CSEC Mathematics: Computation

This comprehensive lesson covers the Computation section of the CXC/CSEC Mathematics syllabus for 2024-2025 examinations.

Introduction to Computation

Computation in mathematics involves performing calculations with numbers and quantities. For the CSEC Mathematics syllabus, this includes working with different types of numbers, applying various operations, and understanding the properties that govern these operations.

1. Number Systems and Types

1.1 Natural Numbers and Integers

Natural numbers are the counting numbers: 1, 2, 3, 4, 5, ...

Integers include natural numbers, zero, and negative whole numbers: ..., -3, -2, -1, 0, 1, 2, 3, ...

Example: Identifying integers in a set
From the set {-5, -1.5, 0, 2, √4, π, 10}, the integers are {-5, 0, 2, 10}

1.2 Rational and Irrational Numbers

Rational numbers can be expressed as fractions (p/q where p and q are integers and q ≠ 0). This includes:

Irrational numbers cannot be expressed as fractions, such as:

Example: Classifying numbers
√9 = 3 (Rational - natural number)
0.25 = 1/4 (Rational - terminating decimal)
0.333... = 1/3 (Rational - recurring decimal)
π = 3.141592... (Irrational - non-terminating, non-recurring)

1.3 Real Numbers

The set of real numbers includes all rational and irrational numbers, which can be represented on a number line.

-4 -3 -2 -1 0 1 2 3 4 Real Numbers Rational Irrational

Figure 1: The Real Number Line showing rational and irrational numbers

2. Basic Arithmetic Operations

2.1 Addition and Subtraction

These operations follow certain properties:

Example: Working with integers
(-7) + 15 = 8
12 - (-5) = 12 + 5 = 17

2.2 Multiplication and Division

Key properties include:

Example: Multiplication and division with fractions
2/3 × 9/4 = (2 × 9)/(3 × 4) = 18/12 = 3/2 = 1.5
8/5 ÷ 2/3 = 8/5 × 3/2 = (8 × 3)/(5 × 2) = 24/10 = 12/5 = 2.4

2.3 Order of Operations (BEDMAS/PEMDAS)

When evaluating expressions with multiple operations, follow this order:

  1. Brackets (or Parentheses)
  2. Exponents (or indices)
  3. Division and Multiplication (from left to right)
  4. Addition and Subtraction (from left to right)
Example: Applying BEDMAS
Calculate: 18 ÷ 3 + 5² - (7 - 2) × 4
= 18 ÷ 3 + 25 - 5 × 4
= 6 + 25 - 20
= 31 - 20
= 11

3. Fractions, Decimals, and Percentages

3.1 Converting Between Forms

These three forms represent parts of a whole and can be converted between each other:

Fraction → Decimal: Divide numerator by denominator
Decimal → Fraction: Write as a fraction and simplify
Fraction → Percentage: Multiply by 100%
Percentage → Fraction: Divide by 100% and simplify
Decimal → Percentage: Multiply by 100%
Percentage → Decimal: Divide by 100%
Example: Conversions
3/4 = 0.75 = 75%
0.125 = 1/8 = 12.5%
45% = 0.45 = 9/20

3.2 Operations with Fractions

To add or subtract fractions, find a common denominator first:

a/b + c/d = (ad + bc)/(bd)
a/b - c/d = (ad - bc)/(bd)

To multiply fractions, multiply numerators and denominators:

a/b × c/d = (a × c)/(b × d)

To divide by a fraction, multiply by its reciprocal:

a/b ÷ c/d = a/b × d/c = (a × d)/(b × c)
Example: Operations with fractions
2/5 + 1/3 = (2 × 3 + 1 × 5)/(5 × 3) = (6 + 5)/15 = 11/15
3/4 - 1/6 = (3 × 6 - 1 × 4)/(4 × 6) = (18 - 4)/24 = 14/24 = 7/12

3.3 Recurring Decimals

Converting recurring decimals to fractions:

Example: Convert 0.363636... to a fraction
Let x = 0.363636...
100x = 36.3636...
100x - x = 36.3636... - 0.3636...
99x = 36
x = 36/99 = 4/11

4. Powers and Roots

4.1 Laws of Indices

When working with expressions in the form an, where a is the base and n is the exponent:

Example: Applying the laws of indices
23 × 24 = 23+4 = 27 = 128
92 ÷ 91/2 = 92-1/2 = 93/2 = (91/2)3 = 33 = 27
(41/2)-2 = 41/2 × (-2) = 4-1 = 1/4

4.2 Square Roots and Cube Roots

Square root (√) and cube root (∛) are common roots we work with:

Example: Simplifying radicals
√48 = √(16 × 3) = √16 × √3 = 4√3
∛54 = ∛(27 × 2) = ∛27 × ∛2 = 3 × ∛2

4.3 Surds

Surds are expressions containing irrational square roots that cannot be simplified further.

Rationalizing the denominator:
a/(b√c) = (a × √c)/(b × c)
a/(√b + √c) = [a × (√b - √c)]/[(√b + √c)(√b - √c)] = [a(√b - √c)]/(b - c)
Example: Rationalizing the denominator
5/√3 = (5 × √3)/(√3 × √3) = 5√3/3
2/(√5 + √2) = [2 × (√5 - √2)]/[(√5 + √2)(√5 - √2)] = [2(√5 - √2)]/(5 - 2) = [2(√5 - √2)]/3

5. Scientific Notation

5.1 Writing Numbers in Scientific Notation

A number in scientific notation has the form a × 10n, where 1 ≤ a < 10 and n is an integer.

Example: Converting to scientific notation
43,500 = 4.35 × 104
0.00728 = 7.28 × 10-3

5.2 Operations with Scientific Notation

When performing operations with numbers in scientific notation:

Example: Operations with scientific notation
(3.6 × 105) × (2.5 × 10-3) = (3.6 × 2.5) × 105 + (-3) = 9.0 × 102 = 900
(8.4 × 106) ÷ (2.1 × 104) = (8.4 ÷ 2.1) × 106 - 4 = 4 × 102 = 400

6. Ratio, Proportion, and Rates

6.1 Ratios

A ratio compares quantities of the same kind and can be written as a:b or a/b.

Equivalent ratios: a:b = c:d if ad = bc
Example: Working with ratios
Divide $350 in the ratio 2:3:5
Total parts = 2 + 3 + 5 = 10
Value of each part = $350 ÷ 10 = $35
First share = 2 × $35 = $70
Second share = 3 × $35 = $105
Third share = 5 × $35 = $175

6.2 Proportion

Direct proportion: y ∝ x or y = kx, where k is the constant of proportionality

Inverse proportion: y ∝ 1/x or y = k/x

Example: Direct proportion
If 5 workers can complete a task in 12 days, how long would it take 8 workers?
Time ∝ 1/(Number of workers)
t1/t2 = w2/w1
12/t2 = 8/5
t2 = (12 × 5)/8 = 7.5 days

6.3 Rates

A rate is a ratio of quantities with different units, such as speed (distance/time) or currency exchange rates.

Example: Currency conversion
Exchange rate: 1 USD = 2.5 BBD (Barbados Dollars)
To convert 75 USD to BBD: 75 × 2.5 = 187.5 BBD
To convert 125 BBD to USD: 125 ÷ 2.5 = 50 USD

7. Approximation and Estimation

7.1 Rounding

Numbers can be rounded to a specified number of:

Example: Rounding numbers
37.846 rounded to:
- 2 d.p. = 37.85
- 1 d.p. = 37.8
- 3 s.f. = 37.8
- 2 s.f. = 38
- Nearest 10 = 40

7.2 Error Calculations

When working with approximations, we can calculate:

Example: Error calculation
If π is approximated as 3.14:
Absolute error = |3.14159... - 3.14| ≈ 0.00159
Relative error ≈ 0.00159/3.14159 ≈ 0.00051
Percentage error ≈ 0.00051 × 100% ≈ 0.051%

7.3 Upper and Lower Bounds

When a value has been rounded to a specific degree of accuracy:

Example: Finding bounds
For a measurement of 5.7 cm (to 1 d.p.):
Lower bound = 5.7 - 0.05 = 5.65 cm
Upper bound = 5.7 + 0.05 = 5.75 cm

8. Consumer Arithmetic

8.1 Percentages in Business

Example: Calculating profit
A smartphone purchased for $250 is sold for $320.
Profit = $320 - $250 = $70
Percentage profit = ($70/$250) × 100% = 28%

8.2 Simple and Compound Interest

Simple Interest formula: I = Prt

where I = interest, P = principal, r = rate (decimal), t = time (years)

Compound Interest formula: A = P(1 + r)t

where A = final amount, P = principal, r = rate (decimal), t = time (years)

Example: Interest calculations
A deposit of $2000 at 5% per annum for 3 years:
Simple interest: I = $2000 × 0.05 × 3 = $300
Final amount = $2000 + $300 = $2300

Compound interest: A = $2000 × (1 + 0.05)3 = $2000 × 1.157625 = $2315.25

8.3 Depreciation

Straight-line depreciation: Equal yearly depreciation amount

Reducing balance method: Value = Original value × (1 - r)t

where r = depreciation rate (decimal), t = time (years)

Example: Calculating depreciation
A car worth $25,000 depreciates at 12% per year using the reducing balance method.
After 3 years, its value = $25,000 × (1 - 0.12)3
= $25,000 × 0.883
= $25,000 × 0.681472
= $17,036.80

9. Applying Computation to Real-World Problems

9.1 Problem-Solving Strategies

  1. Understand the problem: Identify given information and what needs to be found
  2. Plan a solution approach: Select appropriate computational methods
  3. Execute the plan: Perform calculations carefully
  4. Verify the solution: Check if the answer makes sense in context
Example: Integrated problem
A group of friends share a vacation rental costing $1575 for 9 days. If two friends leave after 3 days and the remaining 5 friends stay for the full period, how much should each person pay if costs are split fairly by person-days?

Solution:
Total person-days = (7 people × 3 days) + (5 people × 6 days)
= 21 + 30 = 51 person-days

Cost per person-day = $1575 ÷ 51 ≈ $30.88 per person-day

Amount for each early-leaving friend = 3 days × $30.88 = $92.64
Amount for each staying friend = 9 days × $30.88 = $277.92

Glossary of Terms

Self-Assessment Questions

Section 1: Number Systems

Question 1: From the set {-5, 2.7, 0, √16, -3/4, √5, π}, identify:

a) The integers

b) The rational numbers

c) The irrational numbers

a) Integers: {-5, 0, 4}

b) Rational numbers: {-5, 2.7, 0, 4 (√16), -3/4}

c) Irrational numbers: {√5, π}

Section 2: Basic Arithmetic

Question 2: Evaluate the following expression using the correct order of operations: 24 ÷ (7 - 3) × 2² - 5

24 ÷ (7 - 3) × 2² - 5

= 24 ÷ 4 × 4 - 5

= 6 × 4 - 5

= 24 - 5

= 19

Section 3: Fractions, Decimals, and Percentages

Question 3: Convert the following:

a) 0.375 to a fraction in lowest terms

b) 5/8 to a percentage

c) 42% to a decimal and then to a fraction in lowest terms

a) 0.375 = 375/1000 = 3/8

b) 5/8 = 5/8 × 100% = 62.5%

c) 42% = 0.42 = 42/100 = 21/50

Question 4: Perform the following operations with fractions:

a) 3/4 + 5/6

b) 2/3 × 9/10

c) 5/8 ÷ 2/3

a) 3/4 + 5/6 = (3 × 6 + 5 × 4)/(4 × 6) = (18 + 20)/24 = 38/24 = 19/12

b) 2/3 × 9/10 = (2 × 9)/(3 × 10) = 18/30 = 3/5

c) 5/8 ÷ 2/3 = 5/8 × 3/2 = 15/16

Section 4: Powers and Roots

Question 5: Simplify the following expressions:

a) 23 × 25 ÷ 24

b) (34)2 ÷ 35

c) 163/4

a) 23 × 25 ÷ 24 = 23+5-4 = 24 = 16

b) (34)2 ÷ 35 = 34×2 ÷ 35 = 38 ÷ 35 = 38-5 = 33 = 27

c) 163/4 = (24)3/4 = 24×3/4 = 23 = 8

Question 6: Simplify the following surds:

a) √75

b) 6√8 + 3√32

c) Rationalize the denominator: 7/(3√5)

a) √75 = √(25 × 3) = √25 × √3 = 5√3

b) 6√8 + 3√32 = 6√(4 × 2) + 3√(16 × 2) = 6 × 2√2 + 3 × 4√2 = 12√2 + 12√2 = 24√2

c) 7/(3√5) = (7 × √5)/(3√5 × √5) = 7√5/(3 × 5) = 7√5/15

Section 5: Scientific Notation

Question 7: Convert the following:

a) 0.00325 to scientific notation

b) 7.65 × 108 to standard form

c) Calculate (3.6 × 104) × (2.5 × 10-2) in scientific notation

a) 0.00325 = 3.25 × 10-3

b) 7.65 × 108 = 765,000,000

c) (3.6 × 104) × (2.5 × 10-2) = (3.6 × 2.5) × 104 + (-2) = 9.0 × 102 = 900

Section 6: Ratio, Proportion, and Rates

Question 8: A recipe requires 2 cups of flour, 3 eggs, and 1.5 cups of milk. If you have 5 cups of flour:

a) How many eggs would you need?

b) How much milk would you need?

Original ratio: 2 cups flour : 3 eggs : 1.5 cups milk

New amount of flour = 5 cups

Scaling factor = 5/2 = 2.5

a) Eggs needed = 3 × 2.5 = 7.5 eggs (so 8 eggs would be needed in practice)

b) Milk needed = 1.5 × 2.5 = 3.75 cups

Question 9: If 4 workers can complete a job in 10 days, how long would it take 5 workers to complete the same job, assuming they all work at the same rate?

Number of workers × Days = Constant (inverse proportion)

4 × 10 = 5 × x

40 = 5x

x = 8 days

Section 7: Approximation and Estimation

Question 10: Round the number 86.7249 to:

a) 2 decimal places

b) 3 significant figures

c) The nearest integer

a) 2 decimal places: 86.72

b) 3 significant figures: 86.7

c) Nearest integer: 87

Question 11: The value of π is approximately 3.14. Calculate:

a) The absolute error

b) The percentage error

a) Absolute error = |π - 3.14| ≈ |3.14159... - 3.14| ≈ 0.00159

b) Percentage error = (Absolute error / Exact value) × 100%

= (0.00159 / 3.14159...) × 100%

≈ 0.051%

Section 8: Consumer Arithmetic

Question 12: A laptop originally priced at $1200 is on sale for 15% off. What is the sale price?

Discount = 15% of $1200 = 0.15 × $1200 = $180

Sale price = $1200 - $180 = $1020

Question 13: Jennifer invests $5000 for 3 years. Calculate the final amount if the interest is:

a) 4.5% simple interest per annum

b) 4.5% compound interest per annum

a) Simple interest:

I = P × r × t = $5000 × 0.045 × 3 = $675

Final amount = $5000 + $675 = $5675

b) Compound interest:

A = P(1 + r)^t = $5000(1 + 0.045)^3 = $5000 × 1.141166 = $5705.83

Question 14: A car purchased for $35,000 depreciates by 12% per year. What is its value after 4 years?

Using the reducing balance method:

Value after 4 years = $35,000 × (1 - 0.12)^4

= $35,000 × 0.88^4

= $35,000 × 0.5997

= $20,989.46

Section 9: Comprehensive Problems

Question 15: At a market, Tanya bought 3 kg of carrots at $2.50 per kg and 2 kg of potatoes at $1.80 per kg. She received $0.35 in change after paying with a $10 bill. How much did 1 kg of onions cost if she also bought 1.5 kg of onions?

Cost of carrots = 3 kg × $2.50/kg = $7.50

Cost of potatoes = 2 kg × $1.80/kg = $3.60

Total spent = $10 - $0.35 = $9.65

Cost of onions = $9.65 - $7.50 - $3.60 = $9.65 - $11.10 = -$1.45

This is impossible because the cost cannot be negative.

Let's recalculate:

Total spent = $10 - $0.35 = $9.65

Cost of carrots and potatoes = $7.50 + $3.60 = $11.10

Remaining amount for onions = $9.65 - $11.10 = -$1.45

This negative result suggests an error in the original problem statement.

Let's assume she paid with a $20 bill instead of a $10 bill:

Total spent = $20 - $0.35 = $19.65

Cost of onions = $19.65 - $11.10 = $8.55

Cost per kg of onions = $8.55 ÷ 1.5 kg = $5.70/kg

Extended Practice Problems

Problem 1: A product originally costing $x is marked up by 25%. During a sale, a discount of 20% is applied to the marked-up price. Show that the final price can be expressed as 1.05x. What percentage of the original price is this?

Original price = $x

After 25% markup, new price = x + 0.25x = 1.25x

After 20% discount on marked-up price = 1.25x - 0.2(1.25x) = 1.25x - 0.25x = x

= 1.25x × (1 - 0.2) = 1.25x × 0.8 = x

= 1.00x

This is 100% of the original price.

Note: There was an error in the problem statement. The final price is not 1.05x but x (or 1.00x), which is exactly the original price. The markup and discount effectively cancel each other out.

Problem 2: In an examination, the marks of 120 students were distributed as follows:

Marks (out of 100) Number of Students
0 - 20 12
21 - 40 18
41 - 60 48
61 - 80 30
81 - 100 12

Calculate the mean mark, taking the midpoint of each class as representative.

Using the midpoint of each class:

0-20: midpoint = 10

21-40: midpoint = 30.5

41-60: midpoint = 50.5

61-80: midpoint = 70.5

81-100: midpoint = 90.5

Mean = (10 × 12 + 30.5 × 18 + 50.5 × 48 + 70.5 × 30 + 90.5 × 12) ÷ 120

= (120 + 549 + 2424 + 2115 + 1086) ÷ 120

= 6294 ÷ 120

= 52.45

The mean mark is approximately 52.45 out of 100.

Problem 3: If a car travels 210 km using 15 liters of fuel, and the fuel costs $1.25 per liter, calculate:

a) The fuel efficiency in km/liter

b) The cost of fuel per kilometer

c) The cost of a 350 km journey

a) Fuel efficiency = Distance traveled ÷ Fuel used = 210 km ÷ 15 liters = 14 km/liter

b) Cost per kilometer = Cost per liter ÷ km per liter = $1.25 ÷ 14 = $0.089 per km

c) Cost of 350 km journey = 350 km × $0.089 per km = $31.15

Alternatively: Fuel needed for 350 km = 350 km ÷ 14 km/liter = 25 liters

Cost = 25 liters × $1.25 per liter = $31.25

(The slight difference is due to rounding)

Additional Resources

Summary of Key Points

  1. Understand the different number systems and their properties
  2. Master the four basic operations and their order (BEDMAS/PEMDAS)
  3. Be able to convert between fractions, decimals, and percentages
  4. Know the laws of indices and how to work with powers and roots
  5. Use scientific notation for very large and very small numbers
  6. Apply ratio, proportion, and rates to solve real-world problems
  7. Understand rounding rules and error calculation
  8. Apply mathematics to consumer and business contexts
  9. Develop systematic problem-solving strategies

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