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GRAPHS, FUR1 2002 VCAA 9 MC

A company uses both trainee and qualified inspectors for quality control on an assembly line. The trainee inspectors can examine  `15`  items per hour, while the qualified inspectors can examine  `25`  items per hour.
The company needs at least  `1800`  items inspected in each eight-hour day.
Let  `x`  be the number of trainee inspectors employed by the company, and  `y`  be the number of qualified inspectors employed.
The constraint imposed by the number of items that need to be inspected in an eight-hour day is expressed by the inequality

A.           `8x + 8y ≥ 1800`

B.       `15x + 25y ≤ 1800`

C.       `15x + 25y ≥ 1800`

D.   `120x + 200y ≥ 1800`

E.   `120x + 200y ≤ 1800`

 


Show Answers Only

`D`

Show Worked Solution

`x = text(no. trainees)`

`y = text(no. qualified)`

`text(1 trainee inspects 15 items/hr)`

`text(1 qualified inspector inspects 25 items/hr)`

`text(Hence in 8 hrs)`

`text(No. items trainee inspects)`

`= 15x xx 8text(hrs)`

`= 120x`

`text(No. items qualified inspector inspects)`

`= 25y xx 8text(hrs)`

`= 200y`

`text(At least 1800 items inspected in 8 hrs.)`

`:.\ 120x + 200y ≥ 1800`

`=>  D`

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2002 VCAA 7 MC

The relation  `y = 2x^2`, for  `x > 0`, is represented by

2002VCAA-Graphs-7ab 2002VCAA-Graphs-7cd

2002VCAA-Graphs-7e

 


Show Answers Only

`B`

Show Worked Solution

`y = 2x^2 \ \ \ … \ text{(i)}`

`text(Consider)\ A`

`text(Sub)\ x^2 = 3\ text{into (i)}`

`y` `= 2(3)`
  `= 2 xx 3`
  `= 6`

`text(Does not fit point given in)\ A.`

`:.\ text(Eliminate)\ A`

`text(Consider)\ B`

`text(Sub)\ x^2 = 9\ text{into (i)}`

`y` `= 2(9)`
  `= 18`

`text(Fits point given in)\ B.`

`text(Similarly)\ C, D, E\ text(can be ruled out.)`

`=>  B`

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2002 VCAA 6 MC

For linear programming problems in general, which of the following statements is false?

A.   The objective function is a linear expression.

B.   The expressions defining the constraints are linear inequalities.

C.   There may be more than one optimal solution.

D.   The value of the objective function must be positive.

E.   The constraints are used to define the feasible region.

 


Show Answers Only

`D`

Show Worked Solution

`text(Consider)\ A`

`text(True, as objective function has to be)`

`text(linear for linear programming problems.)`

`:.\ text(Eliminate)\ A`

`text(Consider)\ B`

`text(True, as linear inequalities help define)`

`text(the region or constraints on the graph.)`

`:.\ text(Eliminate)\ B`

`text(Consider)\ C`

`text(True, as any point in the defined region)`

`text(can be an optimal solution.)`

`:.\ text(Eliminate)\ C`

`text(Consider)\ D`

`text(False, as depending on the point given)`

`text(that fits the constraints, the objective)`

`text(function may turn out to be negative)`

`text(when this point is substituted into the)`

`text(objective function.)`

`text(Similarly)\ E\ text(can be ruled out.)`

`=>  D`

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2002 VCAA 5 MC

The feasible region for a particular linear programming problem is shown shaded on the following graph. All relevant vertices are labelled.

2002VCAA-Graphs-5

The minimum value of the expression  `4x -2y`  for this feasible region is

A.   –20

B.   –10

C.       0

D.     22

E.     44

 


Show Answers Only

`A`

Show Worked Solution

`text(Boundary points:)`

`(0, 10), (2, 9), (8, 5), (11, 0), (0, 0)`

`text(Sub each boundary point into)\ 4x − 2y`

`text(Consider)\ (0, 10)`

`4(0) − 2(10) = text(–20)`

`text(This is the minimum value of all the`

`text(choices given in the answers.)`

`text((Similarly all other values of)\ 4x − 2y`

`text(can be calculated.))`

`=>  A`

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2002 VCAA 4 MC

For the pair of simultaneous equations

`4x = 7 - y`

`5x + 7y = 3`

the solution is

A.   `x = –2, y = –1`

B.   `x = –2, y = 1`

C.   `x = 1,   y = 3`

D.   `x = –1, y = 2`

E.   `x = 2,   y = –1`

 


Show Answers Only

`E`

Show Worked Solution
`4x` `= 7 − y`
`4x + y` `= 7`
`7(4x + y)` `= 7(7)`
`28x + 7y`  `= 49 \ \ \ …\ text{(i)}`
`5x + 7y“= 3 \ \ \ \ …\ text{(ii)}`

`text{(i) − (ii)}`

`23x` `= 46`
`x` `= 2`

`text(Sub)\ x = 2\ text{into (ii)}`

`5(2) + 7y` `= 3`
`10 + 7y` `= 3`
`7y` `= text(−7)`
`y` `= text(−1)`

`:.\ x = 2, y = text(−1)`

`=>  E`

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2002 VCAA 3 MC

 

The following graph shows the number of cattle in a particular state during the 1990s.

2002VCAA-Graphs-3

Of the years listed below, in which year did the number of cattle in the state remain relatively unchanged?

A.   1991

B.   1992

C.   1994

D.   1996

E.   1997

 


Show Answers Only

`E`

Show Worked Solution

`text(From an unchanged year, line must)`

`text(be horizontal.)`

`text(Most horizontal year = 1997)`

`=>  E`

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2002 VCAA 2 MC

A courier company’s charges are based on the distance required to deliver a parcel. Within a radius of  `40`  km, charges can be determined from a graph like that shown below.

2002VCAA-Graphs-2

Which one of the following rules could be used to describe this graph?

A.   `C ={(10 text( for ) \ \ 0 < x < 10),(20 text( for ) 10 < x < 20),(40 text( for ) 20 < x < 40) :}`

B.   `C ={(10 text( for ) \ \ 0 ≤ x ≤ 10),(20 text( for ) 10≤ x ≤ 20),(40 text( for ) 20 ≤ x ≤ 40) :}`

C.   `C ={(10 text( for ) \ \ 0 < x ≤ 10),(20 text( for ) 10 < x ≤ 20),(40 text( for ) 20 < x ≤ 40) :}`

D.   `C ={(10 text( for ) \ \ 0 ≤ x < 10),(20 text( for ) 10≤ x < 20),(40 text( for ) 20 ≤ x < 40) :}`

E.   `C ={(10 text( for ) \ \ 0 < x < 10),(20 text( for ) 10≤ x ≤ 20),(40 text( for ) 20 < x < 40) :}`

 


Show Answers Only

`C`

Show Worked Solution

`text(From the graph)`

`text(Open circle means not inclusive)`

`text(Closed circle means inclusive)`

`:.\ ` `10\ \ text(for)\ \ \ \ 0 < x ≤ 10`
  `20\ \ text(for)\ \ 10 < x ≤ 20`
  `40\ \ text(for)\ \ 20 < x ≤ 40`

`=>  C`

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2002 VCAA 1 MC

2002VCAA-Graphs-1

The equation of this straight line is

A.             `x = 7`

B.              `y = 5`

C.   `7x + 5y = 35`

D.   `5x + 7y = 35`

E.   `7x - 5y = 35`

 


Show Answers Only

`D`

Show Worked Solution

`text(Using two point form)`

`text{Sub (0, 5) and (7, 0) into }`

`(y − y_1)/(x − x_1)` `= (y_2 − y_1)/(x_2 − x_1)`
`(y − (5))/(x − (0))` `= (0 − (5))/(7 − (0))`
`(y − 5)/x` `= text(−5)/7`
`7y − 35`  `= text(−5)x`
`5x + 7y` `= 35`

`=>  D`

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2004 VCAA 9 MC

2004VCAA-Graph-9

The graph above represents a relationship  `y = kx^3`  for  `x ≥ 0`.
A graph that shows this relationship when  `y`  is plotted against  `x`  is

2004VCAA-Graph-9ab

2004VCAA-Graph-9cd

2004VCAA-Graph-9e

 


Show Answers Only

`A`

Show Worked Solution

`y = kx^3\ text(for)\ x ≥ 0`

`text(From the graph,)\ (8, 1)\ text(fits this line)`

`:.\ y = 1\ text(when)\ x^3 = 8`

`text(Sub)\ y = 1, x^3 = 8\ text(into)\ y = kx^3`

`1` `= k(8)`
`k` `= 1/8`

`:.\ y = 1/8 x^3\ \ \ … \ text{(i)}`

`text(Consider)\ A`

`text{ Sub (2, 1) into (i)}`

`text(LHS)` `= 1`
`text(RHS)` `= 1/8 (2)^3`
  `= 1/8 (8)`
  `= 1`

`:.\ text(LHS = RHS)`

`text(Similarly)\ B, C, D, E\ text(can be eliminated)`

`=>  A`

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2004 VCAA 8 MC

2004VCAA-Graph-8

The shaded region shown in the graph above (with boundaries included) is described by

A. `3x + 4y ≤ 12`
  `x -y ≤ 1`
  `x ≥ 0`
  `y ≥ 0`
B. `3x + 4y ≤ 12`
  `x -y ≥ 1`
  `x ≥ 0`
  `y ≥ 0`
C. `3x + 4y ≥ 12`
  `x -y ≥ 1`
  `x ≥ 0`
  `y ≥ 0`
D. `4x + 3y ≤ 12`
  `x -y ≤ 1`
  `x ≥ 0`
  `y ≥ 0`
E. `4x + 3y ≤ 12`
  `x -y ≥ 1`
  `x ≥ 0`
  `y ≥ 0`

 


Show Answers Only

`B`

Show Worked Solution

`text(From the graph)`

`text(Horizontal boundary:)\ y >= 0`

`text(For)\ 1^text(st)\ text(line to the left)`

`text(Line passes)\ (0, text(−1))\  text(and)\ (1, 0)`

`text(Using )` `(y − y_1)/(x − x_1)` `= (y_2 − y_1)/(x_2 − x_1)`
  `(y − (text(−1)))/(x − (0))` `= ((0) − (text(−1)))/((1) − (0))`
  `(y + 1)/x` `= 1/1`
  `y + 1` `= x`
  `x − y` `= 1`

`text(From the graph,)\ x − y >= 1`

`text(For)\ 2^text(nd)\ text(line to the right)`

`text{Line passes (4, 0) and (0, 3)}`

`text(Using )` `(y − y_1)/(x − x_1)` `= (y_2 − y_1)/(x_2 − x_1)`
  `(y − (0))/(x − (4))` `= ((3) − (0))/((0) − (4))`
  `y/(x − 4)` `= 3/text(−4)`
  `text(−4)y` `= 3x − 12`
  `text(−3)x − 4y` `= text(−12)`
  `3x + 4y` `= 12`

`text(From the graph)`

`3x + 4y <= 12`

`=>  B`

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2004 VCAA 7 MC

2004VCAA-Graph-7

The shaded region shown in the graph above (with boundaries included) represents the feasible region for a linear programming problem.
The maximum value of the objective function  `y - 2x + 20`, for this feasible region, is

A.   18

B.   23

C.   25

D.   27

E.   33

 


Show Answers Only

`C`

Show Worked Solution

`text(Corner pts of feasible region)`

`(2, 9), (4, 11), (6, 10), (6, 1)`

`text(Let value) = V`

`V = y − 2x + 20 \ \ \ … \ text{(i)}`

`text(Check each corner pt)`

`text(Sub)\ (2, 9)\ text{into (i)}`

`V` `= 9 − 2(2) + 20`
  `= 9 − 4 + 20`
  `= 25`

`text(Sub)\ (4, 11)\ text{into (i)}`

`V` `= 11 − 2(4) + 20`
  `= 11 − 8 + 20`
  `= 23`

`text(Sub)\ (6, 10)\ text{into (i)}`

`V` `= 10 − 2(6) + 20`
  `= 10 − 12 + 20`
  `= 18`

`text(Sub)\ (6, 1)\ text{into (i)}`

`V` `= 1 − 2(6) + 20`
  `= 1 − 12 + 20`
  `= 9`

`:.\ text{Max value when (2,9) at 25.}`

`=>  C`

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2004 VCAA 6 MC

The cost,  `$C`, of hiring a boat for  `x`  hours is given by the equation  `C = ax + b`  where  `a`  is the hourly rate and  `b`  is a fixed booking fee.
When the boat is hired for  `4`  hours the cost is  `$320`.
When the boat is hired for  `6`  hours the cost is  `$450`.
When the boat is hired for one hour the cost is

A.     $65

B.     $75

C.     $77

D.     $80

E.   $125

 


Show Answers Only

`E`

Show Worked Solution

`C = ax + b\ \ \ …\ text{(i)}`

`a = text(hourly rate)`

`b = text(fixed booking fee)`

`text(Sub)\ x = 4, c = 320\ text{into (i)}`

`320` `= a(4) + b`
`4a + b` `= 320 \ \ \ … \ text{(ii)}`

`text(Sub)\ x = 6, c = 450\ text{into (ii)}`

`450 ` `= a(6) + b`
`6a + b` `= 450 \ \ \ … \ text{(iii)}`

`text{(iii) − (ii)}`

`2a` `= 130`
`:.\ a` `= 65`

`text(Sub)\ a = 65\ text{into (ii)}`

`4(65) + b` `= 320`
`b` `= 320 − 260`
`b` `= 60`

`:.\ c = 65x + 60 \ \ \ …\ text{(iv)}`

`text(Sub)\ x = 1\ text{into (iv)}`

`c` `= 65(1) + 60`
  `= 125`

`:.\ text(C)text(ost is $125 for 1 hr)`

`=>  E`

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2004 VCAA 4-5 MC

The graph shows a distance-time graph for a car travelling from home along a long straight road over a  `16`-hour period.

2004VCAA-Graph-4

Part 1

In which one of the time intervals is the speed of the car greatest?

A.   0 to 5 hours

B.   5 to 9 hours

C.   9 to 12 hours

D.   12 to 14 hours

E.   14 to 16 hours

 

Part 2

After twelve hours the car has travelled a total distance of

A.   100 km

B.   350 km

C.   450 km

D.   600 km

E.   700 km

 


Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ D`

Show Worked Solution

`text(Part 1)`

`text(From the graph, speed is seen by slope)`

`text(of line.)`

`text(In)\ 1^text(st)\ text(part of line,)`

`text(gradient)` `= 350/5`
  `= 70\ text(km/hr)`

`text(In)\ 2^text(nd)\ text(and)\ 4^text(th)\ text(part of line)`

`text(gradient = 0 km/hr)`

`text(In)\ 3^text(rd)\ text(part of line,)`

`text(gradient)` `= (350 − 100)/(12 − 9)`
  `= 250/3`
  `= 83.33 …\ text(km/hr)`

`text(In)\ 4^text(th)\ text(part of line,)`

`text(gradient)` `= 100/(16 − 14)`
  `= 100/2`
  `= 50\ text(km/hr)`
`:.\ ` `3^text(rd)\ text(part of line, which is)`
  `text(9 to 12 hours has greatest speed.)`

`=>  C`

 

`text(Part 2)`

`text(From the graph,)`

`text(Distance)` `= 350 + (350 − 100)`
  `= 350 + 250`
  `= 600\ text(km)`

`=>  D` 

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2004 VCAA 3 MC

The lines  `y + 8 = 0`  and `x - 12 = 0`  intersect at the point

A.   (–12, 8)

B.   (–8, 12)

C.   (0, 0)

D.   (8, –12)

E.   (12, –8)

 


Show Answers Only

`E`

Show Worked Solution
`y + 8` `= 0`
 `y` `= −8\ \ \ …\ text{(i)}`
`x − 12` `= 0`
`x` `= 12\ \ \ …\ text{(ii)}`

`text{From (i) and (ii), lines intersect at}`

`(12, −8)`

`=>  E`

 

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2004 VCAA 2 MC

The point  `(2, 1)`  lies on the line  `y = 3x + c`.
The value of  `c`  is

A.   −7

B.   −5

C.   −1

D.     5

E.     7

 


Show Answers Only

`B`

Show Worked Solution

`text(Sub)\ (2, 1)\ text(into)\ y = 3x + c`

`(1)` `= 3(2) + c`
 `1` `= 6 + c`
`c` `= text(−5)`

 `=>  B`

 

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2004 VCAA 1 MC

The graph below shows the cost (dollars) of mobile telephone calls up to  `240`  seconds long.

2004VCAA-Graph-1

The cost of making a  `90`-second call followed by a  `30`-second call is

A.   $1.00

B.   $1.20

C.   $1.25

D.   $1.50

E.   $1.75

 


Show Answers Only

`D`

Show Worked Solution

`text(From the graph)`

`90\ text(second call)` `= $1`
 `30\ text(second call)` `= $0.50`
`:.\ $1 + 0.5` `= $1.5`

 `=>  D`

 

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2005 VCAA 9 MC

The relationship between the intensity of sound, and the distance from the source of the sound, is  `I = k/d^2`, where  `I`  is the intensity of sound measured in watts per square metre (W/m2),  `d`  is the distance in metres from the source, and  `k`  is a constant.

The intensity of sound is  `20`  W/m2 at a distance of  `50`  m from the source.

A graph showing the relationship between intensity of sound and distance from the source is

2005VCAA-Graphs-9ab

2005VCAA-Graphs-9cd

2005VCAA-Graphs-9e

 


Show Answers Only

`A`

Show Worked Solution

`I = k/d^2 \ \ \ …\ text{(i)}`

`text(Sub)\ I = 20, d = 50\ text{into (i)}`

`20` `= k/50^2`
`20` `= k/2500`
`k` `= 20 xx 2500`
  `= 50000`
`:.\ I` `= 50000/d^2`

`text(Eliminate)\ E, D\ text(as graph shows linear)`

`text(relationship.)`

`text(Consider)\ A`

`text(Sub)\ (5, 2000)\ text(into)\ I = 50000/d^2`

`text(LHS)` `= 2000`
`text(RHS)` `= 50000/5^2`
  `= 50000/25`
  `= 2000`
`:.\ text(LHS)` `= text(RHS)`

`text(Similarly)\ B\ text(and)\ C\ text(can be)`

`text(eliminated.)`

`=>  A`

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2005 VCAA 8 MC

2005VCAA-Graphs-8

For the shaded region above (with boundaries included), the value of the objective function  `text(P) = 4x – 3y`  is a
maximum at the point

A.         (0, 0)

B.     (0, 100)

C.   (50, 100)

D.     (90, 60)

E.     (120, 0)

 


Show Answers Only

`E`

Show Worked Solution

`text(Consider)\ A`

`text(Sub)\ (0, 0)\ text(into P) = 4x − 3y`

`text(P)` `= 4(0) − 3(0)`
  `= 0`

`text(Consider)\ B`

`text(Sub)\ (0, 100)\ text(into P) = 4x − 3y`

`text(P)` `= 4(0) − 3(100)`
  `= text(−300)`

`text(Consider)\ C`

`text(Sub)\ (50, 100)\ text(into P) = 4x − 3y`

`text(P)` `= 4(50) − 3(100)`
  `= 200 − 300`
  `= text(−100)`

`text(Consider)\ D`

`text(Sub)\ (90, 60)\ text(into P) = 4x − 3y`

`text(P)` `= 4(90) − 3(60)`
  `= 360 − 180`
  `= 180`

`text(Consider)\ E`

`text(Sub)\ (120, 0)\ text(into P) = 4x − 3y`

`text(P)` `= 4(120) − 3(0)`
  `= 480`

`:.\ E\ text(has max value.)`

`=>  E`

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2005 VCAA 7 MC

2005VCAA-Graphs-7

The relationship between the variables  `a`  and  `b`  as shown in the graph above is

A.   `b = a^2`

B.   `b = 2a^2`

C.   `b = 2a`

D.   `a = 2b`

E.   `a^2 = 2b`

 


Show Answers Only

`B`

Show Worked Solution

`a^2\ text(and)\ b\ text(has a linear relationship)`

`text(from the graph.)`

`:.\ b = ka^2`

`text(Sub)\ (2, 4)\ text(into)\ a^2 = kb`

`text(where)\ a^2 = 2, b = 4`

`4` `= k(2)`
`4` `= 2k`
`k` `= 2`
`:.\ b` `= 2a^2`

 `=>  B`

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2005 VCAA 6 MC

One afternoon at the beach Mr Smith bought four ice creams and three drinks for his family at a cost of  `$21.40`.
Mrs Brown bought five of the same ice creams and two of the same drinks for  `$20.80`.
Based on these prices, the cost of one drink is

A.   $2.80

B.   $2.90

C.   $3.00

D.   $3.30

E.   $3.40

 


Show Answers Only

`E`

Show Worked Solution
`text(Let   I)` `= text(cost for one ice cream)`
`text(D)` `= text(cost for one drink)`

`text(Simultaneous equations)`

`text(4I + 3D)` `= $21.40 \ \ \ …\ text{(i)}`
`text(5I + 2D)` `= $20.80 \ \ \ …\ text{(ii)}`

`text{(i)} xx 2:`

`text(8I + 6D = $ 42.80 )\ \ \ …\ text{(iii)}`

`text{(ii)} xx 3:`

`text(15I + 6D = $62.40 )\ \ \ …\ text{(iv)}`

`text{(iii) − (ii)}`

`text(15I − 8I)` `= 62.40 − 42.80`
`text(7I)` `= 19.6`
`text(I)` `= 2.8`

`text{Sub I = 2.8 into (i)}`

`4(2.8) + text(3D)` `= 21.40`
`11.2 +text(3D)` `= 21.40`
`text(3D)` `= 10.2`
`text(D)` `= 3.4`

`:.\ text(One drink costs $3.40)`

`=>  E`

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2005 VCAA 5 MC

An electrician charges a fixed call-out fee of  `$50`  and then charges  `$65`  per hour for each hour worked.
For  `n`  hours worked, the total charge in dollars is

A.   `115`

B.   `n + 115`

C.   `50n + 65`

D.   `65n + 50`

E.   `115n`

 


Show Answers Only

`D`

Show Worked Solution
`text(Fixed call out fee)` `= $50`
`text(Hourly charge)` `= $65\ text(per hour)`
`:.\ text(Total charge)` `= $50 + $65n`

`(n = text(hours worked))`

`=>  D`

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2005 VCAA 4 MC

2005VCAA-Graphs-4

If the line above has equation  `3x + 2y = 4k`, then the value of  `k`  must be

A.     2

B.     3

C.     6

D.     8

E.   12

 


Show Answers Only

`C`

Show Worked Solution

`3x + 2y = 4k \ \ \ …\ text{(i)}`

`text(From the graph,)\ (0, 12)\ text(and)\ (8, 0)`

`text(passes through this line)`

`text(Sub)\ (0, 12)\ text(into)\ text{(i)}`

`3(0) + 2(12)` `= 4k`
`24` `= 4k`
`k` `= 6`

`=>  C`

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2005 VCAA 3 MC

Which one of the following statements is not true?

A.   The line with equation  `7x - 4y = 0`  passes through the point  `(4, 7)`.

B.   The point  `(3, 5)`  lies in the region defined by  `7x - 4y >= 0`.

C.   The line with equation  `3x + 5y = 0`  has a positive gradient.

D.   The lines  `7x - 4y = 0`  and  `3x + 5y = 0`  meet at the origin.

E.   For the line with the equation  `7x - 4y = 0`,  `y`  increases as  `x`  increases.

 


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`C`

Show Worked Solution

`text(Consider)\ A`

`text(Sub)\ (4, 7)\ text(into)\ 7x − 4y = 0`

`text(LHS)` `= 7(4) − 4(7)`
  `= 28 − 28`
  `= 0`
  `= text(RHS)`

`:. text(True, eliminate)\ A.`

 

`text(Consider)\ B`

`text(Sub)\ (3, 5)\ text(into)\ 7x − 4y >= 0`

`text(LHS)` `= 7(3) − 4(5)`
  `= 21 − 20`
  `= 1`
  `>= 0`

`:. text(True, eliminate)\ B.`

 

`text(Consider)\ C`

`3x + 5y` `= 0`
`5y` `= text(−3)x`
`y` `= −3/5x`
`:.\ text(gradient)` `= −3/5`
  `= text(negative)`

`:.\ text(False)`

`text(Similarly)\ D, E\ text(can be eliminated.)`

`=>  C`

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2005 VCAA 2 MC

Two lines have equations  `y = text(-5)`  and  `y = text(-)x + 5`  respectively.
The point that lies on both of these lines is

A.   (−10, 5)

B.     (−5, 5)

C.     (0, −5)

D.     (5, −5)

E.   (10, −5)

 


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`E`

Show Worked Solution

`y = text(−5) \ \ \ …\ text{(i)}`

`y = text(−x) + 5 \ \ \ …\ text{(ii)}`

`text{(i) − (ii)}`

`0 = text(−5) − (text(−x) + 5)`

`0 = text(−5) + x − 5`

`x = 10`

`:.\ `  `text(From multiple choice given,)`
  `text((10, −5))\ text(is correct.)`

`=>  E`

Filed Under: Geometry FUR1 lost

GRAPHS, FUR1 2005 VCAA 1 MC

The graph below shows the temperature (in degrees Celsius) over a  `24`-hour period.

2005VCAA-Graphs-1

The period of greatest temperature increase was

A.   6 am−9 am

B.   9 am−noon

C.   noon−3 pm

D.   3 pm−6 pm

E.   6 pm−9 pm

 


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`A`

Show Worked Solution

`text(Greatest increase indicated by slope)`

`text(of curve and rising up.)`

`:.\ ` `text(From the graph, 6-9am has)`
  `text(the greatest increase in temperature.)`

`=>  A`

Filed Under: Geometry FUR1 lost

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