4MA0 Higher June 2011 Paper 4 Q21

EdexcelHigherOld spec6 marksForming Equations Solving Quadratics

21 Clare buys some shares for $50$x$.
Later, she sells the shares for $(600 + 5$x$).
She makes a profit of $x$%

(a) Show that $x^2 + 90x - 1200 = 0$ (3)

(b) Solve $x^2 + 90x - 1200 = 0$
Find the value of $x$ correct to 3 significant figures. (3)

Mark scheme (a)

Scheme	Marks
$\dfrac{(600 + 5x) - 50x}{50x} \times 100 = x$ oe or $50x \times \left[1 + \dfrac{x}{100}\right] = 600 + 5x$ oe or $\left(\dfrac{(600 + 5x)}{50x} - 1\right) \times 100 = x$ oe	M1
$100(600 + 5x - 50x) = 50x^2$ oe or $5000x + 50x^2 = 60000 + 500x$ or $(600 + 5x - 50x) \times 100 = 50x^2$	M1 dep
$2(600 - 45x) = x^2$ oe (but not ans) or $x^2 = 1200 - 90x$ or $1200 - 90x = x^2$	A1
	(3)

Notes

M1: $\dfrac{\text{actual profit}}{\text{original}} \times 100 = x$

M1 dep: (removing denominator)

A1: reducing to $1x^2$
dep on M2

Apart from questions 5b, 8, 15d, 20b, 21b, 23, 24b (where the mark scheme states otherwise) the correct answer, unless clearly obtained by an incorrect method, should be taken to imply correct working.

Mark scheme (b)

Scheme	Marks
$x = \dfrac{-90 \pm \sqrt{90^2 - 4 \times 1 \times -1200}}{2}$	M1
$x = \dfrac{-90 \pm \sqrt{8100 + 4800}}{2}$	M1
11.789........	A1
	(3)
	(Total 6 marks)

Notes

M1: condone 1 sign error (working can be seen in part a))
sign error = +90 instead of − 90 or +1200 instead of −1200

A1: dep on M2 awrt 11.8 (ignore negative root).