Primary Mathematics: Balance Scales and Weighing Puzzles | Worked Learning Guide

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A balance can tell us two different things: which quantities are equal, and which possibilities remain after a comparison. The first leads to equations and substitution. The second leads to weighing strategies. Both become clearer when every move has a reason.

Imagine three identical closed boxes and a 12 g mass balancing two of the same boxes and a 37 g mass. Opening the boxes is unnecessary. Remove two boxes from each side. One box and 12 g still balance 37 g. Remove 12 g from each side. The remaining box has mass 25 g. We have discovered something hidden without seeing inside it.

Now imagine several identical-looking tokens, exactly one of which is heavier. Removing equal quantities is no longer the whole problem: we must choose a comparison that separates the possible identities. A successful first guess is not enough. A complete strategy must work whichever token is the heavier one.

This is a worked-learning and enrichment guide. Begin with equal masses and whole-number arithmetic. Use the later strategy sections only when the learner is ready to follow alternative outcomes. These extensions are not presented as compulsory content for every Primary level. The examples use idealised puzzle objects, not the assumption that real fruit or everyday packets have identical masses.

What balance means · Substitution · Weighing strategies · 24 questions · Worked solutions · Transfer and teaching

1. Read the relationship before calculating

Throughout this guide, a level equal-arm balance means equal total mass on the two pans. When it tilts, the lower pan is heavier. The balance is assumed reliable, and empty containers are either identical or already accounted for. These are mathematical conditions of the model. A drawing that merely looks balanced is not evidence unless the question states that it is.

Identical labels mean identical masses: every box labelled A has the same mass. Different labels need not have different masses unless that is stated. Counting three objects on one pan and two on the other does not tell us which pan is heavier when the objects have different masses.

Keep the units attached to the quantities. Three bags is a count. A mass of 30 g is a measurement. An equation may connect them, but they are not interchangeable kinds of information. When we write A + B = 30 g, A and B stand for the masses of the labelled objects, not the number of objects.

Equal actions preserve equality

If the pans balance, adding the same mass to each pan preserves the balance. Removing the same mass from each pan also preserves it, provided that mass is available to remove. Dividing both totals into the same number of equal groups preserves equality between corresponding groups. The point is not to memorise that a symbol moves across an equals sign. It is to preserve the relationship while making the unknown easier to see.

For four equal boxes plus 20 g balancing 100 g, first remove 20 g from each side. Four boxes then balance 80 g. Divide into four equal groups: one box balances 20 g. Check the original statement, not just the last step: four masses of 20 g plus 20 g give 100 g.

Why removing from only one side fails

Suppose two boxes balance 50 g. Removing one box from the left while leaving 50 g on the right does not produce a new equality. It changes the relationship. A learner who writes one box = 50 g has not made an arithmetic slip; the learner has changed one side without making the corresponding change on the other.

A useful repair is to write a brief reason beside each line: remove equal masses, replace by an equal mass, or share into equal groups. The explanation can be spoken rather than written, but it must exist. An unexplained answer may be right by chance; a preserved relationship can be checked.

2. Replace an object by an equal mass

Substitution means replacing a quantity with another quantity known to be equal to it. Suppose one red block balances two blue blocks, and one blue block balances three green blocks. Replace each blue block in the first relationship by three green blocks. One red block therefore balances six green blocks.

This does not tell us a red block’s mass in grams. It gives a relative mass. To obtain grams, we need a measurement such as one green block has mass 4 g. Then the red block has mass 24 g. The distinction between a relative unit and an absolute measurement is essential.

Worked example: bags and counters

Two identical bags and three identical counters balance seventeen of those counters. Remove three counters from each side. Two bags balance fourteen counters, so one bag balances seven counters.

The answer is seven counter-masses per bag, not necessarily seven counters inside the bag. A closed bag could contain another material. The balance reveals total mass, not its contents. This is an example of keeping a conclusion within the information supplied.

Worked example: two linked balances

Two A blocks and one B block balance 28 g. One A block and one B block balance 19 g. The second collection is entirely contained in the first. Removing that equal 19 g collection leaves one A block balancing 9 g. Therefore A = 9 g. Substitute into A + B = 19 g to obtain B = 10 g.

There are two independent conditions to check. The first becomes 2 × 9 + 10 = 28. The second becomes 9 + 10 = 19. Satisfying only the simpler equation would not be enough. A wrong pair such as A = 8 g, B = 11 g satisfies the second but fails the first.

Worked example: three pair totals

Suppose A + B = 36 g, B + C = 44 g, and A + C = 40 g. Add the three statements. Each object appears twice, so twice the total mass of A, B and C is 120 g. One complete collection therefore has mass 60 g.

Subtract B + C = 44 g from the total to find A = 16 g. Subtract A + C = 40 g to find B = 20 g. Subtract A + B = 36 g to find C = 24 g. This method works because the repeated appearances are counted explicitly. It is closely related to the double-counting ideas in the Sets and Venn Diagrams guide.

3. Decide whether the information determines one answer

If A + B = 24 g is the only condition, the mass of A cannot be determined uniquely. A = 8 g and B = 16 g works. A = 10 g and B = 14 g also works. Producing two different valid pairs proves that the stated information does not force a single value of A.

Do not confuse an unanswered question with an impossible question. The relationship A + B = 24 g has many solutions. By contrast, three identical bags plus 80 g balancing the same three bags plus 100 g is inconsistent in this ideal model. Removing the bags from both sides would claim 80 g = 100 g.

A tilt provides an inequality, not an exact mass

If an object together with 12 g is heavier than 30 g, the object must be heavier than 18 g. It might have mass 19 g, 23 g or another value above 18 g. Writing exactly 18 g would incorrectly turn a strict comparison into equality.

For younger learners, the sentence “more than 18 g” is sufficient. The optional symbol A > 18 g names the same relationship. Use symbolic notation to shorten an understood statement, not to replace understanding.

Fractional masses are not fractional objects

Three identical objects balance 25 g. One object’s mass is 25 ÷ 3 = 8⅓ g. That is a valid exact mass even though the number of objects is whole. There is no reason to round to 8 g: three objects of 8 g would total only 24 g.

Likewise, two identical parcels plus 200 g balance 1.4 kg. Convert 1.4 kg to 1400 g before subtracting. The two parcels have combined mass 1200 g, so each has mass 600 g. Mixing 1.4 and 200 without converting units would destroy the meaning of the calculation.

4. A weighing strategy must cover every outcome

We now use an ideal balance that reports only three outcomes: left heavier, right heavier, or level. It does not show a numerical difference. Each token may be placed on at most one pan during a comparison. Genuine tokens all have the same positive mass. Unless stated otherwise, exactly one token is known to be heavier than the genuine tokens.

In this known-heavier model, comparing equal numbers is especially useful. If the pans balance, all tokens on those pans are genuine and the heavier token is outside the comparison. If the left pan is heavier, the unusual token must be among the tokens on the left. The same reasoning applies on the right.

The qualification “known to be heavier” matters. With an unknown lighter-or-heavier token, a left-heavy result could mean an unusually heavy token on the left or an unusually light token on the right. We will handle that as a separate problem rather than quietly carrying over the simpler rule.

Three candidates, one comparison

Label the candidates A, B and C. Weigh A against B. If A is heavier, choose A. If B is heavier, choose B. If the pans balance, choose C. Each possible result identifies a different candidate. The unweighed token has not been ignored: it owns the balanced branch.

Nine candidates, two comparisons

Divide the tokens into three groups of three: A, B and C. Compare Group A with Group B. The heavier group contains the unusual token; if the groups balance, Group C contains it. Exactly three candidates remain in every branch. Within the selected group, compare one token against a second and use the three-candidate procedure.

Every first result leaves a solvable final comparison

First result	Remaining candidates	Next action
A heavier	Three tokens in A	Compare two; balance identifies the third
B heavier	Three tokens in B	Compare two; balance identifies the third
Level	Three tokens in C	Compare two; balance identifies the third

This is a strategy, not a report of one lucky experiment. It says what to do before the outcome is known. The standard three-outcome argument is also explored in the linked NRICH enrichment task listed under sources.

5. Why a lower bound is different from a construction

One comparison has at most three distinguishable results. Two comparisons have at most three choices followed by three choices: nine final result sequences. If each of ten possible heavier-token identities must receive its own distinguishable final result, nine sequences cannot be enough. Thus ten known-heavier candidates cannot be resolved in two comparisons in the worst case.

This counting argument rules out an impossibly short strategy. It does not automatically construct a valid strategy. We must still explain how legal pan comparisons produce useful branches. The distinction is the same as in Optimisation and Minimum Moves: establish a bound, then find a method that attains it.

Eight candidates do not need perfectly equal thirds

Compare three tokens against three, leaving two aside. A tilt leaves three candidates, which one comparison can resolve. A balance leaves two candidates, which can be compared directly. Because exactly one of those two is heavier, that last comparison cannot balance. The strategy uses no more than two comparisons in any branch.

Ten candidates can be resolved in three comparisons

Compare three against three and leave four aside. If one pan is heavier, resolve its three candidates by comparing one against one. If the first comparison balances, take the four unweighed candidates and compare two against two. One pair must be heavier. Compare the two members of that pair to finish.

Some branches finish after two comparisons; the most demanding branch needs three. The guaranteed maximum is three. A statement that a strategy “uses two comparisons” is misleading when that is true only for the convenient branches.

Twenty-seven candidates show the repeating structure

For twenty-seven known-heavier candidates, compare nine against nine, leaving nine aside. Select the relevant nine. Then compare three against three within that group, leaving three aside. Select the relevant three and compare one against one. Three comparisons suffice.

Two comparisons have only nine final result sequences, so they cannot separate twenty-seven identities. The lower bound and the construction agree: three comparisons are necessary and sufficient in this stated model. The optional shorthand is that w comparisons provide at most 3 to the power w outcome sequences.

6. When the unusual token might be lighter

With three tokens A, B and C, suppose exactly one is unusual but we do not know whether it is heavier or lighter. We want both its identity and its direction of difference. There are six possible states: A heavy, A light, B heavy, B light, C heavy and C light.

First compare A with B. If they balance, C is unusual and A is genuine. Compare C against A to determine whether C is heavier or lighter. If A is heavier than B, only two possibilities remain: A is heavy or B is light. C is genuine. Compare A against C. If A is heavier, A is the heavy unusual token. If they balance, B is the light unusual token.

The remaining first branch is the mirror image. If B is heavier than A, the possibilities are B heavy or A light. Compare A against C. If A is lighter, A is unusual and light. If they balance, B is unusual and heavy. We have checked all six original states, not just selected examples.

Why an outcome count is not always sufficient

Consider only two tokens, one genuine and one unusual, with no known genuine reference and no numerical mass reading. A comparison shows A heavier than B. That observation fits either A being the heavy unusual token or B being the light unusual token. Swapping their pans does not distinguish those explanations.

A separate genuine reference token changes the available information. Compare A with the reference. If A differs, its identity and direction are known immediately. If they balance, B is unusual; compare B with the reference to find its direction. What matters is not just how many outcomes a scale can display, but which hypotheses the permitted comparisons can separate.

7. Find the failure in an attractive first move

With nine known-heavier tokens, a learner proposes weighing four against four first. The balanced branch identifies the unweighed token immediately. That sounds excellent. However, either tilted branch leaves four candidates and only one comparison remaining. Three possible outcomes cannot distinguish four identities.

The strategy is excellent for one outcome but fails the guarantee. To repair it, distribute the candidates so that every branch fits the capacity of the remaining comparisons. Three against three with three aside does this. The best first move is not necessarily the one that sometimes gives the most dramatic result.

Another error is to conclude that balanced pans always contain only genuine tokens. That conclusion follows in the exactly-one-unusual-token model. It need not follow when two unusual tokens are allowed: one equally heavier token on each pan could balance. A method inherits the assumptions under which it was justified.

8. Practice: 24 original questions

Use the ideal-balance conditions above. Same-labelled objects have equal masses. Give exact masses unless the question requests a bound. For strategy questions, state what happens in every possible result, not only the result you hope to see.

Questions 1–8: Equalities and substitution

1. Four identical boxes together with 20 g balance 100 g. Find one box’s mass.
2. Three identical packets plus 12 g balance two such packets plus 37 g. Find one packet’s mass.
3. Two identical bags and three counters balance seventeen identical counters. Express one bag’s mass in counter-masses.
4. One red block balances two blue blocks. One blue block balances three green blocks. How many green blocks balance one red block?
5. Two A blocks and one B block balance 28 g. One A and one B balance 19 g. Find both masses.
6. A + B = 36 g, B + C = 44 g and A + C = 40 g. Find A, B and C.
7. Two identical parcels and 200 g balance 1.4 kg. Find each parcel’s mass in grams.
8. A + B = 24 g is the only information. Can A be found uniquely? Support your answer.

Questions 9–16: Conditions and known-heavier strategies

9. An object plus 12 g is heavier than 30 g. What bound follows for the object’s mass?
10. Three identical objects balance 25 g. Find their exact individual mass.
11. Three identical bags plus 80 g are said to balance the same three bags plus 100 g. Is this consistent?
12. A balances twice the mass of B, and A + B = 45 g. Find A and B.
13. Exactly one of three tokens is heavier. Identify it in one comparison.
14. Exactly one of five tokens is heavier. Give a strategy using at most two comparisons.
15. Exactly one of eight tokens is heavier. Explain a three-against-three first comparison and all subsequent branches.
16. Exactly one of nine tokens is heavier. Give a two-comparison strategy and explain why one comparison cannot guarantee identification.

Questions 17–24: Guarantees and limits

17. Can ten known-heavier candidates always be distinguished in two comparisons? Explain.
18. Give a strategy for ten known-heavier candidates using at most three comparisons.
19. Find the minimum guaranteed number of comparisons for twenty-seven known-heavier candidates.
20. Among A, B and C, exactly one token is unusual, either heavier or lighter. Give a two-comparison strategy identifying the token and its direction.
21. Two tokens contain one genuine token and one unusual token of unknown direction. With no reference and only balance comparisons between these tokens, can both identity and direction be determined?
22. Add a known genuine reference to Question 21. Give a strategy using at most two comparisons.
23. For nine known-heavier candidates, explain why four against four is not a guaranteed two-comparison strategy.
24. A learner claims that balanced pans prove every token on them is genuine, even when two unusual tokens may exist. Give a counterexample.

9. Worked solutions

Solutions 1–8

1. 20 g. Remove 20 g from both sides to leave four boxes balancing 80 g. Share 80 g into four equal masses. Rebuild the original total: 4 × 20 + 20 = 100 g.

2. 25 g. Remove two packets from each side, then remove 12 g from each side. The remaining packet balances 37 − 12 = 25 g. Check: both original pans total 87 g.

3. Seven counter-masses. Removing three counters leaves two bags balancing fourteen counters. Halving gives seven counter-masses per bag. This identifies mass equivalence, not the bag’s actual contents.

4. Six green blocks. Replace each of the two blue blocks by its equal mass of three green blocks. The total replacement is 2 × 3 = 6 green blocks.

5. A = 9 g and B = 10 g. Remove the collection A + B, known to equal 19 g, from the 28 g collection. The extra A is 9 g. Then B = 19 − 9 = 10 g. Both original balances check.

6. A = 16 g, B = 20 g, C = 24 g. The three pair totals add to 120 g, counting every object twice. The single total is 60 g. Subtract each pair total to obtain the remaining object.

7. 600 g each. Convert the right side to 1400 g. Remove 200 g to leave 1200 g for two parcels. Divide by two. Combining 600 + 600 + 200 returns 1400 g.

8. No unique value. A = 8 g, B = 16 g and A = 10 g, B = 14 g are two different valid assignments. Another independent condition is needed to determine A.

Solutions 9–16

9. More than 18 g. Remove the same 12 g from the comparison. The object’s mass exceeds 30 − 12 = 18 g. Equality at 18 g would produce level pans, contrary to the given tilt.

10. 8⅓ g. Calculate 25 ÷ 3 exactly. A fractional measurement is allowed; we still have three whole objects. Rounding each to 8 g would fail the total-mass check.

11. Inconsistent. Removing the same three bags from each side would leave 80 g balancing 100 g. Under the reliable ideal-balance model, that is impossible.

12. A = 30 g, B = 15 g. Replace A by two B-masses. The total becomes three B-masses equalling 45 g. One B is 15 g and A is twice that, 30 g.

13. Weigh one against one. A tilt identifies the heavier token on the lower pan. If the pans balance, the third, unweighed token is the unusual one. All three identities have distinct outcomes.

14. Compare two against two, leaving one aside. A balance identifies the outside token immediately. A tilt leaves two candidates on the heavier pan; compare them directly to identify the heavier one.

15. Compare three against three, leaving two aside. A tilt leaves three candidates; compare one against one and use the unweighed candidate for the balanced result. An initial balance leaves two outside candidates; compare those directly.

16. Divide into three groups of three. Compare two groups, select the heavier group or the outside group after balance, then resolve its three tokens by one-against-one. One comparison has only three outcomes, insufficient for nine identities.

Solutions 17–24

17. No. Two comparisons provide at most nine distinct final outcome sequences. Ten possible identities require at least ten distinct endings for certain identification. The counting bound rules out the proposed guarantee.

18. Three against three, with four aside. A tilted branch contains three candidates and takes one more comparison. A balanced branch contains four; compare two against two, then compare the two members of the heavier pair. The worst case uses three.

19. Three. Use groups of nine, then groups of three, then individual tokens. This gives a construction using three comparisons. Two cannot suffice because nine outcome sequences cannot distinguish twenty-seven identities.

20. Compare A with B, then use C as a reference when they differ. If A and B balance, C is unusual; comparing C with A gives its direction. If A is heavier, compare A with C: A heavier means A is unusual and heavy; balance means B is unusual and light. If A is lighter, A lighter than C identifies A as light; balance identifies B as heavy.

21. No. Observing A heavier than B cannot distinguish A being unusually heavy from B being unusually light. Reversing the pans preserves that ambiguity. There is no separately known genuine mass for comparison.

22. Compare A with the genuine reference. A tilt identifies A and its direction. A balance makes A genuine and B unusual; compare B with the reference to determine the direction. The worst branch needs two comparisons.

23. A tilted branch leaves four candidates. Only one comparison remains, with at most three outcomes. The easy balanced branch does not compensate for the unsolved tilted branches. A guarantee must cover all branches.

24. Put one equally heavy unusual token on each pan. For example, two unusual tokens of 11 g balance each other even if genuine tokens have mass 10 g. The original inference depended on exactly one unusual token.

10. Transfer: twelve candidates and an unknown mass

First task: exactly one of twelve tokens is known to be heavier. Find a guaranteed strategy and justify the minimum number of comparisons. Second task: A balances three B-masses, and two A objects plus one B object balance 63 g. Find A and B. These tasks deliberately separate information gathering from mass calculation.

Strategy solution: three comparisons are necessary because two allow at most nine distinct endings. Divide the twelve tokens into three groups of four. Compare two groups and select the heavier group, or the outside group after balance. Compare two against two within the selected four, then compare the two candidates in the heavier pair. Every identity is resolved within three comparisons.

Mass solution: replacing each A by three B-masses makes the total seven B-masses. Therefore B = 63 ÷ 7 = 9 g and A = 27 g. The original total checks as 2 × 27 + 9 = 63 g. Do not import the three-outcome argument into this second task: its difficulty is preserving equality, not identifying a hidden token.

11. Teaching the first decision, not only the answer

For a learner struggling with balance equations, begin with removable paper objects on two labelled sides. Ask which collection can be removed equally. If the learner removes different masses, return to the meaning of equality before introducing a quicker written procedure. The existing Patterns, Early Algebra and Equations guide supplies the broader equality route.

For weighing puzzles, write the remaining possibilities under each branch. “Compare these three against those three” is only the start of an explanation. Ask what a left-heavy result leaves, what a right-heavy result leaves, and what a balanced result leaves. Then ask whether each remaining set fits the next available comparison.

Do not label a child weak at algebra because an advanced weighing puzzle is difficult. These tasks place different demands on representation and branching. A learner may calculate masses confidently but need a smaller candidate set to practise strategy. Another may enjoy strategy while still needing help converting kilograms to grams.

A delayed return

After a gap, give one new equal-mass equation, one underdetermined pair, and a known-heavier puzzle with a different candidate count. Ask the learner to explain the first move without looking at this guide. Record the point at which a prompt becomes necessary. A supported correct answer and an independently chosen strategy are useful but different observations.

The final check is always local: does the mass satisfy every original balance, and does the strategy distinguish every allowed state? Neither an attractive diagram nor one successful trial answers both questions.

Sources, scope and further reading

University of Cambridge NRICH — Spot the Fake provides an external exploration of the known-heavier-token problem and its three-outcome comparison structure. It is an advanced enrichment reference, not evidence that these puzzles are required in Primary school. The mass examples, staged explanations and practice set here are independently written.

For official subject documents, use the MOE Primary curriculum and syllabus page alongside the learner’s school programme. This article is not a complete syllabus, an official assessment or a performance guarantee.

Continue the worked-learning collection

For endpoint reasoning, continue to Intervals, Fence Posts and Cuts. For exhaustive visual counting, use Counting Rectangles, Squares and Triangles. For the difference between detecting a mistake and proving correctness, read Digit Sums and Arithmetic Error Checks.

Return to the BTT Primary Mathematics Learning Hub. The aim is a justified next move: preserve an equality, eliminate an impossible state, or show why no shorter complete strategy can work.