Contemporary Business Mathematics with Canadian Applications . Questions with correct and verified answers.

Chapter 1 Review of Arithmetic 1) Simplify: (14 + 7)/3 Answer: (14 + 7)/3 = 21/3 = 7 Diff: 1 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 2) Simplify: 8 + 6 ∗ 2 Answer: 8 + 12 = 20 Diff: 1 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 3) Simplify: 3 + 8 ∗ 4 Answer: 3 + 32 = 35 Diff: 1 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 4) Simplify: 5(4 + 3) Answer: 5 ∗ 7 = 35 Diff: 1 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 5) Simplify: Answer: 5/20 = .25 Diff: 1 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 6) Simplify: Answer: 40/20 = 2 Diff: 1 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 7) Simplify: 9(8 – 5) + 5(6 + 4) Answer: 9 ∗ 3 + 5 ∗ 10 = 27 + 50 = 77 Diff: 1 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 8) Simplify: 30 + 8 – 6 Answer: 30 + 8 – 6 = 30 + 8 – 6 = 30 + 8 – 6 = 30 + 8 – 6 = 30 + 8(7) – 6 = 30 + 56 – 6 = 80 Diff: 2 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 9) Evaluate: Answer: 268/(4400 ∗ .4262295) = 268/1875.4098 = .1429021 Diff: 2 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 10) Evaluate: 395(2 + .15 ∗ 290/365) Answer: 395 * (2 + .15 ∗ .7945205) = 395 ∗ (2 + .1191781) = 395 ∗ (2.1191781) = 837.075 Diff: 2 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 11) Evaluate: 400(1 + .10 ∗ 100/365) Answer: 400 * (1 + .10 ∗ .2739726) = 400 ∗ (1 + .) = 400 ∗ (1.) = 410.959 Diff: 2 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 12) Evaluate: 8600(1 – .27 * 226/360) Answer: 8600 ∗ (1 – .27 ∗ .6277778) = 8600 ∗ (1 – .1695) = 8600(.8305) = 7142.3 Diff: 2 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 13) Evaluate: Answer: 2424/(1 + .2 ∗ .4547945) = 2424/(1 + .0909589) = 2424/1.0909589 = 2221.899 Diff: 2 Type: SA Page Ref: 6-10 Topic: 1.2 Fractions Objective: 1-2: Determine equivalent fractions and convert fractions to decimals. 14) Evaluate: Answer: 2910/(1 – .015 ∗ .2054795) = 2910/(1 – .0030822) = 2910/.9969178 = 2918.997 Diff: 2 Type: SA Page Ref: 6-10 Topic: 1.2 Fractions Objective: 1-2: Determine equivalent fractions and convert fractions to decimals. 15) Evaluate: Answer: 5000/(1 + .1 ∗ .5) = 5000/(1 + 0.05 ) = 5000/1.05 = 4761.90 Diff: 2 Type: SA Page Ref: 6-10 Topic: 1.2 Fractions Objective: 1-2: Determine equivalent fractions and convert fractions to decimals. 16) Spade Realty sold lots for $23 240 per hectare. What is the total sales value if the lot sizes, in hectares, were 2 , 3 , 4 ? Answer: 23240 ∗ (2 + 3 + 4 ) = 23240 ∗ (2 10/20 + 3 5/20 + 4 4/20) = 23240 ∗ (9 19/20) = 23240 ∗ 9.95 = $231238 Diff: 2 Type: SA Page Ref: 6-10 Topic: 1.2 Fractions Objective: 1-2: Determine equivalent fractions and convert fractions to decimals. 17) Three mechanics worked 15 , 14 , 18 hours respectively. What was the total cost of labor if the mechanics were paid $14.75 per hour? Answer: Total Hours = 15 + 14 + 14 = 15.5 + 14.75 + 18.125 = 48.375 Total cost of labor = 48.375 ∗ 14.75 = $713.53 Diff: 2 Type: SA Page Ref: 6-10 Topic: 1.2 Fractions Objective: 1-2: Determine equivalent fractions and convert fractions to decimals. 18) Three workers worked 10 , 15 , 20 hours respectively. What was the total cost of labour if the workers were paid $20.00 per hour? Answer: Total Hours = 10 + 15 + 20 = 10.5 + 15.60 + 20.25 = 46.35 Total cost of labor = 46.35 ∗ 20.00 = $927.00 Diff: 2 Type: SA Page Ref: 6-10 Topic: 1.2 Fractions Objective: 1-2: Determine equivalent fractions and convert fractions to decimals. 19) A retailer returned 300 defective items to the manufacturer and received a credit for the retail price of $0.75 less a discount of 1/3 of the retail price. What was the amount of the credit received by the retailer? Answer: Retail value = 300($0.75) = $225 Credit = (1-1/3 )∗ $225 = (2/3)*$225 = $150 Diff: 2 Type: SA Page Ref: 12-17 Topic: 1.3 Applications – Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 20) Complete the following inventory sheet and find the total value. Item Quantity Cost per Unit Total 1 69 $.85 2 111 16 2/3 cents 3 155 $2.75 4 350 $1.66 Answer: 69 × 0.85 = $58.65 111 × 0.16 2/3 = 330 × 0.1666667 = 18.50 155 × 2.75 = 426.25 350 × 1.66 = 581.00 $1084.40 Diff: 1 Type: SA Page Ref: 12-17 Topic: 1.3 Applications – Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 21) Extend each of the following and determine the total. Quantity Unit Price 48 $2.45 48 $0.83 16 $2.12 60 $1.33 Answer: Quantity Unit Price Value 48 $2.45 $117.60 48 0.83 1/8 39.90 16 2.12 33.92 60 1.33 1/6 79.90 Total: $271.32 Diff: 1 Type: SA Page Ref: 12-17 Topic: 1.3 Applications – Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 22) Purchases of an inventory item during the last accounting period were as follows: Number of items Unit price 5 $4.00 3 $6.00 7 $9.00 11 $7.00 What was the weighted average price per item? Answer: Number of items Unit price Weighted value 5 × $4.00 = 20.00 3 × 6.00 = 18.00 7 × 9.00 = 63.00 11 × 7.00 = 77.00 Total: 26 178.00 Average price was 178/26 = $6.85 Diff: 2 Type: SA Page Ref: 12-17 Topic: 1.3 Applications – Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 23) Purchases of an inventory item during last month were as follows: Number of items Unit price 5 $5.00 10 $8.00 8 $6.00 15 $3.00 What was the weighted average price per item? Answer: Number of items Unit price Weighted value 5 × $5.00 = 25.00 10 × 8.00 = 80.00 8 × 6.00 = 48.00 15 × 3.00 = 45.00 Total: 38 198.00 Average price was 198/38 = $5.21 Diff: 2 Type: SA Page Ref: 12-17 Topic: 1.3 Applications – Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 24) Noriko’s final mark for her Financial Mathematics course was based on four tests with different weightings. Test one counted for 10% of the final grade, test two for 20%, test three for 30% and test four for 40%. If Clara received 70% on test one, 85% on test two, 64% on test three and 72% on test four, calculate her final mark. Answer: = 70(0.1) + 85(0.2) + 64(0.3) + 72(0.4) = 7 + 17 + 19.2 + 28.8 = 72 Diff: 2 Type: SA Page Ref: 12-17 Topic: 1.3 Applications – Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 25) On a trip, a motorist purchased gasoline as follows: 66 litres at 69.0 cents per litre; 69 litres at 70.5 cents per litres; 80 litres at 71.5 cents per litre; and 57 litres at 74.5 cents per litre. 1. a) What was the average number of litres per purchase? 2. b) What was the average cost per litre? 3. c) If the motorist averaged 9.75 km per litre, what was her average cost of gasoline per kilometre? Answer: 1. a) 66 + 69 + 80 + 57 = 272 Average number of litres = 272 ÷ 4 = 68 1. b) Average cost per litre: Total cost = 66 × 69.0 = 45.54 69 × 70.5 = 48.645 80 × 71.5 = 57.20 57 × 74.5 = 42.465 193.85 cents Average cost = 193.86 ÷ 272 = 71.27 cents 71. c) Average cost per km = 71.27 ÷ 9.75 = 7.3097436 cents Diff: 2 Type: SA Page Ref: 12-17 Topic: 1.3 Applications – Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages.