Foundation June 2025 Paper 2 Q20
20 Show that \(7\dfrac{1}{3} - 3\dfrac{4}{7} = 3\dfrac{16}{21}\)
(3)
| Scheme | Marks |
|---|---|
\(\dfrac{22}{3}(-)\dfrac{25}{7}\) or \((7)\dfrac{7}{21}(-)(3)\dfrac{12}{21}\) or \((7)\dfrac{7a}{21a}(-)(3)\dfrac{12a}{21a}\) | M1 |
\(\dfrac{154}{21} - \dfrac{75}{21}\) or \(\dfrac{22 \times 7}{21} - \dfrac{25 \times 3}{21}\) or \(\dfrac{22 \times 7 - 25 \times 3}{21}\) \(\dfrac{154a}{21a} - \dfrac{75a}{21a}\) or \(7\dfrac{7}{21} - 3\dfrac{12}{21} = 4 - \dfrac{5}{21}\) oe or \(7\dfrac{7}{21} - 3\dfrac{12}{21} = 6\dfrac{28}{21} - 3\dfrac{12}{21}\) | M1 |
\(\dfrac{154}{21} - \dfrac{75}{21} = \dfrac{79}{21} = 3\dfrac{16}{21}\) or \(4 - \dfrac{5}{21} = 3\dfrac{16}{21}\) or \(7\dfrac{7}{21} - 3\dfrac{12}{21} = 6\dfrac{28}{21} - 3\dfrac{12}{21} = 3\dfrac{16}{21}\) Working required Answer: A fully correct solution shown | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for correct improper fractions or fractional part of numbers written correctly over a common denominator
M1: for correct fractions with a common denominator with minus sign or mixed numbers to the stage shown
\(\dfrac{154}{21} - \dfrac{75}{21}\) or \(\dfrac{22 \times 7}{21} - \dfrac{25 \times 3}{21}\) implies the first M1
A1: Dep on M2 for a correct answer from fully correct working
If a student shows that \(3\dfrac{16}{21} = \dfrac{79}{21}\) then they must show correct working to \(\dfrac{79}{21}\) and can gain full marks for this