AQA C1 2014 June — Question 2 4 marks

Exam BoardAQA
ModuleC1 (Core Mathematics 1)
Year2014
SessionJune
Marks4
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Mark schemeDownload PDF ↗
TopicIndices and Surds
TypeRationalize denominator simple
DifficultyModerate -0.8 This is a straightforward rationalizing the denominator question requiring division of surds. Students must divide (15 + 7√3) by (9 + 5√3), multiply by the conjugate, and simplify—a standard C1 technique with no conceptual challenges beyond careful algebraic manipulation.
Spec1.02b Surds: manipulation and rationalising denominators
2 A rectangle has length \(( 9 + 5 \sqrt { 3 } ) \mathrm { cm }\) and area \(( 15 + 7 \sqrt { 3 } ) \mathrm { cm } ^ { 2 }\).
Find the width of the rectangle, giving your answer in the form \(( m + n \sqrt { 3 } ) \mathrm { cm }\), where \(m\) and \(n\) are integers.
[0pt] [4 marks]
Question 2:
AnswerMarks Guidance
AnswerMark Guidance
Width \(= \frac{15 + 7\sqrt{3}}{9 + 5\sqrt{3}}\)M1 Divide area by length
\(\times \frac{9 - 5\sqrt{3}}{9 - 5\sqrt{3}}\)M1 Multiply by conjugate surd
Numerator: \(135 - 75\sqrt{3} + 63\sqrt{3} - 105 = 30 - 12\sqrt{3}\)A1
Denominator: \(81 - 75 = 6\)A1
Width \(= 5 - 2\sqrt{3}\)A1 \(m=5\), \(n=-2\) 
# Question 2:
| Answer | Mark | Guidance |
|--------|------|----------|
| Width $= \frac{15 + 7\sqrt{3}}{9 + 5\sqrt{3}}$ | M1 | Divide area by length |
| $\times \frac{9 - 5\sqrt{3}}{9 - 5\sqrt{3}}$ | M1 | Multiply by conjugate surd |
| Numerator: $135 - 75\sqrt{3} + 63\sqrt{3} - 105 = 30 - 12\sqrt{3}$ | A1 | |
| Denominator: $81 - 75 = 6$ | A1 | |
| Width $= 5 - 2\sqrt{3}$ | A1 | $m=5$, $n=-2$ |

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2 A rectangle has length $( 9 + 5 \sqrt { 3 } ) \mathrm { cm }$ and area $( 15 + 7 \sqrt { 3 } ) \mathrm { cm } ^ { 2 }$.\\
Find the width of the rectangle, giving your answer in the form $( m + n \sqrt { 3 } ) \mathrm { cm }$, where $m$ and $n$ are integers.\\[0pt]
[4 marks]

\hfill \mbox{\textit{AQA C1 2014 Q2 [4]}}
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Q1 13 Q2 4 Q3 12 Q4 7 Q5 7 Q6 12 Q7 14 Q8 6