Foundation November 2018 Paper 1 Q9
9 Work out the fraction that is halfway between \(\dfrac{1}{2}\) and \(1\dfrac{1}{4}\)
[3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\left(1\dfrac{1}{4} =\right) \dfrac{5}{4}\) | M1 | oe improper fraction |
| \(\dfrac{4}{8}\) and \(\dfrac{10}{8}\) or \(\dfrac{2}{4}\) and \(\dfrac{5}{4}\) or \(\dfrac{3.5}{4}\) | M1dep | oe common denominator with at least one correct numerator may be seen as start and end of a list |
| \(\dfrac{7}{8}\) | A1 | oe fraction |
| Alternative method 2 | ||
| \(\left(1\dfrac{1}{4} - \dfrac{1}{2} =\right) \dfrac{3}{4}\) | M1 | oe |
| \(\dfrac{1}{2} +\) their \(\left(\dfrac{3}{4} \div 2\right)\) or \(1\dfrac{1}{4} -\) their \(\left(\dfrac{3}{4} \div 2\right)\) | M1dep | oe |
| \(\dfrac{7}{8}\) | A1 | oe fraction |
| Alternative method 3 | ||
| \(\left(1\dfrac{1}{4} + \dfrac{1}{2} =\right) 1\dfrac{3}{4}\) or \(\dfrac{7}{4}\) | M1 | oe |
| their \(1\dfrac{3}{4} \div 2\) or their \(\dfrac{7}{4} \div 2\) | M1dep | oe |
| \(\dfrac{7}{8}\) | A1 | oe fraction |
| Alternative method 4 | ||
| \((1.25 - 0.5 =)\) 0.75 or \((1.25 + 0.5 =)\) 1.75 | M1 | accept equivalent in percentages but must see % sign |
| \((0.5 + 0.75 \div 2 =)\) 0.875 or \((1.25 - 0.75 \div 2 =)\) 0.875 or \(\left(\dfrac{1.25 + 0.5}{2} =\right)\) 0.875 or 87.5% | M1dep | 0.875 must be correct accept equivalent in percentages but must see % sign |
| \(\dfrac{7}{8}\) | A1 | oe fraction |
| Alternative method 5 | ||
| Positions of \(\dfrac{1}{2}\) and \(1\dfrac{1}{4}\) correctly marked on line or correct midpoint marked on line | M1 | if more points are marked, labels of \(\dfrac{1}{2}\) and \(1\dfrac{1}{4}\) must be given or indicated mark intention in terms of exact position accept decimals or equivalent fractions |
| Correct midpoint marked on line and \(\dfrac{3}{4}\) marked as \(\dfrac{6}{8}\) and 1 marked as \(\dfrac{8}{8}\) | M1dep | oe fractions with common denominator > 4 |
| \(\dfrac{7}{8}\) | A1 | oe fraction |
Additional guidance
In alternative method 5: \(\dfrac{1}{4}\) marked at \(1\dfrac{1}{4}\) is sufficient for \(1\dfrac{1}{4}\)
In all schemes, award of M1dep means that M2 is awarded
Use the scheme that gives the greatest number of marks – ignore errors in the scheme(s) you do not use