Solution Manual For Contemporary Business Mathematics with Canadian Applications, 13th Edition by Si by digitaldownload87

Contents PART ONE Mathematics Fundamentals and Business Applications Chapter 1: Review of Arithmetic

Chapter 2: Review of Basic Algebra

Chapter 3: Ratio, Proportion, and Percent

Chapter 4: Linear Systems

135

Chapter 5: Cost-Volume-Profit Analysis and Break-Even

193

PART TWO Mathematics of Business and Management Chapter 6: Trade Discount, Cash Discount, Markup, and Markdown

225

Chapter 7: Simple Interest

263

Chapter 8: Simple Interest Applications

289

PART THREE Mathematics of Finance and Investment Chapter 9: Compound Interest—Future Value and Present Value

335

Chapter 10: Compound Interest—Further Topics

379

Chapter 11: Ordinary Simple Annuities

415

Chapter 12: Ordinary General Annuities

455

Chapter 13: Annuities Due, Deferred Annuities, and Perpetuities

493

Chapter 14: Amortization of Loans, Including Residential Mortgages

545

Chapter 15: Bond Valuation and Sinking Funds

611

Chapter 16: Investment Decision Applications

669

PART ONE Chapter 1

Mathematics Fundamentals and Business Applications Review of Arithmetic

Exercise 1.1 A. 1.

12  6  3  12  2  14

(3  8  6)  2  (24  6)  2  18  2  9

(7  4)  5  2  11 5  2  55  2  53

5  3  2  4  15  8  23

6(7  2)  3(5  3)  6(5)  3(2)  30  6  24

20  16 4 1    0.2 15  5 20 5

4(8  5)2  5(3  22 )  4(3)2  5(3  4)  4(9)  5(7)  36  35  1

(3  4  2)2  (2  2  72 )  (12  2) 2  (2  2  49)

 102  (2  98)  100  96  4 9.

250(1  0.08)10  250(2.158925)  539.73

10. (1  0.04)4  1  1.169859  1  0.17 11. 30  600  2500  12  600  18,000  2500  7200  8300 12.

1  [(1  0.40)(1  0.25)(1  0.05)]  1  [(0.6)(0.75)(0.95)]  1  [0.4275]  0.5725  0.57 13. 15  7  6(2  3)  3  15  7  6(5)  3

 15  7  30  3  15  7  10  18 14. 16  2  4  6(4  2)

 8  4  6(6)  32  36  68

15. (1  0.7)  4  20  5  (0.3)  4  4  (0.3)  16

 15.7 16. 50[(1  0.2)(1  0.175)(1  0.04)]  50[(0.8)(0.825)(0.96)]  50[(0.6336)]

 31.68 17. 7a  6[4  (3a  6)]  7a  6[4  3a  6]

 7a  6[2  3a]  7a  12  18a  25a  12 18. 6a  4b  2(16  2a  b)  6a  4b  32  4a  2b  2a  6b  32

Exercise 1.2

24 24 / 2 12 12 / 2 6 6 / 3 2       36 36 / 2 18 18 / 2 9 9 / 3 3

also

24 / 12 2  36 / 12 3

28 28 / 2 14 14 / 2 7 7/7 1       56 56 / 2 28 28 / 2 14 14 / 7 2

also

28 / 28 1  56 / 28 2

210 210 / 10 21 21 / 3 7     360 360 / 10 36 36 / 3 12

also

210 / 30 7  360 / 30 12

360 360 / 5 72 72 / 9 8     225 225 / 5 45 45 / 9 5

also

360 / 45 8  225 / 45 5

144 144 / 2 72 72 / 9 8 8/ 4 2 144 / 72 2       also  360 360 / 2 180 180 / 9 20 20 / 4 5 360 / 72 5

25 25 / 5 5   365 365 / 5 73

A. 1.

365 365 / 73 5   73 73 / 73 1

365 365 / 73 5   219 219 / 73 3

B. 1. 6

1 13  2 2

5 29 2. 4  6 6 3 15 3. 3  4 4

2 26 4. 8  3 3 5.

23 1  11 2 2

51 1  5 10 10

31 3  7 4 4

19 5  2 7 7

C. 1.

11  1.375 8

7  1.75 4

5  1.666667  1.6& 3

5  0.833333  0.83& 6

11  1.833333  1.83& 6

7  0.777778  0.7& 9

13  1.083333  1.083& 12

19  1.266667  1.26& 15

D. 1.

3 3  3.375 8

2 3  3.4 5

1 8  8.333333  8.3& 3

2 16  16.666667  16.6& 3

1 33  33.333333  33.3& 3

1 83  83.333333  83.3& 3

7 7  7.777778  7.7& 9

1  7.083333  7.083& 12

E. 1.

$5.63

$17.45

$18

$253.49

$57.70

$3.10

$13

$40

F. 1.

25, 000(15  8)  146, 000  25, 000(7)  146, 000  175, 000  146, 000  29, 000

(300  8000)  (180  8000)  63, 000  2, 400, 000  1, 440, 000  63, 000  897, 000

1  [(1  0.4)(1  0.25)(1  0.08)]  1  [(0.6)(0.75)(0.92)]  1  [0.414]  0.586

1  [(1  0.32)(1  0.15)(1  0.12)]  1  [(0.68)(0.85)(0.88)]  1  [0.50864]  0.49136

1500 

$54 $54 $54    $730 225 0.12  365 0.12  0.616438 0.073973

264 264 264    0.15 146 4400  365 4400  0.4 1760

45   $620 1  0.14    $620(1  0.017260)  $620(1.017260)  $630.70 365  

292   $375 1  0.16    $375(1  0.128)  $375(1.128)  $423 365  

10.

$250, 250 $250, 250 $250, 250    $220,364.90 330 1  0.15  365 1  0.135616 1.135616

11.

$2358 $2358 $2358    $2250 146 1  0.12  365 1  0.048 1.048

1500  1500  30,000  31,500 0.05

 (1  0.03) 24  1  1.032794   1000   1000[34.426470]  $34, 426.47 12. $1000   0.03  0.03   

 (1  0.02) 20  1  0.485947   70(1.02)  13. $70(1  0.02)   0.02  0.02     71.4[24.29737]  $1734.83 14. $50

[1  (1  0.075) 8 ] 50[1  (0.560702)] 50[0.439297]    50[5.857303] 0.075 0.075 0.075  $292.87

Exercise 1.3

64  0.64 100

A. 1.

64% 

300% 

2.5% 

2.5  0.025 100

0.1% 

0.1  0.001 100

0.5% 

0.5  0.005 100

85% 

250% 

4.8% 

4.8  0.048 100

7.5% 

7.5  0.075 100

10. 0.9% 

0.9  0.009 100

300 3 100

85  0.85 100 250  2.5 100

11. 6.25%  12. 99% 

13.

6.25  0.0625 100

99  0.99 100

225% 

225  2.25 100

14. 0.05% 

0.05  0.0005 100

1 8.25 15. 8 %   0.0825 4 100

16.

1 0.5 %  0.005 2 100

1 112.5 17. 112 %   1.125 2 100 3 9.375 18. 9 %   0.09375 8 100 19.

3 0.75 %  0.0075 4 100

1 162.5 20. 162 %   1.625 2 100 21.

2 0.4 %  0.004 5 100

22.

1 0.25 %  0.0025 4 100

23.

1 0.025 %  0.00025 40 100

1 137.5 24. 137 %   1.375 2 100 25.

5 0.625 %  0.00625 8 100

26. 0.875% 

27.

0.875  0.00875 100

1 2.25 2 %  0.0225 4 100

2 16.6& 28. 16 %   0.16& 3 100 2 116.6 29. 116 %   1.16 3 100 1 183.3& 30. 183 %   1.83& 3 100 1 83.3 31. 83 %   0.83 3 100

2 66.6& 32. 66 %   0.6& 3 100

B. 1.

25% 

25 1  100 4

1 62.5 625 5 62 %    2 100 1000 8

175% 

5% 

1 37.5 375 3 37 %    2 100 1000 8

75% 

4% 

4 1  100 25

8% 

8 2  100 25

40% 

175 7  100 4

5 1  100 20

75 3  100 4

40 2  100 5

1 87.5 875 7 10. 87 %    2 100 1000 8 11.

250% 

12.

2% 

250 5  100 2

2 1  100 50

1 12.5 125 1   13. 12 %  2 100 1000 8 14. 60% 

60 3  100 5

15.

2.25% 

16. 0.5% 

17.

2.25 225 9   100 10, 000 400

0.5 5 1   100 1000 200

1 1 1 %  8 8(100) 800

1 100 100 1 18. 33 %  %  3 3 3(100) 3 19.

3 3 3 %  4 4(100) 400

2 200 200 2 20. 66 %  %  3 3 3(100) 3 21. 6.25% 

6.25 625 1   100 10, 000 16

22. 0.25% 

0.25 25 1   100 10, 000 400

2 50 50 1 23. 16 %  %   3 3 3(100) 6 24. 7.5% 

7.5 75 3   100 1000 40

25. 0.75% 

26.

7 7 7 %  8 8(100) 800

27. 0.1% 

28.

0.75 75 3   100 10, 000 400

0.1 1  100 1000

3 3 3 %  5 5(100) 500

29.

2.5% 

2.5 25 1   100 1000 40

1 400 400 4 30. 133 %  %  3 3 3(100) 3 1 550 550 11 31. 183 %  %  3 3 3(100) 6 2 500 500 5 32. 166 %  %  3 3 3(100) 3 C. 1.

3.5  3.5(100)  350%

0.075  0.075(100)  7.5%

0.005  0.005(100)  0.5%

0.375  0.375(100)  37.5%

0.025  0.025(100)  2.5%

2  2(100)  200%

0.125  0.125(100)  12.5%

0.001  0.001(100)  0.1%

0.225  0.225(100)  22.5%

10. 0.008  0.008(100)  0.8% 11. 1.45  1.45(100)  145% 12. 0.0225  0.0225(100)  2.25% 13. 0.0025  0.0025(100)  0.25% 14. 0.995  0.995(100)  99.5% 15.

0.09  0.09(100)  9%

16.

3  3(100)  300%

17.

3  0.75(100)  75% 4

18.

3  0.12(100)  12% 25

19.

5 &  1.666667(100)  166.6% 3

20.

7  0.035(100)  3.5% 200

21.

9  0.045(100)  4.5% 200

22.

5  0.625(100)  62.5% 8

23.

3  0.0075(100)  0.75% 400

24.

5 &  0.833333(100)  83.3% 6

25.

9  0.01125(100)  1.125% 800

26.

7 &  1.166667(100)  116.6% 6

27.

3  0.375(100)  37.5% 8

28.

11  0.275(100)  27.5% 40

29.

4 &  1.333333(100)  133.3% 3

30.

9  0.0225(100)  2.25% 400

31.

13  0.65(100)  65% 20

32.

4  0.8(100)  80% 5

Exercise 1.4 A. 1.

Total weight  1 1 3  2 3 4  1 5 8  3 5 6  1.3& 2.75  1.625  3.83& 9.5416&ounces Total selling value of 4 pieces  $1569  9.5416& $14,970.88

1 3 1 1 3 Total hours  15  13  18  21  22 2 4 2 4 4  15.5  13.75  18.5  21.25  22.75  91.75 Total cost of labour  91.75  25.75  $2362.56

6  224, 400 = $122, 400 11 3.75 Property tax  122, 400   $4590 100

Assessed value 

Retail value  $0.90  2700  $0.90  2700  $2430 3 Discount   2430  $911.25 8 Credit received  2430  911.25  $1518.75

64  $0.75  $ 48.00 1 45.00 54  83 ¢  54  $0.83&  3 27.00 72  $0.375  & & 42  $1.33  42  $1.3  56.00 Total  $176.00

96  $0.875 2 330  16 ¢  330  $0.16& 3 144  $1.75 240  $1.66& 240  $1.6& Total

 $ 84.00 

55.00

 252.00  400.00  $791.00

7. Assessment Quiz 1

Mark 7

Weight 5%

Contribution to Final Grade 3.5

Quiz 2 Quiz 3 Quiz 4 Test 1 Test 2 Test 3 Exam

7.25 9

5% 5% 20% 20% 20% 25% 100%

6.5 38 41 43

50 50 50

79%

3.625 4.5 15.20 16.40 17.20 19.75 80.175

Michael’s final grade in physics is 80% . (His teacher did not count Quiz 4.) B. 1.

1100 1.088  $1196.80 1600 1.197  $1915.20 1400 1.277  $1787.80 Total cost  $4899.80 Average cost per litre 

$4899.80  $1.195073 4100  $1.195

(a) 56  60  70  54  240

Average number of litres  240  4  60 (b) Total cost  56  $2.080  $116.48

60  $1.985  $119.10 70  $2.122  $148.54 54  $2.075  $112.05 $496.17 Average cost per litre  $496.17  240  $2.067375  $2.067 (c) Average cost per km  $2.067375  8.75  $0.236271  $0.236

 3  4  5  2  2  6  4  2  4 1 2  6  12  10  12  8  4  12  58 Total hours  3  5  2  4  4  2  20 58 Grade-point average   2.9 20

Weighted hours

Weighted investment:

January1  February 28 : $7500  2  $15, 000 March1  July 31: 6600  5  33, 000 August1  August 31: 8100 1  8100 September1  December 31: 7800  4  31, 200 $87,300 Average investment balance  $87,300  12  $7275 5.

(a) Simple average of unit prices



10.00  10.60  11.25  9.50  9.20  12.15 62.70   $10.45 6 6

(b) Number of units purchased  Date

Amount invested Unit price

Amount Invested

Unit Price

Number of Units Purchased

February 1

200

10.00

200  20.000 10

March 1

200

10.60

200  18.868 10.60

April 1

200

11.25

200  17.778 11.25

May 1

200

9.50

200  21.053 9.50

June 1

200

9.20

200  21.739 9.20

July 1

200

12.15

200  16.461 12.15

Total number of units purchased (c) Average cost of units purchased 

1200  $10.35 115.899

(d) Value on July 31  115.899(11.90)  $1379.20

Exercise 1.5 A. 1.

(a) Annualsalary  $43, 056

Semi-monthly payment 

43, 056  $1794 24

43, 056  $828 52 828 Hourly rate   $23 36

(b) Weekly pay 

= $ 1794.00 Overtime pay  11 23 1.5  379.50 Gross pay

 $2173.50

115.899

(a) Annual salary  $43,875

Biweekly pay 

43,875  $1687.50 26

1687.50  $843.75 2 843.75 Hourly rate   $22.50 37.5

(b) Weekly pay 

 $1687.50 Overtime pay  8  22.50  1.5  270.00

 $1957.50

Gross pay 3.

(a) Monthly pay  $2657.20

Yearly pay  2657.20 12  $31,886.40 Weekly pay  31,886.40  52  $613.20 Hourly rate of pay  613.20  35  $17.52  $2657.20 Overtime pay  7.75 17.52 1.5  203.67

(b) Regular pay for May

 $2860.87

Gross pay 4.

(a) Semi-monthly pay  $1586

Yearly salary  1586  24  $38, 064 Weekly gross pay  38, 064  52  $732 Hourly rate  732  40  $18.30 (b) Gross pay Regular pay

 $1816.58  $1586.00

Overtime pay  $ 230.58 Number of overtime hours  ($230.58  1.5)  $18.30  8.4 5.

Total hours = 45 Regular hours = 40 Overtime hours = 5 At time-and-a-half, 5 overtime hours are equivalent to 5 1.5  7.5 regular hours

Rate of pay 

$917.70  $19.32 47.5

(a) Biweekly payment  $3942

Annual salary  3942  22  $86, 724 Daily pay  86, 724  200  $433.62 Hourly rate  433.62  7.5  57.816  $57.82  $3942.00 Less: two days  433.62  2  867.24

(b) Regular pay

 $3074.76

Gross pay 7.

Gross sales Less:returns

 $12, 660.00  131.20

Net sales  $12,528.80 Gross commission  12,528.80  0.0975  $1221.56 Less:drawings  720.00  $501.56

Amount due 8.

Net sales  $16, 244 1 Commission: 8 % on first $6000 4 3 9 % on next $6000 4 11.5% on $(16, 244  12, 000) Total commission

Gross sales Less:returns

 $24, 250  855

Net sales

 $23,395



$495.00



585.00



488.06

 $1568.06

Commission: 4.5% on first $10, 000  0.045 10, 000  $450.00 6% on next $5000  0.06  5000  300.00 8% on remaining $8395  0.08  8395  671.60 Total commission

 $1421.60

10. (a) Sales  $8125 Base salary on quota of $8500  $825

(b) Sales  $10,150

Base salary on quota of $8500  $825.00 1 Commission  6 % on $1650  0.065  $1650  107.25 2 Gross earnings  $932.25 11. (a) Commission at 6.5% on sales of $5830 = 0.065 × $5830 = $378.95. This is less than $540 guarantee, therefore weekly salary  $540 (b) Commission at 6.5% on sales of $8830  0.065  $8830  $573.95 This exceeds $540 guarantee, therefore weekly salary = $573.95 12. Gross sales Less: returns  3% of $31, 240

 $30,302.80

Net sales Rate of commission 

Commission: Sales for week Quota: Commission sales Rate of commission

Net sales 

1590.90  0.0525 30,302.80

 5.25%

 $566.25  450.00

13. Gross earnings Less: base salary

14.

 $31, 240.00  937.20

 $116.25  $6550  5000  $1550 116.25   0.075  7.5% 1550

$Commission $2036.88   $18,105.60 Rate 0.1125

Net sales  gross sales  returns 18,105.60  S  0.08S 0.92S  18,105.60 S  19, 680 Gross sales were $19,680

15. Gross earnings Less: Base salary

 $837.50  664.00

 $173.50 173.50 Commission sales   $1982.86 0.0875 Sales for week  $4800  $1982.86  $6782.86 Commission

16. Method A

Method B

Regular hours  40 17.60 Overtime pay  3.5 17.60 1.5 6 17.60  2

 $704.00  92.40  211.20

Gross earnings

 $1007.60

At regular rate: 49.5 17.60 Overtime premium: 3.5 17.60  0.5 6 17.60 1

 $ 871.20  30.80  105.60

Gross earnings

 $1007.60

Exercise 1.6 1. GST collected

GST paid 5% of purchases

GST payable

Month

5% of sales

January

$27,345

$7391.60

$19,953.40

February

12,200

3475.00

8725.00

March

29,400

43,300.00

(13,900.00)

April

32,515

22,500.00

10,015.00

May

7840

4904.90

2935.10

$109,300

$81,571.50

$27,728.50

5-month totals

(GST receivable)

Cook’s owes the government $27, 728.50.

Riza’s revenue of $28,350 includes 5% GST.

GST taxable revenue 

28,350  $27,000 1.05

GST collected  5% of $27,000  $1350 GST paid  5% of $8000  $400 Riza owes the Canada Revenue Agency $(1350  400)  $950

Savings on GST  5% of $780  0.05(780)  $39

Cost of shirt  $15.00 GST in Regina  5% of $15  0.05(15)  0.75 PST  6% of $15  0.06(15)  0.90  $16.65

Consumer pays 5.

At Blackcomb, B.C. Cost of ski pass

GST  5% of $214  0.05(214) PST  7% of $214  0.07(214)

 $214.00  10.70  

Amount paid at Blackcomb, B.C.

 $

At Mont Tremblant, Que. Cost of ski pass  $214.00 GST  5% of $214.00  0.05(214.00)  10.70 PST  9.975% of $214.00  0.09975(214.00)  21.35 Amount paid at Mont Tremblant

 $246.05

Difference  246.05  239.68

 $ 6.37

Total cost in Toronto Retail price HST  13% of $625  0.13(625) Total cost in Toronto

 $625.00  81.25  $706.25

Total cost in Calgary Retail price GST  5% of $625  0.05(625.00) PST 

 $625.00  31.25 nil

Total cost in Calgary

 $656.25

Difference  PST

$ 50.00

Purchase price of the first item = $70.56  0.25 = $17.64 Purchase price of the second item, including 5% GST = 70.56 – 17.64 = $52.92 Purchase price of the second item = $52.92 1.05  $50.40 GST paid = $52.92 – 50.40 = $2.52

 22.751  Property tax  125, 000    $2843.88  1000 

Property tax  307,500(0.019368)  $5955.66

2216  0.004626 479, 000 Semi-annual tax rate  0.004626(1000)  4.626305 The annual tax rate  2(4.626305)  9.252610

10. Semi-annual tax rate 

 9.25 mills 11. (a) Total expenditure  $(3, 050, 000  2, 000, 000  250, 000  700, 000  850, 000)  $6,850, 000

Total residential property tax  0.80(6,850, 000)  $5, 480, 000 (b) Tax rate per $1000 

5, 480,000 (1000)  10.96 500,000,000

 10.96  (c) Property tax  $375, 000    $4110 1000  

Business Math News Box

1. There are 52 weeks per year during which the employee works a 40-hour week. Total hours worked during the year is 52 × 40 = 2080. Hourly Rate Calculations Location Vancouver Calgary Toronto Montreal National Average

Financial Controller Human Resources Manager 99,500/2080 = $47.84 88,324/2080 = $42.46 106,082/2080 = $51.00 88,611/2080 = $42.60 98,500/2080 = $47.36 83,350/2080 = $40.07 99,758/2080 = $47.96 80,641/2080 = $38.77 99,234/2080 = $47.71 78,669/2080 = $37.82

Marketing Manager 78,663/2080 = $37.82 84,836/2080 = $40.79 77,823/2080 = $37.41 76,554/2080 = $36.80 75,450/2080 = $36.27

2. Dollar and percentage differences by job function:

Financial Controller National Average $99,234 National Average $99,234 National Average $99,234 National Average $99,234

–

Vancouver $99,500 Calgary $1,06,082 Toronto $98,500 Montreal $99,758

= = = = = = = =

$ difference $266 $ difference $6848 $ difference ($734) $ difference $524

– – – – – – – –

Vancouver $88,324 Calgary $88,611 Toronto $83,350 Montreal $80,641

= = = = = = = =

$ difference $9655 $ difference $9942 $ difference $4681 $ difference $1972

– – – – – – – –

Vancouver $78,663 Calgary $84,836 Toronto $77,823 Montreal $76,554

= = = = = = = =

$ difference $3213 $ difference $9386 $ difference $2373 $ difference $1104

– – – – – – –

= 266/99,234 = 6848/99,234 = –734/99,234 = 524/99,234

% difference 0.002681 0.27 % difference 0.069009 6.90 % difference –0.007397 –0.74 % difference 0.005280 0.53

Human Resources Manager National Average $78,669 National Average $78,669 National Average $78,669 National Average $78,669

= 9655/78,669 = 9942/78,669 = 4681/78,669 = 1972/78,669

% difference 0.122729 12.2 % difference 0.126378 12.6 % difference 0.059502 5.9 % difference 0.025067 2.5

Marketing Manager National Average $75,450 National Average $75,450 National Average $75,450 National Average $75,450

= 3213/75,450 = 9386/75,450 = 2373/75,450 = 1104/75,450

% difference 0.042584 4.26% % difference 0.124400 12.44% % difference 0.031451 3.15% % difference 0.014632 1.46%

3. Discrepancies between the national averages and specific metropolitan centres might be the result of many factors, including: - National average takes into account data supplied from all geographic locations. - Lack of supply and/or high demand for specific jobs in geographic locations might cause salaries to exceed the national average.

Review Exercise 1.

(a) 32  24  8  32  3  29 (b) (48 18) 15 10  30 15 10  2 10  8 (c) (8  6  4)  (16  4  3)  (48  4)  (16  12)  44  4  11 (d) 9(6  2)  4(3  4)  9(4)  4(7)  36  28  8 (e)

108 108 108    $1520.83 216 0.12  365 0.12  0.591781 0.071014

(f)

288 288 288    0.15 292 2400  365 2400  0.8 1920

225   (g) 320 1  0.10    320(1  0.061644)  320(1.061644)  339.73 365   150   (h) 1000 1  0.12    1000(1  0.049315)  1000(0.950685)  950.68 365  

(i)

660 660 660    625.45 144 1  0.14  365 1  0.055233 1.055233

(j)

1120 1120 1120    1250 292 1  0.13  365 1  0.104 0.896

(a) 185%  1.85 (b) 7.5%  0.075 (c) 0.4%  0.004 (d) 0.025%  0.00025

1 (e) 1 %  1.25%  0.0125 4 (f )

3 %  0.75%  0.0075 4

1 (g) 162 %  162.5%  1.625 2 3 (h) 11 %  11.75%  0.1175 4 1 8.3& (i) 8 %   0.083& 3 100 1 83.3& (j) 83 %   0.83& 3 100 2 266.6& (k) 266 %   2.6& 3 100

3 (l) 10 %  10.375%  0.10375 8 3.

(a) 50% 

50 1  100 2

1 37.5 375 3 (b) 37 %    2 100 1000 8 16 2 503 2 1 (c) 16 %  3  100  3 100 6 1 100  66 23 2 2 5  1  (d) 166 %  3 100 3 3 (e)

1 1 1 1 1 % 2    2 100 2 100 200

(f ) 7.5% 

7.5 75 3   100 1000 40

3 3 (g) 0.75%  %  4 400

(h)

5 5 1 %  8 800 160

(a) 2.25  2.25 100  225% (b) 0.02  0.02 100  2% (c) 0.009  0.009 100  0.9% (d) 0.1275  0.1275 100  12.75%

(e)

5 5  100  125% 4 4

(f )

11  1.375  1.375 100  137.5% 8

(g)

5  0.025  0.025 100  2.5% 200

(h)

7 28   28% 25 100

(a)

150% of 140  1.5 140  210

(b)

3% of 240  0.03  240  7.2

(c)

(d)

3 9 % of 2000 4  0.0975  2000  195 0.9% of 400  0.009  400  3.6

1 3 1 5 (a) 4  3  5  6 3 4 2 8 &  4.3& 3.75  5.5  6.625  20.2083kg (b) 20.2083&1.20  $24.25 (c) 20.2083& 4  5.052083& 5.05 kg

(d) 24.25  4  6.0625  $6.06 7.

56  $0.625



$ 35.00

1 180  83 ¢  180  $0.83& 3 126  $1.16&



$150.00



$147.00

144  $1.75



$252.00

Total



$584.00

(a)

30.45  20.20  16.40  15.50 82.55   20.6375  $20.64 4 4

(b) 30.45  2  20.20  6  16.40  9  15.50 13  30 

$ 60.90 $121.20 $147.60 $201.50 $531.20

Average rate 

$531.20  $17.71 30

January 1  May 31: 15, 000  5  $ 75, 000 June 1  July 31: 13, 000  2  26, 000 August 1  October 31: 11,500  3  34,500 November 1  December 31: 15,500  2  31, 000 12  $166,500 $166,500 Average monthly investment   $13,875 12 Total

10.

January 1  March 31: April 1  May 31: June 1  September 30: October 1  December 31:

12, 000  3  $ 36, 000 14, 400  2  28,800 12,960  4  51,840 15,840  3  47,520

12  $164,160 $164,160 Average monthly investment   $ 13, 680 12 Total

11. (a) Monthly remuneration 

34,944  $2192 12

(b) Weekly pay  34,944  52  $672 Hourly rate  672  35  $19.20

Regular gross pay Overtime pay

 3387.20  2912.00  475.20

Overtime hours  475.20  (19.20 1.5)  16.5

12. (a) Semi-monthly pay  31, 487.04  24  $1311.96 (b) Weekly pay  31, 487.04  52  $605.52 Hourly rate  605.52  36  $16.82

 

Overtime pay  12 16.82 1.5

$1311.96 $302.76

 $1614.72

Gross earnings 13. (a) Gross sales  11,160 Less: returns  120 Net sales  11, 040

Commission: 4% of $6000 8% of $3000 12.5% of $[11, 040  9000]

 $ 240  240  255

Gross earnings



$735

(b) Average hourly rate  735  43  $17.09 14. (a) Regular earnings  44 15.80  $ 695.20

Overtime pay  6.5 15.80 1.5  154.05 Gross earnings

 $849.25

(b) Overtime premium  6.5 15.80  0.5  $51.35 15. (a) Base salary on quota of $8000  $540.00

Commission  4.75% on $3340  158.65  $698.65

Gross earnings

(b) Hourly rate  698.65  35  $19.96 16. Gross earnings

Base salary

 $741.30  $675.00

Commission  $66.30 Commission sales  6560  5000  $1560 Rate of commission  66.30  1560  0.0425  4.25%

17.

Net sales  2101.05  0.105  $20, 010 Net sales  Gross sales  Returns 20, 010  Gross sales  8% of Gross sales 20, 010  92% of Gross sales 20, 010 Gross sales   $21, 750 0.92

18.

Hours worked  47 Regular hours  40 Overtime hours  7 7 overtime hours are equivalent to 7 1.5  10.5 regular hours. Total hours paid at regular rate  40  10.5  50.5 779.72 Hourly rate of pay   $15.44 50.5

19. (a) Annual salary  1413.75  24  $33,930

Weekly pay  33,930  52  $652.50 Hourly rate of pay  652.50  37.5  $17.40 (b) Gross earnings  $1552.55

Regular earnings  1413.75 Overtime pay  $138.80 Overtime hourly rate  17.40  1.5  $26.10 Overtime hours  138.80  26.10  5.318008  5.32 20. Gross earnings  $728.54 Less: base salary  $680.00

Commission  $ 48.54 Commission sales  48.54  0.06  $809 Net sales  5000  809  $5809 Gross sales  5809  136  $5945

 $731.92 21. Gross earnings Regular earnings  35 15.80  553.00 Overtime pay  $178.92 Overtime hours  178.92  (15.80  1.5)  7.549367 Number of hours worked  35  7.549367  42.55

22. GST collected  5% of $76, 000  0.05(76, 000)  $3800

 5% of $14,960  0.05(14,960) 

GST paid

GST remittance

748

$3052

23. GST collected:

Parts : 5% of $175, 000 Labour : 5% of $165, 650 Total : 5% of $340, 650 = 0.05(340, 650) GST paid : Parking fees : 5% of $ 2000 Supplies : 5% of $55, 000 Utilities : 5% of $ 4000 Other : 5% of $ 3300 Total :

 $17, 032.50

5% of $64,300 = 0.05(64,300) = $ 3215.00 GST remittance

$13,817.50

24. Amount paid in Kelowna, B.C.  Retail price  5% GST  7% PST = 1868  0.05(1868)  0.07(1868) = 1868  93.40  130.76  2092.16

Amount paid in Kenora, Ont.  Retail price  13% HST  1868  0.13(1868)  1868  242.84  2110.84 The difference = 2110.84  2092.16 = $18.68

 10.051  25. Property tax in Ripley  350, 000    $3517.85  1000   12.124  Property tax in Amberly  335, 000    $4061.54  1000 

The person in Amberly pays $543.69 more in property tax.

26. (a) Tax rate =

15,567,000 (1000)  15.957970 975,500,000

 15.957970  (b) Property tax = 435, 000    $6941.72  1000 

2, 000, 000 (1000)  2.050231 975,500, 000

 2.050231  (d) Additional property tax = 435, 000    $891.85  1000 

Self-Test 1.

45   (a) 4320 1  0.18    4320(1  0.022192)  4415.87 365   105   (b) 2160  0.15    2160(0.043151)  93.21 365   285   (c) 2880 1  0.12    2880(1  0.093699)  2610.15 365  

(d)

410.40 410.40   4623.33 0.24  135 0.088767 365

(e)

5124 5124   5489.46 270 1  0.09  365 0.933424658

(a) 175% 

(b)

175  1.75 100

3 3 1 3 %    0.00375 8 8 100 800

1 5 5 1 5 1   (a) 2 %  %   2 2 2 100 200 40 50 16 23 2 2 50 1 7 (b) 116 %  100%  16 %  1   1 3  1  1  3 3 100 100 300 6 6

(a) 1.125  1.125  100  112.5% (b)

9  0.0225  0.0225  100  2.25% 400

72  $1.25  $ 90.00 2 84 16 ¢ = 84  $0.16&  $ 14.00 3 40  $0.875  $ 35.00 48  $1.33& 48  $1.3&  $ 64.00 Total

$203.00

5  $9 6  $7 3  $8 6  $6

 $ 45  $ 42  $ 24  $ 36

Total 20

 $147

Average cost 

147  $7.35 20

1 3 5  1 Total size =  5  6  4  3  sq. metres 3 8 6  4 & & sq. metres  (5.25  6.3  4.375  3.83) = 19.7916&sq. metres Sales value = 25,120 19.7916& = $497,166.67

January 1  February 28 : March 1  July 31: August 1  September 30 : October 1  December 31:

7200  2  $14, 400 6720  5  33, 600 7320  2  14, 640 7440  3  22,320

Total

Average monthly balance = 9.

$84,960

84,960  $7080 12

Annual salary  2080  24 = $49,920 Weekly pay  49,920  52  $960 Hourly rate of pay  960  40  $24

10.

Net sales  0.885  5880  $5203.80 806.59 Commission rate   0.155  15.5% 5203.80

11.

Weekly pay  52,956.80  52 = $1018.40 Hourly pay  1018.40  38 = $26.80 Regular monthly pay  52,956.80  12  $4413.07 Overtime earnings  26.80  8.75 1.5  351.75  $4764.82

Gross pay

12. Total hours  8.25  8.25  9.5  11.5  7.25  44.75

Regular hours  8  8  8  8  7.25  39.25 Overtime hours  0.25  0.25  1.5  3.5  5.50 Regular pay  39.25  16.60  $651.55 Overtime pay  5.5  16.60  1.5  136.95  $788.50

Gross earnings 13.

Total hours  52.5 Regular hours  44.0 Overtime hours  8.5

At time-and-a-half, 8.5 overtime hours are equivalent to 8.5 1.5  12.75 regular hours

Hourly rate of pay  14.

983.15  $17.32 56.75

Base salary on first $4500  $600 Commission on next $2000 = 0.11 2000  220 Commission on additional sales = (8280  6500)  0.15  1780  0.15  267  $1087

Gross earnings

$6400  $20 GST 5% of $6420 $321.00 Manitoba PST 7% of $6420 449.40

15. Total value

Total purchase price

16.

Purchase price Less discount Net price Add shipping charge Total cost before taxes

 $6420.00 770.40 $7190.40

$ 17.95 2.50 $15.45 1.45 $16.90

HST

15% of $16.90  2.535

Final purchase price is 17.

18.

 $2.54 $19.44

= Assessed Value  Tax Rate 18 4502.50  Assessed Value  1000 4502.50(1000) Assessed Value =  $250,139 18 Property Tax

2 Assessed value   $390, 000  $260, 000 3 12.5 Property tax  $260, 000   $3250 1000

Challenge Problems 1.

Purchase price of the first item = $821.40  0.29 = $238.206 Purchase price of the second item, including 5% GST and 7% PST = $821.40 – 238.206 = $583.194 Purchase price of the second item = $583.194 / 1.12 = $520.708929 Total GST paid = $520.708929(0.05) = $26.035446 = $26.04 BC PST paid on second item = $520.708929(0.07) = $36.449625 = $36.45 BC PST paid on first item = ($238.206 / 1.07)(0.07) = $15.583570 = $15.58 Total BC PST paid = $36.45 + $15.58 = $52.03

2. Test score

Weight

Final grade contribution

Test 1

30%

60(0.30) = 18

Test 2

30%

50(0.30)  15

Final exam

40%

Final mark

Final exam contribution to final mark  70  (18  15)  70  33  37 37 Final examination mark required   92.5% 0.40 Case Study 1.

HST collected

13% of $280,000

$36,400

HST paid HST remittance

13% of $ 40,000

$ 5200 $31,200

(a) HST by Quick Method HST on sales = 185,000  0.13 = $24,050 Purchases: Goods for resale (185,000 × 47%) × 1.13 = $98,253.50 Other expenses (48,000 – 42,000) × 1.13 = Total taxable goods and expenses

6,780.00 105,033.50

Input tax credits = 13/113 × 105,033.50 =

$12,083.50

Remittance by Quick Method: $24,050.00 – 12,083.50 = $11,966.50 (b) HST by Standard Method HST collected 13% of $185,000

$24,050.50

HST paid on purchases and taxable services 13% of (47% of $185,000)

$11,303.50

13% of ($48,000 – $42,000)

780.00

Remittance by Standard Method

12,083.50 $11,966.50

Line 101 Line 103 Line 104 Line 105 Line 106 Line 107 Line 108 Line 109 Line 110 Line 111 Line 112 Line 113 Line 114 Line 115

13% of $486,530

13% of $239,690

63,248.90 – 31,159.70 3120 × (12)

32,089.20 – 37,440

$486,530.00 63,248.90 0.00 63,248.90 31,159.70 0.00 31,159.70 32,089.20 37,440.00 0.00 37,440.00 –5350.80 5350.80 0

Refund Claimed is $5350.80

Chapter 2

Review of Basic Algebra

Exercise 2.1 A. 1.

19a

a  10

3a 14

2 x  4 y

3p  q

14 f  4v

2c  3d

0.8x

10. 1.06x 11. 1.4x 12. 0.98x 13. 2.79x 14. 4.05 y 15.  x2  x  8 16. ax  x  2 17. 2 x  3 y  x  4 y  x  7 y 18. 4  5a  2  3a  2a  2 19. 12b  4c  9  8  8b  2c 15  4b  2c  2

20. a 2  ab  b2  3a 2  5ab  4b2  2a 2  6ab  5b2 21. 3m2  4m  5  4  2m  2m2  m2  6m  1 22. 6  4 x  3 y  1  5 x  2 y  9  14  9 x  y 23. 7a  5b  3a  4b  5b  10a 14b 24. 3 f  f 2  fg  f  3 f 2  2 fg  2 f  2 f 2  3 fg 25. 4b4 d  2ac 7  (5b 4 d )  3ac

 9b4 d  2ac7  3ac 26. (8t 2  6t  9)  (7t 2  6t  7)

 8t 2  6t  9  7t 2  6t  7  t 2  16 27.

18 y 12 1  3 y 2 5 4

 9 y  2.4  3.25 y  12.25 y  2.4 x 28. 1.3x  x 2   2 x  4 2

 x 2  (1.3  0.5  2) x  4  x 2  0.2 x  4 29.

k k  (1  0.05) (1  0.05)2  0.952381k  0.907029k  1.859410k

142  x  30. x 1  0.052   91  365    1  0.052   365  

B. 1.

12x

56a

10ax

27ab

2x2

24m 2

60xy

24abc

2 x  4 y

10. 10 x  20 11. 2ax 2  3ax  a 12. 24 x  12bx  6b2 x 13. 20 x  24  6  15x  35x  30 14. 24a  3b  14a  18b  10a  15b 15. 15ax  3a  5a  2ax  3ax  3a  20ax  5a 16. 24 y  32  4 y  2  1  y  21y  31 17. 3x 2  x  6 x  2  3x 2  5x  2 18. 5m2  2mn  15mn  6n2  5m2  17mn  6n2 19. x3  x2 y  xy 2  x2 y  xy 2  y3  x3  y3 20. a3  2a 2  a  a 2  2a  1  a3  3a 2  3a  1 21. 10 x 2  8x  5x  4  3x 2  21x  5x  35  7 x 2  3x  39

22. 2(2a 2  2a  3a  3)  3(3a 2  2a  3a  2)

 4a 2  10a  6  9a 2  3a  6  5a 2  13a  12 23. 3x 2 ( x 2  2 x  3)  4 x( x 2  2 x  3)  ( x 2  2 x  3)

 3x 4  6 x3  9 x 2  4 x3  8 x 2  12 x  x 2  2 x  3  3x 4  2 x3  16 x 2  14 x  3 24. (5b2  5b  5)(b3  4b  2)  5b2 (b3  4b  2)  5b(b3  4b  2)  5(b3  4b  2)  5b5  20b3  10b 2  5b 4  20b 2  10b  5b3  20b  10  5b5  5b 4  15b3  30b 2  10b  10

25. 4ab 26. 5 y 27. 4x 28. 6 29. 10m  4 30. 2 x  3 31. 2 x 2  3x  6 32. a 2  4a  3 C. 1. 3x  2 y  3  3(4)  2(5)  3  12  10  3  5 2.

1 1 (3x2  x  1)  (5  2 x  x 2 ) 2 4

1 1  [3(3) 2  (3)  1]  [5  2(3)  (3) 2 ] 2 4 1 1  (27  3  1)  (5  6  9) 2 4 1 1  (29)  (2) 2 4  14.5  0.5  14 3. ( pq  vq)  f  ( p  v)q  f  (12  7)2000  4500  10,000  4500  5500 4. F /C  13,000/0.65  20,000 5. (1  d1 )(1  d 2 )(1  d3 )  (1  0.35)(1  0.08)(1  0.02)  (0.65)(0.92)(0.98)  0.58604 6. C  0.38C  0.24C  (1  0.38  0.24)C  1.62C  1.62 ($25)  $40.50 7.

RP(n  1) 0.21 $1200  (77  1)   $378 2N 2  26

I 63 63    0.125 219 Pt 840  365 840  0.60

I $198 $198    $3000 146 rt 0.165  365 0.165  0.40

10.

2NC 2  52  60 2  52  60    0.13 P(n  1) 1800(25  1) 1800  26

76   11. P(1  rt )  $880 1  0.12   365  

 $880(1  0.024986)  $880(1.024986)  $901.99 256   12. FV(1  rt )  $1200 1  0.175   365  

 $1200(1  0.122740)  $1200(0.877260)  $1052.71 13.

P $1253 $1253 $1253     $1400.06 284 1  dt 1  0.135  365 1  0.105041 0.894959

14.

S $1752 $1752 $1752     $1600.08 228 1  rt 1  0.152  365 1  0.094948 1.094948

t   15. S 1  r  365  

for

S  3240, r  0.125, t  290

  290    3240 1  (0.125)    365     3240 (1.099315)  3561.780822 16. (SP  X )  FC  (VC  X ) for SP  13, X  125, FC  875, VC  L

 (13 125)  875  (4 125)  1625  875  500  250 17. (1  i)m  1 for i  0.0275, m  2  (1  0.0275) 2  1  1.055756  1

 0.055756  (1  i ) n  1  PmT 18.   for PmT  500, i  0.025, n  2 i  

 (1  0.025) 2  1   500   0.025    0.050625   500   0.025   500 (2.025)  1012.50 19. 1  [(1  d1 )(1  d 2 )] for d1  0.15, d 2  0.04

 1  [(1  0.15)(1  0.04)]  1  [(0.85)(0.96)]  1  (0.816)  0.184

 ( FV )(i )  20.   for n  (1  i)  1 

FV  10, 000, i  0.0075, n  20

 (10, 000)(0.0075)    20  (1  0.0075)  1  75      0.161184   465.306319 Exercise 2.2 A. 1.

16 81

625 1296



1 64



8 27

0.25

10. 113.379904 11. 0.001 12. 335.544320 13. 1

14. 1 15.

1 9

16. 512 17. 

18.

1 125

1 167.9616

19. 125 20.

81 16

21.

1 1.01

22. 1 23. 11.526683 24.

1 1  1 0 (1.07) 1

25.

1 1   0.781198 10 (1  0.025) 1.280085

26. 100(1  0.0225)7  100(1.168539)  116.853901 27. 425(1  0.16)4  425(0.552291)  234.723717 0.5

 1500  28.    1  2.738613  1  1.738613  200  (1  0.03) 29. 0.03



2.093778  69.792598 0.03

1  (1.01) 20  1  0.819544 0.180456    18.045553 30.  0.01  0.01 0.01 

B. 1. 25  23  253  28 2. (4)3  (4)  (4)31  (4)4 3. 47  44  474  43 4. (3)9  (3)7  (3)97  (3) 2 5. (23 )5  235  215 6

6. (4)3   (4)36  (4)18 7. a 4  a10  a 410  a14 8. m12  m7  m127  m5 9. 34  36  3  3461  311 10. (1)3 (1)7 (1)5  (1)375  (1)15

67  63 11.  6 7  3 9  6 9 6 12.

( x 4 )( x5 )  x 4 57  x 2 7 x 4

3 3 3 13.        5 5 5 5

47

1 1 1 14.         6 6 6 6

311  11 5

5 3

1 62



1 6  4

 3  3   3   3  15.              2  2   2   2  8

87

 3  3  3 16.             4  4  4

 

(3)11  211

3 4

17. (1.025)80 (1.025)70  (1.025)8070  1.025150 18. 1.005240 1.005150  1.005240150  1.00590

19. 1.0420   1.04204  1.0480 3

 3 5   3 53 315 20.          15 7  7    7  21. (1  i)100 (1  i)100  (1  i)100100  (1  i)200 22. (1  r )2 (1  r )2 (1  r )2  (1  r )222  (1  r )6 2

23. (1  i)80   (1  i)802  (1  i)160 3

24. (1  r )40   (1  r )403  (1  r )120 25. (ab)5  a5b5 26. (2 xy)4  16 x4 y 4 27. (m3n)8  m24 n8 4

 a3b2  a12b8 28.    x4  x  29. 23  25  24  2354  24 30. 52  53  52( 3)  55 8

b8 a 31.    8 a b

 1 i  32.    i 

n

in  (1  i)n

Exercise 2.3

5184  72.0000

A. 1.

205.9225  14.3500

2. 3.

2187  3.0000

1.1046221  1.0100

4.3184  1.075886  1.0759

0.00001526  0.500002  0.5000

1.0825  1.0133

1.15  1.011715  1.0117 1

B. 1. 3025 2  55 1

2. 24014  7 2

3. 525.21875 5  12.25 4

4. 21.6 3  60.154991 5.

1.1257  1.071122

1.095  1.015241  1  

7. 4 3  

1 4

8. 1.06

1 3

 1    12 



1  0.629961 1.587401



1 1.06

1 12



1  0.995156 1.004868

1.0360  1 5.891603  1   163.053437 0.03 0.03

10.

1  1.0536 1  0.172657   16.546852 0.05 0.05

11. 2.158925 12. 0.589664  3.536138  1  13. 26.50(1.043)    26.50(1.043)(58.979962)  1630.176673 0.043  

 2.653298  1  14. 350 (1.05)    350 (1.05)(33.065954)  12,151.73813 0.05  

 1  0.520035  15. 133    133(8.570795)  1139.915716 0.056    1  0.759412  16. 270    270 (6.873956)  1855.967995 0.035    1  0.581251  17. 5000 (0.581251)  137.50    0.0275 

 2906.252832  137.50(15.227252)  2906.252832  2093.747168  5000  1  0.623167  18. 1000 (0.623167)  300   0.03  

 623.166939  300(12.561102)  623.166939  3768.330608  4391.497547 19. 112.55  100(1  i) 4

(1  i ) 4  1.1255 (1  i )  1.12550.25 (1  i )  1.029998 i  0.029998 20. 380.47  300(1  i)12

(1  i )12  1.268233 (1  i )  1.2682330.083 (1  i )  1.019999 i  0.019999

21. 3036.77  2400(1  i)6

(1  i )6  1.265321 (1  i )  1.2653210.16 (1  i )  1.04 i  0.04 22. 1453.36  800(1  i)60

(1  i )60  1.8167 (1  i )  1.8167 0.016 (1  i )  1.01 i  0.01 Exercise 2.4 A. 1. 29  512 9  log 2 512

2. 37  2187 7  log 3 2187

3. 53 

1 125

3  log5

1 125

4. 105  0.00001 5  log10 0.00001

5. e2 j  18

2 j  log e 18 or 2 j  ln 18 6. e3 x  12

 3x  log e 12 or 3x  ln 12 B. 1. log 2 32  5

25  32

2. log3

1  4 81 34 

1 81

3. log10 10  1

101  10 4. ln e 2  2

e2  e2 C. 1. ln 2  0.693147 2. ln 200  5.298317 3. ln 0.105  2.253795 4. ln 300(1.1015 )   ln 300  ln 1.1015

 ln 300  15(ln 1.10)  5.703782  15(0.095310)  5.703782  1.429653  7.133435  2000   ln 2000  ln 1.099 5. ln  9 1.09 

 ln 2000  9(ln 1.09)  7.600902  9(0.086178)  7.600902  0.775599  6.825303 1.01120   ln 850  ln 1.01120  ln 0.01 6. ln 850    0.01 

 ln 850  120(ln 1.01)  ln 0.01  6.745236  120(0.009950)  (4.605170)  6.745236  1.194040  4.605170  10.156367

Business Math News Box 1.

Total amount invested  $1200 Total number of shares purchased  10 + 10.225 + 9.615 + 10.395 + 9.524 + 9.302 + 10.132 + 9.302  9.009  8.696  8.849  8.888  113.937 Average cost per share 

1200  $10.53 113.937

$10.53 is less than the current $11.25 cost per share. 2. Number of shares purchased  Share price  Amount invested 10 shares  $17  $170 a  16 15 shares   240  b 16.50  20 shares  330 c $740 Average cost per share 

740  $16.44 45

3. Amount invested  Share price  Number of shares purchased $5000  $25  200 156.25 5000  32  5000





250 606.25

Average cost per share 

15, 000  $24.74 606.25

The first $5000 allocation purchased 200 shares at $25 per share. The second $5000 allocation only bought 156.25 shares because the price rose to $32 per share in the second month. The third $5000 allocation bought 250 shares at $20 per share. After three months, the couple owned 606.25 shares at an average cost of $24.74. Their investment is worth $15,156.25. (i.e., 606.25 shares  $25 current value). If they had invested $15,000 all at once, they would only have 600 shares. At the current share price, their investment would only be worth $15,000, the same as the original lump sum. 4. Answers will vary. However, markets tend to go up in the long term.

Exercise 2.5 A. 1. 15 x  45 x  3

2. 7 x  35

x  5 3. 0.9 x  72

x  80 4. 0.02 x  13

x  650 5.

1 x3 6 x  18

1 6.  x  7 8

x  56 7.

3 x  21 5 1 x  7 5 x  35

4 8.  x  32 3 1 x8 3 x  24

9. x  3  7

x  4 10. 2 x  7  3x

x7 11. x  6  2

x  8

12. 3x  9  2 x

x9 13. 4  x  9  2 x

x5 14. 2 x  7  x  5

x  12 15. x  0.6 x  32

1.6 x  32 x  20 16. x  0.3x  210

0.7 x  210 x  300 17. x  0.04 x  192

0.96 x  192 x  200 18. x  0.07 x  64.20

1.07 x  64.20 x  60 B. 1. 3x  5  7 x 11

4 x  16 x4 LS: 3 x  5  3(4)  5  12  5  17 RS: 7 x  11  7(4)  11  28  11  17

2. 5  4 x  4  x

 3x  9 x3 LS: 5  4 x  5  (4)(3)  5  12  7 RS:  4  x  4  3  7

3. 2  3x  9  2 x  7  3x

 3x  7  5 x  7  8x  0 x0 LS:  2  3x  9  2  3(0)  9  7 RS:  2 x  7  3x  2(0)  7  3(0)  7 4. 4 x  8  9 x  10  2 x  4

 5x  8  6  2 x  7 x  14 x  2 LS:  4 x  8  9 x  4(2)  8  9(2)  8  8  18  2 RS:  10  2 x  4  10  2(2)  4  10  4  4  2 5. 3x  14  4 x  9

 x  5 x5 LS:  3x  14  3(5)  14  15  14  29 RS:  4 x  9  4(5)  9  20  9  29

6. 16 x 12  6 x  32

10 x  20 x  2 LS:  16 x  12  16(2)  12  32  12  44 RS:  6 x  32  6(2)  32  12  32  44 7. 5  3  4 x  5x  12  25

4 x  5 x  12  25  5  3  x  21 x  21 LS:  5  3  4 x  8  4(21)  8  84  92 RS:  5 x  12  25  5(21)  13  105  13  92 8. 3  2 x  5  5x  36  14

 2 x  5 x  36  14  3  5  3x  24 x 8 LS:  3  2 x  5  3  2(8)  5  3  16  5  18 RS:  5 x  36  14  5(8)  36  14  40  36  14  18

x  50  100  0.34 x  0.21x x  50  100  0.55 x x  0.55 x  150 (1  0.55) x  150 0.45 x  150

x  333.3 CHECK: L.S.

R.S.

333.3  50

100  0.34(333.3)  0.21(333.3)

 283.3

 100  113.3  70  283.3

10.



23 x 32

6 12 x 8  x 1.125 9 0.8 x  0.8 x  0

1  0.25 

x  all real numbers CHECK: L.S. 1

1  0.25 

R.S. 23 (1) 32 8  (1) 9  0.8

6 12

1 1.125  0.8 

Exercise 2.6 A. 1. 12 x  4(9 x  20)  320 12 x  36 x  80  320

 24 x  240 x  10 LS  12(10)  4[9(10)  20]  120  4[90  20]  120  440  320 RS  320

2. 5( x  4)  3(2  3x)  54 5 x  20  6  9 x  54

14 x  26  54 14 x  28 x  2

LS  5[2  4]  3[2  3(2)]  5(6)  3(2  6)  30  24  54 RS  54 3. 3(2 x  5)  2(2 x  3)  15

6 x  15  4 x  6  15 2 x  9  15 2 x  6 x  3 LS  3[2(3)  5]  2[2(3)  3]  3[65]  2[6  3]  3(11)  2(9)  33  18  15 RS  15 4. 17  3(2 x  7)  7 x  3(2 x  1)

17  6 x  21  7 x  6 x  3  6 x  38  x  3  7 x  35 x5 LS  17  3[2(5)  7]  17  3[10  7]  17  9  8 RS  7(5)  3[2(5)  1]  35  3[10  1]  35  27  8 5. 4 x  2(2 x  3)  18

4 x  4 x  6  18 8 x  24 x3

LS  4(3)  2[2(3)  3]  12  2[6  3]  12  6  18 RS  18 6. 3(1  11x)  (8 x  15)  187

 3  33x  8 x  15  187 33x  8 x  187  3  15 41x  205 x5 LS  3[(1  11(5)]  [8(5)  15]  3[54]  25  162  25  187 RS  187 7. 10 x  4(2 x  1)  32

10 x  8 x  4  32 2 x  28 x  14 LS  10(14)  4[2(14)  1]  140  4[27]  140  108  32 RS  32

8. 2( x  4)  12(3  2 x)  8

 2 x  8  36  24 x  8  2 x  24 x  8  8  36  26 x  52 x2 LS  2(2  4)  12[3  2(2)]  4  8  36  48  8 RS  8

65   9. x 1  0.12    1225.64 365   1.021370 x  1225.64 x  1199.996245

x  1200 CHECK:

L.S.

65   1200 1  0.12   365    1200 1.021370 

R.S. 1225.64

 1225.643836 10. x 

x x 1000   3148  2 1.25 (1.25) (1.25)3

x  0.80 x  0.64 x  3148  512 2.44 x  3660

x  1500 CHECK: L.S. 1500 1500 1500   1.25 (1.25) 2  1500  1200  960

3660

 3660 B. 1.

1 x  x  15 4 4 x  x  60 3 x  60 x  20

5 x  x  26 8 8 x  5 x  208 13x  208 x  16

2 1 7 5 x   x 3 4 4 6

8 x  3  21  10 x 18 x  18 x  1 4.

5 2 1 1  x  x 3 5 6 30

50  12 x  5 x  1  17 x  51 x3 5.

3 113 2 x4  x 4 24 3

18 x  96  113  16 x 34 x  17 x

1 2

3 2 31 2 x  x 2 3 9

36  27 x  12 x  62  39 x  26 x

2 3

1 2 1  x  15  x 3 3 2   1   3 1  x   3 15  x  3   3   3  x  45  2 x 42  3x x  14

3x  2 2 x  1  5 3 3(3 x  2)  5(2 x  1) 9 x  6  10 x  5

x  1 9.

21 2 11 1  x  x 8 5 4 10 1  21 2  11 40   x   40  x   10  8 5  4 105  16 x  110 x  4 109  126 x x

10.

109 126

2 1 3 1  x x   x 3 12 4 24 1   2 3 1  24   x  x   24   x  12   3  4 24  16 x  2 x  18  x 18 x  18  x 18  19 x x

C. 1.

18 19

3 1 55 (2 x  1)  (5  2 x)   4 3 12 9(2 x  1)  4(5  2 x)  55 18 x  9  20  8 x  55 26 x  29  55 26 x  26 x  1

4 53 3 7 (4  3x)   x  (2 x  3) 5 40 10 8

32(4  3x)  53  12 x  35(2 x  3) 128  96 x  53  12 x  70 x  105  96 x  181  58 x  105  38 x  76 x2

2 3 20 (2 x  1)  (3  2 x)  2 x  3 4 9

24(2 x  1)  27(3  2 x)  72 x  80 48 x  24  81  54 x  72 x  80 102 x  105  72 x  80 30 x  25 x

5 6

4 3 11 (3x  2)  (4 x  3)   3x 3 5 60

80(3x  2)  36(4 x  3)  11  180 x 240 x  160  144 x  108  11  180 x 96 x  52  11  180 x  84 x  63 x

3 4

2 3 (5 x  1)   ( x  2) 3 5 2   3  15  (5 x  1)   15   ( x  2)  3   5  10(5 x  1)  9( x  2) 50 x  10  9 x  18 59 x  8 x

8 59

y  mx  b

D. 1.

y  b  mx x

r

y b m

M S

Sr  M S

M r

PV 

PMT i

PMT  PVi 4.

I  P rt t

I Pr

S  P(1  rt ) S  1  rt P S  1  rt P S 1 r P t SP r P t SP r Pt

PV  FV(1  i) n

PV  (1  i )  n FV 

 PV  n  FV   1  i 1

 FV  n  PV   1  i 1

 FV  n i 1  PV 

S for t (1  rt ) P(1  rt )  S P  Prt  S Prt  S  P SP t Pr

7. P 

N  L(1  d ) for d N  L  dL dL  L  N

d

LN L

f  (1  i ) m  1 for i (1  f )  (1  i) m 1

(1  f ) m  ((1  i) m ) m 1

(1  f ) m  1  i 1

i  (1  f ) m  1 10. FV  PV (1  i)n

for n

FV  (1  i ) n PV  FV  ln    n ln (1  i )  PV 

 FV  ln   PV  n  ln 1  i  Exercise 2.7 1. Let the cost be $x. 3   Selling price  $  x  x  4  

3 x  49.49 4 4 x  3x  197.96 7 x  197.96 x  28.28

x 

The cost was $28.28.

2. Let the regular selling price be $x. 1   Sale price  $  x  x  3  

1  x  x  576 3 3x  x  1728 2 x  1728 x  864 The regular selling price was $864. 3. Let the price be $x. Total  $ x  0.05x  x  0.05 x  $57.75 1.05 x  $57.75 x  55

The price was $55. 4. Let the regular price be $x. Sale price  $( x  0.40 x)  x  0.40 x  11.34 0.60 x  11.34 x  18.90

The regular selling price was $18.90. 5. Let the last month’s index be x. This month’s index  x 

1 x 12

1 x  176 12 12 x  x  2112 11x  2112 x  192

x 

Last month the index was 192.

6. Let the original hourly wage be $x. 1   New hourly wage  $  x  x  8  

1  x  x  15.75 8 8 x  x  126 9 x  126 x  14 The hourly wage before the increase was $14. 7. Let Vera’s sales be $x. Tai’s sales  $(3x  140) Total sales  $( x  3x  140)  x  3 x  140  940 4 x  1080 x  270

Tai’s sales  3  270   140  $670. 8. Let the shorter piece be x cm. Length of longer piece  (2 x  15) cm. Total length  ( x  2 x  15) cm  x  2 x  15  90 3 x  75 x  25

The longer piece is 2(25) cm + 15 cm  65 cm. 9. Let the cost of a ticket be $x. Total  $( x  18.40) 1.05  2

 ( x  18.40) 1.05  2  345.24 ( x  18.40)  2.10  345.24 ( x  18.40)  164.40 x  146

The cost per ticket is $146.

10. Let Ken’s investment be $x. 2  Martina’s investment  $  x  2500  3  2   Total investment  $  x  x  2500  3   x 

2 x  2500  55, 000 3 5x  52,500 3 x  31,500

Martina’s investment is

2  31,500  2500  $23,500. 3

11. Let the number of chairs produced by the first shift be x. Number of chairs produced by the second shift 

4 x  60. 3

4 Total production  x  x  60  2320. 3 x 

4 x  60  2320 3 7 x  2380 3 x  1020

Production by the second shift is

4 1020  60  1300. 3

12. Let the number of type A lights be x. Number of type B lights  60  x. Value of type A lights  $40 x. Value of type B lights  $(60  x)50.

 40 x  50(60  x)  2580 40 x  3000  50 x  2580  10 x  420 x  42

The number of type B lights is 18.

13. Let the number of units of Product A be x; then the number of units of Product B is 60  x. The number of hours for Product A is 4x; The number of hours for Product B is 3(60  x).  4 x  3(60  x)  200 4 x  180  3 x  200 x  20

Production of Product A is 20 units. 14. Let the number of dimes be x. Number of nickels  3x  4 Number of quarters 

3 x 1 4

Value of the dimes  10x cents Value of nickels  5(3x  4) cents 3  Value of quarters  25  x  1 cents 4 

3  10 x  5(3 x  4)  25  x  1  880 4  75 10 x  15 x  20  x  25  880 4 75 25 x  x  875 4 175 x  3500 x  20 Alick has 20 dimes, 56 nickels, and 16 quarters.

15. Let the number of $12 tickets be x. Number of $8 tickets  3x  10 Number of $15 tickets 

4 x 3 5

Value of the $12 tickets  $12x Value of the $8 tickets  $8(3x  10) 4  Value of the $15 tickets  $15  x  3  5 

4  12 x  8(3x  10)  15  x  3   1475 5  12 x  24 x  80  12 x  45  1475 48 x  1440 x  30 30 $12 tickets, 100 $ 8 tickets, Sales were and 21 $15 tickets. 16. Let the number of medium pizzas be x. Number of large pizzas  3x  1 Number of small pizzas  2 x  1 Value of medium pizzas  $15x Value of large pizzas  $18(3x  1) Value of small pizzas  $11(2 x  1)

15 x  18(3x  1)  11(2 x  1)  539 15 x  54 x  18  22 x  11  539 91x  546 x6 6 medium pizzas, 17 large pizzas, Sales were and 13 small pizzas.

17. Let the taxable income (in dollars) be x. Then x – 50,197 is the amount by which his income exceeds $50,197. 7529.55 + 0.205 (x – 50,197 ) = 10,121.16 7529.55 + 0.205x – 10,290.385 = 10,121.16 0.205x = 12,881.995 x = $62,839 His taxable income is $62,839. 18. Let the amount invested at 3% be $x. Then the amount invested at 4.5% is (3000 – x). 0.03x + 0.045 (3000 – x) = 128.25 0.03x + 135 – 0.045x = 128.25 –0.015x  – 6.75 x  $450 at 3% 3000 – 450 = $2550 at 4.5% 19. Let x be the number on the second shift. Then 2x is the number on the second shift. And x  12 is the number on the third shift. x + 2x + (x – 12) = 196 4x – 12 = 196 4x = 208 x = 52 on the second shift 2x = 2(52) = 104 on the first shift x – 12 = 52 – 12 = 40 on the third shift 20. Let x be the number of options received by each employee. Then 1.5x is the number received by each team leader. And 3x is the number received by each senior manager. 421x + 22(1.5x) + 7(3x) = 171,000 Copyright © 2025 Pearson Canada Inc.

421x + 33x + 21x = 171,000 475x = 171,000 x = 360 options for each employee 1.5x = 1.5(360) = 540 options for each team leader (2)(540) = 1080 options for each senior manager Check: 421(360) + 22(540) + 7(1080) = 171,000

21. Let the amount of money spent on recreational players be $x. If twice as much money was spent on rep players, then the amount spent on recreational players can be determined by $x + $2x = $4320 $3x = $4320 $x = $1440 And therefore, the amount spent on rep players was $4320  $1440 = $2880. Let the number of Youth Large shirts purchased for recreational players be y. $10y + $8(50) + $8(50) = $1440 $10y + $400 + $400 = $1440 $10y = $640 y = 64 64 Youth Large shirts were purchased for recreational players. Let the number of Adult Small and Adult Medium shirts be z. For rep players, the cost of shirts is given by $8(50 – 10) + $10(3 × 64) + $16z + $16z = $2880 Therefore, the number of shirts of each Adult size ordered can be calculated as $320 + $1920 + $16(2z) = $2880 $2240 + $16(2z) = $2880 $16(2z) = $640 2z = 40 z = 20 20 Adult Small and 20 Adult Medium shirts were purchased for rep players. (50 + 50 + 64) + (40 + 192 + 20 + 20) = 436 There are a total of 436 players in the organization.

Review Exercise 1. (a) 3 x  4 y  3 y  5 x  2 x  7 y (b) 2 x  0.03x  1.97 x (c) (5a  4)  (3  a)  5a  4  3  a  6a  7 (d) (2 x  3 y )  (4 x  y )  ( y  x)  2 x  3 y  4 x  y  y  x  x  3 y (e) (5a 2  2b  c)  (3c  2b  4a 2 )

 5a 2  2b  c  3c  2b  4a 2  9a 2  4b  4c (f ) (2 x  3)  ( x 2  5 x  2)  2 x  3  x 2  5 x  2   x 2  3x  1 2. (a) 3(5a)  15a (b) 7m(4 x)  28mx (c) 14m  (2m)  7 (d) (15a 2b)  (5a)  3ab (e) 6(3 x)(2 y )  36 xy (f ) 4(3a)(b)(2c)  24abc (g) 4(3 x  5 y  1)  12 x  20 y  4 (h) x(1  2 x  x 2 )  x  2 x 2  x3 (i) (24 x  16)  (4)  6 x  4 (j) (21a2 12a)  3a  7a  4 (k)

4(2a  5)  3(3  6a)  8a  20  9  18a  26a  29

(l)

2a( x  a)  a(3x  2)  3a(5 x  4)

 2ax  2a 2  3ax  2a  15ax  12a  14ax  2a 2  10a (m)

(m  1)(2m  5)  2m2  2m  5m  5  2m 2  7 m  5

(n)

(3a  2)(a 2  2a  3)  3a3  2a 2  6a 2  4a  9a  6  3a3  8a 2  5a  6

(o)

3(2 x  4)( x  1)  4( x  3)(5 x  2)  3(2 x 2  4 x  2 x  4)  4(5 x 2  15 x  2 x  6)  6 x 2  18 x  12  20 x 2  52 x  24  14 x 2  34 x  36

(p)

 2a(3m  1)(m  4)  5a(2m  3)(2m  3)  2a(3m2  m  12m  4)  5a(4m2  6m  6m  9)  6am2  26am  8a  20am 2  45a  26am2  26am  37a

(a) for x  2, y  5,

3xy  4 x  5 y  3(2)(5)  4(2)  5(5)  30  8  25  47

1 2 (b) for a   , b  , 4 3  5(2a  3b)  2(a  5b)  10a  15b  2a  10b  12a  5b 1 1  1 2  12     5    3  3  6 3 3  4 3

2NC 2 12  432 12  48 16     0.16 P(n  1) 1800  (35  1) 100  36 100 (d) for I  600, r  0.15, P  7300,

365 I 365  600 2    200 rP 0.15  7300 0.01

(e) for A  $720, d  0.135, t 

280 , 365

280   A(1  dt )  $720 1  0.135    $720(1  0.103562)  645.435616  $645.44  365 

(f ) for S  2755, r  0.17, t 

219 , 365

S 2755 2755 2755     2500 219 1  rt 1  0.17  365 1  0.034  3 1  0.102

4. (a) (3)5  243 4

16 2 (b)    81 3 (c) (5)0  1 (d) (3)1  

1 3

4

625 2 5 (e)       16 5 2 (f ) (1.01)0  1 (g) (3)5 (3)4  (3)9  19,683 (h) 47  42  45  1024 5

(i) (3)2   (3)10  59,049 (j) (m3 )4  m12 3

6

16 2 2 2 2 (k)           81 3 3 3 3 5

25  5  5  5 (l)             16  4  4  4

(m) (1.0350 )(1.03100 )  1.03150 (n) (1  i)180  (1  i)100  (1  i)80 5

(o) (1.05)30   1.05150 (p) (2 xy)4  16 x4 y 4 4

 a 2b  81  3  (q)    2   8 4 ab a b  3  (r) (1  i)  n 

5. (a)

1 (1  i) n

0.9216  0.96

(b) 6 1.075  1.012126 (c) 14.9744581/ 40  1.07 (d) 1.085/12 

1  0.968442 1.085/12

(e) ln 3  1.098612 (f ) ln 0.05  2.995732 (g) ln(5.1) / ln(1.015)  1.629241 / 0.014889  109.428635

 5500   ln 5500  ln 1.1016 (h) ln  16  1.10  

 ln 5500  16ln 1.10  8.612503  16(0.095310)  8.612503  1.524963  7.087540

  1  1.0172   72 (i) ln 375(1.01)     ln 375  ln 1.01  ln (1  1.01 )  ln 0.01  0.01   

 ln 375  ln 1.01  ln (1  0.488496)  ln 0.01  ln 375  ln 1.01  ln 0.511504  ln 0.01  5.926926  0.009950  0.670400  (4.605170)  9.871647

6. (a) 9 x  63

x  7 (b) 0.05x  44

5 x  4400 x  880 1 (c)  x  3 7

 x  21 x  21 (d)

5 x  15 6 1 x  3 6 x  18

(e)

x  8  5 x  8  8  5  8 x3

(f )

x  9  2 x  9  9  2  9 x  11

(g) x  0.02 x  255

1.02 x  255 x  250 (h) x  0.1x  36

0.9 x  36 9 x  360 x  40 (i) 4 x  3  9 x  2

 5x  5 x  1 (j) 9 x  6  3x  15  4 x  7

6x  6  8  4x 2 x  14 x7 1 (k) x  x  26 3

2 x  26 3 1 x  13 3 x  39 3 (l) x  x  77 8

11 x  77 8 1 x7 8 x  56 7. (a) 9(3x  8)  8(9  7 x)  5  4(9 x  11)

 27 x  72  72  56 x  5  36 x  44 29 x  49  36 x  7 x  49 x  7 Check LS  9[3(7)  8]  8[9  7(7)]  9(29)  8(58)  203 RS  5  4[9(7)  11]  5  4(52)  5  208  203 (b) 21x  4  7(5 x  6)  8 x  4(5 x  7)

21x  4  35 x  42  8 x  20 x  28  14 x  38  12 x  28  2 x  10 x5

Check LS  21(5)  4  7[5(5)  6]  105  4  7(19)  101 133  32 RS  8(5)  4[5(5)  7]  40  4(18)  40  72  32

(c)

5 1 5 2 x   x 7 2 14 3 5  1  5 2  42  x   42    42    42  x  7  2  14  3  6(5 x)  21(1)  3(5)  14(2 x) 30 x  21  15  28 x 2 x  6 x  3 5 1 30  7 23 Check LS  (3)    7 2 14 14 5 2 5 23 RS   (3)   2   14 3 14 14

(d)

4x 9 x 2  3 8 6

8(4 x)  24(2)  3(9)  4( x) 32 x  48  27  4 x 36 x  21 x

7 12

4 7  28 7 18 11 Check LS      2    2     3  12  36 9 9 9 9 1  7  9 7 81  7 88 11 RS           8 6  12  8 72 72 72 9 (e)

7 3 (6 x  7)  (7 x  15)  25 5 8 56(6 x  7)  15(7 x  15)  40(25) 336 x  392  105 x  225  1000 231x  617  1617 x7 7 3 Check LS  [6(7)  7]  [7(7)  15] 5 8 7 3  (35)  (64)  7(7)  24  49  24  25 5 8 RS  25

(f )

5 3 1 1 (7  6 x)  (3 15 x)  (3 x  5)  9 4 12 2

20(7  6 x)  27(3  15 x)  3(3 x  5)  18 140  120 x  81  405 x  9 x  15  18 285 x  59  9 x  33 276 x  92 x

Check LS 

1 3

5  1  3   1  7  6      3  15      9  3  4   3 

5 3  (7  2)  (3  5) 9 4  5  6  1 RS 

1   1  1 3    5  12   3   2

1 1 (6)  12 2 1 1     1 2 2 

(g)

5 2 16 (4 x  3)  (3x  4)  5 x  (1  3x) 6 5 15

25(4 x  3)  12(3 x  4)  150 x  32(1  3 x) 100 x  75  36 x  48  150 x  32  96 x 64 x  123  246 x  32  182 x  91 x

1 2

Check LS 

5  1  2  1  4     3  3    4 6   2   5   2  

5 2 3   (2  3)     4  6 5 2  5 25 25 31  (5)       1   6 52 6 6  1  16   1  RS  5     1  3      2  15   2  5 16  3  5 16  5     1        2 15  2  2 15  2  5 8 15  16 31     2 3 6 6 8. (a) I  P rt I Pt

r

(b)

S  P(1  rt ) S  1  rt P S  1  rt P S 1 P t r SP t P r SP t Pr

r

D L

é(1  p) n  1ù ú (d) FV  PMT ê ê ú p ë û é FVp ù ú PMT  ê êë(1  p) n  1ú û

9. Let the size of the workforce be x. Number laid off 

1 x 6

1 Number after the layoff  x  x 6

1  x  x  690 6 5 x  690 6 5 x  4140 x  828  the number laid off is

1  828  138. 6

10. Let last year’s average property value be $x. 2   Current average value  $  x  x  7  

2  x  x  346,162.50 7 9 x  346,162.50 7 1 x  38, 462.50 7 x  269, 237.50  Last year’s average value was $269, 237.50.

11. Let the quoted price be $x.

x 

1 x  $12,957 20 21 x  12,957 20 1 x  617 20

 The gratuities were $617.

12. Let the value of the building be $x.

1 Value of the land  $ x  2000 3 1 Total value of the property  $ x  x  2000 3

1  x  x  2000  790, 000 3 4 x  792, 000 3 1 x  198, 000 3 x  594, 000 The value assigned to land is $(790,000  594,000)  $196,000. 13. Let the cost of power be $x. 3  Cost of heat  $  x  22  4  1  Cost of water  $  x  11 3 

3 1 Total cost  x  x  22  x  11  2010  10% of 2010. 4 3 12 x  9 x  4 x  12(2010  201  11) 25 x  26 400 x  1056

3 Cost of heat  1056  22  $814 4 Cost of power  $1056

1 Cost of water  1056  11  $341 3

14. Let the amount allocated to newspaper advertising be $x. Amount allocated to TV advertising  $(3x  1000)

3 Amount allocated to direct selling  [ x  3x  1000] 4

3  x  3x  1000  [4 x  1000]  87,500 4 3 4 x  [4 x  1000]  86,500 4 16 x  12 x  3000  346, 000 28 x  343, 000 x  12, 250 The amount allocated to newspaper advertising is $12,250; the amount allocated to TV advertising is $37,750; the amount allocated to direct selling is $37,500. 15. Let the number of minutes on Machine B be x. Time on Machine A 

4 x  3 minutes 5

5 4  Time on Machine C   x  x  3  minutes 6 5 

Total time  x 

x

4 5 4  x  3   x  x  3  minutes 5 6 5 

4 5 4  x  3   x  x  3   77 5 6 5 

4   30 x  24 x  90  25  x  x  3   30(77) 5   54 x  90  25 x  20 x  75  2310 99 x  165  2310 99 x  2475 x  25 Time on Machine B is 25 minutes; time on Machine A is time on Machine C is

5 (25  17)  35 minutes. 6

4 (25)  3  17 minutes; 5

16. Let the number of pairs of superlight poles be x. Number of pairs of ordinary poles  72  x Value of superlight poles  $130x Value of ordinary poles  $56(72  x) Total value of all poles  $130 x  56(72  x)

130 x  56(72  x)  $6030 130 x  4032  56 x  6030 74 x  1998 x  27 The number of pairs of superlight poles is 27; the number of pairs of ordinary poles is 45. 17. Let the number of $2 coins be x. Number of $1 coins 

3 x 1 5

 

 

3 5

Number of quarters  4  x  x  1 Value of the $2 coins  $2x

3 5

 

Value of the $1 coins  $  x  1

1 4

 

3 5

 

3 5

Value of the quarters  $ (4)  x  x  1  x  x  1 3 3 Total value  2 x  x  1  x  x  1  107 5 5 10 x  3x  5  5 x  3x  5  535 21x  10  535 21x  525 x  25

3 5

 

The number of $2 coins is 25; the number of $1 coins is   25  1  16; the number of quarters is 4(25  16)  164.

18. Let $x represent Jaime’s monthly savings. $2975 ¸ 2 = $1487.50 Jaime has $1487.50 after paying for school and transportation. 0.30(1487.50) + 900 + x  1487.50 1346.25 + x  1487.50 x  141.25 Jaime has $141.25 left over for savings each month. 19. Let x represent the total valuation of Baldwin Industries. Then Inspire Inc.’s stake is 0.49x and Crown Company’s stake is 0.24x. 0.80(0.49x) = $19,600,000 0.392x = $19,600,000 x = $50,000,000 0.24(50,000,000) = $12,000,000 Crown Company’s stake in Baldwin Industries is worth $12 Million. Self-Test 1. (a) 4  3x  6  5x  2  8x (b) (5x  4)  (7 x  5)  5x  4  7 x  5  2 x  9 (c)

2(3a  4)  5(2a  3)

 6a  8  10a  15  16a  7 (d)

 6( x  2)( x  1)

 6( x 2  2 x  x  2)  6( x 2  x  2)  6 x 2  6 x  12

2. (a) For x  3, y  5

2 x 2  5 xy  4 y 2  2(3)2  5(3)(5)  4(5)2  18  75  100  7 2 3

(b) For a  , b  

3 4

3(7 a  4b)  4(5a  3b)  21a  12b  20a  12b  a  24b 2  3  24    3  4 2   18 3 2  18 3 

2NC (2)(12)(400) 2(12)(400)    0.192 P(n  1) 2000(24  1) 2000(25) (d) For I = 324, P = 5400, r = 0.15 I 324   0.4 P r 5400  0.15

(e) For S = 1606, d = 0.125, t =

240 365

240   S(1  dt )  1606 1  0.125   365    1606(1  0.082192)  1606(0.917808)  1474

(f ) For S = 1566, r = 0.10, t =

292 365

S 1566  292 1  rt 1  0.10  365 1566 1  0.08  1450 

3. (a) (2)  8 3

2 4 (b)     

3

(d) (3) (3)  (3)  2187 2

2

1 1 9 4   (e)    2 16 16 3 4   9 3 (f) ( x3 )5   x15

4. (a) 10 1.35  1.3510  1.35

0.10

 1.030465

1  1.0340 1  0.306557 0.693443    23.114772 (b) 0.03 0.03 0.03 (c) ln 1.025  0.024693 (d) ln (3e0.2 )

 ln 3  ln e0.2  ln 3  0.2ln e  1.098612  0.2  0.898612

 600  11   1.06 

(e) ln 

 ln 600  ln 1.0611  ln 600  11ln 1.06  6.396930  11(0.058269)  6.396930  0.640958  5.755972

  1.075 1  ln (f ) 250     0.07   ln 250  ln (1.075  1)  ln 0.07  ln 250  ln 0.402552  ln 0.07  5.521461  0.909932  ( 2.659260)  5.521461  0.909932  2.659260  7.270789 5. (a)

1 1   81  3 

n 2

1 1   34  3  4

n2

1 1     3 3

n2

Since the bases are common

4 n2 n6

5 1  40   2 2

(b)

1 1  8  2 2 1 1   16  2 

n1

n 1

1 1     2 2 4  n 1 n5 2 3

6. (a)  x  24

 3 x  24     2 x  36 (b) x  0.06 x  8.46

0.94 x  8.46 x9 (c) 0.2 x  4  6  0.3x

0.5 x  10 x  20 (d) (3  5x)  (8 x  1)  43

3  5 x  8 x  1  43  13x  39 x  3 (e) 4(8 x  2)  5(3x  5)  18

32 x  8  15 x  25  18 17 x  33  18 17 x  51 x3

(f ) x 

3 1 3 x   x  x  1  103 10 2 5

3 3 3 x  x   103 10 5 2 20 x  3x  6 x  15  1030 29 x  1015

2x 

x  35

4 5 4  x  x  3   x  x  3   77 5 6 5 

(g)

4   30 x  24 x  90  25  x  x  3   30(77) 5   54 x  90  25 x  20 x  75  2310 99 x  165  2310 99 x  2475 x  25 (h)

2 3 9 5 (3x  1)  (5 x  3)  x  (7 x  9) 3 4 8 6

16(3x  1)  18(5 x  3)  27 x  20(7 x  9) 48 x  16  90 x  54  27 x  140 x  180  42 x  38  113x  180 71x  142 x2 7. (a) I  P rt P

I rt

(b) S 

P 1  dt

S 1  P 1  dt P  1  dt S dt  1 

P S

1

P S

d

S P d S t

d

S P St

8. Let the regular selling price be $x. 1 5

Reduction in price  $ x x 

1 x  1920 5 4 x  1920 5 x  2400

The regular selling price is $2400. 9. Let the floor space occupied by shipping be x. Floor space occupied by weaving  2 x  400 Total floor space  x  2 x  400  x  2 x  400  6700 3 x  6300 x  2100

The floor space occupied by weaving is 2(2100)  400  4600 square metres.

10. Let the number of units of Product A be x. Number of units of Product B  95  x Number of hours for Product A  3x Number of hours for Product B  5(95  x)

 3x  5(95  x)  395 3x  475  5 x  395  2 x  80 x  40 11. The number of units of Product B is 95  40  55. Let the sum of money invested in the bank be $x. 2 3

Sum of money invested in the credit union  $ x  500 Yield on the bank investment  $

1 x 12

12 93

 

Yield on the credit union investment  $  x  500 



1 12  x   x  500   1000 12 93  2  3 x  4  x  500   36, 000 3  8 3 x  x  2000  36, 000 3 17 x  34, 000 3 17 x  102, 000 x  6000

The sum of money invested in the credit union certificate is

2  $   6000  500   $4500. 3 

Challenge Problems 1. Counting a nickel as a quarter overstates the total by $0.20; for x nickels, the total must be reduced by $0.20x. Counting a toonie as a loonie understates the total by $1; for x toonies, the total must be increased by $1x. The total adjustment  0.20 x  1x  $0.80 x The clerk must increase the total by $0.80x. 2. There are 5 tires, so each tire is idle at some point. Therefore, the number of rotations is 5. The distance per rotation 

4000  800 km; each tire will be used for four rotations 5

for a total distance of 3200 km. (See table below.) Distance Rotation

Tire A

Tire B

Tire C

Tire D

Tire E

travelled

800

—

800

—

800

—

800

—

800

—

800

Total

3200

4000

3. The lowest possible two-digit number is 10; the highest possible two-digit number is 99. For a difference in value of $17.82, the two-digit numbers must differ by 18, such as 10 and 28, 11 and 29, etc. The lowest possible correct value of the cheque is $10.28; the largest possible correct value of the cheque is $81.99. In either case the difference between is $17.82. (a) FALSE than 70.

In the possible correct cheque value $81.99, the x-value 81 is greater

(b) TRUE

In the possible correct cheque value $18.36, the y-value 36 equals 2x.

A cheque cannot have zero cents.

(d) FALSE

Let the correct amount be $A;