Fractions – Year 7 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. The highest common factor of 36 and 48 is 12. Divide top and bottom by 12: \(36\div12=3\) and \(48\div12=4\), so the simplest form is \(\dfrac{3}{4}\). (1 mark)
   2. Divide the numerator by the denominator: 6 goes into 23 three times (18) with remainder 5. The whole number is 3 and the remainder stays over 6: \(3\dfrac{5}{6}\). (1 mark)
   3. Write all three fractions with the common denominator 10: \(\dfrac{3}{5}=\dfrac{6}{10}\), \(\dfrac{7}{10}\) stays the same and \(\dfrac{1}{2}=\dfrac{5}{10}\) (1 mark). Now compare the numerators 5, 6, 7: \(\dfrac{1}{2},\ \dfrac{3}{5},\ \dfrac{7}{10}\) (1 mark).

   Check (b): \(3\times6+5=23\), so the mixed number converts back correctly.

2. Answer to Question 2 (4 marks)

   1. Change quarters into eighths: \(\dfrac{1}{4}=\dfrac{2}{8}\). Add the numerators only: \(\dfrac{3}{8}+\dfrac{2}{8}=\)\(\dfrac{5}{8}\). (1 mark)
   2. Change fifths into tenths: \(\dfrac{2}{5}=\dfrac{4}{10}\). Then \(\dfrac{7}{10}-\dfrac{4}{10}=\)\(\dfrac{3}{10}\). (1 mark)
   3. Convert the mixed numbers to improper fractions: \(2\dfrac{1}{3}=\dfrac{7}{3}\) and \(1\dfrac{1}{2}=\dfrac{3}{2}\). Use the common denominator 6: \(\dfrac{14}{6}+\dfrac{9}{6}\) (1 mark). This gives \(\dfrac{23}{6}=\)\(3\dfrac{5}{6}\) (1 mark).

   Another method for (c): add the wholes, \(2+1=3\), then the parts, \(\dfrac{2}{6}+\dfrac{3}{6}=\dfrac{5}{6}\). Never add denominators together.

3. Answer to Question 3 (4 marks)

   1. To find \(\dfrac{5}{6}\) of 72, divide by the denominator then multiply by the numerator: \(72\div6=12\) and \(12\times5=\)60. (1 mark)
   2. Multiply the numerators and the denominators: \(\dfrac{4\times3}{9\times8}=\dfrac{12}{72}\). Divide top and bottom by 12 to get \(\dfrac{1}{6}\). (1 mark)
   3. Dividing by a fraction is the same as multiplying by its reciprocal: \(\dfrac{5}{6}\times\dfrac{12}{5}=\dfrac{60}{30}=\)2. (1 mark)
   4. Change to improper fractions first: \(\dfrac{3}{2}\times\dfrac{8}{3}=\dfrac{24}{6}=\)4. (1 mark)

   Cancelling before multiplying saves work: in (d) the 3s cancel and \(8\div2=4\). Check (c): \(\dfrac{5}{12}\times2=\dfrac{5}{6}\).

4. Answer to Question 4 (4 marks)

   1. One fifth of £180 is \(180\div5=36\), so two fifths are \(2\times36=\)£72 for the charity. (1 mark)
   2. The whole amount is 1. Use the common denominator 15: \(\dfrac{2}{5}=\dfrac{6}{15}\) and \(\dfrac{1}{3}=\dfrac{5}{15}\) (1 mark). The fraction saved is \(\dfrac{15}{15}-\dfrac{6}{15}-\dfrac{5}{15}=\)\(\dfrac{4}{15}\) (1 mark).
   3. One fifteenth of £180 is \(180\div15=12\), so four fifteenths are \(4\times12=\)£48. (1 mark)

   Check: the sports kit gets \(180\div3=£60\). Then \(72+60+48=180\), so the three shares make up the whole £180 and nothing is lost or counted twice.

5. Answer to Question 5 (4 marks)

   1. Write \(1\dfrac{1}{2}\) km as \(\dfrac{6}{4}\) km so both distances are in quarters: \(\dfrac{3}{4}+\dfrac{6}{4}=\dfrac{9}{4}=\)\(2\dfrac{1}{4}\) km. (1 mark)
   2. On Friday he swims \(\dfrac{2}{3}\) of \(\dfrac{9}{4}\) km: \(\dfrac{2}{3}\times\dfrac{9}{4}=\dfrac{18}{12}=\dfrac{3}{2}=\)\(1\dfrac{1}{2}\) km. (1 mark)
   3. Total for the week: \(\dfrac{9}{4}+\dfrac{6}{4}=\dfrac{15}{4}=3\dfrac{3}{4}\) km (1 mark). This is less than the 4 km target, so no, Leo misses it by \(4-3\dfrac{3}{4}=\dfrac{1}{4}\) km (1 mark).

   Leo must change both distances to quarters before adding; \(\dfrac{3}{4}+1\dfrac{1}{2}\) is not \(1\dfrac{4}{6}\). In decimals: \(0.75+1.5+1.5=3.75\) km.

Total : 20 marks