Foundation June 2022 Paper 2 Q20
20 Show that \(5\dfrac{1}{3} - 2\dfrac{6}{7} = 2\dfrac{10}{21}\)
(3)
| Scheme | Marks |
|---|---|
| \(\dfrac{16}{3} - \dfrac{20}{7}\) or \((5)\dfrac{7}{21} - (2)\dfrac{18}{21}\) or \((5)\dfrac{7a}{21a} - (2)\dfrac{18a}{21a}\) | M1 |
\(\dfrac{112}{21} - \dfrac{60}{21}\) or \(\dfrac{112a}{21a} - \dfrac{60a}{21a}\) or \(5\dfrac{7}{21} - 2\dfrac{18}{21} = 3 - \dfrac{11}{21}\) oe or \(5\dfrac{7}{21} - 2\dfrac{18}{21} = 4\dfrac{28}{21} - 2\dfrac{18}{21}\) | M1 |
\(\dfrac{112}{21} - \dfrac{60}{21} = \dfrac{52}{21} = 2\dfrac{10}{21}\) oe or \(3 - \dfrac{11}{21} = 2\dfrac{10}{21}\) or \(5\dfrac{7}{21} - 2\dfrac{18}{21} = 4\dfrac{28}{21} - 2\dfrac{18}{21} = 2\dfrac{10}{21}\) Working required Answer: 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
A1: Dep on M2 for a correct answer from fully correct working
If all 3 fractions turned into improper fractions on the first line \(\dfrac{16}{3} - \dfrac{20}{7} = \dfrac{52}{21}\) then the student clearly needs to show that the LHS \(= \dfrac{52}{21}\)