Mathematics

Home > Mathematics > General Mathematics > Preliminary Course > Algebraic Modelling > AM2: Modelling linear relationships > Quick quiz: A sample of multiple choice questions for this topic.

AM2

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  1. When applying a function, the procedure follows the idea of Input – Process – Output.

    The cost of staging a preschool concert was calculated using the formula C = 80 + 3n, where C is the cost and n is the number of students involved.

    Which variable is considered to be the input for this scenario?
C, the cost.
n, the number of students.
The answers to each of the calculations.

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  1. The dependent variable in a linear model is w and the independent variable is b. If the gradient of the line representing this model is 2, the equation could be:
b = 2w
w = 2b
w = b + 2

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  1. The linear function representing the cost, C, of an excursion for a number of students, n, was given as C = 4.5n + 50.

    To graph this by first creating a table of values, the table could be:
n
0
1
2
3
C
50
95
140
185

n
0
1
2
3
C
50
54.5
109
163.5

n
0
1
2
3
C
50
54.5
59
63.5

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  1. To find the gradient of the line AB, Rita measured AC as 3 mm and BC as 4 mm.

Rise and Run between 2 points on line AB


The gradient would be:

3/4

4/3

5

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  1. This graph shows the cost of producing boxes of fudge to sell for charity.

graph depicting cost of producing boxes of fudge

The gradient of this graph is 5. This gradient represents:

The cost per box of fudge.
The cost of the box to put the fudge in.
The cost of the fudge for each box.

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  1. This graph shows the cost of producing boxes of fudge to sell for charity.

graph showing cost of producing boxes of fudge.

The graph meets the vertical axis at 13 in an open circle. This value of 13 represents:

The cost of a box with no fudge in it must be $13.
There are 13 pieces of fudge in each box.
It would cost $13 to get started with the production.

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  1. A sketch of the graph y = -2x + 1 would look like:
     

Line y=-2, x=1

Line y=1, x=-2

Line y=1, x=0.5

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  1. This graph shows that the cost, C, of a holiday varies directly with the distance, d, from home.

Graph depicting C=$90. d=$630km


If the cost of travelling 630 km is $90 the equation of this graph

C = 7d

C = 7/d

C = d/7

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  1. This graph shows that the cost, C, of a holiday varies directly with the distance, d, from home. The cost of travelling 630 km is $90.

Graph depicting C=$90, d=630km

For a cost of $840 you could travel:

5880 km from home.
120 km from home.
1200 km from home.

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  1. This is a graph of the distance travelled by Priya on her way to school.

The section AB represents:

Priya went back towards home (maybe she had dropped something).
Priya reached a down-hill section of the road.
Priya stopped for a short period of time.

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  1. A suburban car park has the following parking rates:
Time Cost
Up to 1 hour
Free
More than 1 hour and up to 2 hours
$5
More than 2 hours and up to 3 hours
$9
More than 3 hours
$12


When using this table of data to graph Cost against Time, the type of graph produced is called:

A scatter or dot graph.
A step graph.
A pie or sector graph.

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  1. When representing some physical phenomena as a linear function(such as shoe size against age) there are often limitations to the model produced because:
The graph may be too big to fit on a page.
The graph may be a vertical line.
The graph may only be a straight line for a limited section.

 

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  1. This represents a little pocket graph that a group of year 12 students took with them to Bali after the HSC. It converts the currency in Bali (Rupiah) to Australian dollars.

Line graph depicting currency conversion

This graph uses the conversion rate of 5000 Rupiah to the dollar which is a reasonable approximation.

The students are offered a great deal on an excursion to other islands for 425 000 Rupiah.

What is this approximately equivalent to in Australian dollars?

$90
$85
$850

 

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  1. This is a travel graph representing two cars travelling between Taree and Maitland.



The point where the two lines cross shows where:

The two cars collided.
Car A overtook Car B which was going slower.
Cars A and B pass each other travelling in opposite directions.

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  1. When using a pencil and ruler to sketch a line of best fit for the scatter plot below, you should just use your ruler to get the general trend of the gradient of the line and:

Scatter plot

Draw a line that has about the same number of points on either side of it.
Find a straight line that goes through any four points.
Join the lowest and highest points.

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