The equation of the line that passes through the points (0, 4) and (2, 4) is
- `x = 4`
- `y = 4`
- `y = 4x`
- `y = 4x + 2`
- `y = 2x + 4`
Aussie Maths & Science Teachers: Save your time with SmarterEd
The equation of the line that passes through the points (0, 4) and (2, 4) is
`B`
`m = (4 – 4)/(2 – 0) = 0`
`ytext(-intercept = 4)`
`:. text(Equation is:)\ y = 4`
`=> B`
Three circles of radius 50 mm are placed so that they just touch each other.
The region enclosed by the circles is shaded in the diagram below.
The area of the shaded region, in square millimetres, is closest to
`A`
A triangle `ABC` has:
• one side, `bar(AB)`, of length 4 cm
• one side, `bar(BC)`, of length 7 cm
• one angle, `∠ACB`, of 26°.
Which one of the following angles, correct to the nearest degree, could not be another angle in triangle `ABC`?
`E`
`text(Using the sine rule,)`
`(sin x)/7` | `= (sin 26)/4` |
`sin x` | `= (7 xx sin 26)/4` |
`= 0.767…` | |
`:. x` | `= 50° text{or 130° (nearest degree)}` |
`text(If)\ \ x = 50°,\ text(other angle = 104°)`
`text(If)\ \ x = 130°,\ text(other angle = 24°)`
`=> E`
A hemispherical bowl of radius 10 cm is shown in the diagram below.
The bowl contains water with a maximum depth of 2 cm.
The radius of the surface of the water, in centimetres, is
`B`
A segment is formed by an angle of 75° in a circle of radius 112 mm.
This segment is shown shaded in the diagram below.
Which one of the following calculations will give the area of the shaded segment?
`C`
`text(Area)` | `=\ text(Area small sector − Area of triangle)` |
`= pi xx 112^2 xx 75/360 – 1/2 xx 112^2 xx sin(75°)` |
`=> C`
A grain storage silo in the shape of a cylinder with a conical top is shown in the diagram below.
The volume of this silo, in cubic metres, is closest to
`B`
`V` | `=\ text(volume of cylinder + volume of cone)` |
`= pir^2h + 1/3pir^2h` | |
`= pi xx 5^2 xx 16 + 1/3 xx pi xx 5^2 xx 1.8` | |
`= 1303.76…\ text(m³)` |
`=> B`
The locations of four cities are given below.
Adelaide (35° S, 139° E) | Buenos Aires (35° S, 58° W) |
Melilla (35° N, 3° W) | Heraklion (35° N, 25° E) |
In which order, from first to last, will the sun rise in these cities on New Year’s Day 2018?
`B`
`text(Sun rises in the East.)`
`text(Adelaide is further east than Heraklion.)`
`text(Melilla is further east (i.e. less west))`
`text(than Buenos Aires.)`
`:.\ text{Order is: Adelaide (139°E) , Heraklion (25°E),}`
`text{Melilla (3°W), Buenos Aires (58°W)}`
`=> B`
A right-angled triangle, `XYZ`, has side lengths `XY = 38.5\ text(cm)` and `YZ = 24.0\ text(cm)`, as shown in the diagram below.
The length of `XZ`, in centimetres, is closest to
`D`
`text(By Pythagoras:)`
`XZ` | `= sqrt(38.5^2 + 24^2)` |
`= sqrt(2058.25)` | |
`= 45.36…\ text(cm)` |
`=> D`
A wheel has five spokes equally spaced around a central hub, as shown in the diagram below.
The angle `theta` between two spokes is labelled on the diagram.
What is the angle `theta`?
`D`
`theta` | `= 360/5qquad(360^@\ text(about a point))` |
`= 72^@` |
`=> D`
Kim lives in Perth (32°S, 115°E). He wants to watch an ice hockey game being played in Toronto (44°N, 80°W) starting at 10.00 pm on Wednesday.
What is the time in Perth when the game starts?
A. 9.00 am on Wednesday
B. 7.40 pm on Wednesday
C. 9.00 pm on Wednesday
D. 12.20 am on Thursday
E. 11.00 am on Thursday
`E`
`text(Perth is East of Toronto) =>\ text(Ahead)`
`text(Longitudinal difference)`
`= 115 + 80`
`= 195^@`
`text(Time difference)` | `= 195/15` |
`= 13\ text(hours)` |
`:.\ text(Time in Perth)`
`=\ text{10 pm (Wed) + 13 hours}`
`=\ text(11 am on Thursday)`
`=> E`
Makoua and Macapá are two towns on the equator.
The longitude of Makoua is 16°E and the longitude of Macapá is 52°W.
How far apart are these two towns if the radius of Earth is approximately 6400 km?
A. `4000\ text(km)`
B. `7600\ text(km)`
C. `1\ 367\ 200\ text(km)`
D. `1\ 447\ 600\ text(km)`
E. `2\ 734\ 400\ text(km)`
`B`
Stockholm is located at 59°N 18°E and Darwin is located at 13°S 131°E.
What is the time difference between Stockholm and Darwin? (Ignore time zones and daylight saving.)
`D`
`text(Stockholm is 59°N 18°E, Darwin is 13°S 131°E)`
`text{Angular difference (longitude)}`
`= 131^@− 18^@`
`= 113^@`
`:.\ text(Time difference)` | `= 113 xx 4` |
`= 452\ text(minutes)` |
`⇒ D`
Which expression will give the shortest distance, in kilometres, between Mount Isa (20°S 140°E) and Tokyo (35°N 140°E)?
`B`
In this diagram of the Earth, `O` represents the centre and `B` lies on both the Equator and the Greenwich Meridan.
What is the latitude and longitude of point `A`?
A. `text(30°N 110°E)`
B. `text(70°N 30°W)`
C. `text(110°N 60°E)`
D. `text(30°N 110°W)`
E. `text(60°N 110°W)`
`A`
`text(S)text(ince A is) 30^circ\ text(North of the Equator)`
`→ text(Latitude is) 30^circ text(N)`
`text(S)text(ince A is) 110^circ\ text(East of Greenwich)`
`→ text(Longitude is) 110^circ text(E)`
`:. A\ text(coordinates are:) \ 30^circ text(N) 110^circ text(E)`
`=> A`
Two cities lie on the same meridian of longitude. One is 40° north of the other.
What is the distance between the two cities, correct to the nearest kilometre? (2 marks)
`4468\ text{km (nearest km)}`
A ship sails due South from Channel-Port-aux-Basques, Canada, 47°N 59°W to Barbados, 13°N 59°W.
How far did the ship sail, to the nearest kilometre? Assume that the radius of Earth is 6400 km. (2 marks)
`3798\ text{km (nearest km)}`
`text(Angular difference in latitude)` | `=47` `-13` |
`=34^@` |
`text{(No difference in longitude)}`
`:.\ text(Distance)` | `=34/360` `xx 2 pi r` |
`=34/360` `xx 2` `xx pi` `xx 6400` | |
`= 3797.836…` | |
`= 3798` `text{km (nearest km)}` |
Osaka is at 34°N, 135°E, and Denver is at 40°N, 105°W.
What time and day is it in Osaka then? (1 mark)
What was the time and day in Denver when John received the text? (1 mark)
a. | `text(Longitude difference)` | `= 135 + 105` |
`= 240^@` |
`text(Time difference)` | `= 240/15` |
`= 16\ text(hours … as required)` |
b. `text(Denver is behind Osaka time because it is further west.)`
`:.\ text(Time in Osaka)` | `= 9\ text(pm Monday plus 16 hours)` |
`= 1\ text(pm Tuesday)` |
c. `text(Denver is 16 hours behind Osaka)`
`:.\ text(John will receive the text at 10 am Thursday less 16)`
`text{plus 14 hours (i.e. 8 am Thursday.)}`
Pontianak has a longitude of 109°E, and Jarvis Island has a longitude of 160°W.
Both places lie on the Equator
a. | `text(Longitude difference)` | `= 109 + 160` |
`= 269^@` |
`=> text(Shortest distance)\ text{(by degree)}` | `= 360 – 269` |
`= 91^@` |
`:.\ text(Shortest distance)` | `= 91/360 xx 2 pi r` |
`= 91/360 xx 2 xx pi xx 6400` | |
`= 10\ 164.79…` | |
`=10\ 165\ text(km)\ text{(nearest km)}` |
b. | `text(Latitude)` |
`4^@\ text(South of Jarvis Island)` | |
`text(S)text(ince Jarvis Island is on equator)` | |
`=>\ text(Latitude is)\ 4^@ text(S)` | |
`text(Longitude)` | |
`text(Jarvis Island is)\ 160^@ text(W)` | |
`text(Rubail is)\ 48^@\ text(West of Jarvis Island, or 208° West)` | |
`text(which is)\ 28^@\ text{past meridian (180°)}` |
`=>\ text(Longitude)` | `= (180\ -28)^@ text(E)` |
`= 152^@ text(E)` |
`:.\ text(Position is)\ (4^@text{S}, 152^@text{E})`
Melbourne is located at (38°S, 145°E) and Dubai is located at (24°N, 55°E).
What will be the time and the day in Dubai when the plane is due to land? (1 marks)
a. `text(Difference in longitude)`
`= 145 – 55`
`= 90^@`
b. `text(S)text(ince)\ 1^@ = 4\ text(min time difference,)`
`text(Time difference)`
`= 90 xx 4`
`= 360\ text(mins)`
`= 6\ text(hours … as required)`
c. | `text(Arrival time)` | `= 11:30 + 15\ text(hours)` |
`= 14:30\ text{(Melbourne time)}` |
`:.\ text(Arrival time in Dubai time)`
`= 14:30 – 6\ text(hours)`
`= 8:30\ text(am on Saturday)`
This diagram represents Earth. `O` is at the centre, and `A` and `B` are points on the surface.
Find the shortest great circle distance from `A` to `B`.
Give your answer in to the nearest km. (2 marks)
`text{4803 km (nearest km)}`
An aircraft travels at an average speed of 913 km/h. It departs from a town in Kenya (0°, 38°E) on Tuesday at 10 pm and flies east to a town in Borneo (0°, 113°E).
a. | `text(Angular difference in longitude)` |
`= 113 – 38`
`= 75^@`
`text(Arc length)` | `= 75/360 xx 2 xx pi xx 6400` |
`= 8377.58…` | |
`= 8378\ text(km)\ text{(nearest km)}` |
`:.\ text(The distance between the two towns is 8378 km.)`
b. | `text(Flight time)` | `= text(Distance)/text(Speed)` |
`= 8378/913` | ||
`= 9.176…` | ||
`= 9\ text(hours)\ text{(nearest hr)}` |
c. | `text(Time Difference)` | `= 75 xx 4` |
`= 300\ text(minutes)` | ||
`= 5\ text(hours)` |
`text(S)text(ince Kenya is further East,)`
`=>\ text(Kenya is +5 hours)`
`:.\ text(Arrival time in Kenya)`
`= text{10 pm (Tues) + 5 hrs + 9 hrs}\ text{(flight)}`
`= 12\ text(midday on Wednesday)`
The area under the curve `y = sin (x)` between `x = 0` and `x = pi/2` is approximated by two rectangles as shown.
This approximation to the area is
A. `1`
B. `pi/2`
C. `((sqrt 3 + 1) pi)/12`
D. `0.5`
E. `((sqrt 3 + 1) pi)/6`
`C`
`text(Rectangle width) = pi/6`
`text(Area)` | `~~ pi/6 [sin (pi/6) + sin (pi/3)]` |
`~~pi/6(sqrt3/2 + 1/2)` | |
`~~ (pi (sqrt 3 + 1))/12` |
`=> C`
A part of the graph of `f: R -> R, f(x) = x^2` is shown below. Zoe finds the approximate area of the shaded region by drawing rectangles as shown in the second diagram.
Zoe's approximation is `ptext(%)` more than the exact value of the area.
The value of `p` is closest to
A. `10`
B. `15`
C. `20`
D. `25`
E. `30`
`=> D`
`A_text(exact)` | `= int_0^6 x^2 dx` |
`= 72` |
`A_text(approx)` |
`= 1[f(1) + f(2) + f(3) + …` `+ f(4) + f(5) + f(6)]` |
`= 91` |
`p text(%)` | `= text(increase)/text(exact area) xx 100` |
`= (91 – 72)/72 xx 100` | |
`= 26.4 text(%)` |
`=> D`
A company produces two types of hockey stick, the ‘Flick’ and the ‘Jink’.
Let `x` be the number of Flick hockey sticks that are produced each month.
Let `y` be the number of Jink hockey sticks that are produced each month.
Each month, up to 500 hockey sticks in total can be produced.
The inequalities below represent constraints on the number of each hockey stick that can be produced each month.
Constraint 1 | `x >= 0` | Constraint 2 | `y >= 0` |
Constraint 3 | `x + y <= 500` | Constraint 4 | `y <= 2x` |
There is another constraint, Constraint 5, on the number of each hockey stick that can be produced each month.
Constraint 5 is bounded by Line `A`, shown on the graph below.
The shaded region of the graph contains the points that satisfy constraints 1 to 5.
The profit, `P`, that the company makes from the sale of the hockey sticks is given by
`P = 62x + 86y`
The profit made on the Flick hockey sticks is `m` dollars per hockey stick.
The profit made on the Jink hockey sticks is `n` dollars per hockey stick.
The maximum profit of $42 000 is made by selling 400 Flick hockey sticks and 100 Jink hockey sticks.
What are the values of `m` and `n`? (2 marks)
a. `text(Constraint 4 means the number of Jink sticks)`
`text(produced each month is less than or equal to)`
`text(twice the number of flick sticks produced each)`
`text(month.)`
b. `y <= 300`
c. `text(From the equation, 1 Jink stick produces a)`
`text(higher profit than 1 Flick stick.)`
`text(Maximum profit at)\ (200,300)`
`P` | `= (62 xx 200) + (86 xx 300)` |
`= $38\ 200` |
d. `Q = mx + ny`
`text(Max profit at)\ (400,100)`
`(400,100)\ text(lies on)\ x + y = 500`
`=>\ text(Max profit equation has the same)`
`text(gradient as the profit line.)`
`=> m = −1, m = n`
`text(Using the maximum profit = $42 000,)`
`400m + 100n` | `= 42\ 000` |
`500m` | `= 42\ 000` |
`m` | `= (42\ 000)/500` |
`= 84` |
`:. m = n = 84`
The bonus money is provided by a company that manufactures and sells hockey balls.
The cost, in dollars, of manufacturing a certain number of balls can be found using the equation
cost = 1200 + 1.5 × number of balls
Find the selling price of one hockey ball. (1 mark)
Maria is a hockey player. She is paid a bonus that depends on the number of goals that she scores in a season.
The graph below shows the value of Maria’s bonus against the number of goals that she scores in a season.
Another player, Bianca, is paid a bonus of $125 for every goal that she scores in a season.
How many goals did Maria and Bianca each score in the season? (1 mark)
A golf course has a sprinkler system that waters the grass in the shape of a sector, as shown in the diagram below.
A sprinkler is positioned at point `S` and can turn through an angle of 100°.
The shaded area on the diagram shows the area of grass that is watered by the sprinkler.
Round your answer to the nearest metre. (1 mark)
This sprinkler will water a section of grass as shown in the diagram below.
The section of grass that is watered is 4.5 m wide at all points.
Water can reach a maximum of 12 m from the sprinkler at `L`.
What is the area of grass that this sprinkler will water?
Round your answer to the nearest square metre. (2 marks)
a. | `text(Area of sector)` | `= theta/360 xx pi xx r^2` |
`147.5` | `= 100/360 xx pi xx d^2` | |
`d^2` | `= 147.5/pi xx 360/100` | |
`= 169.02…` | ||
`:.d` | `= 13.00…` | |
`= 13\ text{m (nearest m)}` |
b. | `text(Area)` | `= ((360 – theta))/360 xx pi xx R^2 – ((360 – theta))/360 xx pi xx r^2` |
`= 260/360 xx pi (12^2 – 7.5^2)` | ||
`= 199.09…` | ||
`= 199\ text{m² (nearest m²)}` |
During a game of golf, Salena hits a ball twice, from `P` to `Q` and then from `Q` to `R`.
The path of the ball after each hit is shown in the diagram below.
After Salena’s first hit, the ball travelled 80 m on a bearing of 130° from point `P` to point `Q`.
After Salena’s second hit, the ball travelled 100 m on a bearing of 054° from point `Q` to point `R`.
Use the cosine rule to find the distance travelled by this ball.
Round your answer to the nearest metre. (2 marks)
Round your answer to the nearest degree. (1 mark)
a. |
`PR^2` | `= PQ^2 + QR^2 – 2 xx PQ xx QR xx cos104^@` |
`= 80^2 + 100^2 – 2 xx 80 xx 100 xx cos104^@` | |
`= 20\ 270.75…` | |
`= 142.37…` | |
`= 142\ text{m (nearest m)}` |
b. `text(Find)\ angle RPQ.`
`text(Using the sin rule),`
`(sin angleRPQ)/100` | `= (sin104^@)/142` |
`sin angle RPQ` | `= (100 xx sin104^@)/142` |
`= 0.683…` | |
`angle RPQ` | `= 43.1^@\ \ (text(1 d.p.))` |
`:. text(Bearing of)\ R\ text(from)\ P`
`= 130 – 43.1`
`= 086.9`
`= 087^@\ \ (text(nearest degree))`
A golf tournament is played in St Andrews, Scotland, at location 56° N, 3° W.
Find the shortest great circle distance to the equator from St Andrews.
Round your answer to the nearest kilometre. (1 mark)
Many people in Melbourne will watch the tournament live on television.
Assume that the time difference between Melbourne (38° S, 145° E) and St Andrews (56° N, 3° W) is 10 hours.
On what day and at what time will the tournament begin in Melbourne? (1 mark)
a. |
`text(Great circle distance to equator)`
`= theta/360 xx 2pir`
`= 56/360 xx 2 xx pi xx 6400`
`= 6255.26…`
`= 6255\ text{km (nearest km)}`
b. `text(Melbourne is East of St Andrews)`
`=>\ text(Melbourne is 10 hours ahead.)`
`:.\ text(Time in Melbourne)`
`= 6:32\ text(am + 10 hours)`
`= 4:32\ text(pm on Thursday)`
Salena practises golf at a driving range by hitting golf balls from point `T`.
The first ball that Salena hits travels directly north, landing at point `A`.
The second ball that Salena hits travels 50 m on a bearing of 030°, landing at point `B`.
The diagram below shows the positions of the two balls after they have landed.
a. `text(Let)\ \ d\ text(= distance apart)`
`sin30^@` | `= d/50` |
`:. d` | `= 50 xx sin 30^@` |
`= 25\ text(m)` |
b. `text(Let)\ \ x^@\ text(= angle of elevation from)\ T`
`tanx` | `= 16.8/200` |
`= 0.084` | |
`:. x` | `= 4.801…` |
`= 5^@\ text{(nearest degree)}` |
A golf ball is spherical in shape and has a radius of 21.4 mm, as shown in the diagram below.
Assume that the surface of the golf ball is smooth.
a. | `text(Surface Area)` | `= 4pir^2` |
`= 4 xx pi xx 21.4^2` | ||
`= 5754.89…` | ||
`= 5755\ text(mm²)` |
b. | `text(Minimum length)` | `= 21.4 xx 10` |
`= 214\ text(mm)` |
Megan walks from her house to a shop that is 800 m away.
The equation for the relationship between the distance, in metres, that Megan is from her house `t` minutes after leaving is
`text(distance) = {(100t, 0 <= t <= 6),(\ 600, 6 < t <= a),(quadkt, a < t <= 10):}`
If Megan reaches the shop 10 minutes after leaving her house, the value of `a` is
`B`
`text(When)\ t = 10, d = 800`
`800` | `= 10k` |
`k` | `= 80` |
`text(From the equations and using)\ \ k=80,`
`80a` | `= 600` |
`:. a` | `= 600/80` |
`= 7.5` |
`=> B`
Simon grows cucumbers and zucchinis.
Let `x` be the number of cucumbers that are grown.
Let `y` be the number of zucchinis that are grown.
For every two cucumbers that are grown, Simon grows at least three zucchinis.
An inequality that represents this situation is
`E`
`text(Inequality can be restated:)`
`text(For every one cucumber grown, Simon)`
`text(grows at least)\ 3/2\ text(zucchinis.)`
`:. y >= (3x)/2`
`=> E`
The point (2, 12) lies on the graph of `y = kx^n`, as shown below.
Another graph that represents this relationship between `y` and `x` could be
`E`
`text(Consider each option,)`
`text(Option)\ A:\ text(graph between)\ x\ text(and)\ y\ text(cannot be linear.)`
`text(Option)\ B:\ y = 6x^2`
`text(Test)\ (2,12):\ 6 xx 2^2 = 24 != 12`
`text(Option)\ C:\ y = 3/2 x^2`
`text(Test)\ (2,12):\ 3/2 xx 2^2 = 6 != 12`
`text(Option)\ D:\ y = 6x^3`
`text(Test)\ (2,12):\ 6 xx 2^3 = 48 != 12`
`text(Option)\ E:\ y = 3/2 x^3`
`text(Test)\ (2,12):\ 3/2 xx 2^3 = 12`
`=> E`
The feasible region for a linear programming problem is shaded in the diagram below.
The equation of the objective function for this problem is of the form
`P = ax + by`, where `a > 0` and `b > 0`
The dotted line in the diagram has the same slope as the objective function for this problem.
The maximum value of the objective function can be determined by calculating its value at
`C`
`text(By applying the sliding rule technique)`
`text{(moving the objective function out in}`
`text{a series of parallel lines), the maximum}`
`text(value occurs at point)\ C.`
`=> C`
The point (3, –2) satisfies the inequality
`D`
`text(Consider option)\ D,`
`y` | `< 5 – x` |
`−2` | `< 5 – 3` |
`−2` | `< 2\ \ \ text{(correct)}` |
`=> D`
The graph below shows a straight line that passes through the points (6, 6) and (8, 9).
The coordinates of the point where the line crosses the `x`-axis are
`D`
`m` | `= (y_2 – y_1)/(x_2 – x_1)` |
`= (9 – 6)/(8 – 6)` | |
`= 3/2` |
`text(Using the point gradient formula,)`
`y – y_1` | `= m(x – x_1)` |
`y – 6` | `= 3/2(x – 6)` |
`y` | `= 3/2 x – 3` |
`text(When)\ y = 0,`
`0` | `=3/2 x – 3` |
`x` | `= 2` |
`:.\ text(Crosses)\ xtext{-axis at (2,0)}`
`=> D`
A phone company charges a fixed, monthly line rental fee of $28 and $0.25 per call.
Let `n` be the number of calls that are made in a month.
Let `C` be the monthly phone bill, in dollars.
The equation for the relationship between the monthly phone bill, in dollars, and the number of calls is
`A`
`C = 28 + 0.25n`
`=> A`
The graph below shows the temperature, in degrees Celsius, between 6 am and 6 pm on a given day.
From 2 pm to 6 pm, the temperature decreased by
`B`
`text(Temperature)` | `= 23 – 12` |
`= 11°text(C)` |
`=> B`
A string of seven flags consisting of equilateral triangles in two sizes is hanging at the end of a racetrack, as shown in the diagram below.
The edge length of each black flag is twice the edge length of each white flag.
For this string of seven flags, the total area of the black flags would be
`D`
`text(Shapes are similar.)`
`text(Scale factor of sides = 2)`
`text(Scale factor of areas) = 2^2 = 4`
`text(S)text(ince there are 4 black and 3)`
`text(white flags,)`
`:.\ text{Total area of black flags (in white flags)}`
`= 4/3 xx 4`
`= 16/3 xx text(Area of white flags)`
`=> D`
The diagram below shows a rectangular-based right pyramid, ABCDE.
In this pyramid, AB = DC = 24 cm, AD = BC = 10 cm and AE = BE = CE = DE = 28 cm.
The height, OE, of the pyramid, in centimetres, is closest to
`C`
`text(Find)\ AC,\ text(using Pythagoras,)`
`AC^2` | `= 24^2 + 10^2` |
`= 676` | |
`AC` | `= 26` |
`text(Consider)\ DeltaAEO,`
`AO` | `= 1/2AC` |
`= 13` |
`text(Using Pythagoras,)`
`OE^2 + AO^2` | `= AE^2` |
`OE^2` | `= 28^2 – 13^2` |
`= 615` | |
`:. OE` | `= 24.79…` |
`=> C`
Marcus is on the opposite side of a large lake from a horse and its stable. The stable is 150 m directly east of the horse. Marcus is on a bearing of 170° from the horse and on a bearing of 205° from the stable.
The straight-line distance, in metres, between Marcus and the horse is closest to
`E`
A water tank in the shape of a cylinder with a hemispherical top is shown below.
The volume of water that this tank can hold, in cubic metres, is closest to
`A`
`V` | `=\ text(cylinder + half sphere)` |
`= pir^2h + 1/2 xx 4/3 xx pir^3` | |
`= pi xx 2^2 xx 5 + 1/2 xx 4/3 xx pi xx 2^3` | |
`= 79.587…` | |
`~~ 80\ text(m³)` |
`=> A`
All towns in the state of Victoria are in the same time zone.
Mallacoota (38°S, 150°E) and Portland (38°S, 142°E) are two coastal towns in the state of Victoria.
On one day in January, the sun rose in Mallacoota at 6.03 am.
Assuming that 15° of longitude equates to a one-hour time difference, the time that the sun was expected to rise in Portland is
`E`
`text(Angular difference)` | `= 150 – 142` |
`= 8^@` |
`text(Time difference)` | `= 8/15 xx 60` |
`= 32\ text(minutes)` |
`text(S)text(ince Portland is West of Mallacoota,)`
`text(it’s sunrise will be later.)`
`:.\ text(Portland)` | `= 6:03 + 32\ text(minutes)` |
`= 6:35\ text(am)` |
`=> E`
The diagram above shows the position of three cities, A, B and C.
According to the diagram
`E`
`A\ text(and)\ C\ text(are on the same line)`
`text(of longitude.)`
`text(All other statements can be shown)`
`text(to be false.)`
`=> E`
A continuous random variable, `X`, has a probability density function given by
`f(x) = {{:(1/5e^(−x/5),x >= 0),(0, x < 0):}`
The median of `X` is `m`.
Find `text(Pr)(X < 1 | X <= m)`. (2 marks)
a. | `1/5 int_0^m e^(−x/5)dx` | `= 1/2` |
`1/5 xx (−5)[e^(−x/5)]_0^m` | `= 1/2` | |
`[-e^(- x/5)]_0^m` | `= 1/2` | |
`-e^(−m/5) + 1` | `= 1/2` | |
`e^(−m/5)` | `= 1/2` | |
`- m/5` | `= log_e(1/2)` |
`:. m = −5log_e(1/2)\ \ \ (text(or)\ \ 5log_e(2),\ text(or)\ \ log_e 32)`
b. `text(Using Conditional Probability:)`
`text(Pr)(X < 1 | X <= m)` | `= (text(Pr)(X < 1))/(text(Pr)(X <= m))` |
`= (1/5 int_0^1 e^(−x/5)dx)/(1/2)` | |
`= (1/5(−5)[e^(−x/5)]_0^1)/(1/2)` | |
`= −2[(e^(−1/5)) – e^0]` | |
`= 2(1 – e^(−1/5))` |
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.
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` |
Tiffany’s pulse rate (in beats/minute) during the first 60 minutes of a long-distance run is shown in the graph below.
Write your answer in beats/minute. (1 mark)
The target zone for aerobic exercise is between 60% and 75% of a person’s maximum pulse rate.
Tiffany is 20 years of age.
Write your answers correct to the nearest whole number. (1 mark)
a. `text(110 beats/min)`
b. `text(Increase of pulse rate)`
`=\ text(Rate at 60 min − initial rate)`
`= 150 – 70`
`= 80 \ text(beats/min)`
c.i. `text(Maximum pulse rate = 220 − age)`
c.ii. `text(Tiffany’s maximum pulse rate)`
`= 200 – 20`
`= 200\ text(beats/min)`
`=>\ text(Lower range) = 60text(%) xx 200 = 120`
`=>\ text(Higher range) = 75text(%) xx 200 = 150`
`:.\ text(Target range is 120 − 150 beats/min.)`
Cheapstar Airlines wishes to find the optimum number of flights per day on two of its most popular routes: Alberton to Bisley and Alberton to Crofton.
Let `x` be the number of flights per day from Alberton to Bisley
`y` be the number of flights per day from Alberton to Crofton
Table 4 shows the constraints on the number of flights per day and the number of crew per flight.
The lines `x + y = 10` and `3x + 5y = 41` are graphed below.
A profit of $1300 is made on each flight from Alberton to Bisley and a profit of $2100 is made on each flight from Alberton to Crofton.
Determine the maximum total profit that Cheapstar Airlines can make per day from these flights. (2 marks)
`$17\ 300`
`P = 1300x + 2100y`
`text(Within the feasible area, profit is)`
`text(maximised at)\ (2,7).`
`:.\ text(Maximum profit)`
`= 1300 xx 2 + 2100 xx 7`
`= $17\ 300`
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` |
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` |