# IBDP Maths AI: Topic : SL 3.1: The distance between two points: IB style Questions SL Paper 1

## Question

P (4, 1) and Q (0, –5) are points on the coordinate plane.

Determine the

(i) coordinates of M, the midpoint of P and Q.

(ii) gradient of the line drawn through P and Q.

(iii) gradient of the line drawn through M, perpendicular to PQ.

a.

The perpendicular line drawn through M meets the y-axis at R (0, k).

Find k.

b.

## Markscheme

(i) (2, – 2) parentheses not required.     (A1)

(ii) gradient of PQ $$= \left( {\frac{{ – 5 – 1}}{{0 – 4}}} \right) = \frac{6}{4} = \frac{3}{2}(1.5)$$     (M1)(A1)

(M1) for gradient formula with correct substitution

Award (A1) for $$y = \frac{3}{2}x – 5$$ with no other working

(iii) gradient of perpendicular is $$– \frac{2}{3}$$     (A1)(ft)     (C4)[4 marks]

a.

$$\left( {\frac{{k + 2}}{{0 – 2}}} \right) = – \frac{2}{3}$$, $$k = – \frac{2}{3}$$ or $$y = – \frac{2}{3}x + c$$, $$c = – \frac{2}{3}\therefore k = – \frac{2}{3}$$     (M1)(A1)(ft)

Allow ($$0, – \frac{2}{3}$$)

(M1) is for equating gradients or substituting gradient into $$y = mx + c$$     (C2)[2 marks]

b.

## 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.

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.

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.

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

The mid-point, M, of the line joining A(s , 8) to B(−2, t) has coordinates M(2, 3).

Calculate the values of s and t.

a.

Find the equation of the straight line perpendicular to AB, passing through the point M.

b.

## Markscheme

$$s = 6$$     (A1)

$$t = – 2$$     (A1)     (C2)[2 marks]

a.

$${\text{gradient of AB}} = \frac{{ – 2 – 8}}{{ – 2 – 6}} = \frac{{ – 10}}{{ – 8}} = \frac{5}{4}$$     (A1)(ft)

(A1) for gradient of AM or BM $$= \frac{5}{4}$$

$${\text{Perpendicular gradient}} = – \frac{4}{5}$$     (A1)(ft)

Equation of perpendicular bisector is

$$y = – \frac{4}{5}x + c$$

$$3 = – \frac{4}{5}(2) + c$$     (M1)

$$c = 4.6$$

$$y = -0.8x + 4.6$$

or $$5y = -4x + 23$$     (A1)(ft)     (C4)[4 marks]

b.

## Question

Triangle $${\text{ABC}}$$ is such that $${\text{AC}}$$ is $$7{\text{ cm}}$$, angle $${\text{ABC}}$$ is $${65^ \circ }$$ and angle $${\text{ACB}}$$ is $${30^ \circ }$$.

Sketch the triangle writing in the side length and angles.

a.

Calculate the length of $${\text{AB}}$$.

b.

Find the area of triangle $${\text{ABC}}$$.

c.

## Markscheme (A1)     (C1)

Note: (A1) for fully labelled sketch.
[1 mark]

a.

Unit penalty (UP) may apply in this question.

$$\frac{{{\text{AB}}}}{{\sin 30}} = \frac{7}{{\sin 65}}$$     (M1)

(UP)     $${\text{AB}} = 3.86{\text{ cm}}$$     (A1)(ft)     (C2)

Note: (M1) for use of sine rule with correct values substituted.[2 marks]

b.

Unit penalty (UP) may apply in this question.

$${\text{Angle BAC}} = {85^ \circ }$$     (A1)

$${\text{Area}} = \frac{1}{2} \times 7 \times 3.86 \times \sin {85^ \circ }$$     (M1)

(UP)     $$= 13.5{\text{ }}{{\text{cm}}^2}$$     (A1)(ft)     (C3)[3 marks]

c.

## Question

A race track is made up of a rectangular shape $$750{\text{ m}}$$ by $$500{\text{ m}}$$ with semi-circles at each end as shown in the diagram. Michael drives around the track once at an average speed of $$140{\text{ km}}{{\text{h}}^{ – 1}}$$.

Calculate the distance that Michael travels.

a.

Calculate how long Michael takes in seconds.

b.

## Markscheme

Unit penalty (UP) may apply in this question.

$${\text{Distance}} = \pi \times 500 + 2 \times 750$$     (M1)

(UP)     $$= 3070{\text{ m}}$$     (A1)     (C2)[2 marks]

a.

Unit penalty (UP) may apply in this question.

$${\text{140 km}}{{\text{h}}^{ – 1}} = \frac{{140 \times 1000}}{{60 \times 60}}{\text{ m}}{{\text{s}}^{ – 1}}$$     (M1)

$$= 38.9{\text{ m}}{{\text{s}}^{ – 1}}$$     (A1)

$${\text{Time}} = \frac{{3070}}{{38.889}}$$     (M1)

(UP)     $$= 78.9{\text{ seconds}}$$ (accept $$79.0$$ seconds)     (A1)(ft)     (C4)[4 marks]

b.

## Question

Triangle $${\text{ABC}}$$ is drawn such that angle $${\text{ABC}}$$ is $${90^ \circ }$$, angle $${\text{ACB}}$$ is $${60^ \circ }$$ and $${\text{AB}}$$ is $$7.3{\text{ cm}}$$.

(i) Sketch a diagram to illustrate this information. Label the points $${\text{A, B, C}}$$. Show the angles $${90^ \circ }$$, $${60^ \circ }$$ and the length $$7.3{\text{ cm}}$$ on your diagram.

(ii) Find the length of $${\text{BC}}$$.

a.

Point $${\text{D}}$$ is on the straight line $${\text{AC}}$$ extended and is such that angle $${\text{CDB}}$$ is $${20^ \circ }$$.

(i) Show the point $${\text{D}}$$ and the angle $${20^ \circ }$$ on your diagram.

(ii) Find the size of angle $${\text{CBD}}$$.

b.

## Markscheme

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

(i) (A1)

For $${\text{A}}$$, $${\text{B}}$$, $${\text{C}}$$, $$7.3$$, $${60^ \circ }$$, $${90^ \circ }$$, shown in correct places     (A1)

Note: The $${90^ \circ }$$ should look like $${90^ \circ }$$ (allow $$\pm {10^ \circ }$$)

(ii) Using $$\tan 60$$ or $$\tan 30$$     (M1)

(UP)     $$4.21{\text{ cm}}$$     (A1)(ft)

Note: (ft) on their diagram

Or

Using sine rule with their correct values     (M1)

(UP)     $$= 4.21{\text{ cm}}$$     (A1)(ft)

Or

Using special triangle $$\frac{{7.3}}{{\sqrt 3 }}$$     (M1)

(UP)     $$4.21{\text{ cm}}$$     (A1)(ft)

Or

Any other valid solution

Note: If A and B are swapped then $${\text{BC}} = 8.43{\text{ cm}}$$     (C3)[3 marks]

a.

(i) For $${\text{ACD}}$$ in a straight line and all joined up to $${\text{B}}$$, for $${20^ \circ }$$ shown in correct place and $${\text{D}}$$ labelled. $${\text{D}}$$ must be on $${\text{AC}}$$ extended.     (A1)

(ii) $${\text{B}}\hat {\text{C}}{\text{D}} = {120^ \circ }$$     (A1)

$${\text{C}}\hat {\text{B}}{\text{D}} = {40^ \circ }$$     (A1)     (C3)[3 marks]

b.

## Question

The diagram shows a pyramid $${\text{VABCD}}$$ which has a square base of length $$10{\text{ cm}}$$ and edges of length $$13{\text{ cm}}$$. $${\text{M}}$$ is the midpoint of the side $${\text{BC}}$$. Calculate the length of $${\text{VM}}$$.

a.

Calculate the vertical height of the pyramid.

b.

## Markscheme

Unit penalty (UP) applies in this question.

$${\text{VM}}^{2} = {13^2} – {5^2}$$     (M1)

UP     $$= 12{\text{ cm}}$$     (A1)     (C2)[2 marks]

a.

Unit penalty (UP) applies in this question.

$${h^2} = {12^2} – {5^2}$$ (or equivalent)     (M1)

UP     $$= 10.9{\text{ cm}}$$     (A1)(ft)     (C2)[2 marks]

b.

## Question

A rectangle is 2680 cm long and 1970 cm wide.

Find the perimeter of the rectangle, giving your answer in the form $$a \times {10^k}$$, where $$1 \leqslant a \leqslant 10$$ and $$k \in \mathbb{Z}$$.

a.

Find the area of the rectangle, giving your answer correct to the nearest thousand square centimetres.

b.

## Markscheme

Note: Unit penalty (UP) applies in this part

(2680 + 1970) × 2     (M1)

(UP)     = 9.30 × 103 cm     (A1)(A1)     (C3)

Notes: Award (M1) for correct formula.

(A1) for 9.30 (Accept 9.3).

(A1) for 103.[3 marks]

a.

2680 × 1970     (M1)

= 5279600     (A1)

= 5,280,000 (5280 thousand)     (A1)(ft)     (C3)

Note: Award (M1) for correctly substituted formula.

Accept 5.280 × 106.

Note: The last (A1) is for specified accuracy, (ft) from their answer.

The (AP) for the paper is not applied here.[3 marks]

b.

## Question

Consider the statement p:

“If a quadrilateral is a square then the four sides of the quadrilateral are equal”.

Write down the inverse of statement p in words.

a.

Write down the converse of statement p in words.

b.

Determine whether the converse of statement p is always true. Give an example to justify your answer.

c.

## Markscheme

If a quadrilateral is not a square (then) the four sides of the quadrilateral are not equal.     (A1)(A1)     (C2)

Note: Award (A1) for “if…(then)”, (A1) for the correct phrases in the correct order.[2 marks]

a.

If the four sides of the quadrilateral are equal (then) the quadrilateral is a square.     (A1)(A1)(ft)     (C2)

Note: Award (A1) for “if…(then)”, (A1)(ft) for the correct phrases in the correct order.

Note: Follow through in (b) if the inverse and converse in (a) and (b) are correct and reversed.[2 marks]

b.

The converse is not always true, for example a rhombus (diamond) is a quadrilateral with four equal sides, but it is not a square.     (A1)(R1)     (C2)

Note: Do not award (A1)(R0).[2 marks]

c.

## Question

A line joins the points A(2, 1) and B(4, 5).

Find the gradient of the line AB.

a.

Let M be the midpoint of the line segment AB.

Write down the coordinates of M.

b.

Let M be the midpoint of the line segment AB.

Find the equation of the line perpendicular to AB and passing through M.

c.

## Markscheme

$${\text{Gradient}} = \frac{{(5 – 1)}}{{(4 – 2)}}$$     (M1)

Note: Award (M1) for correct substitution in the gradient formula.

$$= 2$$     (A1)     (C2)[2 marks]

a.

Midpoint = (3, 3) (accept x = 3, y = 3 )     (A1)     (C1)

[1 mark]

b.

$${\text{Gradient of perpendicular}} = -\frac{{1}}{{2}}$$     (A1)(ft)

$$y = – \frac{{1}}{{2}} x + c$$     (M1)

$$3 = – \frac{{1}}{{2}} \times 3 + c$$

$$c = 4.5$$

$$y = -0.5x + 4.5$$     (A1)(ft)

OR

$$y – 3 = -0.5(x – 3)$$     (A1)(A1)(ft)

Note: Award (A1) for –0.5, (A1) for both threes.

OR

$$2y + x = 9$$     (A1)(A1)(ft)     (C3)

Note: Award (A1) for 2, (A1) for 9.[3 marks]

c.

## Question

The diagram shows a triangle ABC in which AC = 17 cm. M is the midpoint of AC.
Triangle ABM is equilateral. Write down the size of angle MCB.

a.1.

Write down the length of BM in cm.

a.i.

Write down the size of angle BMC.

a.ii.

Calculate the length of BC in cm.

b.

## Markscheme

30°     (A1)     (C3)[1 mark]

a.1.

8.5 (cm)     (A1)[1 mark]

a.i.

120°     (A1)[1 mark]

a.ii.

$$\frac{{{\text{BC}}}}{{\sin 120}} = \frac{{8.5}}{{\sin 30}}$$     (M1)(A1)(ft)

Note: Award (M1) for correct substituted formula, (A1) for correct substitutions.

$${\text{BC}} = {\text{14}}{\text{.7}}\left( {\frac{{17\sqrt 3 }}{2}} \right)$$     (A1)(ft)[3 marks]

b.

## Question

A rectangular cuboid has the following dimensions.

Width      0.50 metres     (DG)

Height     1.80 metres     (DC) Calculate the length of AG.

a.

Calculate the length of AF.

b.

Find the size of the angle between AF and AG.

c.

## Markscheme

$${\text{AG}} = \sqrt {{{0.8}^2} + {{0.5}^2}}$$     (M1)

AG = 0.943 m     (A1)     (C2)[2 marks]

a.

$${\text{AF}} = \sqrt {{\text{A}}{{\text{G}}^2} + {{1.80}^2}}$$     (M1)

= 2.03 m     (A1)(ft)     (C2)

b.

$$\cos {\rm{G\hat AF}} = \frac{{0.943(39 \ldots )}}{{2.03(22 \ldots )}}$$     (M1)

$$\operatorname{G\hat AF} = 62.3^\circ$$     (A1)(ft)     (C2)

Notes: Award (M1) for substitution into correct trig ratio.

Accept alternative ratios which give 62.4° or 62.5°.

c.

## Question

The base of a prism is a regular hexagon. The centre of the hexagon is O and the length of OA is 15 cm. Write down the size of angle AOB.

a.

Find the area of the triangle AOB.

b.

The height of the prism is 20 cm.

Find the volume of the prism.

c.

## Markscheme

60°     (A1)     (C1)[1 mark]

a.

$$\frac{{15 \times \sqrt {{{15}^2} – {{7.5}^2}} }}{2} = 97.4{\text{ c}}{{\text{m}}^2}$$     (97.5 cm2)     (A1)(M1)(A1)

Notes: Award (A1) for correct height, (M1) for substitution in the area formula, (A1) for correct answer.

Accept 97.5 cm2 from taking the height to be 13 cm.

OR

$$\frac{1}{2} \times {15^2} \times \sin 60^\circ = 97.4{\text{ c}}{{\text{m}}^2}$$     (M1)(A1)(A1)(ft)     (C3)

Notes: Award (M1) for substituted formula of the area of a triangle, (A1) for correct substitution, (A1)(ft) for answer.

If radians used award at most (M1)(A1)(A0).[3 marks]

b.

97.4 × 120 = 11700 cm3     (M1)(A1)(ft)     (C2)

Notes: Award (M1) for multiplying their part (b) by 120.[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.

a.

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

b.

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

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

A satellite travels around the Earth in a circular orbit $$500$$ kilometres above the Earth’s surface. The radius of the Earth is taken as $$6400$$ kilometres.

Write down the radius of the satellite’s orbit.

a.

Calculate the distance travelled by the satellite in one orbit of the Earth. Give your answer correct to the nearest km.

b.

Write down your answer to (b) in the form $$a \times {10^k}$$ , where $$1 \leqslant a < 10{\text{, }}k \in \mathbb{Z}$$ .

c.

## Markscheme

$$6900$$ km     (A1)     (C1)[1 mark]

a.

$$2\pi (6900)$$     (M1)(A1)(ft)

Notes: Award (M1) for substitution into circumference formula, (A1)(ft) for correct substitution. Follow through from part (a).

$$= 43354$$     (A1)(ft)     (C3)

Notes: Follow through from part (a). The final (A1) is awarded for rounding their answer correct to the nearest km. Award (A2) for $$43 400$$ shown with no working.[3 marks]

b.

$$4.3354 \times {10^4}$$     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(ft) for $$4.3354$$, (A1)(ft) for $$\times {10^4}$$ . Follow through from part (b). Accept $$4.34 \times {10^4}$$ .[2 marks]

c.

## Question

A room is in the shape of a cuboid. Its floor measures $$7.2$$ m by $$9.6$$ m and its height is $$3.5$$ m. Calculate the length of AC.

a.

Calculate the length of AG.

b.

Calculate the angle that AG makes with the floor.

c.

## Markscheme

$${\text{A}}{{\text{C}}^2} = {7.2^2} + {9.6^2}$$     (M1)

Note: Award (M1) for correct substitution in Pythagoras Theorem.

$${\text{AC}} = 12{\text{ m}}$$     (A1)     (C2)[2 marks]

a.

$${\text{A}}{{\text{G}}^2} = {12^2} + {3.5^2}$$     (M1)

Note: Award (M1) for correct substitution in Pythagoras Theorem.

$${\text{AG}} = 12.5{\text{ m}}$$     (A1)(ft)     (C2)

b.

$$\tan \theta = \frac{{3.5}}{{12}}$$ or $$\sin \theta = \frac{{3.5}}{{12.5}}$$ or $$\cos \theta = \frac{{12}}{{12.5}}$$     (M1)

Note: Award (M1) for correct substitutions in trig ratio.

$$\theta = {16.3^ \circ }$$     (A1)(ft)     (C2)

Notes: Follow through from parts (a) and/or part (b) where appropriate. Award (M1)(A0) for use of radians (0.284).[2 marks]

c.

## Question

The diagram shows the straight lines $${L_1}$$ and $${L_2}$$ . The equation of $${L_2}$$ is $$y = x$$ . Find
(i)     the gradient of $${L_1}$$ ;
(ii)    the equation of $${L_1}$$ .

a.

Find the area of the shaded triangle.

b.

## Markscheme

(i)     $$\frac{{0 – 2}}{{6 – 0}}$$     (M1)
$$= – \frac{1}{3}{\text{ }}\left( { – \frac{2}{6}{\text{, }} – 0.333} \right)$$     (A1)     (C2)

(ii)    $$y = – \frac{1}{3}x + 2$$     (A1)(ft)     (C1)

Notes: Follow through from their gradient in part (a)(i). Accept equivalent forms for the equation of a line.[3 marks]

a.

$${\text{area}} = \frac{{6 \times 1.5}}{2}$$     (A1)(M1)

Note: Award (A1) for $$1.5$$ seen, (M1) for use of triangle formula with $$6$$ seen.

$$= 4.5$$     (A1)     (C3)[2 marks]

b.

## Question

The planet Earth takes one year to revolve around the Sun. Assume that a year is 365 days and the path of the Earth around the Sun is the circumference of a circle of radius $$150000000{\text{ km}}$$. Calculate the distance travelled by the Earth in one day.

a.

Give your answer to part (a) in the form $$a \times {10^k}$$ where $$1 \leqslant a \leqslant 10$$ and $$k \in \mathbb{Z}$$ .

b.

## Markscheme

$$2\pi \frac{{150000000}}{{365}}$$     (M1)(A1)(M1)

Notes: Award (M1) for substitution in correct formula for circumference of circle.

Award (A1) for correct substitution.

Award (M1) for dividing their perimeter by $$365$$.

Award (M0)(A0)(M1) for $$\frac{{150000000}}{{365}}$$ .

$$2580000{\text{ km}}$$     (A1)     (C4)[4 marks]

a.

$$2.58 \times {10^6}$$     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(ft) for $$2.58$$, (A1)(ft) for $${10^6}$$ . Follow through from their answer to part (a). The follow through for the index should be dependent not only on the answer to part (a), but also on the value of their mantissa. No (AP) penalty for first (A1) provided their value is to 3 sf or is all their digits from part (a).[2 marks]

b.

## Question

The straight line $$L$$ passes through the points $${\text{A}}( – 1{\text{, 4}})$$ and $${\text{B}}(5{\text{, }}8)$$ .

Calculate the gradient of $$L$$ .

a.

Find the equation of $$L$$ .

b.

The line $$L$$ also passes through the point $${\text{P}}(8{\text{, }}y)$$ . Find the value of $$y$$ .

c.

## Markscheme

$$\frac{{8 – 4}}{{5 – ( – 1)}}$$     (M1)

Note: Award (M1) for correct substitution into the gradient formula.

$$\frac{2}{3}{\text{ }}\left( {\frac{4}{6}{\text{, }}0.667} \right)$$     (A1)     (C2)[2 marks]

a.

$$y = \frac{2}{3}x + c$$     (A1)(ft)

Note: Award (A1)(ft) for their gradient substituted in their equation.

$$y = \frac{2}{3}x + \frac{{14}}{3}$$     (A1)(ft)     (C2)

Notes: Award (A1)(ft) for their correct equation. Accept any equivalent form. Accept decimal equivalents for coefficients to 3 sf.

OR

$$y – {y_1} = \frac{2}{3}(x – {x_1})$$     (A1)(ft)

Note: Award (A1)(ft) for their gradient substituted in the equation.

$$y – 4 = \frac{2}{3}(x + 1)$$     OR     $$y – 8 = \frac{2}{3}(x – 5)$$     (A1)(ft)     (C2)

Note: Award (A1)(ft) for correct equation.[2 marks]

b.

$$y = \frac{2}{3} \times 8 + \frac{{14}}{3}$$     OR     $$y – 4 = \frac{2}{3}(8 + 1)$$     OR     $$y – 8 = \frac{2}{3}(8 – 5)$$     (M1)

Note: Award (M1) for substitution of $$x = 8$$ into their equation.

$$y = 10$$ ($$10.0$$)     (A1)(ft)     (C2)

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.

a.

Solve the quadratic equation found in part (a).

b.

Write down the value of the perimeter of the triangle.

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

The diagram shows points A(2, 8), B(14, 4) and C(4, 2). M is the midpoint of AC. Write down the coordinates of M.

a.

Calculate the gradient of the line AB.

b.

Find the equation of the line parallel to AB that passes through M.

c.

## Markscheme

$$\left( {\frac{{2 + 4}}{2},\frac{{8 + 2}}{2}} \right)$$     (M1)

Note: Award (M1) for a correct substitution into the midpoint formula.

$$= (3, 5)$$     (A1)     (C2)

Note: Brackets must be present for final (A1) to be awarded.

Note: Accept $$x = 3$$, $$y = 5$$ .[2 marks]

a.

$$\frac{{8 – 4}}{{2 – 14}}$$     (M1)

Note: Award (M1) for correctly substituted formula.

$$= – \frac{1}{3}$$ $$\left( {\frac{{ – 4}}{{12}}, – 0.333} \right)$$    $$( – 0.333333 \ldots )$$     (A1)     (C2)[2 marks]

b.

$$(y – 5) = – \frac{1}{3}(x – 3)$$     (M1)(A1)(ft)

OR

$$5 = – \frac{1}{3}(3) + c$$     (M1)

$$y = – \frac{1}{3}x + 6$$     (A1)(ft)     (C2)

Notes: Award (M1) for substitution of their gradient into equation of line with their values from (a) correctly substituted.

Accept correct equivalent forms of the equation of the line. Follow through from their parts (a) and (b).[2 marks]

c.

## Question

The coordinates of point A are (−4, p) and the coordinates of point B are (2, −3) .

The mid-point of the line segment AB, has coordinates (q, 1) .

Find the value of

(i) q ;

(ii) p .

a.

Calculate the distance AB.

b.

## Markscheme

(i) $$\frac{{ – 4 + 2}}{2} = q$$     (M1)

Note: Award (M1) for correct substitution in the correct formula.

q = –1     (A1)

(ii) $$\frac{{p + ( – 3)}}{2} = 1$$     (M1)

Note: Award (M1) for correct substitution into the correct formula or consistent with their equation in (i).

p = 5     (A1)     (C4)

Notes: Award A marks for integer values. Penalise if answers left as a fraction the first time a fraction is seen.[4 marks]

a.

$${\text{AB}} = \sqrt {{{(2 + 4)}^2} + {{( – 3 – 5)}^2}}$$     (M1)

Note: Award (M1) for the correct substitution of their coordinates for A and B in the correct formula.

AB = 10     (A1)(ft)     (C2)

b.

## Question

Line L is given by the equation 3y + 2x = 9 and point P has coordinates (6 , –5).

Explain why point P is not on the line L.

a.

Find the gradient of line L.

b.

(i) Write down the gradient of a line perpendicular to line L.

(ii) Find the equation of the line perpendicular to L and passing through point P.

c.

## Markscheme

3 × (–5) + 2 × 6 ≠ 9     (A1)     (C1)

Note: Also accept 3 × (–5) + 2x = 9 gives x =12 ≠ 6 or 3y + 2 × (6) = 9 gives y = –1 ≠ –5.[1 mark]

a.

3y = –2x + 9     (M1)

Note: Award (M1) for 3y = –2x + 9 or $$y = \frac{{ – 2}}{3}x + 3$$ or $$y = \frac{{( – 2x + 9)}}{3}$$.

$${\text{gradient}} = – \frac{2}{3}( – 0.667)( – 0.666666…)$$     (A1)     (C2)[2 marks]

b.

(i) gradient of perpendicular line $$= \frac{3}{2}(1.5)$$     (A1)(ft)

(ii) $$y = \frac{3}{2}x + c$$

$$– 5 = \frac{3}{2} \times 6 + c$$     (M1)

Note: Award (M1) for substitution of their perpendicular gradient and the point (6, –5) into the equation of their line.

$$y = \frac{3}{2}x – 14$$     (A1)(ft)

OR

$$y + 5 = \frac{3}{2}(x – 6)$$     (M1)(A1)(ft)     (C3)

Notes: Award (M1) for substitution of their perpendicular gradient and the point (6, –5) into the equation of their line. Follow through from their perpendicular gradient.[3 marks]

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.

a.

Write down the size of angle CAB.

b.

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

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

The area of a circle is equal to 8 cm2.

Find the radius of the circle.

a.

This circle is the base of a solid cylinder of height 25 cm.

Write down the volume of the solid cylinder.

b.

This circle is the base of a solid cylinder of height 25 cm.

Find the total surface area of the solid cylinder.

c.

## Markscheme

πr2 = 8     (M1)

Note: Award (M1) for correct area formula.

r = 1.60 (cm)     (1.59576…)     (A1)     (C2)[2 marks]

a.

200 cm3     (A1)(ft)     (C1)

Notes: Units are required. Follow through from their part (a). Accept 201 cm3 (201.061…) for use of r = 1.60 .[1 mark]

b.

Surface area = 16 + 2π(1.59576…)25     (M1)(M1)

Note: Award (M1) for correct substitution of their r into curved surface area formula, (M1) for adding 16 or 2 × π × (their answer to part (a))2

267 cm2     (266.662…cm2)     (A1)(ft)     (C3)

Note: Follow through from their part (a).[3 marks]

c.

## Question

The diagram shows a right triangular prism, ABCDEF, in which the face ABCD is a square.

AF = 8 cm, BF = 9.5 cm, and angle BAF is 90°. Calculate the length of AB .

a.

M is the midpoint of EF .

Calculate the length of BM .

b.

M is the midpoint of EF .

Find the size of the angle between BM and the face ADEF .

c.

## Markscheme

9.52 = 82 + AB2     (M1)

Note: Award (M1) for correct substitution into Pythagoras’ theorem.

AB = 5.12 (cm) (5.12347…)     (A1)     (C2)[2 marks]

a.

$${\text{BM = }}\sqrt {{{9.5}^2} + {{\left( {\frac{{5.12347…}}{2}} \right)}^2}}$$     (M1)

Note: Award (M1) for correct substitution into Pythagoras’ theorem.

= 9.84 (cm) (9.83933…)     (A1)(ft)     (C2)

Notes: Accept alternative methods. Follow through from their answer to part[2 marks]

b.

sin $${\text{A}}\hat {\rm M}{\text{B = }}\frac{{5.12347…}}{{9.83933…}}$$     (M1)

Note: Award (M1) for a correctly substituted trigonometrical equation using $${\text{A}}\hat {\rm M}{\text{B}}$$ .

= 31.4 (31.3801…)     (A1)(ft)     (C2)

Notes: If radians used, the answer will be 0.5476… award (M1)(A0)(ft). Degree symbol ° not required. Follow through from their answers to part (a) and to part (b).[2 marks]

c.

## Question

The straight line, L1, has equation $$2y − 3x =11$$. The point A has coordinates (6, 0).

Give a reason why L1 does not pass through A.

a.

b.

L2 is a line perpendicular to L1. The equation of L2 is $$y = mx + c$$.

Write down the value of m.

c.

L2 does pass through A.

Find the value of c.

d.

## Markscheme

$$2 \times 0 – 3 \times 6 \ne 11$$     (R1)

Note: Stating $$2 \times 0 – 3 \times 6 = – 18$$ without a conclusion is not sufficient.

OR

Clear sketch of L1 and A. (R1)

OR

Point A is (6, 0) and $$2y – 3x = 11$$ has x-intercept at $$- \frac{11}{3}$$ or the line has only one x-intercept which occurs when x is negative.     (R1)     (C1)

a.

$$2y = 3x + 11$$ or $$y – \frac{3}{2}x = \frac{{11}}{2}$$     (M1)

Note: Award (M1) for a correct first step in making y the subject of the equation.

$$({\text{gradient equals}}) = \frac{3}{2}(1.5)$$     (A1)     (C2)

Note: Do not accept 1.5x.

b.

$$(m = ) – \frac{2}{3}$$     (A1)(ft)     (C1)

Notes: Follow through from their part (b).

c.

$$0 = – \frac{2}{3}(6) + c$$     (M1)

Note: Award (M1) for correct substitution of their gradient and (6, 0) into any form of the equation.

(c =) 4     (A1)(ft)     (C2)

Note: Follow through from part (c).

d.

## Question

The diagram below represents a rectangular flag with dimensions 150 cm by 92 cm. The flag is divided into three regions A, B and C. Write down the total area of the flag.

a.

Write down the value of y.

b.

The areas of regions A, B, and C are equal.

Write down the area of region A.

c.

Using your answers to parts (b) and (c), find the value of x.

d.

## Markscheme

Units are required in this question for full marks to be awarded.

13800 cm2     (A1)     (C1)

a.

75     (A1)     (C1)

b.

Units are required in this question for full marks to be awarded.

4600 cm2     (A1)(ft)     (C1)

Notes: Units are required unless already penalized in part (a). Follow through from their part (a).

c.

$$0.5(x + 92) \times 75 = 4600$$     (M1)(A1)(ft)

OR

$$0.5 \times 150 \times (92 – x) = 4600$$     (M1)(A1)(ft)

Note: Award (M1) for substitution into area formula, (A1)(ft) for their correct substitution.

(= 30.7 (cm)(30.6666…(cm))     (A1)(ft)     (C3)

Note: Follow through from their parts (b) and (c).

d.

## Question

In triangle $${\text{ABC}}$$, $${\text{AC}} = 20 {\text{ cm}}$$, $${\text{BC}} = 12 {\text { cm}}$$ and $${\rm{A\hat BC}} = 90^\circ$$. diagram not to scale

Find the length of $${\text{AB}}$$.

a.

$${\text{D}}$$ is the point on $${\text{AB}}$$ such that $$\tan ({\rm{D\hat CB}}) = 0.6$$.

Find the length of $${\text{DB}}$$.

b.

$${\text{D}}$$ is the point on $${\text{AB}}$$ such that $$\tan ({\rm{D\hat CB}}) = 0.6$$.

Find the area of triangle $${\text{ADC}}$$.

c.

## Markscheme

$$({\text{A}}{{\text{B}}^2}) = {20^2} – {12^2}$$     (M1)

Note: Award (M1) for correctly substituted Pythagoras formula.

$${\text{AB}} = 16{\text{ cm}}$$     (A1)     (C2)[2 marks]

a.

$$\frac{{{\text{DB}}}}{{12}} = 0.6$$     (M1)

Note: Award (M1) for correct substitution in tangent ratio or equivalent ie seeing $$12 \times 0.6$$.

$${\text{DB}} = 7.2{\text{ cm}}$$     (A1)     (C2)

Note: Award (M1)(A0) for using $$\tan 31$$ to get an answer of $$7.21$$.

Award (M1)(A0) for $$\frac{{12}}{{\sin 59}} = \frac{{{\text{DB}}}}{{\sin 31}}$$ to get an answer of $$7.2103…$$ or any other incorrect answer.[2 marks]

b.

$$\frac{1}{2} \times 12 \times (16 – 7.2)$$     (M1)

Note: Award (M1) for their correct substitution in triangle area formula.

OR

$$\frac{1}{2} \times 12 \times 16 – \frac{1}{2} \times 12 \times 7.2$$     (M1)

Note: Award (M1) for subtraction of their two correct area formulas.

$$= 52.8{\text{ c}}{{\text{m}}^2}$$     (A1)(ft)     (C2)

Notes: Follow through from parts (a) and (b).

Accept alternative methods.[2 marks]

c.

## Question

The diagram shows the points M(a, 18) and B(24, 10) . The straight line BM intersects the y-axis at A(0, 26). M is the midpoint of the line segment AB. Write down the value of $$a$$.

a.

Find the gradient of the line AB.

b.

c.

## Markscheme

12     (A1)     (C1)

Note: Award (A1) for $$\left( {12,18} \right)$$.[1 mark]

a.

$$\frac{{26 – 10}}{{0 – 24}}$$     (M1)

Note: Accept $$\frac{{26 – 18}}{{0 – 12}}$$   or   $$\frac{{18 – 10}}{{12 – 24}}$$   (or equivalent).

$$= – \frac{2}{3}{\text{ }}\left( { – \frac{{16}}{{24}},{\text{ }} – 0.666666 \ldots } \right)$$     (A1)     (C2)

Note: If either of the alternative fractions is used, follow through from their answer to part (a).

The answer is now (A1)(ft).[2 marks]

b.

gradient of $${\text{OM}} = \frac{3}{2}$$     (A1)(ft)

$$– \frac{2}{3} \times \frac{3}{2}$$     (M1)

Note: Award (M1) for multiplying their gradients.

Since the product is $$-1$$, OAM is a right-angled triangle     (R1)(ft)

Notes: Award the final (R1) only if their conclusion is consistent with their answer for the product of the gradients.

The statement that OAM is a right-angled triangle without justification is awarded no marks.

OR

$${(26 – 18)^2} + {12^2}$$ and $${12^2} + {18^2}$$     (A1)(ft)

$$\left( {{{(26 – 18)}^2} + {{12}^2}} \right) + ({12^2} + {18^2}) = {26^2}$$     (M1)

Note: This method can also be applied to triangle OMB.

Hence a right angled triangle     (R1)(ft)

Note: Award the final (R1) only if their conclusion is consistent with their (M1) mark.

OR

$$OA = OB = 26$$ (cm) an isosceles triangle     (A1)

Note:     Award (A1) for $$OA = 26$$ (cm) and $$OB = 26$$ (cm).

Line drawn from vertex to midpoint of base is perpendicular to the base     (M1)

Conclusion     (R1)     (C3)

Note: Award, at most (A1)(M0)(R0) for stating that OAB is an isosceles triangle without any calculations.[3 marks]

c.

## Question

The average radius of the orbit of the Earth around the Sun is 150 million kilometres. The average radius of the orbit of the Earth around the Sun is 150 million kilometres. Write down this radius, in kilometres, in the form $$a \times {10^k}$$, where $$1 \leqslant a < 10,{\text{ }}k \in \mathbb{Z}$$.

a.

The Earth travels around the Sun in one orbit. It takes one year for the Earth to complete one orbit.

Calculate the distance, in kilometres, the Earth travels around the Sun in one orbit, assuming that the orbit is a circle.

b.

Today is Anna’s 17th birthday.

Calculate the total distance that Anna has travelled around the Sun, since she was born.

c.

## Markscheme

$$1.5 \times {10^8}{\text{ (km)}}$$     (A2)     (C2)

Notes: Award (A2) for the correct answer.

Award (A1)(A0) for 1.5 and an incorrect index.

Award (A0)(A0) for answers of the form $$15 \times {10^7}$$.[2 marks]

a.

$$2\pi 1.5 \times {10^8}$$     (M1)

$$= 942\,000\,000{\text{ (km) (942}}\,{\text{477}}\,{\text{796.1}} \ldots {\text{, }}3 \times {10^8}\pi ,{\text{ }}9.42 \times {10^8})$$     (A1)(ft)     (C2)

Notes: Award (M1) for correct substitution into correct formula. Follow through from part (a).

Do not accept calculator notation $$9.42{\text{E}}8$$.

Do not accept use of $$\frac{{22}}{7}$$ or$$3.14$$ for $$\pi$$.[2 marks]

b.

$$17 \times 942\,000\,000$$     (M1)

$$= 1.60 \times {10^{10}}{\text{ (km) }}\left( {{\text{1.60221}} \ldots \times {{10}^{10}}{\text{, 1.6014}} \times {{10}^{10}},{\text{ 16}}\,{\text{022}}\,{\text{122}}\,{\text{530, }}(5.1 \times {{10}^9})\pi } \right)$$     (A1)(ft)     (C2)

Note: Follow through from part (b).[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.

a.

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

Calculate the value of $$x$$.

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.

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

The four points A(−6, −11) , B(−2, 7) , C(4, 9) and D(6, 3) define the vertices of a kite. Calculate the distance between points $${\text{B}}$$ and $${\text{D}}$$.

a.

The distance between points $${\text{A}}$$ and $${\text{C}}$$ is $$\sqrt {500}$$.

Calculate the area of the kite $${\text{ABCD}}$$.

b.

## Markscheme

$${\text{BD}} = \sqrt {\left( {{4^2} + {8^2}} \right)}$$     (M1)

Note:     Award (M1) for correct substitution into the distance formula.

$$= 8.94\left( {8.94427 \ldots ,{\text{ }}\sqrt {80} ,{\text{ }}4\sqrt 5 } \right)$$     (A1)     (C2)

a.

Area $${\text{ABCD}} = 2 \times \left( {0.5 \times \frac{{{\text{their BD}}}}{2} \times \sqrt {500} } \right)$$     (M1)(M1)(M1)

Note: Award (M1) for dividing their BD by 2, (M1) for correct substitution into the area of triangle formula, (M1) for adding two triangles (or multiplied by 2).

Accept alternative methods:

Area of kite $$= 0.5 \times \sqrt {500} \times$$ their part (a).

Award (M1) for stating kite formula.

Award (M1) for correctly substituting in $$\sqrt {500}$$.

Award (M1) for correctly substituting in their part (a).

$$= 100$$     (A1)     (C4)

Note: Accept 99.9522 if 3 sf answer is used from part (a).

b.

## Question

Tuti has the following polygons to classify: rectangle (R), rhombus (H), isosceles triangle (I), regular pentagon (P), and scalene triangle (T).

In the Venn diagram below, set $$A$$ consists of the polygons that have at least one pair of parallel sides, and set $$B$$ consists of the polygons that have at least one pair of equal sides. Complete the Venn diagram by placing the letter corresponding to each polygon in the appropriate region. For example, R has already been placed, and represents the rectangle.

a.

State which polygons from Tuti’s list are elements of

(i)     $$A \cap B$$;

(ii)     $$(A \cup B)’$$.

b.

## Markscheme (A3)     (C3)

Note: Award (A3) if all four letters placed correctly,

(A2) if three letters are placed correctly,

(A1) if two letters are placed correctly.

a.

(i)     Rhombus and rectangle OR H and R     (A1)(ft)

(ii)     Scalene triangle OR T     (A2)(ft)     (C3)

Notes: Award (A1) for a list R, H, I, P seen (identifying the union).

Follow through from their part (a).

b.

## Question

A hotel has a rectangular swimming pool. Its length is $$x$$ metres, its width is $$y$$ metres and its perimeter is $$44$$ metres.

Write down an equation for $$x$$ and $$y$$.

a.

The area of the swimming pool is $${\text{112}}{{\text{m}}^2}$$.

Write down a second equation for $$x$$ and $$y$$.

b.

Use your graphic display calculator to find the value of $$x$$ and the value of $$y$$.

c.

An Olympic sized swimming pool is $$50$$ m long and $$25$$ m wide.

Determine the area of the hotel swimming pool as a percentage of the area of an Olympic sized swimming pool.

d.

## Markscheme

$$2x + 2y = 44$$     (A1)     (C1)

Note: Accept equivalent forms.

a.

$$xy = 112$$     (A1) (C1)

b.

$$8$$, $$14$$     (A1)(ft)(A1)(ft)     (C2)

Notes: Accept $$x = 8$$, $$y = 14$$   OR   $$x = 14$$, $$y = 8$$

Follow through from their answers to parts (a) and (b) only if both values are positive.

c.

$$\frac{{112}}{{1250}} \times 100$$     (M1)

Note: Award (M1) for $$112$$ divided by $$1250$$.

$$= 8.96$$     (A1)     (C2)

Note: Do not penalize if percentage sign seen.

d.

## Question

The diagram shows the straight line $${L_1}$$, which intersects the $$x$$-axis at $${\text{A}}(6,{\text{ }}0)$$ and the $$y$$-axis at $${\text{B}}(0,{\text{ }}2)$$ . Write down the coordinates of M, the midpoint of line segment AB.

a.

Calculate the gradient of $${L_1}$$.

b.

The line $${L_2}$$ is parallel to $${L_1}$$ and passes through the point $$(3,{\text{ }}2)$$.

Find the equation of $${L_2}$$. Give your answer in the form $$y = mx + c$$.

c.

## Markscheme

$$(3,{\text{ }}1)$$     (A1)(A1)     (C2)

Note: Accept $$x = 3$$, $$y = 1$$. Award (A0)(A1) if parentheses are missing.

a.

$$\frac{{2 – 0}}{{0 – 6}}$$     (M1)

Note: Award (M1) for correct substitution into gradient formula.

$$= – \frac{1}{3}( – 0.333333 \ldots )$$     (A1)     (C2)

Note: Accept $$– \frac{2}{6}$$.

b.

$$(y – 2) = – \frac{1}{3}(x – 3)$$     (M1)

OR

$$2 = – \frac{1}{3}(3) + c$$     (M1)

Note: Award (M1) for substitution of their gradient from part (b).

$$y = – \frac{1}{3}x + 3$$     (A1)(ft)     (C2)

Note: Follow through from part (b).

The answer must be an equation in the form $$y = mx + c$$ for the (A1)(ft) to be awarded.

c.

## Question

A right pyramid has apex $${\text{V}}$$ and rectangular base $${\text{ABCD}}$$, with $${\text{AB}} = 8{\text{ cm}}$$, $${\text{BC}} = 6{\text{ cm}}$$ and $${\text{VA}} = 13{\text{ cm}}$$. The vertical height of the pyramid is $${\text{VM}}$$. Calculate $${\text{VM}}$$.

a.

Calculate the volume of the pyramid.

b.

## Markscheme

$${\text{A}}{{\text{C}}^2} = {8^2} + {6^2}$$     (M1)

Note: Award (M1) for correct substitution into Pythagoras, or recognition of Pythagorean triple.

$${\text{AC}} = 10$$     (A1)

Note: Award (A2) for $${\text{AC}} = 10\;\;\;$$OR$$\;\;\;{\text{AM}} = 5$$ with no working seen.

$${\text{V}}{{\text{M}}^2} = {13^2} – {5^2}$$     (M1)

Note: Award (M1) for correct second use of Pythagoras, using the result from the first use of Pythagoras.

$${\text{VM}} = 12{\text{ (cm)}}$$     (A1)     (C4)

Notes: Accept alternative methods and apply the markscheme as follows: Award (M1)(A1) for first correct use of Pythagoras with lengths from the question, (M1) for a correct second use of Pythagoras, consistent with the method chosen, (A1) for correct height.

a.

$$\frac{1}{3} \times 8 \times 6{\kern 1pt} \times 12$$     (M1)

Note: Award (M1) for their correct substitutions into volume formula.

$$= 192{\text{ c}}{{\text{m}}^3}$$     (A1)(ft)     (C2)

Notes: Follow through from part (a), only if working seen.

b.

## Question

The distance $$d$$ from a point $${\text{P}}(x,{\text{ }}y)$$ to the point $${\text{A}}(1,{\text{ }} – 2)$$ is given by $$d = \sqrt {{{(x – 1)}^2} + {{(y + 2)}^2}}$$

Find the distance from $${\text{P}}(100,{\text{ }}200)$$ to $${\text{A}}$$. Give your answer correct to two decimal places.

a.

Write down your answer to part (a) correct to three significant figures.

b.

Write down your answer to part (b) in the form $$a \times {10^k}$$, where $$1 \leqslant a < 10$$ and $$k \in \mathbb{Z}$$.

c.

## Markscheme

$$\sqrt {{{(100 – 1)}^2} + {{(200 + 2)}^2}}$$     (M1)

$$\sqrt {50605} \;\;\;( = 224.955 \ldots )$$     (A1)

Note: Award (M1)(A1) if $$\sqrt {50605}$$ seen.

$${\text{224.96}}$$     (A1)     (C3)

Note: Award (A1) for their answer given correct to 2 decimal places.

a.

$$225$$     (A1)(ft)     (C1)

Note: Follow through from their part (a).

b.

$$2.25 \times {10^2}$$     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(A0) for $$2.25$$ and an incorrect index value.

Award (A0)(A0) for answers such as $$22.5 \times {10^1}$$.

c.

## Question

Fabián stands on top of a building, T, which is on a horizontal street.

He observes a car, C, on the street, at an angle of depression of 30°. The base of the building is at B. The height of the building is 80 metres.

The following diagram indicates the positions of T, B and C. Show, in the appropriate place on the diagram, the values of

(i)     the height of the building;

(ii)     the angle of depression.

a.

Find the distance, BC, from the base of the building to the car.

b.

Fabián estimates that the distance from the base of the building to the car is 150 metres. Calculate the percentage error of Fabián’s estimate.

c.

## Markscheme (A1)(A1)     (C2)

Notes: Award (A1) for 80 m in the correct position on diagram.

Award (A1) for 30° in a correct position on diagram.

a.

$$\tan 30^\circ = \frac{{80}}{{{\text{BC}}}}\;\;\;$$OR$$\;\;\;\tan 60^\circ = \frac{{{\text{BC}}}}{{80}}\;\;\;$$OR$$\;\;\;\frac{{80}}{{\sin 30^\circ }} = \frac{{{\text{BC}}}}{{\sin 60^\circ }}$$     (M1)

Note: Award (M1) for a correct trigonometric or Pythagorean equation for BC with correctly substituted values.

$$({\text{BC}} = ){\text{ 139 (m)}}\;\;\;\left( {138.564 \ldots {\text{ (m)}}} \right)$$     (A1)(ft)     (C2)

Notes: Accept an answer of $$80\sqrt 3$$ which is the exact answer.

b.

$$\left| {\frac{{150 – 138.564 \ldots }}{{138.564 \ldots }}} \right| \times 100$$     (M1)

Notes: Award (M1) for their correct substitution into the percentage error formula.

$$= 8.25(\% )\;\;\;(8.25317 \ldots \% )$$     (A1)(ft)     (C2)

Notes: Accept $$7.91(\%)$$ ($$7.91366…$$ if $$139$$ is used.

Accept $$8.23(\%)$$ ($$8.22510…$$ if $$138.6$$ is used.

If answer to part (b) is $$46.2$$, answer to part (c) is $$225\%$$, award (M1)(A1)(ft) with or without working seen. If answer to part (b) is negative, award at most (M1)(A0).

c.

## Question

The diagram shows a triangle $${\rm{ABC}}$$. The size of angle $${\rm{C\hat AB}}$$ is $$55^\circ$$ and the length of $${\rm{AM}}$$ is $$10$$ m, where $${\rm{M}}$$ is the midpoint of $${\rm{AB}}$$. Triangle $${\rm{CMB}}$$ is isosceles with $${\text{CM}} = {\text{MB}}$$. Write down the length of $${\rm{MB}}$$.

a.

Find the size of angle $${\rm{C\hat MB}}$$.

b.

Find the length of $${\rm{CB}}$$.

c.

## Markscheme

$$10$$ m     (A1)(C1)

a.

$${\rm{A\hat MC}} = 70^\circ \;\;\;$$OR$$\;\;\;{\rm{A\hat CM}} = 55^\circ$$     (A1)

$${\rm{C\hat MB}} = 110^\circ$$     (A1)     (C2)

b.

$${\text{C}}{{\text{B}}^2} = {10^2} + {10^2} – 2 \times 10 \times 10 \times \cos 110^\circ$$     (M1)(A1)(ft)

Notes: Award (M1) for substitution into the cosine rule formula, (A1)(ft) for correct substitution. Follow through from their answer to part (b).

OR

$$\frac{{{\text{CB}}}}{{\sin 110^\circ }} = \frac{{10}}{{\sin 35^\circ }}$$     (M1)(A1)(ft)

Notes: Award (M1) for substitution into the sine rule formula, (A1)(ft) for correct substitution. Follow through from their answer to part (b).

OR

$${\rm{A\hat CB}} = 90^\circ$$     (A1)

$$\sin 55^\circ = \frac{{{\text{CB}}}}{{55}}\;\;\;$$OR$$\;\;\;\cos 35^\circ = \frac{{{\text{CB}}}}{{20}}$$     (M1)

Note: Award (A1) for some indication that $${\rm{A\hat CB}} = 90^\circ$$, (M1) for correct trigonometric equation.

OR

Perpendicular $${\rm{MN}}$$ is drawn from $${\rm{M}}$$ to $${\rm{CB}}$$.     (A1)

$$\frac{{\frac{1}{2}{\text{CB}}}}{{10}} = \cos 35^\circ$$     (M1)

Note: Award (A1) for some indication of the perpendicular bisector of $${\rm{BC}}$$, (M1) for correct trigonometric equation.

$${\text{CB}} = 16.4{\text{ (m)}}\;\;\;\left( {16.3830 \ldots {\text{ (m)}}} \right)$$     (A1)(ft)(C3)

Notes: Where a candidate uses $${\rm{C\hat MB}} = 90^\circ$$ and finds $${\text{CB}} = 14.1{\text{ (m)}}$$ award, at most, (M1)(A1)(A0).

Where a candidate uses $${\rm{C\hat MB}} = 60^\circ$$ and finds $${\text{CB}} = 10{\text{ (m)}}$$ award, at most, (M1)(A1)(A0).

c.

## Question

In the following diagram, ABCD is the square base of a right pyramid with vertex V. The centre of the base is O. The diagonal of the base, AC, is 8 cm long. The sloping edges are 10 cm long. Write down the length of $${\text{AO}}$$.

a.

Find the size of the angle that the sloping edge $${\text{VA}}$$ makes with the base of the pyramid.

b.

Hence, or otherwise, find the area of the triangle $${\text{CAV}}$$.

c.

## Markscheme

$${\text{AO}} = 4{\text{ (cm)}}$$     (A1)     (C1)

a.

$$\cos {\rm{O\hat AV}} = \frac{4}{{10}}$$     (M1)

Note: Award (M1) for their correct trigonometric ratio.

OR

$$\cos {\rm{O\hat AV}} = \frac{{{{10}^2} + {8^2} – {{10}^2}}}{{2 \times 10 \times 8}}\;\;\;$$OR$$\;\;\;\frac{{{{10}^2} + {4^2} – {{(9.16515 \ldots )}^2}}}{{2 \times 10 \times 4}}$$     (M1)

Note: Award (M1) for correct substitution into the cosine rule formula.

$${\rm{O\hat AV}} = 66.4^\circ \;\;\;(66.4218 \ldots )$$     (A1)(ft)     (C2)

b.

$${\text{area}} = \frac{{8 \times 10 \times \sin (66.4218 \ldots ^\circ)}}{2}\;\;\;$$OR$$\;\;\;\frac{1}{2} \times 8 \times \sqrt {{{10}^2} – {4^{\text{2}}}}$$

OR$$\;\;\;\frac{1}{2} \times 10 \times 10 \times \sin (47.1563 \ldots ^\circ )$$     (M1)(A1)(ft)

Notes: Award (M1) for substitution into the area formula, (A1)(ft) for correct substitutions. Follow through from their answer to part (b) and/or part (a).

$${\text{area}} = 36.7{\text{ c}}{{\text{m}}^2}\;\;\;(36.6606 \ldots {\text{ c}}{{\text{m}}^2})$$     (A1)(ft)     (C3)

Notes: Accept an answer of $$8\sqrt {21} {\text{ c}}{{\text{m}}^2}$$ which is the exact answer.

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.

a.

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

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.

c.

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

Hence write down the value of $$m$$ and of $$n$$.

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)

c.

$$m = 3600\;\;\;$$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 $$3600 \leqslant A \leqslant 10\,000$$.

d.

## Question

The diagram shows a rectangular based right pyramid VABCD in which $${\text{AD}} = 20{\text{ cm}}$$, $${\text{DC}} = 15{\text{ cm}}$$ and the height of the pyramid, $${\text{VN}} = 30{\text{ cm}}$$. Calculate
(i)     the length of AC;
(ii)    the length of VC.

a.

Calculate the angle between VC and the base ABCD.

b.

## Markscheme

(i)     $$\sqrt {{{15}^2} + {{20}^2}}$$     (M1)

Note: Award (M1) for correct substitution in Pythagoras Formula.

$${\text{AC}} = 25{\text{ (cm)}}$$     (A1)     (C2)

(ii)    $$\sqrt {{{12.5}^2} + {{30}^2}}$$     (M1)

Note: Award (M1) for correct substitution in Pythagoras Formula.

$${\text{VC}} = 32.5{\text{ (cm)}}$$     (A1)(ft)     (C2)

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

a.

$$\sin {\text{VCN}} = \frac{{30}}{{32.5}}$$     OR     $$\tan {\text{VCN}} = \frac{{30}}{{12.5}}$$     OR     $$\cos {\text{VCN}} = \frac{{12.5}}{{32.5}}$$     (M1)
$${ = 67.4^ \circ }$$ ($$67.3801 \ldots$$)     (A1)(ft)     (C2)

Note: Accept alternative methods. Follow through from part (a) and/or part (b).

b.

## Question

A child’s toy consists of a hemisphere with a right circular cone on top. The height of the cone is $$12{\text{ cm}}$$ and the radius of its base is $$5{\text{ cm}}$$ . The toy is painted red. Calculate the length, $$l$$, of the slant height of the cone.

a.

Calculate the area that is painted red.

b.

## Markscheme

$$\sqrt {{5^2} + {{12}^2}}$$     (M1)

Note: Award (M1) for correct substitution in Pythagoras Formula.

=$$13{\text{ (cm)}}$$     (A1)     (C2)

a.

$${\text{Area}} = 2\pi {(5)^2} + \pi (5)(13)$$     (M1)(M1)(M1)

Notes: Award (M1) for surface area of hemisphere, (M1) for surface of cone, (M1) for addition of two surface areas. Follow through from their answer to part (a).

$$= 361{\text{ c}}{{\text{m}}^2}$$ ($$361.283 \ldots$$)     (A1)(ft)     (C4)

Note: The answer is $$361{\text{ c}}{{\text{m}}^2}$$ , the units are required.

b.

## Question

Tennis balls are sold in cylindrical tubes that contain four balls. The radius of each tennis ball is 3.15 cm and the radius of the tube is 3.2 cm. The length of the tube is 26 cm.

Find the volume of one tennis ball.

a.

Calculate the volume of the empty space in the tube when four tennis balls have been placed in it.

b.

## Markscheme

Unit penalty (UP) applies

$${\text{Volume of tennis ball}} = \frac{{4}}{{3}} \pi 3.15^3$$     (M1)

Note: Award (M1) for correct substitution into correct formula.

(UP)     Volume of tennis ball = 131 cm3     (A1)     (C2)[2 marks]

a.

Unit penalty (UP) applies

$${\text{Volume of empty space}} = \pi 3.2^2 \times 26 – 4 \times 130.9$$     (M1)(M1)(M1)

Note: Award (M1) for correct substitution into cylinder formula, (M1) 4 × their (a), (M1) for subtracting appropriate volumes.

(UP)     Volume of empty space = 313 cm3     (A1)(ft)     (C4)

Note: Accept 312 cm3 with use of 131. [4 marks]

b.

## Question

The volume of a sphere is $$V{\text{ = }}\sqrt {\frac{{{S^3}}}{{36\pi }}}$$, where $$S$$ is its surface area.

The surface area of a sphere is 500 cm2 .

Calculate the volume of the sphere. Give your answer correct to two decimal places.

a.

b.

Write down your answer to (b) in the form $$a \times {10^n}$$, where $$1 \leqslant a < 10$$ and $$n \in \mathbb{Z}$$.

c.

## Markscheme

$$V{\text{ = }}\sqrt {\frac{{{500^3}}}{{36\pi }}}$$     (M1)
Note: Award (M1) correct substitution into formula.

V = 1051.305 …     (A1)
V = 1051.31 cm3     (A1)(ft)    (C3)

Note: Award last (A1)(ft) for correct rounding to 2 decimal places of their answer. Unrounded answer must be seen so that the follow through can be awarded.[3 marks]

a.

1051     (A1)(ft)[1 mark]

b.

$$1.051 \times {10^3}$$     (A1)(ft)(A1)(ft)     (C2)

Note: Award (A1) for 1.051 (accept 1.05) (A1) for $$\times {10^3}$$.[2 marks]

c.

## Question

In a television show there is a transparent box completely filled with identical cubes. Participants have to estimate the number of cubes in the box. The box is 50 cm wide, 100 cm long and 40 cm tall.

Find the volume of the box.

a.

Joaquin estimates the volume of one cube to be 500 cm3. He uses this value to estimate the number of cubes in the box.

Find Joaquin’s estimated number of cubes in the box.

b.

The actual number of cubes in the box is 350.

Find the percentage error in Joaquin’s estimated number of cubes in the box.

c.

## Markscheme

$$50 \times 100 \times 40 = 200\,000{\text{ c}}{{\text{m}}^3}$$     (M1)(A1)     (C2)

Note: Award (M1) for correct substitution in the volume formula.[2 marks]

a.

$$\frac{{200\,000}}{{500}} = 400$$     (M1)(A1)(ft)     (C2)

Note: Award (M1) for dividing their answer to part (a) by 500.[2 marks]

b.

$$\frac{{400 – 350}}{{350}} \times 100 = 14.3{\text{ }}\%$$     (M1)(A1)(ft)     (C2)

Notes: Award (M1) for correct substitution in the percentage error formula.

Accept –14.3 %.

% sign not necessary.[2 marks]

c.

## Question

The base of a prism is a regular hexagon. The centre of the hexagon is O and the length of OA is 15 cm. Write down the size of angle AOB.

a.

Find the area of the triangle AOB.

b.

The height of the prism is 20 cm.

Find the volume of the prism.

c.

## Markscheme

60°     (A1)     (C1)[1 mark]

a.

$$\frac{{15 \times \sqrt {{{15}^2} – {{7.5}^2}} }}{2} = 97.4{\text{ c}}{{\text{m}}^2}$$     (97.5 cm2)     (A1)(M1)(A1)

Notes: Award (A1) for correct height, (M1) for substitution in the area formula, (A1) for correct answer.

Accept 97.5 cm2 from taking the height to be 13 cm.

OR

$$\frac{1}{2} \times {15^2} \times \sin 60^\circ = 97.4{\text{ c}}{{\text{m}}^2}$$     (M1)(A1)(A1)(ft)     (C3)

Notes: Award (M1) for substituted formula of the area of a triangle, (A1) for correct substitution, (A1)(ft) for answer.

If radians used award at most (M1)(A1)(A0).[3 marks]

b.

97.4 × 120 = 11700 cm3     (M1)(A1)(ft)     (C2)

Notes: Award (M1) for multiplying their part (b) by 120.[2 marks]

c.

## Question

$$75$$ metal spherical cannon balls, each of diameter $$10{\text{ cm}}$$, were excavated from a Napoleonic War battlefield.

Calculate the total volume of all $$75$$ metal cannon balls excavated.

a.

The cannon balls are to be melted down to form a sculpture in the shape of a cone. The base radius of the cone is $$20{\text{ cm}}$$.

Calculate the height of the cone, assuming that no metal is wasted.

b.

## Markscheme

$$75 \times \frac{4}{3}\pi \times {5^3}$$     (M1)(M1)

Notes: Award (M1) for correctly substituted formula of a sphere. Award (M1) for multiplying their volume by $$75$$. If $$r = 10$$ is used, award (M0)(M1)(A1)(ft) for the answer 314000 cm3 .

$$39300{\text{ c}}{{\text{m}}^3}$$ .     (A1)     (C3)

a.

$$\frac{1}{3}\pi \times {20^2} \times h = 39300$$     (M1)(M1)

Notes: Award (M1) for correctly substituted formula of a cone. Award (M1) for equating their volume to their answer to part (a).

$$h = 93.8{\text{ cm}}$$     (A1)(ft)     (C3)

Notes: Accept the exact value of $$93.75$$ . Follow through from their part (a).[3 marks]

b.

## Question

An observatory is built in the shape of a cylinder with a hemispherical roof on the top as shown in the diagram. The height of the cylinder is 12 m and its radius is 15 m. Calculate the volume of the observatory.

a.

The hemispherical roof is to be painted.

Calculate the area that is to be painted.

b.

## Markscheme

$$V = \pi {(15)^2}(12) + 0.5 \times \frac{{4\pi {{(15)}^3}}}{3}$$     (M1)(M1)(M1)

Note: Award (M1) for correctly substituted cylinder formula, (M1) for correctly substituted sphere formula, (M1) for dividing the sphere formula by 2.

$$= 15550.8 \ldots$$

$$= 15600{\text{ }}{{\text{m}}^3}{\text{ }}(4950\pi {\text{ }}{{\text{m}}^3})$$     (A1)     (C4)

Notes: The final answer is $$15600{\text{ }}{{\text{m}}^3}$$; the units are required. The use of $$\pi = 3.14$$ which gives a final answer of $$15 500$$ ($$15 543$$) is premature rounding; the final (A1) is not awarded.[4 marks]

a.

$$SA = 0.5 \times 4\pi {\left( {15} \right)^2}$$     (M1)

$$= 1413.71 \ldots$$

$$= 1410{\text{ }}{{\text{m}}^2}{\text{ }}(450\pi {\text{ }}{{\text{m}}^2})$$     (A1)     (C2)

Notes: The final answer is $$1410{\text{ }}{{\text{m}}^2}$$ ; do not penalize lack of units if this has been penalized in part (a).[2 marks]

b.

## Question

The area of a circle is equal to 8 cm2.

Find the radius of the circle.

a.

This circle is the base of a solid cylinder of height 25 cm.

Write down the volume of the solid cylinder.

b.

This circle is the base of a solid cylinder of height 25 cm.

Find the total surface area of the solid cylinder.

c.

## Markscheme

πr2 = 8     (M1)

Note: Award (M1) for correct area formula.

r = 1.60 (cm)     (1.59576…)     (A1)     (C2)[2 marks]

a.

200 cm3     (A1)(ft)     (C1)

Notes: Units are required. Follow through from their part (a). Accept 201 cm3 (201.061…) for use of r = 1.60 .[1 mark]

b.

Surface area = 16 + 2π(1.59576…)25     (M1)(M1)

Note: Award (M1) for correct substitution of their r into curved surface area formula, (M1) for adding 16 or 2 × π × (their answer to part (a))2

267 cm2     (266.662…cm2)     (A1)(ft)     (C3)

Note: Follow through from their part (a).[3 marks]

c.

## Question

A cuboid has the following dimensions: length = 8.7 cm, width = 5.6 cm and height = 3.4 cm.

Calculate the exact value of the volume of the cuboid, in cm3.

a.

(i) one decimal place;

(ii) three significant figures.

b.

Write your answer to part (b)(ii) in the form $$a \times 10^k$$, where $$1 \leqslant a < 10 , k \in \mathbb{Z}$$.

c.

## Markscheme

$${\text{V}} = 8.7 \times 5.6 \times 3.4$$     (M1)

Note: Award (M1) for multiplication of the 3 given values.

$$=165.648$$     (A1)     (C2)

a.

(i) 165.6     (A1)(ft)

(ii) 166     (A1)(ft)     (C2)

b.

$$1.66 \times 10^2$$     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(ft) for 1.66, (A1)(ft) for $$10^2$$. Follow through from their answer to part (b)(ii) only. The follow through for the index should be dependent on the value of the mantissa in part (c) and their answer to part (b)(ii).

c.

## Question

A child’s wooden toy consists of a hemisphere, of radius 9 cm , attached to a cone with the same base radius. O is the centre of the base of the cone and V is vertically above O.

Angle OVB is $${27.9^ \circ }$$.

Diagram not to scale. Calculate OV, the height of the cone.

a.

Calculate the volume of wood used to make the toy.

b.

## Markscheme

$$\tan 27.9^\circ = \frac{9}{{{\text{OV}}}}$$     (M1)

Note: Award (M1) for correct substitution in trig formula.

$${\text{OV}} = 17.0\left( {{\text{cm}}} \right)\left( {16.9980 \ldots } \right)$$     (A1)     (C2)[2 marks]

a.

$$\frac{{\pi {{(9)}^2}(16.9980 \ldots )}}{3} + \frac{1}{2} \times \frac{{4\pi {{(9)}^3}}}{3}$$     (M1)(M1)(M1)

Note: Award (M1) for correctly substituted volume of the cone, (M1) for correctly substituted volume of a sphere divided by two (hemisphere), (M1) for adding the correctly substituted volume of the cone to either a correctly substituted sphere or hemisphere.

$$= 2970{\text{ c}}{{\text{m}}^3}{\text{ (2968.63}} \ldots {\text{)}}$$     (A1)(ft)     (C4)

Note: The answer is $$2970{\text{ c}}{{\text{m}}^3}$$, the units are required.[4 marks]

b.

## Question

Chocolates in the shape of spheres are sold in boxes of 20.

Each chocolate has a radius of 1 cm.

Find the volume of 1 chocolate.

a.

Write down the volume of 20 chocolates.

b.

The diagram shows the chocolate box from above. The 20 chocolates fit perfectly in the box with each chocolate touching the ones around it or the sides of the box. Calculate the volume of the box.

c.

The diagram shows the chocolate box from above. The 20 chocolates fit perfectly in the box with each chocolate touching the ones around it or the sides of the box. Calculate the volume of empty space in the box.

d.

## Markscheme

The first time a correct answer has incorrect or missing units, the final (A1) is not awarded.

$$\frac{4}{3}\pi {(1)^3}$$     (M1)

Notes: Award (M1) for correct substitution into correct formula.

$$= 4.19{\text{ }}\left( {{\text{4.18879}} \ldots ,{\text{ }}\frac{4}{3}\pi } \right){\text{ c}}{{\text{m}}^3}$$     (A1)     (C2)[2 marks]

a.

The first time a correct answer has incorrect or missing units, the final (A1) is not awarded.

$$83.8{\text{ }}\left( {{\text{83.7758}} \ldots ,{\text{ }}\frac{{80}}{3}\pi } \right){\text{ c}}{{\text{m}}^3}$$     (A1)(ft)     (C1)

b.

The first time a correct answer has incorrect or missing units, the final (A1) is not awarded.

$$10 \times 8 \times 2$$     (M1)

Note: Award (M1) for correct substitution into correct formula.

$$= 160{\text{ c}}{{\text{m}}^3}$$     (A1)     (C2)[2 marks]

c.

The first time a correct answer has incorrect or missing units, the final (A1) is not awarded.

$$76.2{\text{ }}\left( {{\text{76.2241}} \ldots ,{\text{ }}\left( {160 – \frac{{80}}{3}\pi } \right)} \right){\text{ c}}{{\text{m}}^3}$$     (A1)(ft)     (C1)

Note: Follow through from their part (b) and their part (c).[1 mark]

d.

## Question

Assume the Earth is a perfect sphere with radius 6371 km.

Calculate the volume of the Earth in $${\text{k}}{{\text{m}}^3}$$. Give your answer in the form $$a \times {10^k}$$, where $$1 \leqslant a < 10$$ and $$k \in \mathbb{Z}$$.

a.

The volume of the Moon is $$2.1958 \times {10^{10}}\;{\text{k}}{{\text{m}}^3}$$.

Calculate how many times greater in volume the Earth is compared to the Moon.

b.

## Markscheme

$$\frac{4}{3}\pi {(6371)^3}$$     (M1)

Note: Award (M1) for correct substitution into volume formula.

$$= 1.08 \times {10^{12}}\;\;\;(1.08320 \ldots \times {10^{12}})$$     (A2)     (C3)

Notes: Award (A1)(A0) for correct mantissa between 1 and 10, with incorrect index.

Award (A1)(A0) for $$1.08\rm{E}12$$

Award (A0)(A0) for answers of the type: $$108 \times {10^{10}}$$.

a.

$$\frac{{1.08320 \ldots \times {{10}^{12}}}}{{2.1958 \times {{10}^{10}}}}$$     (M1)

Note: Award (M1) for dividing their answer to part (a) by $$2.1958 \times {10^{10}}$$.

$$= 49.3308 \ldots$$     (A1)(ft)

Note: Accept $$49.1848…$$ from use of 3 sf answer to part (a).

$$= 49$$     (A1)     (C3)

Notes: Follow through from part (a).

The final (A1) is awarded for their unrounded non-integer answer seen and given correct to the nearest integer.

Do not award the final (A1) for a rounded answer of 0 or if it is incorrect by a large order of magnitude.

b.

## Question

A right pyramid has apex $${\text{V}}$$ and rectangular base $${\text{ABCD}}$$, with $${\text{AB}} = 8{\text{ cm}}$$, $${\text{BC}} = 6{\text{ cm}}$$ and $${\text{VA}} = 13{\text{ cm}}$$. The vertical height of the pyramid is $${\text{VM}}$$. Calculate $${\text{VM}}$$.

a.

Calculate the volume of the pyramid.

b.

## Markscheme

$${\text{A}}{{\text{C}}^2} = {8^2} + {6^2}$$     (M1)

Note: Award (M1) for correct substitution into Pythagoras, or recognition of Pythagorean triple.

$${\text{AC}} = 10$$     (A1)

Note: Award (A2) for $${\text{AC}} = 10\;\;\;$$OR$$\;\;\;{\text{AM}} = 5$$ with no working seen.

$${\text{V}}{{\text{M}}^2} = {13^2} – {5^2}$$     (M1)

Note: Award (M1) for correct second use of Pythagoras, using the result from the first use of Pythagoras.

$${\text{VM}} = 12{\text{ (cm)}}$$     (A1)     (C4)

Notes: Accept alternative methods and apply the markscheme as follows: Award (M1)(A1) for first correct use of Pythagoras with lengths from the question, (M1) for a correct second use of Pythagoras, consistent with the method chosen, (A1) for correct height.

a.

$$\frac{1}{3} \times 8 \times 6{\kern 1pt} \times 12$$     (M1)

Note: Award (M1) for their correct substitutions into volume formula.

$$= 192{\text{ c}}{{\text{m}}^3}$$     (A1)(ft)     (C2)

Notes: Follow through from part (a), only if working seen.

b.

## Question

A cuboid has a rectangular base of width $$x$$ cm and length 2$$x$$ cm . The height of the cuboid is $$h$$ cm . The total length of the edges of the cuboid is $$72$$ cm. The volume, $$V$$, of the cuboid can be expressed as $$V = a{x^2} – 6{x^3}$$.

Find the value of $$a$$.

a.

Find the value of $$x$$ that makes the volume a maximum.

b.

## Markscheme

$$72 = 12x + 4h\;\;\;$$(or equivalent)     (M1)

Note: Award (M1) for a correct equation obtained from the total length of the edges.

$$V = 2{x^2}(18 – 3x)$$     (A1)

$$(a = ){\text{ }}36$$     (A1)     (C3)

a.

$$\frac{{{\text{d}}V}}{{{\text{d}}x}} = 72x – 18{x^2}$$     (A1)

$$72x – 18{x^2} = 0\;\;\;$$OR$$\;\;\;\frac{{{\text{d}}V}}{{{\text{d}}x}} = 0$$     (M1)

Notes: Award (A1) for  $$– 18{x^2}$$  seen. Award (M1) for equating derivative to zero.

$$(x = ){\text{ 4}}$$     (A1)(ft)     (C3)

Note: Follow through from part (a).

OR

Sketch of $$V$$ with visible maximum     (M1)

Sketch with $$x \geqslant 0,{\text{ }}V \geqslant 0$$ and indication of maximum (e.g. coordinates)     (A1)(ft)

$$(x = ){\text{ 4}}$$     (A1)(ft)     (C3)

Notes: Follow through from part (a).

Award (M1)(A1)(A0) for $$(4,{\text{ }}192)$$.

Award (C3) for $$x = 4,{\text{ }}y = 192$$.

b.

## Question

FreshWave brand tuna is sold in cans that are in the shape of a cuboid with length $$8\,{\text{cm}}$$, width $${\text{5}}\,{\text{cm}}$$ and height $${\text{3}}{\text{.5}}\,{\text{cm}}$$. HappyFin brand tuna is sold in cans that are cylindrical with diameter $${\text{7}}\,{\text{cm}}$$ and height $${\text{4}}\,{\text{cm}}$$. Find the volume, in $${\text{c}}{{\text{m}}^3}$$, of a can of

i)    FreshWave tuna;

ii)   HappyFin tuna.

a.

The price of tuna per $${\text{c}}{{\text{m}}^3}$$ is the same for each brand. A can of FreshWave tuna costs $$90$$ cents.

Calculate the price, in cents, of a can of HappyFin tuna.

b.

## Markscheme

i)    $$8\,\, \times \,\,5\,\, \times \,3.5$$        (M1)

$$= 140$$        (A1)

Note: Award (M1) for correct substitution in volume formula.

ii)   $$\pi \,\, \times \,\,{3.5^2}\,\, \times \,\,4$$        (M1)

$$= 154\,\,\,\,(153.938…,\,\,49\pi )$$        (A1) (C4)

Note: Award (M1) for correct substitution in volume formula.

a.

$$\frac{{90\,\, \times \,\,{\text{their}}\,154}}{{{\text{their}}\,140}}$$       (M1)

Note: Award (M1) for multiplying the given $$90$$ by their part (a)(ii) and dividing by their part (a)(i). Follow through from (a). Accept correct alternative methods.

$$= 99$$       (A1)(ft) (C2)

Note: Award a maximum of (M1)(A0) if the final answer is not an integer.

b.

## Question

Assume that the Earth is a sphere with a radius, $$r$$ , of $$6.38 \times {10^3}\,{\text{km}}$$ . i)     Calculate the surface area of the Earth in $${\text{k}}{{\text{m}}^2}$$.

ii)    Write down your answer to part (a)(i) in the form $$a \times {10^k}$$ , where $$1 \leqslant a < 10$$ and $$k \in \mathbb{Z}$$ .

a.

The surface area of the Earth that is covered by water is approximately $$3.61 \times {10^8}{\text{k}}{{\text{m}}^2}$$ .

Calculate the percentage of the surface area of the Earth that is covered by water.

b.

## Markscheme

i)     $$4\pi {(6.38 \times {10^3})^2}$$       (M1)

Note: Award (M1) for correct substitution into the surface area of a sphere formula.

$$= 512\,000\,000\,\,\,(511506576,\,\,162\,817\,600\pi )$$       (A1)    (C2)

Note: Award at most (M1)(A0) for use of $$3.14$$ for $$\pi$$, which will give an answer of $$511\,247\,264$$.

ii)    $$5.12 \times {10^8}\,\,\,(5.11506… \times {10^8},\,\,1.628176\pi \times {10^8})$$       (A1)(ft)(A1)(ft)    (C2)

Note: Award (A1) for $$5.12$$ and (A1) for $$\times {10^8}$$.
Award (A0)(A0) for answers of the type: $$5.12 \times {10^7}$$.

a.

$$\frac{{3.61 \times {{10}^8}}}{{5.11506…\,\, \times {{10}^8}}} \times 100$$  OR $$\frac{{3.61}}{{5.11506…\,}} \times 100$$  OR $$0.705758… \times 100$$        (M1)

Note: Award (M1) for correct substitution. Multiplication by $$100$$ must be seen.

$$= 70.6\,(\% )\,\,\,\,(70.5758…\,(\% ))$$        (A1)(ft)   (C2)

Note: Follow through from part (a). Accept the use of $$3$$ sf answers, which gives a final answer of $$70.5\,(\% )\,\,\,\,(70.5758…\,(\% ))$$ .

b.

## Question

A snack container has a cylindrical shape. The diameter of the base is $$7.84\,{\text{cm}}$$. The height of the container is $$23.4\,{\text{cm}}$$. This is shown in the following diagram. Write down the radius, in $${\text{cm}}$$, of the base of the container.

a.

Calculate the area of the base of the container.

b.

Dan is going to paint the curved surface and the base of the snack container.

Calculate the area to be painted.

c.

## Markscheme

$$3.92\,({\text{cm}})$$        (A1)    (C1)

a.

$$\pi \times {3.92^2}$$             (M1)

$$= 48.3\,{\text{c}}{{\text{m}}^2}\,\,\,(15.3664\,\pi \,{\text{c}}{{\text{m}}^2},\,\,48.2749…\,{\text{c}}{{\text{m}}^2})$$              (A1)(ft)         (C2)

Note: Award (M1) for correct substitution in area of circle formula. Follow through from their part (a). The answer is $$48.3\,{\text{c}}{{\text{m}}^2}$$, units are required.

b.

$$2 \times \pi \times 3.92 \times 23.4 + 48.3$$        (M1)(M1)

$$625\,{\text{c}}{{\text{m}}^2}\,\,\,(624.618…\,{\text{c}}{{\text{m}}^2})$$                  (A1)(ft)    (C3)

Note: Award (M1) for correct substitution in curved surface area formula, (M1) for adding their answer to part (b). Follow through from their parts (a) and (b). The answer is $$625\,{\text{c}}{{\text{m}}^2}$$, units are required.

c.

## Question

A balloon in the shape of a sphere is filled with helium until the radius is 6 cm.

The volume of the balloon is increased by 40%.

Calculate the volume of the balloon.

a.

Calculate the radius of the balloon following this increase.

b.

## Markscheme

Units are required in parts (a) and (b).

$$\frac{4}{3}\pi \times {6^3}$$    (M1)

Note:     Award (M1) for correct substitution into volume of sphere formula.

$$= 905{\text{ c}}{{\text{m}}^3}{\text{ }}(288\pi {\text{ c}}{{\text{m}}^3},{\text{ }}904.778 \ldots {\text{ c}}{{\text{m}}^3})$$    (A1)     (C2)

Note:     Answers derived from the use of approximations of $$\pi$$ (3.14; 22/7) are awarded (A0).[2 marks]

a.

Units are required in parts (a) and (b).

$$\frac{{140}}{{100}} \times 904.778 \ldots = \frac{4}{3}\pi {r^3}$$ OR $$\frac{{140}}{{100}} \times 288\pi = \frac{4}{3}\pi {r^3}$$ OR $$1266.69 \ldots = \frac{4}{3}\pi {r^3}$$     (M1)(M1)

Note:     Award (M1) for multiplying their part (a) by 1.4 or equivalent, (M1) for equating to the volume of a sphere formula.

$${r^3} = \frac{{3 \times 1266.69 \ldots }}{{4\pi }}$$ OR $$r = \sqrt{{\frac{{3 \times 1266.69 \ldots }}{{4\pi }}}}$$ OR $$r = \sqrt{{(1.4) \times {6^3}}}$$ OR $${r^3} = 302.4$$     (M1)

Note:     Award (M1) for isolating $$r$$.

$$(r = ){\text{ }}6.71{\text{ cm }}(6.71213 \ldots )$$     (A1)(ft)     (C4)

Note:     Follow through from part (a).[4 marks]

b.

## Question

A type of candy is packaged in a right circular cone that has volume $${\text{100 c}}{{\text{m}}^{\text{3}}}$$ and vertical height 8 cm. Find the radius, $$r$$, of the circular base of the cone.

a.

Find the slant height, $$l$$, of the cone.

b.

Find the curved surface area of the cone.

c.

## Markscheme

$$100 = \frac{1}{3}\pi {r^2}(8)$$     (M1)

Note:     Award (M1) for correct substitution into volume of cone formula.

$$r = 3.45{\text{ (cm) }}\left( {3.45494 \ldots {\text{ (cm)}}} \right)$$     (A1)     (C2)[2 marks]

a.

$${l^2} = {8^2} + {(3.45494 \ldots )^2}$$     (M1)

Note:     Award (M1) for correct substitution into Pythagoras’ theorem.

$$l = 8.71{\text{ (cm) }}\left( {8.71416 \ldots {\text{ (cm)}}} \right)$$     (A1)(ft)     (C2)

Note:     Follow through from part (a).[2 marks]

b.

$$\pi \times 3.45494 \ldots \times 8.71416 \ldots$$     (M1)

Note:     Award (M1) for their correct substitutions into curved surface area of a cone formula.

$$= 94.6{\text{ c}}{{\text{m}}^2}{\text{ }}(94.5836 \ldots {\text{ c}}{{\text{m}}^2})$$     (A1)(ft)     (C2)

Note:     Follow through from parts (a) and (b). Accept $$94.4{\text{ c}}{{\text{m}}^2}$$ from use of 3 sf values.[2 marks]

c.

## Question

A cylindrical container with a radius of 8 cm is placed on a flat surface. The container is filled with water to a height of 12 cm, as shown in the following diagram. A heavy ball with a radius of 2.9 cm is dropped into the container. As a result, the height of the water increases to $$h$$ cm, as shown in the following diagram. Find the volume of water in the container.

a.

Find the value of $$h$$.

b.

## Markscheme

$$\pi \times {8^2} \times 12$$     (M1)

Note:     Award (M1) for correct substitution into the volume of a cylinder formula.

$$2410{\text{ c}}{{\text{m}}^3}{\text{ }}(2412.74 \ldots {\text{ c}}{{\text{m}}^3},{\text{ }}768\pi {\text{ c}}{{\text{m}}^3})$$     (A1)     (C2)[2 marks]

a.

$$\frac{4}{3}\pi \times {2.9^3} + 768\pi = \pi \times {8^2}h$$     (M1)(M1)(M1)

Note:     Award (M1) for correct substitution into the volume of a sphere formula (this may be implied by seeing 102.160…), (M1) for adding their volume of the ball to their part (a), (M1) for equating a volume to the volume of a cylinder with a height of $$h$$.

OR

$$\frac{4}{3}\pi \times {2.9^3} = \pi \times {8^2}(h – 12)$$     (M1)(M1)(M1)

Note:     Award (M1) for correct substitution into the volume of a sphere formula (this may be implied by seeing 102.160…), (M1) for equating to the volume of a cylinder, (M1) for the height of the water level increase, $$h – 12$$. Accept $$h$$ for $$h – 12$$ if adding 12 is implied by their answer.

$$(h = ){\text{ }}12.5{\text{ (cm) }}\left( {12.5081 \ldots {\text{ (cm)}}} \right)$$     (A1)(ft)     (C4)

Note:     If 3 sf answer used, answer is 12.5 (12.4944…). Follow through from part (a) if first method is used.[4 marks]

b.

## Question

A solid right circular cone has a base radius of 21 cm and a slant height of 35 cm.
A smaller right circular cone has a height of 12 cm and a slant height of 15 cm, and is removed from the top of the larger cone, as shown in the diagram. Calculate the radius of the base of the cone which has been removed.

a.

Calculate the curved surface area of the cone which has been removed.

b.

Calculate the curved surface area of the remaining solid.

c.

## Markscheme

$$\sqrt {{{15}^2} – {{12}^2}}$$     (M1)

Note: Award (M1) for correct substitution into Pythagoras theorem.

OR

$$\frac{{{\text{radius}}}}{{21}} = \frac{{15}}{{35}}$$     (M1)

Note: Award (M1) for a correct equation.

= 9 (cm)     (A1) (C2)[2 marks]

a.

$$\pi \times 9 \times 15$$      (M1)

Note: Award (M1) for their correct substitution into curved surface area of a cone formula.

$$= 424\,\,{\text{c}}{{\text{m}}^2}\,\,\,\,\,\left( {135\pi ,\,\,424.115…{\text{c}}{{\text{m}}^2}} \right)$$     (A1)(ft) (C2)

Note: Follow through from part (a).[2 marks]

b.

$$\pi \times 21 \times 35 – 424.115…$$     (M1)

Note: Award (M1) for their correct substitution into curved surface area of a cone formula and for subtracting their part (b).

$$= 1880\,\,{\text{c}}{{\text{m}}^2}\,\,\,\,\,\left( {600\pi ,\,\,1884.95…{\text{c}}{{\text{m}}^2}} \right)$$     (A1)(ft) (C2)

Note: Follow through from part (b).[2 marks]

c.

## Question

Julio is making a wooden pencil case in the shape of a large pencil. The pencil case consists of a cylinder attached to a cone, as shown.

The cylinder has a radius of r cm and a height of 12 cm.

The cone has a base radius of r cm and a height of 10 cm. Find an expression for the slant height of the cone in terms of r.

a.

The total external surface area of the pencil case rounded to 3 significant figures is 570 cm2.

Using your graphic display calculator, calculate the value of r.

b.

## Markscheme

(slant height2 =) 102 + r 2     (M1)

Note: For correct substitution of 10 and r into Pythagoras’ Theorem.

$$\sqrt {{{10}^2} + {r^2}}$$     (A1) (C2)[2 marks]

a.

$$\pi {r^2} + 2\pi r \times 12 + \pi r\sqrt {100 + {r^2}} = 570$$     (M1)(M1)(M1)

Note: Award (M1) for correct substitution in curved surface area of cylinder and area of the base, (M1) for their correct substitution in curved surface area of cone, (M1) for adding their 3 surface areas and equating to 570. Follow through their part (a).

= 4.58   (4.58358…)      (A1)(ft) (C4)

Note: Last line must be seen to award final (A1). Follow through from part (a).[4 marks]

b.

## Question

A child’s toy consists of a hemisphere with a right circular cone on top. The height of the cone is $$12{\text{ cm}}$$ and the radius of its base is $$5{\text{ cm}}$$ . The toy is painted red. Calculate the length, $$l$$, of the slant height of the cone.

a.

Calculate the area that is painted red.

b.

## Markscheme

$$\sqrt {{5^2} + {{12}^2}}$$     (M1)

Note: Award (M1) for correct substitution in Pythagoras Formula.

=$$13{\text{ (cm)}}$$     (A1)     (C2)

a.

$${\text{Area}} = 2\pi {(5)^2} + \pi (5)(13)$$     (M1)(M1)(M1)

Notes: Award (M1) for surface area of hemisphere, (M1) for surface of cone, (M1) for addition of two surface areas. Follow through from their answer to part (a).

$$= 361{\text{ c}}{{\text{m}}^2}$$ ($$361.283 \ldots$$)     (A1)(ft)     (C4)

Note: The answer is $$361{\text{ c}}{{\text{m}}^2}$$ , the units are required.

b.