Math // Fractions
Fraction Word Problems
Solve word problems involving addition and subtraction of fractions with unlike denominators, using visual models and benchmark fractions to estimate and assess reasonableness
Quick check
If the learner walks 1 and 1/2 miles to school and 3/4 of a mile to a friend's house, can they work out the total distance — and check whether their answer makes sense?
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Topic // Mastery
What mastery looks like
- Solve: 'Tara ate 2/5 of a pizza and Sam ate 1/3. How much did they eat together?'
- Recognise that 2/5 + 1/2 ≠ 3/7 because 3/7 < 1/2 but the sum should exceed 1/2
- Estimate a fraction sum using benchmarks (0, 1/2, 1) before computing exactly
Topic // Map
Where this fits
- RequiredMust come first
- HelpfulHelps, but not essential
Before this
2 topicsFractions // Ages 10 to 11
Word problems require the computational skill of adding/subtracting unlike-denominator fractions
Fractions // Ages 9 to 10
Extends Y5 fraction word problems to unlike denominators
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Fraction Word Problems
Ages 10 to 11 // Method
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Topic // Research-informed
Teaching notes
From the Marble Skill Taxonomy. These notes have not been validated in classrooms, so treat them as a starting point for your own judgement.
The approach
Pupils (age 10–11) solve word problems adding and subtracting fractions with unlike denominators, but the defining skill is using benchmark estimation (0, 1/2, 1) as a reasonableness gate. They predict the sum's region BEFORE computing with common denominators, then check the exact answer against that prediction so an implausible result (e.g. an add-across sum that falls below 1/2 when it must exceed it) is caught rather than accepted.
Common misconceptions
- Adds across (2/5 + 1/3 → 3/8), producing a sum smaller than an addend
- Accepts an implausible answer that violates a benchmark (a sum below 1/2 when it must exceed it) because no estimate was made
- Partial-scaling to the common denominator — scales one numerator but not the other (2/5 + 1/3 → 6/15 + 1/15)
- Treats 'estimate' as optional and computes blind, missing the reasonableness check
Useful representations
- Benchmark number line (0, 1/2, 1) with each fraction placed
- Bar models recut to a common denominator via equivalent fractions
- Estimate-then-compute two-column layout
- The comparison 3/7 < 1/2 shown against the true sum region
Teaching practices
- Estimate with benchmarks first — is each fraction near 0, 1/2, or 1? — and predict whether the sum is above/below 1/2 and 1 before computing (IES)
- Compute exactly with common denominators, then CHECK against the benchmark estimate and reject implausible answers (e.g. a sum below 1/2 when it must exceed it)
- Model on a number line marked with 0, 1/2, 1 so 'add-across gives too small an answer' is visible
- Scale BOTH fractions to the common denominator (not just one) and keep the value unchanged
- Use the estimate to explain WHY 2/5 + 1/2 ≠ 3/7, turning reasonableness into a habit
Activities to try
- 1
Predict the region on a benchmark number line, then compute
Why it works: Benchmark estimation predicts a region for the sum; the exact answer must land inside it. Here add-across produces a sum (3/8) smaller than the addend 2/5, which the benchmark exposes as impossible.
- Read the problem: 'Tara ate 2/5 of a pizza and Sam ate 1/3 of a same-size pizza. How much did they eat together?'
- On a 0-to-1 number line marked at 1/4, 1/2 and 3/4, place each fraction: note that both 2/5 and 1/3 sit below 1/2 but above 1/4.
- Before computing, shade the region where you predict the TOTAL lands (above 1/2 but below a whole) and say why: each amount is more than 1/4, so together they are more than 1/2; each is less than 1/2, so together they are less than 1.
- Rename BOTH fractions to a common denominator (fifteenths): 2/5 = 6/15 and 1/3 = 5/15, then add to get 11/15.
- Place 11/15 on the line and check it falls inside your predicted region.
- 2
Argue why an answer is impossible
Why it works: Reasonableness can be used to reject an answer before or without full computation: adding 1/2 to a positive amount must give more than 1/2, and benchmarks make that concrete.
- A friend proposes that 2/5 + 1/2 = 3/7. Decide whether this can be right.
- Place 3/7 and 1/2 on a benchmark line and state where each sits relative to 1/2.
- Explain in your own words why 3/7 cannot be the sum: 3/7 is less than 1/2, but adding 1/2 to a positive amount must give more than 1/2.
- Give the correct method: rename both to a common denominator and compute the true sum.
- Repeat the argument for another proposed value, 2/7: show it is below 1/2, but since one addend here is exactly 1/2, the total must exceed 1/2, so 2/7 cannot be the sum.
- 3
Rename both fractions to the common unit
Why it works: Both fractions must be renamed to the common denominator and each equivalent fraction must keep its original value; changing only one numerator breaks the equivalence.
- For 2/5 + 1/3, find a common denominator (15) and recut both fractions to fifteenths.
- Write each equivalent fraction and verify it equals the original: 2/5 = 6/15 (×3/3), 1/3 = 5/15 (×5/5).
- Spot the error in a given partial-scaling attempt: 2/5 + 1/3 → 6/15 + 1/15 = 7/15. Identify that 1/15 does not equal 1/3.
- Correct it to 6/15 + 5/15 = 11/15.
- Check the result against a benchmark estimate (between 1/2 and 1).
Topic // Standards
Curriculum standards
- Common Core State Standards for Mathematics
- 5.NF.2
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