Foundation November 2021 Paper 1 Q10
10
(a) Show that \(\quad \dfrac{3}{10} \div \dfrac{1}{4} = \dfrac{6}{5}\) (2)
(b) Show that \(\quad \dfrac{5}{6} - \dfrac{3}{4} = \dfrac{1}{12}\) (2)
| Scheme | Marks |
|---|---|
| eg \(\dfrac{3}{10} \times \dfrac{4}{1}\left(= \dfrac{12}{10}\right)\) or \(\dfrac{6}{20} \div \dfrac{5}{20}\) or \(\dfrac{12}{40} \div \dfrac{10}{40}\) | M1 |
eg \(\dfrac{3}{10} \times \dfrac{4}{1} = \dfrac{12}{10} = \dfrac{6}{5}\) or \(\dfrac{6}{20} \div \dfrac{5}{20} = \dfrac{6}{5}\) or eg \(\dfrac{3}{\cancel{10}_{\,5}} \times \dfrac{\cancel{4}^{\,2}}{1} = \dfrac{6}{5}\) Answer: shown | A1 |
| (2) |
Notes
M1: Inverting \(\tfrac{1}{4}\) and changing to multiply or writing both fractions with the same denominator.
A1: Conclusion to \(\dfrac{6}{5}\) from correct working – either sight of the result of the multiplication eg \(\dfrac{12}{10}\) must be seen or correct cancelling prior to multiplication.
NB use of decimals scores no marks.
| Scheme | Marks |
|---|---|
eg \(\dfrac{10}{12} - \dfrac{9}{12}\) or \(\dfrac{20}{24} - \dfrac{18}{24}\) oe or eg \(\dfrac{10 - 9}{12}\) | M1 |
eg \(\dfrac{10}{12} - \dfrac{9}{12} = \dfrac{1}{12}\) or \(\dfrac{20}{24} - \dfrac{18}{24} = \dfrac{2}{24} = \dfrac{1}{12}\) oe Answer: clearly shown | A1 |
| (2) | |
| (4 marks) |
Notes
M1: for correct fractions with a common denominator of 12 or a multiple of 12.
A1: dep on M1 for a correct answer from fully correct working.