Singapore School Mathematics Operating Manual · Chapter 11
Students are often taught to read a Mathematics question from the beginning and work forward. That is necessary, but it is not always sufficient.
Longer or unfamiliar questions may provide many facts without announcing the next operation. In those situations, a strong solver often looks at the target and asks a reverse question: what would I need to know immediately before I could answer this?
That creates a subgoal. Then the learner asks what would be needed to reach that subgoal, and continues until the chain meets information already available.
This chapter develops backward reasoning as a planning tool and forward reasoning as the final justification. The distinction matters. Working backwards can reveal a route. It should not become circular reasoning in which the desired answer is assumed to be true.
Read the target · Decompose the target · Backward proof planning · Word problems · Practice · Worked answers
1. The target tells you what kind of object must be built
“Find the area” asks for an area. “Find the equation of the line” asks for a relationship between coordinates. “Find the probability” asks for a value between zero and one under a specified random model. “Prove the triangles are similar” asks for a valid similarity condition.
Before searching for methods, identify the final mathematical object.
If the target is area, ask what quantities determine that area. If it is a tangent equation, ask what determines a line. If it is a percentage increase, ask for the old value, new value and correct reference base.
This is target decomposition: replace a large question with smaller requirements.
2. Area problems demonstrate backward planning clearly
Suppose a triangle’s area is required.
The most familiar formula is A = 1/2 bh. Backward reasoning asks: do I know a base and its perpendicular height?
If not, perhaps I know two sides and the included angle, allowing A = 1/2 ab sin C. If not, perhaps the geometry can produce a missing side or height.
The target “area” therefore opens several possible dependency routes. The learner chooses the route that matches the information actually available.
This is more flexible than scanning for a formula whose letters resemble the labels in the diagram.
3. A target tree can expose missing prerequisites
Suppose the final target is the cost of tiling a floor.
Cost may require number of tiles multiplied by price per tile. Number of tiles may require floor area divided by usable tile area, followed by a whole-number decision. Floor area may require dimensions found from earlier geometry.
The dependency tree becomes:
cost ← number of tiles + unit price ← floor area + tile area ← dimensions.
The learner can now see which unknowns matter and in what order.
Not every printed number needs to enter the tree. Irrelevant facts remain outside unless another branch needs them.
4. Reverse planning is especially useful when the first move is unclear
Consider an invented question: a rectangle has perimeter 54 cm and length 3 cm more than twice its width. Find its area.
Starting from the target, area needs length and width.
The relationship gives length in terms of width: L = 2w + 3.
The perimeter gives 2L + 2w = 54, so L + w = 27.
Substitute the relationship: 2w + 3 + w = 27, giving 3w = 24 and w = 8. Then L = 19.
Area = 19 × 8 = 152 cm².
The reverse plan was “area needs dimensions; dimensions can be determined from the two relationships.” The final working proceeds forward from the givens.
5. Reverse planning should stop when it reaches known information
A backward chain can continue indefinitely if the learner keeps asking “why?” without recognising when a sufficient starting point has been reached.
If a question directly gives the radius of a circle, there is no need to derive it from the circumference unless the target requires some independent relationship.
A useful stopping rule is: once every required subgoal is supplied by a given fact or a valid result already established, switch to forward execution.
Backward planning is for locating dependencies, not for making every solution longer.
6. Algebraic targets often tell you which form is useful
If the target asks for the minimum of x² − 8x + 20, completing the square is attractive because minimum structure becomes visible.
Write x² − 8x + 20 = (x − 4)² + 4. Since the square is non-negative, the minimum is 4.
If the target instead asks for the roots of x² − 8x + 15 = 0, factorisation into (x − 3)(x − 5) is immediately useful.
The underlying topic is quadratic algebra, but the target helps choose the representation.
7. “What would make this easy?” is a strong backward question
Suppose a difficult expression must be simplified. Ask what intermediate form would make the target obvious.
For example, to show that 1/(x − 1) − 1/(x + 1) = 2/(x² − 1), the target denominator x² − 1 suggests the factorisation (x − 1)(x + 1).
That reveals the common denominator required by the left-hand side.
The final derivation should still begin with the left side and simplify to the right. The target was used to plan, not to assume.
8. Backward proof planning asks what would be sufficient
Suppose a geometry task asks to prove two triangles congruent.
Working backwards, the learner asks which congruence condition could apply: SSS, SAS, ASA or another allowed criterion in the relevant syllabus.
If SAS looks promising, ask: which two side equalities and included angle equality would be needed?
Then inspect the givens and previously proved facts to see whether those conditions can be established.
This is not assuming congruence. It is identifying a sufficient set of subgoals.
9. A desired statement can guide factorisation without being used as a premise
To prove n² − n is even for every integer n, the target “even” suggests looking for a factor 2.
Factor n² − n = n(n − 1). The two factors are consecutive integers, so one is even. Therefore the product is even.
The target property shaped the representation. The proof itself still proceeds from the expression to the conclusion.
This is one of the most productive uses of backward reasoning in proof.
10. Circular reasoning happens when the target is smuggled into the givens
Suppose the goal is to show x = 5 from an equation. Writing “Let x = 5” and then verifying the equation only proves that 5 works, not that the original equation forces 5.
Likewise, to prove two lines are parallel, beginning with the assumption that they are parallel and deriving an angle equality does not establish the original claim unless the reasoning is explicitly structured as a valid converse argument.
Backward planning is private route discovery. Forward proof must rest on accepted premises and valid equivalences.
11. Reversible steps can allow a backward derivation to become a proof
Sometimes every step in a backward chain is logically reversible.
Suppose we want to solve 2x + 3 = 11. Working backwards from x = 4 might produce 2x = 8 and then 2x + 3 = 11. Because adding 3 and dividing by 2 are reversible, the chain can be rewritten as a valid equivalence argument.
In more complicated equations, not every step is reversible. Squaring or cancelling variable expressions can change the solution set.
The companion chapter Reversible Steps, Lost Information and Extraneous Solutions is essential when converting reverse exploration into rigorous algebra.
12. Word problems become easier when you work backward from the requested quantity
Suppose an invented school fair asks for total revenue from adult and child tickets.
Revenue needs number of adult tickets, adult price, number of child tickets and child price.
If the prices are given but the counts are not, the dependency problem becomes finding the counts.
If total attendance and an adult:child ratio are given, those determine the counts.
The reverse plan is therefore:
revenue ← ticket counts ← total + ratio.
This keeps the learner from combining all printed numbers indiscriminately.
13. Percentage questions require the reference quantity to be identified backward
Suppose the target is percentage increase.
The formula is change divided by original value, multiplied by 100%.
Backward reasoning asks: what is the change, and which value is the original reference?
If a price rises from $80 to $100, change = 20 and original = 80, so increase = 25%.
To return from 100 to 80, the decrease is still 20 but the original for that second movement is 100, so decrease = 20%.
The target decomposition prevents the common mistake of using whichever denominator happens to be nearby.
14. Rates can be decomposed into numerator and denominator quantities
If the target is speed, we need distance and time.
If distance is missing but coordinates are given, geometry may supply it. If time is given in minutes but speed is required in km/h, unit conversion becomes a subgoal.
The dependency tree may therefore be:
speed ← distance + time ← coordinate distance + unit conversion.
Each branch can be solved and checked independently before recombination.
15. Coordinate geometry naturally supports reverse planning
Suppose the target is the equation of a perpendicular bisector of AB.
A line equation needs a point and a gradient.
The required point is the midpoint of AB. The required gradient is the negative reciprocal of the gradient of AB, assuming the line is neither vertical nor horizontal in a way requiring special handling.
Thus:
perpendicular bisector ← midpoint + perpendicular gradient ← coordinates of A and B.
The target tells the learner exactly which sibling calculations to perform.
16. Trigonometric synthesis becomes manageable through subgoals
A triangle problem may ask for area while giving no perpendicular height.
Backward reasoning asks whether two sides and included angle can be found.
Perhaps the included angle comes from an angle sum, and a missing side comes from the sine or cosine rule.
The learner does not need to know the whole route instantly. It is enough to identify the next necessary subgoal.
This is one reason sophisticated problem solving often feels like local planning rather than one giant insight.
17. Calculus targets often reveal which derivative information matters
If the target is a stationary point, we need where dy/dx = 0 and then the corresponding y-coordinate.
If the target is a tangent equation, we need the point on the curve and the derivative value there.
If the target is a local maximum or minimum, we need a stationary point plus classification evidence, such as a sign change or second derivative where appropriate.
Different calculus tasks share derivative machinery but require different final dependency trees.
18. A final target may branch into parallel subgoals
Not every route is a single chain.
To find the equation of a circle, the learner may need both its centre and radius. Those may be derived from different pieces of information.
To calculate total revenue, adult and child revenues can be computed in parallel then added.
To prove two shapes congruent, several independent equalities may need to be established before the criterion can be invoked.
Backward planning therefore creates a graph of dependencies, not always a straight line.
19. A subgoal can be useful even if it is not explicitly asked
A problem asks for the surface area of a cylinder but gives volume and height. The radius is not requested, yet it may be an essential intermediate quantity.
The student should not confuse “not explicitly asked” with “unnecessary”.
Subgoals exist because they unlock the target.
Conversely, not every intermediate quantity that could be calculated is worth calculating. The target tree separates useful subgoals from optional facts.
20. Reverse planning helps decide when enough information has been found
Once every input to the target relationship is known, additional calculation may be unnecessary.
If area needs base and height and both are now known, there is no need to calculate every other side and angle in the diagram.
This connects backward reasoning to the Data Sufficiency chapter. The key question becomes not “have I used everything?” but “do I now have enough to determine the target?”
21. Reverse planning can improve checking
After solving, walk the dependency tree backward again.
Did the final answer use all necessary inputs? Was each subgoal correctly established? Did any branch depend on a value that was only assumed? Did a unit conversion occur before the quantities were combined?
This gives a structured audit rather than a vague rereading of the page.
For a long problem, the check can focus on handoff points where one subgoal became the next input.
22. Build a route with the smallest necessary subgoals
More subgoals are not automatically better.
If a direct relationship already gives the target, do not manufacture a long route. If two methods exist, prefer one whose intermediate quantities are stable, interpretable and easy to verify.
For example, if a rectangle’s area can be found directly from known dimensions, computing its diagonal first adds work without serving the target.
Efficient backward reasoning is selective.
23. Independent practice
1. A rectangle has perimeter 44 cm and length 4 cm more than width. Find its area. Write the dependency chain first.
2. A line passes through A(2, 3) and B(8, 15). Find the equation of the perpendicular bisector of AB by identifying the two subgoals.
3. A quantity rises from 120 to 150. Find the percentage increase by decomposing the target.
4. A triangle has base 10 cm and area 35 cm². Find its perpendicular height. Work backward from the area formula.
5. Show that n² + n is even for every integer n. Use the target “even” to choose a useful algebraic form.
6. Find the tangent to y = x² + 1 at x = 2. State the two quantities needed for a line equation.
7. A circle has area 49π cm². Find its circumference by identifying the intermediate quantity needed.
8. An invented event has 90 tickets sold in the ratio adult:child = 2:3. Adult tickets cost $14 and child tickets $8. Find total revenue using a target tree.
9. Explain why working backwards from a desired result is not automatically a proof.
10. A problem asks for average speed in km/h. It gives distance 7.2 km and time 24 minutes. Identify and solve the unit-conversion subgoal.
24. Worked answers
1. Area needs length and width. Let width = w and length = w + 4. Perimeter gives 2w + 2(w + 4) = 44, so 4w + 8 = 44, w = 9 and length = 13. Area = 117 cm².
2. A perpendicular bisector needs midpoint and perpendicular gradient. Midpoint M = ((2 + 8)/2, (3 + 15)/2) = (5, 9). Gradient AB = 12/6 = 2, so perpendicular gradient = −1/2. Equation: y − 9 = −1/2(x − 5), or x + 2y = 23.
3. Percentage increase needs change and original value. Change = 30. Original = 120. Increase = 30/120 × 100% = 25%.
4. A = 1/2 bh. Therefore h = 2A/b = 70/10 = 7 cm.
5. Factor n² + n = n(n + 1). Consecutive integers contain one even factor, so the product is even.
6. A tangent equation needs point and gradient. Point at x = 2 is (2, 5). Derivative dy/dx = 2x, so gradient = 4. Hence y − 5 = 4(x − 2), giving y = 4x − 3.
7. Circumference needs radius. πr² = 49π gives r = 7 cm. Circumference = 2πr = 14π cm.
8. Revenue needs adult and child counts. Five ratio parts represent 90 tickets, so one part = 18. Adults = 36, children = 54. Revenue = 36(14) + 54(8) = 504 + 432 = $936.
9. Backward reasoning can reveal which conditions would be sufficient, but the final argument must show those conditions follow from accepted givens. Assuming the desired conclusion would be circular.
10. Convert 24 minutes to 0.4 hours. Then average speed = 7.2/0.4 = 18 km/h.
25. Continue through the operating manual
Use Given, Deduced, Assumed and Conjectured Information to ensure each subgoal rests on legitimate facts. Use Invariants and Conservation Checks when a preserved quantity suggests the route. Continue to Parameter Sensitivity, Stability and What-If Analysis to test how the route behaves when inputs change.
Return to the BTT Mathematics Hub for Batch 03.