Arithmetic — Multiples, Factors, PEMDAS, Fractions & Decimals - Exam answer templates
Proper answer writing technique is crucial for maximizing your scores in mathematics exams. Examiners look for specific formats, clear working, and logical presentation of solutions. These templates show you exactly how to structure your answers to earn full marks at different question levels. Remember: even if your final answer is incorrect, you can still earn partial marks by showing proper working and method.
Answer templates
Find the LCM of 12 and 18.
Examiner tip
Show alternative method (listing multiples) for bonus appreciation
Model answer
Method 1 (Prime Factorization):
12 = 2² × 3
18 = 2 × 3²
LCM = 2² × 3² = 4 × 9 = 36
Therefore, LCM of 12 and 18 = 36
Answer structure
- Line 1: Write prime factorization of both numbers [1 mark]
- Line 2: Apply LCM formula and calculate [1 mark]
Scoring breakdown
Criteria
Correct prime factorization of both numbers
Criteria
Correct application of LCM method and final answer
Common mark deductions
- Incomplete prime factorization
- Calculation errors
- Missing final statement
Key phrases to include
- Prime factorization
- LCM
- Therefore
Check if 847 is divisible by 7 using divisibility rule.
Examiner tip
Always verify your final step calculation to ensure accuracy
Model answer
Given: Number = 847
To check: Divisibility by 7
Using divisibility rule for 7:
Step 1: Double the last digit: 7 × 2 = 14
Step 2: Subtract from remaining digits: 84 - 14 = 70
Step 3: Check if 70 is divisible by 7: 70 ÷ 7 = 10
Since 70 is divisible by 7, therefore 847 is divisible by 7.
Answer structure
- Line 1: State the divisibility rule [1 mark]
- Line 2-3: Apply the rule step by step [1 mark]
- Line 4: State conclusion clearly [1 mark]
Scoring breakdown
Criteria
Correct statement of divisibility rule for 7
Criteria
Correct application of the rule with calculations
Criteria
Correct conclusion about divisibility
Common mark deductions
- Wrong divisibility rule
- Calculation errors
- No clear conclusion
Key phrases to include
- Divisibility rule
- Step by step
- Therefore
Simplify: 2 + 3 × (4² - 2 × 6) ÷ 2
Examiner tip
Label each step clearly to show you understand the order of operations
Model answer
Given: 2 + 3 × (4² - 2 × 6) ÷ 2
Using PEMDAS:
Step 1: Solve brackets first
4² - 2 × 6 = 16 - 12 = 4
Step 2: Expression becomes: 2 + 3 × 4 ÷ 2
Step 3: Multiplication and Division (left to right):
3 × 4 = 12
12 ÷ 2 = 6
Step 4: Addition:
2 + 6 = 8
Therefore, the answer is 8.
Answer structure
- Line 1: Identify PEMDAS order [1 mark]
- Line 2-4: Solve step by step following order [1 mark]
- Line 5: State final answer [1 mark]
Scoring breakdown
Criteria
Correct identification and application of PEMDAS
Criteria
Accurate calculations at each step
Criteria
Correct final answer
Common mark deductions
- Wrong order of operations
- Arithmetic errors
- Not showing working
Key phrases to include
- PEMDAS
- Step by step
- Therefore
Convert 3.75 to a fraction in simplest form.
Examiner tip
Always check if your fraction can be simplified further
Model answer
Given: 3.75
Step 1: Write as fraction
3.75 = 375/100
Step 2: Simplify by dividing by GCF
GCF of 375 and 100 = 25
375 ÷ 25 = 15
100 ÷ 25 = 4
Therefore, 3.75 = 15/4
Answer structure
- Line 1: Convert decimal to fraction [1 mark]
- Line 2: Simplify to lowest terms [1 mark]
Scoring breakdown
Criteria
Correct conversion to fraction form
Criteria
Correct simplification to lowest terms
Common mark deductions
- Not simplifying completely
- Wrong GCF calculation
Key phrases to include
- Simplest form
- GCF
- Therefore
A class has 24 boys and 18 girls. If they are to be divided into groups with equal number of boys and girls in each group, what is the maximum number of groups possible?
Examiner tip
Always verify your answer by checking if the division works out correctly
Model answer
Given:
Number of boys = 24
Number of girls = 18
To Find: Maximum number of groups with equal boys and girls in each group
Solution:
For equal distribution, we need to find GCF of 24 and 18.
Finding GCF using prime factorization:
24 = 2³ × 3 = 8 × 3
18 = 2 × 3² = 2 × 9
GCF = 2¹ × 3¹ = 2 × 3 = 6
Therefore, maximum number of groups = 6
Verification:
Boys per group = 24 ÷ 6 = 4
Girls per group = 18 ÷ 6 = 3
Total students per group = 4 + 3 = 7
Answer: Maximum 6 groups can be formed, each having 4 boys and 3 girls.
Answer structure
- Line 1: Write Given and To Find [1 mark]
- Line 2-3: Identify that GCF is needed [1 mark]
- Line 4-5: Calculate GCF correctly [2 marks]
- Line 6: Verify and state final answer [1 mark]
Scoring breakdown
Criteria
Correct identification of given data and requirement
Criteria
Understanding that GCF gives maximum groups
Criteria
Correct calculation of GCF of 24 and 18
Criteria
Correct final answer with verification
Common mark deductions
- Not writing Given/To Find
- Confusing GCF with LCM
- No verification
- Wrong prime factorization
Key phrases to include
- Given
- To Find
- GCF
- Prime factorization
- Verification
- Therefore
Find all factors of 36.
Examiner tip
List factors in ascending order to avoid missing any
Model answer
Given: Number = 36
To find factors, we find all numbers that divide 36 exactly:
36 ÷ 1 = 36 ✓
36 ÷ 2 = 18 ✓
36 ÷ 3 = 12 ✓
36 ÷ 4 = 9 ✓
36 ÷ 6 = 6 ✓
Therefore, factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18, 36
Answer structure
- Line 1: Show systematic division method [1 mark]
- Line 2: List all factors correctly [1 mark]
Scoring breakdown
Criteria
Systematic approach to finding factors
Criteria
Complete and correct list of all factors
Common mark deductions
- Missing factors
- Including non-factors
- No systematic method
Key phrases to include
- Factors
- Divide exactly
- Therefore
Is 91 prime or composite? Justify your answer.
Examiner tip
Always show the factorization when proving a number is composite
Model answer
Given: Number = 91
To check if 91 is prime or composite:
Step 1: Find √91 ≈ 9.5
So we check prime factors up to 9
Step 2: Check divisibility by prime numbers 2, 3, 5, 7:
91 ÷ 2 = 45.5 (not exact)
91 ÷ 3 = 30.33... (not exact)
91 ÷ 5 = 18.2 (not exact)
91 ÷ 7 = 13 (exact!)
Step 3: Since 91 = 7 × 13, and both 7 and 13 are primes
Therefore, 91 is composite because it has factors other than 1 and itself.
Answer structure
- Line 1: Apply systematic method to check primality [1 mark]
- Line 2: Show division tests correctly [1 mark]
- Line 3: State conclusion with justification [1 mark]
Scoring breakdown
Criteria
Correct method for testing primality
Criteria
Accurate division calculations
Criteria
Correct conclusion with proper justification
Common mark deductions
- Not checking systematically
- Calculation errors
- Wrong conclusion
Key phrases to include
- Prime
- Composite
- Factors
- Therefore
Add: 2/3 + 5/8
Examiner tip
Always check if the final fraction can be simplified
Model answer
Given: 2/3 + 5/8
Step 1: Find LCM of denominators 3 and 8
LCM of 3 and 8 = 24
Step 2: Convert to equivalent fractions
2/3 = (2 × 8)/(3 × 8) = 16/24
5/8 = (5 × 3)/(8 × 3) = 15/24
Step 3: Add the fractions
16/24 + 15/24 = 31/24
Therefore, 2/3 + 5/8 = 31/24
Answer structure
- Line 1: Find common denominator [1 mark]
- Line 2: Add correctly and simplify if possible [1 mark]
Scoring breakdown
Criteria
Correct LCM and equivalent fractions
Criteria
Correct addition and final answer
Common mark deductions
- Wrong LCM
- Calculation errors
- Not in simplest form
Key phrases to include
- LCM
- Equivalent fractions
- Therefore
A number when divided by 15 leaves remainder 7, and when divided by 25 leaves remainder 17. Find the smallest such positive number.
Examiner tip
Show several values to demonstrate systematic approach, then verify your answer
Model answer
Given:
Number ÷ 15 = quotient + remainder 7
Number ÷ 25 = quotient + remainder 17
To Find: Smallest positive number satisfying both conditions
Solution:
Let the number be N.
Then: N = 15q₁ + 7 = 25q₂ + 17
From first condition: N = 15q₁ + 7
This means N can be: 7, 22, 37, 52, 67, 82, 97, 112, 127, 142, 157, 172, 187, 202, 217, 232, 247, 262, 277, 292, 307, 322, 337, 352, 367, 382, 397, 412, 427, 442, 457, 472, 487, 502, 517, 532...
Checking which of these satisfies second condition:
517 ÷ 25 = 20 remainder 17 ✓
Verification:
517 ÷ 15 = 34 remainder 7 ✓
517 ÷ 25 = 20 remainder 17 ✓
Therefore, the smallest such number is 517.
Answer structure
- Line 1: Set up the problem using remainder theorem [1 mark]
- Line 2-3: Generate possible values from first condition [2 marks]
- Line 4: Test values against second condition [1 mark]
- Line 5: Verify and state final answer [1 mark]
Scoring breakdown
Criteria
Correct understanding and setup of remainder problem
Criteria
Systematic generation of possible values
Criteria
Correct testing against second condition
Criteria
Verification and correct final answer
Common mark deductions
- No systematic approach
- Calculation errors
- No verification
- Missing working
Key phrases to include
- Remainder theorem
- Systematic checking
- Verification
- Therefore
Express 0.375 as a fraction.
Examiner tip
Remember to simplify by dividing by GCF
Model answer
0.375 = 375/1000 = 3/8
Answer structure
- Line 1: Direct conversion and simplification [1 mark]
Scoring breakdown
Criteria
Correct conversion to simplest fraction form
Common mark deductions
- Not simplifying
- Wrong conversion
Key phrases to include
- Simplest form
Find the value of: 3² + 2³ - 4 × 5 ÷ 2
Examiner tip
Show each operation clearly to demonstrate understanding of order
Model answer
Given: 3² + 2³ - 4 × 5 ÷ 2
Using PEMDAS:
Step 1: Exponents first
3² = 9, 2³ = 8
Expression becomes: 9 + 8 - 4 × 5 ÷ 2
Step 2: Multiplication and Division (left to right)
4 × 5 = 20, 20 ÷ 2 = 10
Expression becomes: 9 + 8 - 10
Step 3: Addition and Subtraction (left to right)
9 + 8 = 17, 17 - 10 = 7
Therefore, the answer is 7.
Answer structure
- Line 1: Apply PEMDAS systematically [1 mark]
- Line 2: Calculate correctly and state answer [1 mark]
Scoring breakdown
Criteria
Correct application of order of operations
Criteria
Accurate calculations and final answer
Common mark deductions
- Wrong order
- Arithmetic errors
Key phrases to include
- PEMDAS
- Step by step
- Therefore
Two numbers are in the ratio 3:4. If their LCM is 60, find the numbers.
Examiner tip
Always verify your answer by checking both the ratio and LCM conditions
Model answer
Given:
Two numbers are in ratio 3:4
LCM of the numbers = 60
To Find: The two numbers
Solution:
Let the numbers be 3x and 4x where x is a positive integer.
For LCM calculation:
3x = 3 × x
4x = 4 × x = 2² × x
LCM of 3x and 4x = LCM of (3 × x) and (2² × x)
= 3 × 2² × x = 12x
Given that LCM = 60:
12x = 60
x = 60 ÷ 12 = 5
Therefore:
First number = 3x = 3 × 5 = 15
Second number = 4x = 4 × 5 = 20
Verification:
Ratio check: 15:20 = 3:4 ✓
LCM of 15 and 20 = 60 ✓
Answer: The two numbers are 15 and 20.
Answer structure
- Line 1: Set up variables based on ratio [1 mark]
- Line 2-3: Express LCM in terms of variable [2 marks]
- Line 4: Solve for the variable [1 mark]
- Line 5: Find numbers and verify [1 mark]
Scoring breakdown
Criteria
Correct setup using ratio relationship
Criteria
Correct LCM calculation in terms of variable
Criteria
Correct solution for the variable
Criteria
Correct final answer with verification
Common mark deductions
- Wrong variable setup
- LCM calculation errors
- No verification
Key phrases to include
- Let
- Ratio
- LCM
- Verification
- Therefore
Multiply: 2¼ × 1⅗
Examiner tip
Always convert mixed numbers to improper fractions before multiplying
Model answer
Given: 2¼ × 1⅗
Step 1: Convert mixed numbers to improper fractions
2¼ = (2×4+1)/4 = 9/4
1⅗ = (1×5+3)/5 = 8/5
Step 2: Multiply the fractions
9/4 × 8/5 = (9×8)/(4×5) = 72/20
Step 3: Simplify
72/20 = 18/5 = 3⅗
Therefore, 2¼ × 1⅗ = 3⅗
Answer structure
- Line 1: Convert to improper fractions [1 mark]
- Line 2: Multiply and simplify [1 mark]
Scoring breakdown
Criteria
Correct conversion to improper fractions
Criteria
Correct multiplication and final answer
Common mark deductions
- Wrong conversion
- Not simplifying
- Arithmetic errors
Key phrases to include
- Improper fractions
- Simplify
- Therefore
State the multiplicative inverse property and give an example.
Examiner tip
Always mention that the number must be non-zero for multiplicative inverse to exist
Model answer
Multiplicative Inverse Property:
The product of any non-zero number and its reciprocal equals 1.
Mathematically: a × (1/a) = 1, where a ≠ 0
Example: 5 × (1/5) = 1
Or: ¾ × 4/3 = 1
Therefore, every non-zero number has a multiplicative inverse which is its reciprocal.
Answer structure
- Line 1: State the property clearly [1 mark]
- Line 2: Give a correct example [1 mark]
Scoring breakdown
Criteria
Correct statement of multiplicative inverse property
Criteria
Appropriate example demonstrating the property
Common mark deductions
- Incomplete definition
- Wrong example
- Not mentioning non-zero condition
Key phrases to include
- Multiplicative inverse
- Reciprocal
- Non-zero
- Example
Find three consecutive even numbers whose sum is 78.
Examiner tip
Always define your variable clearly and verify your final answer
Model answer
Given: Three consecutive even numbers with sum = 78
To Find: The three numbers
Solution:
Let the three consecutive even numbers be x, x+2, and x+4
According to given condition:
x + (x+2) + (x+4) = 78
3x + 6 = 78
3x = 78 - 6 = 72
x = 72 ÷ 3 = 24
Therefore:
First number = x = 24
Second number = x+2 = 26
Third number = x+4 = 28
Verification: 24 + 26 + 28 = 78 ✓
Answer: The three consecutive even numbers are 24, 26, and 28.
Answer structure
- Line 1: Set up algebraic equation [1 mark]
- Line 2: Solve the equation correctly [1 mark]
- Line 3: Find all three numbers and verify [1 mark]
Scoring breakdown
Criteria
Correct algebraic setup with consecutive even numbers
Criteria
Accurate solution of the equation
Criteria
Correct final answer with verification
Common mark deductions
- Wrong algebraic setup
- Calculation errors
- No verification
Key phrases to include
- Let
- Consecutive even
- Verification
- Therefore
Mark wise strategy
Write direct answer with minimal working for basic recall questions
Do
- Write answer clearly
- Use proper mathematical notation
- Be precise
Don't
- Over-explain simple concepts
- Show unnecessary working
- Waste time on elaboration
Show clear method and calculation steps
Do
- Show key steps
- State method used
- Box final answer
- Check calculation
Don't
- Skip important steps
- Make careless errors
- Forget to simplify
Demonstrate understanding through systematic approach
Do
- Show complete working
- Label steps clearly
- State conclusions
- Verify when possible
Don't
- Rush through steps
- Miss intermediate calculations
- Forget proper notation
Provide comprehensive solution with Given/To Find/Solution format
Do
- Write Given and To Find
- Show all working
- Verify answer
- Use proper mathematical language
- Conclude clearly
Don't
- Skip the setup
- Miss verification
- Provide unclear explanations
- Make computational errors
General answer writing tips
- Always write 'Given:', 'To Find:', and 'Solution:' for word problems to earn method marks
- Show all working steps clearly - never skip intermediate calculations
- Box or underline your final answer to make it stand out for the examiner
- Use proper mathematical notation and symbols consistently throughout
- For factor/multiple problems, list systematically to avoid missing any values
- When using PEMDAS, show each step of the order of operations
- For fraction problems, always express final answers in simplest form
- Include units in your final answer when dealing with word problems
Diagrams
Arithmetic Concepts Overview
Complete overview of all arithmetic topics covered in this chapter, showing the relationship between different concepts
Problem Solving Strategy for Word Problems
Step-by-step flowchart showing how to approach different types of arithmetic word problems systematically
PEMDAS Order of Operations
Visual representation of the order of operations showing the step-by-step process for evaluating mathematical expressions
Distribution of Question Types in UPCAT Math
Shows the typical distribution of arithmetic question types in UPCAT mathematics section, helping students prioritize their study time