Higher November 2019 Paper 2 Q18

AQACurrent spec4 marksForming Equations Solving Quadratics

18 Kate has the following question for homework.

Homework question: The net of a box is made by cutting four squares from a piece of cardboard. The cardboard is a rectangle with width x cm and length (x + 5) cm. Each square has side length 10 cm. The area of the net is 1000 cm². Work out the value of x. The diagram shows the rectangle with a 10 cm square cut from each corner

(a) Show that Kate can form the equation \(\quad x^2 + 5x - 1400 = 0\) [3 marks]

(b) Kate correctly factorises the equation to get \(\quad (x + 40)(x - 35) = 0\)

Her answer to the homework question is \(\quad x = -40\) or \(x = 35\)

Is her answer correct?

Tick a box.

Give a reason for your answer. [1 mark]

Mark scheme (a)

Answer	Mark	Comments
Alternative method 1: large rectangle − 4 squares
\(x(x + 5)\)	M1
\(x^2 + 5x - 400 = 1000\) or \(x^2 + 5x - 400 - 1000 = 0\) or \(x^2 + 5x = 1000 + 400\) with M1 seen	M1dep	400 may be seen as \(4 \times 10^2\) or \(4 \times 100\) oe equation with brackets expanded and 400 and 1000 seen
\(x^2 + 5x - 1400 = 0\) with M2 seen	A1	must have = 0
Alternative method 2: three vertical rectangles
\((x + 5)(x - 20)\) or \((2 \times)10(x - 15)\)	M1	\((x - 20)\) may be seen as \((x - 10 - 10)\) \((x - 15)\) may be seen as \((x + 5 - 10 - 10)\)
\(x^2 - 20x + 5x - 100 + 20x - 300 = 1000\) or \(x^2 - 15x - 100 + 20x - 300 = 1000\) with M1 seen	M1dep	oe equation with brackets expanded and 100 and 300 and 1000 seen allow 150 seen twice for 300
\(x^2 + 5x - 1400 = 0\) with M2 seen	A1	must have = 0
Alternative method 3: three horizontal rectangles
\(x(x - 15)\) or \((2 \times)10(x - 20)\)	M1	\((x - 20)\) may be seen as \((x - 10 - 10)\) \((x - 15)\) may be seen as \((x + 5 - 10 - 10)\)
\(x^2 - 15x + 20x - 400 = 1000\) with M1 seen	M1dep	oe equation with brackets expanded and 400 and 1000 seen allow 200 seen twice for 400
\(x^2 + 5x - 1400 = 0\) with M2 seen	A1	must have = 0
Alternative method 4: central rectangle + four outer rectangles
\((x - 15)(x - 20)\) or \((2 \times)10(x - 15)\) or \((2 \times)10(x - 20)\)	M1	\((x - 20)\) may be seen as \((x - 10 - 10)\) \((x - 15)\) may be seen as \((x + 5 - 10 - 10)\)
\(x^2 - 20x - 15x + 300 + 20x - 300 + 20x - 400 = 1000\) or \(x^2 - 35x + 300 + 20x - 300 + 20x - 400 = 1000\) with M1 seen	M1dep	oe equation with brackets expanded and 300 seen twice and 400 and 1000 seen allow 150 seen twice for one of the 300s allow 200 seen twice for 400
\(x^2 + 5x - 1400 = 0\) with M2 seen	A1	must have = 0

Additional guidance

If 1st M1 seen award M1 even if expression is not subsequently used
For M1 allow multiplication signs eg \(x \times (x + 5)\)	M1
\(x(x + 5) = x^2 + 5x\) \(1000 + 400 = 1400\) \(x^2 + 5x = 1400\) (previous line shows 1000 and 400) \(x^2 + 5x - 1400 = 0\)	M1 M1 A1
\(x(x + 5) = x^2 + 5x\) \(x^2 + 5x = 1400\) (equation does not have 1000 and 400) \(x^2 + 5x - 1400 = 0\)	M1 M0 A0
Only equation seen is \(x^2 + 5x - 1400 = 0\) the maximum mark is M1

Mark scheme (b)

Answer	Mark	Comments
No and valid reason	B1	eg No and \(x\) cannot be negative (in this context)

Additional guidance

If neither box is ticked condone if No is clearly stated in working lines
Yes or both boxes ticked	B0
Allow ‘it’ to represent \(x\)
No and \(x\) is (only) 35	B1
No and it cannot be −40	B1
No and the width would be negative	B1
No and the width should be positive	B1
No she put −40	B1
No and you can’t have two answers	B0
No and the answers are too big	B0
No and it should be 40 (and −35)	B0