Constant-Difference Problems: Comparing Quantities Across Change

Two quantities can change dramatically and still keep exactly the same gap between them.

That is the small idea behind a large family of Primary Mathematics problems.

If A and B both gain the same amount, the difference between them does not change.

If A and B both lose the same amount, the difference between them does not change either.

Same additive change → same difference.

This is the mechanism behind constant-difference problems.

The arithmetic is usually not the hardest part. The harder part is recognising that a fixed gap is still present after the story has changed the two quantities.

What exactly is being preserved?

Suppose Maya has $24 more than Leo.

If both receive $20, Maya still has $24 more than Leo.

If both spend $15, Maya still has $24 more than Leo.

Why?

Because adding the same amount to both sides preserves subtraction:

(Maya + 20) − (Leo + 20) = Maya − Leo.

And subtracting the same amount from both sides does the same:

(Maya − 15) − (Leo − 15) = Maya − Leo.

The values move. The gap does not.

Singapore’s Primary Mathematics syllabus places mathematical problem solving at the centre of learning and includes reasoning, connections, modelling, thinking skills and heuristics among the processes that support it. “Constant difference” is a useful classroom heuristic for seeing an invariant; it should not be mistaken for a separate official syllabus topic. See the MOE Primary Mathematics Syllabus, updated December 2024.

A worked problem from start to finish

Maya had $24 more than Leo. They each received $20. After that, the ratio of Maya’s money to Leo’s money was 7 : 5. How much money did each have at first?

The phrase “they each received $20” is the key structural clue.

Both amounts increase by the same number. Therefore the difference remains $24.

Step 1: connect the final ratio to the unchanged difference

After both receive $20:

Maya : Leo = 7 : 5.

The difference is:

7 − 5 = 2 units.

Those 2 units represent the same $24 difference that existed before the change.

So:

2 units = $24.

1 unit = $12.

Step 2: find the final amounts

Maya: 7 × $12 = $84.
Leo: 5 × $12 = $60.

Step 3: reverse the equal change

Each had received $20, so subtract $20 from each final amount.

Maya originally had $84 − $20 = $64.
Leo originally had $60 − $20 = $40.

Check:

$64 − $40 = $24.
After both receive $20: $84 and $60.
$84 : $60 simplifies to 7 : 5.

Every condition is satisfied.

Why the ratio changes even though the difference does not

This is one of the most useful conceptual points in the entire topic.

Originally, Maya and Leo had $64 and $40.

Their ratio was:

64 : 40 = 8 : 5.

After each receives $20, they have $84 and $60.

The ratio becomes:

84 : 60 = 7 : 5.

The difference stayed at $24, yet the ratio changed.

This happens because ratios respond to scale. Adding the same amount to two unequal quantities changes each quantity by a different proportion of its starting value.

For the smaller quantity, +20 is a larger percentage increase than it is for the larger quantity. The two quantities therefore become relatively closer, even though their absolute gap is unchanged.

This is why constant-difference problems often combine naturally with changing ratios.

The bar-model view

A bar model can make the fixed gap visible.

Imagine two horizontal bars. Maya’s bar is longer than Leo’s by one fixed segment labelled $24.

Now attach an equal $20 segment to the end of each bar.

The bars both grow.

The extra $24 segment between their endpoints remains exactly the same length.

When the final ratio 7 : 5 is drawn, the difference between the bars is 2 ratio units. Those 2 units must therefore represent $24.

The model is not a decorative picture. It is a visual proof of what stayed invariant.

The algebra underneath the heuristic

Let the original quantities be M and L, with:

M − L = 24.

After both receive 20:

(M + 20) : (L + 20) = 7 : 5.

The difference after the change is:

(M + 20) − (L + 20) = M − L = 24.

This identity explains the entire Primary-school shortcut.

Secondary algebra makes the rule explicit. The Primary method makes the same invariant visible through units and models.

When the difference really is constant

Change	Difference preserved?	Reason
Add 15 to both quantities	Yes	The same amount cancels in subtraction
Subtract 8 from both quantities	Yes	The same amount cancels in subtraction
Double both quantities	No	The difference also doubles
Add 10 to one and 5 to the other	No	The gap changes by 5
Transfer 10 from one to the other	No	The gap changes by 20

The phrase “both changed” is not enough.

The same additive change must happen to both quantities for the difference to remain fixed.

Constant difference is not constant ratio

If both quantities are multiplied by the same factor, their ratio stays constant but their difference usually changes.

For example:

6 and 4 have ratio 3 : 2 and difference 2.
12 and 8 still have ratio 3 : 2, but the difference is now 4.

So learners should separate two families:

same additive change → difference invariant;
same multiplicative scaling → ratio invariant.

This distinction becomes important later in algebra, similarity, direct proportion and functions.

Constant difference is not constant total either

Constant-total problems usually involve a transfer between parts.

If A gives 10 to B:

A decreases by 10;
B increases by 10;
A + B stays constant;
the difference changes by 20.

Constant-difference problems behave differently.

If A and B both receive 10:

A increases by 10;
B increases by 10;
A − B stays constant;
A + B increases by 20.

The learner should therefore ask:

Is the story moving quantity between the people, or applying the same change to both people?

A second worked example with subtraction

Two containers held different amounts of water. The first container held 18 litres more than the second. Then 6 litres were removed from each container. The remaining amounts were in the ratio 5 : 3. How much water was in each container originally?

Removing the same 6 litres from both containers preserves the 18-litre difference.

The final ratio has a difference of:

5 − 3 = 2 units.

2 units = 18 litres.

1 unit = 9 litres.

Final amounts:

first container: 5 × 9 = 45 litres;
second container: 3 × 9 = 27 litres.

Add back the 6 litres removed from each:

original first amount = 51 litres;
original second amount = 33 litres.

Check: 51 − 33 = 18.

Again, the final ratio and the invariant difference work together.

Why learners often misread these problems

They focus on the change instead of the invariant

The problem may mention birthdays, savings, prizes, spending, adding water or losing tokens. Those actions attract attention because they are visible verbs.

But the useful relationship may be the invisible one: the gap stayed fixed.

They treat ratio units as permanent objects

A ratio such as 7 : 5 describes a state. After equal addition or subtraction, the same people can have a different ratio. The units belong to that moment, not forever.

They preserve the wrong quantity

Some students preserve the total because they remember a previous heuristic. But if both people receive money from outside the pair, the total increases. The difference, not the total, is the invariant.

A diagnostic decision test

Give the learner several short stories without asking for calculations.

A and B each receive 12.
A gives 12 to B.
A and B are both doubled.
A receives 12 while B receives 7.
A and B each spend 9.

Ask what, if anything, stays constant:

difference?
total?
ratio?
none of these?

This is a better diagnostic than giving five almost-identical constant-difference questions. The learner must choose the invariant rather than imitate the previous page.

How this connects to PSLE Mathematics

The 2026 PSLE Mathematics syllabus assesses more than recall and routine computation. Its assessment objectives include interpreting information, applying concepts in varied contexts, reasoning mathematically, making inferences and selecting appropriate strategies. See the SEAB 2026 PSLE Mathematics syllabus.

A constant-difference problem reveals exactly that distinction. A learner may be perfectly capable of subtracting, simplifying ratios and multiplying units. The learner can still fail because the wrong relationship was selected.

In difficult problems, method selection is often the mathematics.

From model method to algebraic thinking

Constant-difference reasoning is also a bridge towards formal algebra.

The invariant:

A − B = d

remains true after adding the same quantity k to both:

(A + k) − (B + k) = d.

This is not merely a PSLE trick. It is an instance of a broader algebraic principle: applying the same additive translation to two values preserves their distance on the number line.

Seen geometrically, both points move the same distance in the same direction. Their separation stays unchanged.

Seen algebraically, the added terms cancel.

Seen through a bar model, the same block is attached to or removed from both bars.

Three representations. One relationship.

How to practise for transfer

Do not practise only money questions.

Use different surfaces:

ages increasing by the same number of years;
two containers losing the same volume;
two runners moving the same additional distance;
two accounts receiving the same credit;
two stacks having the same number removed.

Then change the structure deliberately:

give different amounts to each;
transfer from one to the other;
multiply both by the same factor.

The learner’s job is to say which invariant survives before doing any arithmetic.

For parents and teachers: ask for the mechanism

A useful explanation sounds like this:

“Both amounts increased by the same $20, so their difference is still $24. In the final 7 : 5 ratio, the 2-unit difference must represent that same $24.”

That sentence contains the entire method.

If the learner cannot say why the difference is unchanged, a correct answer may still be fragile.

Where the idea goes next

Constant difference is an early encounter with invariance under transformation.

On a number line, equal translations preserve distance.
In coordinate geometry, translating two points by the same vector preserves the vector between them.
In algebra, adding the same term to two expressions preserves their difference.
In data analysis, shifting every observation by the same constant changes the mean and median but leaves measures such as range and standard deviation unchanged.

The PSLE problem may involve two wallets.

The mathematical habit is much larger:

When two things change together, ask whether their relationship moved with them.

Related eduKateSG Mathematics routes

Official references

Constant-difference problems are not solved by finding a magic keyword.

They are solved by noticing that equal change can leave a gap untouched.