Higher June 2025 Paper 5 Q5
5
\[1\frac{1}{5} + \frac{22}{d} = 1\frac{3}{4}\]Find the value of \(d\). [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 40 | 3 | M2 for \(\frac{11}{20}\) or \(\frac{22}{40}\) or \([1]\frac{15}{20} - [1]\frac{4}{20}\) or \([1]\frac{4}{20} + \frac{22}{d} = [1]\frac{15}{20}\) or \(\left[\frac{22}{d} =\right] \frac{35}{20} - \frac{24}{20}\) oe or \(\frac{24}{20} + \frac{22}{d} = \frac{35}{20}\) oe | Accept other equivalent fractions with a common denominator for M2 or M1 M2 implied by \(\frac{22}{0.55}\) For M2 accept e.g. \(\frac{7 \times 5}{4 \times 5} - \frac{6 \times 4}{4 \times 5}\) Accept e.g. \(\left[-\frac{22}{d} =\right] \frac{24}{20} - \frac{35}{20}\) for M2 |
| or M1 for \([1]\frac{3}{4} - [1]\frac{1}{5}\) oe or \(\frac{35}{20}\) and \(\frac{k}{20}\) or \(\frac{k}{20}\) and \(\frac{24}{20}\) or \([1]\frac{15}{20}\) and \([1]\frac{p}{20}\) or \([1]\frac{p}{20}\) and \([1]\frac{4}{20}\) | M1 implied by 0.55 or [1].75 – [1].2 \(p \lt 20\) | ||
| OR | |||
| M2 for \(7 \times 5 \times d - 6 \times 4 \times d = 22 \times 5 \times 4\) or better | \(35d - 24d = 440\) | ||
| or M1 for \(\frac{6 \times d + 22 \times 5}{5d} = \frac{7}{4}\) oe | \(\frac{6d + 110}{5d} = \frac{7}{4}\), accept e.g. \(\frac{6d}{5d} + \frac{110}{5d} = \frac{7}{4}\) | ||
| or \(\frac{4 \times 6 \times d + 22 \times 5 \times 4}{20d} = \frac{7 \times 5 \times d}{20d}\) oe | \(\frac{24d + 440}{20d} = \frac{35d}{20d}\) | ||