Higher June 2025 Paper 1R Q15
15
(a) Solve \(\dfrac{5a + 8}{3} - \dfrac{2a + 5}{4} = 23\)
Show clear algebraic working. (4)
Show clear algebraic working. (4)
(b) Express \(\left(\dfrac{\sqrt{y}}{3}\right)^{-1}\) in the form \(cy^n\) where \(c\) and \(n\) are numbers to be found. (2)
| Scheme | Marks |
|---|---|
eg \(12 \times \dfrac{5a + 8}{3} - 12 \times \dfrac{2a + 5}{4} = 12 \times 23\) or eg \(4(5a + 8) - 3(2a + 5) = 12 \times 23 (= 276)\) or eg \(\dfrac{4(5a + 8)}{12} - \dfrac{3(2a + 5)}{12}(= 23)\) or eg \(\dfrac{4(5a + 8) - 3(2a + 5)}{12}(= 23)\) | M1 |
| eg \(20a + 32 - 6a - 15 = 12 \times 23 (= 276)\) oe or \(14a + 17 = 276\) | M1 |
| eg \(20a - 6a = 276 - 32 + 15\) oe or \(14a = 259\) | M1 |
| Working required Answer: 18.5 | A1 |
| (4) |
Notes
M1: for clear intention to multiply all terms by 12 or a multiple of 12
or
to express LHS as two fractions over 12 or a multiple of 12
or as a single fraction with a denominator of 12 or a multiple of 12
(If expanded numerator, allow one sign error or one numerical error but not both)
Accept
\(\dfrac{20a + 32}{12} - \dfrac{6a + 15}{12}(= 23)\) or \(\dfrac{20a + 32 - 6a + 15}{12}(= 23)\) or
\(\dfrac{20a + 32 - 6a + 15}{12}(= 23)\)
M1: ft for expanding brackets and multiplying both sides by denominator with no more than one error in total leading to a linear equation
Accept a linear equation leading to
\(14a - 17 = 276\) oe or \(14a + 47 = 276\) oe or
\(26a + 17 = 276\)
This mark implies the previous M mark if not already awarded
M1: ft dep on previous M1 for correctly rearranging terms in \(a\) on one side and number terms on the other side
A1: oe dep on M2 eg \(\dfrac{259}{14}\) or \(\dfrac{37}{2}\)
| Scheme | Marks |
|---|---|
eg \(\dfrac{3}{\sqrt{y}}\left(= \dfrac{3\sqrt{y}}{y}\right)\) or \(\dfrac{3}{y^{0.5}}\) or \(\dfrac{3}{y^{\frac{1}{2}}}\) or \(\left(\dfrac{y^{0.5}}{3}\right)^{-1}\) or \(\left(\dfrac{y^{\frac{1}{2}}}{3}\right)^{-1}\) oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: \(3y^{-0.5}\) | A1 |
| (2) | |
| (6 marks) |
Notes
M1: for a correct first step by applying one of the following index rules
\(\sqrt{x} = x^{\frac{1}{2}} = x^{0.5}\) or \(\left(\dfrac{a}{b}\right)^{-1} = \left(\dfrac{b}{a}\right)\)
A1: oe eg \(3y^{-\frac{1}{2}}\), accept \(c = 3\) and \(n = -0.5\) oe