# IBDP Maths AI: Topic : SL 1.8: Systems of linear equations: IB style Questions SL Paper 1

## Question

A store sells bread and milk. On Tuesday, 8 loaves of bread and 5 litres of milk were sold for $21.40. On Thursday, 6 loaves of bread and 9 litres of milk were sold for$23.40.

If $$b =$$ the price of a loaf of bread and $$m =$$ the price of one litre of milk, Tuesday’s sales can be written as $$8b + 5m = 21.40$$.

Using simplest terms, write an equation in b and m for Thursday’s sales.[2]

a.

Find b and m.[2]

b.

Draw a sketch, in the space provided, to show how the prices can be found graphically.

[2]

c.

## Markscheme

Thursday’s sales, $$6b + 9m = 23.40$$     (A1)

$$2b + 3m = 7.80$$     (A1)     (C2)[2 marks]

a.

$$m = 1.40$$     (accept 1.4)     (A1)(ft)

$$b = 1.80$$     (accept 1.8)     (A1)(ft)

Award (A1)(d) for a reasonable attempt to solve by hand and answer incorrect.     (C2)[2 marks]

b.

(A1)(A1)(ft)

(A1) each for two reasonable straight lines. The intersection point must be approximately correct to earn both marks, otherwise penalise at least one line.

Note: The follow through mark is for candidate’s line from (a).     (C2)[2 marks]

c.

## Question

The number of cells, C, in a culture is given by the equation $$C = p \times 2^{0.5t} + q$$, where t is the time in hours measured from 12:00 on Monday and p and q are constants.

The number of cells in the culture at 12:00 on Monday is 47.

The number of cells in the culture at 16:00 on Monday is 53.

Use the above information to write down two equations in p and q ;[2]

a.

Use the above information to calculate the value of p and of q ;[2]

b.

Use the above information to find the number of cells in the culture at 22:00 on Monday.[2]

c.

## Markscheme

p + q = 47     (A1)

4p + q = 53     (A1)     (C2)[2 marks]

a.

Reasonable attempt to solve their equations     (M1)

p = 2, q = 45     (A1)     (C2)

Note: Accept only the answers p = 2, q = 45.[2 marks]

b.

C = 2 × 20.5(10) + 45     (M1)

C = 109     (A1)(ft)     (C2)

Note: Award (M1) for substitution of 10 into the formula with their values of p and q.[2 marks]

c.

## Question

Consider the function $$f(x) = p{(0.5)^x} + q$$ where p and q are constants. The graph of f (x) passes through the points $$(0,\,6)$$ and $$(1,\,4)$$ and is shown below.

Write down two equations relating p and q.[2]

a.

Find the value of p and of q.[2]

b.

Write down the equation of the horizontal asymptote to the graph of f (x).[2]

c.

## Markscheme

p + q = 6     (A1)

0.5p + q = 4     (A1)     (C2)

Note: Accept correct equivalent forms of the equations.[2 marks]

a.

p = 4, q = 2     (A1)(A1)(ft)     (C2)

Notes: If both answers are incorrect, award (M1) for attempt at solving simultaneous equations.[2 marks]

b.

y = 2     (A1)(A1)(ft)     (C2)

Notes: Award (A1) for “y = a constant”, (A1)(ft) for 2. Follow through from their value for q as long as their constant is greater than 2 and less than 6.

An equation must be seen for any marks to be awarded.[2 marks]

c.

## Question

$$10 000$$ people attended a sports match. Let $$x$$ be the number of adults attending the sports match and $$y$$ be the number of children attending the sports match.

Write down an equation in $$x$$ and $$y$$ .[1]

a.

The cost of an adult ticket was $$12$$ Australian dollars (AUD). The cost of a child ticket was $$5$$ Australian dollars (AUD).

Find the total cost for a family of 2 adults and 3 children.[2]

b.

The total cost of tickets sold for the sports match was $$108800{\text{ AUD}}$$.

Write down a second equation in $$x$$ and $$y$$ .[1]

c.

Write down the value of $$x$$ and the value of $$y$$ .[2]

d.

## Markscheme

$$x + y = 10000$$     (A1)     (C1)[1 mark]

a.

$$2 \times 12 + 3 \times 5$$     (M1)

$$39{\text{ }}(39.0{\text{, }}39.00)$$ (AUD)     (A1)     (C2)[2 marks]

b.

$$12x + 5y = 108800$$     (A1)     (C1)[1 mark]

c.

$$x = 8400$$, $$y = 1600$$     (A1)(ft)(A1)(ft)     (C2)

Notes: Follow through from their equations. If $$x$$ and $$y$$ are both incorrect then award (M1) for attempting to solve simultaneous equations.[2 marks]

d.

## Question

A function f (x) = p×2x + q is defined by the mapping diagram below.

Find the value of

(i) p ;

(ii) q .[3]

a.

Write down the value of r .[1]

b.

Find the value of s .[2]

c.

## Markscheme

(i) 2p + q = 11 and 4p + q = 17     (M1)

Note: Award (M1) for either two correct equations or a correct equation in one unknown equivalent to 2p = 6 .

p = 3     (A1)

(ii) q = 5     (A1)     (C3)

Notes: If only one value of p and q is correct and no working shown, award (M0)(A1)(A0).[3 marks]

a.

r = 8     (A1)(ft)     (C1)

Note: Follow through from their answers for p and q irrespective of whether working is seen.[1 mark]

b.

3 × 2s + 5 = 197     (M1)

Note: Award (M1) for setting the correct equation.

s = 6     (A1)(ft)     (C2)

Note: Follow through from their values of p and q.[2 marks]

c.

## Question

The equation of a line L1 is $$2x + 5y = −4$$.

Write down the gradient of the line L1.[1]

a.

A second line L2 is perpendicular to L1.

Write down the gradient of L2.[1]

b.

The point (5, 3) is on L2.

Determine the equation of L2.[2]

c.

Lines L1 and L2 intersect at point P.

Using your graphic display calculator or otherwise, find the coordinates of P.[2]

d.

## Markscheme

$$\frac{-2}{{5}}$$     (A1)     (C1)

a.

$$\frac{5}{{2}}$$     (A1)(ft)     (C1)

b.

$$3 = \frac{5}{2} \times 5 + c$$     (M1)

Notes: Award (M1) for correct substitution of their gradient into equation of line. Follow through from their answer to part (b).

$$y = \frac{5}{2}x – \frac{19}{2}$$     (A1)(ft)

OR

$$y – 3 = \frac{5}{2}(x – 5)$$     (M1)(A1)(ft)     (C2)

Notes: Award (M1) for correct substitution of their gradient into equation of line. Follow through from their answer to part (b).

c.

(3, −2)     (A1)(ft)(A1)(ft)     (C2)

Notes: If parentheses not seen award at most (A0)(A1)(ft). Accept x = 3, y = −2. Follow through from their answer to part (c), even if no working is seen. Award (M1)(A1)(ft) for a sensible attempt to solve $$2x + 5y = −4$$ and their $$y = \frac{5}{2}x – \frac{19}{2}$$ or equivalent, simultaneously.

d.

## Question

A quadratic function $$f:x \mapsto a{x^2} + b$$, where $$a$$ and $$b \in \mathbb{R}$$ and $$x \geqslant 0$$, is represented by the mapping diagram.

Using the mapping diagram, write down two equations in terms of $$a$$ and $$b$$.[2]

a.

Solve the equations to find the value of

(i)     $$a$$;

(ii)     $$b$$.[2]

b.

Find the value of $$c$$.[2]

c.

## Markscheme

$$a{(1)^2} + b = – 9$$     (A1)

$$a{(3)^2} + b = 119$$     (A1)     (C2)

Note: Accept equivalent forms of the equations.[2 marks]

a.

(i)     $$a = 16$$     (A1)(ft)

(ii)     $$b = – 25$$     (A1)(ft)     (C2)

Note: Follow through from part (a) irrespective of whether working is seen.

If working is seen follow through from part (i) to part (ii).[2 marks]

b.

$$16{c^2} – 25 = 171$$     (M1)

Note: Award (M1) for correct quadratic with their $$a$$ and $$b$$ substituted.

$$c = 3.5$$     (A1)(ft)     (C2)

Note: Accept $$x$$ instead of $$c$$.

Award (A1) only, for an answer of $$\pm 3.5$$ with or without working.[2 marks]

c.

## Question

The graph of the quadratic function $$f(x) = a{x^2} + bx + c$$ intersects the y-axis at point A (0, 5) and has its vertex at point B (4, 13).

Write down the value of $$c$$.[1]

a.

By using the coordinates of the vertex, B, or otherwise, write down two equations in $$a$$ and $$b$$.[3]

b.

Find the value of $$a$$ and of $$b$$.[2]

c.

## Markscheme

5     (A1)     (C1)[1 mark]

a.

at least one of the following equations required

$$a{(4)^2} + 4b + 5 = 13$$

$$4 = – \frac{b}{{2a}}$$

$$a{(8)^2} + 8b + 5 = 5$$     (A2)(A1)     (C3)

Note: Award (A2)(A0) for one correct equation, or its equivalent, and (C3) for any two correct equations.

The equation $$a{(0)^2} + b(0) = 5$$ earns no marks.[3 marks]

b.

$$a = – \frac{1}{2},{\text{ }}b = 4$$     (A1)(ft)(A1)(ft)     (C2)

Note: Follow through from their equations in part (b), but only if their equations lead to unique solutions for $$a$$ and $$b$$.[2 marks]

c.

## Question

An iron bar is heated. Its length, $$L$$, in millimetres can be modelled by a linear function, $$L = mT + c$$, where $$T$$ is the temperature measured in degrees Celsius (°C).

At 150°C the length of the iron bar is 180 mm.

Write down an equation that shows this information.[1]

a.

At 210°C the length of the iron bar is 181.5 mm.

Write down an equation that shows this second piece of information.[1]

b.

At 210°C the length of the iron bar is 181.5 mm.

Hence, find the length of the iron bar at 40°C.[4]

c.

## Markscheme

$$180 = 150m + c\;\;\;$$(or equivalent)     (A1)     (C1)

a.

$$181.5 = 210m + c\;\;\;$$(or equivalent)     (A1)     (C1)

b.

$$m = 0.25,{\text{ }}c = 176.25$$     (A1)(A1)(ft)

Note: Follow through from part (a) and part (b), irrespective of working shown.

$$L = 0.025(4) + 176.25$$     (M1)

Note: Award (M1) for substitution of their $$m$$, their $$c$$ and 40 into equation.

$$L = 177\;\;\;(177.25){\text{ (mm)}}$$     (A1)(ft)     (C4)

Note: Follow through, within part (c), from their $$m$$ and $$c$$ only if working shown.

c.

## Question

A liquid is heated so that after $$20$$ seconds of heating its temperature, $$T$$ , is $${25^ \circ }{\text{C}}$$ and after $$50$$ seconds of heating its temperature is $${37^ \circ }{\text{C}}$$ .

The temperature of the liquid at time $$t$$ can be modelled by $$T = at + b$$ , where $$t$$ is the time in seconds after the start of heating.

Using this model one equation that can be formed is $$20a + b = 25$$ .

Using the model, write down a second equation in $$a$$ and $$b$$ .[2]

a.

Using your graphic display calculator or otherwise, find the value of $$a$$ and of $$b$$ .[2]

b.

Use the model to predict the temperature of the liquid $$60{\text{ seconds}}$$ after the start of heating.[2]

c.

## Markscheme

$$50a + b = 37$$ (A1)(A1)     (C2)

Note: Award (A1) for $$50a + b$$ , (A1) for $$= 37$$ .

a.

$$a = 0.4$$, $$b = 17$$     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (M1) for attempt to solve their equations if this is done analytically. If the GDC is used, award (ft) even if no working seen.

b.

$$T = 0.4(60) + 17$$     (M1)

Note: Award (M1) for correct substitution of their values and $$60$$ into equation for $$T$$.

$$T = 41{\text{ }}{{\text{(}}^ \circ }{\text{C}})$$     (A1)(ft)     (C2)

Note: Follow through from their part (b).

c.

## Question

The length of one side of a rectangle is 2 cm longer than its width.

If the smaller side is x cm, find the perimeter of the rectangle in terms of x.[1]

a.

The length of one side of a rectangle is 2 cm longer than its width.

The perimeter of a square is equal to the perimeter of the rectangle in part (a).

Determine the length of each side of the square in terms of x.[1]

b.

The length of one side of a rectangle is 2 cm longer than its width.

The perimeter of a square is equal to the perimeter of the rectangle in part (a).

The sum of the areas of the rectangle and the square is $$2x^2 + 4x +1$$ (cm2).

(i) Given that this sum is 49 cm2, find x.

(ii) Find the area of the square.[4]

c.

## Markscheme

Unit penalty (UP) is applicable where indicated in the left hand column.

(UP) $${\text{P (rectangle)}} = 2x + 2(x + 2) = 4x + 4{\text{ cm}}$$     (A1)     (C1)

(UP) Simplification not required[1 mark]

a.

Unit penalty (UP) is applicable where indicated in the left hand column.

(UP) Side of square = (4x + 4)/4 = x + 1 cm     (A1)(ft)     (C1)[1 mark]

b.

(i) $$2x^2 + 4x + 1 = 49$$ or equivalent     (M1)

$$(x + 6)(x – 4) = 0$$

$$x = – 6$$ and $$4$$     (A1)

Note: award (A1) for the values or for correct factors

Choose $$x = 4$$     (A1)(ft)

Award (A1)(ft) for choosing positive value.     (C3)

(ii) $${\text{Area of square}} = 5 \times 5 = 25{\text{ c}}{{\text{m}}^2}$$     (A1)(ft)

Note: Follow through from both (b) and (c)(i).     (C1)[4 marks]

c.

## Question

Solve the following equation for x

$$3(2x +1) − 2(3 − x)=13$$.[2]

a.

Factorize the expression $$x^2 + 2x − 3$$.[2]

b.

Find the positive solution of the equation

$$x^2 + 2x − 6 = 0$$.[2]

c.

## Markscheme

$$6x+ 3 – 6 + 2x = 13$$     (M1)

$$8x = 16$$

$$x = 2$$     (A1)     (C2)[2 marks]

a.

$$(x + 3) (x – 1)$$     (A1)(A1)     (C2)[2 marks]

b.

$$x = 1.64575…$$

$$x = 1.65$$     (A2)

If formula is used award (M1)(A1) for correct solution. If graph is sketched award (M1)(A1) for correct solution.     (C2)[2 marks]

c.

## Question

Factorise the expression $${x^2} – kx$$ .[1]

a.

Hence solve the equation $${x^2} – kx = 0$$ .[1]

b.

The diagram below shows the graph of the function $$f(x) = {x^2} – kx$$ for a particular value of $$k$$.

Write down the value of $$k$$ for this function.[1]

c.

The diagram below shows the graph of the function $$f(x) = {x^2} – kx$$ for a particular value of $$k$$.

Find the minimum value of the function $$y = f(x)$$ .[3]

d.

## Markscheme

$$x(x – k)$$     (A1)     (C1)[1 mark]

a.

$$x = 0$$ or $$x = k$$     (A1)     (C1)

Note: Both correct answers only.[1 mark]

b.

$$k = 3$$     (A1)     (C1)[1 mark]

c.

$${\text{Vertex at }}x = \frac{{ – ( – 3)}}{{2(1)}}$$     (M1)

Note: (M1) for correct substitution in formula.

$$x = 1.5$$     (A1)(ft)

$$y = – 2.25$$     (A1)(ft)

OR

$$f'(x) = 2x – 3$$     (M1)

Note: (M1) for correct differentiation.

$$x = 1.5$$     (A1)(ft)
$$y = – 2.25$$     (A1)(ft)

OR

for finding the midpoint of their 0 and 3     (M1)
$$x = 1.5$$     (A1)(ft)
$$y = – 2.25$$     (A1)(ft)

Note: If final answer is given as $$(1.5{\text{, }}{- 2.25})$$ award a maximum of (M1)(A1)(A0)[3 marks]

d.

## Question

Let $$f (x) = x^2 – 6x + 8$$.

Factorise $$x^2 – 6x + 8$$.[2]

a.

Hence, or otherwise, solve the equation $$x^2 – 6x + 8 = 0$$.[2]

b.

Let $$g(x) = x + 3$$.

Write down the solutions to the equation $$f (x) = g(x)$$.[2]

c.

## Markscheme

$$(x – 2)(x – 4)$$     (A1)(A1)     (C2)[2 marks]

a.

x = 2, x = 4     (A1)(ft)(A1)(ft)     (C2)[2 marks]

b.

x = 0.807, x = 6.19     (A1)(A1)     (C2)

Note: Award maximum of (A0)(A1) if coordinate pairs given.

OR

(M1) for an attempt to solve $$x^2 – 7x + 5 = 0$$ via formula with correct values substituted.     (M1)

$$x = \frac{{7 \pm \sqrt {29} }}{2}$$     (A1)     (C2)[2 marks]

c.

## Question

The length of a square garden is (x + 1) m. In one of the corners a square of 1 m length is used only for grass. The rest of the garden is only for planting roses and is shaded in the diagram below.

The area of the shaded region is A .

Write down an expression for A in terms of x.[1]

a.

Find the value of x given that A = 109.25 m2.[3]

b.

The owner of the garden puts a fence around the shaded region. Find the length of this fence.[2]

c.

## Markscheme

(x + 1)2 – 1  or  x2 + 2x     (A1)     (C1)[1 mark]

a.

(x + 1)2 – 1 = 109.25     (M1)

x2 + 2x – 109.25 = 0     (M1)

Notes: Award (M1) for writing an equation consistent with their expression in (a) (accept equivalent forms), (M1) for correctly removing the brackets.

OR

(x + 1)2 – 1 = 109.25     (M1)

x + 1 = $$\sqrt {110.25}$$     (M1)

Note: Award (M1) for writing an equation consistent with their expression in (a) (accept equivalent forms), (M1) for taking the square root of both sides.

OR

(x + 1)2 – 10.52 = 0     (M1)

(x – 9.5) (x + 11.5) = 0     (M1)

Note: Award (M1) for writing an equation consistent with their expression in (a) (accept equivalent forms), (M1) for factorised left side of the equation.

x = 9.5     (A1)(ft)     (C3)

Note: Follow through from their expression in part (a).

The last mark is lost if x is non positive.

If the follow through equation is not quadratic award at most (M1)(M0)(A1)(ft).[3 marks]

b.

4 × (9.5 + 1) = 42 m     (M1)(A1)(ft)     (C2)

Notes: Award (M1) for correct method for finding the length of the fence. Accept equivalent methods.[2 marks]

c.

## Question

In the diagram, $${\text{B}}\hat {\text{A}}{\text{C}} = {90^ \circ }$$ . The length of the three sides are $$x{\text{ cm}}$$, $$(x + 7){\text{ cm}}$$ and $$(x + 8){\text{ cm}}$$.

Write down and simplify a quadratic equation in $$x$$ which links the three sides of the triangle.[3]

a.

Solve the quadratic equation found in part (a).[2]

b.

Write down the value of the perimeter of the triangle.[1]

c.

## Markscheme

$${(x + 8)^2} = {(x + 7)^2} + {x^2}$$     (A1)

Note: Award (A1) for a correct equation.

$${x^2} + 16x + 64 = {x^2} + 14x + 49 + {x^2}$$     (A1)

Note: Award (A1) for correctly removed parentheses.

$${x^2} – 2x – 15 = 0$$     (A1)     (C3)

Note: Accept any equivalent form.[3 marks]

a.

$$x = 5$$, $$x = – 3$$     (A1)(ft)(A1)(ft)     (C2)

Notes: Accept (A1)(ft) only from the candidate’s quadratic equation.[2 marks]

b.

$$30{\text{ cm}}$$     (A1)(ft)     (C1)

Note: Follow through from a positive answer found in part (b).[1 mark]

c.

## Question

In the diagram, triangle ABC is isosceles. AB = AC and angle ACB is 32°. The length of side AC is x cm.

Write down the size of angle CBA.

[1]
a.

Write down the size of angle CAB.[1]

b.

The area of triangle ABC is 360 cm2. Calculate the length of side AC. Express your answer in millimetres.[4]

c.

## Markscheme

32°     (A1)     (C1)[1 mark]

a.

116°     (A1)     (C1)[1 mark]

b.

$$360 = \frac{1}{2} \times {x^2} \times \sin 116^\circ$$     (M1)(A1)(ft)

Notes: Award (M1) for substitution into correct formula with 360 seen, (A1)(ft) for correct substitution, follow through from their answer to part (b).

x = 28.3 (cm)     (A1)(ft)

x = 283 (mm)     (A1)(ft)     (C4)

Notes: The final (A1)(ft) is for their cm answer converted to mm. If their incorrect cm answer is seen the final (A1)(ft) can be awarded for correct conversion to mm.[4 marks]

c.

## Question

f (x) = 5x3 − 4x2 + x

Find f‘(x).[3]

a.

(i) the local maximum point;

(ii) the local minimum point.[3]

b.

## Markscheme

15x2 – 8x + 1     (A1)(A1)(A1)     (C3)

Note: Award (A1) for each correct term.[3 marks]

a.

15x2 – 8x +1 = 0     (A1)(ft)

Note: Award (A1)(ft) for setting their derivative to zero.

(i) $$(x =)\frac{1}{5}(0.2)$$     (A1)(ft)

(ii) $$(x =)\frac{1}{3}(0.333)$$     (A1)(ft)     (C3)

b.

## Question

The graph of the quadratic function $$f (x) = c + bx − x^2$$ intersects the y-axis at point A(0, 5) and has its vertex at point B(2, 9).

Write down the value of c.[1]

a.

Find the value of b.[2]

b.

Find the x-intercepts of the graph of f .[2]

c.

Write down $$f (x)$$ in the form $$f (x) = −(x − p) (x + q)$$.[1]

d.

## Markscheme

5     (A1)     (C1)

a.

$$\frac{{ – b}}{{2( – 1)}} = 2$$     (M1)

Note: Award (M1) for correct substitution in axis of symmetry formula.

OR

$$y = 5 + bx – x^2$$

$$9 = 5 + b (2) – (2)^2$$     (M1)

Note: Award (M1) for correct substitution of 9 and 2 into their quadratic equation.

$$(b =) 4$$     (A1)(ft)     (C2)

Note: Follow through from part (a).

b.

5, −1     (A1)(ft)(A1)(ft)     (C2)

Notes: Follow through from parts (a) and (b), irrespective of working shown.

c.

$$f (x) = -(x – 5)(x + 1)$$     (A1)(ft)     (C1)

Notes: Follow through from part (c).

d.

## Question

A quadratic function $$f:x \mapsto a{x^2} + b$$, where $$a$$ and $$b \in \mathbb{R}$$ and $$x \geqslant 0$$, is represented by the mapping diagram.

Using the mapping diagram, write down two equations in terms of $$a$$ and $$b$$.[2]

a.

Solve the equations to find the value of

(i)     $$a$$;

(ii)     $$b$$.[2]

b.

Find the value of $$c$$.[2]

c.

## Markscheme

$$a{(1)^2} + b = – 9$$     (A1)

$$a{(3)^2} + b = 119$$     (A1)     (C2)

Note: Accept equivalent forms of the equations.[2 marks]

a.

(i)     $$a = 16$$     (A1)(ft)

(ii)     $$b = – 25$$     (A1)(ft)     (C2)

Note: Follow through from part (a) irrespective of whether working is seen.

If working is seen follow through from part (i) to part (ii).[2 marks]

b.

$$16{c^2} – 25 = 171$$     (M1)

Note: Award (M1) for correct quadratic with their $$a$$ and $$b$$ substituted.

$$c = 3.5$$     (A1)(ft)     (C2)

Note: Accept $$x$$ instead of $$c$$.

Award (A1) only, for an answer of $$\pm 3.5$$ with or without working.[2 marks]

c.

## Question

The surface of a red carpet is shown below. The dimensions of the carpet are in metres.

Write down an expression for the area, $$A$$, in $${{\text{m}}^2}$$, of the carpet.[1]

a.

The area of the carpet is $${\text{10 }}{{\text{m}}^2}$$.

Calculate the value of $$x$$.[3]

b.

The area of the carpet is $${\text{10 }}{{\text{m}}^2}$$.

Hence, write down the value of the length and of the width of the carpet, in metres.[2]

c.

## Markscheme

$$2x(x – 4)$$   or   $$2{x^2} – 8x$$     (A1)     (C1)

Note: Award (A0) for $$x – 4 \times 2x$$.[1 mark]

a.

$$2x(x – 4) = 10$$     (M1)

Note: Award (M1) for equating their answer in part (a) to $$10$$.

$${x^2} – 4x – 5 = 0$$     (M1)

OR

Sketch of $$y = 2{x^2} – 8x$$ and $$y = 10$$     (M1)

OR

Using GDC solver $$x = 5$$ and $$x = – 1$$     (M1)

OR

$$2(x + 1)(x – 5)$$     (M1)

$$x = 5{\text{ (m)}}$$     (A1)(ft)     (C3)

Award at most (M1)(M1)(A0) if both $$5$$ and $$-1$$ are given as final answer.

Final (A1)(ft) is awarded for choosing only the positive solution(s).[3 marks]

b.

$$2 \times 5 = 10{\text{ (m)}}$$     (A1)(ft)

$$5 – 4 = 1{\text{ (m)}}$$     (A1)(ft)     (C2)

Do not accept negative answers.[2 marks]

c.

## Question

A building company has many rectangular construction sites, of varying widths, along a road.

The area, $$A$$, of each site is given by the function

$A(x) = x(200 – x)$

where $$x$$ is the width of the site in metres and $$20 \leqslant x \leqslant 180$$.

Site S has a width of $$20$$ m. Write down the area of S.[1]

a.

Site T has the same area as site S, but a different width. Find the width of T.[2]

b.

When the width of the construction site is $$b$$ metres, the site has a maximum area.

(i)     Write down the value of $$b$$.

(ii)     Write down the maximum area.[2]

c.

The range of $$A(x)$$ is $$m \leqslant A(x) \leqslant n$$.

Hence write down the value of $$m$$ and of $$n$$.[1]

d.

## Markscheme

$$3600{\text{ (}}{{\text{m}}^2})$$     (A1)(C1)

a.

$$x(200 – x) = 3600$$     (M1)

Note: Award (M1) for setting up an equation, equating to their $$3600$$.

$$180{\text{ (m)}}$$     (A1)(ft)     (C2)

b.

(i)     $$100{\text{ (m)}}$$     (A1)     (C1)

(ii)     $$10\,000{\text{ (}}{{\text{m}}^2})$$     (A1)(ft)(C1)

$$m = 3600\;\;\;$$and$$\;\;\;n = 10\,000$$     (A1)(ft)     (C1)
Notes: Follow through from part (a) and part (c)(ii), but only if their $$m$$ is less than their $$n$$. Accept the answer $$3600 \leqslant A \leqslant 10\,000$$.