Make complex deductions from linked totals
Combine several summaries, track what is counted more than once and verify every deduced quantity.
Section 26-2.3 · 2026–2027 · 2.3 · Deduce and compare outcomes
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Before you begin
- Read tables and instructions in English
- Use arithmetic, ratios and whole-number constraints
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Before calculating, what should you identify?
What you will learn
- Recover individual quantities from linked aggregates
- Check consistency and non-negative whole-number requirements
Label unknowns and overlapping totals
Totals can conceal repeated counting. If a+b, a+c and b+c are known, adding all three counts each variable twice. Half that sum gives a+b+c; subtracting a pair then isolates the third quantity. This method uses structure rather than trying random triples.
Fictional scores satisfy A+B=13, A+C=15 and B+C=18. Their sum is 46, so A+B+C=23. Hence C=10, B=8 and A=5.
Verify before interpreting
Substitute the values into every original relation. A calculation that gives a negative number of people or a fractional number of whole items may reveal inconsistent information, a mistaken equation or a model that needs a different domain.
For a long problem, keep intermediate conclusions labelled so they can be reused without confusing them with raw source data. If several solutions survive the constraints, state the range or set instead of presenting one as uniquely forced.
Sources for this lesson
Pause and explain
What does (a+b)+(a+c)+(b+c) count?
Worked example
Recover the three scores from the pair totals.
Show the worked solution
- 46÷2=23 is the total score.
- C=23−13=10, B=23−15=8, A=23−18=5.
- Check 5+8=13, 5+10=15 and 8+10=18.
Answer The uniquely determined scores are A=5, B=8 and C=10. Each pair sum matches, and the values are permissible non-negative whole numbers.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Guided: Verify A=5, B=8 and C=10 against every original pair total.
Give me a hint
Use all three equations.
Compare my reasoning
- A+B=5+8=13.
- A+C=5+10=15.
- B+C=8+10=18.
All three pair totals match. The verification shows the recovered values satisfy the combined data, not merely the first equation.
Look for these in your work
- Checks all pairs
- Uses correct labels
- Confirms consistency
Independent: Box counts satisfy X+Y=11, X+Z=14 and Y+Z=17. Find each count.
Give me a hint
First recover the combined total.
Compare my reasoning
- The pair sums total 42, so X+Y+Z=21.
- X=21−17=4; Y=21−14=7.
- Z=21−11=10 and all sums check.
The counts are X=4, Y=7 and Z=10. They give pair sums 11,14 and 17, with no fractional or negative box count.
Look for these in your work
- Corrects double counting
- Finds all values
- Checks the domain
Application: Three whole-item counts have pair sums 5,6 and 8. Can all the data be correct?
Give me a hint
Look at the parity of the sum.
Compare my reasoning
- The pair sums total nineteen.
- This must equal twice the total item count.
- Nineteen cannot be twice an integer.
No. The data would require a total of 19÷2=9.5 items, impossible for whole-item counts. At least one pair total or the whole-item assumption must be wrong.
Look for these in your work
- Uses the invariant
- Detects inconsistency
- Names the incompatible assumption
Common mistakes
- Forgetting duplicated quantities in summaries
- Checking only one relation
- Ignoring the allowed domain
What can you explain now?
What does (a+b)+(a+c)+(b+c) count?
Compare with the explanation
Twice the total a+b+c
Every variable appears in two pair sums.
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Make it stick
What does (a+b)+(a+c)+(b+c) count?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.