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Higher June 2025 Paper 1 Q16
16
(a) Rationalise the denominator of \(\dfrac{35}{\sqrt{7}}\)
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)
Mark scheme (a)| 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 \(\sqrt{175}\) |
Mark scheme (b)| 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 \(7 + 5\sqrt{3}\)
Higher November 2024 Paper 1 Q16
16
(a) Rationalise the denominator of \(\dfrac{15}{\sqrt{5}}\)
Give your answer in its simplest form. (2)
(b) Write \(\dfrac{\sqrt{75} - 2}{1 + 2\sqrt{3}}\) in the form \(\dfrac{a - b\sqrt{3}}{c}\) where \(a\), \(b\) and \(c\) are integers. (4)
Mark scheme (a)| Answer | Mark | Mark scheme |
|---|
| \(3\sqrt{5}\) | M1 | for \(\dfrac{15}{\sqrt{5}} \times \dfrac{\sqrt{5}}{\sqrt{5}}\) or \(\dfrac{15}{\sqrt{5}} \times \dfrac{-\sqrt{5}}{-\sqrt{5}}\) |
| A1 | for \(3\sqrt{5}\) or \(\sqrt{45}\) |
Mark scheme (b)| Answer | Mark | Mark scheme |
|---|
| \(\dfrac{32 - 9\sqrt{3}}{11}\) | M1 | (indep) for writing \(\sqrt{75}\) as \(5\sqrt{3}\) |
| M1 | for method to rationalise the denominator,
eg \(\dfrac{\sqrt{75} - 2}{1 + 2\sqrt{3}} \times \dfrac{1 - 2\sqrt{3}}{1 - 2\sqrt{3}}\) or \(\dfrac{5\sqrt{3} - 2}{1 + 2\sqrt{3}} \times \dfrac{1 - 2\sqrt{3}}{1 - 2\sqrt{3}}\) |
| M1 | (dep on previous M1) for expanding terms, condone one error in numerator or denominator
eg \(\dfrac{\sqrt{75} - 2\sqrt{75}\sqrt{3} - 2 + 4\sqrt{3}}{1 - 2\sqrt{3} + 2\sqrt{3} - 4\sqrt{3}\sqrt{3}}\) or \(\dfrac{5\sqrt{3} - 10\sqrt{3}\sqrt{3} - 2 + 4\sqrt{3}}{1 - 2\sqrt{3} + 2\sqrt{3} - 4\sqrt{3}\sqrt{3}}\) |
| A1 | for \(\dfrac{32 - 9\sqrt{3}}{11}\) oe eg \(\dfrac{-32 + 9\sqrt{3}}{-11}\) |
Additional guidance
This mark can be awarded whenever this is seen, which might be later in the process.
Accept \(a = 32\), \(b = 9\), \(c = 11\)
No questions match these filters.
