1 × 1 = | 1 |

1 × 2 = | 2 |

1 × 3 = | 3 |

1 × 4 = | 4 |

1 × 5 = | 5 |

1 × 6 = | 6 |

1 × 7 = | 7 |

1 × 8 = | 8 |

1 × 9 = | 9 |

1 × 10 = | 10 |

1 × 11 = | 11 |

1 × 12 = | 12 |

2 × 1 = | 2 |

2 × 2 = | 4 |

2 × 3 = | 6 |

2 × 4 = | 8 |

2 × 5 = | 10 |

2 × 6 = | 12 |

2 × 7 = | 14 |

2 × 8 = | 16 |

2 × 9 = | 18 |

2 × 10 = | 20 |

2 × 11 = | 22 |

2 × 12 = | 24 |

3 × 1 = | 3 |

3 × 2 = | 6 |

3 × 3 = | 9 |

3 × 4 = | 12 |

3 × 5 = | 15 |

3 × 6 = | 18 |

3 × 7 = | 21 |

3 × 8 = | 24 |

3 × 9 = | 27 |

3 × 10 = | 30 |

3 × 11 = | 33 |

3 × 12 = | 36 |

4 × 1 = | 4 |

4 × 2 = | 8 |

4 × 3 = | 12 |

4 × 4 = | 16 |

4 × 5 = | 20 |

4 × 6 = | 24 |

4 × 7 = | 28 |

4 × 8 = | 32 |

4 × 9 = | 36 |

4 × 10 = | 40 |

4 × 11 = | 44 |

4 × 12 = | 48 |

5 × 1 = | 5 |

5 × 2 = | 10 |

5 × 3 = | 15 |

5 × 4 = | 20 |

5 × 5 = | 25 |

5 × 6 = | 30 |

5 × 7 = | 35 |

5 × 8 = | 40 |

5 × 9 = | 45 |

5 × 10 = | 50 |

5 × 11 = | 55 |

5 × 12 = | 60 |

6 × 1 = | 6 |

6 × 2 = | 12 |

6 × 3 = | 18 |

6 × 4 = | 24 |

6 × 5 = | 50 |

6 × 6 = | 36 |

6 × 7 = | 42 |

6 × 8 = | 48 |

6 × 9 = | 54 |

6 × 10 = | 60 |

6 × 11 = | 66 |

6 × 12 = | 72 |

7 × 1 = | 7 |

7 × 2 = | 14 |

7 × 3 = | 21 |

7 × 4 = | 28 |

7 × 5 = | 35 |

7 × 6 = | 42 |

7 × 7 = | 49 |

7 × 8 = | 56 |

7 × 9 = | 63 |

7 × 10 = | 70 |

7 × 11 = | 77 |

7 × 12 = |

8 × 1 = | 8 |

8 × 2 = | 16 |

8 × 3 = | 24 |

8 × 4 = | 32 |

8 × 5 = | 40 |

8 × 6 = | 48 |

8 × 7 = | 56 |

8 × 8 = | 64 |

8 × 9 = | 72 |

8 × 10 = | 80 |

8 × 11 = | 88 |

8 × 12 = | 96 |

9 × 1 = | 9 |

9 × 2 = | 18 |

9 × 3 = | 27 |

9 × 4 = | 36 |

9 × 5 = | 45 |

9 × 6 = | 54 |

9 × 7 = | 63 |

9 × 8 = | 72 |

9 × 9 = | 81 |

9 × 10 = | 90 |

9 × 11 = | 99 |

9 × 12 = | 108 |

10 × 1 = | 10 |

10 × 2 = | 20 |

10 × 3 = | 30 |

10 × 4 = | 40 |

10 × 5 = | 50 |

10 × 6 = | 60 |

10 × 7 = | 70 |

10 × 8 = | 80 |

10 × 9 = | 90 |

10 × 10 = | 100 |

10 × 11 = | 110 |

10 × 12 = | 120 |

11 × 1 = | 11 |

11 × 2 = | 22 |

11 × 3 = | 33 |

11 × 4 = | 44 |

11 × 5 = | 55 |

11 × 6 = | 66 |

11 × 7 = | 77 |

11 × 8 = | 88 |

11 × 9 = | 99 |

11 × 10 = | 110 |

11 × 11 = | 121 |

11 × 12 = | 132 |

12 × 1 = | 12 |

12 × 2 = | 24 |

12 × 3 = | 36 |

12 × 4 = | 48 |

12 × 5 = | 60 |

12 × 6 = | 72 |

12 × 7 = | 84 |

12 × 8 = | 96 |

12 × 9 = | 108 |

12 × 10 = | 120 |

12 × 11 = | 132 |

12 × 12 = | 144 |

× | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|

0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |

2 | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 | 20 | 22 | 24 |

3 | 0 | 3 | 6 | 9 | 12 | 15 | 18 | 21 | 24 | 27 | 30 | 33 | 36 |

4 | 0 | 4 | 8 | 12 | 16 | 20 | 24 | 28 | 32 | 36 | 40 | 44 | 48 |

5 | 0 | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 45 | 50 | 55 | 60 |

6 | 0 | 6 | 12 | 18 | 24 | 30 | 36 | 42 | 48 | 54 | 60 | 66 | 72 |

7 | 0 | 7 | 14 | 21 | 28 | 35 | 42 | 49 | 56 | 63 | 70 | 77 | 84 |

8 | 0 | 8 | 16 | 24 | 32 | 40 | 48 | 56 | 64 | 72 | 80 | 88 | 96 |

9 | 0 | 9 | 18 | 27 | 36 | 45 | 54 | 63 | 72 | 81 | 90 | 99 | 108 |

10 | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 100 | 110 | 120 |

11 | 0 | 11 | 22 | 33 | 44 | 55 | 66 | 77 | 88 | 99 | 110 | 121 | 132 |

12 | 0 | 12 | 24 | 36 | 48 | 60 | 72 | 84 | 96 | 108 | 120 | 132 | 144 |

- An electrician used switch outlet boxes on eight different jobs. The number of boxes used on each job is 56, 9, 36, 93, 105, 42, 86, and 56. The total number of outlet boxes used is boxes.
- For a residential job, a reel containing 1050 feet of cable is delivered. Three 45 foot lengths and three 65 foot lengths are used. There are feet left.
- Two thousand five hundred feet of plastic tubing are ordered. The tubing is shipped in 250 foot coils. There are coils shipped.
- The materials charged to a wiring job are as follows: 100 amp distribution panel, $36; meter switch, $8; conduit, $28; number 2 wire, $43; BX cable, $25; conduit fittings, $9; outlet boxes. $92; switches, $35; fixtures, $65; and $37 for wire nuts, grounding clips, staples and pipe clamps. The total amount charged for these materials is dollars.
- An HVAC company has a payroll of 27 people. Seven people earn $8/hr, eleven people earn $10/hr, and nine people earn $6/hr. If all the employees work 40 hours during the week, the total amount of money earned in one week is dollars.
- A large room contains forty fluorescent lamps. Twenty three of the lamps are 40 watt lamps, and the remaining number are 60 watt lamps. The lamps use a total of Watts.
- On a certain job, a sum of $438 is spent for materials. Of this amount, $76 is spent for 1 fittings, and $105 is spent for miscellaneous materials. dollars is spent on other materials.
- Three plumbers install 180 fixtures in 4 working days. The average number of fixtures installed by each plumber in one day is fixtures.
- On a 124 foot length of pipe, 32 supports are used. The supports are equally spaced. If one support is placed at the beginning and one at the end of the pipe, there will be a support every feet.
- A 7 floor apartment building has an average of 7 electrical circuits per apartment, and there are 8 apartments per floor. There are circuits in the building.

- 483
- 720
- 10
- 378
- 8800
- 1940
- 257
- 15
- 4
- 392

- 11.25 − 3.875 =
- 7.15
- 7.875
- 7.375
- 15.125

- Reduce 24/64 to lowest terms.
- 12/32
- 6/16
- 3/8
- 1.5/4

- Reduce 150/275 to lowest terms.
- 6/11
- 3/5.5
- 3/11
- 12/22

- Change 34/8 into a mixed number.
- 3 4/8
- 4 1/4
- 4 1/8
- 4 1/16

- Express 5 7/16 as an improper fraction.
- 103/16
- 57/16
- 87/16
- 35/16

- 1/4 + 5/16 + 7/32 =
- 7/16
- 13/32
- 25/32
- 13/16

- 5/8 + 9/32 − 1/16 =
- 7/8
- 27/32
- 13/16
- 7/16

- 7/8 × 5 =
- 4 3/4
- 3 5/40
- 6 3/8
- 4 3/8

- 3/16 × 5/8 =
- 9/32
- 9/16
- 80/24
- 15/128

- 4 1/2 × 4 =
- 18
- 8 1/2
- 16 1/2
- 16 1/8

- 3/8 ÷ 1/16 =
- 6
- 12
- 3/16
- 3/128

- 5/16 ÷ 4 =
- 20/16
- 1 1/4
- 5/16
- 5/64

- 17/32 + 1/2 × 1/4 =
- 8/32
- 21/32
- 23/32
- 1 9/32

- On the sheet metal layout in the drawing, the distance between the centres of hole A and hole D, measuring lengthwise, is 16 centimetres. All dimensions are in centimetres. True or false?
- A total of 256 bronze castings were machined. Each casting had a mass of 793 g, and 158 g were removed from each by machining. The total mass of bronze removed by machining was 162,560 g. True or false?
- A counter top is made of 3/4'' particleboard and is covered with 1/32'' plastic laminate. The width of metal edging required to finish off the edge is 27/32''. True or false?
- In one minute, a large reduction gear turns 61/3 revolutions. How many revolutions will it turn in 27 1/2 minutes?
- 174 1/6
- 162 1/6
- 168 1/3
- 170 1/3

- 12 billets of cold drawn steel 1½'' long can be cut from a bar 20'' long, if 1/8'' is allowed for each saw cut. True or false?
- A 3/16 sheet of metal is thicker than a 13/64'' fastener. True or false?
- Eight pins, each 43/8'' long, have to be cut from a piece of drill rod. There is a 3/32'' waste for each cut. What is the minimum length of rod you will need?
- 35''
- 35 3/4''
- 35 21/32
- 35 43/64''

- C
- C
- A
- B
- C
- C
- A
- D
- D
- A
- A
- D
- B
- True
- False
- False
- A
- True
- False
- C

- Round 5237.02046 to the nearest thousandth.
- 5000
- 5237.02
- 5237.021
- 5237.02106

- Round 12465.078 to the nearest thousand.
- 12465
- 12000
- 10000
- 12465.078

- 13.963 + 335.021 + 2267.123 =
- 2616.17
- 2616.107
- 2616.296
- 2616.182

- Add 2.72, 0.6, 110, and 17.225.
- 130.545
- 135.945
- 136.545
- 131.545

- 11.25 − 3.875 =
- 7.15
- 7.875
- 7.375
- 15.125

- Multiply 19.5 by 7.65.
- 14917.5
- 14.9175
- 1491.75
- 149.175

- 18.753 ÷ 0.87 =
- 210.55
- 21.555
- 2.1555
- 0.21555

- Divide 0.0028 by 0.95. Express your answer to the nearest thousandth.
- 0.003
- 0.002
- 0.0029
- 0.00295

- 178 + 18 × 6 =
- 286
- 295
- 1176
- 19224

- 158.58 − 33 × 4 =
- −26.58
- 502.32
- 26.58
- −502.32

- 1890.33 – 543.48 + 101 =
- 1447.52
- 2534.81
- 2534.85
- 1447.85

- C
- B
- B
- A
- C
- D
- B
- A
- A
- C
- D

- 40% of 24 is what?
- 14.4%
- 9.6%
- 8.4%
- 10.2%

- What percent of 1.32 is 0.6?
- 46%
- 55%
- 45%
- 10%

- 3 is 15% of 22. True or false?
- One part acid and four parts water are mixed as an electrolyte for a storage battery. What percent of the electrolyte is acid?
- 25%
- 20%
- 75%
- 80%

- The efficiency of a motor is 90%. If the motor delivers 13.5 horsepower, what is the input? (efficiency=output/input)
- 15 hp
- 15.85 hp
- 12.15 hp
- 15.28 hp

- What percent is wasted when 3.6 of every 120 sheets of metal are spoiled?
- 3%
- 0.3%
- 30%
- 300%

- A floor joist is allowed to vary 5% of its stated width of 184 mm. The maximum width would be what?
- 200 mm
- 188.4 mm
- 193.2 mm
- 197.9 mm

- A contractor wants to make a profit of 12% on a job. If the break-even cost is $2345, what should the total job price be, including the profit?
- $2626
- $2600.40
- $2600
- $2626.40

- The minimum idle speed for a particular engine is 630 rpm. This is 94% of the maximum idle speed of 668 rpm. True or false?
- The operating spindle speed of a lathe spindle that rotates freely is 354 rpm. When 16% is lost through slippage and cutting pressure, the new speed is 297 rpm. True or false?

- B
- C
- False
- B
- A
- A
- C
- D
- False
- True

- An electrician used switch outlet boxes on eight different jobs. The number of boxes used on each job is 56, 9, 36, 93, 105, 42, 86, and 56. The total number of outlet boxes used is boxes.
- For a residential job, a reel containing 1050 feet of cable is delivered. Three 45 foot lengths and three 65 foot lengths are used. There are feet left.
- Two thousand five hundred feet of plastic tubing are ordered. The tubing is shipped in 250 foot coils. There are coils shipped.
- The materials charged to a wiring job are as follows: 100 amp distribution panel, $36; meter switch, $8; conduit, $28; number 2 wire, $43; BX cable, $25; conduit fittings, $9; outlet boxes. $92; switches, $35; fixtures, $65; and $37 for wire nuts, grounding clips, staples and pipe clamps. The total amount charged for these materials is dollars.
- An HVAC company has a payroll of 27 people. Seven people earn $8/hr, eleven people earn $10/hr, and nine people earn $6/hr. If all the employees work 40 hours during the week, the total amount of money earned in one week is dollars.
- A large room contains forty fluorescent lamps. Twenty three of the lamps are 40 watt lamps, and the remaining number are 60 watt lamps. The lamps use a total of Watts.
- On a certain job, a sum of $438 is spent for materials. Of this amount, $76 is spent for 1 fittings, and $105 is spent for miscellaneous materials. dollars is spent on other materials.
- Three plumbers install 180 fixtures in 4 working days. The average number of fixtures installed by each plumber in one day is fixtures.
- On a 124 foot length of pipe, 32 supports are used. The supports are equally spaced. If one support is placed at the beginning and one at the end of the pipe, there will be a support every feet.
- A 7 floor apartment building has an average of 7 electrical circuits per apartment, and there are 8 apartments per floor. There are circuits in the building.

- 483
- 720
- 10
- 378
- 8800
- 1940
- 257
- 15
- 4
- 392

- 11.25 − 3.875 =
- 7.15
- 7.875
- 7.375
- 15.125

- Reduce [latex]\tfrac{24}{64}[/latex] to lowest terms.
- [latex]\tfrac{12}{32}[/latex]
- [latex]\tfrac{6}{16}[/latex]
- [latex]\tfrac{3}{8}[/latex]
- [latex]\tfrac{1.5}{4}[/latex]

- Reduce [latex]\tfrac{150}{275}[/latex] to lowest terms.
- [latex]\tfrac{6}{11}[/latex]
- [latex]\tfrac{3}{5.5}[/latex]
- [latex]\frac{3}{11}[/latex]
- [latex]\frac{12}{22}[/latex]

- Change [latex]\frac{34}{8}[/latex] into a mixed number.
- [latex]3\frac{4}{8}[/latex]
- [latex]4\frac{1}{4}[/latex]
- [latex]4\frac{1}{8}[/latex]
- [latex]4\frac{1}{16}[/latex]

- Express [latex]5\frac{7}{16}[/latex] as an improper fraction.
- [latex]\frac{103}{16}[/latex]
- [latex]\frac{57}{16}[/latex]
- [latex]\frac{87}{16}[/latex]
- [latex]\frac{35}{16}[/latex]

- [latex]\frac{1}{4}+\frac{5}{16}+\frac{7}{32}=[/latex]
- [latex]\frac{7}{16}[/latex]
- [latex]\frac{13}{32}[/latex]
- [latex]\frac{25}{32}[/latex]
- [latex]\frac{13}{16}[/latex]

- [latex]\frac{5}{8}+\frac{9}{32}-\frac{1}{16}=[/latex]
- [latex]\frac{7}{8}[/latex]
- [latex]\frac{27}{32}[/latex]
- [latex]\frac{13}{16}[/latex]
- [latex]\frac{7}{16}[/latex]

- [latex]\frac{7}{8}\times5=[/latex]
- [latex]4\frac{3}{4}[/latex]
- [latex]3\frac{5}{40}[/latex]
- [latex]6\frac{3}{8}[/latex]
- [latex]4\frac{3}{8}[/latex]

- [latex]\frac{3}{16}\times\frac{5}{8}=[/latex]
- [latex]\frac{9}{32}[/latex]
- [latex]\frac{9}{16}[/latex]
- [latex]\frac{80}{24}[/latex]
- [latex]\frac{15}{128}[/latex]

- [latex]4\frac{1}{2}\times4=[/latex]
- [latex]18[/latex]
- [latex]8\frac{1}{2}[/latex]
- [latex]16\frac{1}{2}[/latex]
- [latex]16\frac{1}{8}[/latex]

- [latex]\frac{3}{8}\div\frac{1}{16}=[/latex]
- [latex]6[/latex]
- [latex]12[/latex]
- [latex]\frac{3}{16}[/latex]
- [latex]\frac{3}{128}[/latex]

- [latex]\frac{5}{16}\div4=[/latex]
- [latex]\frac{20}{16}[/latex]
- [latex]1\frac{1}{4}[/latex]
- [latex]\frac{5}{16}[/latex]
- [latex]\frac{5}{64}[/latex]

- [latex]\frac{17}{32}+\frac{1}{2}\times\frac{1}{4}=[/latex]
- [latex]\frac{8}{32}[/latex]
- [latex]\frac{21}{32}[/latex]
- [latex]\frac{23}{32}[/latex]
- [latex]1\frac{9}{32}[/latex]

- On the sheet metal layout in the drawing, the distance between the centres of hole A and hole D, measuring lengthwise, is 16 centimetres. All dimensions are in centimetres. True or false?
- A total of 256 bronze castings were machined. Each casting had a mass of 793 g, and 158 g were removed from each by machining. The total mass of bronze removed by machining was 162,560 g. True or false?
- A counter top is made of [latex]\frac{3}{4}''[/latex] particleboard and is covered with [latex]\frac{1}{32}''[/latex] plastic laminate. The width of metal edging required to finish off the edge is [latex]\frac{27}{32}''[/latex]. True or false?
- In one minute, a large reduction gear turns [latex]\frac{61}{3}[/latex] revolutions. How many revolutions will it turn in [latex]27\frac{1}{2}[/latex] minutes?
- [latex]174\frac{1}{6}[/latex]
- [latex]162\frac{1}{6}[/latex]
- [latex]168\frac{1}{3}[/latex]
- [latex]170\frac{1}{3}[/latex]

- 12 billets of cold drawn steel [latex]1\frac{1}{2}''[/latex] long can be cut from a bar [latex]20''[/latex] long, if [latex]\frac{1}{8}''[/latex] is allowed for each saw cut. True or false?
- A [latex]\frac{3}{16}''[/latex] sheet of metal is thicker than a [latex]\frac{13}{64}''[/latex] fastener. True or false?
- Eight pins, each [latex]\frac{43}{8}''[/latex] long, have to be cut from a piece of drill rod. There is a [latex]\frac{3}{32}''[/latex] waste for each cut. What is the minimum length of rod you will need?
- [latex]35''[/latex]
- [latex]35\frac{3}{4}''[/latex]
- [latex]35\frac{21}{32}''[/latex]
- [latex]35\frac{43}{64}''[/latex]

- C
- C
- A
- B
- C
- C
- A
- D
- D
- A
- A
- D
- B
- True
- False
- False
- A
- True
- False
- C

- Round 5237.02046 to the nearest thousandth.
- 5000
- 5237.02
- 5237.021
- 5237.02106

- Round 12465.078 to the nearest thousand.
- 12465
- 12000
- 10000
- 12465.078

- 13.963 + 335.021 + 2267.123 =
- 2616.17
- 2616.107
- 2616.296
- 2616.182

- Add 2.72, 0.6, 110, and 17.225.
- 130.545
- 135.945
- 136.545
- 131.545

- 11.25 − 3.875 =
- 7.15
- 7.875
- 7.375
- 15.125

- Multiply 19.5 by 7.65.
- 14917.5
- 14.9175
- 1491.75
- 149.175

- 18.753 ÷ 0.87 =
- 210.55
- 21.555
- 2.1555
- 0.21555

- Divide 0.0028 by 0.95. Express your answer to the nearest thousandth.
- 0.003
- 0.002
- 0.0029
- 0.00295

- 178 + 18 × 6 =
- 286
- 295
- 1176
- 19224

- 158.58 − 33 × 4 =
- −26.58
- 502.32
- 26.58
- −502.32

- 1890.33 – 543.48 + 101 =
- 1447.52
- 2534.81
- 2534.85
- 1447.85

- C
- B
- B
- A
- C
- D
- B
- A
- A
- C
- D

- 40% of 24 is what?
- 14.4%
- 9.6%
- 8.4%
- 10.2%

- What percent of 1.32 is 0.6?
- 46%
- 55%
- 45%
- 10%

- 3 is 15% of 22. True or false?
- One part acid and four parts water are mixed as an electrolyte for a storage battery. What percent of the electrolyte is acid?
- 25%
- 20%
- 75%
- 80%

- The efficiency of a motor is 90%. If the motor delivers 13.5 horsepower, what is the input? ([latex]\text{efficiency}=\frac{\text{output}}{\text{input}}[/latex])
- 15 hp
- 15.85 hp
- 12.15 hp
- 15.28 hp

- What percent is wasted when 3.6 of every 120 sheets of metal are spoiled?
- 3%
- 0.3%
- 30%
- 300%

- A floor joist is allowed to vary 5% of its stated width of 184 mm. The maximum width would be what?
- 200 mm
- 188.4 mm
- 193.2 mm
- 197.9 mm

- A contractor wants to make a profit of 12% on a job. If the break-even cost is $2345, what should the total job price be, including the profit?
- $2626
- $2600.40
- $2600
- $2626.40

- The minimum idle speed for a particular engine is 630 rpm. This is 94% of the maximum idle speed of 668 rpm. True or false?
- The operating spindle speed of a lathe spindle that rotates freely is 354 rpm. When 16% is lost through slippage and cutting pressure, the new speed is 297 rpm. True or false?

- B
- C
- False
- B
- A
- A
- C
- D
- False
- True

[latex]\LARGE\dfrac{1}{2}\ \dfrac{3}{8}\ \dfrac{5}{16}\ \dfrac{4}{9}\ \dfrac{7}{15}\ \dfrac{77}{145}[/latex]

[latex]\LARGE\dfrac{8}{8}=1[/latex]

Now let’s say that one of the three of them has another slice from the second pizza. They will have now eaten 1 whole pizza plus 1 slice.[latex]\LARGE1+\dfrac{1}{8}=1\dfrac{1}{8}\longleftarrow\text{Mixed number}[/latex]

This is what is known as a mixed number. A mixed number can be defined as the following:[latex]\LARGE\dfrac{9}{8}[/latex]

Now, we want to change a mixed number into an improper fraction, and then do the reverse and take an improper fraction and change it back to a mixed number.Example

Change the following mixed number into an improper fraction:

[latex]\LARGE1\dfrac{3}{4}[/latex]

**Step 1**: Change the whole number into a fraction, with the denominator being 4.

[latex]\LARGE1=\dfrac{4}{4}[/latex]

**Step 2**: Add the two fractions together. Now, we'll have to jump ahead a little here, as we haven't covered adding fractions yet. I'll give you the cheap and easy version here. As long as the denominators are the same, we are all good. When adding fractions, we simply keep the denominators the same and add the numerators. (We'll go through adding fractions thoroughly in the next chapter.)

[latex]\LARGE\dfrac{4}{4}+\dfrac{3}{4}=\dfrac{7}{4}[/latex]

So:[latex]\LARGE1\dfrac{3}{4}=\dfrac{7}{4}[/latex]

Another way to find your answer would be as follows: That may look a little confusing, but follow me through it. With the mixed number 1 ¾, take the 4 and multiply it by the 1. Then add 3, and you end up with 7. It's the same answer — just a different way of getting there.Practice Question A

Change the following into an improper fraction. Check the video answer to see how you did.

[latex]\LARGE3\dfrac{3}{8}[/latex]

https://video.bccampus.ca/id/0_6zl7fv1c?width=608&height=402&playerId=23448552Example

Change the following improper fraction into a mixed number:

[latex]\LARGE\dfrac{27}{6}[/latex]

**Step 1**: Find out how many times 6 goes into 27. We can do this using long division. The good news here is that we already went through long division in the first chapter. If you need to review it, take a look back to see how it's done (see Dividing Whole Numbers).

What we end up with is 6 going into 27 four times with 3 left over. So our mixed number becomes:

[latex]\LARGE4\dfrac{3}{6}[/latex]

Practice Question B

Change the following improper fraction into a mixed number. Check out the video answer to see how you did.

[latex]\LARGE\dfrac{17}{3}[/latex]

https://video.bccampus.ca/id/0_iepaq6aw?width=608&height=402&playerId=23448552[latex]\LARGE\dfrac{1}{2}[/latex]

We could then conclude that the two fractions represent the same thing mathematically, and they are just two different ways to represent the same thing. You could look at it this way: I cut two pieces of wood. One is 12 inches in length, and the other is 1 foot in length. They are the same length — their lengths are just expressed in different ways. So in the end, we end up with this:[latex]\LARGE\dfrac{4}{8}=\dfrac{1}{2}[/latex]

What we’ve done is reduced the fraction from 4 over 8 to 1 over 2 without changing the actual value represented. How this was done mathematically is we took the original numerator of 4 and divided it by 4. What is done to one part of the fraction must also be done to the other, so we also divided the denominator of 8 by 4, resulting in a fraction of 1 over 2. Doing the same thing to both the numerator and the denominator guarantees that the original fraction and the final fraction are equal in value. We reduce fractions when we can evenly divide the same number into both the numerator and the denominator. In our example, 4 can be divided into both. Note that the number 2 can also be divided into both the numerator and the denominator. If we divided both by 2, we would get: Although this still works mathematically, we often want to get a fraction into its lowest terms, meaning to a point where it can no longer be reduced. The fraction 2 over 4 could be reduced even further to 1 over 2, so there is further work we could do, if we chose to.Example

Let's go through the thought pattern when reducing fractions. Take the following fraction and reduce it to its lowest terms:
**Step 1**: What we want to do here is take a look at both the numerator and the denominator and determine if there is a number that can go into both of them. It might be easier if you write down numbers starting from 1 and then decide which numbers can go into both 8 and 12.
From this, we can conclude that the largest number that can go into both 8 and 12 is 4.
**Step 2**: Divide both the numerator and the denominator by 4.
There you have it: the fraction has now been reduced to its lowest terms. Always take a look at the answer when you are done, just to make sure that there definitely isn't another number that can go into the numerator and denominator, as this would mean the fraction could be reduced even further.

[latex]\LARGE\dfrac{8}{12}[/latex]

Example

Reduce the following fraction into its lowest terms:
**Step 1**: Determine if there is a number that can go into both the numerator and the denominator. If there is more than one number, then use the larger number.
This is a bit tougher than the first question, as the numbers are a lot larger and harder to work with. Going back to our times tables, we can see that 6, 8, and 12 all go into 24. We could also say that 24 goes into 24. But what about 168? What goes into that?
One thing we know for sure is that 2 goes into both, so why don't we start by taking each part of the fraction and dividing it by 2. If you have trouble dividing 168 by 2 in your head, go ahead and use your calculator.
**Step 2**: Determine if the fraction can be reduced any further. We can see that, once again, we can divide both numbers by 2.
**Step 3**: Repeat step 2 and determine if the fraction can be reduced any further. What we note this time is that 6 can go into both 6 and 42, so we divide both the numerator and the denominator by 6.

[latex]\LARGE\dfrac{24}{168}[/latex]

Question 1

[latex]\LARGE\dfrac{15}{18}[/latex]

https://video.bccampus.ca/id/0_7z8qtt2o?width=608&height=402&playerId=23448552Question 2

[latex]\LARGE\dfrac{24}{36}[/latex]

https://video.bccampus.ca/id/0_ocni4bjy?width=608&height=402&playerId=23448552[latex]\LARGE\dfrac{6}{16}+\dfrac{5}{16}+\dfrac{4}{16}=?[/latex]

What you’ll note is that the numerators are all different, while the denominators are all the same (16). When adding or subtracting fractions, the denominators must be the same. We refer to this as having a common denominator. So, in order to get the answer to the above question, you just add all the numerators. Adding fractions is very simple in this respect. Notice that the denominator in the final answer is the same as that in the fractions being added. By the end, the apprentices will have gone through 15 of the 16 chapters separately, and then they will go through the last chapter together. The concept of adding fractions with common denominators is easy enough, and we did enough adding whole numbers that going through examples at this point might not be worth it (but if you need a review, see Adding Whole Numbers). What we will do instead is write down some examples of adding fractions so you can see the idea.[latex]\LARGE\dfrac{1}{8}+\dfrac{2}{8}=\dfrac{3}{8}[/latex]

[latex]\LARGE\dfrac{5}{16}+\dfrac{6}{16}=\dfrac{11}{16}[/latex]

[latex]\LARGE\dfrac{13}{32}+\dfrac{11}{32}=\dfrac{24}{32}[/latex]

Do you notice anything about the answer to the last one? It can be reduced.[latex]\LARGE\dfrac{24}{32}\longrightarrow\dfrac{2}{3}[/latex]

Before we get going any further with work on fractions, this might be a good time to state that, when working with fractions, we generally want to put the answer in lowest terms.[latex]\LARGE\dfrac{5}{8}-\dfrac{2}{8}=\dfrac{3}{8}[/latex]

[latex]\LARGE\dfrac{9}{16}-\dfrac{5}{16}=\dfrac{4}{16}\longrightarrow\dfrac{1}{4}[/latex]

[latex]\LARGE\dfrac{27}{32}-\dfrac{14}{32}=\dfrac{13}{32}[/latex]

[latex]\LARGE\dfrac{1}{2}+\dfrac{3}{8}=?[/latex]

We can’t simply add up the numerators and the denominators, as it just won’t work. Take a look at the two circles drawn below. One is split into 2 parts, and one is split into 8 parts. Do you notice anything about the sizes of the parts? You’ll note that the parts in the 2-part circle are much larger than those in the 8-part circle. If we were to add up the parts in each of the circles, it would be like adding apples and oranges. So the idea becomes making it so that the parts we are adding are the same size. If we can somehow get to that point, then we are good to go, and we can add the two fractions. This is referred to as finding a common denominator, and most often, we try and find the[latex]\LARGE\dfrac{1}{2}+\dfrac{3}{8}=?[/latex]

The process here is similar to when we were putting fractions into their lowest terms in the last section, only this time, we will be increasing at least one of the denominators, and sometimes, we will be increasing both until we find one that is common. What we are looking for is a number that both denominators can go into evenly. In this example, we see that 2 can go into 8 and 8 can go into 8. This leaves us with a common denominator of 8. We have determined that 8 is going to be our common denominator, which means that one of the fractions is already good to go. But what about 1 over 2, or one-half? We have to turn the one-half into a fraction with 8 as the denominator. As we calculated above, 2 goes into 8 four times.[latex]\LARGE2\times4=8[/latex]

That's good for the denominator, but what about the numerator? Well, whatever we do to the one part of the fraction, we must do the same to the other part. This leaves the fraction with the same value. We then have to also multiply the 1 by 4.[latex]\LARGE1\times4=4[/latex]

If we wanted to do it all in one step, it would look something like this: Now we have something we can work with. Go back to the original equation and replace the [latex]\dfrac{1}{2}[/latex] with [latex]\dfrac{4}{8}[/latex].[latex]\LARGE\dfrac{4}{8}+\dfrac{3}{8}=\dfrac{7}{8}[/latex]

Okay, so that works for adding fractions, but what about subtracting fractions? Well, subtracting fractions follows the same principle: if the denominators are not the same, then we have to find a common denominator first before subtracting the two fractions.Example

Calculate the following:
**Step 1**: Find the common denominator. This can get a little tricky when the numbers begin to get larger. As you get more familiar with the patterns in numbers, the answers will come easier. The question we are asking right now is, "What number can both 8 and 16 go into evenly?"
We might even start by seeing if the smaller denominator can go into the larger denominator. In this case, it does.
The fraction with the common denominator of 16 is already good to go, but we have to work with the fraction with a denominator of 8.
**Step 2**: Multiply both the numerator and the denominator of ⅞ by 2 to give the fraction the common denominator of 16.
**Step 3**: Subtract the new versions of the fractions.

[latex]\LARGE\dfrac{7}{8}-\dfrac{13}{16}=[/latex]

[latex]\LARGE\dfrac{14}{16}-\dfrac{13}{16}=\dfrac{1}{16}[/latex]

Question 1

[latex]\LARGE\dfrac{3}{16}+\dfrac{5}{8}=[/latex]

https://video.bccampus.ca/id/0_535fcxuh?width=608&height=402&playerId=23448552Question 2

[latex]\LARGE\dfrac{5}{8}-\dfrac{5}{16}=[/latex]

https://video.bccampus.ca/id/0_1765fg0o?width=608&height=402&playerId=23448552Question 3

[latex]\LARGE\dfrac{1}{2}+\dfrac{7}{8}=[/latex]

https://video.bccampus.ca/id/0_8s54ppy3?width=608&height=402&playerId=23448552Question 4

[latex]\LARGE2\dfrac{1}{2}+1\dfrac{7}{8}=[/latex]

https://video.bccampus.ca/id/0_clcs9eo0?width=608&height=402&playerId=23448552[latex]\LARGE4\dfrac{3}{8}+3\dfrac{2}{8}=7\dfrac{5}{8}[/latex]

Pretty straightforward, right? You simply add the two whole numbers, and then add the fractions. It works out quite nicely. But what about a situation like the next example?[latex]\LARGE4\dfrac{5}{8}+3\dfrac{4}{8}=?[/latex]

Do you see the issue?

The issue (it's not really an issue) is that, when we add the fractions, we end up with a bigger number in the numerator than in the denominator.[latex]\LARGE4\dfrac{5}{8}+3\dfrac{4}{8}=7\dfrac{9}{8}[/latex]

The solution is to change the improper fraction part of the answer into a mixed number, and then add that to the whole number part of the answer.[latex]\LARGE\dfrac{9}{8}\longrightarrow1\dfrac{1}{8}[/latex]

Take the 7 and add it to the mixed number to get our final answer.[latex]\LARGE7+1\dfrac{1}{8}=8\dfrac{1}{8}[/latex]

Okay, that seemed pretty straightforward, but what about subtraction? Well, we follow the same rules. Take a look at the following example:Example

[latex]\LARGE8\dfrac{7}{8}-6\dfrac{3}{8}=?[/latex]

The procedure is similar to that for adding fractions, but instead of adding, we are subtracting. We can break it down into two parts. We start by subtracting the whole numbers, and then follow that up by subtracting the fractions portion.[latex]\LARGE8-6=2[/latex]

[latex]\LARGE\dfrac{7}{8}-\dfrac{3}{8}=\dfrac{4}{8}\rightarrow\dfrac{1}{2}[/latex]

[latex]\LARGE8\dfrac{7}{8}-6\dfrac{3}{8}=2\dfrac{4}{8}\rightarrow2\dfrac{1}{2}[/latex]

Example

[latex]\LARGE5\dfrac{2}{8}-3\dfrac{7}{8}=?[/latex]

The problem emerges not when you are subtracting the whole numbers, but when you are subtracting the fractions.[latex]\LARGE\dfrac{2}{8}-\dfrac{7}{8}=?[/latex]

We would end up with an answer less than zero. This is not going to work for us. So how do we solve the problem? Well, the answer lies in borrowing, and what we are borrowing from is the whole number, 5. Let's just say we borrow 1 from the 5. This would leave us with 4, and then what? Take a look at the following logic.[latex]\LARGE5=4+1[/latex]

[latex]\LARGE1=\dfrac{8}{8}[/latex]

If we go ahead and break the 5 down into 4 and 1, and then split that 1 down into parts of 8, we have a lot more eighths to work with. We can now put everything together to get the following:[latex]\LARGE5\dfrac{2}{8}=4+\dfrac{8}{8}+\dfrac{2}{8}=4\dfrac{10}{8}[/latex]

We now have numbers we can work with in our original question.[latex]\LARGE4\dfrac{10}{8}-3\dfrac{7}{8}=?[/latex]

We now follow the same steps as before.[latex]\LARGE4-3=1[/latex]

[latex]\LARGE\dfrac{10}{8}-\dfrac{7}{8}=\dfrac{3}{8}[/latex]

[latex]\LARGE4\dfrac{10}{8}-3\dfrac{7}{8}=1\dfrac{3}{8}[/latex]

Question 1

[latex]\LARGE7\dfrac{3}{16}+4\dfrac{5}{16}=[/latex]

https://video.bccampus.ca/id/0_s8dew6mc?width=608&height=402&playerId=23448552Question 2

[latex]\LARGE2\dfrac{7}{16}+3\dfrac{7}{8}=[/latex]

https://video.bccampus.ca/id/0_nxy48mwt?width=608&height=402&playerId=23448552Question 3

[latex]\LARGE8\dfrac{27}{32}-1\dfrac{15}{32}=[/latex]

https://video.bccampus.ca/id/0_tbjhhjze?width=608&height=402&playerId=23448552Question 4

[latex]\LARGE6\dfrac{5}{16}-5\dfrac{5}{8}=[/latex]

https://video.bccampus.ca/id/0_zomyq5wz?width=608&height=402&playerId=23448552[latex]\LARGE\dfrac{1}{4}\times\dfrac{1}{2}=?[/latex]

We’ll take a look at this visually, using a circle cut into parts to work this out. To start, we'll divide the circle into 4 equal parts. One of those parts would equal one-quarter of the circle. If we were to multiply that ¼ by ½, what we would be doing mathematically is taking ½ of the ¼ piece, or essentially splitting that ¼ into two equal parts. That would end up representing ⅛ of the circle. Mathematically, it's done like this:**Multiply the numerators together**

[latex]1\times1=1[/latex]

**AND**

**Multiply the denominators together**

[latex]2\times4=8[/latex]

What we end up with is this:[latex]\LARGE\dfrac{1}{2}\times\dfrac{1}{4}=\dfrac{1}{8}[/latex]

Example

Let’s go back to Abigail, Hanna, and Naomi. They’ve now completed another level of their schooling and are getting to the end of their apprenticeships. All three are working on the same job, which is a three-storey wood-frame building, and each is responsible for roughing-in 30 suites. They are required to wire ⅙ of those suites every week. One week, Hanna had to miss two days. Therefore, she only worked 3 out of the 5 days, or ⅗ of the time. What fraction of suites would she have been able to rough-in that week, taking into consideration her time away?
Start by writing down the fractions we are going to work with in this situation.

[latex]\dfrac{1}{6}\text{ The amount of suites needed to be completed during a 5-day work week.}[/latex] [latex]\dfrac{3}{5}\text{ The fraction of time worked during the week, 3 out of 5 days.}[/latex]Then multiply the two fractions together, sticking to our formula of multiplying the numerators together and then multiplying the denominators together.

[latex]\LARGE\text{numerators }1\times3=3[/latex]

[latex]\LARGE\text{denominators }6\times5=30[/latex]

And that makes the answer:[latex]\LARGE\dfrac{1}{6}\times\dfrac{3}{5}=\dfrac{3}{30}[/latex]

Which can then be reduced to its lowest terms:[latex]\LARGE\dfrac{3}{30}\rightarrow\dfrac{1}{10}[/latex]

Example

[latex]\LARGE\dfrac{5}{8}\times\dfrac{3}{4}=?[/latex]

[latex]\LARGE5\times3=15[/latex]

[latex]\LARGE8\times4=32[/latex]

[latex]\LARGE\dfrac{5}{8}\times\dfrac{3}{4}=\dfrac{15}{32}[/latex]

[latex]\LARGE\text{Final answer }=\dfrac{15}{32}[/latex]

Example

[latex]\LARGE4\dfrac{2}{5}\times2\dfrac{1}{4}=?[/latex]

Before you start, do you see a problem? The problem is that you are now trying to multiply two mixed numbers together. How does that work? Can you just go ahead and try to multiply them as they are? The answer is NO, but the solution to the problem is not that difficult: you just have to take one extra step before going through the process. The first thing you have to do is change each of the mixed numbers into improper fractions. From that point on, the process is the same.[latex]\LARGE4\dfrac{2}{5}=\dfrac{22}{5}[/latex]

(5 × 4 + 2 = 22)

[latex]\LARGE2\dfrac{1}{4}=\dfrac{9}{4}[/latex]

(4 × 2 + 1 = 9)

[latex]\LARGE22\times9=198[/latex]

[latex]\LARGE5\times4=20[/latex]

[latex]\LARGE\dfrac{22}{5}\times\dfrac{9}{4}=\dfrac{198}{20}[/latex]

[latex]\LARGE\dfrac{99}{10}=9\dfrac{9}{10}\text{ Mixed number}[/latex]

Question 1

[latex]\LARGE\dfrac{4}{7}\times\dfrac{3}{8}=[/latex]

https://video.bccampus.ca/id/0_102s3rf3?width=608&height=402&playerId=23448552Question 2

[latex]\LARGE\dfrac{6}{11}\times\dfrac{5}{9}=[/latex]

https://video.bccampus.ca/id/0_uwfhiu9a?width=608&height=402&playerId=23448552Question 3

[latex]\LARGE5\dfrac{1}{2}\times6\dfrac{3}{8}=[/latex]

https://video.bccampus.ca/id/0_iismlgjd?width=608&height=402&playerId=23448552Question 4

[latex]\LARGE7\dfrac{5}{9}\times8\dfrac{5}{7}=[/latex]

https://video.bccampus.ca/id/0_x03fh0rj?width=608&height=402&playerId=23448552[latex]\LARGE20÷10=2[/latex]

You end up with 2 groups of 10. Now divide those 20 screwdrivers by 5 (or into groups of 5).[latex]\LARGE20÷5=4[/latex]

You end up with 4 groups of 5. Now divide those 20 screwdrivers by 2 (or into groups of 2).[latex]\LARGE20÷2=10[/latex]

You end up with 10 groups of 2. Take a look at the math here. Do you see a pattern? What did you come up with? Did you notice that, when you take your original amount (in this case, 20) and divide it by a number that continues to get smaller (10, then 5, then 2), we end up with an answer that gets larger.[latex]\LARGE20÷10=2[/latex]

[latex]\LARGE20÷5=4[/latex]

[latex]\LARGE20÷2=10[/latex]

Follow this logic into fractions, keeping in mind that fractions are not only less than 10, 5, and 2, but 1. Using this pattern, we determine that dividing the 20 screwdrivers by a number less than 1 would get us a larger answer than if we divided the 20 by 10, 5, or 2. Try this. Take the 20 screwdrivers, and divide them by ½. What do you think your answer will be?[latex]\LARGE20÷\dfrac{1}{2}=?[/latex]

Following our logic, the answer should be more than 10, and in fact, it is.[latex]\LARGE20÷\dfrac{1}{2}=40[/latex]

This does not mean that we end up with 40 screwdrivers, though. What it means is that we end up with 40 parts of screwdrivers. You have to imagine that each of the screwdrivers were split into 2. Twenty screwdrivers split in half would give us 40 pieces in the end. The question now becomes, how do we do this mathematically? The answer lies in using what is known as a reciprocal. Here is the definition.[latex]\LARGE5=\dfrac{5}{1}[/latex]

Using our definition of reciprocal, we need to find a number that, when multiplied by[latex]\LARGE\dfrac{5}{1}\times\dfrac{1}{5}=1[/latex]

In the end, to find the reciprocal of a fraction, we simply take the numerator and make it the denominator and take the denominator and make it the numerator. Essentially, we are just flipping the fraction around. Here are some more examples of reciprocals.[latex]\LARGE\dfrac{3}{8}\text{ and }\dfrac{8}{3}[/latex]

[latex]\LARGE\dfrac{2}{9}\text{ and }\dfrac{9}{2}[/latex]

[latex]\LARGE\dfrac{24}{17}\text{ and }\dfrac{17}{24}[/latex]

Okay, so now that we have the reciprocal issue out of the way, the question then becomes, why do we need reciprocals in the first place? Well, the answer lies in the rule for dividing fractions. The rule for dividing fractions is you take the first fraction and multiply it by the reciprocal of the second fraction. Yes, you heard that right: to divide, you end up multiplying, but only after first flipping the second fraction around. Flipping the second fraction around (finding its reciprocal) changes the value of the equation. In order to keep the equation mathematically the same, we have to change the division question into a multiplication question. Take a look at the following example to see how this is done.Example

[latex]\LARGE\dfrac{1}{2}÷\dfrac{3}{8}=?[/latex]

[latex]\LARGE\text{The reciprocal of }\dfrac{3}{8}\text{ is }\dfrac{8}{3}[/latex]

[latex]\LARGE\text{Check: }\dfrac{3}{8}\times\dfrac{8}{3}=\dfrac{24}{24}=1[/latex]

So we end up with:[latex]\LARGE\dfrac{1}{2}÷\dfrac{3}{8}\text{ becomes }\dfrac{1}{2}\times\dfrac{8}{3}=?[/latex]

**Multiply the numerators together**

[latex]\LARGE1\times8=8[/latex]

**Multiply the denominators together**

[latex]\LARGE2\times3=6[/latex]

[latex]\LARGE\dfrac{1}{2}\times\dfrac{8}{3}=\dfrac{8}{6}[/latex]

[latex]\LARGE\dfrac{4}{3}=1\dfrac{1}{3}\text{ Mixed number}[/latex]

Final answer:[latex]\LARGE\dfrac{1}{2}÷\dfrac{3}{8}=1\dfrac{1}{3}[/latex]

Example

[latex]\LARGE\dfrac{5}{9}÷\dfrac{7}{4}=?[/latex]

[latex]\LARGE\dfrac{5}{9}÷\dfrac{7}{4}\text{ becomes }\dfrac{5}{9}\times\dfrac{4}{7}[/latex]

**Multiply numerators together**

[latex]\LARGE5\times4=20[/latex]

**Multiply denominators together**

[latex]\LARGE9\times7=63[/latex]

[latex]\LARGE\dfrac{5}{9}\times\dfrac{4}{7}=\dfrac{20}{63}[/latex]

[latex]\LARGE\dfrac{5}{9}÷\dfrac{7}{4}=\dfrac{20}{63}[/latex]

Question 1

[latex]\LARGE\dfrac{7}{8}÷\dfrac{7}{16}=[/latex]

https://video.bccampus.ca/id/0_f64q56vn?width=608&height=402&playerId=23448552Question 2

[latex]\LARGE4\dfrac{7}{8}÷2\dfrac{3}{4}=[/latex]

Note: This one is slightly different from what we have been doing, as it involves dividing mixed numbers. What do you think you are going to have to do when dealing with this?A number of years ago, a student named Harpreet went through a plumbing apprenticeship program. After completing his schooling and receiving his Red Seal plumbing ticket, he has decided to open up his own plumbing company.

Like most new entrepreneurs, he experienced a steep learning curve, especially when it came to ordering material for jobs and organizing and scheduling when he had a number of jobs on the go. Although he might not be thinking about counting numbers or whole numbers, he has ended up working with these when ordering materials and during the day-to-day operations of the company. If we want to start with the very basics, we must start withThe discovery of the number zero was a big step in the history of mathematics. Don’t ask me why, but apparently it was. As for Harpreet, let’s hope that he bids jobs properly and that his profit doesn’t amount to zero.Counting numbers: 1, 2, 3, 4, 5, …Whole numbers: 0, 1, 2, 3, 4, 5, …

I would have:

[latex]\LARGE2\times$100=$200[/latex]

[latex]\LARGE3\times$20=$60[/latex]

[latex]\LARGE1\times$5 =$5[/latex]

Adding those up, I would have:[latex]\LARGE$200+$60+$5=$265[/latex]

Notice that each of the original values has a specific place in the total. Positioning the digits in the reverse order would make it appear as if we had a lot more money than we actually have:[latex]\LARGE$562[/latex]

This is the importance of the place value system. Another aspect of this system is known as the5 = ones place

6 = tens place

2 = hundreds place

Each of the digits in the place value system can have a value anywhere between zero and nine. When the value of a digit increases past nine, we start again at zero but add one to the value of the digit in the next highest place value.The simplest way to think about this is to go from 9 to 10:

In this case, the ones place goes back to zero and the tens place increases by one. The ones place has gone from nine to zero. There was no value in the tens place when we started, but now, the tens place has increased by one. Now, you might be thinking that this really is far too basic to be reading, but these are the building blocks of the language of algebra. Although the example used here is quite simple, when you get to larger and more complex numbers, the mathematical grammar that you are learning here is the basis for understanding more complicated mathematical grammar in the chapters to follow. Now let's look at a bigger number. One way to understand the place value system is to take a number such as:[latex]\LARGE5,217,364,958[/latex]

If we were to sound out this number, we would say:Ones | 8 |

Tens | 5 |

Hundreds | 9 |

Thousands | 4 |

Ten thousands | 6 |

Hundred thousands | 3 |

Millions | 7 |

Ten millions | 1 |

Hundred millions | 2 |

Billions | 5 |

Ten billions | – |

Hundred billions | – |

- The digit 5 is in the billions place. Its value is 5,000,000,000.
- The digit 2 is in the hundred millions place. Its value is 200,000,000.
- The digit 1 is in the ten millions place. Its value is 10,000,000.
- The digit 7 is in the millions place. Its value is 7,000,000.
- The digit 3 is in the hundred thousands place. Its value is 300,000.
- The digit 6 is in the ten thousands place. Its value is 60,000.
- The digit 4 is in the thousands place. Its value is 4,000.
- The digit 9 is in the hundreds place. Its value is 900.
- The digit 5 is in the tens place. Its value is 50.
- The digit 8 is in the ones place. Its value is 8.

Example

As you go through this question, think of the actual value of each of the digits. For example, the 4 in the leftmost place of the number below represents 40,000,000. Doing this will help you to visualize the value each number actually stands for.
Find the place value of each of the following digits:

[latex]\LARGE45,837,249[/latex]

Use the following layout as a guideline to help you answer the question. 4 = ___________________ 5 = ____________________ 8 =____________________ 3 =____________________ 7 =____________________ 2 = ____________________ 4 = ____________________ 9 = ____________________ Inserting the numbers into the table below is also a good place to start:Question 1

[latex]\LARGE385,922,102[/latex]

https://video.bccampus.ca/id/0_hsf5qwsc?width=608&height=402&playerId=23448552Question 2

[latex]\LARGE20,433,876,011[/latex]

https://video.bccampus.ca/id/0_63724jxg?width=608&height=402&playerId=23448552First, look at 53, which has 5 tens and 3 ones. Then, look at 34, which has 3 tens and 4 ones. Add the tens up: 5 tens + 3 tens = 8 tens. Add the ones up: 3 ones + 4 ones = 7 ones.When we put the original numbers in the place value system and add, we get a total of 87. Separating this number into each of its digits and positioning them in the place value system can help us visualize the value of each of the digits. Follow the steps outlined below to see how this process looks mathematically.

Example

We’ll use the same numbers from the previous example:
**Step 1**: Put the question into a formula that is easy to work with.
**Step 2**: Add up the ones. In this case, we have 3 and 4. Together, they add up to 7.
**Step 3**: Add up the tens. In this case, we have 5 and 3. Together, they add up to 8.

The word "sum" is math language for adding numbers together. When we say, "What is the sum of 27 and 45?" what we are saying is, "What is the total of those numbers if we added them together?"

Example

Find the sum of 27 and 45.
To get the sum of these numbers, we will go back to our three steps.
**Step 1**: Put the question into a formula that is easy to work with.
**Step 2**: Add up the ones. In this case, we have 7 and 5. Together, they add up to 12. This is where we do what is called “carrying the one.” Take a look at the following image to see this in action:
**Note:** In this case, we do not put the number 12 at the bottom of the equation. Instead, we “carry the one” into the next spot (the tens) in the place value system.
**Step 3**: Add up the tens. In this case, we have 2 and 4. Together, they add up to 6. We also have to take into account the one that we carried over. We add that in as well to get a total of 7 in the tens place.
**Note:** If it turned out that the tens added up to more than 9, we could carry the one again into the hundreds column. Take a look below to see how this works.
As there are no other values in the hundreds column in this question, we just go ahead and place the one in the hundreds column.

Question 1

https://video.bccampus.ca/id/0_sp8cz8lj?width=608&height=402&playerId=23448552

Question 2

https://video.bccampus.ca/id/0_zafmjqkz?width=608&height=402&playerId=23448552

[latex]\LARGE17-5=12[/latex]

OR

[latex]\LARGE17-2=5[/latex]

We'll use the first way it is written down as our example, and then what we'll do is change that into another format that will be easier to work with. What you might note is that the way the question is written is similar to how we worked the equation when we were adding whole numbers. Writing it this way gives us a better representation of the ones and the tens columns, which we'll need to use when working through the question. Subtracting, like adding, requires us to work through each of the columns one by one until we reach our final answer. We'll answer this question visually in order to get the picture. Start with 7 apples in the ones column. Remove (or subtract) 5 of those apples, and you are left with 2 apples. That takes care of the ones columns. We started with 7 apples, then we subtracted 5 apples, leaving us with 2 apples in the ones column. We would say that 2 is the difference between 7 and 5. Now off to the tens column. What is interesting to note here is that there is only one number in the tens column, and that happens to be the number 1. This makes things easy, as there is no work for us to do. We just move the 1 down into the tens column of the answer, and we then have our final answer. Okay, that was pretty straightforward. Now we'll try something a little more challenging.Example

Imagine things worked out differently for Harpreet and Jamieson. Let’s say that, of those 17 jobs they bid on and successfully got, they had to turn down 9 of them. How many of those jobs would they have taken? Before you continue reading and see the answer, try and visualize what it would look like when we put those numbers into the formula. Do you see the problem?
If we were to start with the ones column as we did in the last example, the problem would show up right away. The problem is that, if you try to subtract 9 from 7, you would end up below zero. So we have to come up with some method of subtracting that compensates for that.

[latex]\LARGE7-9=\text{less than zero}[/latex]

What we end up doing is borrowing from the tens column. We would end up with something that looked like this: When we borrow from the tens column, we are borrowing a value of ten and adding it to the ones column to help the ones. We end up with 17 in the ones column, which is now more than enough to deal with the 9 being subtracted. Also, whatever value we had in the tens column (in this case, it was a 1) is reduced by 1 to account for the fact that it has been borrowed.Example

We’ll go through another example in which we return to our plumbers, Harpreet and Jamieson. The three employees they hired are Dixon, Kavanir, and Arman, and it's a good thing they hired them. Those 17 jobs they originally bid on involved the installation of 246 fixtures, including bathtubs, toilets, and sinks. Harpreet and Jamieson could not do all that work by themselves.
But, due to the fact that they turned down 5 jobs, they won’t need to install 75 of those fixtures, so with the help of the three new employees, they should be able to complete all the jobs. The question is, "How many fixtures will they have to install?"
As usual, start with the mathematical formula that allows us to properly answer the question. But this time, we’ll go through the process using steps, so that when you look back, you can see how it breaks down.
You may have noticed in the first couple of sections that we often go through examples using steps. This is done to break down a large question into manageable parts. If you follow this idea when working through math problems, it can help keep you on track.
Okay, back to the problem:
**Step 1**: Put the question into a format that is easy to work with.
**Step 2**: Subtract the ones. In this case, we have 6 minus 5, which equals 1.
**Step 3**: Subtract the tens. Here, we have the issue wherein the number on top (4) is less than the number on the bottom (7), so we have to borrow from the hundreds. The 2 in the hundreds column has to be reduced by 1, and then that 1 is added to the tens. We end up with 14 minus 7, which equals 7.
**Step 4**: Subtract the hundreds. In this case, we only have the 1, so 1 minus 0 is 1.

Question 1

https://video.bccampus.ca/id/0_sbor44zz?width=608&height=402&playerId=23448552

Question 2

https://video.bccampus.ca/id/0_51r50q8w?width=608&height=402&playerId=23448552

[latex]\LARGE\text{Four times five OR }4\times5[/latex]

If we looked at it from a more visual perspective, it would look like this: In this case we could physically count up all the wrenches, and we would end up with 20. We could also count up each row of 5 and add them as follows:[latex]\LARGE5+5+5+5=20[/latex]

Or we could take one row of 5 and multiply it 4 times because there are 4 rows. The equation would look like this:[latex]\LARGE4\times5=20[/latex]

Each method should get us the same answer, but going 4 times 5 gets us to the answer quicker. If we were to multiply larger numbers such as 8 time 9 (8 × 9), we could spend a long time counting wrenches or a long time adding numbers. Multiplication simplifies the process. Before moving on to multiplying larger numbers, why don't we go through another example of multiplying smaller numbers and use visuals to help us with our answer.Example

Harpreet decides to buy some screwdrivers for the van and decides that each van needs 7 screwdrivers. He also decides to buy 2 extra sets for the shop. In total he decides to buy 6 sets of screwdrivers.
The first thing we should do is write the question in a form that we can work with.
Remember you can also think of it like this:

[latex]\LARGE7+7+7+7+7+7=[/latex]

And once again we could look at it visually and count up the screwdrivers. If we were to count them up, we would find that we have 42 screwdrivers. If we took our calculator and plugged in 6 times 7, we would get the same answer. If we added 7 plus 7 plus 7 and so on, we would also end up with 42.
The word "product" in math terms means the multiplying of two numbers together.

One thing to keep in mind is that the process we are going to work our way through here takes some time and is a lot of work. If during the process you are asking yourself, Example

[latex]\LARGE437\times392=?[/latex]

[latex]\LARGE437\times392 =171,304[/latex]

Question 1

https://video.bccampus.ca/id/0_u8edz3lv?width=608&height=402&playerId=23448552

Question 2

https://video.bccampus.ca/id/0_g37moqi9?width=608&height=402&playerId=23448552

[latex]\LARGE12÷4=?[/latex]

How would we go about solving this? Well let’s look at it visually to start. We have 12 jobs to work with. What we do now when dividing is take the number we have (12) and split it into the number of groups we are going to have. In this case we have 4 employees so we will split it into 4 even groups. You can now count that there are 3 jobs in every group. So what we get in the end is:[latex]\LARGE12÷4=3[/latex]

Like multiplying, dividing small whole numbers is usually pretty straight forward. But what if we had a case where we have larger numbers. How would you go about doing this calculation without the use of a calculator? What we use is a system called “LONG DIVISION” and it looks something like the following:Example

We’ll use this process of long division to answer the following question. How many times does 5 go into 90?

**Step 1:** Set up the equation in a workable form.

**Step 2:** Take the 5 and divide it into the first digit in the number to be divided into. In this case this is the 9. We have to figure out how many times 5 goes into 9 without going over.

This is where knowing our times tables really comes in handy. We know that:

[latex]\LARGE5\times1=5[/latex]

and

[latex]\LARGE5\times2=10[/latex]

Therefore we know that 5 goes into 9 one complete time with 4 left over.

To complete this step we need to bring down the next number in the equation. In this case, it’s the 0. What we have now is 40 and the next step becomes how many times 5 goes into 40.

**Step 3:** Figure out how many times 5 goes into 40 without going over. If we go back to our multiplying, we might be able to remember that 5 goes into 40 exactly 8 times.

Example

How many times does 7 go into 167?
**Step 1:** Set up the equation in a workable form.
**Step 2:** Take the 7 and divide it into the first digit in the number to be divided into. In this case this is the 1, and we can see that this is not going to work. As such, we just move one step to the right and divide the 7 into 16.
To complete this step we need to bring down the next number in the equation. In this case it’s the 7. We now have 27 and the next step becomes how many times 7 goes into 27.
**Step 3:** Figure out how many times 7 goes into 27 without going over.
We end up with a little different scenario here than in the first question. When we get to the end of the question we have 6 left over. What this is telling us is that 7 goes into 167 twenty three times with 6 left over.

Question 1

[latex]\LARGE130÷8=[/latex]

https://video.bccampus.ca/id/0_oh503eff?width=608&height=402&playerId=23448552Question 2

[latex]\LARGE684÷12=[/latex]

https://video.bccampus.ca/id/0_zar81rgd?width=608&height=402&playerId=23448552[latex]\LARGE10\times10=100[/latex]

We would end up with 100 parts, or we could look at it like 100 cents. Now, take the fifth box out, divide that box into 10 parts, and fill up 7 of them. Imagine that each of the original 10 boxes are split into 10 even parts. What gets crazier is that we could then take each of those 100 boxes and split them into another 10 parts each. What this would leave us with is a number that takes us to the thousandths place, because we would be splitting things up into 1000 parts. What we are really doing with decimals is taking a whole number and breaking it down into parts, similar to fractions. In fact, if we were to look at decimals like fractions, this is how we would think about it:[latex]\LARGE\dfrac{1}{10}=0.1[/latex]

[latex]\LARGE\dfrac{1}{100}=0.01[/latex]

[latex]\LARGE\dfrac{1}{1000}=0.001[/latex]

Now that we have the first three digits after the decimal point figured out, let's work our way even further. Take a look at the number below. To the left of the decimal point is the whole number, and to the right is the decimal or part of a whole number.[latex]\LARGE123.456789[/latex]

Note: 1, 2, and 3 are all part of the whole number, and 4, 5, 6, 7, 8, and 9 are all part of the decimal. Using the place value system for both whole numbers and decimals, we would end up with the following:- 1 = hundreds place
- 2 = tens place
- 3 = ones place
- (.) = decimal
- 4 = tenths place
- 5 = hundredths place
- 6 = thousandths place
- 7 = ten thousandths place
- 8 = hundred thousandths place
- 9 = millionths place

Example

In the number below, indicate which digit is in the thousandths place.

[latex]\LARGE57.29652[/latex]

The easiest way to do this is to write out the number, and then, starting from the left, indicate the place value of each of the digits.- 5 = tens place
- 7 = ones place
- 2 = tenths place
- 9 = hundredths place
**6 = thousandths place**- 5 = ten thousandths place
- 2 = hundred thousandths place

Example

In the number below, indicate which digit is in the hundred thousandths place.

[latex]\LARGE369.246813[/latex]

Once again, to find the answer, write down the number, and then indicate the place value of each of the digits.- 3 = hundreds place
- 6 = tens place
- 9 = ones place
- 2 = tenths place
- 4 = hundredths place
- 6 = thousandths place
- 8 = ten thousandths place
**1 = hundred thousandths place**- 3 = millionths place

- 8.16
- 8.151
- 8.1513
- 8.15138

Example

Which of the following is the largest number?
**Step 1**: Start with the whole number. Which number has the largest whole number? As they are all equal, this will not be the determining factor.
**Step 2**: Go to the first digit to the right of the decimal. It works out that the digit is a 3 in all cases. Therefore, this is not the deciding factor, either.
**Step 3**: Go one place to the right again. This would be the hundredths place. Here, we find the first difference. Answer a gives us the largest number in this column. Therefore, answer a is the largest number.

- 7.34
- 7.332
- 7.323
- 7.3234

**The whole number in all four cases is a 7. This is not the deciding factor.**

**The number in the tenths column in all four cases is a 3. This is not the deciding factor.**

**The digit in the hundredths column is different for each number. In this case, the number with the highest value in the hundredths column is the largest number.**

[latex]\LARGE7.34[/latex]

Question 1

- 27.5
- 27.54
- 27.539
- 27.5392

Question 2

- 14.001
- 13.099
- 13.999
- 13.001

**How many did you count?**

- 12
- 27
- 45
- 98

- 12 rounds down to 10
- 27 rounds up to 30
- 45 rounds up to 50
- 98 rounds up to 100

Example

Round the following number to the nearest hundredth.
**Step 1**: Identify which number is in the hundredths column.
**Step 2**: Look at the number directly to the right of the hundredths column. In this case, it would be the digit in the thousandths column, which is 6.
**Step 3**: Follow the rules of rounding.
**RULE**: If the number is 4 or less, round down. If the number is 5 or greater, round up.
As we have 6 in the thousandths column, we round up.
Therefore, 123.4567 rounded to the nearest hundredth is:

[latex]\LARGE123.45678[/latex]

[latex]\LARGE123.46[/latex]

Question 1

Round to the nearest tenth.

[latex]\LARGE27.71536[/latex]

https://video.bccampus.ca/id/0_rpzfqn1j?width=608&height=402&playerId=23448552Question 2

Round to the nearest thousandth.

[latex]\LARGE14.75638[/latex]

https://video.bccampus.ca/id/0_ll2glhtt?width=608&height=402&playerId=23448552Question 3

Round to the nearest hundredth. This is a new concept. Before checking the video answer, give this one a try and see how you do.

[latex]\LARGE75.9999[/latex]

https://video.bccampus.ca/id/0_77elprhp?width=608&height=402&playerId=23448552Example

We’ll start with an easy decimal like 0.25 and work towards changing this into a fraction.
**Step 1**: Put the question into fraction form such that the decimal is over 1.
**Step 2**: Take both the numerator and the denominator and multiply each by 10 for every digit to the right of the decimal point. In this case, we have 2 digits to the right. Therefore, we multiply each by 100 (10 × 10).
**Step 3**: Reduce the fraction to its lowest terms, and we end up with the final answer.

[latex]\LARGE\dfrac{0.25}{1}[/latex]

Example

Change the decimal 0.729 into a fraction.
**Step 1**: Put the equation into fraction form such that the decimal is over 1.
**Step 2**: Take both the numerator and the denominator and multiply each by 10 for every digit to the right of the decimal point. In this case, we have 3 digits to the right. Therefore, we multiply each by 1000 (10 × 10 × 10).
**Step 3**: Reduce the fraction to get our final answer.
This is a tough one to reduce. My suggestion would be to do this in steps. Start with small numbers like 2 and 3. Does the number 2 go into both the numerator and the denominator? The answer would be no. How about 3? Also no. Continue this process until you find a number that works.
It is possible that you will not find a number and we are already in lowest terms. This is the case here. Our final answer is simply:

[latex]\LARGE\dfrac{0.729}{1}[/latex]

[latex]\LARGE\dfrac{729}{1000}[/latex]

Question 1

[latex]\LARGE0.362[/latex]

https://video.bccampus.ca/id/0_z81n8xao?width=608&height=402&playerId=23448552Question 2

[latex]\LARGE0.963[/latex]

https://video.bccampus.ca/id/0_wqbz70ju?width=608&height=402&playerId=23448552Example

[latex]\LARGE1\text{ foot}=12\text{ inches}[/latex]

Take the decimal of a foot and multiply it by 12.[latex]\LARGE0.9384\times12=11.2608\text{ inches}[/latex]

[latex]\LARGE0.2608\times16=4.1728[/latex]

When we are looking at 4.1728, what we are actually looking at is:[latex]\LARGE\dfrac{4.1728}{16}[/latex]

If we had taken the 0.2608 inches and multiplied it by 8, our answer would be in eighths of an inch. If we had multiplied by 4, our answer would be in fourths (or quarters) of an inch.[latex]\LARGE\dfrac{4.1728}{16}\rightarrow\dfrac{4}{16}\rightarrow\dfrac{1}{4}[/latex]

Our final answer then becomes:Example

Change the following feet and decimals of a foot into feet, inches, and sixteenths of an inch.
**Step 1**: Change the decimal of a foot into inches.
**Step 2**: We are left with 8 inches and decimals of an inch. The 8 inches is good the way it is, but we need to change the decimal of an inch into fractions of an inch. We need to change the decimal of an inch into sixteenths, so multiply it by 16.
**Step 3**: Round the answer to the nearest fraction of an inch, and then reduce the fraction if necessary.

[latex]\LARGE7.6939\text{ feet}[/latex]

[latex]\LARGE0.6939\times12=8.3268\text{ inches}[/latex]

[latex]\LARGE0.3268\times16=5.2288[/latex]

[latex]\LARGE\dfrac{5.2288}{16}\rightarrow\dfrac{5}{16}[/latex]

Our final answer becomes:Question 1

[latex]\LARGE9.1234\text{ feet}[/latex]

https://video.bccampus.ca/id/0_6iseqgh6?width=608&height=402&playerId=23448552Question 2

[latex]\LARGE0.058\text{ feet}[/latex]

https://video.bccampus.ca/id/0_6iseqgh6?width=608&height=402&playerId=23448552[latex]\LARGE\dfrac{1}{2}\text{ to }0.5[/latex]

Once again, the quick and easy way to do this is to go through a few examples. Start with a plain old fraction and work towards changing this into a decimal.[latex]\LARGE\dfrac{3}{8}[/latex]

This can be done a number of ways, with the simplest being:[latex]\LARGE3÷8=0.375[/latex]

You could plug the numbers into your calculator, or you could use long division to come up with the answer.That’s it. You’re done!

Okay, that was pretty easy, but now let’s move on to an example that is a bit more difficult. We are going to change a number written in feet, inches, and fractions of an inch into a decimal.Example

Change this number into a decimal:
**Step 1**: Identify any part of the number that is already in a form that works. In this case, the 7 feet is good to go and nothing needs to be done with it.
**Step 2**: Take the fraction of an inch and turn it into a decimal.
**Step 3**: Change the inches and decimals of an inch into decimals of a foot using the fact that 1 foot = 12 inches.
In the last section, when we went from decimals to fractions, we multiplied by 12 to get inches. As we are doing the reverse calculation here by going from fractions to decimals, we end up dividing by 12 inches to get decimals of a foot.

[latex]\LARGE\dfrac{5}{16}\text{ OR }5÷16=0.3125[/latex]

Pause here for a moment and ask yourself what unit the 0.3125 is in. All we have done is change the fraction of an inch into decimals of an inch. We now have turned the 9 inches and fractions of an inch into 9 inches and decimals of an inch.[latex]\LARGE9\dfrac{5}{16}\text{ inches into }9.3125\text{ inches}[/latex]

[latex]\LARGE9.3125÷12=0.776[/latex]

Once again, ask yourself what units you end up with here. When we take the inches and divide it by 12, we end up with feet or, in this case, decimals of a foot. So our final answer becomes:Example

Change the following number written in feet, inches, and fractions of an inch into feet and decimals of a foot.
**Step 1**: Identify any part of the number that is already in a form that works. In this case, the 3 feet is good to go and nothing needs to be done with it.
**Step 2**: Take the fraction of an inch and turn it into a decimal.
**Step 3**: Change the inches and decimals of an inch into decimals of a foot using the fact that 1 foot = 12 inches.

[latex]\LARGE\dfrac{3}{8}\text{ OR }3÷8=0.375[/latex]

Put the inches and decimals of an inch together.[latex]\LARGE6\dfrac{3}{8}\text{ inches into }6.375\text{ inches}[/latex]

[latex]\LARGE6.375÷12=0.531[/latex]

So our final answer becomes:Question 1

https://video.bccampus.ca/id/0_n3ejv8mt?width=608&height=402&playerId=23448552

Question 2

https://video.bccampus.ca/id/0_612e14f7?width=608&height=402&playerId=23448552

92%

The symbol % means percent and can be used in place of writing the actual word percent. You can find this symbol on your computer keyboard, generally on the same key as one of the numbers. On my keyboard, the percent symbol is on the same key as the number 5. To get the % symbol, you would press the [Shift] and [5] keys at the same time.

The following is an example of percent.
According to a recent survey by BCIT Piping Foundation Instructors, 25% of students coming into the Piping Foundation Program would like to get into the plumbing trade, 20% would like to get into the steamfitting trade, and 10% would like to get into the gasfitting trade. The remaining 45% of students are undecided when they begin the program. If you add up those percentages, they equal 100.
Piping Trade | Percentage |
---|---|

Plumbing | 25% |

Gasfitting | 10% |

Undecided | 45% |

Total | 100% |

[latex]\LARGE3:8[/latex]

Notice how the ratio is written. It has its own style, just like a fraction does. A ratio of 3:8 means that we have eaten 3 of the 8 pieces in the pizza. We could also write a ratio that identified how many pieces of the pizza were eaten and how many were not eaten. The ratio for that would look like:[latex]\LARGE3:5[/latex]

In this case, 3 pieces of the pizza were eaten while 5 pieces were not. Regardless of which way we describe our pizza eating, what we are dealing with is the relationship between two numbers. Here is another example: let’s say we are on a job site, and we have to install some pipe. We have 42 feet of plastic pipe and 79 feet of steel pipe to install. What is the ratio of plastic pipe to steel pipe that we need to install? The answer:[latex]\LARGE42:79[/latex]

Another question might be how much of the pipe is plastic in relation to how much total pipe we have. In this case, the ratio would look like:[latex]\LARGE42:121[/latex]

In this case, the 121 is derived from adding the plastic pipe and the steel pipe lengths together.[latex]\LARGE42+79=121[/latex]

At this point, you might be thinking that this looks and sounds familiar, and you would be correct. Ratios are similar to fractions, and each ratio can be written as a fraction.[latex]\LARGE\dfrac{42}{121}[/latex]

We would say that 42 feet out of a total of 121 feet of pipe is plastic. All right: we’ve added a few new things to our math library here, but you might be asking right about now, “Aren’t we dealing with percentages in this chapter? How does all this ratio and fraction stuff relate to percentages?" The idea is we can take these ratios or fractions and turn them into percentages by making the ratio or fraction out of 100. In truth, we don’t even need to make a ratio or fraction out of 100 to do this, but it’s a good place to start our understanding of the math behind the whole process. We are at a point where we can finally introduce you to some new people in the story. For the story of percentage, we’re going to use apprentices from carpentry, electrical, and plumbing, in addition to statistics.[latex]\LARGE97+123+80=300[/latex]

What we can do now is put each of the apprentice totals into both a ratio and a fraction of the whole. If percentages are based on 100, then we have to translate the ratio and the fraction into forms based on 100. In other words, we have to make it so that the number on the right in the ratio is 100 and the denominator in the fraction is 100. For this, we have to go back to our work with fractions. We’ll start with the 97 carpentry apprentices. We have 97 apprentices out of a total of 300. Our goal here is to get this fraction down to a point where the denominator is 100. We are essentially reducing the fraction. Luckily for us, going from 300 down to 100 is quite easy. We just divide the denominator by 3, and then also divide the numerator by 3. We could now write this number as a ratio.[latex]\LARGE32.33:100[/latex]

What we end up with is the fact that 32.33% of the trades apprentices are carpentry apprentices. Once we find out the number of carpentry students per 100 students, we automatically have our percentage. In just a bit, we’ll talk about working with numbers that don’t give us a nice even number of 100 and simply translate to percentage in one step. But for now, let’s continue on with our apprentices.Example

Find the percentage of electrical students.
**Step 1**: Put the numbers into an equation that we can work with. In this case, put the numbers into a fraction.
**Step 2**: Turn the fraction into one with a denominator of 100.
As a ratio, we would have:

[latex]\LARGE\dfrac{123}{300}[/latex]

[latex]\LARGE41:100[/latex]

And finally, as a percentage, we would get:[latex]\LARGE41%[/latex]

Practice Question

Now calculate the percentage of plumbing apprentices. Check the video to see if your answer is correct.
https://video.bccampus.ca/id/0_dloiiah6?width=608&height=402&playerId=23448552

[latex]\LARGE57÷97=0.59[/latex]

Now take the 0.59 and turn that into a percentage by multiplying the 0.59 by 100. This moves the decimal point 2 spots to the right and leaves us with a whole number. It's as simple as that.[latex]\LARGE0.59\times100=59%[/latex]

Example

Using the same method we just used with the carpentry apprentices, find the percentage of plumbing students in the trades college.
**Step 1**: Write down the equation in a format we can use.
**Step 2**: Work this out in your head. Just kidding: grab your calculator and plug the numbers in.
**Step 3**: Multiply the answer by 100 to put it into a percentage.

[latex]\LARGE80÷300=?[/latex]

[latex]\LARGE80÷300=0.267[/latex]

[latex]\LARGE0.267\times100 = 26.7%[/latex]

Question 1

The electrical students are wiring an electrical project as part of their practical mark. The students must receive a minimum of 35 out of a possible 50 marks. What percentage is that?
https://video.bccampus.ca/id/0_hardemes?width=608&height=402&playerId=23448552

Question 2

The plumbing students have been asked to move a bunch of pipe from one side of the shop to the other. There are 279 pieces of cast iron pipe that have to be moved. At the end of the day, they have moved 222 pieces. What percentage of the pipe have they moved?
https://video.bccampus.ca/id/0_ihhz5g2j?width=608&height=402&playerId=23448552

Question 3

Referring to question 2, what percentage of the pipe do the students still have to move in relation to the original amount?
https://video.bccampus.ca/id/0_1nyxdvv8?width=608&height=402&playerId=23448552

[latex]\LARGE\dfrac{12}{100}[/latex]

What this will eventually tell us is that, for every 100 candies there are in the jar, Patrick will get 12 of them. Now turn that fraction into a decimal by dividing 12 by 100.[latex]\LARGE12÷100=0.12[/latex]

Have you noticed a pattern when it comes to decimals? Here is the relationship showing a few different ways in which we can end up with 12%:[latex]\LARGE12\%=\dfrac{12}{100}=0.12[/latex]

All three of these numbers represent the same amount. Learning to work between them is important in math, and it's also important to begin to see the relationships between numbers. Okay, back to Patrick. The final step in this situation is to take the 0.12, which is really 12%, and multiply it by the number of candies in the jar.[latex]\LARGE0.12\times117=14.04[/latex]

Now, slicing off 0.04 of a candy is hard to do, so we’ll use our rounding skills and round down to 14 candies.Example

Now move on to Bryce’s next friend, Matt. Matt gets 25% of the candy.
**Step 1**: Turn the percentage into a fraction with a denominator of 100.
**Step 2**: Take the 25 and divide it by the 100.
**Step 3**: Take the 0.25 and multiply it by the number of candies in the jar.

[latex]\LARGE\dfrac{25}{100}[/latex]

[latex]\LARGE25÷100=0.25[/latex]

[latex]\LARGE0.25\times117=29.25[/latex]

We would again round down to get our final answer of 29.Example

We won’t worry about how many candies the instructor gets, but what we will calculate is how many candies Bryce keeps for himself. We want to calculate the percentage of candies Bryce ends up with, and then calculate how many candies that is.
**Step 1**: Add up the percentage of the candies that the other three people have, and then subtract the sum from 100.
**Step 2**: At this point, we’ve done a couple of examples in which we first took the percentage and turned it into a fraction over 100. Then, we took the percentage and divided it by 100. What if we just skipped that step? What we end up doing for a percentage is just taking the decimal point and moving it over 2 places to the left (this represents dividing by 100). Try that.
**Step 3**: Take the 0.46 and multiply it by the number of candies in the jar.

- Patrick gets 12%
- Matt gets 25%
- Instructor gets 17%

[latex]\LARGE12\%+25\%+17\%=54\%[/latex]

[latex]\LARGE\text{Bryce gets }100\%-54\%=46\%[/latex]

[latex]\LARGE46\%=0.46[/latex]

[latex]\LARGE0.46\times117=53.82[/latex]

This can be rounded up to 54. Bryce still gets to eat a lot of candy, which will satisfy his sweet tooth, but he won’t eat so much that he’ll end up getting sick.Question 1

Find 37% of 229

https://video.bccampus.ca/id/0_702d0kqu?width=608&height=402&playerId=23448552Question 2

Find 78% of 1928

https://video.bccampus.ca/id/0_ey9ruvvk?width=608&height=402&playerId=23448552Version | Date | Change | Details |
---|---|---|---|

1.00 | April 17, 2020 | Added to the B.C. Open Textbook Collection. | |

1.01 | July 24, 2020 | Correction to Whole Numbers Quiz and Fractions Quiz. | Whole Numbers: Added "86" to the list of numbers in the switch outlet boxes question. Fractions: Changed 11/2 to 1½ in the steel question. |

Flinn, C. & Overgard, M. (2020). *Math for Trades: Volume 1*. Victoria, B.C.: BCcampus. Retrieved from https://pressbooks.bccampus.ca/mark/.

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Outcomes

- Identify and describe different types of fractions.
- Add and subtract fractions.
- Multiply fractions.
- Divide fractions.

Outcomes

- Identify counting numbers and whole numbers.
- Locate digits in the place value system.
- Add whole numbers.
- Subtract whole numbers.
- Multiply whole numbers.
- Divide whole numbers.

Outcomes

- Understand decimal notation.
- Compare and round decimals.
- Convert fractions to decimals.
- Convert decimals to fractions.

Outcomes

- Define percent
- Understand the similarities between percent, ratios and fractions
- Calculate the percent of a number
- Calculate change as a percent and percent change

[latex]\text{Percentage change}=\dfrac{\text{Actual increase or decrease}}{\text{Original amount}}\times100[/latex]

There are a couple of things to note here. One is that this formula works if you have either an increase or a decrease from the original amount. If Melissa’s wage went down, you could also express that as a percentage. The second thing to note is that there is the number 100 at the end of the formula. This is due to the fact that the calculation would give us a decimal, and multiplying this number by 100 makes it a percentage. So, to follow through on our calculation, let’s put the numbers in and see what we get.[latex]\LARGE\text{Percentage change}=\dfrac{\$2}{\$14}\times100[/latex]

[latex]\LARGE\text{Percentage change}=0.1429\times100[/latex]

[latex]\LARGE\text{Percentage change}=14.29\%[/latex]

This is telling us that Melissa will get a 14.29% increase in her wage after the first six months. Not bad!Example

During her foundation program, Melissa had to keep her job working at a coffee shop in order to pay the bills. On her first day of work, she served 78 cups of coffee. During her six months on the job, the most cups of coffee she made in one day was 201. How can the difference between these amounts be expressed as a percentage increase?
**Step 1**: Calculate the increase in cups served.
**Step 2**: Write down the formula you are going to work with.
**Step 3**: Plug the numbers into the formula.
**Step 4**: Work through the answer.

[latex]\LARGE201-78=\text{123 more cups of coffee}[/latex]

[latex]\text{Percentage change}=\dfrac{\text{Actual increase or decrease}}{\text{Original amount}}\times100[/latex]

[latex]\LARGE\text{Percentage change}=\dfrac{123}{78}\times100[/latex]

[latex]\LARGE\text{Percentage change}=1.58\times100[/latex]

[latex]\LARGE\text{Percentage change}=158\%[/latex]

This answer tells us that, from the time Melissa started to the time she had her most productive day, she had a one-day increase of 158%. If she were to increase her output by 100%, that would mean mathematically that she would be making twice as much coffee as before. Increasing by 158% indicates that she is making MORE than twice the amount of coffee as when she first started.Question 1

In its first year of operation, a trade school has 57 students. In its second year of operation, its student body increased to 104 students. What percent increase is this?
https://video.bccampus.ca/id/0_d76pd4vn?width=608&height=402&playerId=23448552

Question 2

Referring back to the trade school in question 1, their third year of operation did not go so well. In their third year, the number of students went down to 86. What percent decrease is this from year two?
https://video.bccampus.ca/id/0_4ow2w0um?width=608&height=402&playerId=23448552

[latex]\LARGE\dfrac{20}{100}=0.2[/latex]

Multiply 27 by 0.2 to get our increase.[latex]\LARGE27\times0.2=5.4[/latex]

What this answer indicates is that the company wants to increase their number of apprentices by 5.4. Now, we can’t have 0.4 of an apprentice, so we have to round up or down. Following the rules we established before, we round down to 5. In the end, the company wants to increase the number of apprentices by 5. But we are not quite done yet: we still have to calculate how many apprentices they want to have in the future. To do this, we simply add the 5 extra students to our starting number, and we end up with 32 students.[latex]\LARGE27+5=32[/latex]

Example

We’re going to change it up here and use sports as our example.
A professional hockey team scored 219 goals in the 2018–2019 season and finished in 26th place in the league. In the 2019–2020 season, they want to improve on that total and are looking for a 25% increase in goal production. How many goals would they have to score to achieve their goal?
**Step 1**: Calculate 25% of 219.
**Step 2**: Add 55 to our original number, and we get:

[latex]\LARGE\dfrac{25}{100}=0.25\text{, so }219\times0.25=54.75[/latex]

As we can’t score 0.75 of a goal, we need to round this up to 1 complete goal and make it 55 more goals.[latex]\LARGE219+55=274\text{ goals}[/latex]

In the end, to achieve the 25% increase in goal production that they are looking for, the team would have to score 274 goals. If they scored that many in the 2018–2019 season, they would end up in the top 5 or 6 teams for goal production.Question 1

Melissa, the carpentry apprentice, likes to garden during the spring and summer. This year, she planted 42 plants. Next year, she wants to increase the total number of plants by 15%. How many plants will she need to plant in order to achieve her goal?
https://video.bccampus.ca/id/0_ra48wkew?width=608&height=402&playerId=23448552

Question 2

Let’s go back to sports for this one. A quarterback threw 47 touchdown passes in one year. The next year, the number of touchdown passes he threw dropped by 35%. How many touchdown passes did he throw that year?
https://video.bccampus.ca/id/0_z2oxbpct?width=608&height=402&playerId=23448552