Lesson 8: Bar Models — Before-After

Before-after problems describe change: money spent, water poured out, pupils joining or leaving. "Ahmad had some money. He spent S$25 and had S$40 left. How much at first?" The before-after model draws two bars — before and after — aligned to show what changed. This model is the closest to algebra: it handles unknowns at the start, transfers between people, and the constant-total and constant-difference patterns that unlock the PSLE's hardest problems.

The Before-After Idea

The before-after model draws the "before" bar and the "after" bar, aligned so that unchanged portions line up and changed portions stand out. Problem: "Ahmad had some money. He spent S$25 and had S$40 left." Before-bar: ? (whole). After-bar: S$40, with a marked-off S$25 section showing what left. The alignment shows: before = 25 + 40 = S$65. The picture is a timeline of the quantity.

The model's power is making time visible. Words describe change sequentially; bars show before and after simultaneously, so the relationship between them can be inspected. "What was removed?" "What stayed the same?" "What is the new total?" — the bars answer by display.

Before-after combines with part-whole (Lesson 6) and comparison (Lesson 7): the before-bar is often a part-whole structure, and transfers between people create comparisons. The three models form a complete toolkit — most PSLE problems yield to some combination.

Worked Example 1 — Simple Change

Problem: "A tank contained some water. 120 litres were pumped out, leaving 280 litres. How much was there at first?" Before-bar: ?; after-bar: 280, with 120 marked as removed. Before = 120 + 280 = 400 litres. The "at first" unknown sits at the start of the timeline; the model reads it directly.

Reverse: "A tank had 400 litres. Some was pumped out, leaving 280 litres. How much was pumped out?" Before-bar: 400; after-bar: 280 with ? marked as removed. Pumped out = 400 − 280 = 120 litres. Same picture, different unknown — the model adapts to whichever quantity is missing.

The key discipline: label which bar is before and which is after, and mark the change explicitly. Students who skip this labelling mix up the bars and add when they should subtract. The model only works if the timeline is clear.

Worked Example 2 — Transfer Between People

Problem: "Ahmad had S$120 and Bala had S$60. Ahmad gave some money to Bala. Now they have the same amount. How much did Ahmad give?" Before-bars: Ahmad 120, Bala 60. After-bars: equal, equal. The total is constant (120 + 60 = 180), so each has 90 after. Ahmad gave 120 − 90 = S$30. The before-after model with the constant-total insight solves it.

The constant total is the key pattern: when money (or items) moves between people, the total does not change. Draw before-bars (unequal), draw after-bars (equal, same combined length), and the transfer is the difference. This pattern appears in countless problems: card trades, marble exchanges, water poured between tanks.

Generalize: if Ahmad gives x to Bala, Ahmad has 120 − x, Bala has 60 + x; equality gives 120 − x = 60 + x, so 2x = 60, x = 30. The bar model is this algebra, drawn. Pupils who master the visual will find the algebra natural later.

Worked Example 3 — Constant Difference

Problem: "Ahmad is 12 years older than Bala. In 5 years' time, the sum of their ages will be 50. How old is Bala now?" Ages before (now) and after (+5 years): both increase by 5, so the difference (12) is constant. After-bars: total 50, difference 12. Equalize: (50 − 12) ÷ 2 = 19 — that is Bala's age in 5 years. Now: 19 − 5 = 14. Answer: Bala is 14.

The constant-difference pattern: when both quantities change by the same amount (both age 5 years, both gain the same bonus), their difference never changes. The before-after model shows this — the extra 12-year section persists in both diagrams — while the total shifts. Recognizing "difference is constant" versus "total is constant" is the mark of an advanced modeller.

Check: now, Bala 14, Ahmad 26 (difference 12 ✓). In 5 years: 19 and 31; sum 50 ✓. The model's predictions verify cleanly.

Fractions in Before-After — The Classic PSLE Type

Problem: "Ahmad had some stamps. He gave 1/3 of them to Bala and had 40 left. How many did he have at first?" Before-bar: whole (?), divided into 3 units. After: 1 unit given away (marked), 2 units left = 40. So 1 unit = 20, whole = 60 stamps. The before-after timeline plus fraction units cracks the classic "fraction of remainder" family.

Harder: "Ahmad gave 1/3 of his stamps to Bala, then gave 1/4 of the remainder to Chen, and had 30 left. How many at first?" Work backwards with bars: after giving 1/4 of remainder, 3/4 of remainder = 30, so remainder = 40. That remainder was 2/3 of the original (after giving 1/3 away), so original = 40 × 3/2 = 60. Each "after" becomes the next "before" — the timeline chains.

These layered fraction problems are the PSLE's signature challenge. The method is always: draw the timeline, identify what each fraction refers to (the original? the remainder?), use the known final quantity to find units, work backwards. Patience with bars beats brilliance without them.

Conclusion — Why This Matters

The before-after model is the crown of primary problem solving: it handles change, unknowns at any point in time, transfers, and layered fractions. Combined with part-whole and comparison, it completes a toolkit that solves virtually every word problem through the PSLE.

The deeper lesson is temporal thinking: quantities have histories, and modelling the history reveals the present. This is the same thinking behind algebra's equations (which relate before and after symbolically) and calculus's rates of change. The bar model is early training in the mathematics of change.

From here, the course turns to geometry, measurement, and then algebra — where the bar model's visual wisdom becomes symbolic power.

Before-After in Science — Conservation Laws

The before-after model's deepest application is in science, where it becomes the principle of conservation. Water poured between beakers: the total volume is constant (before = after). A chemical reaction: atoms are conserved — the before-bar (reactants) and after-bar (products) contain the same atoms, rearranged. The bar model is conservation of mass, drawn.

Energy follows the same pattern: a roller coaster's potential energy (before, at the top) becomes kinetic energy (after, at the bottom) — the total energy bar keeps its length while its sections relabel. Singapore's science curriculum teaches energy conservation from primary level; the before-after bar is its mathematical shadow.

Even money problems are conservation problems: transfers between people conserve the total (Lesson 8's constant total). The universe, it seems, runs on before-after models — quantities change form but not amount. The pupil who masters the model has a head start in physics, chemistry, and economics.

The philosophical payoff: mathematics' bar models and science's conservation laws are one idea. The best problem-solvers see these connections — and the Singapore curriculum, with its unified CPA philosophy across subjects, is designed to reveal them.

Debate — Does Before-After Handle Every Change Problem?

Enthusiasts claim the before-after bar solves all change problems: money spent, water poured, ages advancing. The timeline picture — before-bar, after-bar, marked change — seems universal. Combined with constant-total and constant-difference patterns, it covers the PSLE syllabus completely.

Cautious voices note the limits: multi-agent transfers (three people exchanging simultaneously), continuous change (water flowing in and out at once), and percentage changes on changing bases can tangle the bars beyond readability. At some complexity, algebra's compactness beats the picture's clarity. The model is a bridge, not a destination.

The wise position: use before-after bars as far as they illuminate, switch to algebra when they clutter. Knowing when to abandon a tool is as important as knowing how to use it. The bar model earns its place by handling 95% of change problems beautifully — and by teaching the temporal thinking that algebra later formalizes.

Worked Example — Bars, Working Backwards, or an Equation?

Problem: "After giving away 24 marbles, Ahmad had 3 times as many marbles as Bala. Bala has 18 marbles. How many marbles did Ahmad have at first?" Three routes reach the answer. Before-after bars: Ahmad's "after" bar is 3 units against Bala's 1 unit = 18, so Ahmad has 54 after giving; the "before" bar is longer by the 24 given away: 54 + 24 = 78. Working backwards: undo the story from the end — 3 × 18 = 54 now; before giving, 54 + 24 = 78. Algebra: let Ahmad start with x; then x − 24 = 3 × 18, so x = 78.

The three routes share one insight: each forward step must be undone in reverse order. Choose the route that keeps the story clearest — for chained changes, the bars usually win.

Common Mistakes

• Taking a fraction of the wrong whole. "Gave 1/4 of the remainder" is not "gave 1/4 of the original". Mark on the bars which quantity each fraction refers to before computing.
• Assuming the total changes in a transfer. When Ahmad gives Bala marbles, the combined total is unchanged — only the split changes.
• Assuming the difference changes when both grow. When both ages increase by 5 years, the age difference is unchanged. Constants are clues: name what stays the same before touching what changes.

Exam Experience

Question 1 [2 marks]. After spending S$18 on a book, Mei had S$47 left. How much money did she have at first?

Question 2 [3 marks]. Ahmad gave 2/5 of his savings to his brother and had S$96 left. How much money did he give to his brother?

Question 3 [6 marks]. Boxes A and B contain 230 balls altogether. 25 balls are moved from Box A to Box B. After the move, the two boxes contain the same number of balls. (a) How many balls are in each box after the move? (b) How many balls were in Box A at first? (c) How many balls were in Box B at first?

Marking scheme. Q1: M1 for 47 + 18 (working backwards); A1 for S$65. Q2: M1 for recognising S$96 is 3/5 of the savings (whole = 96 ÷ 3 × 5 = 160); A1 for whole = S$160; A1 for amount given = S$64. Q3: (a) M1 for using the constant total, 230 ÷ 2; B1 for 115 balls each. (b) M1 for reversing the transfer, 115 + 25; A1 for 140. (c) M1 for 115 − 25; A1 for 90.