Business Accounting
Business Accounting
RBA1002 & RBA10x2
Chapter 3:
Straight Line Graphs
And
Break-even Analysis
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 1
Business Accounting
3.1
GRAPHS
Exercise 1
1.
C
13
10
7
4
1
0
1
2
3
4
5
Q
2.
Cost
8
7
6
5
4
3
0
50
100
150
200
250
Quantity
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 2
Business Accounting
3.
Price of Pizza
900
800
700
600
500
400
0
50
55
60
65
70
Number of pizzas sold
10
20
30
40
50
Quantity
4.a)
Price
100
90
80
70
60
50
0
b) Negative linear relationship
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 3
Business Accounting
Exercise 2
1(a)
y = 3x + 6
Determine the y-intercept. Let x = 0 and solve for y.
y = 3(0) + 6
Therefore the y-intercept is 6.
Coordinates (0;6)
Determine the x-intercept. Let y = 0 and solve for x.
0 = 3x + 6
3x = -6
x = -2
Therefore the x-intercept is -2
Coordinates (-2;0)
y
(0;6)
6
5
4
3
2
(-2;0)
-4
-3
1
-2
-1
x
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 4
Business Accounting
(b)
Determine the slope of the graph.
Coordinates (0;6)
Coordinates (-2;0)
change in y
Slope =
=
or
change in x
y
x
6−0
0−(−2)
=
6
2
= 3 =
3
1
2(a)
Therefore:
2x + y = 10
y = -2x + 10
Determine the y-intercept. Let x = 0 and solve for y.
y = -2(0) + 10
y = 10
Therefore the y-intercept is 10.
Coordinates (0;10)
Determine the x-intercept. Let y = 0 and solve for x.
0 = -2x + 10
2x = 10
x=5
Therefore the x-intercept is 5
Coordinates (5;0)
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 5
Business Accounting
y
12
10 (0,10)
8
6
2
(5,0)
-4
-3
-2
-1
1
2
3
4
5
x
x + 3y = 12
Therefore:
3y = -x +12
y = -
𝟏
𝟑
x+ 4
Determine the y-intercept. Let x = 0 and solve for y.
y=-
1
3
(0) + 4
Therefore the y-intercept is 4.
Coordinates (0;4)
Determine the x-intercept. Let y = 0 and solve for x.
0=1
3
1
3
x+4
x=4
x = 12
Therefore the x-intercept is 12
Coordinates (12;0)
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 6
Business Accounting
y
12
10
8
6
4 (0,4)
2
-4
b)
-3
-2
-1
(12,0)
3
6
9
12
15
x
2x + y = 10 or y = -2x + 10
Coordinates (0;10)
Coordinates (5;0)
Slope =
=
=
change in y
y
or
change in x
x
10−0
0−5
10
−5
= -2 = -
x + 3y = 12 or y = -
1
3
2
1
x+ 4
Coordinates (0;4)
Coordinates (12;0)
Slope =
=
=
change in y
y
or
change in x
x
4−0
0−12
4
−12
= -
1
3
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 7
Business Accounting
3.
y = -2x + 5
Determine the y-intercept. Let x = 0 and solve for y.
y = -2(0) + 5
Therefore the y-intercept is 5.
Coordinates (0;5)
Determine the x-intercept. Let y = 0 and solve for x.
0 = -2x + 5
2x = 5
x=2
1
2
Therefore the x-intercept is 2
1
Coordinates (2
2
𝟏
𝟐
;0)
y
6
5
(0,5)
4
3
2
1
-4
-3
-2
-1
(2
1
2
1
2
, 0)
3
4
5
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
x
Page 8
Business Accounting
4
A(4;6) B(12;2)
Slope =
=
=
change in y
y
or
change in x
x
6−2
4−12
4
−8
= -
1
2
Equation of a graph: y = mx + c, using the point A(4,6)
1
6 = - (4) + c
2
6 = -2 + c
8 = c
c = 8
1
Equation: y = - x + 8
2
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 9
Business Accounting
Exercise 3
1. a)
y
6
4
2
-2
-1
80
100
120
140
160
180
x
-2
-4
-6
b) A(80;-6) and B(100;-4)
slope =
=
=
=
change in y
y
or
change in x
x
−4−(−6)
100−80
2
20
1
10
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 10
Business Accounting
C(120;-2) and D(140;0)
slope =
=
=
=
2.
change in y
y
or
change in x
x
0−(−2)
140−120
2
20
1
10
3y + 4x = 12
3y = -4x + 12
Div by 3
y= -
4
3
+4
y-intercept = 4
3.
y = 2x + 3
Determine the y-intercept. Let x = 0 and solve for y.
y = 2(0) + 3
Therefore the y-intercept is 3.
Coordinates (0;3)
Determine the x-intercept. Let y = 0 and solve for x.
0 = 2x + 3
2x = -3
x=
−3
2
x=-1
1
2
Therefore the x-intercept is - 1
1
2
Coordinates (- 1
𝟏
𝟐
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
; 0)
Page 11
Business Accounting
.
y
6
5
4
3 (0,3)
2
1
(-1 2 ,0)
-4
4.
-3
-2
1
-1
1
2
3
4
5
x
A(1;-4) B(-1;4)
Slope =
=
=
change in y
y
or
change in x
x
4−(−4)
−1−1
8
−2
= -4
= -
4
1
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 12
Business Accounting
5.a)
Ice-cream sold is a dependent variable as the number of ice creams sold depends on
the temperature and the temperature is an independent variable.
Ice cream sold
350
300
250
200
150
100
0
26
28
30
b)
Positive linear relationship
c)
(26;100) (28;150)
slope =
=
=
32
34
36
Temperature
change in y
y
or
change in x
x
150−100
28−26
50
2
= 25
=
25
1
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 13
Business Accounting
6.
a) A negative linear relationship
b) 1980(6.2 ; 143) and 1981(5.7 ; 156)
slope =
=
=
change in y
y
or
change in x
x
156−143
5.7−6.2
13
−0.5
= - 26
= -
26
1
1981(5.7 ; 156) and 1982(5.2 ; 169)
Slope =
=
=
change in y
y
or
change in x
x
169−156
5.2−5.7
13
−0.5
= -26
= -
26
1
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 14
Business Accounting
7.
y = 5x - 3
Determine the y -intercept. Let x = 0 and solve for y.
y = 5(0) - 3
Therefore the y-intercept is -3.
Coordinates (0;-3)
Determine the x-intercept. Let y = 0 and solve for x.
0 = 5x - 3
5x = 3
x=
3
5
3
Therefore the x-intercept is
.
Coordinates (
5
𝟑
𝟓
; 0)
y
3
1
-4
-3
-2
-1
3
( ,0)
5
1
2
3
4
5
x
-3 (0,-3)
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 15
Business Accounting
8.
3y = -6x +15
Div by 3:
y = -2x + 5
Determine the y -intercept. Let x = 0 and solve for y.
y = -2(0) + 5
Therefore the y-intercept is 5.
Coordinates (0;5)
Determine the x-intercept. Let y = 0 and solve for x.
0 = -2x + 5
2x = 5
x=
5
2
= 2
1
2
Therefore the x-intercept is 2
.
1
Coordinates (2
2
𝟏
𝟐
; 0)
y
5 (0,5)
3
1
-4
-3
-2
-1
(2
1
2
1
2
3
,0)
4
5
x
-1
-3
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 16
Business Accounting
Exercise 4
1.
Price
25
(400,25)
20
(300,20)
15
(200,15)
10
(100,10)
5
(0,5)
0
100
200
300
400
500
Quantity
2.
y
15
(5,15)
9
(3.9)
3
(0,0)
0
(1,3)
1
3
5
x
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 17
Business Accounting
a) First quadrant
b) The points (2,6) and (4,12) also lie on the line (and many others)
c) There is a positive / direct linear relationship between the x- and y-coordinates.
If x increases by 1 unit, y increases by 3 units.
The y-coordinate is 3 times the x-coordinate.
3.
a)
Use any two points on the graph
(4;12) (8;9)
Slope =
=
=
b)
change in y
y
OR
change in x
x
12−9
4−8
3
−4
Determine c (y-intercept)
Use any point on the straight line graph and substitute in place of x and y, (12,6)
y = mx + c
6=
3
(12) + c
−4
6 = -9 + c
c = 15
Therefore: y =
3
−4
x + 15
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 18
Business Accounting
c) Determine the x-intercept. Let y = 0 and solve for x.
y=
∴
∴
∴
0=
3
4
3
−4
x + 15
3
−4
x + 15
x = 15
x = 20
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 19
Business Accounting
3.2. Breakeven Analysis
1.
P
45000
45000
45000
45000
45000
45000
45000

QBE
SBE
2a)
QBE
SBE
Q
0
40
80
120
160
200
240
FC
4 200 000
4 200 000
4 200 000
4 200 000
4 200 000
4 200 000
4 200 000
VC
10 000
10 000
10 000
10 000
10 000
10 000
10 000
TVC
0
400 000
800 000
1 200 000
1 600 000
2 000 000
2 400 000
TC
4 200 000
4 600 000
5 000 000
5 400 000
5 800 000
6 200 000
6 600 000
SALES
0
1 800 000
3 600 000
5 400 000
7 200 000
9 000 000
10 800 000
PROFIT
-4 200 000
-2 800 000
-1 400 000
0
1 400 000
2 800 000
4 200 000
= 120
(break-even units)
= R 5 400 000
(break-even in Rands)
9000
= 3 000
( 4  1)
= 3 000  4 = R 12 000
=
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 20
Business Accounting
(‘000 Rands)
Total Sales Income
Profit
Total Costs (TC)
Variable Costs (VC)
12
Fixed Costs (FC)
9
Loss
0
3
Quantity
(‘000)
3.
P
15
15
15
15
15
15
Q
4 000
4 400
4 800
5 200
5 600
6 000
FC
65 000
65 000
65 000
65 000
65 000
65 000
VC
2.50
2.50
2.50
2.50
2.50
2.50
TVC
10 000
11 000
12 000
13 000
14 000
15 000
TC
75 000
76 000
77 000
78 000
79 000
80 000
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
SALES
60 000
66 000
72 000
78 000
84 000
90 000
PROFIT
-15 000
-10 000
-5 000
0
5 000
10 000
Page 21
Business Accounting

QBE
SBE
4.
a)
QBE
SBE
= 5 200
= R 78 000
200000
= 800
(300  50)
= 800  300 = R 240 000
=
(‘000 Rands)
Total Sales Income
Profit
Total Costs
TC)
Variable Costs (VC)
240
Break-even
200
Fixed Costs (FC)
Loss
0
5.1
Quantity (‘000)
800
Contribution ratio
=
=
=
Total contribution in Rands
Sales in Rands
R1 050 000
R1 800 000
x
x
100
100
58.33% of sales
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 22
Business Accounting
5.2
Break-even units
=
=
Break-even sales
5.3
Contribution per unit
R700 000
R35
=
20 000 units
=
Break-even units x price
=
20 000 x R60
=
R1 200 000
Margin of safety units
Margin of safety rands
6.1
Fixed cost
=
sales in units - breakeven in units
=
30 000 – 20 000
=
10 000 units
=
margin of safety units x selling price per unit
=
10 000 x R60
=
R600 000
Product A
Break-even units
=
=
Break-even sales
Fixed cost
Contribution per unit
R12 000
R2.50−R1.50
=
12 000 units
=
Break-even units x price
=
12 000 x R2.50
=
R30 000
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 23
Business Accounting
Product B
Break-even units
=
=
Break-even sales
Fixed cost
Contribution per unit
R12 000
R2.00−R1.40
=
20 000 units
=
Break-even units x price
=
20 000 x R2.00
=
R40 000
Product A
Margin of safety units
Margin of safety rands
=
sales in units - breakeven in units
=
24 000 – 12 000
=
12 000 units
=
margin of safety units x selling price per unit
=
12 000 x R2.50
=
R30 000
=
sales in units - breakeven in units
=
40 000 – 20 000
=
20 000 units
=
margin of safety units x selling price per unit
=
20 000 x R2.00
=
R40 000
Product B
Margin of safety units
Margin of safety rands
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 24
Business Accounting
7.1
Break-even units
=
=
Break-even sales
7.2
R600 000
R2 000−R1 200
750 units
=
Break-even units x price
=
750 x R2 000
=
R1 500 000
Margin of safety units
Contribution ratio
Contribution per unit
=
Margin of safety rands
7.3
Fixed cost
=
=
=
=
sales in units - breakeven in units
=
930 – 750
=
180 units
=
margin of safety units x selling price per unit
=
180 x R2 000
=
R360 000
Total contribution in Rands
Sales in Rands
R800 x 930
R2 000 x 930
x
x
100
100
40% of sales
Chapter 3 – Straight Line Graphs and Break-even Analysis Solutions
Page 25