GCSE Maths Practice: fractions

Question 1 of 11

This question tests compound fraction skills: you must evaluate two products and then subtract. Use BIDMAS to structure your working, reduce fractions early by cross-cancelling, and present your final answer in simplest form.

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\( \begin{array}{l} \text{Evaluate: }\left(\frac{3}{5}\times\frac{4}{7}\right)-\left(\frac{1}{5}\times\frac{11}{7}\right).\end{array} \)

Choose one option:

Multiply first, then subtract. Cross-cancel within each product to simplify, and check the final fraction is in lowest terms.

At Higher GCSE level, fraction questions often combine several operations. The key is to structure your work using BIDMAS: do the multiplications first, then the subtraction. Only at the end should you simplify or convert forms if needed.

Method Blueprint (no spoilers for the quiz)

  1. Multiply first. Evaluate each product separately. Use cross-cancellation before multiplying to keep numbers small.
  2. Align denominators. Once both products are simplified, you usually have like denominators already (if not, find an LCM).
  3. Subtract numerators. Keep the common denominator the same.
  4. Final tidy-up. Reduce your result to lowest terms.

Worked Examples

Example 1: \( \left(\tfrac{5}{6}\times\tfrac{3}{4}\right)-\left(\tfrac{1}{3}\times\tfrac{1}{8}\right) \)

  1. First product: \( \tfrac{5}{6}\times\tfrac{3}{4}=\tfrac{15}{24}=\tfrac{5}{8} \).
  2. Second product: \( \tfrac{1}{3}\times\tfrac{1}{8}=\tfrac{1}{24} \).
  3. Subtract: \( \tfrac{5}{8}-\tfrac{1}{24}=\tfrac{15}{24}-\tfrac{1}{24}=\tfrac{14}{24}=\tfrac{7}{12} \).
  4. Answer: \( \tfrac{7}{12} \).

Example 2: \( \left(\tfrac{7}{9}\times\tfrac{3}{7}\right)-\left(\tfrac{2}{3}\times\tfrac{1}{9}\right) \)

  1. First product: \( \tfrac{7}{9}\times\tfrac{3}{7}=\tfrac{3}{9}=\tfrac{1}{3} \).
  2. Second product: \( \tfrac{2}{3}\times\tfrac{1}{9}=\tfrac{2}{27} \).
  3. Subtract: \( \tfrac{1}{3}-\tfrac{2}{27}=\tfrac{9}{27}-\tfrac{2}{27}=\tfrac{7}{27} \).
  4. Answer: \( \tfrac{7}{27} \).

Example 3: \( \left(\tfrac{4}{5}\times\tfrac{5}{14}\right)-\left(\tfrac{3}{10}\times\tfrac{2}{7}\right) \)

  1. First: cross-cancel 5s → \( \tfrac{4}{14}=\tfrac{2}{7} \).
  2. Second: \( \tfrac{3}{10}\times\tfrac{2}{7}=\tfrac{6}{70}=\tfrac{3}{35} \).
  3. Subtract: \( \tfrac{2}{7}-\tfrac{3}{35}=\tfrac{10}{35}-\tfrac{3}{35}=\tfrac{7}{35}=\tfrac{1}{5} \).
  4. Answer: \( \tfrac{1}{5} \).

Common Mistakes

  • Adding/subtracting before multiplying. In expressions like \( (A\times B)-(C\times D) \), products come first.
  • Forgetting cross-cancellation. Reducing before multiplying avoids unwieldy numbers and reduces error.
  • Changing both denominator and numerator incorrectly during subtraction: once denominators match, only the numerators change.
  • Dropping factors. Keep intermediate fractions in simplest form to see structure clearly.

Why This Matters

Compound fraction expressions appear across GCSE topics: ratio scaling (difference of two scaled amounts), enlargements (subtracting a second scaled copy), and mixture/dilution calculations (net change after two fractional operations). Mastering the structure means you can navigate multi-step fraction tasks confidently.

Quick FAQs

  • Do I always need an LCM? Not if your two products already share a denominator (common when denominators match or simplify to the same base).
  • Is \( (A\times B)-(C\times D) \) the same as \( (A-C)\times(B-D) \)? No — that’s a classic algebra error. Multiply first, then subtract results.
  • Should I convert to decimals? Fractions keep exactness and are preferred unless the question asks for decimals.

Study Tip

Lightly pencil a “spine” in your rough work: multiply → multiply → match denoms → subtract → simplify. This keeps you from skipping steps under time pressure.

Try These Yourself (no answers shown)

  • \( \left(\tfrac{2}{3}\times\tfrac{5}{8}\right)-\left(\tfrac{1}{6}\times\tfrac{3}{4}\right) \)
  • \( \left(\tfrac{9}{10}\times\tfrac{2}{9}\right)-\left(\tfrac{1}{5}\times\tfrac{1}{2}\right) \)
  • \( \left(\tfrac{7}{12}\times\tfrac{3}{7}\right)-\left(\tfrac{5}{18}\times\tfrac{1}{2}\right) \)