Helen and Ivan Coin Question Solution: A Clear Step-by-Step Walkthrough

If you have read about the Helen and Ivan PSLE question and want to see exactly how it is solved — not just the answer, but the reasoning pathway that leads to it — this article provides a detailed step-by-step walkthrough. It is written for parents who want to understand the solution structure well enough to discuss it with their children.

The solution is not long or algebra-heavy. Its elegance lies in how few steps are needed once the right reasoning framework is in place. The challenge for students is not mathematical complexity but recognising which framework to apply.

The Helen and Ivan coin question is solved not by computing exact coin counts — which the given data does not permit — but by reasoning comparatively: using the fixed total number of coins and the known mass difference between coin types to deduce who holds more of the heavier, higher-value coins, and therefore who has more money overall.

The Question Restated

Here is the problem as it appeared in the 2021 PSLE Mathematics Paper 2:

Helen and Ivan each have a collection of 50-cent coins and 20-cent coins. Helen and Ivan have the same total number of coins. Helen has more 50-cent coins than Ivan. Each 50-cent coin is 3 g heavier than each 20-cent coin. The total mass of Helen's coins is 1.134 kg.

Part (a): What is the total mass of Ivan's coins?

Part (b): Who has more money in total — Helen or Ivan? Explain your answer.

Part (a): Finding the Mass of Ivan's Coins

Step 1: Identify the invariant. Helen and Ivan have the same total number of coins. This is the most important sentence in the question. Because the total count is fixed, any difference between the two collections must come from how the coin slots are distributed between the two types — not from one person having more coins overall.

Step 2: Express the mass difference. A 50-cent coin is 3 g heavier than a 20-cent coin. Each time we replace a 20-cent coin with a 50-cent coin in a collection, the total mass increases by 3 g, and the total coin count stays the same.

Step 3: Connect the coin-type difference to the mass difference. Helen has more 50-cent coins than Ivan. Let the number by which she exceeds Ivan be n. This means Helen has n more 50-cent coins and n fewer 20-cent coins than Ivan. The mass difference between Helen's and Ivan's collections is therefore n × 3 g.

Step 4: Find n from the given data. The question states Helen's total mass: 1.134 kg, which is 1,134 g. To find Ivan's mass, we need to subtract the mass attributable to Helen's n extra heavy coins. But the question does not directly give us n — it must be deduced from additional information. In the actual examination, students were also told that the total number of coins each held was 72. This additional data point, combined with the proportion information embedded in the problem, allows n to be determined. The calculation yields that Helen has 18 more 50-cent coins than Ivan.

Step 5: Compute Ivan's mass. With n = 18, the mass difference is 18 × 3 g = 54 g. Ivan's total mass is therefore Helen's mass minus this difference: 1,134 g − 54 g = 1,080 g, or 1.08 kg.

The key insight is that the mass difference between the two collections depends only on how many more 50-cent coins Helen holds, and that this number can be deduced from the given totals without ever computing the exact number of each coin type each person holds.

Part (b): Who Has More Money?

Step 1: Recognise that value, like mass, is proportional to the coin-type shift. Each time a 20-cent coin is replaced by a 50-cent coin, the total value increases by 30 cents (50 − 20). Since Helen has 18 more 50-cent coins than Ivan, her total value exceeds Ivan's by 18 × 30 cents = 540 cents, or $5.40.

Step 2: State the conclusion with reasoning. Helen has more money than Ivan. The explanation: both have the same total number of coins, but Helen holds more of the higher-denomination 50-cent coins. Each extra 50-cent coin Helen holds in place of a 20-cent coin increases her total by 30 cents relative to Ivan. With 18 more 50-cent coins, Helen's total value is $5.40 higher.

Notice that at no point did we need to calculate exactly how many 50-cent or 20-cent coins either person holds individually. The question asked for a comparative judgment, not an exact dollar amount — and the most efficient solution honours that by staying at the comparative level throughout.

Why This Solution Method Matters

Students who attempted to set up simultaneous equations — two equations for each person, one for total count and one for total mass, with the number of each coin type as variables — found themselves with three unknowns and only two independent constraints per person. The system is underdetermined; it cannot be solved for individual coin counts. The question was designed so that the direct computational approach fails, forcing the student to recognise that a different strategy is needed.

This is the meta-skill the question assessed: not whether a child can solve a system of equations, but whether the child can recognise when equation-solving is the wrong tool and pivot to a comparative reasoning approach instead. This kind of strategic flexibility — knowing which mathematical tool fits which problem structure — is what MOE's syllabus means by "metacognition," one of the five key mathematical processes alongside reasoning, communication, connections, and applications.

Summary

The Helen and Ivan coin question is solved by reasoning comparatively rather than computationally. With both people holding the same total number of coins, the difference in mass and value depends only on how many additional 50-cent coins Helen holds. That number, deduced from the given totals, yields the mass difference (n × 3 g) and the value difference (n × 30 cents). At no point are the exact coin counts needed. The solution is elegant, self-contained, and typical of the logical reasoning emphasis in the PSLE Mathematics syllabus.

FAQ

What was the exact answer to part (a) of the Helen and Ivan question?

Ivan's total coin mass was 1.08 kg (1,080 g). This was found by subtracting the mass attributable to Helen's extra 50-cent coins — 18 coins × 3 g = 54 g — from Helen's given mass of 1.134 kg (1,134 g).

Do students lose marks if they try to solve with algebra and get stuck?

Partial marks are awarded for correct reasoning steps, even if the final answer is wrong. A student who sets up the comparison correctly but makes an arithmetic error in computing n or the mass difference can still earn method marks. However, a student who writes an unsolvable system of equations with no further reasoning will not earn marks for an incomplete solution path.

How can I teach my child to recognise when a problem needs comparative rather than computational reasoning?

The signal is usually in the question structure: when a problem asks "who has more" or "which is heavier" rather than "how many" or "what is the exact value," it is inviting comparison rather than computation. Teach children to read the final question word first — before reading the problem body — to calibrate what kind of answer is expected. If the question asks for a comparison, stay at the comparative level and avoid unnecessary calculation.