Business Mathematics In Canada by F. Ernest Jerome Solution Manual |ISBN: 9781260065992| Guide A+

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COMPLETE SOLUTION MANUAL FOR
Business Mathematics in Canada 10th
Edition F. Ernest Jerome & Tracy Worswick


1 Review and Applications of
Basic Mathematics
SO

Exercise 1.1
a. 10 + 10 x 0 = 10 + 0 = 10
b. 2x2+4–8=4+4–8=0
LU
c. (10 + 10) x 0 = 20 x 0 = 0
d. 2 x (2 + 4) – 8 = 2 x 6 – 8 = 12 – 8 = 4
e. 0 + 3 x 3 – 32 + 10 = 0 + 9 – 9 + 10 = 10
f. 12 – 2 x 5 + 22 x 0 = 12 – 10 + 4 x 0 = 12 – 10 + 0 = 2
TI
g. 0 + 3 x 3 – (32 + 10) = 0 + 9 – 19 = -10
h. (12 – 2) x (5 + 22) x 0 = 10 x 9 x 0 = 0
22 −4 4 −4 0
i. = = =0
O
(4−2)2 22 4

(2 −4)2 (−2)2 4
= = =4
5−22 5 −4 1
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j.
1. 20 − 4  2 − 8 = 20 − 8 − 8 = 4
2. 18  3 + 6  2 = 6 + 12 = 18
S
3. (20 − 4)  2 − 8 = 16  2 − 8 = 32 − 8 = 24
4. 18  (3 + 6)  2 = 18  9  2 = 2  2 = 4

5. 20 − (4  2 − 8) = 20 − (8 − 8) = 20

6. (18  3 + 6)  2 = (6 + 6)  2 = 24
7. 54 − 36  4 + 22 = 54 − 9 + 4 = 49

8. (5 + 3)2 − 32  9 + 3 = 82 − 9  9 + 3 = 64 − 1+ 3 = 66

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9. (54 − 36)  (4 + 2)2 = 18  62 = 18  36 = 0.5
(
10. 5 + 32 − 3 )2  (9 + 3) = 5 + (9 − 3)2  12 = 5 + 36  12 = 5 + 3 = 8
82 − 42 64 − 16 = 48 = 6
11. =
(4 − 2)3 23 8
2
12.
(8 − 4)2 =
4
=
16
=−4
4 − 23 4−8 −4
13. 3(6 + 4)2 − 5(17 − 20)2 = 3 10 2 − 5(− 3)2 = 3 100 − 5  9 = 300 − 45 = 255

14. (4  3 − 2)2  (4 − 3  22 ) = (12 − 2)2  (4 − 3  4) = 102  (4 − 12) = 100  (− 8) = − 12.5
15. (20 + 8  5) − 7  (− 3) 9 = (20 + 40 + 21)  9 = 81 9 = 9
SO

( )  
519 + 52 − 16 2  = 5 19 + (25 − 16 )2 2 = 5(19 + 81)2 = 5  100 2 = 50,000
2
16.
 
17. $100(1+ 0.06  365
45
) = $100(1+ 0.00739726) = $100.74
$200 $200 $200
= = = $194.17
LU
18.
1+ 0.09  12 1+ 0.03 1.03
4


$500 = $500 $500
=
19.
(1 + 0.05) 1.05 1.1025 = $453.51
2 2


( )
20. $1000 (1+ 0.02) = $1000 1.023 = $1000 (1.061208 ) = $1061.21
3
TI
 (1+ 0.04) − 1
2
 1.042 − 1  0.0816 
21. $100 0.04  = $100 0.04  = $100 0.04  = $204.00
     
 1   1 − 1 
O
1 − (1 + 0.03)2  1.0609  1 − 0.942596 
22. $300  = $300  = $300  = $574.04
 0.03   0.03   0.03 
   
N
   

Concept Questions (Section 1.2)
S
1. You must retain at least one more figure than you require in the answer. To achieve four-
figure accuracy in the answer, you must retain a minimum of five figures in the values
used in the calculations. B)
2. We want six-figure accuracy in the answer. Therefore, values used in the calculations
must be accurate to at least seven figures. B)
3. We want seven-figure accuracy in the answer. Therefore, values used in the calculations
must retain at least eight figures. C)
4. To be accurate to the nearest 0.01%, an interest rate greater than 10% must have four-
figure accuracy. Therefore, five figures must be retained in numbers used in the
calculations. C)

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Exercise 1.2
a. 1
= 0.10 = 10%
10
b. 2 = 0.40 = 40%
5
c. 1 = 0.25 = 25%
4
d. 3 = 0.75 = 75%
4
1
e. 1 = 1.50 = 150%
2
1
f. 2 = 2.3333 = 233.33%
3
g. 10 = 2.00 = 200%
5
h. 52 = 5.6667 = 566.67%
SO
3
i. 0.25 x 80 = 20
j. 0.20 x 120 = 24
k. Money in Savings = 0.20 x $1000 = $200
Money in TFSA = 0.50 x $200 = $100
1. 7
8
= 0.87500 = 87.500%
LU
2. 65
104
= 0.62500 = 62.500%
3. 47
20
= 2.3500 = 235.00%
4. −916=−0.56250 =−56.250%
TI
5. −35 = −1.4000 = −140.00%
25


7 = 1.2800 = 128.00%
6. 1 25
O
7. 1000
25 = 0.025000 = 2.5000%

8. 1000 = 40.000 = 4000.0%
N
25

2 = 2.0200 = 202.00%
9. 2 100
S
10. − 1 32
11 = −1.3438 = −134.38%


11. 37.5
50
= 0.75000 = 75.000%
12. 22.5
−12
= − 1.8750 = − 187.50%

13. 5
6 = 0.83 = 83.3%
14. − 8 = −2.6 = −266.6%
3

15. 7 79 = 7.7 = 777.7%
1 = 1.09 = 109.09%
16. 1 11

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17. 10
9 = 1.1= 111.1%
4
18. − = −0.004̅ = −0. 4̅%
900

19. − 270
7
= − 0.0259 = − 2.592%

20. 37
27 = 1.370 = 137.037%
21. 11.3845  11.38
22. 9.6455  9.646
23. 0.5545454  0.5545
24. 1000.49  1000
25. 1.0023456  1.002
SO
26. 0.030405  0.03041
27. 40.09515  40.10
28. 0.0090909  0.009091

29. 1
= 0.16667 = 16.667%
LU
6

30. 7
6
= 1.1667 = 116.67%

31. 1
60
= 0.016667 = 1.6667%

32. 2 59 = 2.5556 = 255.56%
TI
33. 250
365
= 0.68493 = 68.493%

34. 15
365
= 0.041096 = 4.1096%
O
35. 0.11
12
= 0.0091667 = 0.91667%

36. 0.095
12
= 0.0079167 = 0.79167%
N
37. (
$92 1+ 0.095  112) = $92  1.02915 = $94.68
365

38. $100(1+ 0.11 5 ) = $100  1.04583 = $104.58
S
12

39. $454.76(1− 0.105  11) = $454.76  0.903750 = $410.99
12
40. 1
= 0.550707 = 0.55
(1+0.22)3


0.18 3 0.21 2
41. $1447 (1 + 2
) (1 + 3
) = $1447 (1.295029) (1.1449) = $2145.44
$790.84 $790.84
42. 1+0.13311
365
= =$711.98
1.110767