Higher June 2022 Unit 1 Q13
13. Marian is competing in a race.
The race is \((6x + 5)\) miles long.
She completes the race in \(x\) hours.
Her average speed during the race is \((2x + 3)\) miles per hour.
Calculate how long Marian takes to complete the race.
You must use an algebraic method (not trial and improvement). [6]
From the formula list: The solutions of \(ax^2 + bx + c = 0\) where \(a \neq 0\) are given by \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
| Scheme | Marks | Comments |
|---|---|---|
| \(\dfrac{6x + 5}{x} = 2x + 3\) or \(\dfrac{6x + 5}{2x + 3} = x\) or \(6x + 5 = x\,(2x + 3)\) | B1 | Correct use of ‘speed = distance / time’, using three expressions. May be implied by further working. |
| \(6x + 5 = 2x^2 + 3x\) or \(6x + 5 - 2x^2 - 3x\ [= 0]\) or \(2x^2 + 3x - 6x - 5\ [= 0]\) or equivalent | M1 | Expanding brackets FT ‘their equation’ if of equivalent difficulty |
| \(2x^2 - 3x - 5\ (=0)\) | A1 | Collecting like terms and re-arranging quadratic equation. Ignore presence of a denominator (provided correct). |
| \((x + 1)(2x - 5)\ (= 0)\) | B2 | B1 for \((x\ldots..1)(2x\ldots.5)\) FT their quadratic equation, provided of equivalent difficulty. |
| (Marian takes) 2.5 (hours) or equivalent | B1 | Mark final answer. FT provided first B1 awarded and an algebraic method used to solve quadratic equation. Ignore negative solution (\(x = -1\)). Strict FT ‘their derived brackets’. No marks for a trial and improvement method. No marks for starting with \((2x + 3)(6x + 5)[ = 0]\). |
Alternative method to solve quadratic equation
| Scheme | Marks | Comments |
|---|---|---|
| (\(x\) =) \(\dfrac{3 \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)}\) \(x = \dfrac{3 \pm \sqrt{49}}{4}\) | M1 A1 | FT their quadratic equation, provided of equivalent difficulty. Allow one error, in sign or substitution, but not in formula |
| (Marian takes) 2.5 (hours) or equivalent | A1 | Mark final answer. FT provided first B1 awarded. Ignore negative solution (\(x = -1\)). |