CIMA - C3 Fundamentals Of Business Mathematics
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360 Answer bank
21 Y = aX
b
= 9
× 7
– 0.24
= 5.64
22 3 (X + 4) < 2X + 1
3X + 12 < 2X + 1 (multiply out the brackets)
3X < 2X – 11 (subtract 12 from both sides)
X < – 11 (subtract 2X from both sides)
23 x =
–5
y =
4
Workings
x – y = –9 (1)
9x – 13y = –97 (2)
9x – 9y = –81 (3) (1)
× 9
4y = 16 (3) – (2)
y = 4
Substituting in (1) we get
x – 4 = –9
x = –9 + 4
x = –5
24
x
2
– 4x – 437 = 0
or
y
2
+ 4y – 437 = 0
Workings
Let the two positive numbers be x and y.
x
2
+ y
2
= 890 (1)
x – y = 4 (2)
y = x – 4 (3) (from (2)).
Substituting in (1) we get
x
2
+ (x – 4)
2
= 890
x
2
+ (x
2
– 8x + 16) = 890
2x
2
– 8x – 874 = 0
x
2
– 4x – 437 = 0
Alternatively x = y + 4 (3)
and hence (y + 4)
2
+ y
2
= 890
y
2
+ 8y + 16 + y
2
= 890
2y
2
+ 8y – 874 = 0
y
2
+ 4y – 437 = 0
25
–b
2
+ 22b – 112 = 0
or
–a
2
+ 22a – 112 = 0
Answer bank 361
Workings
Let the length and breadth of the rectangle be a and b.
2a + 2b = 44
a + b = 22 (1)
ab = 112 (2)
From (1) we get a = 22 – b
Substituting in (2) we get
b(22 – b) = 112
–b
2
+ 22b – 112 = 0
By symmetry the same equation:
–a
2
+ 22a – 112 = 0
can be solved for 'a'.
26 x =
–2
or
+4
Workings
– 3x
2
+ 6x + 24 = 0
This can be solved using:
x =
a2
ac4bb
2
−±−
Where a = –3, b = 6, c = 24
x =
()
32
)24346(6
2
−×
××+±−
=
)6(
3246
−
±−
=
)6(
]186[
−
±−
= –2 or +4.
27 B The number of goals scored is an example of
quantitative data as they can be measured. Since the
number of goals scored cannot take on any value, they can only score 1, 2, 3 or any
whole number
of goals (they cannot score 2½ goals) the data are said to be
discrete.
You should have been able to eliminate options C and D immediately since qualitative data are data
that cannot be measured but which reflect some quality of what is being observed.
28 B
Systematic sampling is a sampling method which works by selecting every n
th
item (or m
th
in this
case) after a random start.
362 Answer bank
29 B A sampling frame is a numbered list of all items in a population (not a sample). Therefore
Statement III is not true.
Cluster sampling involves selecting one definable subsection of the population which therefore
makes the potential for
bias considerable. Therefore Statement II is not true.
Statements I and IV are true and the correct answer is therefore B.
30
Sampling method Sampling frame required (
9
)
Random
9
Stratified
9
Quota
Systematic
9
31 C Height of histogram =
3
2
20
= 30
If you selected option A, you multiplied 20 by 2/3 instead of dividing it by 2/3.
Option B is incorrect because it represents the score in the class under consideration.
Option D represents the score in the class (20) plus 2/3 of 20 = 20 + 13.33 = 33.33.
32 B A histogram is a chart that looks like a bar chart except that the bars are joined together. On a
histogram, frequencies are presented by the
area covered by the bars.
33 C An ogive represents a cumulative frequency distribution which shows the cumulative number of
items with a value less than or equal to, or alternatively greater than or equal to, a certain amount.
This is a 'less than' ogive.
34 A =
6.5
B =
12
C =
16
D =
4.5
Workings
The first and last intervals have widths of $10, ie twice the standard width. Their frequencies must therefore
be halved in order to calculate the frequency density. In all other cases, the frequency density is the same
as the frequency (or number of employees).
Weekly average wage Number of employees Frequency density
$
51 and <61 13 6.5
61 and <66 12 12
66 and <71 16 16
71 and <76 21 21
76 and <81 25 25
81 and <86 20 20
86 and <91 14 14
91 and <101 9 4.5
Answer bank 363
35
36 E =
62
F =
87
G =
130
Workings
Average weekly wage Number of employees Cumulative frequency
$
51 and <61 13 13
61 and <66 12 25
66 and <71 16 41
71 and <76 21 62
76 and <81 25 87
81 and <86 20 107
86 and <91 14 121
91 and <101 9 130
364 Answer bank
37
38 A
x
fx
f
$
$
$5.00
800
800
÷ 5.00
160
$6.25
1,000
1,000
÷ 6.25
160
$5.50
1,100
1,100
÷ 5.50
200
$6.00
1,200
1,200
÷ 6.00
200
4,100
720
∴
720
100,4
f
fx
x
=
∑
∑
= = $5.69
39 C In order of magnitude, scores are
16 18 22 24 28 30 34 42 48 68
Position of median is
2
110
+
= 5
2
1
and therefore the median is the average of the 5
th
and 6
th
items,
ie of 28 and 30. The median is therefore 29.
40 D
Arithmetic mean =
n
x
∑
∴ 10 =
19
x
∑
∑x = 10 × 19
= 190
Let x = twentieth number
190 + x = 20
× 11
190 + x = 220
x = 220 – 190
= 30
Answer bank 365
41
Daily sales
Frequency
Midpoint
f
x
fx
> 0
≤ 3
8
2
16
> 3
≤ 6
16
5
80
> 6
≤ 9
12
8
96
> 9
≤ 12
4
11
44
40
236
The variables in the table are
discrete. The arithmetic mean of the data =
40
236
f
fx
=
∑
∑
= 5.9 televisions per day
42
32.67
minutes
Workings
Times
Midpoint
Frequency
x
f
fx
0 – 10
5
14
70
10 – 20
15
26
390
20 – 30
25
64
1,600
30 – 40
35
46
1,610
40 – 50
45
28
1,260
50 – 60
55
16
880
60 – 80
70
8
560
80 – 100
90
4
360
206
6,730
Mean
206
730,6
f
fx
x
=
∑
∑
= = 32.67 minutes
43
6.22
Workings
fX
X
f
Σ
=
Σ
2
2
2
fX 1161 79
X 50.478 11.798
f2323
Σ
⎛⎞
σ= − = − = −
⎜⎟
Σ
⎝⎠
=
38.68 6.22=
44 C The formula for the coefficient of variation is
Mean
deviationStandard
Distribution Coefficient of variation
1 0.24
2 0.25
3 0.35 largest
4 0.21
Option C is therefore correct.
366 Answer bank
If you chose any of the other options you did not calculate the coefficient of variation using the
correct formula.
45 C The arithmetic mean of the five numbers,
x , is equal to
5
1512864 ++++
=
5
45
= 9
We can now calculate the variance:
x
x – x
()
2
xx −
4
–5
25
6
–3
9
8
–1
1
12
3
9
15
6
36
∑(x – x )
2
=
80
Variance =
n
)xx(
2
−∑
=
5
80
= 16
The correct answer is therefore C.
If you selected option A, you calculated the standard deviation.
If you selected option B, you calculated the arithmetic mean.
If you selected option D, you forgot to divide ∑
(
)
2
xx −
by n (or 5).
46 C To find the percentage increase since year 9, we must take the increase as a percentage of the year 9
value. The increase in the index of 18 points between year 9 and year 14 is therefore a percentage
increase of (18/180 × 100%) = 10%.
The correct answer is therefore C.
If you selected option D you interpreted the increase of 18 points as an increase of 18% which is
incorrect.
47
Year Index Real wage
$
20X0 88 170
20X1 90 179
20X2 93 181
20X3 95 188
20X4 96 193
20X5 98 195
20X6 100 197
20X7 102 199
20X8 104 199
20X9 106 201
20Y0 108 214
The index number for each year with 20X6 as the base year will be the original index number divided by
1.14, and the real wages for each year will be (money wages × 100)/index number for the year.
Answer bank 367
48 (a)
101.3
Workings
Weighted average of price relative index =
W
P/PW
01
∑
×∑
× 100
where P
1
= price in 20X1
P
0
= price in 20X0
Price relatives
Price relative P
1
/P
0
A 100/100 = 1.00
B 108/100 = 1.08
C 90/100 = 0.90
Weightings (W)
Q
0
A
5,000
B
6,000
C
3,000
14,000
Index
W × P
1
/P
0
A
1.00 × 5,000 =
5,000
B
1.08 × 6,000 =
6,480
C
0.90 × 3,000 =
2,700
14,180
Index =
000,14
180,14
× 100 = 101.3 (to 1 decimal place)
(b)
110.0
Workings
Weighted average of price relative index =
W
P/PW
01
∑
×∑
× 100
Price relatives
Price relative P
1
/P
0
A 110/100 = 1.1
B 115/100 = 1.15
C 100/100 = 1.0
Weightings (W) = 14,000 (as calculated in (a))
Index
W × P
1
/P
0
A
1.1 × 5,000 =
5,500
B
1.15 × 6,000 =
6,900
C
1.00 × 3,000 =
3,000
15,400
Index =
000,14
400,15
× 100 = 110.0
368 Answer bank
49 C A demand of 53 units occurred on 42 days out of a total of 200.
P(demand = 53 units) =
200
42
× 100%
= 21%
If you selected option A, you calculated the probability that daily demand was 51 units instead of 53.
If you selected option B, you calculated the probability that daily demand was 52 units instead of 53.
If you selected option D, you calculated the probability that daily demand was 54 units instead of 53.
50 D Probability of A or B occurring = P(A) + P(B) provided A and B cannot both be true.
Pr(sales remain same or rise) = P(same) + P(rise)
= 0.61 + 0.22
= 0.83
You must make sure that you understand the laws of probability so that you can apply them
correctly to objective test questions such as this.
51 Probability of visiting area x = 0.50
Probability of visiting area y = 0.25
Probability of visiting area z = 0.25
EV of probability of a sale in area x = 0.20
EV of probability of a sale in area y = 0.0875
EV of probability of a sale in area z = 0.0625
Workings
The probabilities of visiting each area are obtained from the ratios in which he visits them.
Area
Ratio
Probability
x
2
2/4 = 0.50
y
1
1/4 = 0.25
z
1
1/4 =
0.25
4
1.00
The expected value (EV) of probability of a sale = P(sale) × P(visiting area).
Area P(sale) P(visiting area) EV of probability of a sale
x 0.40 0.50 0.20
y 0.35 0.25 0.0875
z 0.25 0.25 0.0625
52 (a) With replacement
169
1
Without replacement
221
1
Answer bank 369
Workings
Assuming replacement
P(Ace) =
13
1
52
4
=
P(Ace on second draw) =
13
1
P(Ace followed by Ace with replacement) =
169
1
13
1
13
1
=
⎟
⎠
⎞
⎜
⎝
⎛
×
⎟
⎠
⎞
⎜
⎝
⎛
Without replacement
P(Ace) =
13
1
P(Ace on second draw) =
17
1
51
3
=
P(Ace followed by Ace without replacement) =
221
1
17
1
13
1
=
⎟
⎠
⎞
⎜
⎝
⎛
×
⎟
⎠
⎞
⎜
⎝
⎛
(b)
704,2
1
Workings
P(Drawing ace of Hearts) =
52
1
P(Drawing ace of Spades) =
52
1
(if replacement is assumed)
P(Ace of hearts followed by ace of Spades) =
704,2
1
52
1
52
1
=
⎟
⎠
⎞
⎜
⎝
⎛
×
⎟
⎠
⎞
⎜
⎝
⎛
(c) (i)
0.3
(ii)
0.27
(iii)
0.02
Workings
P(Own Home) = P(H) = 0.5
P(Own Car) = P(C) = 0.6
P(Own TV) = P(TV) = 0.9
(i) P(Home and Car) = P(Home)
× P(Car)
= 0.5 × 0. 6
= 0.3
(ii) P(all three owned) = P(Home)
× P(car) × P(TV)
= 0.5 × 0.6 × 0.9
= 0.27