Test Bank for Contemporary Business Mathematics with Canadian
Applications, 12th edition By S. A. Hummelbrunner
[email protected]
Contemporary
... [Show More] Business Mathematics, 12e (Hummelbrunner)
Chapter 1 Review of Arithmetic
1) Simplify: (28 - 4)/3
Answer: (28 - 4)/3 = 24/3 = 8
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: 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.
4) 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.
5) 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.
6) 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.
7) 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.
1
Copyright © 2021 Pearson Education, Inc.
[email protected]
8) Evaluate and round to 3 decimal places: 125(6 + 0.35 ∗ 142/678)
Answer: 125 ∗ (6 + 0.35 ∗ 0.2094395) = 125 ∗ (6 + 0.0733038) = 125 ∗ (6.0733038) = 759.163
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: 400(1 + .10 ∗ 100/365)
Answer: 400 ∗ (1 + .10 ∗ .2739726) = 400 ∗ (1 + .02739726) = 400 ∗ (1.02739726) = 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.
10) Evaluate and round to one decimal place: 9210(5 - 1.38 ∗ 169/420)
Answer: 9210 ∗ (5 - 1.38 ∗ 0.40238095) = 9210 ∗ (5 - 0.5552857) = 9210(4.444714) = 40,935.8
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:
Answer: 2424/(1 + .2 ∗ .4547945) = 2424/(1 + .0909589) = 2424/1.0909589 = 2221.899
Diff: 2 Type: SA Page Ref: 6-11
Topic: 1.2 Fractions
Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice
versa.
12) Evaluate and round to three decimal places:
Answer: 3140/(2 + 0.35 * 0.2123288) = 3140/(2 + 0.0743151) = 2910/2.074315068 = 1513.753
Diff: 2 Type: SA Page Ref: 7-13
Topic: 1.2 Fractions
Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice
versa.
13) Evaluate:
Answer: 5000/(1 + .1 ∗ .5) = 5000/(1 + 0.05 ) = 5000/1.05 = 4761.90
Diff: 2 Type: SA Page Ref: 6-11
Topic: 1.2 Fractions
Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice
versa.
2
Copyright © 2021 Pearson Education, Inc.
[email protected]
14) 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 ∗ ( + 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: 6-11
Topic: 1.2 Fractions
Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice
versa.
15) Three mechanics worked 15 , 14 , 18 hours respectively. What was the total cost of labour if the
mechanics were paid $14.75 per hour?
Answer: Total Hours
= 15 + 14 + 18
= 15.5 + 14.75 + 18.125
= 48.375
Total cost of labor = 48.375 ∗ 14.75 = $713.53
Diff: 2 Type: SA Page Ref: 6-11
Topic: 1.2 Fractions
Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice
versa.
16) Ana, Aamir and Charlotte worked 11 , 14 , and 22 hours respectively. What was the total cost of
labour if they were paid $18.00 per hour?
Answer: Total Hours
= 11 + 14 + 22
= 11.75 + 14.65 + 22.80
= 49.20
Total cost of labor = 49.20 ∗ 18.00 = $885.60
Diff: 2 Type: SA Page Ref: 7-13
Topic: 1.2 Fractions
Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice
versa.
17) 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: 16-21
Topic: 1.4 Applications - Averages
Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages.
3
Copyright © 2021 Pearson Education, Inc.
[email protected]
18) 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:
1161616161669 × 0.85 = $58.65
111 × 0.16 2/3 = 330 × 0.1666667 = 18.50
155 × 2.75 = 426.25
350 × 1.66 = 581.00
Diff: 1 Type: SA Page Ref: 16-21
$1084.40
Topic: 1.4 Applications - Averages
Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages.
19) 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: 16-21
Topic: 1.4 Applications - Averages
Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages.
4
Copyright © 2021 Pearson Education, Inc.
[email protected]
20) Purchases of packs of printing paper during the last accounting period were as follows:
Number of items Unit price
8 $13.00
4 $12.00
15 $10.00
10 $10.50
What was the weighted average price per item?
Answer:
Number of items Unit price Weighted value
8 × $13.00 = 104.00
4 × $12.00 = 48.00
15 × $10.00 = 150.00
10 × $10.50 = 105.00
Total: 37 407.00
Average price was 407/37 = $11.00
Diff: 2 Type: SA Page Ref: 17-24
Topic: 1.4 Applications - Averages
Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages.
21) 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: 16-21
Topic: 1.4 Applications - Averages
Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages.
5
Copyright © 2021 Pearson Education, Inc.
[email protected]
22) 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: 16-21
Topic: 1.4 Applications - Averages
Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages.
23) 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.
a) What was the average number of litres per purchase?
b) What was the average cost per litre?
c) If the motorist averaged 9.75 km per litre, what was her average cost of gasoline per kilometre?
Answer:
a) 66 + 69 + 80 + 57 = 272
Average number of litres = 272 ÷ 4 = 68
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
c) Average cost per km = 71.27 ÷ 9.75 = 7.3097436 cents
Diff: 2 Type: SA Page Ref: 16-21
Topic: 1.4 Applications - Averages
Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages.
24) Dorian Frump invested $12 500 in a business on January 1. She withdrew $850 on April 3, reinvested
$1920 on August 1, and withdrew $700 on September 1. What is Don's average monthly investment
balance for the year?
Answer:
Weighted investment:
January 1 — March 31: 12500 × 3/12 = 3125.0000
April 1 - July 31: 11650 × 4/12 = 3883.3333
August 1 - August 31: 13570 × 1/12 = 1130.8333
September 1 - December 31: 12870 × 4/12 = 4290.0000
Average investment balance = $12429.17
Diff: 2 Type: SA Page Ref: 17-24
Topic: 1.4 Applications - Averages
Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. [Show Less]