Higher June 2024 Paper 4 Q3
3 Four numbers are written, in ascending order, as algebraic expressions.
\(a \qquad a + b \qquad a + 2b \qquad 3a - b\)
The mean of these four numbers is 27.
The range of these four numbers is 24.
Find the value of \(a\) and the value of \(b\).
You must show your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [\(a =\)] 15.6 [\(b =\)] 7.2 with correct working | 5 | B4 for one correct answer with correct working | “Correct working” requires evidence of at least M1M1 for the two equations and M1 for some evidence of solving them |
| OR | |||
| M1 for \(\dfrac{a + a + b + a + 2b + 3a - b}{4} = 27\) oe | e.g. \(6a + 2b = 108\) | ||
| M1 for \(3a - b - a = 24\) oe | e.g. \(2a - b = 24\) | ||
| M1 for equating coefficients of one variable for their linear equations | dependant on two linear equations e.g. \(6a - 3b = 72\) or \(4a - 2b = 48\) or \(3a + b = 54\) | ||
| M1 for correct method to eliminate one variable for their original linear equations | dependant on two linear equations e.g. \(5b = 36\) Allow one numerical error in each step of solving their equations A sign error is not an arithmetic error | ||
| If 0, M1 or M2 scored, instead award SC3 for answers 15.6 and 7.2 with no working or insufficient working If 0 or M1 scored, instead award SC2 for \(a = 7.2\) and \(b = 15.6\) with no working or insufficient working If 0 scored, instead award SC1 for two answers which satisfy one of the original conditions | Substitution method: M1 and M1 for the two equations M1 dependant on two linear equations and for rearranging one equation to make one variable the subject e.g. \(a = \frac{24 + b}{2}\) M1 for substituting it into their other equation Trials (need to see the mean or total, and the range evaluated for each): M1 for each correct trial up to a maximum of 3 After three correct trials, correct final answers score 5 | ||