ࡱ>  N_5@ 0pbjbj22 pXXd,,,@$$$8\p@-XX( h6;6;6;(W*W*W*W*W*W*W$YRk[NW,>>/ 6;>>NW hLWHIHIHI>Z ^,h(WHI>(WHI HIhI:N,.,(OhL ų$B2O V4W0-XO [,Eb[(O@@[,(O6;;|HI`<d<6;6;6;NWNW@@$G@@$Chapter 4 Understanding Percent Section 4.1 Representing Percents Section 4.1 Page 128 Question 4 a) The first grid is entirely shaded, so it represents 100%. The second grid has 12 out of 100 squares shaded. There are a total of 112 squares shaded, which corresponds to 112%. b) Since only a portion of one square of a hundred grid is shaded, the percent represented is between 0% and 1%. The enlarged diagram of the partially shaded square shows two out of ten parts shaded. The diagram represents  EMBED Equation.3  or 0.2%. c) The diagram shows 85 squares shaded plus a portion of another square shaded. When you zoom in on the partially shaded square, it is apparent that  EMBED Equation.3  of it is shaded. The shading represents  EMBED Equation.3  of 1% of the whole diagram. The diagram represents  EMBED Equation.3 . Section 4.1 Page 128 Question 5  a) The diagram has only one square shaded, so it represents a percent between 0% and 1%. The enlarged image of the partially shaded square indicates that  EMBED Equation.3  of 1% is shaded. The diagram represents  EMBED Equation.3 . b) A total of 125 squares are shaded on the two hundred grids. There is also one square that is half shaded. The diagram represents  EMBED Equation.3 . c) Two complete hundred grids are shaded, which represents 200%. On the third hundred grid, 82 squares are shaded. The diagram represents 282%. The answer could also be found by subtracting the number of non-shaded squares from the three hundred grids: 300 12 = 282; the diagram represents 282%. Section 4.1 Page 128 Question 6  a) Since 125% is greater than 100%, more than one hundred grid is needed. You can represent 100% by shading one grid. You can represent 25% by shading 25 squares of a second hundred grid. b)  EMBED Equation.3  is a fractional percent between 1% and 100%, so you use one hundred grid. Shade ten squares to represent 10%. Shade  EMBED Equation.3  of another square to represent  EMBED Equation.3   c) 0.4% is a fractional percent between 0% and 1%. Zoom in on one square of a hundred grid. Since 0.4 represents  EMBED Equation.3 , divide the enlarged square into ten equal sections. Shade four of the ten sections.  d) Shade two hundred grids to represent 200%. Then shade 62 squares on a third hundred grid to represent 62%. e)  EMBED Equation.3  is a fractional percent between 0% and 1%. Zoom in on one square of a hundred grid. Divide the enlarged square into eight equal sections and shade seven sections. f) 45.6% is a fractional percent between 1% and 100%, so use one hundred grid. Shade 45 squares. Then shade  EMBED Equation.3  of another square to represent 0.6%. Section 4.1 Page 129 Question 7 a) For 3.2%, shade three squares and  EMBED Equation.3  of another square. b) For 0.13%, shade  EMBED Equation.3  of one square. c) Since 230% is greater than 100%, more than one hundred grid is needed. Shade two hundred grids to represent 200%, and shade 30 squares on a third hundred grid to represent 30%. Section 4.1 Page 129 Question 8 a) Three completely shaded hundred grids are needed to show 300%. b) To show 466%, five hundred grids are needed. Four grids will be shaded entirely and the fifth grid will have 66 squares shaded. c) Twelve completely shaded hundred grids are needed to represent 1200%. Section 4.1 Page 129 Question 9 Answers may vary. Example: Two situations where the percent will be greater than 100% are a mothers mass compared to that of her newborn child, and the volume of water in the Pacific Ocean in relation to a lake in Canada. In both cases, the first measure (mothers mass; volume of water in the Pacific Ocean) is larger than the second measure (babys mass; volume of water in a lake in Canada). Section 4.1 Page 129 Question 10 Answers will vary. Example: The percent concentration of different pollutants in a water sample will most likely be between 0% and 1%. Section 4.1 Page 129 Question 11  Represent the land area of Saskatchewan as 100%, which is equal to one hundred grid. Then represent the area of Albertas land with one hundred grid and 13 shaded squares on a second hundred grid, which represents 113%. Section 4.1 Page 129 Question 12 Answers may vary. Example: A single hundred grid represents the 100% calcium that is needed by the body each day. Each glass of milk (30% of recommended daily calcium) is represented by 30 shaded squares on the hundred grid. You would need to drink  EMBED Equation.DSMT4  glasses of milk to get 100% of the daily value of calcium. Section 4.1 Page 129 Question 13 a)  EMBED Equation.3  is a fractional percent between 0% and 1%. Zoom in on one square and divide it into three equal regions. Shade one of these regions. b) Answers may vary. Example: Fractional percents with repeating decimals can be difficult to show on a hundred grid if you do not know how to convert the decimal to a proper fraction. Section 4.1 Page 129 Question 14 a) Two squares represent 1%. Therefore, 0.5 or  EMBED Equation.DSMT4  of a square will represent 0.25%.  b) Four squares represent 1%. Therefore, three squares will represent 0.75%.  Section 4.1 Page 129 Question 15 Since one square on a hundred grid is equal to 1%, then one square on a thousand grid (ten hundred grids placed together) would equal 0.1%. If this pattern is continued, then very small percents can be expressed on larger grids. If a ten million grid was used, then 0.000 012 5% would be represented by 1.25 squares.  Section 4.1 Page 129 Question 16 a) If the square is broken into 1000 smaller equal-sized squares, then each square is worth 0.1% (100 1000). Therefore, 17 pieces will be equivalent to 1.7% (17 0.1). b) The two large squares represent 200%. If each square is divided into ten smaller equal-sized squares, then each square is equivalent to 10%. Thirteen of these squares will be equivalent to 130%. c) Determine the value of one of the eight equal-sized squares. 100 8 = 12.5 Find the number of squares that will represent  EMBED Equation.3 . 87.5 12.5 = 7 Seven squares will represent  EMBED Equation.3 . Find the number of squares that will represent  EMBED Equation.3 . 56.25 12.5 = 4.5 A total of 4.5 squares will represent  EMBED Equation.3 . Section 4.2 Fractions, Decimals, and Percents Section 4.2 Page 135 Question 4 a)  EMBED Equation.3 =  EMBED Equation.DSMT4   EMBED Equation.3 =  EMBED Equation.DSMT4  x = 0.4 That is 0.4% or 0.004. b)  EMBED Equation.3 =  EMBED Equation.DSMT4   EMBED Equation.3 =  EMBED Equation.DSMT4  x = 40.5 That is 40.5% or 0.405. c)  EMBED Equation.3 can be expressed as  EMBED Equation.DSMT4  +  EMBED Equation.DSMT4  = 1 +  EMBED Equation.DSMT4  One whole represents 100%.  EMBED Equation.DSMT4  represents 20%. So,  EMBED Equation.DSMT4  represents 40%. So,  EMBED Equation.3 = 100% + 40% = 140%.  EMBED Equation.3 = 7 ( 5 = 1.4 Section 4.2 Page 135 Question 5 a)  EMBED Equation.3 =  EMBED Equation.DSMT4   EMBED Equation.3 =  EMBED Equation.DSMT4  x = 170 That is 170% or 1.7. b)  EMBED Equation.3 =  EMBED Equation.DSMT4   EMBED Equation.3 =  EMBED Equation.DSMT4  x = 10.5 That is 10.5% or 0.105. c)  EMBED Equation.3 =  EMBED Equation.DSMT4   EMBED Equation.3 =  EMBED Equation.DSMT4  x = 0.6 That is 0.6% or 0.006. Section 4.2 Page 135 Question 6 a) 0.0072 =  EMBED Equation.DSMT4  0.0072 =  EMBED Equation.DSMT4 = 0.72%  EMBED Equation.3  b) 0.548 =  EMBED Equation.DSMT4  0.548 =  EMBED Equation.DSMT4  = 54.8%  EMBED Equation.3  c) 3.45 =  EMBED Equation.DSMT4 = 345%  EMBED Equation.3  Section 4.2 Page 136 Question 7 a) 0.256 =  EMBED Equation.DSMT4  0.256 =  EMBED Equation.DSMT4  = 25.6%  EMBED Equation.3  b) 0.0005 =  EMBED Equation.DSMT4  0.0005 =  EMBED Equation.DSMT4 = 0.05%  EMBED Equation.3  c) 6.5 =  EMBED Equation.DSMT4  6.5 =  EMBED Equation.DSMT4  = 650%  EMBED Equation.3  Section 4.2 Page 136 Question 8 a) 248% =  EMBED Equation.DSMT4  =  EMBED Equation.DSMT4   EMBED Equation.DSMT4 =  EMBED Equation.DSMT4  248 ( 100 = 2.48 b) 0.56% =  EMBED Equation.DSMT4   EMBED Equation.DSMT4  = 0.56 ( 100 = 0.0056 0.0056 =  EMBED Equation.DSMT4 =  EMBED Equation.DSMT4  c)  EMBED Equation.DSMT4  = 75% +  EMBED Equation.DSMT4  75% = 0.75 =  EMBED Equation.DSMT4   EMBED Equation.DSMT4 = 3 4 = 0.75  EMBED Equation.DSMT4  = 0.75 100 = 0.0075  EMBED Equation.DSMT4 = 0.75 + 0.0075 = 0.7575 0.7575 =  EMBED Equation.DSMT4   EMBED Equation.DSMT4  =  EMBED Equation.DSMT4  =  EMBED Equation.DSMT4  Section 4.2 Page 136 Question 9 a)  EMBED Equation.DSMT4  = 5% +  EMBED Equation.DSMT4  5% = 0.05 =  EMBED Equation.DSMT4   EMBED Equation.DSMT4 = 1 20 = 0.05  EMBED Equation.DSMT4  = 0.9 100 = 0.009  EMBED Equation.DSMT4  = 0.05 + 0.009 = 0.059 0.059 =  EMBED Equation.DSMT4   EMBED Equation.DSMT4  =  EMBED Equation.DSMT4  b) 550% =  EMBED Equation.DSMT4  =  EMBED Equation.DSMT4   EMBED Equation.3  550 100 = 5.5 c) 0.8% =  EMBED Equation.DSMT4   EMBED Equation.DSMT4 = 0.8 ( 100 = 0.008 0.008 =  EMBED Equation.3  Section 4.2 Page 136 Question 10  Section 4.2 Page 136 Question 11 a) There are 17 shaded squares on the 25 square grid:  EMBED Equation.3  Convert to a decimal. 17 25 = 0.68 Convert to a percent. 0.68 100 = 68% b) There are 9 shaded squares on the 24 square grid:  EMBED Equation.3  Convert to a decimal. 3 8 = 0.375 Convert to a percent. 0.375 100 = 37.5% Section 4.2 Page 136 Question 12 a) One whole grid is shaded ( EMBED Equation.3 ), and eight squares are shaded on the second grid ( EMBED Equation.3 ). Determine the combined value.  EMBED Equation.3  Convert to a decimal. 33 25 = 1.32 Convert to a percent. 1.32 100 = 132% b) Two whole grids are shaded ( EMBED Equation.3 ), and seven squares are shaded on the third grid ( EMBED Equation.3 ). Determine the combined value.  EMBED Equation.3  Convert to a decimal. 47 20 = 2.35 Convert to a percent. 2.35 100 = 235% Section 4.2 Page 136 Question 13 Write as a fraction and then convert to a percent.  EMBED Equation.3  The comic book is worth 2000% of its initial value. Section 4.2 Page 136 Question 14 Write the fraction that the snack (0.9g) represents of his total fat intake (40g):  EMBED Equation.3  Multiply by  EMBED Equation.3  to acquire an equivalent proper fraction.  EMBED Equation.3  Convert to a decimal. 9 400 = 0.0225 Convert to a percent. 0.0225 100 = 2.25% The snack represented 2.25% of his total fat intake for the day. Section 4.2 Page 136 Question 15  0.6%: smallest  0.6%,  EMBED Equation.3 (0.625%), 33.5%, 0.65 (65%), 1.32 (132%), 145% 145%: largest Section 4.2 Page 136 Question 16 Display as a fraction and reduce.  EMBED Equation.3  Convert to a decimal. 1 225 =  EMBED Equation.3  Convert to a percent.  EMBED Equation.3  100 =  EMBED Equation.3 % Section 4.2 Page 137 Question 17 Answers may vary. Example: a) Ticket sales are  EMBED Equation.3  of what they were this time last year. The number 1.3 sounds like a smaller number than the fraction EMBED Equation.3 . b) We are already at 0.605 of our target and we just started! The decimal 0.605 is easily recognizable as more than half. It sounds like it is a large number because there are 3 decimal places. c) We have managed to cut our costs by  EMBED Equation.3 . The large denominator makes this number appear large. Section 4.2 Page 137 Question 18 Add the numbers in the column titled Number in order to find the total number of fish. 143 + 122 + 2 = 267 Express the fractions of the total using 267 and complete the column Fraction of Total. Convert each fraction into a decimal using division, and then convert to percents.  Section 4.2 Page 137 Question 19 Express as a fraction in reduced form.  EMBED Equation.3  Express as a decimal: 6.0. Express as a percent. 6.0 100 = 600% The new circulation is 600% of the circulation five years ago (or 6.0, or  EMBED Equation.DSMT4 ). Section 4.2 Page 137 Question 20 fraction decimal percent For 90 beats per minute:  EMBED Equation.3  6 5 = 1.2 1.2 100 = 120% For 125 beats per minute:  EMBED Equation.3  5 3 =  EMBED Equation.3   EMBED Equation.3  100 =  EMBED Equation.3  For 150 beats per minute:  EMBED Equation.3  2 1 = 2.0 2.0 100 = 200% Section 4.2 Page 137 Question 21 Divide each value in a row by 2 to get the values for the row directly below it.  Section 4.3 Percent of a Number Section 4.3 Page 142 Question 3 a) 300% is 3 100%. 100% of 2000 is 2000. 3 2000 = 6000 300% of 2000 is 6000. b)  EMBED Equation.3  is 1% +  EMBED Equation.3 . 100% of 60 is 60. 10% of 60 is 6. 1% of 60 is 0.6.  EMBED Equation.3  of 60 is 0.6 2 = 0.3.  EMBED Equation.3  of 60 is 0.3 2 = 0.15. 0.6 + 0.15 = 0.75  EMBED Equation.3  of 60 is 0.75. c) 100% of 40 is 40. 10% of 40 is 4. 1% of 40 is 0.4. 0.1% of 40 is 0.04. Section 4.3 Page 142 Question 4 a) 100% of 60 is 60. 10% of 60 is 6. 20% is 2 10%. 2 6 = 12 20% of 60 is 12. b) 250% = 100% + 100% + 50% 100% of 400 is 400. 50% of 400 is 400 2 = 200. 400 + 400 + 200 = 1000 250% of 400 is 1000. c)  EMBED Equation.3  10.5% of 100 is 10.5. Section 4.3 Page 142 Question 5 a)  EMBED Equation.3  Divide by 100 to write the percent as a decimal. 0.4 100 = 0.004 0.004 325 = 1.3  EMBED Equation.3  of 325 is 1.3. b)  EMBED Equation.3  Divide by 100 to write the percent as a decimal. 15.25 100 = 0.1525 0.1525 950 = 144.875 c) Divide 175% by 100 to write the percent as a decimal. 175 100 = 1.75 1.75 125.5 = 219.625 175% of $125.50 is $219.63. Section 4.3 Page 142 Question 6 a)  EMBED Equation.3  Divide by 100 to write the percent as a decimal. 0.625 100 = 0.006 25 0.006 25 520 = 3.25 b)  EMBED Equation.3  Divide by 100 to write the percent as a decimal. 75.4 100 = 0.754 0.754 200 = 150.8 c) Divide 250% by 100 to write the percent as a decimal. 250 100 = 2.5 2.5 76.5 = 191.25 250% of $76.50 is $191.25. Section 4.3 Page 142 Question 7 a) Convert the chance of winning from a fraction to a percent.  EMBED Equation.3  The chance of winning is 0.5%. b) Divide the chance of winning as a percent into the desired percent, 2.5%. 2.5 0.5 = 5 You would need to purchase five tickets in order to have a 2.5% chance of winning. Section 4.3 Page 142 Question 8 Calculate  EMBED Equation.3  of $84.00. 25.5% = 0.255 0.255 84 = 21.42 The price was reduced by $21.42. Section 4.3 Page 142 Question 9 Determine 159% of 3747. 1.59 3747 = 5957.73 The height of Mount Logan is 5957.73 m. Section 4.3 Page 142 Question 10 a) Determine 10% of 750 mL. 0.1 750 = 75.0 The volume of water increases by 75 mL. b) Add the initial amount of water to the increase from part a). 750 + 75 = 825 The volume of ice is 825 mL. Section 4.3 Page 143 Question 11 Determine  EMBED Equation.3  of 9 984 670 km2. 0.065 9 984 670 = 649 003.55 The area of Manitoba is approximately 649 004 km2. Section 4.3 Page 143 Question 12 Calculate 200% of 550 km. 2.0 550 = 1100 The hybrid version will travel 1100 km on a tank of gas. Section 4.3 Page 143 Question 13 a) Commission is the portion of the sale price that the real estate agent earns. b) Calculate 5% of the first $200 000. 0.05 200 000 = 10 000 Calculate 6% on the remaining $145 000 of the selling price. 0.06 145 000 = 8700 Find the sum of the two portions of the commission. 10 000 + 8700 = 18 700 The real estate agent earns $18 700 on the sale of the house. Section 4.3 Page 143 Question 14 Find 4% of 100. 0.04 100 = 4 Determine what number, x, will equal 4 when 8% is taken of it. 0.08x = 4 x = 4 0.08 x = 50 4% is half of 8%, and 50 is half of 100. Section 4.3 Page 143 Question 15 Find the second bid by calculating 135% of 100. 1.35 100 = 135 Find the third bid by calculating 257% of $135. 2.57 135 = 346.95 Find the next five bids by determining 110.5% of each previous bid. Fourth bid: 1.105 346.95 = 383.379 75 Fifth bid: 1.105 383.38 = 423.6349 Sixth bid: 1.105 423.63 = 468.111 15 Seventh bid: 1.105 468.11 = 517.261 55 Eighth bid: 1.105 517.26 = 571.5723 Find the winning bid by calculating 100.1% of the eighth bid. 1.001 571.57 = 572.14 The final bid was $572.14. This answer involved rounding each bid to the nearest hundredth. Without rounding at each step, the answer is $572.15. The repeated multiplication that occurs for bids four through eight could be solved more quickly in exponential form. 346.95 (1.1055) 1.001 = 572.15 Section 4.3 Page 143 Question 16 Determine the total number of shots for the two games. 30 + 10 = 40 Find the number of baskets she made for both games. 0.5 40 = 20 Find the number of baskets she made in the second game. 20 12 = 8 Josephine made eight of her ten shots in the second game. Section 4.4 Combining Percents Section 4.4 Page 148 Question 4 Find the combined tax rate. 5% + 7% = 12% Find the cost of the all items before tax. 2 4.99 + 3.99 + 19.99 = 33.96 Calculate the tax, 12%, on $33.96. 0.12 33.96 = 4.0752 Find the total cost including tax. 33.96 + 4.08 = 38.04 The total cost is $38.04. Section 4.4 Page 148 Question 5 Find the total cost before taxes. 3 19.99 = 59.97 Combine the cost and tax percents. 100 + 6 + 5 = 111 Find 111% of the cost before taxes. 1.11 59.97 = 66.5667 The total cost with taxes for the DVDs is $66.57. Section 4.4 Page 148 Question 6 Determine the reduced price after the first week. 0.5 85 = 42.5 Determine the amount of the discount in the second week. 42.5 = 4.25 Find the sale price in the second week. 42.5 4.25 = 38.25 The reduced price in the second week is $38.25. Section 4.4 Page 149 Question 7 a) Determine the increase in population after the first year. 100 = 10 Determine the increase in population after the second year. (100 + 10) = 22 The population of caribou after two years is 132. b) The increase is not 30% because the 20% increase in the second year is based on the new population after the 10% increase in the first year. Section 4.4 Page 149 Question 8 Answers may vary based on the PST rate in your province. Example: Based on a total tax of 12% (GST = 5% and PST = 7%), Calculate the amount of tax. Then, calculate the total cost. a) 0.12 119.99 = 14.40 119.99 + 14.40 = 134.39 b) 0.12 89.99 = 10.80 89.99 + 10.80 = 100.79 c) 0.12 39.99 = 4.80 39.99 + 4.80 = 44.79 d) 0.12 189.99 = 22.80 189.99 + 22.80 = 212.79  Section 4.4 Page 149 Question 9 a) Find 103.2% of $23 000. 1.032 23 000 = 23 736 The price of the car was $23 736. b) Find the combined cost and tax percents. 100 + 5 + 5 = 110 Find 110% of $23 736 1.1 23 736 = 26 109.6 The total cost of the car with taxes was $26 109.60. Section 4.4 Page 149 Question 10 Find the cost before taxes for four tires. 4 85 = 340 Find the combined cost and tax percents. 100 + 5 + 1.5 = 106.5 Calculate the total cost for the four tires. 1.065 340 = 362.1 The total cost for the four tires is $362.10. Section 4.4 Page 149 Question 11 a) Find the amount of interest after the first year. 0.03 1000 = 30 Find the value of the scholarship after the first year. 1000 + 30 = 1030 The value of the scholarship after the first year is $1030. Find the amount of interest accumulated in the second year. 0.03 1030 = 30.90 Find the value of the scholarship after the second year. 1030 + 30.90 = 1060.90 The value of the scholarship after two years is $1060.90. b) The original value of the scholarship is $1000. The value of the scholarship after two years is $1060.90. Subtract to find the total increase in value. 1060.90 1000 = 60.90 The total increase in value is $60.90. Determine the percent the increase is of the original value.  EMBED Equation.DSMT4  = 0.0609 The single percent increase in value is 6.09% for the two-year period. Section 4.4 Page 149 Question 12 a) Find the total distance of the race. 1.5 + 40 + 10 = 51.5 The total distance of the triathlon is 51.5 km. Determine the percent of the total distance that each event represents. swim: 1.5 51.5 H" 0.029 0.029 100 = 2.9 bike ride: 40 51.5 H" 0.777 0.777 100 = 77.7 run: 10 51.5 H" 0.194 0.194 100 = 19.4 The swim component represents approximately 2.9% of the race. The bike ride represents approximately 77.7% of the race. The run represents approximately 19.4% of the race. b) 77.7 + 19.4 = 97.1 97.1% of the race is spent on land. This value could also be calculated by subtracting the only water component from 100%: 100% 2.9% = 97.1%. Section 4.4 Page 149 Question 13 Find the overall percent of the original cost that must be paid after all discounts. 100 20 = 80 After the first discount, 80% of the cost still applies. (100% 25%) 80% = 0.75 0.8 = 0.6 After the second discount, 60% of the original cost still applies. After the third discount, 50% of 60% of the cost still applies. 0.5 0.6 = 0.3 30% of the original cost must be paid. Find the overall single discount as a percent. 100 30 = 70 The overall percent saved is 70%. Section 4.4 Page 149 Question 14 As a selling price, the DVD is 135% of the cost to the store. The sale at a discount of 20% indicates that the DVD is being sold for 80% (100 20) of the selling price, 135%. 0.8 1.35 = 1.08 Therefore, the store is selling the DVD for 108% of the cost to the store. The store gains 8% on every DVD sold. Chapter 4 Review Chapter 4 Review Page 150 Question 1 PERCENT means out of 100. Chapter 4 Review Page 150 Question 2 A FRACTIONAL percent is a percent that includes a portion of 1%. Chapter 4 Review Page 150 Question 3 Percents that are added together are COMBINED percents. Chapter 4 Review Page 150 Question 4 a) Two hundred grids are needed to show 101%. One complete grid will be shaded to represent 100%, and one square of a second grid will be shaded to represent 1%. b) Six hundred grids will be required to show 589%. The first five grids will be shaded completely to represent 500%. The sixth grid will have 89 squares shaded to represent 89%. c) Fifteen hundred grids are needed to show 1450%. Fourteen grids will be shaded completely to represent 1400%. Half of the fifteenth grid (50 squares) will be shaded to represent 50%. Chapter 4 Review Page 150 Question 5 a) A portion of one square is shaded, so the percent is between 0% and 1%. In the enlarged image of the one square, seven of the ten sections are shaded. The diagram represents  EMBED Equation.3  or 0.7%. b) A portion of one square is shaded, so the percent is between 0% and 1%. In the enlarged image of the one square, three of the five sections are shaded. The diagram represents  EMBED Equation.3  or 0.6%. c) Fifty squares on the grid are shaded. One additional square is partially shaded. In the enlarged image of the one square, one of the four sections is shaded. The diagram represents  EMBED Equation.3  d) There are three hundred grids. Each of the two completely shaded grids represents 100%. The third grid has 45 squares shaded. The diagram represents 245%. Chapter 4 Review Page 150 Question 6  a) 110% = 100% + 10% b)  EMBED Equation.3  c)  EMBED Equation.3  = 7% +  EMBED Equation.DSMT4 % d) 172.5% = 100% + 72% + 0.5% e) 0.75% f) 500% = 100% + 100% + 100% + 100% + 100%  Chapter 4 Review Page 150 Question 7 a) 0.4% b) 12% c) 115% = 100% + 15% Chapter 4 Review Page 150 Question 8 a) 79.1% = 79% + 0.1% b) 223% = 100% + 100% + 23% c) 0.8% Chapter 4 Review Page 150 Question 9 a) Find the percent. 0.115 100 = 11.5% Find the fraction. 0.115 =  EMBED Equation.3  b) Find the decimal.  EMBED Equation.DSMT4  = 23% +  EMBED Equation.DSMT4  23% = 0.23 =  EMBED Equation.DSMT4   EMBED Equation.DSMT4  = 0.75 100 = 0.0075  EMBED Equation.DSMT4  = 0.0075 + 0.23 = 0.2375 Find the fraction. 0.2375 =  EMBED Equation.DSMT4   EMBED Equation.DSMT4  =  EMBED Equation.DSMT4  c) Find the decimal. 3 200 = 0.015 Find the percent. 0.015 100% = 1.5% d) Find the percent. 3.85 100% = 385% Find the fraction.  EMBED Equation.3  Chapter 4 Review Page 151 Question 10 a) Convert 110% to a decimal. 110% =  EMBED Equation.DSMT4  =  EMBED Equation.DSMT4   EMBED Equation.DSMT4 = 110 100 = 1.1 Express as a fraction and reduce.  EMBED Equation.3  b) Answers may vary. Example: It means that you must give more of an effort than you would normally. Chapter 4 Review Page 151 Question 11 a) Convert 95.5% to decimal form. 95.5% = 95% + 0.5% 95% = 0.95 0.5% = 0.005 95.5% = 0.95 + 0.005 = 0.955 Convert to fraction form.  EMBED Equation.3  Kyle scored  EMBED Equation.3 on a practice test. b) Convert 140% to decimal form. 140% =  EMBED Equation.DSMT4  =  EMBED Equation.DSMT4   EMBED Equation.DSMT4  = 140 100 = 1.4 Convert to fraction form.  EMBED Equation.3  The stores sales increased by a factor of 1.4. c) Convert  EMBED Equation.3  to a decimal. 0.9 100 = 0.009 Convert to fraction form.  EMBED Equation.3  By getting your car tuned up, you can reduce emissions by 0.009 times the original amount. Chapter 4 Review Page 151 Question 12 Convert each percent to decimal form and then multiply. a) 1.15 230 = 264.5 b) 0.8075 50 = 40.375 H" 40.4 c) 5.0 0.02 = 0.1 d) 0.001 800 = 0.8 e) 0.638 12 000 = 7656 f) 0.0005 1 000 000 = 500 Chapter 4 Review Page 151 Question 13 Find 250% of 2.5 cm. 2.5 2.5 = 6.25 The length of the enlarged diagram is 6.25 cm. Chapter 4 Review Page 151 Question 14 Calculate 5.5% of $100. 0.055 100 = 5.5 Julia had to pay $5.50 in interest to her brother. Chapter 4 Review Page 151 Question 15 a) Find the sum of all the counted trees. 567 + 324 + 156 + 89 + 678 = 1814 There were 1814 trees recorded. b) For each species, determine the decimal portion of all the trees and then convert to a percent. Fir: 567 1814 H" 0.31 0.31 100% = 31% Pine: 324 1814 H" 0.18 0.18 100% = 18% Larch: 156 1814 H" 0.09 0.09 100% = 9% Cedar: 89 1814 H" 0.05 0.05 100% = 5% Hemlock: 678 1814 H" 0.37 0.37 100% = 37% Chapter 4 Review Page 151 Question 16 Determine the combined taxes. 5 + 7 + 1 + 0.75 = 13.75 The total taxes amount to 13.75%. Combine the cost and taxes as a percent: 100 + 13.75 = 113.75 Calculate 113.75% of $289.50. 1.1375 289.5 = 329.306 25 The total cost of the ticket is $329.31. Chapter 4 Review Page 151 Question 17 a) No, the populations did not increase by the same amount. In the second year, the 7% increase in Cedarville is applied to the new population after the initial year increase of 7%. In Pinedale, the 15% increase is applied to the initial population of 1200. b) Determine the populations of each town. Cedarville: In the first year, the population increased by 7%. 1.07 1200 = 1284 In the second year, the population increased a further 8%. 1.08 1284 = 1386.72 The population of Cedarville after two years is 1387. Pinedale: The percent increase over two years was 15%. Find 115% of 1200. 1.15 1200 = 1380 The population of Pinedale after two years is 1380. Chapter 4 Practice Test Chapter 4 Practice Test Page 152 Question 1 Answer: A Convert 0.0235 to a percent. 0.0235 100 = 2.35 Chapter 4 Practice Test Page 152 Question 2 Answer: B Convert 135% to a decimal. 135 100 = 1.35 Chapter 4 Practice Test Page 152 Question 3 Answer: C Convert  EMBED Equation.3  to a fraction.  EMBED Equation.3  =  EMBED Equation.3  =  EMBED Equation.3  Chapter 4 Practice Test Page 152 Question 4 Answer: D Convert  EMBED Equation.3  to a decimal, and then to a percent. 1 8 = 0.125 0.125 100 = 12.5  EMBED Equation.3  = 12.5% Chapter 4 Practice Test Page 152 Question 5 Answer: B Find the price of the bike after the first discount. 0.1 420 = 42 The original discount was $42.00. 420 42 = 378 The price of the bike after the 10% discount was $378.00. Find the price of the bike after the further 5% discount. 0.05 378 = 18.9 The second discount is $18.90. 378 18.9 = 359.1 The final sale price is $359.10. Chapter 4 Practice Test Page 152 Question 6  The hundred grids shown represent 130%. The first grid is entirely shaded, so it represents 100%. The second grid has 30 squares shaded, so it represents 30%. 100% + 30% = 130% Chapter 4 Practice Test Page 152 Question 7 The hundred grid shown represents 0.4% or  EMBED Equation.3  One square of the hundred grid is partially shaded, so the diagram represents a percent between 0% and 1%. In the enlarged image of the single square, two of the five sections are shaded. Chapter 4 Practice Test Page 152 Question 8 a) 0.1% b) 35% c) 102% = 100% + 2%  Chapter 4 Practice Test Page 152 Question 9 a) Convert 15% to a decimal. 15% =  EMBED Equation.DSMT4  15 100 = 0.15 Convert 15% to a fraction and reduce.  EMBED Equation.3  b) Convert 1.24 to a percent. 1.24 =  EMBED Equation.DSMT4 = 124% Convert 124% to a fraction and reduce.  EMBED Equation.3  c) Convert  EMBED Equation.3  to a decimal.  EMBED Equation.3 =  EMBED Equation.DSMT4   EMBED Equation.3 =  EMBED Equation.DSMT4  x = 52 That is 52% or 0.52. Chapter 4 Practice Test Page 152 Question 10 Calculate 5% of 250 000. 0.05 250 000 = 12 500 Calculate 7% on the difference between $423 000 and $250 000 ($173 000). 0.07 173 000 = 12 110 Find the total commission. 12 500 + 12 110 = 24 610 The real estate agent makes $24 610 on the sale of the house. Chapter 4 Practice Test Page 152 Question 11 If the population increased by 0.7% each year, then calculate 100.7% of the current population to find the population for the next year. Year 1: 1.007 50 000 = 50 350 Year 2: 1.007 50 350 = 50 702.45 The population after two years was 50 702. Chapter 4 Practice Test Page 153 Question 12 a) Determine the combined tax rate. 5 + 7 = 12 Calculate the tax on the scooter. 0.12 64.98 = 7.7976 Helen paid $7.80 in tax altogether. b) Add the calculated tax to the sale price. 7.80 + 64.98 = 72.78 The total cost of the scooter was $72.78. Chapter 4 Practice Test Page 153 Question 13 a) Calculate 25% of $5000. 0.25 5000 = 1250 The club will earn $1250 in commission. b) Determine 125% of the commission. 1.25 1250 = 1562.5 The parent committee will donate $1562.50 to the club. c) Determine the sum of the amounts from parts a) and b). 1250 + 1562.5 = 2812.5 The club will receive a total of $2812.50. Chapter 4 Practice Test Page 153 Question 14 a) Determine the percent of ice that will be retained after one year. 100 6 = 94 Calculate 94% of the current ice-covered percent, 70%. 0.94 70 = 65.8 Ice will cover 65.8% of the region after the first year. b) Find the percent after year 2. 0.94 65.8 = 61.852 Find the percent after year 3. 0.94 61.852 = 58.140 88 Ice will cover 58.1% of the region after the third year. c) Continue calculating the coverage each year until the percent drops below 50%. Year 4: 0.94 58.140 88 H" 54.652 43 Year 5: 0.94 54.652 43 H" 51.373 Year 6: 0.94 51.373 H" 48.29 After six years, less than one half of the region will be covered by ice. Chapter 1 4 Review Chapter 1 Representing Data Chapter 1 4 Review Page 156 Question 1  a) A bar graph is used to compare data across categories. For example, the number of jewellery items sold per week over a four-week period. b) A double bar graph compares two sets of data across categories. For example, the sales of four types of music in two distinct time periods.  c) A circle graph compares categories to the whole using percents. For example, the relative sales of four types of music in a given year. d) A line graph reveals changes in data over time. For example, the number of items of jewellery sold each week over a four-week period.  Chapter 14 Review Page 156 Question 2 a) A pictograph provides a visual picture of the difference between the numbers. It uses symbols to compare the number of people who prefer different types of food. It would show more clearly that respondents prefer Italian food and Chinese food. b) Let each circle represent 100 people who preferred a specific food.  c) A line graph would not be appropriate for this survey because the data do not show changes over time. Chapter 14 Review Page 156 Question 3 a) Answers may vary. Example: More game systems were sold in December than in any previous month in the six-month period; more game system 1s were sold than game system 2s in each month; the sales of both systems increase throughout the six-month period. b) c) Answers may vary. Example: Sales of game system 1 are increasing faster than sales of game system 2; sales of game systems 1 and 2 both increased from July to December. d) Answers may vary. Example: From the bar graph, it can be quickly determined that December represents the largest increase in sales for both game systems. e) Answers may vary. Example: A bar graphs strength is that it is easy to compare two sets of data; a bar graphs limitation is that it is harder to see that one set of data is increasing faster than the other. A line graphs strength is that it is easy to see changes over time; a line graphs limitation is that it is harder to compare sales in a particular month. Chapter 14 Review Page 156 Question 4 a) Answers may vary. Example: This graph is misleading because computers appear to be the favourite; the line for computers is the longest one and the symbols for computers and books are much larger. b) Ensure that each symbol is similar in size.  c) Convert each number into a percent of the whole. Display the data in proportion in the circle using the proper central angles. Colour and label each sector.  d) Answers may vary. Example: One advantage of using a circle graph is that each section can be easily compared so you know which items are the most and least popular. Chapter 14 Review Page 156 Question 5  a) Answers may vary. A sample graph is shown. b) His pulse rate increases for the first three minutes and then begins to level off. c) A line graph shows change over time, so you can see how Calvins pulse rate changes over 5 min. Chapter 2 Ratios, Rates, and Proportional Reasoning Chapter 14 Review Page 157 Question 6 a) Determine three eighths of 32. =  EMBED Equation.3  = 12 There are 12 boys in the class. b) Find the number of girls. 32 12 = 20 There are 20 girls. Express the ratio of girls to the total number of students.  EMBED Equation.3  or 62.5% The ratio of girls to total students is  EMBED Equation.3  or 62.5%. c) The ratio of girls to boys is  EMBED Equation.3  Chapter 14 Review Page 157 Question 7 Find the unit price for 1 kg of Happy Kitty cat food. 12.99 5 = 2.598 Multiply the unit price by 4 to find the equivalent price for a 4-kg bag. 2.598 4 = 10.392 Purr n Chew must charge less than $10.39 for their bag of cat food. Chapter 14 Review Page 157 Question 8  a) Answers may vary. Example: Pasta Supreme appears to be the better buy because it is a much larger quantity for just a little bit more cost. b) Find the unit price for each product. Super Choice:  EMBED Equation.3  0.99 7 =  EMBED Equation.3  The unit cost is $0.14/100 g.   Pasta Supreme:  EMBED Equation.3  1.29 12.5 = 0.1032 The unit cost is $0.10/100 g. c) Pasta Supreme is a better buy because it costs less per 100 g (about 4 cents less). d) Answers may vary. Example: Estimating unit costs is useful because it can help you determine the cheapest brand and help you save money. Chapter 14 Review Page 157 Question 9 a) Find the unit rate for each vehicles fuel consumption.   Vehicle 1:  EMBED Equation.3  20.2 1.9 H" 10.63  The fuel consumption for Vehicle 1 is 10.63 L/100 km.   Vehicle 2:  EMBED Equation.3  44.7 4.6 H" 9.72 The fuel consumption for Vehicle 2 is 9.72 L/100 km.  Vehicle 3:  EMBED Equation.3  85 8 = 10.625 The fuel consumption for Vehicle 3 is 10.63 L/100 km. b) Vehicle 2 has the lowest fuel consumption. It requires the least amount of fuel (9.72 L) to travel 100 km. Chapter 14 Review Page 157 Question 10 a) Let x be the cost of 8 lemons.    EMBED Equation.3  x =  EMBED Equation.3   = 2.56 The cost of 8 lemons is $2.56. b) Let x represent the distance on the map that corresponds to 550 km.    EMBED Equation.3  x = 4.4  A distance of 550 km is represented by 4.4 cm on the map. Chapter 14 Review Page 157 Question 11 Let n represent the number of nickels that corresponds to the value of 5 quarters.     EMBED Equation.3  n = 20 1.25 = 25 The value of 5 quarters is equivalent to the value of 25 nickels. Chapter 3 Pythagorean Relationship Chapter 14 Review Page 157 Question 12 a) 82 = 8 8 = 64 b) 132 = 13 13 = 169 c) 172 = 17 17 = 289 d) 802 = 80 80 = 6400 Chapter 14 Review Page 157 Question 13 a)  EMBED Equation.3  = 11 b)  EMBED Equation.3  = 30 c)  EMBED Equation.3  = 7 d)  EMBED Equation.3  = 16 Chapter 14 Review Page 158 Question 14 a) Find the square root of 42.  EMBED Equation.3  Find the squares of the whole number on either side of 6.48. 62 = 36 and 72 = 49 The perfect squares on either side of 42 are 36 and 49. b) Find the square root of 139.  EMBED Equation.3  Find the squares of the whole number on either side of 11.79. 112 = 121 and 122 = 144 The perfect squares on either side of 139 are 121 and 144. c) Find the square root of 200.  EMBED Equation.3  Find the squares of the whole number on either side of 14.14. 142 = 196 and 152 = 225 The perfect squares on either side of 200 are 196 and 225. Chapter 14 Review Page 158 Question 15 a) Answers may vary. Example: EMBED Equation.3  b) Answers may vary. Example: EMBED Equation.3  c) Answers may vary. Example: EMBED Equation.3  d) Answers may vary. Example: EMBED Equation.3  Chapter 14 Review Page 158 Question 16 The perfect squares on either side of 90 are 81 and 100. Since 90 is almost exactly halfway between 81 and 100, then EMBED Equation.3 should be almost halfway between 9 and 10. Therefore, the best estimate of  EMBED Equation.3  is 9.5. Chapter 14 Review Page 158 Question 17 Substitute the values into the Pythagorean relationship, where 61 cm represents the hypotenuse.  EMBED Equation.3  121 + 3600 = 3721 Since the equation is true, these three values represent the sides of a right triangle. Chapter 14 Review Page 158 Question 18 a) Use the Pythagorean relationship to find the missing side length of the corral, x. x2 + 282 = 422  x2 + 784 = 1764 x2 + 784 784 = 1764 784 x2 = 980  EMBED Equation.3   EMBED Equation.3  The side length of the rectangular corral is 31.3 m. Find the perimeter of the corral. P = 2 31.3 + 2 28 = 62.6 + 56 = 118.6 The perimeter of the corral is 118.6 m. b) pre-tax cost = 118.6 15 = 1779 The cost for the fencing is $1779. Chapter 14 Review Page 158 Question 19 a) Find the distance from B to C.  EMBED Equation.3    EMBED Equation.3   EMBED Equation.3  BC = 20 m Find the distance from A to B.  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  AB = 25 m Chapter 4 Understanding Percent Chapter 14 Review Page 158 Question 20 The rear sprocket is represented by one complete hundred grid. The larger front sprocket is represented by one complete hundred grid and 55 parts of a second hundred grid. front sprocket  Chapter 14 Review Page 158 Question 21 Determine the percent 9 grams is of 1000 grams. 9 1000 = 0.009 0.009 100% = 0.9% The percent, 0.9%, represents nine tenths of one cell on a hundred grid.  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T0'$MT&. t tS| t5 to tq!tK~ tta3 tt$ tqt7[ tt7 t qt t +6t'tKGtt"tut ZAtt'zt!Kt(U"t$NWt) s" t1"{ T2wT*CT+pE@,Iz oT@ b  D$bL7$b ?$bL=$b L?$b >$b  @$b =$ b@$!b B$"bL;$#bLE$NNCC7<==ly֩   TTJJB<==qy֩ _ *urn:schemas-microsoft-com:office:smarttagscountry-regionhttp://www.5iantlavalamp.com/i*urn:schemas-microsoft-com:office:smarttagsState0http://www.5iamas-microsoft-com:office:smarttagsV *urn:schemas-microsoft-com:office:smarttagsplacehttp://www.5iantlavalamp.com/V*urn:schemas-microsoft-com:office:smarttags PersonNamehttp://www.microsoft.com     ./124Wnrtuv 679:;mO f g i j k y     " B Y Z \ ] ` Mdeghi,.234"/;/()0) &(@ACDKLSTZ[abz{ũƩʩΩҩ֩.Wn 6mO f y  B Y Md,"/9/T4k4t4488PPttYutu:vUv(?by}ĩ֩ Helen Masonu;}Whh^h`o(hh^h`o(.0^`0o(..0^`0o(... 88^8`o( .... 88^8`o( ..... `^``o( ...... `^``o(....... ^`o(........hh^h`o(hh^h`o(.0^`0o(..0^`0o(... 88^8`o( .... 88^8`o( ..... `^``o( ...... `^``o(....... ^`o(........}Wu; :IK pW  . t |x O] t  a t Yl)tA^ky'+Bcc@WzX+ ^dje.6uRf;I+6Bl~1$tG=QGv#'b h"D""Y"r"/#IB#;$``$ %w%>&&'(2(f(<)p)~)*?*n*+DF+SF+u , ,+-d- ;.Na/f/J0e0:1|2{2B}233 l3{34D*4y4D 55$58z566}6.8e:bH;<%<f9<CG<^=s> >D>O>T>*?>?O?N@A nByB'CkCDD\D&sUozUu*VAAW6XMvXZj,Z3Z6ZUZ~.[W[-\FX\s]k_X `#aZGaud e&eeefg/g5PhUh!0imBiVi$j.j:k=kkEkA%lRl)"mOHnVknL2opFqpqzqqrv s6svsd"t'quB+w>wWw8,xIy8zYDzF{P|w|My;3'6T9KkI >WS,Y!Kt{I~+?X?Ycn{_8u>{_ }$65gx!^]z~sRJSa?B=EMRYtm{;-.Vq|J7bISN{k'1O|8aJ~O *Miq7H[jvvHF?=qUyWRp_cFDq0>?UH=`weITQ+9`AdQa~CAS.i?EF4X %oB&[cJw%U0:2\| #+_sA8]35gZRcCD,S-V`}x\0I[a`4ZS na W]&2E.u\ 49:x2Dm"8}U'a_j t2af5j+~6IVeG&ta3[2jl1m@%Oq\;`P @'Ps<7Bh^OwiC~[ lga}*hsG?] V>2qHX ?; Chapter 5Sentinel Secondary School Helen Mason  CompObjj