Numbers can be grouped in to a number of sets. From the diagram you see that all rational numbers are also real numbers; i.e. ℚ is a subset of ℝ.
Positive integers
Positive integers and zero
Integers
Rational numbers
Numbers
Our detailed guide on **Number** for IB Exams. Whether you're a student preparing for the International Baccalaureate (IB) Mathematics exams or seeking to strengthen your foundational number skills, this guide offers thorough explanations, properties, methods, IB-style exam questions with answers, strategies, common mistakes, practice questions, combined exercises, and additional resources to help you excel in the Number section of your IB Mathematics assessments.
Introduction
The **Number** topic in IB Mathematics covers a wide range of fundamental concepts that form the basis for higher-level mathematical studies. Mastery of these concepts is essential not only for performing well in IB exams but also for developing critical thinking and problem-solving skills applicable in various real-world scenarios.
Importance of Number in IB Problem Solving
The Number topic is pivotal in IB Mathematics for several reasons:
- It lays the groundwork for more advanced topics such as algebra, calculus, and statistics.
- Understanding number properties and operations is essential for solving complex mathematical problems.
- It enhances logical reasoning and analytical skills, which are crucial for the IB’s rigorous assessment standards.
- Real-world applications of number concepts make them relevant beyond the classroom.
Proficiency in Number concepts ensures a strong mathematical foundation, enabling students to tackle higher-level mathematics with confidence.
Basic Concepts of Number
Before delving into advanced applications, it's essential to grasp the foundational elements of the Number topic in IB Mathematics.
Types of Numbers
Natural Numbers (N): Counting numbers starting from 1, 2, 3, ...
Whole Numbers: Natural numbers including zero: 0, 1, 2, 3, ...
Integers (Z): Whole numbers and their negatives: ..., -3, -2, -1, 0, 1, 2, 3, ...
Rational Numbers (Q): Numbers that can be expressed as a fraction of two integers, where the denominator is not zero.
Irrational Numbers: Numbers that cannot be expressed as a simple fraction, with non-repeating, non-terminating decimal expansions.
Real Numbers: All rational and irrational numbers.
Complex Numbers: Numbers in the form a + bi, where a and b are real numbers, and i is the imaginary unit.
Prime Numbers and Factors
Prime Numbers: Natural numbers greater than 1 that have no positive divisors other than 1 and themselves.
Composite Numbers: Natural numbers greater than 1 that are not prime.
Factors: Numbers that divide another number without leaving a remainder.
Prime Factorization: Expressing a number as a product of its prime factors.
Divisibility Rules
Quick methods to determine if one number is divisible by another without performing full division.
- Divisible by 2: If the last digit is even.
- Divisible by 3: If the sum of the digits is divisible by 3.
- Divisible by 5: If the last digit is 0 or 5.
- Divisible by 9: If the sum of the digits is divisible by 9.
- Divisible by 10: If the last digit is 0.
Properties of Numbers
Understanding the properties that govern different types of numbers is crucial for simplifying problems and performing accurate calculations.
Properties of Integers
- Closure Property: The sum and product of any two integers is an integer.
- Commutative Property: a + b = b + a; a × b = b × a.
- Associative Property: (a + b) + c = a + (b + c); (a × b) × c = a × (b × c).
- Distributive Property: a × (b + c) = (a × b) + (a × c).
Rational and Irrational Numbers
- Rational Numbers: Can be expressed as a fraction where both numerator and denominator are integers.
- Irrational Numbers: Cannot be expressed as a simple fraction; their decimal expansions are non-repeating and non-terminating.
Methods in Number
Various methods are employed in the Number topic to solve problems efficiently and accurately.
1. Prime Factorization
Breaking down a number into its prime factors to simplify calculations and solve problems related to multiples and divisors.
Example: Find the prime factorization of 60.
Solution:
60 ÷ 2 = 30
30 ÷ 2 = 15
15 ÷ 3 = 5
5 ÷ 5 = 1
Prime factors: 2 × 2 × 3 × 5 = 2² × 3 × 5
2. Divisibility Tests
Applying divisibility rules to quickly determine if a number is divisible by another without performing long division.
Example: Determine if 1,254 is divisible by 3.
Solution:
Sum of digits = 1 + 2 + 5 + 4 = 12
12 is divisible by 3, so 1,254 is divisible by 3.
3. Greatest Common Divisor (GCD) and Least Common Multiple (LCM)
Finding the largest number that divides two or more numbers (GCD) and the smallest number that is a multiple of two or more numbers (LCM).
Example: Find the GCD and LCM of 12 and 18.
Solution:
Prime factors of 12: 2² × 3
Prime factors of 18: 2 × 3²
GCD = 2¹ × 3¹ = 6
LCM = 2² × 3² = 36
4. Exponents and Roots
Understanding and manipulating powers and roots to simplify expressions and solve equations.
Example: Simplify \( (2^3) \times (2^4) \).
Solution:
Using the rule \( a^m \times a^n = a^{m+n} \):
\( 2^3 \times 2^4 = 2^{7} = 128 \)
Calculations in Number
Performing calculations accurately is essential for solving Number-related problems in IB Exams. This involves applying the correct formulas and methods to obtain precise results.
1. Operations with Integers
Adding, subtracting, multiplying, and dividing integers while following the rules of signs.
Example: Calculate \( -7 + 12 - (-5) \).
Solution:
Apply the rules of signs:
\( -7 + 12 = 5 \)
\( 5 - (-5) = 5 + 5 = 10 \)
2. Simplifying Expressions with Exponents
Applying exponent rules to simplify expressions.
Example: Simplify \( (3^2)^3 \).
Solution:
Using the power of a power rule \( (a^m)^n = a^{m \times n} \):
\( (3^2)^3 = 3^{6} = 729 \)
3. Calculating GCD and LCM
Finding the greatest common divisor and least common multiple using prime factorization.
Example: Find the GCD and LCM of 24 and 36.
Solution:
Prime factors of 24: 2³ × 3
Prime factors of 36: 2² × 3²
GCD = 2² × 3 = 12
LCM = 2³ × 3² = 72
4. Working with Rational and Irrational Numbers
Performing operations and simplifying expressions involving rational and irrational numbers.
Example: Simplify \( \frac{3}{4} + \sqrt{2} \).
Solution:
This expression cannot be simplified further as it involves a rational number and an irrational number.
Examples of Problem Solving with Number
Understanding through examples is key to mastering the Number topic. Below are a variety of IB-style problems ranging from easy to hard, each with detailed solutions.
Example 1: Prime Factorization
Problem: Find the prime factorization of 84.
Solution:
84 ÷ 2 = 42
42 ÷ 2 = 21
21 ÷ 3 = 7
7 ÷ 7 = 1
Prime factors: 2 × 2 × 3 × 7 = 2² × 3 × 7
Therefore, the prime factorization of 84 is \( 2^2 \times 3 \times 7 \).
Example 2: Divisibility Rule
Problem: Determine if 2,346 is divisible by 3.
Solution:
Sum of digits = 2 + 3 + 4 + 6 = 15
15 is divisible by 3, so 2,346 is divisible by 3.
Therefore, 2,346 is divisible by 3.
Example 3: GCD and LCM
Problem: Find the GCD and LCM of 16 and 24.
Solution:
Prime factors of 16: 2⁴
Prime factors of 24: 2³ × 3
GCD = 2³ = 8
LCM = 2⁴ × 3 = 48
Therefore, the GCD is 8 and the LCM is 48.
Example 4: Operations with Exponents
Problem: Simplify \( 5^3 \times 5^2 \).
Solution:
Using the rule \( a^m \times a^n = a^{m+n} \):
\( 5^3 \times 5^2 = 5^{5} = 3125 \)
Therefore, \( 5^3 \times 5^2 = 3125 \).
Example 5: Rational and Irrational Numbers
Problem: Simplify \( \frac{7}{8} - \sqrt{3} \).
Solution:
This expression cannot be simplified further as it involves a rational number and an irrational number.
Therefore, \( \frac{7}{8} - \sqrt{3} \) remains as it is.
Word Problems: Application of Number in IB Exams
Applying Number concepts to real-life scenarios enhances understanding and demonstrates their practical utility. Here are several IB-style word problems that incorporate these concepts, along with their solutions.
Example 1: Budget Allocation
Problem: A student has saved $480 from part-time jobs. They want to allocate this money equally to buy books, stationery, and a new backpack. How much money will each category receive?
Solution:
Total categories = 3
Money per category = $480 ÷ 3 = $160
Therefore, each category will receive $160.
Example 2: Divisibility in Grouping
Problem: A teacher has 672 pencils and wants to distribute them equally among 12 students. Is this distribution possible without any leftover pencils?
Solution:
Divide 672 by 12:
672 ÷ 12 = 56
Since 672 is divisible by 12, each student will receive 56 pencils with no leftovers.
Therefore, the distribution is possible without any leftover pencils.
Example 3: Prime Factorization in Simplifying Fractions
Problem: Simplify the fraction \( \frac{56}{98} \) by finding the GCD using prime factorization.
Solution:
Prime factors of 56: 2³ × 7
Prime factors of 98: 2 × 7²
GCD = 2 × 7 = 14
Simplified fraction: \( \frac{56 ÷ 14}{98 ÷ 14} = \frac{4}{7} \)
Therefore, the simplified fraction is \( \frac{4}{7} \).
Example 4: Exponents in Compound Interest
Problem: If you invest $1,000 at an annual interest rate of 5%, compounded yearly, what will be the amount after 3 years? Use exponents to calculate.
Solution:
Compound Interest Formula: \( A = P \times (1 + r)^n \)
P = $1,000
r = 0.05
n = 3
A = $1,000 × (1 + 0.05)^3 = $1,000 × 1.157625 ≈ $1,157.63
Therefore, the amount after 3 years is approximately $1,157.63.
Example 5: Rational and Irrational Numbers in Measurements
Problem: The diagonal of a square garden is 10 meters. Express the side length of the garden in meters using rational and irrational numbers.
Solution:
For a square, \( \text{Diagonal} = \text{Side} \times \sqrt{2} \)
Let the side length be s:
10 = s × \( \sqrt{2} \)
s = \( \frac{10}{\sqrt{2}} \) = \( 5\sqrt{2} \) meters (irrational number)
Therefore, the side length of the garden is \( 5\sqrt{2} \) meters.
Strategies and Tips for Number in IB Exams
Enhancing your skills in the Number topic involves employing effective strategies and consistent practice. Here are some tips to help you improve:
1. Master Prime Factorization
Being proficient in prime factorization aids in simplifying fractions, finding GCDs and LCMs, and solving divisibility problems.
Example: Quickly factorize numbers like 84, 56, and 120 to identify common factors.
2. Utilize Divisibility Rules
Apply divisibility rules to determine if numbers are divisible by 2, 3, 5, 9, and 10, which can simplify calculations and problem-solving.
Example: Determine divisibility of large numbers without performing full division.
3. Practice Operations with Integers
Ensure you are comfortable with adding, subtracting, multiplying, and dividing positive and negative integers, following the correct rules for signs.
Example: Solve problems like \( -8 + 12 - (-5) \).
4. Understand Exponent Rules
Familiarize yourself with the laws of exponents to simplify expressions and solve exponential equations efficiently.
Example: Simplify \( (3^2)^4 \) or \( 2^3 \times 2^5 \).
5. Strengthen GCD and LCM Skills
Be adept at finding the greatest common divisor and least common multiple using both prime factorization and other methods.
Example: Find GCD and LCM of pairs like (24, 36) or (18, 30).
6. Differentiate Between Rational and Irrational Numbers
Understand the characteristics that distinguish rational numbers from irrational ones to simplify expressions and solve equations correctly.
Example: Identify if \( \sqrt{5} \) or \( \frac{7}{9} \) is rational or irrational.
7. Apply Real-World Contexts
Relate Number concepts to real-life situations to enhance comprehension and retention.
Example: Use prime factorization in budgeting or GCD/LCM in scheduling events.
8. Consistent Practice with IB-Style Questions
Regularly attempt IB-style questions to familiarize yourself with the exam format and question types.
Example: Practice past IB exam questions and timed quizzes.
Common Mistakes in Number and How to Avoid Them
Being aware of common errors can help you avoid them and improve your calculation accuracy.
1. Misapplying Exponent Rules
Mistake: Incorrectly simplifying expressions involving exponents.
Solution: Familiarize yourself with exponent rules such as \( a^m \times a^n = a^{m+n} \), \( (a^m)^n = a^{m \times n} \), and \( \frac{a^m}{a^n} = a^{m-n} \).
Example:
Incorrect: \( (2^3)^2 = 2^5 \)
Correct: \( (2^3)^2 = 2^{6} \)
2. Incorrect Prime Factorization
Mistake: Skipping steps or misidentifying prime factors, leading to incorrect GCD or LCM.
Solution: Carefully divide the number by prime numbers starting from the smallest (2, 3, 5, etc.) until you reach 1.
Example:
Incorrect: Prime factors of 60 = 2 × 3 × 10
Correct: Prime factors of 60 = 2² × 3 × 5
3. Overlooking Negative Signs
Mistake: Ignoring negative signs when performing operations with integers.
Solution: Always pay attention to the signs and apply the correct rules for addition, subtraction, multiplication, and division.
Example:
Incorrect: \( -5 + 3 = -2 \) (Incorrect reasoning)
Correct: \( -5 + 3 = -2 \) (Actual result, but ensure to understand why)
4. Confusing GCD and LCM
Mistake: Mixing up greatest common divisor with least common multiple.
Solution: Remember that GCD is the largest number that divides two or more numbers, while LCM is the smallest number that is a multiple of two or more numbers.
Example:
Incorrect: GCD of 12 and 18 is 36
Correct: GCD of 12 and 18 is 6; LCM is 36
5. Rounding Errors in Calculations
Mistake: Rounding numbers too early in multi-step calculations, leading to cumulative errors.
Solution: Maintain precision throughout calculations and round only the final result unless instructed otherwise.
Example:
Incorrect: \( (2.5 × 3.6) ≈ (3 × 4) = 12 \)
Correct: \( 2.5 × 3.6 = 9 \)
Practice Questions: Test Your Number Skills
Practicing with a variety of problems is key to mastering the Number topic. Below are IB-style practice questions categorized by difficulty level, along with their solutions.
Level 1: Easy
- Round 456 to the nearest hundred.
- Find the number of milligrams in 3 grams.
- Compare the GCD of 18 and 24 with the GCD of 20 and 30.
- Simplify \( 2^3 \times 2^2 \).
- Is 1,225 divisible by 5? Explain your reasoning.
Solutions:
-
Solution:
456 rounded to the nearest hundred is 500 (since 56 ≥ 50) -
Solution:
3 grams × 1,000 milligrams/gram = 3,000 milligrams -
Solution:
GCD of 18 and 24: 6
GCD of 20 and 30: 10
Compare: 6 < 10 -
Solution:
\( 2^3 \times 2^2 = 2^{3+2} = 2^5 = 32 \) -
Solution:
Yes, 1,225 is divisible by 5 because its last digit is 5.
Level 2: Medium
- Find the prime factorization of 84.
- Calculate the GCD and LCM of 12 and 18.
- Simplify \( (3^2)^4 \).
- Determine if 1,536 is divisible by both 4 and 9.
- Round 7.856 to two decimal places.
Solutions:
-
Solution:
84 ÷ 2 = 42
42 ÷ 2 = 21
21 ÷ 3 = 7
7 ÷ 7 = 1
Prime factorization: \( 2^2 \times 3 \times 7 \) -
Solution:
Prime factors of 12: \( 2^2 \times 3 \)
Prime factors of 18: \( 2 \times 3^2 \)
GCD = \( 2^1 \times 3^1 = 6 \)
LCM = \( 2^2 \times 3^2 = 36 \) -
Solution:
\( (3^2)^4 = 3^{2 \times 4} = 3^8 = 6,561 \) -
Solution:
Divisible by 4: Last two digits (36) are divisible by 4.
Divisible by 9: Sum of digits = 1 + 5 + 3 + 6 = 15, which is divisible by 3 but not by 9.
Therefore, 1,536 is divisible by 4 but not by 9. -
Solution:
7.856 rounded to two decimal places is 7.86 (since the third decimal digit is 6 ≥ 5)
Level 3: Hard
- Simplify the expression \( \frac{1200}{t} = 4 \times 100 \) and solve for t.
- Find four equivalent estimates for the number 847 by rounding to the nearest hundred.
- Compare the GCD of 56 and 98 with the GCD of 64 and 96.
- Simplify \( 5^4 \times 5^{-2} \).
- Determine if 2,745 is divisible by 3 and 5.
Solutions:
-
Solution:
\( \frac{1200}{t} = 400 \)
Solve for t: \( t = \frac{1200}{400} = 3 \) -
Solution:
Round 847 to the nearest hundred: 800
Equivalent estimates:
800, 850, 790, 810 -
Solution:
Prime factors of 56: \( 2^3 \times 7 \)
Prime factors of 98: \( 2 \times 7^2 \)
GCD of 56 and 98: \( 2^1 \times 7^1 = 14 \)
Prime factors of 64: \( 2^6 \)
Prime factors of 96: \( 2^5 \times 3 \)
GCD of 64 and 96: \( 2^5 = 32 \)
Compare: 14 < 32 -
Solution:
\( 5^4 \times 5^{-2} = 5^{4-2} = 5^2 = 25 \) -
Solution:
Divisible by 3: Sum of digits = 2 + 7 + 4 + 5 = 18, which is divisible by 3.
Divisible by 5: Last digit is 5.
Therefore, 2,745 is divisible by both 3 and 5.
IB Exam Questions: Number
Below are IB-style exam questions on the Number topic, complete with detailed answers to help you understand the application of concepts in exam settings.
Question 1: Prime Factorization and GCD
Problem: Find the greatest common divisor (GCD) of 360 and 504 using prime factorization.
Solution:
Prime factorization of 360:
360 ÷ 2 = 180
180 ÷ 2 = 90
90 ÷ 2 = 45
45 ÷ 3 = 15
15 ÷ 3 = 5
5 ÷ 5 = 1
Prime factors: \( 2^3 \times 3^2 \times 5 \)
Prime factorization of 504:
504 ÷ 2 = 252
252 ÷ 2 = 126
126 ÷ 2 = 63
63 ÷ 3 = 21
21 ÷ 3 = 7
7 ÷ 7 = 1
Prime factors: \( 2^3 \times 3^2 \times 7 \)
GCD = \( 2^3 \times 3^2 = 8 \times 9 = 72 \)
Therefore, the GCD of 360 and 504 is **72**.
Question 2: Simplifying Exponential Expressions
Problem: Simplify the expression \( \frac{4^5 \times 4^{-2}}{4^3} \).
Solution:
Apply exponent rules:
\( 4^5 \times 4^{-2} = 4^{5-2} = 4^3 \)
Now, \( \frac{4^3}{4^3} = 4^{3-3} = 4^0 = 1 \)
Therefore, the simplified expression is **1**.
Question 3: Divisibility and Factorization
Problem: Determine if 2,598 is divisible by both 2 and 9. Provide your reasoning.
Solution:
Divisible by 2: Check if the last digit is even.
Last digit of 2,598 is 8 (even), so it is divisible by 2.
Divisible by 9: Sum the digits and check if the sum is divisible by 9.
Sum = 2 + 5 + 9 + 8 = 24
24 is not divisible by 9 (since 9 × 2 = 18 and 9 × 3 = 27, 24 falls in between).
Therefore, 2,598 is divisible by 2 but not by 9.
Hence, 2,598 is **not** divisible by both 2 and 9.
Question 4: Rational and Irrational Numbers
Problem: Express \( \sqrt{50} \) in its simplest form and state whether it is rational or irrational.
Solution:
Simplify \( \sqrt{50} \):
\( \sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2} \)
Since \( \sqrt{2} \) is an irrational number, \( 5\sqrt{2} \) is also irrational.
Therefore, \( \sqrt{50} = 5\sqrt{2} \) and it is **irrational**.
Question 5: Operations with Integers
Problem: Calculate \( -15 + 22 - (-7) + (-10) \).
Solution:
Apply the rules of addition and subtraction with integers:
\( -15 + 22 = 7 \)
\( 7 - (-7) = 7 + 7 = 14 \)
\( 14 + (-10) = 4 \)
Therefore, \( -15 + 22 - (-7) + (-10) = **4**.
Additional Practice Questions: Number
Further practice with IB-style questions can help solidify your understanding of the Number topic. Below are additional practice questions categorized by difficulty level, along with their solutions.
Level 1: Easy
- Round 589 to the nearest hundred.
- Find the number of grams in 2.5 kilograms.
- Compare the GCD of 14 and 21 with the GCD of 28 and 35.
- Simplify \( 6^2 \times 6^{-1} \).
- Is 1,240 divisible by 10? Explain your reasoning.
Solutions:
-
Solution:
589 rounded to the nearest hundred is 600 (since 89 ≥ 50) -
Solution:
2.5 kilograms × 1,000 grams/kilogram = 2,500 grams -
Solution:
GCD of 14 and 21: 7
GCD of 28 and 35: 7
Compare: 7 = 7 -
Solution:
\( 6^2 \times 6^{-1} = 6^{2-1} = 6^1 = 6 \) -
Solution:
Yes, 1,240 is divisible by 10 because its last digit is 0.
Level 2: Medium
- Find the prime factorization of 180.
- Calculate the GCD and LCM of 24 and 36.
- Simplify \( (4^3) \times (4^{-1}) \).
- Determine if 3,888 is divisible by both 4 and 9.
- Round 12.437 to two decimal places.
Solutions:
-
Solution:
180 ÷ 2 = 90
90 ÷ 2 = 45
45 ÷ 3 = 15
15 ÷ 3 = 5
5 ÷ 5 = 1
Prime factorization: \( 2^2 \times 3^2 \times 5 \) -
Solution:
Prime factors of 24: \( 2^3 \times 3 \)
Prime factors of 36: \( 2^2 \times 3^2 \)
GCD = \( 2^2 \times 3 = 12 \)
LCM = \( 2^3 \times 3^2 = 72 \) -
Solution:
\( (4^3) \times (4^{-1}) = 4^{3-1} = 4^2 = 16 \) -
Solution:
Divisible by 4: Last two digits (88) are divisible by 4.
Divisible by 9: Sum of digits = 3 + 8 + 8 + 8 = 27, which is divisible by 9.
Therefore, 3,888 is divisible by both 4 and 9. -
Solution:
12.437 rounded to two decimal places is 12.44 (since the third decimal digit is 7 ≥ 5)
Level 3: Hard
- Simplify the expression \( \frac{1500}{t} = 5 \times 100 \) and solve for t.
- Find four equivalent estimates for the number 923 by rounding to the nearest hundred.
- Compare the GCD of 45 and 60 with the GCD of 30 and 75.
- Simplify \( 7^4 \times 7^{-2} \).
- Determine if 4,320 is divisible by 8 and 9.
Solutions:
-
Solution:
\( \frac{1500}{t} = 500 \)
Solve for t: \( t = \frac{1500}{500} = 3 \) -
Solution:
Round 923 to the nearest hundred: 900
Equivalent estimates:
900, 950, 920, 930 -
Solution:
Prime factors of 45: \( 3^2 \times 5 \)
Prime factors of 60: \( 2^2 \times 3 \times 5 \)
GCD of 45 and 60: \( 3 \times 5 = 15 \)
Prime factors of 30: \( 2 \times 3 \times 5 \)
Prime factors of 75: \( 3 \times 5^2 \)
GCD of 30 and 75: \( 3 \times 5 = 15 \)
Compare: 15 = 15 -
Solution:
\( 7^4 \times 7^{-2} = 7^{4-2} = 7^2 = 49 \) -
Solution:
Divisible by 8: Last three digits (320) are divisible by 8.
Divisible by 9: Sum of digits = 4 + 3 + 2 + 0 = 9, which is divisible by 9.
Therefore, 4,320 is divisible by both 8 and 9.
IB Exam Questions with Answers: Number
Below are sample IB-style exam questions on the Number topic, complete with detailed answers to help you prepare effectively.
Question 1: Prime Factorization and Simplification
Problem: Simplify the fraction \( \frac{150}{225} \) by finding the GCD using prime factorization.
Solution:
Prime factorization of 150:
150 ÷ 2 = 75
75 ÷ 3 = 25
25 ÷ 5 = 5
5 ÷ 5 = 1
Prime factors: \( 2 \times 3 \times 5^2 \)
Prime factorization of 225:
225 ÷ 3 = 75
75 ÷ 3 = 25
25 ÷ 5 = 5
5 ÷ 5 = 1
Prime factors: \( 3^2 \times 5^2 \)
GCD = \( 3 \times 5^2 = 75 \)
Simplified fraction: \( \frac{150 ÷ 75}{225 ÷ 75} = \frac{2}{3} \)
Therefore, \( \frac{150}{225} \) simplifies to \( \frac{2}{3} \).
Question 2: Exponent Rules
Problem: Simplify the expression \( \frac{5^6}{5^2} \times 5^{-1} \).
Solution:
First, simplify \( \frac{5^6}{5^2} = 5^{6-2} = 5^4 \)
Then, \( 5^4 \times 5^{-1} = 5^{4-1} = 5^3 = 125 \)
Therefore, the simplified expression is **125**.
Question 3: Divisibility and Prime Factors
Problem: Determine if 1,050 is divisible by both 5 and 6. Provide your reasoning.
Solution:
Divisible by 5: Check if the last digit is 0 or 5.
Last digit of 1,050 is 0, so it is divisible by 5.
Divisible by 6: A number is divisible by both 2 and 3.
- Divisible by 2: Last digit is 0 (even), so divisible by 2.
- Divisible by 3: Sum of digits = 1 + 0 + 5 + 0 = 6, which is divisible by 3.
Since 1,050 is divisible by both 2 and 3, it is divisible by 6.
Therefore, 1,050 is divisible by both 5 and 6.
Question 4: Operations with Integers
Problem: Calculate \( -12 + 7 - (-5) \).
Solution:
Apply the rules of addition and subtraction with integers:
\( -12 + 7 = -5 \)
\( -5 - (-5) = -5 + 5 = 0 \)
Therefore, \( -12 + 7 - (-5) = **0**.
Question 5: Rational and Irrational Numbers
Problem: Express \( \frac{16}{64} \) in its simplest form and determine if it is rational or irrational.
Solution:
Simplify the fraction:
\( \frac{16}{64} = \frac{1}{4} \)
Since \( \frac{1}{4} \) is a ratio of two integers, it is a rational number.
Therefore, \( \frac{16}{64} \) simplifies to \( \frac{1}{4} \) and is **rational**.
Combined Exercises: Examples and Solutions
Many mathematical problems require the use of Number concepts in conjunction with other operations. Below are additional IB-style examples that incorporate these concepts alongside logical reasoning and application to real-world scenarios.
Example 1: Budgeting and GCD
Problem: You have $600 to allocate equally to three departments: Marketing, Sales, and Development. Find the greatest common divisor (GCD) of 600 and 3 to determine how much each department should receive.
Solution:
Prime factorization of 600:
600 ÷ 2 = 300
300 ÷ 2 = 150
150 ÷ 2 = 75
75 ÷ 3 = 25
25 ÷ 5 = 5
5 ÷ 5 = 1
Prime factors: \( 2^3 \times 3 \times 5^2 \)
Prime factorization of 3: \( 3 \)
GCD = 3
Amount per department = \( \frac{600}{3} = 200 \)
Therefore, each department should receive $200.
Example 2: Exponential Growth
Problem: A population of bacteria doubles every 3 hours. If the initial population is 150 bacteria, express the population after 9 hours using exponents.
Solution:
Number of doubling periods = 9 ÷ 3 = 3
Population after 9 hours = \( 150 \times 2^3 = 150 \times 8 = 1,200 \)
Therefore, the population after 9 hours is **1,200** bacteria.
Example 3: Divisibility and Scheduling
Problem: A teacher has 1,080 minutes of class time to allocate equally to 6 subjects. Determine if this allocation is possible without any leftover minutes.
Solution:
Divide total minutes by number of subjects:
1,080 ÷ 6 = 180
Since 1,080 is divisible by 6, each subject will receive 180 minutes with no leftovers.
Therefore, the allocation is possible without any leftover minutes.
Example 4: Exponents in Compound Interest
Problem: You invest $2,000 at an annual interest rate of 4%, compounded yearly. Calculate the amount after 5 years using exponents.
Solution:
Compound Interest Formula: \( A = P \times (1 + r)^n \)
P = $2,000
r = 0.04
n = 5
A = $2,000 × (1 + 0.04)^5 ≈ $2,000 × 1.2166529 ≈ $2,433.31
Therefore, the amount after 5 years is approximately **$2,433.31**.
Example 5: Rational Numbers in Measurements
Problem: A recipe requires \( \frac{3}{4} \) cup of sugar. If you want to make 2.5 times the recipe, how many cups of sugar are needed?
Solution:
Multiply the amount of sugar by 2.5:
\( \frac{3}{4} \times 2.5 = \frac{3}{4} \times \frac{5}{2} = \frac{15}{8} = 1 \frac{7}{8} \) cups
Therefore, you need **1 7/8** cups of sugar.
IB Exam Questions: Number with Answers
Below are additional IB-style exam questions on the Number topic, complete with detailed answers to aid your preparation.
Question 1: Simplifying Fractions
Problem: Simplify the fraction \( \frac{480}{600} \) by finding the GCD.
Solution:
Prime factorization of 480:
480 ÷ 2 = 240
240 ÷ 2 = 120
120 ÷ 2 = 60
60 ÷ 2 = 30
30 ÷ 2 = 15
15 ÷ 3 = 5
5 ÷ 5 = 1
Prime factors: \( 2^5 \times 3 \times 5 \)
Prime factorization of 600:
600 ÷ 2 = 300
300 ÷ 2 = 150
150 ÷ 2 = 75
75 ÷ 3 = 25
25 ÷ 5 = 5
5 ÷ 5 = 1
Prime factors: \( 2^3 \times 3 \times 5^2 \)
GCD = \( 2^3 \times 3 \times 5 = 120 \)
Simplified fraction: \( \frac{480 ÷ 120}{600 ÷ 120} = \frac{4}{5} \)
Therefore, \( \frac{480}{600} \) simplifies to \( \frac{4}{5} \).
Question 2: Exponent Manipulation
Problem: Simplify the expression \( \frac{9^4}{9^2} \times 9^{-1} \).
Solution:
Simplify \( \frac{9^4}{9^2} = 9^{4-2} = 9^2 = 81 \)
Then, \( 81 \times 9^{-1} = \frac{81}{9} = 9 \)
Therefore, the simplified expression is **9**.
Question 3: Divisibility and Prime Factors
Problem: Determine if 2,520 is divisible by both 6 and 15.
Solution:
Divisible by 6: A number is divisible by both 2 and 3.
- Divisible by 2: Last digit is 0 (even).
- Divisible by 3: Sum of digits = 2 + 5 + 2 + 0 = 9, which is divisible by 3.
Divisible by 15: A number is divisible by both 3 and 5.
- Divisible by 3: Sum of digits = 9 (as above).
- Divisible by 5: Last digit is 0.
Therefore, 2,520 is divisible by both 6 and 15.
Hence, **2,520** is divisible by both **6 and 15**.
Question 4: Rational and Irrational Numbers
Problem: Express \( \frac{49}{7} \) and \( \sqrt{81} \) in their simplest forms and classify them as rational or irrational.
Solution:
\( \frac{49}{7} = 7 \) (Rational Number)
\( \sqrt{81} = 9 \) (Rational Number)
Therefore, both \( \frac{49}{7} \) and \( \sqrt{81} \) are **rational numbers**.
Question 5: Operations with Integers
Problem: Calculate \( -25 + 40 - 15 + (-10) \).
Solution:
Step 1: \( -25 + 40 = 15 \)
Step 2: \( 15 - 15 = 0 \)
Step 3: \( 0 + (-10) = -10 \)
Therefore, \( -25 + 40 - 15 + (-10) = **-10**.
Summary
Mastering the Number topic is essential for success in IB Mathematics exams. By understanding the different types of numbers, their properties, and the methods to manipulate them, you can tackle a wide range of mathematical problems with confidence. Remember to:
- Understand and apply the different types of numbers: natural, whole, integers, rational, irrational, and complex numbers.
- Master prime factorization to simplify fractions and find GCDs and LCMs.
- Apply exponent rules correctly to simplify expressions.
- Use divisibility rules to determine if numbers are divisible by specific integers.
- Perform operations with integers accurately, paying close attention to the signs.
- Differentiate between rational and irrational numbers to simplify expressions involving them.
- Consistently practice IB-style questions to familiarize yourself with the exam format and question types.
- Avoid common mistakes by carefully following problem-solving steps and verifying your answers.
- Utilize real-world applications to enhance your understanding and retention of Number concepts.
With dedication and consistent practice, you can excel in the Number section of your IB Mathematics exams, laying a strong foundation for more advanced mathematical studies.
Additional Resources
Enhance your learning by exploring the following resources: