1
Construction and interpretation of graphs Solutions
Name __________________________
1
Albert is a salesperson who is paid $300
retainer and 2% commission on his sales.
Complete this table to show the relationship
between Earnings (E) and Sales (S).
(a)
Sales
$0
$5000
$15 000 $35 000
Earnings
(b) Write an equation that relates Earnings (E)
to Sales (S).
(a)
1
Sales
$0
$5000
Earnings
$300
$400
$15 000 $35 000
$600
$1000
(b) E = 300 + 0.02S
2
Using the information in question 1, construct a
graph of the relationship between Earnings (E)
and Sales (S).
3
y y
The points P(3, 2) and Q(5, 14) lie on a line.
m = 2 1 where (x1, y1) = (3, 2) and
Calculate the gradient of the line.
x2  x1
(x2, y2) = (5, 14)
14  2
m=
5  3
16
m=
8
m=2
1
2
1
1
2
4
Consider the equation 8x  9y = 72.
(a) Find the value of y when x = 0.
(b) Find the value of x when y = 0.
(c) Sketch the graph of 8x  9y = 72.
5
6
A plumber charges these rates:
$55 call out fee
$40 per half hour or part thereof
Calculate the charges for these service calls
12 minutes, 23 minutes, 44 minutes,
56 minutes, 73 minutes, 87 minutes.
Construct the graph of charges (C) versus time
of the service call in minutes (t) for calls of up
to 120 minutes.
(a) Put x = 0 in 8x  9y = 72,
8  0  9y = 72
9y = 72
y = 8
(0, 8)
(b) Put y = 0 in 8x  9y = 72,
8x  9  0 = 72
8x = 72
x=9
(9, 0)
(c)
1
1
1
2
Time
12 min
23 min
44 min
56 min
73 min
87 min
Charge
$55 + $40 = $95
$55 + $40 = $95
$55 + 2 x $40 = $135
$55 + 2 x $40 = $135
$55 + 3 x $40 = $175
$55 + 3 x $40 = $175
2
3
7
A telephone company charges users at a rate of
25 cents for each completed 30 seconds. This
implies a call of less than 30 seconds is free.
(a) Copy and complete this table for the calls
(a)
shown:
Length of Call
(seconds)
15
30
45
60
75
90
105
120
Cost
(cents)
(b) Construct the graph of Cost versus Length
for calls up to 120 seconds.
(b)
2
Length of Call
(seconds)
15
30
45
60
75
90
105
120
Cost
(cents)
0
25
25
50
50
75
75
100
2
4
8
Solve these simultaneous equations:
(a) y = 10x  7
y = 2x + 1
by substitution method
(b) 6x  11y = 2
5x  9y = 1 by elimination method.
9
(a) y = 10x  7 . . . . [1]
y = 2x + 1 . . . . . [2]
10x  7 = 2x + 1
10x  2x = 1 + 7
8x = 8
x=1
Substitute x = 1 in [2]
y=21+1
y=3
Solution is (1, 3)
(b)
2
6x  11y = 2 . . . . . [1]
5x  9y = 1 . . . . . [2]
_____________________
[1]  9:
54x  99y = 18 . . . . [3]
[2]  11:
55x  99y = 11 . . . . [4]
_____________________
[3]  [4]:
1x
=7
x
= 7
Substitute x = 7 into [1]:
6  –7  11y = 2
42  11y = 2
11y = 44
y = 4
Solution is (7, 4)
The cost of manufacturing basketballs (C) is
related to the number of basketballs produced
(n), by the formula C = 2800 + 4n.
The revenue (R) made from selling n
basketballs is R = 14n.
Copy and complete this table:
Number of
basketballs
150
200
250
300
350
Cost
Revenue
3
2
Number of
basketballs
150
200
250
300
350
Cost
$3400
$3600
$3800
$4000
$4200
Revenue
$2100
$2800
$3500
$4200
$4900
5
10
(a) Using the information in question 9 above,
construct the graph of Cost (C) versus
number (n) and the graph of Revenue (R)
versus number (n) on the same set of axes.
(a)
2
(b) Using the graph, write down the number of
basketballs produced before Revenue
equals Cost to ‘break-even’.
(b) Number of basketballs n = 280
1