Foundation June 2025 Paper 1 Q13
13 Show that \(\dfrac{3}{7} + \dfrac{1}{4} = \dfrac{19}{28}\)
(2)
| Scheme | Marks |
|---|---|
eg \(\dfrac{12}{28}\) and \(\dfrac{7}{28}\) or \(\dfrac{12n}{28n}\) and \(\dfrac{7n}{28n}\) or 4 × 3 (= 12) and (1 ×) 7 and 4 × 7 (= 28) or \(\dfrac{3 \times 4}{4 \times 7}\) and \(\dfrac{(1 \times)7}{4 \times 7}\) | M1 |
\(\dfrac{12}{28} + \dfrac{7}{28} = \dfrac{19}{28}\) or \(\dfrac{12n}{28n} + \dfrac{7n}{28n} = \dfrac{19n}{28n} = \dfrac{19}{28}\) or 12 + 7 = 19 and \(\dfrac{19}{28}\) or 4 × 3 + (1 ×) 7 (= 19) and 4 × 7 (= 28) and \(\dfrac{19}{28}\) or \(\dfrac{3 \times 4 + (1 \times)7}{4 \times 7} = \dfrac{19}{28}\) or \(\dfrac{3 \times 4}{4 \times 7} + \dfrac{(1 \times)7}{4 \times 7} = \dfrac{19}{28}\) Working required Answer: Shown | A1 |
| (2) | |
| (2 marks) |
Notes
M1: for finding a common denominator with at least one fraction correct
or
for calculations that would lead to 12 and 7 and 28
A1: dep on M1, for a complete correct method leading to \(\dfrac{19}{28}\)