Higher June 2025 Paper 1 Q16
16
(a) Rationalise the denominator of \(\dfrac{35}{\sqrt{7}}\)
Give your answer in its simplest form. (2)
Give your answer in its simplest form. (2)
\(\dfrac{\sqrt{27} - 1}{2 - \sqrt{3}}\) can be written in the form \(a + b\sqrt{3}\) where \(a\) and \(b\) are integers.
(b) Work out the value of \(a\) and the value of \(b\). (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(5\sqrt{7}\) | M1 | for \(\dfrac{35}{\sqrt{7}} \times \dfrac{\sqrt{7}}{\sqrt{7}}\ \left(= \dfrac{35\sqrt{7}}{7}\right)\) or \(\dfrac{35}{\sqrt{7}} \times \dfrac{-\sqrt{7}}{-\sqrt{7}}\ \left(= \dfrac{-35\sqrt{7}}{-7}\right)\) |
| A1 | for \(5\sqrt{7}\) or \(\sqrt{175}\) |
| Answer | Mark | Mark scheme |
|---|---|---|
| \(a = 7\), \(b = 5\) | B1 | for \(\sqrt{27} = 3\sqrt{3}\) or \(2\sqrt{27} = 6\sqrt{3}\) |
| P1 | for process to rationalise the denominator, eg \(\dfrac{\sqrt{27} - 1}{2 - \sqrt{3}} \times \dfrac{2 + \sqrt{3}}{2 + \sqrt{3}}\) or \(\dfrac{3\sqrt{3} - 1}{2 - \sqrt{3}} \times \dfrac{2 + \sqrt{3}}{2 + \sqrt{3}}\) oe | |
| P1 | (dep on previous P1) for expanding terms, condone one error in numerator or denominator, eg \(\dfrac{2\sqrt{27} + \sqrt{27}\sqrt{3} - 2 - \sqrt{3}}{4 + 2\sqrt{3} - 2\sqrt{3} - \sqrt{3}\sqrt{3}}\) or \(\dfrac{6\sqrt{3} + 3\sqrt{3}\sqrt{3} - 2 - \sqrt{3}}{4 + 2\sqrt{3} - 2\sqrt{3} - \sqrt{3}\sqrt{3}}\) or \(6\sqrt{3} + 9 - 2 - \sqrt{3}\) oe | |
| A1 | for \(a = 7\), \(b = 5\) |
Additional guidance
B1 can be awarded whenever this is seen, which might be later in the process.
Accept \(7 + 5\sqrt{3}\)