Higher June 2019 Paper 4 Q5
5 Bennie is 7 years older than Ayesha.
Chloe is twice as old as Bennie.
The sum of their three ages is 57.
Work out the ages of Ayesha, Bennie and Chloe. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Ayesha 9 Bennie 16 Chloe 32 | 6 | allow any letter providing use is consistent this method assumes Ayesha’s age = \(a\) B4 for \(a + a + 7 + 2(a + 7) = 57\) or better OR B1 for [\(b\) =] \(a + 7\) oe e.g. \(a = b - 7\) B1 for \(c = 2b\) oe e.g. \(\frac{c}{2} = b\) or [\(c\) =] \(2(a + 7)\) B1 for their ‘\(a\)’ + their ‘\(b\)’ + their ‘\(c\)’ = 57 e.g. \(a + b + c = 57\) must be algebraic AND M1FT for correctly solving their linear equation in one variable e.g. \(4a = 36\) and \(a = 9\) AND M1 for substituting their \(a\) into \(b = a + 7\) and \(c = 2b\) e.g. \(a = 8\), \(b = 15\) and \(c = 30\) implied by their answer which must be integers see appendix for other methods mark working first, if 0 scored then SC2 for 2 answers correct in the correct place or SC1 for 1 answer correct in the correct place or if 1 scored then award the better of 1 or SC2 for 2 answers correct in the correct place to a maximum of 5 marks | |
Appendix: Question 5 alternatives
e.g. assumes Bennie’s age = \(b\)
B4 for \(b - 7 + b + 2b = 57\) or better
OR
B1 for [\(a\) =] \(b - 7\) oe e.g. \(b = a + 7\)
B1 for [\(c\) =] \(2b\) oe e.g. \(\frac{c}{2} = b\) or \(c = 2(a + 7)\)
B1 for their ‘\(a\)’ + their ‘\(b\)’ + their ‘\(c\)’ = 57 e.g. \(a + b + c = 57\) must be algebraic
AND
M1FT for correctly solving their linear equation in one variable e.g. \(4b = 64\) and \(b = 16\)
AND
M1 for substituting their \(b\) into \(a = b - 7\) and \(\frac{c}{2} = b\) e.g. \(a = 8\), \(b = 15\) and \(c = 30\) implied by their answer which must be integers
e.g. assumes Chloe’s age = \(c\)
B4 for \(\frac{c}{2} - 7 + \frac{c}{2} + c = 57\) or better
OR
B1 for [\(a\) =] \(\frac{c}{2} - 7\) oe e.g. \(c = 2(a + 7)\)
B1 for [\(b\) =] \(\frac{c}{2}\) oe e.g. \(c = 2b\) or \(b = (a + 7)\)
B1 for their ‘\(a\)’ + their ‘\(b\)’ + their ‘\(c\)’ = 57 e.g. \(a + b + c = 57\) must be algebraic
AND
M1FT for correctly solving their linear equation in one variable e.g. \(2c = 64\) and \(c = 32\)
AND
M1 for substituting their \(c\) into \(b = \frac{c}{2}\) and \(a = \frac{c}{2} - 7\) e.g. \(a = 8\), \(b = 15\) and \(c = 30\) implied by their answer which must be integers