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Maths I

About the exam

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Example - 2010, Paper 1, Question 7

Question

A straight line has equation y=mx + c, where m and c are constants

a) The point (2,7) lies on the line. Write down an equation in m and c to illustrate this information.

b) A second point (4, 17) also lies on this line. Write down another equation in m and c to illustrate this information.

c) Hence calculate the values of m and c.

d) Write down the gradient of this line.

Answer
  • Interpret: the style of the question, in four parts, usually indicates that the results of part a) can be used to solve part b) and so on.
  • Interpret: Parts a) and b) and Knowledge and Understanding questions. The answers to a) and b) will be used to solve part c) and d), which are Reasoning and Enquiry.
  • Select: The expected method involves "solving simaltaneous equations". The answer to part d) is obtained from the solution to part c).
  • a) 2m + c = 7
  • b) 4m + c = 17
  • Note – Marks can only be awarded for equations in terms of m and c.

c) By subtracting the equation obtained in a) from the equation obtained in b)

  • 2m = 10 (1 mark RE)
  • m = 5 (1 mark RE)

And by substituting m=5 into the equation in (a) or (b).

  • c = -3 (1 mark RE)

Note – Other methods are possible and acceptable.

  • In part c) it was found that m = 5, and as m is the gradient, the gradient of the straight line must be 5.

Example, paper 1, question 8

Question

a) Simplify √2 x √18

b) Simplify √2 + √18

c) Hence show that {{\sqrt 2 \times \sqrt 18} \over \sqrt 2 + \sqrt 18}{=}{{3 \sqrt 2} \over 4}

Answer
  • Interpret: this is a three part Knowledge and Understanding question. Parts a) and b) are used to simplify c). The topic area is Surds.
  • Select: The techniques used are "simplifying surds" and "rationalising a surd denominator".
  • Implement and communicate: See the solution below.
  • a) √2 x √18
  • = √36
  • = 6 (1 mark KU)
  • b) √2 + √18
  • = √2 + √9 x 2
  • = √2 + 3√2
  • = 4√2 (1 mark KU)
  • {{\sqrt 2 \times \sqrt 18} \over \sqrt 2 + \sqrt 18}
  • = {6 \over {4 \sqrt 2}}
  • = {3 \over {2 \sqrt 2}} {\times} {\sqrt 2 \over \sqrt 2}}
  • = {{3 \sqrt 2} \over 4}
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