1. Bar Modeling Method

The bar modeling method, often called the Singapore Math Model Method, is perhaps the most powerful heuristic in the PSLE toolkit. This visual approach represents quantities and their relationships using rectangular bars, making abstract relationships concrete and visible.

This method is particularly effective for problems involving:

• Part-whole relationships
• Comparison of quantities
• Fractions, ratios, and percentages
• Before-after scenarios
• Age problems

Example Problem: John and Peter have $560 altogether. After John spends 1/4 of his money and Peter spends $85, they have the same amount left. How much did John have at first?

Solution using Bar Modeling:

Step 1: Draw bars to represent John’s and Peter’s initial amounts.
Step 2: Show what happens after John spends 1/4 of his money and Peter spends $85.
Step 3: Since they have the same amount left, the remaining portions must be equal.

If we let John’s initial amount be J and Peter’s initial amount be P:

We know that J + P = $560 (total money)
After spending, John has: J – J/4 = 3J/4 left
After spending, Peter has: P – $85 left

Since they have the same amount left: 3J/4 = P – $85

Also, J + P = $560, so P = $560 – J

Substituting: 3J/4 = ($560 – J) – $85
3J/4 = $475 – J
3J/4 + J = $475
3J/4 + 4J/4 = $475
7J/4 = $475
J = ($475 × 4) ÷ 7
J = $1900 ÷ 7
J = $271.43, which rounds to $271 (since we can’t have partial cents)

Therefore, John had $271 at first.

The bar model provides a visual representation that helps students understand the relationship between quantities, making complex word problems more accessible. At EduFirst, we emphasize this method because of its versatility and how it bridges concrete and abstract thinking.

2. Working Backwards

The working backwards heuristic is particularly useful for problems where the final result is known, but the initial value needs to be found. Instead of starting from the beginning, students start with the end result and reverse the operations to find the original value.

This approach works well for problems involving a series of operations, transformations, or changes over time.

Example Problem: Maya had some stickers. She gave 15 stickers to her friend and then bought 12 more. After giving 1/3 of her remaining stickers to her sister, she had 26 stickers left. How many stickers did Maya have initially?

Solution using Working Backwards:

Step 1: Identify the final state – Maya has 26 stickers left.
Step 2: Work backwards through each operation:

After giving 1/3 of her stickers to her sister, Maya had 26 stickers left.
This means these 26 stickers represent 2/3 of what she had before giving to her sister.
So before giving to her sister, she had: 26 ÷ (2/3) = 26 × (3/2) = 39 stickers.

Before this, Maya bought 12 stickers, so before buying these, she had: 39 – 12 = 27 stickers.

And before that, she gave 15 stickers to her friend, so initially, she had: 27 + 15 = 42 stickers.

Therefore, Maya initially had 42 stickers.

Working backwards is an elegant approach that simplifies problems by reversing the sequence of events. Students at EduFirst learn to identify problem cues that suggest this heuristic, such as phrases like “ended up with,” “finally had,” or “resulting in.”

3. Guess and Check

The guess and check method involves making educated guesses, testing them, and then refining subsequent guesses based on the results. While it might seem unsophisticated, when applied systematically, it’s a powerful problem-solving tool that develops logical reasoning.

This heuristic is particularly useful for:

• Problems with unknown variables that can be tested
• Situations where algebraic approaches might be too complex
• Problems involving integers or discrete values

Example Problem: Jia Ming thinks of a number. When she multiplies it by 4, then subtracts 18, she gets twice the original number. What is her number?

Solution using Guess and Check:

Let’s try some values and see if they meet the condition:

Try: n = 6
4 × 6 = 24
24 – 18 = 6
But this is equal to the original number, not twice the original number (which would be 12).

Try: n = 9
4 × 9 = 36
36 – 18 = 18
This equals twice the original number (2 × 9 = 18). ✓

Therefore, Jia Ming’s number is 9.

Guess and check teaches students persistence and logical thinking. At EduFirst, we emphasize the importance of recording each guess and its outcome so students can identify patterns and refine their subsequent guesses. We also teach them to start with simpler values (like 10 or 5) to make calculations easier.

4. Looking for Patterns

Pattern recognition is a fundamental mathematical skill that helps solve problems involving sequences, series, or repeated structures. By identifying patterns, students can predict subsequent values or find general rules.

This heuristic is especially valuable for:

• Number sequences and series
• Geometric patterns
• Cyclical events
• Problems involving multiple iterations

Example Problem: A pattern of dots is formed as shown below. How many dots will be in Figure 10?

Figure 1: 3 dots
Figure 2: 6 dots
Figure 3: 9 dots
Figure 4: 12 dots

Solution using Pattern Recognition:

Step 1: Analyze the given values to identify the pattern.
Figure 1: 3 dots
Figure 2: 6 dots
Figure 3: 9 dots
Figure 4: 12 dots

Step 2: Recognize that each figure has 3 more dots than the previous one, or that the number of dots equals 3 times the figure number.

The pattern can be expressed as: Number of dots in Figure n = 3n

Step 3: Apply the pattern to find Figure 10.
Number of dots in Figure 10 = 3 × 10 = 30 dots

Pattern recognition develops students’ abilities to make connections and generalizations. At EduFirst, we teach students to organize information in tables and look for differences, ratios, or other relationships between consecutive terms. This systematic approach makes pattern identification more manageable.

5. Making a Systematic List

Systematic listing involves organizing possibilities in a structured way to ensure all outcomes are accounted for. This methodical approach prevents overlooking possibilities and helps identify patterns or solutions.

This heuristic is particularly effective for:

• Combinatorial problems (arrangements, combinations)
• Counting problems
• Problems with multiple possible cases
• Probability scenarios

Example Problem: How many different 3-digit numbers can be formed using only the digits 1, 3, 5, and 7, if each digit can be used only once in a number?

Solution using Systematic Listing:

Step I: Organize our approach to ensure we count all possibilities.
For a 3-digit number, we need to select 3 digits from the 4 available (1, 3, 5, 7).

Step 2: List all possibilities systematically by fixing the first digit, then varying the second and third digits:

When the first digit is 1:
135, 137, 153, 157, 173, 175

When the first digit is 3:
315, 317, 351, 357, 371, 375

When the first digit is 5:
513, 517, 531, 537, 571, 573

When the first digit is 7:
713, 715, 731, 735, 751, 753

Step 3: Count the total number of different 3-digit numbers.
Total count: 6 + 6 + 6 + 6 = 24 different 3-digit numbers

Systematic listing teaches students to be methodical and exhaustive in their approach. At EduFirst, we emphasize creating organized tables or trees to visualize all possibilities clearly. This technique is particularly important for the data analysis and probability sections of the PSLE Mathematics syllabus.

6. Using Before-After Concept

The before-after concept is a powerful heuristic for problems involving changes in quantities over time or after certain operations. This approach compares the initial state with the final state to determine relationships or values.

This method works particularly well for:

• Problems involving increases or decreases in quantities
• Scenarios with multiple transformations
• Rate and time problems
• Problems involving comparisons of states

Example Problem: A shop sold notebooks at $3.50 each. After increasing the price by 20%, sales volume decreased by 15%. If the shop initially sold 240 notebooks per week, find the new weekly revenue.

Solution using Before-After Concept:

Step 1: Identify the before and after states for both price and sales volume.

Before:
Price per notebook = $3.50
Weekly sales volume = 240 notebooks
Weekly revenue = $3.50 × 240 = $840

After:
Price per notebook = $3.50 + 20% of $3.50 = $3.50 + $0.70 = $4.20
Weekly sales volume = 240 – 15% of 240 = 240 – 36 = 204 notebooks
New weekly revenue = $4.20 × 204 = $856.80

Therefore, the new weekly revenue is $856.80.

The before-after concept helps students organize information clearly and track changes systematically. At EduFirst, we teach students to create before-after tables that display all relevant quantities, making it easier to see transformations and calculate final values.

7. Drawing a Diagram

Beyond the bar model method, drawing appropriate diagrams can significantly simplify problems, especially those involving spatial relationships, geometry, or movement. Visualizing the problem often reveals relationships that aren’t immediately apparent from the text.

This heuristic is especially useful for:

• Geometric problems
• Distance, speed, and time questions
• Spatial arrangement problems
• Angle relationships
• Area and perimeter questions

Example Problem: A rectangular garden is 12m long and 8m wide. A path of uniform width surrounds the garden. If the total area of the garden and the path is 168m², find the width of the path.

Solution using Drawing a Diagram:

Step 1: Draw the rectangle representing the garden and the surrounding path.

Step 2: Label the dimensions.
Garden: 12m × 8m
Outer rectangle (garden + path): (12 + 2w)m × (8 + 2w)m, where w is the width of the path

Step 3: Use the area information to solve for the unknown.
Area of garden = 12 × 8 = 96m²
Total area of garden and path = 168m²
Area of outer rectangle = (12 + 2w)(8 + 2w) = 168

Step 4: Expand and solve the equation.
(12 + 2w)(8 + 2w) = 168
96 + 24w + 16w + 4w² = 168
96 + 40w + 4w² = 168
4w² + 40w + 96 – 168 = 0
4w² + 40w – 72 = 0
w² + 10w – 18 = 0

Using the quadratic formula: w = (-10 ± √(100 + 72))/2 = (-10 ± √172)/2

w = (-10 + 13.11)/2 or (-10 – 13.11)/2
w = 1.56 or -11.56

Since width cannot be negative, w = 1.56m, which rounds to 1.6m.

The path is 1.6m wide.

Drawing diagrams helps students visualize problems and identify key relationships. At EduFirst, we emphasize accurate labeling and proportional representations to avoid misinterpretations. Students learn to include all relevant information in their diagrams while keeping them clear and uncluttered.