In the PSLE Mathematics syllabus, word problems in Paper 2 account for more than 55% of the total score. To solve these challenging problems, students must go beyond standard operations and apply problem-solving heuristics. This guide breaks down the top 5 heuristics with step-by-step worked examples.
Standard operations work when a problem directly states what values to add or multiply. However, PSLE Paper 2 questions often present scenarios where variables are hidden or relationships change midway. By categorizing questions into recognizable heuristic patterns, students can instantly retrieve the correct structured strategy.
The Assumption Method is a highly efficient strategy for solving questions that involve two groups of variables with distinct values (e.g., number of chickens/cows and legs, or types of coins and total value). While many students default to the time-consuming "Guess and Check" table, Singapore primary schools teach the Assumption Method because it reaches the answer in just four logical calculations.
A bag contains 40 coins consisting of 20-cent coins and 50-cent coins. The total value of the coins in the bag is $14.60. How many 50-cent coins are there?
Step 1: Assume all coins belong to one category. Let's assume all 40 coins are 20-cent coins.
40 × $0.20 = $8.00 (Assumed total value)
Step 2: Find the difference between the actual total value and our assumed value.
$14.60 - $8.00 = $6.60 (Total value gap)
Step 3: Find the difference in value between the two individual items.
$0.50 - $0.20 = $0.30 (Individual value gap)
Step 4: Divide the total gap by the individual gap to find the number of 50-cent coins.
$6.60 ÷ $0.30 = 22 (Number of 50-cent coins)
Step 5: Check your work. Deduct from total coins to find 20-cent coins ($40 - 22 = 18$). Then calculate: $(22 \times \$0.50) + (18 \times \$0.20) = \$11.00 + \$3.60 = \$14.60$. The solution is correct.
💡 Key Rule: To find the quantity of item A, assume all items are item B. In the example above, assuming all coins were 20-cent coins directly yielded the number of 50-cent coins.
Before-After heuristics analyze what happens when elements change. The key is finding the constant factor—the value that remains unchanged throughout the transition.
Only one category is added to or subtracted from. The other category remains constant. We make the ratio units of the constant category equal before and after the change.
The ratio of the number of boys to girls in a hall was 3 : 5. After 21 girls left, the ratio of boys to girls became 2 : 1. How many boys were in the hall?
Since girls left, the number of boys remained constant. We adjust the boys' ratio units to be equal in both ratios (lowest common multiple of 3 and 2 is 6):
Before ratio (Boys : Girls) = 3 : 5 → Multiply by 2 → 6 : 10
After ratio (Boys : Girls) = 2 : 1 → Multiply by 3 → 6 : 3
Now that the boys' units are equal (6 units), we can compare the girls' units change:
Change in girls' units = 10u - 3u = 7u
7u = 21 girls
1u = 21 ÷ 7 = 3 girls
Boys (6u) = 6 × 3 = 18 boys
When items are shifted internally (e.g., from Person A to Person B), the individual amounts change, but the total sum remains constant. Equate the sum of ratio units before and after the change.
Ahmad and Bala had stamps in the ratio 3 : 5. After Bala gave 12 stamps to Ahmad, they both had the same number of stamps (ratio 1 : 1). How many stamps did they have altogether?
Because it is an internal transfer, the total number of stamps remains constant. Find the total sum of units:
Before (Ahmad : Bala) = 3 : 5 (Total = 8 units)
After (Ahmad : Bala) = 1 : 1 (Total = 2 units)
To equate the total units (LCM of 8 and 2 is 8), multiply the After ratio by 4:
After (Ahmad : Bala) = 4 : 4 (Total = 8 units)
Now track the change for Bala:
Bala went from 5u to 4u → Decrease of 1u
1u = 12 stamps
Total stamps (8u) = 8 × 12 = 96 stamps
When both categories increase or decrease by the same amount (e.g., in age word problems where years pass equally for everyone, or when both people spend the same amount of money), the difference remains constant. Equate the difference in ratio units before and after.
Mrs Tan is 38 years old and her daughter is 10 years old. In how many years' time will Mrs Tan be 3 times as old as her daughter?
The difference in age between mother and daughter is constant:
Age difference = 38 - 10 = 28 years
In the future, the ratio of mother to daughter will be 3 : 1. Compare units:
Future ratio (Mother : Daughter) = 3 : 1 (Difference = 2 units)
2u = 28 years
1u = 14 years (Daughter's future age)
Calculate the years that must pass:
Future age (14) - Current age (10) = 4 years
Repeated Identity occurs when a common entity connects two different ratio relationships. By identifying the repeated element and equating its units across both ratios, you can combine them into a single three-part ratio.
The ratio of Peter's money to Quincy's money is 2 : 3. The ratio of Quincy's money to Ryan's money is 4 : 5. If Ryan has $35 more than Peter, how much money do they have altogether?
Quincy is the repeated identity. We must equate Quincy's units in both ratios (LCM of 3 and 4 is 12):
Ratio 1 (Peter : Quincy) = 2 : 3 → Multiply by 4 → 8 : 12
Ratio 2 (Quincy : Ryan) = 4 : 5 → Multiply by 3 → 12 : 15
Combine into a single ratio:
Combined (Peter : Quincy : Ryan) = 8 : 12 : 15
Solve using Ryan and Peter's difference:
Ryan (15u) - Peter (8u) = 7u
7u = $35
1u = $35 ÷ 7 = $5
Total units = 8u + 12u + 15u = 35u
Total money = 35 × $5 = $175
This heuristic is deployed when the end state of a sequence of transactions is known, and you need to determine the starting amount. Work from the end back to the beginning, reversing every operation (e.g., addition becomes subtraction, division becomes multiplication).
Janice had some sweets. She gave half of them plus 3 sweets to her brother. She then gave half of the remaining sweets plus 2 sweets to her sister. She was left with 5 sweets. How many sweets did she have at first?
Step 1: Start with the final amount.
Leftover sweets = 5
Step 2: Reverse the sister's transaction. She gave "half of remaining plus 2 sweets" to her sister. Reverse this: add the 2 sweets back, then multiply by 2.
Add 2: 5 + 2 = 7 sweets
Multiply by 2: 7 × 2 = 14 sweets (Before sister's share)
Step 3: Reverse the brother's transaction. She gave "half of initial sweets plus 3 sweets" to her brother. Reverse this: add 3 sweets back, then multiply by 2.
Add 3: 14 + 3 = 17 sweets
Multiply by 2: 17 × 2 = 34 sweets
Janice had 34 sweets at first.
Grouping is used when objects are grouped into sets with fixed quantities and values. Instead of dealing with individual objects, calculate the value of one full "set," find the number of sets, and multiply back to find individual quantities.
Apples are sold in bags of 5 for $4. Pears are sold in bags of 3 for $2. Mr Lee bought an equal number of apples and pears. He spent $6 more on apples than on pears. How many pears did he buy?
Step 1: Find the common quantity for one standard set. Since he bought an equal number of apples and pears, find the LCM of 5 and 3, which is 15. One set contains 15 apples and 15 pears.
Step 2: Calculate the cost of apples in one set.
Cost of 15 apples = (15 ÷ 5) × $4 = $12
Step 3: Calculate the cost of pears in one set.
Cost of 15 pears = (15 ÷ 3) × $2 = $10
Step 4: Find the cost difference in one set.
Difference in one set = $12 - $10 = $2
Step 5: Find the total number of sets bought to accumulate the total difference of $6.
Total sets = $6 ÷ $2 = 3 sets
Pears bought = 3 sets × 15 pears per set = 45 pears
💡 Summary Check: Regular revision of these 5 heuristics helps students recognize them in exam questions instantly. In the PSLE Math exam, identifying the heuristic in the first 30 seconds is key to finishing the paper on time.
Q: What are Maths Heuristics in the PSLE syllabus?
Heuristics are problem-solving strategies used when standard arithmetic operations (addition, subtraction, multiplication, division) are not immediately obvious. Common heuristics include the Assumption Method, Before-After concepts, Repeated Identity, Working Backwards, and Grouping.
Q: Why is the Assumption Method preferred over Guess and Check?
The Assumption Method solves two-variable problem sums (e.g., animals/legs or coins/value) in just 4-5 mathematical steps. Guess and Check relies on trial-and-error, which takes significantly longer and carries a high risk of careless calculation errors under exam time pressure.
Q: How does the Before-After concept work in ratios?
The Before-After concept identifies what remains unchanged after a transition. If one side changes, the other remains a 'Constant Individual'. If an internal transfer occurs, the 'Total Sum' remains constant. If both change by equal amounts, the 'Difference' remains constant. Equating the unchanged ratio units solves the problem.
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