account monthly.
Suppose the $40 000 loan is to be fully repaid in equal monthly instalments over five years.
Determine the monthly interest payment, correct to the nearest cent. (1 mark)
a. `text(Find monthly payment by TVM solver:)`
`N` | `= 5 xx 12 = 60` |
`I(text(%))` | `= 7.8` |
`PV` | `= 40\ 000` |
`PMT` | `= ?` |
`FV` | `= 0` |
`text(P/Y)` | `= text(C/Y) = 12` |
`=> PMT = −807.232…`
`:.\ text{Monthly payment is $807.23 (nearest cent)}`
b. `text(Find)\ n\ text(when loan fully paid:)`
`N` | `= ?` |
`I(text(%))` | `= 7.8` |
`PV` | `= 40\ 000` |
`PMT` | `= −1000` |
`FV` | `= 0` |
`text(P/Y)` | `= text(C/Y) = 12` |
`=> N = 46.47…`
`:.\ text(Loan will be fully repaid after 47 months.)`
c.i. `text(Loan balance after 12 months)`
`= 40\ 000 xx (1 + (7.8)/(12 xx 100))^12`
`= 43\ 233.99…`
`= $43\ 234\ \ text{(nearest $) … as required}`
c.ii. `text(Interest paid each month)`
`= 43\ 234 xx (7.8)/(12 xx 100)`
`= 281.021`
`= $281.02\ \ text{(nearest cent)}`
The value of the equipment will be depreciated using the unit cost method.
The initial value of the equipment is $8360. It will depreciate by 22 cents per hour of use.
On average, the equipment will be used for 3800 hours each year.
The initial value of the equipment is $8360. It will depreciate at a rate of 14% per annum of the reducing balance.
Find, correct to the nearest dollar, the depreciated value of the equipment after ten years. (1 mark)
a. `text(Unit cost depreciation per year)`
`= 3800 xx 0.22`
`= $836`
`:.\ text(After 3 years, depreciated value)`
`= 8360 – (3 xx 836)`
`= $5852`
b. `text(10% Depreciation per year)`
`= 10text(%) xx 8360`
`= $836\ \ …text(as required.)`
c. `text(Depreciated value of $0 occurs when)`
`8360 – 836n` | `= 0` |
`836n` | `= 8360` |
`n` | `= 10\ text(years)` |
d. `text(After 1 year,)`
`V_1` | `= 8360(1 – 0.14)` |
`= 8360(0.86)` |
`text(After 10 years,)`
`V_10` | `= 8360(0.86)^10` |
`= 1850.08…` | |
`= $1850\ \ text{(nearest $)}` |
A club purchased new equipment priced at $8360. A 15% deposit was paid.
The amount owing will be fully repaid in 12 instalments of $650.
a. | `text(Deposit)` | `= 15text(%) xx 8360` |
`= $1254` |
b.i. | `text(Amount still owed)` | `= 8360 − 1254` |
`= $7106` |
b.ii. | `text(Total repayments)` | `= 12 xx 650` |
`= $7800` |
`:.\ text(Total interest paid)`
`= 7800 – 7106`
`= $694`
The network below shows the activities that are needed to finish a particular project and their completion times (in days).
Part 1
The earliest start time for Activity `K`, in days, is
A. 7
B. 15
C. 16
D. 19
E. 20
Part 2
This project currently has one critical path.
A second critical path, in addition to the first, would be created by
A. increasing the completion time of `D` by 7 days.
B. increasing the completion time of `G` by 1 day.
C. increasing the completion time of `I` by 2 days.
D. decreasing the completion time of `C` by 1 day.
E. decreasing the completion time of `H` by 2 days.
`text(Part 1:)\ C`
`text(Part 2:)\ A`
`text(Part 1)`
`text(EST for Activity)\ K`
`=\ text(Duration)\ ACFI`
`= 2 + 5 + 6 + 3`
`= 16`
`=> C`
`text(Part 2)`
`text(Original critical path is)`
`ACFHJL\ text{(22 days)}`
`text(Consider option)\ A,`
`text(New critical path is created)`
`ABDJL\ text{(22 days)}`
`=> A`
In the network below, the values on the edges give the maximum flow possible between each pair of vertices. The arrows show the direction of flow. A cut that separates the source from the sink in the network is also shown.
Part 1
The capacity of this cut is
A. `14`
B. `18`
C. `23`
D. `31`
E. `40`
Part 2
The maximum flow between source and sink through the network is
A. `7`
B. `10`
C. `11`
D. `12`
E. `20`
`text(Part 1:)\ C`
`text(Part 2:)\ B`
Part 1
`text(Capacity of the cut)`
`= 11 + 5 + 7`
`= 23`
`=> C`
Part 2
`text(The maximum flow)`
`=\ text{minimum cut (see above)}`
`= 4 + 2 + 3 + 1`
`= 10`
`=> B`
A board game consists of nine labelled squares as shown.
A player must start at square `J` and, moving one square at a time, aim to finish at square `R`.
Each move may only be to the right one square or down one square.
A player who lands on square `N` must stay there and cannot move again.
A player can only stop moving when they reach `N` or `R`.
A digraph that shows all the possible moves that a player could make to reach `N` or `R` from `J` is
`E`
`=> E`
A complete graph with six vertices is drawn.
This network would best represent
`C`
`text(A complete graph has all vertices connected)`
`text(directly to all other vertices without any)`
`text(parallel edges or loops.)`
`rArr C`
There are five teams, `A, B, C, D` and `E`, in a volleyball competition. Each team played each other team once in 2007.
The results are summarised in the directed graph below. An arrow from `A` to `E` signifies that `A` defeated `E.`
In 2007, the team that had the highest number of two-step dominances was
A. team `A`
B. team `B`
C. team `C`
D. team `D`
E. team `E`
`B`
`text(2-step dominance matrix)`
`{: (\ quad A quad B quad C quad D quad E), ([(0, 0, 1, 2, 1), (3, 0, 0, 2, 1), (0, 3, 0, 0, 0), (0, 0, 1, 0, 1), (0, 0, 1, 0, 0)]{:(A – 4), (B – 6), (C – 3), (D – 2), (E – 1):}\ \ \ \ {:text(“2-step” wins):}):}`
`=> B`
The minimal spanning tree for the network below includes two edges with weightings `x` and `y.`
The length of the minimal spanning tree is 19.
The values of `x` and `y` could be
A. `x = 1 and y = 7`
B. `x = 2 and y = 5`
C. `x = 3 and y = 5`
D. `x = 4 and y = 5`
E. `x = 5 and y = 6`
`C`
The following network shows the distances, in kilometres, along a series of roads that connect town `A` to town `B.`
The shortest distance, in kilometres, to travel from town `A` to town `B` is
A. `9`
B. `10`
C. `11`
D. `12`
E. `13`
`B`
`text(Shortest path)`
`= 4 + 1 + 3 + 2`
`= 10`
`=> B`
Four workers, Anna, Bill, Caitlin and David, are each to be assigned a different task.
The table below gives the time, in minutes, that each worker takes to complete each of the four tasks.
The tasks are allocated so as to minimise the total time taken to complete the four tasks.
This total time, in minutes, is
A. `21`
B. `28`
C. `31`
D. `34`
E. `38`
`C`
The network shows the activities that are needed to complete a particular project.
Part 1
The total number of activities that need to be completed before activity `L` may begin is
A. `2`
B. `4`
C. `6`
D. `7`
E. `8`
Part 2
The duration of every activity is initially 5 hours. For an extra cost, the completion times of both activity `F` and activity `K` can be reduced to 3 hours each.
If this is done, the completion time for the project will be
A. decreased by 2 hours.
B. decreased by 3 hours.
C. decreased by 4 hours.
D. decreased by 6 hours.
E. unchanged.
`text(Part 1:)\ D`
`text(Part 1:)\ E`
`text(Part 1)`
`A, B, C, D, E, H\ text(and)\ I\ text(must be)`
`text(completed before)\ L.`
`=> D`
`text(Part 2)`
`F\ text(and)\ K\ text(are not on any critical path and a)`
`text(reduction of 3 hours on either activity will not change)`
`text(the completion time for the project.)`
`=>E`
A connected planar graph has 10 edges and 10 faces.
The number of vertices for this graph is
`A`
`v+f` | `=e+2` |
`:. v` | `= e-f + 2` |
`= 10-10 + 2` | |
`= 2` |
`=> A`
The diagram shows the tasks that must be completed in a project.
Also shown are the completion times, in minutes, for each task.
The critical path for this project includes activities
A. `B and I.`
B. `C and H.`
C. `D and E.`
D. `F and K.`
E. `G and J.`
`D`
`text(The critical path is)\ \ ACFIK.`
`=> D`
The map of Australia shows the six states, the Northern Territory and the Australian Capital Territory (ACT).
In the network diagram below, each of the vertices `A` to `H` represents one of the states or territories shown on the map of Australia. The edges represent a border shared between two states or between a state and a territory.
Part 1
In the network diagram, the order of the vertex that represents the Australian Capital Territory (ACT) is
A. `0`
B. `1`
C. `2`
D. `3`
E. `4`
Part 2
In the network diagram, Queensland is represented by
A. vertex A.
B. vertex B.
C. vertex C.
D. vertex D.
E. vertex E.
`text(Part 1:)\ B`
`text(Part 2:)\ A`
`text(Part 1)`
`text {ACT has 1 border (with NSW)}`
`:.\ text(Its vertex will be one degree.)`
`=> B`
`text(Part 2)`
`text(NSW is Vertex)\ B`
`:.\ text(Queensland is vertex)\ A\ text(as it is)`
`text(connected to)\ B\ text(and has degree)`
`text{3 (}C\ text{is Victoria as it has degree 2)}`
`=> A`
The graph below shows the one-step dominances between four farm dogs, Kip, Lab, Max, and Nim.
In this graph, an arrow from Lab to Kip indicates that Lab has a one-step dominance over Kip.
From this graph, it can be concluded that Kip has a two-step dominance over
A. Max only.
B. Nim only.
C. Lab and Nim only.
D. all of the other three dogs.
E. none of the other three dogs.
`C`
`=> C`
In the digraph above, all vertices are reachable from every other vertex.
All vertices would still be reachable from every other vertex if we remove the edge in the direction from
A. `Q` to `U`
B. `R` to `S`
C. `S` to `T`
D. `T` to `R`
E. `V` to `U`
`A`
`rArr A`
`{:({:qquadqquadPquadQquadRquadS:}),({:(P),(Q),(R),(S):}[(0,0,2,1),(0,0,1,1),(2,1,0,1),(1,1,1,0)]):}`
The adjacency matrix above represents a planar graph with four vertices.
The number of faces (regions) on the planar graph is
A. `1`
B. `2`
C. `3`
D. `4`
E. `5`
`D`