GCSE Maths Practice: fractions

Question 7 of 11

This Higher GCSE question involves a compound fractional expression requiring both multiplication and addition. You must follow BIDMAS and find common denominators to combine the results accurately.

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\( \begin{array}{l}\text{Evaluate: }\frac{5}{6}+\left(\frac{2}{3}\times\frac{3}{4}\right)+\frac{7}{9}.\end{array} \)

Choose one option:

In compound fraction expressions, perform multiplication first. Then find a common denominator and add or subtract step by step, simplifying fully at the end.

At Higher GCSE level, fraction questions frequently combine multiple operations in a single expression. This tests understanding of both the rules for adding fractions with different denominators and the BIDMAS structure (Brackets, Indices, Division/Multiplication, Addition/Subtraction).

Core Principles

  1. Always perform multiplications or divisions before addition or subtraction.
  2. Use the least common multiple (LCM) of denominators when adding or subtracting fractions.
  3. After each operation, simplify whenever possible — especially if numerators and denominators share common factors.

Worked Example 1 – Mixed operations

\( \tfrac{5}{6}+\tfrac{2}{3}\times\tfrac{3}{4}+\tfrac{7}{9} \)

  1. Multiply first: \( \tfrac{2}{3}\times\tfrac{3}{4}=\tfrac{6}{12}=\tfrac{1}{2} \).
  2. Add the fractions: \( \tfrac{5}{6}+\tfrac{1}{2}=\tfrac{5}{6}+\tfrac{3}{6}=\tfrac{8}{6}=\tfrac{4}{3} \).
  3. Then add \( \tfrac{7}{9} \): convert to common denominator 9 → \( \tfrac{4}{3}=\tfrac{12}{9} \).
  4. Add: \( \tfrac{12}{9}+\tfrac{7}{9}=\tfrac{19}{9}=2\tfrac{1}{9} \).

This layered approach ensures consistency and accuracy. Directly combining all three at once often leads to arithmetic slips.

Worked Example 2 – Subtraction in the mix

\( \tfrac{7}{8}-\tfrac{2}{3}\times\tfrac{3}{4}+\tfrac{1}{6} \)

  1. Multiply first: \( \tfrac{2}{3}\times\tfrac{3}{4}=\tfrac{6}{12}=\tfrac{1}{2} \).
  2. Subtract: \( \tfrac{7}{8}-\tfrac{1}{2}=\tfrac{7}{8}-\tfrac{4}{8}=\tfrac{3}{8} \).
  3. Add \( \tfrac{1}{6} \): LCM of 8 and 6 is 24 → \( \tfrac{3}{8}=\tfrac{9}{24},\ \tfrac{1}{6}=\tfrac{4}{24} \).
  4. Final: \( \tfrac{13}{24} \).

Worked Example 3 – Different signs and denominators

\( \tfrac{3}{5}+\tfrac{4}{9}-\tfrac{2}{15} \)

  1. LCM of 5, 9, and 15 is 45.
  2. Convert: \( \tfrac{27}{45}+\tfrac{20}{45}-\tfrac{6}{45}=\tfrac{41}{45} \).
  3. Answer: \( \tfrac{41}{45} \).

Common Mistakes

  • Ignoring BIDMAS. Multiplication must come before addition or subtraction.
  • Forgetting to adjust numerators correctly when finding equivalent fractions.
  • Adding denominators directly instead of creating equivalent fractions.
  • Not simplifying the final answer — always check for common factors.

Why This Matters

Multi-step fraction arithmetic mirrors how GCSE problems are structured in real assessments — they often mix operations to test full understanding. This foundation also supports algebraic manipulation later, when coefficients and terms are fractional.

Quick FAQs

  • Q: Can I find a single common denominator from the start?
    A: Yes, but it’s usually simpler to handle multiplication first, then find a shared base for the remaining additions or subtractions.
  • Q: Do I always need to convert to a mixed number?
    A: No — improper fractions are perfectly acceptable final answers in GCSE unless specified.
  • Q: Should I check by estimating?
    A: Definitely — rough estimates help confirm the result’s reasonableness.

Study Tip

Write intermediate answers line by line. After every major step, simplify before continuing. This helps avoid losing track of denominators during multi-operation questions.

Try These Yourself

  • \( \tfrac{4}{7}+\tfrac{3}{5}\times\tfrac{2}{3} \)
  • \( \tfrac{2}{3}\times\tfrac{3}{4}-\tfrac{1}{2} \)
  • \( \tfrac{5}{8}-\tfrac{3}{10}+\tfrac{2}{15} \)