IB DP Mathematical Studies 5.0 Paper 2

 

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Question

Jenny has a circular cylinder with a lid. The cylinder has height 39 cm and diameter 65 mm.

An old tower (BT) leans at 10° away from the vertical (represented by line TG).

The base of the tower is at B so that \({\text{M}}\hat {\rm B}{\text{T}} = 100^\circ \).

Leonardo stands at L on flat ground 120 m away from B in the direction of the lean.

He measures the angle between the ground and the top of the tower T to be \({\text{B}}\hat {\rm L}{\text{T}} = 26.5^\circ \).

Calculate the volume of the cylinder in cm3. Give your answer correct to two decimal places.[3]

i.a.

The cylinder is used for storing tennis balls. Each ball has a radius of 3.25 cm.

Calculate how many balls Jenny can fit in the cylinder if it is filled to the top.[1]

i.b.

(i) Jenny fills the cylinder with the number of balls found in part (b) and puts the lid on. Calculate the volume of air inside the cylinder in the spaces between the tennis balls.

(ii) Convert your answer to (c) (i) into cubic metres.[4]

i.c.

(i) Find the value of angle \({\text{B}}\hat {\rm T}{\text{L}}\).

(ii) Use triangle BTL to calculate the sloping distance BT from the base, B to the top, T of the tower.[5]

ii.a.

Calculate the vertical height TG of the top of the tower.[2]

ii.b.

Leonardo now walks to point M, a distance 200 m from B on the opposite side of the tower. Calculate the distance from M to the top of the tower at T.[3]

ii.c.
Answer/Explanation

Markscheme

\(\pi \times 3.25^2 \times 39\)     (M1)(A1)

(= 1294.1398)

Answer 1294.14 (cm3)(2dp)     (A1)(ft)(G2)

(UP) not applicable in this part due to wording of question. (M1) is for substituting appropriate numbers from the problem into the correct formula, even if the units are mixed up. (A1) is for correct substitutions or correct answer with more than 2dp in cubic centimetres seen. Award (G1) for answer to > 2dp with no working and no attempt to correct to 2dp. Award (M1)(A0)(A1)(ft) for \(\pi  \times {32.5^2} \times 39{\text{ c}}{{\text{m}}^3}\) (= 129413.9824) = 129413.98

Use of \(\pi = \frac{22}{7}\) or 3.142 etc is premature rounding and is awarded at most (M1)(A1)(A0) or (M1)(A0)(A1)(ft) depending on whether the intermediate value is seen or not. For all other incorrect substitutions, award (M1)(A0) and only follow through the 2 dp correction if the intermediate answer to more decimal places is seen. Answer given as a multiple of \(\pi\) is awarded at most (M1)(A1)(A0). As usual, an unsubstituted formula followed by correct answer only receives the G marks.[3 marks]

i.a.

39/6.5 = 6     (A1)[1 mark]

i.b.

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

(UP) (i) Volume of one ball is \(\frac{4}{3} \pi \times 3.25^3 {\text{ cm}}^3\)     (M1)

\({\text{Volume of air}} = \pi  \times {3.25^2} \times 39 – 6 \times \frac{4}{3}\pi  \times {3.25^3} = 431{\text{ c}}{{\text{m}}^3}\)     (M1)(A1)(ft)(G2)

Award first (M1) for substituted volume of sphere formula or for numerical value of sphere volume seen (143.79… or 45.77… \( \times \pi\)). Award second (M1) for subtracting candidate’s sphere volume multiplied by their answer to (b). Follow through from parts (a) and (b) only, but negative or zero answer is always awarded (A0)(ft)(UP) (ii) 0.000431m3 or  4.31×10−4 m3     (A1)(ft)[4 marks]

i.c.

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

(i) \({\text{Angle B}}\widehat {\text{T}}{\text{L}} = 180 – 80 – 26.5\) or \(180 – 90 – 26.5 – 10\)     (M1)

\(= 73.5^\circ\)     (A1)(G2)

(ii) \(\frac{{BT}}{{\sin (26.5^\circ )}} = \frac{{120}}{{\sin (73.5^\circ )}}\)     (M1)(A1)(ft)

(UP) BT = 55.8 m (3sf)     (A1)(ft) [5 marks]


If radian mode has been used throughout the question, award (A0) to the first incorrect answer then follow through, but

negative lengths are always awarded (A0)(ft).

The answers are (all 3sf)

(ii)(a)     – 124 m (A0)(ft)

(ii)(b)     123 m (A0)

(ii)(c)     313 m (A0)

If radian mode has been used throughout the question, award (A0) to the first incorrect answer then follow through, but negative lengths are always awarded (A0)(ft)

ii.a.

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

TG = 55.8sin(80°) or 55.8cos(10°)     (M1)

(UP) = 55.0 m (3sf)     (A1)(ft)(G2)

Apply (AP) if 0 missing[2 marks]

If radian mode has been used throughout the question, award (A0) to the first incorrect answer then follow through, but
negative lengths are always awarded (A0)(ft).

The answers are (all 3sf)

(ii)(a)     – 124 m (A0)(ft)

(ii)(b)     123 m (A0)

(ii)(c)     313 m (A0)

If radian mode has been used throughout the question, award (A0) to the first incorrect answer then follow through, but negative lengths are always awarded (A0)(ft)

ii.b.

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

\({\text{MT}}^2 = 200^2 + 55.8^2 – 2 \times 200 \times 55.8 \times \cos(100^\circ)\)     (M1)(A1)(ft)

(UP) MT = 217 m  (3sf)     (A1)(ft)

Follow through only from part (ii)(a)(ii). Award marks at discretion for any valid alternative method.[3 marks]


If radian mode has been used throughout the question, award (A0) to the first incorrect answer then follow through, but

negative lengths are always awarded (A0)(ft).

The answers are (all 3sf)

(ii)(a)     – 124 m (A0)(ft)

(ii)(b)     123 m (A0)

(ii)(c)     313 m (A0)

If radian mode has been used throughout the question, award (A0) to the first incorrect answer then follow through, but negative lengths are always awarded (A0)(ft)

ii.c.

Question

ABCDV is a solid glass pyramid. The base of the pyramid is a square of side 3.2 cm. The vertical height is 2.8 cm. The vertex V is directly above the centre O of the base.

Calculate the volume of the pyramid.[2]

a.

The glass weighs 9.3 grams per cm3. Calculate the weight of the pyramid.[2]

b.

Show that the length of the sloping edge VC of the pyramid is 3.6 cm.[4]

c.

Calculate the angle at the vertex, \({\text{B}}{\operatorname {\hat V}}{\text{C}}\).[3]

d.

Calculate the total surface area of the pyramid.[4]

e.
Answer/Explanation

Markscheme

Unit penalty (UP) is applicable in question parts (a), (b) and (e) only.

\({\text{V}} = \frac{1}{3} \times {3.2^2} \times 2.8\)     (M1)

(M1) for substituting in correct formula

(UP) = 9.56 cm3     (A1)(G2)[2 marks]

a.

Unit penalty (UP) is applicable in question parts (a), (b) and (e) only.

\(9.56 \times 9.3\)     (M1)

(UP) = 88.9 grams     (A1)(ft)(G2)[2 marks]

b.

\(\frac{1}{2} {\text{base}} = 1.6 {\text{ seen}}\)     (M1)

award (M1) for halving base

\({\text{OC}}^2 = 1.6^2 + 1.6^2 = 5.12\)     (A1)

award (A1) for one correct use of Pythagoras

\(5.12 + 2.8^2 = 12.96 = {\text{VC}}^2\)     (M1)

award (M1) for using Pythagoras again to find VC2

VC = 3.6 AG

award (A1) for 3.6 obtained from 12.96 only (not 12.95…)     (A1)

OR

\({\text{AC}}^2 = 3.2^2 + 3.2^2 = 20.48\)     (A1)

award (A1) for one correct use of Pythagoras

({\text{OC}} = \frac{1}{2} \sqrt{20.48}\) ( = 2.26…)     (M1)

award (M1) for halving AC

\(2.8^2 + (2.26…)^2 = {\text{VC}}^2 = 12.96\)     (M1)

award (M1) for using Pythagoras again to find VC2

VC = 3.6 AG     (A1)

award (A1) for 3.6 obtained from 12.96 only (not 12.95…)[4 marks]

c.

\(3.2^2 = 3.6^2 + 3.6^2 – 2 \times (3.6) (3.6) \cos\) \({\text{B}}{\operatorname {\hat V}}{\text{C}}\)     (M1)(A1)

\({\text{B}}{\operatorname {\hat V}}{\text{C}}\) \( = {52.8^\circ }\) (no (ft) here)     (A1)(G2)

award (M1) for substituting in correct formula, (A1) for correct substitution

OR

\(\sin\) \({\text{B}}{\operatorname {\hat V}}{\text{M}}\) \( = \frac{{1.6}}{{3.6}}\) where M is the midpoint of BC     (M1)(A1)

\({\text{B}}{\operatorname {\hat V}}{\text{C}}\) \( = {52.8^\circ}\) (no (ft) here)     (A1)[3 marks]

d.

Unit penalty (UP) is applicable in question parts (a), (b) and (e) only.

\(4 \times \frac{1}{2}{(3.6)^2} \times \sin (52.8^\circ ) + {(3.2)^2}\)     (M1)(M1)(M1)

award (M1) for \( \times 4\), (M1) for substitution in relevant triangle area, (\(\frac{1}{2}(3.2)(2.8)\) gets (M0))

(M1) for \(+ {(3.2)^2}\)

(UP) = 30.9 cm2 ((ft) from their (d))     (A1)(ft)(G2)[4 marks]

e.

Question

Mal is shopping for a school trip. He buys \(50\) tins of beans and \(20\) packets of cereal. The total cost is \(260\) Australian dollars (\({\text{AUD}}\)).

The triangular faces of a square based pyramid, \({\text{ABCDE}}\), are all inclined at \({70^ \circ }\) to the base. The edges of the base \({\text{ABCD}}\) are all \(10{\text{ cm}}\) and \({\text{M}}\) is the centre. \({\text{G}}\) is the mid-point of \({\text{CD}}\).

Write down an equation showing this information, taking \(b\) to be the cost of one tin of beans and \(c\) to be the cost of one packet of cereal in \({\text{AUD}}\).[1]

i.a.

Stephen thinks that Mal has not bought enough so he buys \(12\) more tins of beans and \(6\) more packets of cereal. He pays \(66{\text{ AUD}}\).

Write down another equation to represent this information.[1]

i.b.

Stephen thinks that Mal has not bought enough so he buys \(12\) more tins of beans and \(6\) more packets of cereal. He pays \(66{\text{ AUD}}\).

Find the cost of one tin of beans.[2]

i.c.

(i)     Sketch the graphs of the two equations from parts (a) and (b).

(ii)    Write down the coordinates of the point of intersection of the two graphs.[4]

i.d.

Using the letters on the diagram draw a triangle showing the position of a \({70^ \circ }\) angle.[1]

ii.a.

Show that the height of the pyramid is \(13.7{\text{ cm}}\), to 3 significant figures.[2]

ii.b.

Calculate

(i)     the length of \({\text{EG}}\);

(ii)    the size of angle \({\text{DEC}}\).[4]

ii.c.

Find the total surface area of the pyramid.[2]

ii.d.

Find the volume of the pyramid.[2]

ii.e.
Answer/Explanation

Markscheme

\(50b + 20c = 260\)     (A1)[1 mark]

i.a.

\(12b + 6c = 66\)     (A1)[1 mark]

i.b.

Solve to get \(b = 4\)     (M1)(A1)(ft)(G2)

Note: (M1) for attempting to solve the equations simultaneously.[2 marks]

i.c.

(i)

     (A1)(A1)(A1)


Notes: Award (A1) for labels and some idea of scale, (A1)(ft)(A1)(ft) for each line.
The axis can be reversed.

(ii)    \((4,3)\) or \((3,4)\)     (A1)(ft)

Note: Accept \(b = 4\), \(c = 3\)[4 marks]

i.d.

     (A1)[1 mark]

ii.a.

\(\tan 70 = \frac{h}{5}\)     (M1)

\(h = 5\tan 70 = 13.74\)     (A1)

\(h = 13.7{\text{ cm}}\)     (AG)[2 marks]

ii.b.

Unit penalty (UP) is applicable in this part of the question where indicated in the left hand column.

(i)     \({\text{E}}{{\text{G}}^2} = {5^2} + {13.7^2}\) OR \({5^2} + {(5\tan 70)^2}\)     (M1)

(UP)     \({\text{EG}} = 14.6{\text{ cm}}\)     (A1)(G2)

(ii)    \({\text{DEC}} = 2 \times {\tan ^{ – 1}}\left( {\frac{5}{{14.6}}} \right)\)     (M1)

\( = {37.8^ \circ }\)     (A1)(ft)(G2)[4 marks]

ii.c.

Unit penalty (UP) is applicable in this part of the question where indicated in the left hand column.

\({\text{Area}} = 10 \times 10 + 4 \times 0.5 \times 10 \times 14.619\)     (M1)

(UP)     \( = 392{\text{ c}}{{\text{m}}^2}\)     (A1)(ft)(G2)[2 marks]

ii.d.

Unit penalty (UP) is applicable in this part of the question where indicated in the left hand column.

\({\text{Volume}} = \frac{1}{3} \times 10 \times 10 \times 13.7\)     (M1)

(UP)     \( = 457{\text{ c}}{{\text{m}}^3}\) (\(458{\text{ c}}{{\text{m}}^3}\))     (A1)(G2)[2 marks]

ii.e.

Question

Amir needs to construct an isosceles triangle \({\text{ABC}}\) whose area is \(100{\text{ cm}}^2\). The equal sides, \({\text{AB}}\) and \({\text{BC}}\), are \(20{\text{ cm}}\) long.

Sylvia is making a square-based pyramid. Each triangle has a base of length \(12{\text{ cm}}\) and a height of \(10{\text{ cm}}\).

Angle \({\text{ABC}}\) is acute. Show that the angle \({\text{ABC}}\) is \({30^ \circ }\).[2]

i.a.

Find the length of \({\text{AC}}\).[3]

i.b.

Show that the height of the pyramid is \(8{\text{ cm}}\).[2]

ii.a.

\({\text{M}}\) is the midpoint of the base of one of the triangles and \({\text{O}}\) is the apex of the pyramid.

Find the angle that the line \({\text{MO}}\) makes with the base of the pyramid.[3]

ii.b.

Calculate the volume of the pyramid.[2]

ii.c.

Daniel wants to make a rectangular prism with the same volume as that of Sylvia’s pyramid. The base of his prism is to be a square of side \(10{\text{ cm}}\). Calculate the height of the prism.[2]

ii.d.
Answer/Explanation

Markscheme

\(\frac{1}{2}{20^2}\sin B = 100\)     (M1)(A1)

\(B = {30^ \circ }\)    (AG)

Note: (M1) for correct substituted formula and (A1) for correct substitution. \(B = {30^ \circ }\) must be seen or previous (A1) mark is lost.[2 marks]

i.a.

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

\({\overline {{\text{AC}}} ^2} = 2 \times {20^2} – 2 \times {20^2} \times \cos {30^ \circ }\)     (M1)(A1)

(UP)     \(\overline {{\text{AC}}}  = 10.4{\text{ cm}}\)     (A1)(G2)

Note: (M1) for using cosine rule, (A1) for correct substitution. Last (A1) is for the correct answer. Accept use of sine rule or any correct method e.g. \({\text{AC}} = 2 \times 20\sin {15^ \circ }\) .[3 marks]

i.b.

\({x^2} + {6^2} = {10^2}\)     (A1)(M1)

\(x = 8{\text{ cm}}\)     (AG)

Note: (A1) for \(6\) (or \(36\)) seen and (M1) for using Pythagoras with correct substitution. \(x = 8\) must be seen or previous (M1) mark is lost.[2 marks]

ii.a.

\(\cos \beta  = \frac{6}{{10}}\)     (M1)(A1)

\(\beta  = {53.1^ \circ }\)     (A1)(G2)

OR equivalent

Note: (M1) for use of trigonometric ratio with numbers from question. (A1) for use of correct numbers, and (A1) for correct answer.[3 marks]

ii.b.

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

\(vol = \frac{{{{12}^2} \times 8}}{3}\)     (M1)

(UP)     \( = 384{\text{ c}}{{\text{m}}^3}\)     (A1)(G2)

Note: (M1) for correct formula and correct substitution, (A1) for correct answer.[2 marks]

ii.c.

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

Let h be the height

\({10^2}h = 384\)     (M1)

(UP)     \( = 3.84{\text{ cm}}\)     (A1)(ft)(G2)

Note: (M1) for correct formula and correct substitution, (A1) for correct answer. (ft) from answer to part (c).[2 marks]

ii.d.

Question

On the coordinate axes below, \({\text{D}}\) is a point on the \(y\)-axis and \({\text{E}}\) is a point on the \(x\)-axis. \({\text{O}}\) is the origin. The equation of the line \({\text{DE}}\) is \(y + \frac{1}{2}x = 4\).

Write down the coordinates of point \({\text{E}}\).[2]

a.

\({\text{C}}\) is a point on the line \({\text{DE}}\). \({\text{B}}\) is a point on the \(x\)-axis such that \({\text{BC}}\) is parallel to the \(y\)-axis. The \(x\)-coordinate of \({\text{C}}\) is \(t\).

Show that the \(y\)-coordinate of \({\text{C}}\) is \(4 – \frac{1}{2}t\).[2]

b.

\({\text{OBCD}}\) is a trapezium. The \(y\)-coordinate of point \({\text{D}}\) is \(4\).

Show that the area of \({\text{OBCD}}\) is \(4t – \frac{1}{4}{t^2}\).[3]

c.

The area of \({\text{OBCD}}\) is \(9.75\) square units. Write down a quadratic equation that expresses this information.[1]

d.

(i) Using your graphic display calculator, or otherwise, find the two solutions to the quadratic equation written in part (d).

(ii) Hence find the correct value for \(t\). Give a reason for your answer.[4]

e.
Answer/Explanation

Markscheme

\({\text{E}}(8{\text{, }}0)\)    (A1)(A1)

Notes: Brackets required but do not penalize again if mark lost in Q4 (i)(d). If missing award (A1)(A0).
Accept \(x = 8\), \(y = 0\)
Award (A1) for \(x = 8\)

a.

\(y + \frac{1}{2}t = 4\)     (M1)(M1)

Note: (M1) for the equation of the line seen. (M1) for substituting \(t\).

\(y = 4 – \frac{1}{2}t\)     (AG)

Note: Final line must be seen or previous (M1) mark is lost.[2 marks]

b.

\({\text{Area}} = \frac{1}{2} \times (4 + 4 – \frac{1}{2}t) \times t\)     (M1)(A1)

Note: (M1) for substituting in correct formula, (A1) for correct substitution.

\( = \frac{1}{2} \times (8 – \frac{1}{2}t) \times t = \frac{1}{2}(8t – \frac{1}{2}{t^2})\)     (A1)
\( = 4t – \frac{1}{4}{t^2}\)     (AG)

Note: Final line must be seen or previous (A1) mark is lost.[3 marks]

c.

\(4t – \frac{1}{4}{t^2} = 9.75\) or any equivalent form.     (A1)[1 mark]

d.

(i) \(t = 3\) or \(t =13\)     (A1)(ft)(A1)(ft)(G2)

Note: Follow through from candidate’s equation to part (d). Award (A0)(A1)(ft) for \((3{\text{, }}0)\) and \((13{\text{, }}0)\).

(ii) \(t\) must be a value between \(0\) and \(8\) then \(t = 3\)

Note: Accept \({\text{B}}\) is between \({\text{O}}\) and \({\text{E}}\). Do not award (R0)(A1).

e.

Question

The quadrilateral ABCD shown below represents a sandbox. AB and BC have the same length. AD is \(9{\text{ m}}\) long and CD is \(4.2{\text{ m}}\) long. Angles ADC and ABC are \({95^ \circ }\) and \({130^ \circ }\) respectively.

Find the length of AC.[3]

a.

(i)     Write down the size of angle BCA.

(ii)    Calculate the length of AB.[4]

b.

Show that the area of the sandbox is \(31.1{\text{ }}{{\text{m}}^2}\) correct to 3 s.f.[4]

c.

The sandbox is a prism. Its edges are \(40{\text{ cm}}\) high. The sand occupies one third of the volume of the sandbox. Calculate the volume of sand in the sandbox.[3]

d.
Answer/Explanation

Markscheme

\({\text{A}}{{\text{C}}^2} = {9^2} + {4.2^2} – 2 \times 9 \times 4.2 \times \cos {95^ \circ }\)     (M1)(A1)
\({\text{AC}} = 10.3{\text{ m}}\)     (A1)(G2)

Note: (M1) for correct substituted formula and (A1) for correct substitution. If radians used answer is \(6.59\). Award at most (M1)(A1)(A0).

Note: The final A1 is only awarded if the correct units are present; only penalize once for the lack of units or incorrect units.

a.

(i)     \({\text{B}}\hat{\text{C}}{\text{A}} = {25^ \circ }\)     (A1)

(ii)    \(\frac{{{\text{AB}}}}{{\sin {{25}^ \circ }}} = \frac{{10.258 \ldots }}{{\sin {{130}^ \circ }}}\)     (M1)(A1)

\({\text{AB}} = 5.66{\text{ m}}\)     (A1)(ft)(G2)

Note: (M1) for correct substituted formula and (A1) for correct substitution. (A1) for correct answer.

Follow through with angle \({\text{B}}\)\(\hat{\text{C}}\)\({\text{A}}\) and their AC. Allow \({\text{AB}} = 5.68\) if \({\text{AC}} = 10.3\) used. If radians used answer is \(0.938\) (unreasonable answer). Award at most (M1)(A1)(A0)(ft).

OR

Using that ABC is isosceles

\({\text{cos2}}{{\text{5}}^ \circ } = \frac{{\frac{1}{2} \times 10.258 \ldots }}{{{\text{AB}}}}\) (or equivalent)     (A1)(M1)(ft)

\({\text{AB}} = 5.66{\text{ m}}\)     (A1)(ft)(G2)

Note: (A1) for \(\frac{1}{2}\) of their AB seen, (M1) for correct trigonometric ratio and correct substitution, (A1) for correct answer. If \(\frac{1}{2}{\text{AB}}\) seen and correct answer is given award (A1)(G1). Allow \({\text{AB}} = 5.68\) if \({\text{AC}} = 10.3\) used. If radians used answer is \(3.32\). Award (A1)(M1)(A1)(ft). If \(\sin 65\) and radians used answer is \(3.99\). Award (A1)(M1)(A1)(ft).

Note: The final A1 is only awarded in (ii) if the correct units are present; only penalize once for the lack of units or incorrect units.

b.

Area \( = \frac{1}{2} \times 9 \times 4.2 \times \sin {95^ \circ } + \frac{1}{2} \times {(5.6592 \ldots )^2} \times \sin {130^ \circ }\)    (M1)(M1)(ft)(M1)

\( = 31.095 \ldots  = 31.1{\text{ }}{{\text{m}}^2}\) (correct to 3 s.f.)     (A1)(AG)

Note: (M1)(M1) each for correct substitution in the formula of the area of each triangle, (M1) for adding both areas. (A1) for unrounded answer. Follow through with their length of AB but last mark is lost if they do not reach the correct answer.

c.

Volume of sand \( = \frac{1}{3}(31.09 \ldots  \times 0.4)\)     (M1)(M1)

\( = 4.15{\text{ }}{{\text{m}}^3}\)     (A1)(G2)

Note: (M1) for correct formula of volume of prism and for correct substitution, (M1) for multiplying by \(\frac{1}{3}\) and last (A1) for correct answer only.

Note: The final A1 is only awarded if the correct units are present; only penalize once for the lack of units or incorrect units.

d.

Question

The vertices of quadrilateral ABCD as shown in the diagram are A (3, 1), B (0, 2), C (–2, 1) and D (–1, –1).

Calculate the gradient of line CD.[2]

a.

Show that line AD is perpendicular to line CD.[2]

b.

Find the equation of line CD. Give your answer in the form \(ax + by = c\) where \(a,{\text{ }}b,{\text{ }}c \in \mathbb{Z}\).[3]

c.

Lines AB and CD intersect at point E. The equation of line AB is \(x + 3y = 6\).

Find the coordinates of E.[2]

d.

Lines AB and CD intersect at point E. The equation of line AB is \(x + 3y = 6\).

Find the distance between A and D.[2]

e.

The distance between D and E is \(\sqrt{20}\).

Find the area of triangle ADE.[2]

f.
Answer/Explanation

Markscheme

\({\text{Gradient of CD}} = \frac{{1 – ( – 1)}}{{ – 2 – ( – 1)}}\)     (M1)

\( =  – 2\)     (A1)(G2)

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

a.

\({\text{Gradient of AD}} = \frac{1}{2}\)     (A1)

\( – 2 \times \frac{1}{2} =  – 1\) or \(\frac{1}{2}\) is negative reciprocal of –2     (M1)

Hence AD is perpendicular for CD.     (AG)

Note: Last line must be seen for the (M1) to be awarded.[2 marks]

b.

\(y = -2x – 3\)     (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for their (a), (A1)(ft) for –3.

If part (a) incorrect award (A1)(ft) for their y-intercept only if working is seen.

OR

\(y – 1 = -2(x + 2)\)     (A1)(ft)(A1)

OR

\(y + 1 = -2(x + 1)\)     (A1)(ft)(A1)

Note: Award (A1)(ft) for their (a), (A1) for correct substitution of point.

\(2x + y = -3\)     (A1)(ft) 

Note: The final (A1)(ft) is for their equation in the stated form.[3 marks]

c.

E (−3, 3) (Accept x = −3, y = 3)     (G2)

OR

Award (M1) for solving the pair of simultaneous equations by hand. (A1)(ft) for correct answer, (ft) from their (c).     (M1)(A1)(ft)

OR

Award (M1) for having extended the lines in their own graph seen drawn on answer paper. (A1) for correct answer.     (M1)(A1)

Note: Missing coordinate brackets receive (G1)(G0) or (M1)(A0).[2 marks]

d.

Distance between A and D = \(\sqrt {4^2 + 2^2}\)     (M1)

\( = \sqrt{20}\) OR \(2 \sqrt {5}\) OR 4.47 (3 s.f.)     (A1)(G2)

Note: Award (M1) for correct substitution into the distance formula, (A1) for correct answer.[2 marks]

e.

Area of ADE = \(\frac{1}{2}\sqrt {20}  \times \sqrt {20} \)     (M1)

= 10     (A1)(ft)(G2)

Follow through from (e).[2 marks]

f.

Question

A farmer has a triangular field, ABC, as shown in the diagram.

AB = 35 m, BC = 80 m and BÂC = 105°, and D is the midpoint of BC.

Find the size of BĈA.[3]

a.

Calculate the length of AD.[5]

b.

The farmer wants to build a fence around ABD.

Calculate the total length of the fence.[2]

c.

The farmer wants to build a fence around ABD.

The farmer pays 802.50 USD for the fence. Find the cost per metre.[2]

d.

Calculate the area of the triangle ABD.[3]

e.

A layer of earth 3 cm thick is removed from ABD. Find the volume removed in cubic metres.[3]

f.
Answer/Explanation

Markscheme

\(\frac{{\sin {\text{BCA}}}}{{35}} = \frac{{\sin 105^\circ }}{{80}}\)     (M1)(A1)

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

\({\text{B}}{\operatorname{\hat C}}{\text{A}} = 25.0^{\circ}\)     (A1)(G2)[3 marks]

a.

Note: Unit penalty (UP) applies in parts (b)(c) and (e)

Length BD = 40 m     (A1)

Angle ABC = 180° − 105° − 25° = 50°     (A1)(ft)

Note: (ft) from their answer to (a).

AD2 = 352 + 402 − (2 × 35 × 40 × cos 50°)     (M1)(A1)(ft)

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

(UP)     AD = 32.0 m     (A1)(ft)(G3)

Notes: If 80 is used for BD award at most (A0)(A1)(ft)(M1)(A1)(ft)(A1)(ft) for an answer of 63.4 m.

If the angle ABC is incorrectly calculated in this part award at most (A1)(A0)(M1)(A1)(ft)(A1)(ft).

If angle BCA is used award at most (A1)(A0)(M1)(A0)(A0).[5 marks]

b.

Note: Unit penalty (UP) applies in parts (b)(c) and (e)

length of fence = 35 + 40 + 32     (M1)

(UP)     = 107 m     (A1)(ft)(G2)

Note: (M1) for adding 35 + 40 + their (b).[2 marks]

c.

cost per metre \( = \frac{802.50}{107}\)     (M1)

Note: Award (M1) for dividing 802.50 by their (c).

cost per metre = 7.50 USD (7.5 USD) (USD not required)     (A1)(ft)(G2)[2 marks]

d.

Note: Unit penalty (UP) applies in parts (b)(c) and (e)

Area of ABD \( = \frac{1}{2} \times 35 \times 40 \times \sin 50^\circ \)     (M1)

= 536.2311102     (A1)(ft)

(UP)     = 536 m2     (A1)(ft)(G2)

Note: Award (M1) for correct substituted formula, (A1)(ft) for correct substitution, (ft) from their value of BD and their angle ABC in (b).[3 marks]

e.

Volume = 0.03 × 536     (A1)(M1)

= 16.08

= 16.1     (A1)(ft)(G2)

Note: Award (A1) for 0.03, (M1) for correct formula. (ft) from their (e).

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

f.

Question

The diagram below shows a square based right pyramid. ABCD is a square of side 10 cm. VX is the perpendicular height of 8 cm. M is the midpoint of BC.

In a mountain region there appears to be a relationship between the number of trees growing in the region and the depth of snow in winter. A set of 10 areas was chosen, and in each area the number of trees was counted and the depth of snow measured. The results are given in the table below.

A path goes around a forest so that it forms the three sides of a triangle. The lengths of two sides are 550 m and 290 m. These two sides meet at an angle of 115°. A diagram is shown below.

Write down the length of XM.[1]

A, a.

Use your graphic display calculator to find the standard deviation of the number of trees.[1]

A, a, ii.

Calculate the length of VM.[2]

A, b.

Calculate the angle between VM and ABCD.[2]

A, c.

Calculate the length of the third side of the triangle. Give your answer correct to the nearest 10 m.[4]

B, a.

Calculate the area enclosed by the path that goes around the forest.[3]

B, b.

Inside the forest a second path forms the three sides of another triangle named ABC. Angle BAC is 53°, AC is 180 m and BC is 230 m.

Calculate the size of angle ACB.[4]

B, c.
Answer/Explanation

Markscheme

UP applies in this question

(UP)     XM = 5 cm     (A1)[1 mark]

A, a.

16.8     (G1)[1 mark]

A, a, ii.

UP applies in this question

VM2 = 52 + 82     (M1)

Note: Award (M1) for correct use of Pythagoras Theorem.

(UP)     VM = \(\sqrt{89}\) = 9.43 cm     (A1)(ft)(G2)[2 marks]

A, b.

\(\tan {\text{VMX}} = \frac{8}{5}\)     (M1)

Note: Other trigonometric ratios may be used.

\({\rm{V\hat MX}} = 58.0^\circ \)     (A1)(ft)(G2)[2 marks]

A, c.

UP applies in this question

l2 = 2902 + 5502 − 2 × 290 × 550 × cos115°     (M1)(A1)

Note: Award (M1) for substituted cosine rule formula, (A1) for correct substitution.

l = 722     (A1)(G2)

(UP)     = 720 m     (A1)

Note: If 720 m seen without working award (G3).

The final (A1) is awarded for the correct rounding of their answer.[4 marks]

B, a.

UP applies in this question

\({\text{Area}} = \frac{1}{2} \times 290 \times 550 \times \sin 115\)     (M1)(A1)

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

(UP)     = \(72\,300{\text{ }}{{\text{m}}^2}\)     (A1)(G2)[3 marks]

B, b.

\(\frac{{180}}{{\sin {\text{B}}}} = \frac{{230}}{{\sin 53}}\)     (M1)(A1)

Note: Award (M1) for substituted sine rule formula, (A1) for correct substitution.

B = 38.7°     (A1)(G2)

\({\operatorname{A\hat CB}} = 180 – (53^\circ  + 38.7^\circ )\)

\( = 88.3^\circ \)     (A1)(ft) [4 marks]

B, c.

Question

The diagram shows triangle ABC. Point C has coordinates (4, 7) and the equation of the line AB is x + 2y = 8.

Find the coordinates of A.[1]

a.i.

Find the coordinates of B.[1]

a.ii.

Show that the distance between A and B is 8.94 correct to 3 significant figures.[2]

b.

N lies on the line AB. The line CN is perpendicular to the line AB.

Find the gradient of CN.[3]

c.i.

N lies on the line AB. The line CN is perpendicular to the line AB.

Find the equation of CN.[2]

c.ii.

N lies on the line AB. The line CN is perpendicular to the line AB.

Calculate the coordinates of N.[3]

d.

It is known that AC = 5 and BC = 8.06.

Calculate the size of angle ACB.[3]

e.

It is known that AC = 5 and BC = 8.06.

Calculate the area of triangle ACB.[3]

f.
Answer/Explanation

Markscheme

A(0, 4)     Accept x = 0, y = 4     (A1)[1 mark]

a.i.

B(8, 0)     Accept x = 8, y = 0     (A1)(ft)

Note: Award (A0) if coordinates are reversed in (i) and (A1)(ft) in (ii).[1 mark]

a.ii.

\({\text{AB}} = \sqrt {{8^2} + {4^2}} {\text{  }} = \sqrt {80} \)     (M1)

AB = 8.944     (A1)

= 8.94     (AG)[2 marks]

b.

y = –0.5x + 4     (M1)
Gradient AB = –0.5     (A1)

Note: Award (A2) if –0.5 seen.

OR

Gradient \({\text{AB}} = \frac{{(0 – 4)}}{{(8 – 0)}}\)     (M1)
\( = -\frac{1}{2}\)     (A1)

Note: Award (M1) for correct substitution in the gradient formula. Follow through from their answers to part (a).

Gradient CN = 2     (A1)(ft)(G2)

Note: Special case: Follow through for gradient CN from their gradient AB.[3 marks]

c.i.

CN: y = 2x + c

7 = 2(4) + c     (M1)

Note: Award (M1)for correct substitution in equation of a line.

y = 2x – 1     (A1)(ft)(G2)

Note: Accept alternative forms for the equation of a line including y – 7 = 2(x – 4) . Follow through from their gradient in (i).

Note: If c = –1 seen but final answer is not given, award (A1)(d).[2 marks]

c.ii.

x + 2(2x – 1) = 8     or equivalent     (M1)

N(2, 3) (x = 2, y = 3)     (A1)(A1)(ft)(G3)

Note: Award (M1) for attempt to solve simultaneous equations or a sketch of the two lines with an indication of the point of intersection.[3 marks]

d.

Cosine rule: \(\cos ({\rm{A\hat CB)}} = \frac{{{5^2} + {{8.06}^2} – {{8.944}^2}}}{{2 \times 5 \times 8.06}}\)     (M1)(A1)

Note: Award (M1) for use of cosine rule with numbers from the problem substituted, (A1) for correct substitution.

\({\rm{A\hat CB  =  82.9^\circ }}\)     (A1)(G2)

Note: If alternative right-angled trigonometry method used award (M1) for use of trig ratio in both triangles, (A1) for correct substitution of their values in each ratio, (A1) for answer.

Note: Accept 82.8° with use of 8.94.[3 marks]

e.

Area \({\text{ACB}} = \frac{{5 \times 8.06\sin (82.9)}}{2}\)     (M1)(A1)(ft)

Note: Award (M1) for substituted area formula, (A1) for correct substitution. Follow through from their angle in part (e).

OR

Area \({\text{ACB}} = \frac{{{\text{AB}} \times {\text{CN}}}}{2} = \frac{{8.94 \times \sqrt {{{(4 – 2)}^2} + {{(7 – 3)}^2}} }}{2}\)     (M1)(M1)(ft)

Note: Award (M1) substituted area formula with their values, (M1) for substituted distance formula. Follow through from
coordinates of N.

Area ACB = 20.0     (A1)(ft)(G2)

Note: Accept 20[3 marks]

f.

Question

The diagram shows an office tower of total height 126 metres. It consists of a square based pyramid VABCD on top of a cuboid ABCDPQRS.

V is directly above the centre of the base of the office tower.

The length of the sloping edge VC is 22.5 metres and the angle that VC makes with the base ABCD (angle VCA) is 53.1°.

Write down the length of VA in metres.[1]

a.i.

Sketch the triangle VCA showing clearly the length of VC and the size of angle VCA.[1]

a.ii.

Show that the height of the pyramid is 18.0 metres correct to 3 significant figures.[2]

b.

Calculate the length of AC in metres.[3]

c.

Show that the length of BC is 19.1 metres correct to 3 significant figures.[2]

d.

Calculate the volume of the tower.[4]

e.

To calculate the cost of air conditioning, engineers must estimate the weight of air in the tower. They estimate that 90 % of the volume of the tower is occupied by air and they know that 1 m3 of air weighs 1.2 kg.

Calculate the weight of air in the tower.[3]

f.
Answer/Explanation

Markscheme

22.5 (m)     (A1)[1 mark]

a.i.

     (A1)[1 mark]

a.ii.

h = 22.5     sin 53.1°     (M1)
= 17.99     (A1)
= 18.0     (AG)

Note: Unrounded answer must be seen for (A1) to be awarded.

Accept 18 as (AG).[2 marks]

b.

\({\text{AC}} = 2\sqrt {{{22.5}^2} – {{17.99…}^2}} \)     (M1)(M1)

Note: Award (M1) for multiplying by 2, (M1) for correct substitution into formula.

OR

AC = 2(22.5)cos53.1°     (M1)(M1)

Notes: Award (M1) for correct use of cosine trig ratio, (M1) for multiplying by 2.

OR

AC2 = 22.52 + 22.52 – 2(22.5)(22.5) cos73.8°     (M1)(A1)

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

OR

\(\frac{{{\text{AC}}}}{{\sin (73.8^\circ )}} = \frac{{22.5}}{{\sin (53.1^\circ )}}\)     (M1)(A1)

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

AC = 27.0     (A1)(G2)[3 marks]

c.

\({\text{BC}} = \sqrt {{{13.5}^2} + {{13.5}^2}} \)     (M1)

= 19.09     (A1)

= 19.1     (AG)

OR

x2 + x2 = 272     (M1)

2x2 = 272     (A1)

BC = 19.09…     (A1)

= 19.1     (AG)

Notes: Unrounded answer must be seen for (A1) to be awarded.[2 marks]

d.

Volume = Pyramid + Cuboid

\( = \frac{1}{3}(18)({19.1^2}) + (108)({19.1^2})\)     (A1)(M1)(M1)

Note: Award (A1) for 108, the height of the cuboid seen. Award (M1) for correctly substituted volume of cuboid and (M1) for correctly substituted volume of pyramid.

= \(41\,588\)     (41\(\,\)553 if 2(13.52) is used)

= \(41\,600\) m3     (A1)(ft)(G3)[4 marks]

e.

Weight of air = \(41\,600 \times 1.2 \times 0.9\)     (M1)(M1)

= \(44\,900{\text{ kg}}\)     (A1)(ft)(G2)

Note: Award (M1) for their part (e) × 1.2, (M1) for × 0.9.

Award at most (M1)(M1)(A0) if the volume of the cuboid is used.[3 marks]

f.

Question

A gardener has to pave a rectangular area 15.4 metres long and 5.5 metres wide using rectangular bricks. The bricks are 22 cm long and 11 cm wide.

The gardener decides to have a triangular lawn ABC, instead of paving, in the middle of the rectangular area, as shown in the diagram below.

The distance AB is 4 metres, AC is 6 metres and angle BAC is 40°.

In another garden, twelve of the same rectangular bricks are to be used to make an edge around a small garden bed as shown in the diagrams below. FH is the length of a brick and C is the centre of the garden bed. M and N are the midpoints of the long edges of the bricks on opposite sides of the garden bed.

The garden bed has an area of 5419 cm2. It is covered with soil to a depth of 2.5 cm.

It is estimated that 1 kilogram of soil occupies 514 cm3.

Calculate the total area to be paved. Give your answer in cm2.[3]

a.i.

Write down the area of each brick.[1]

a.ii.

Find how many bricks are required to pave the total area.[2]

a.iii.

Find the length of BC.[3]

b.i.

Hence write down the perimeter of the triangular lawn.[1]

b.ii.

Calculate the area of the lawn.[2]

b.iii.

Find the percentage of the rectangular area which is to be lawn.[3]

b.iv.

Find the angle FCH.[2]

c.i.

Calculate the distance MN from one side of the garden bed to the other, passing through C.[3]

c.ii.

Find the volume of soil used.[2]

d.

Find the number of kilograms of soil required for this garden bed.[2]

e.
Answer/Explanation

Markscheme

15.4 × 5.5     (M1)

84.7 m2     (A1)

= 847000 cm2     (A1)(G3)

Note: Award (G2) if 84.7 m2 seen with no working.

OR

1540 × 550     (A1)(M1)

= 847000 cm2     (A1)(ft)(G3)

Note: Award (A1) for both dimensions converted correctly to cm, (M1) for multiplication of both dimensions. (A1)(ft) for correct product of their sides in cm.[3 marks]

a.i.

242 cm2 (0.0242 m2)     (A1)[1 marks}

a.ii.

\(\frac {15.4}{0.22} = 70\)     (M1)

\(\frac{5.5}{0.11} = 50\)

\(70 \times 50 = 3500\)     (A1)(G2)

OR

\(\frac {847000}{242} = 3500\)     (M1)(A1)(ft)(G2)

Note: Follow through from parts (a) (i) and (ii).[2 marks]

a.iii.

\({\text{B}}{{\text{C}}^2} = {4^2} + {6^2}-2 \times 4 \times 6 \times \cos 40^\circ \)     (M1)(A1)

\({\text{BC}} = 3.90{\text{ m}}\)     (A1)(G2)

Note: Award (M1) for correct substituted formula, (A1) for correct substitutions, (A1) for correct answer.[3 marks]

b.i.

perimeter = 13.9 m     (A1)(ft)(G1)

Notes: Follow through from part (b) (i).[1 mark]

b.ii.

\({\text{Area}} = \frac{1}{2} \times 4 \times 6 \times \sin 40^\circ \)     (M1)

= 7.71 m2     (A1)(ft)(G2)

Notes: Award (M1) for correct formula and correct substitution, (A1)(ft) for correct answer.[2 marks]

b.iii.

\(\frac{{7.713}}{{84.7}} \times 100{\text{ }}\%  = 9.11{\text{ }}\% \)     (A1)(M1)(A1)(ft)(G2)

Notes: Accept 9.10 %.

Award (A1) for both measurements correctly written in the same unit, (M1) for correct method, (A1)(ft) for correct answer.

Follow through from (b) (iii) and from consistent error in conversion of units throughout the question.[3 marks]

b.iv.

\(\frac{{360^\circ }}{{12}}\)     (M1)

\( = 30^\circ\)     (A1)(G2)[2 marks]

c.i.

\(\text{MN} = 2 \times \frac{11}{\tan 15} \)     (A1)(ft)(M1)

OR

\(\text{MN} = 2 \times 11 \tan 75^\circ \)

\({\text{MN}} = 82.1{\text{ cm}}\)     (A1)(ft)(G2)

Notes: Award (A1) for 11 and 2 seen (implied by 22 seen), (M1) for dividing by tan15 (or multiplying by tan 75).

Follow through from their angle in part (c) (i).[3 marks]

c.ii.

volume = 5419 × 2.5     (M1)

= 13500 cm3     (A1)(G2)[2 marks]

d.

\(\frac{{13547.34 \ldots }}{{514}} = 26.4\)     (M1)(A1)(ft)(G2)

Note: Award (M1) for dividing their part (d) by 514.

Accept 26.3.[2 marks]

e.

Question

The points A (−4, 1), B (0, 9) and C (4, 2) are plotted on the diagram below. The diagram also shows the lines AB, L1 and L2.

Find the gradient of AB.[2]

a.

L1 passes through C and is parallel to AB.

Write down the y-intercept of L1.[1]

b.

L2 passes through A and is perpendicular to AB.

Write down the equation of L2. Give your answer in the form ax + by + d = 0 where a, b and d \( \in \mathbb{Z}\).[3]

c.

Write down the coordinates of the point D, the intersection of L1 and L2.[1]

d.

There is a point R on L1 such that ABRD is a rectangle.

Write down the coordinates of R.[2]

e.

The distance between A and D is \(\sqrt {45} \).

(i) Find the distance between D and R .

(ii) Find the area of the triangle BDR .[4]

f.
Answer/Explanation

Markscheme

\(\frac{{9 – 1}}{{0 – ( – 4)}}\)     (M1)

= 2    (A1)(G2)

Notes: Award (M1) for correct substitution into the gradient formula.[2 marks]

a.

–6     (A1)

Note: Accept (0, –6) .[1 mark]

b.

\(y =  – \frac{1}{2}x – 1\)  (or equivalent)     (A1)(ft)(A1)

Notes: Award (A1)(ft) for gradient, (A1) for correct y-intercept. Follow through from their gradient in (a).

x + 2y + 2 = 0     (A1)(ft)

Notes: Award (A1)(ft) from their gradient and their y-intercept. Accept any multiple of this equation with integer coefficients.

OR

\(y – 1 =  – \frac{1}{2}(x + 4)\) (or equivalent)     (A1)(ft)(A1)

Note: Award (A1)(ft) for gradient, (A1) for any point on the line correctly substituted in equation.

x + 2y + 2 = 0     (A1)(ft)

Notes: Award (A1)(ft) from their equation. Accept any multiple of this equation with integer coefficients.[3 marks]

c.

D(2, –2) or x = 2, y = –2     (A1)

Note: Award (A0) if brackets not present.[1 mark]

d.

R(6, 6) or x = 6, y = 6     (A1)(A1)

Note: Award at most (A0)(A1)(ft) if brackets not present and absence of brackets has not already been penalised in part (d).[2 marks]

e.

(i) \({\text{DR}} = \sqrt {{8^2} + {4^2}} \)     (M1)

\({\text{DR}} = \sqrt {80} \)  (8.94)     (A1)(ft)(G2)

Note: Award (M1) for correct substitution into the distance formula. Follow through from their D and R.

(ii) \({\text{Area}} = \frac{{\sqrt {80}  \times \sqrt {45} }}{2}\)     (M1)

= 30 (30.0)     (A1)(ft)(G2)

Note: Award (M1) for correct substitution in the area of triangle formula. Follow through from their answer to part (f) (i).[4 marks]

f.

Question

In the diagram below A, B and C represent three villages and the line segments AB, BC and CA represent the roads joining them. The lengths of AC and CB are 10 km and 8 km respectively and the size of the angle between them is 150°.

Find the length of the road AB.[3]

a.

Find the size of the angle CAB.[3]

b.

Village D is halfway between A and B. A new road perpendicular to AB and passing through D is built. Let T be the point where this road cuts AC. This information is shown in the diagram below.

Write down the distance from A to D.[1]

c.

Show that the distance from D to T is 2.06 km correct to three significant figures.[2]

d.

A bus starts and ends its journey at A taking the route AD to DT to TA.

Find the total distance for this journey.[3]

e.

The average speed of the bus while it is moving on the road is 70 km h–1. The bus stops for 5 minutes at each of D and T .

Estimate the time taken by the bus to complete its journey. Give your answer correct to the nearest minute.[4]

f.
Answer/Explanation

Markscheme

AB2 = 102 + 82 – 2 × 10 × 8 × cos150°     (M1)(A1)

AB = 17.4 km     (A1)(G2)

Note: Award (M1) for substitution into correct formula, (A1) for correct substitution, (A1) for correct answer.[3 marks]

a.

\(\frac{8}{{\sin {\text{C}}\hat {\rm A}{\text{B}}}} = \frac{{17.4}}{{\sin 150^\circ }}\)     (M1)(A1)

\({\text{C}}\hat {\rm A}{\text{B}} = 13.3^\circ \)     (A1)(ft)(G2)

Notes: Award (M1) for substitution into correct formula, (A1) for correct substitution, (A1) for correct answer. Follow through from their answer to part (a).[3 marks]

b.

AD = 8.70 km (8.7 km)     (A1)(ft)

Note: Follow through from their answer to part (a).[1 mark]

c.

DT = tan (13.29…°) × 8.697… = 2.0550…     (M1)(A1)

= 2.06     (AG)

Notes: Award (M1) for correct substitution in the correct formula, award (A1) for the unrounded answer seen. If 2.06 not seen award at most (M1)(AO).[2 marks]

d.

\(\sqrt {{{8.70}^2} + {{2.06}^2}}  + 8.70 + 2.06\)     (A1)(M1)

= 19.7 km     (A1)(ft)(G2)

Note: Award (A1) for AT, (M1) for adding the three sides of the triangle ADT, (A1)(ft) for answer. Follow through from their answer to part (c).[3 marks]

e.

\(\frac{{19.7}}{{70}} \times 60 + 10\)     (M1)(M1)

= 26.9     (A1)(ft)

Note: Award (M1) for time on road in minutes, (M1) for adding 10, (A1)(ft) for unrounded answer. Follow through from their answer to (e).

= 27  (nearest minute)     (A1)(ft)(G3)

Note: Award (A1)(ft) for their unrounded answer given to the nearest minute.[4 marks]

f.

Question

The diagram represents a small, triangular field, ABC , with \({\text{BC}} = 25{\text{ m}}\) , \({\text{angle BAC}} = {55^ \circ }\) and \({\text{angle ACB}} = {75^ \circ }\) .

Write down the size of angle ABC.[1]

a.

Calculate the length of AC.[3]

b.

Calculate the area of the field ABC.[3]

c.

N is the point on AB such that CN is perpendicular to AB. M is the midpoint of CN.

Calculate the length of NM.[3]

d.

A goat is attached to one end of a rope of length 7 m. The other end of the rope is attached to the point M.

Decide whether the goat can reach point P, the midpoint of CB. Justify your answer.[5]

e.
Answer/Explanation

Markscheme

\({\text{Angle ABC}} = {50^ \circ }\)     (A1)[1 mark]

a.

\(\frac{{{\text{AC}}}}{{\sin {{50}^ \circ }}} = \frac{{25}}{{\sin {{55}^ \circ }}}\)     (M1)(A1)(ft)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for correct substitution. Follow through from their angle ABC.

\({\text{AC}} = 23.4{\text{ m}}\)     (A1)(ft)(G2)[3 marks]

b.

\({\text{Area of }}\Delta {\text{ ABC}} = \frac{1}{2} \times 23.379 \ldots  \times 25 \times \sin {75^ \circ }\)     (M1)(A1)(ft)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for correct substitution. Follow through from their AC.

OR

\({\text{Area of triangle ABC}} = \frac{{29.479 \ldots  \times 19.151 \ldots }}{2}\)     (A1)(ft)(M1)

Note: (A1)(ft) for correct values of AB (29.479…) and CN (19.151…). Follow through from their (a) and /or (b). Award (M1) for substitution of their values of AB and CN into the correct formula.

\({\text{Area of }}\Delta {\text{ ABC}} = 282{\text{ }}{{\text{m}}^2}\)     (A1)(ft)(G2)

Note: Accept \(283{\text{ }}{{\text{m}}^2}\) if \(23.4\) is used.[3 marks]

c.

\({\text{NM}} = \frac{{25 \times \sin {{50}^ \circ }}}{2}\)     (M1)(M1)

Note: Award (M1) for \({25 \times \sin {{50}^ \circ }}\) or equivalent for the length of CN. (M1) for dividing their CN by \(2\).

\({\text{NM}} = 9.58{\text{ m}}\)     (A1)(ft)(G2)

Note: Follow through from their angle ABC.

Notes: Premature rounding of CN leads to the answers \(9.60\) or \(9.6\). Award at most (M1)(M1)(A0) if working seen. Do not penalize with (AP). CN may be found in (c).

Note: The working for this part of the question may be in part (b).[3 marks]

d.

\({\text{Angle NCB}} = {40^ \circ }\) seen     (A1)(ft)

Note: Follow through from their (a).

From triangle MCP:

\({\text{M}}{{\text{P}}^2} = {(9.5756 \ldots )^2} + {12.5^2} – 2 \times 9.5756 \ldots  \times 12.5 \times \cos ({40^ \circ })\)     (M1)(A1)(ft)

\({\text{MP}} = 8.034 \ldots {\text{ m}}\)     (A1)(ft)(G3)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for their correct substitution. Follow through from their d). Award (G3) for correct value of MP seen without working.

OR

From right triangle MCP

\({\text{CP}} = 12.5{\text{ m}}\) seen     (A1)

\({\text{M}}{{\text{P}}^2} = {(12.5)^2} – {(9.575 \ldots )^2}\)     (M1)(A1)(ft)

\({\text{MP}} = 8.034 \ldots {\text{ m}}\)     (A1)(G3)(ft)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for their correct substitution. Follow through from their (d). Award (G3) for correct value of MP seen without working.

OR

From right triangle MCP

\({\text{Angle MCP}} = {40^ \circ }\) seen     (A1)(ft)

\(\frac{{{\text{MP}}}}{{12.5}} = \sin ({40^ \circ })\) or equivalent     (M1)(A1)(ft)

\({\text{MP}} = 8.034 \ldots {\text{ m}}\)     (A1)(G3)(ft)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for their correct substitution. Follow through from their (a). Award (G3) for correct value of MP seen without working.

The goat cannot reach point P as \({\text{MP}} > 7{\text{ m}}\) .     (A1)(ft)

Note: Award (A1)(ft) only if their value of MP is compared to \(7{\text{ m}}\), and conclusion is stated.[5 marks]

e.

Question

The diagram shows triangle ABC in which \({\text{AB}} = 28{\text{ cm}}\), \({\text{BC}} = 13{\text{ cm}}\), \({\text{BD}} = 12{\text{ cm}}\) and \({\text{AD}} = 20{\text{ cm}}\).

Calculate the size of angle ADB.[3]

a.

Find the area of triangle ADB.[3]

b.

Calculate the size of angle BCD.[4]

c.

Show that the triangle ABC is not right angled.[4]

d.
Answer/Explanation

Markscheme

\(\cos {\text{ADB}} = \frac{{{{12}^2} + {{20}^2} – {{28}^2}}}{{2(12)(20)}}\)     (M1)(A1)

Notes: Award (M1) for substituted cosine rule formula, (A1) for correct substitutions.

\(\angle {\text{ADB}} = 120\)     (A1)(G2)[3 marks]

a.

\({\text{Area}} = \frac{{(12)(20)\sin {{120}^ \circ }}}{2}\)     (M1)(A1)(ft)

Notes: Award (M1) for substituted area formula, (A1)(ft) for their correct substitutions.

\( = 104{\text{ c}}{{\text{m}}^2}\) (\(103.923 \ldots {\text{ c}}{{\text{m}}^2}\))     (A1)(ft)(G2)

Note: The final answer is \(104{\text{ c}}{{\text{m}}^2}\) , the units are required. Accept \(100{\text{ c}}{{\text{m}}^2}\) .[3 marks]

b.

\(\frac{{\sin {\text{BCD}}}}{{12}} = \frac{{\sin {{60}^ \circ }}}{{13}}\)     (A1)(ft)(M1)(A1)

Note: Award (A1)(ft) for their 60 seen, (M1) for substituted sine rule formula, (A1) for correct substitutions.

\({\text{BCD}} = {53.1^ \circ }\) (\(53.0736 \ldots \))     (A1)(G3)

Note: Accept \(53\), do not accept \(50\) or \(53.0\).[4 marks]

c.

Using triangle ABC

\(\frac{{\sin {\text{BAC}}}}{{13}} = \frac{{\sin {{53.1}^ \circ }}}{{28}}\)     (M1)(A1)(ft)

OR

Using triangle ABD

\(\frac{{\sin {\text{BAD}}}}{{12}} = \frac{{\sin {{120}^ \circ }}}{{28}}\)     (M1)(A1)(ft)

Note: Award (M1) for substituted sine rule formula (one of the above), (A1)(ft) for their correct substitutions. Follow through from (a) or (c) as appropriate.

\({\text{BAC}} = {\text{BAD}} = {21.8^ \circ }\) (\(21.7867 \ldots \))     (A1)(ft)(G2)

Notes: Accept \(22\), do not accept \(20\) or \(21.7\). Accept equivalent methods, for example cosine rule.

\({180^ \circ } – ({53.1^ \circ } + {21.8^ \circ }) \ne {90^ \circ }\), hence triangle ABC is not right angled     (R1)(AG)

OR

\(\frac{{{\text{CD}}}}{{\sin {{66.9}^ \circ }}} = \frac{{13}}{{\sin {{60}^ \circ }}}\)     (M1)(A1)(ft)

Note: Award (M1) for substituted sine rule formula, (A1)(ft) for their correct substitutions. Follow through from (a) and (c).

\({\text{CD}} = 13.8{\text{ }}(13.8075 \ldots )\)     (A1)(ft)

\({13^3} + {28^2} \ne {33.8^2}\), hence triangle ABC is not right angled.     (R1)(ft)(AG)

Note: The complete statement is required for the final (R1) to be awarded.[4 marks]

d.

Question

The diagram shows a Ferris wheel that moves with constant speed and completes a rotation every 40 seconds. The wheel has a radius of \(12\) m and its lowest point is \(2\) m above the ground.

Initially, a seat C is vertically below the centre of the wheel, O. It then rotates in an anticlockwise (counterclockwise) direction.

Write down

(i)     the height of O above the ground;

(ii)    the maximum height above the ground reached by C .[2]

a.

In a revolution, C reaches points A and B , which are at the same height above the ground as the centre of the wheel. Write down the number of seconds taken for C to first reach A and then B .[2]

b.

The sketch below shows the graph of the function, \(h(t)\) , for the height above ground of C, where \(h\) is measured in metres and \(t\) is the time in seconds, \(0 \leqslant t \leqslant 40\) .

Copy the sketch and show the results of part (a) and part (b) on your diagram. Label the points clearly with their coordinates.[4]

c.
Answer/Explanation

Markscheme

(i)     \(14\) m     (A1)

(ii)    \(26\) m     (A1)[2 marks]

a.

A:\(10\), B:\(30\)     (A1)(A1)[2 marks]

b.

     (A1)(ft)(A1)(ft)(A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for coordinates of each point clearly indicated either by scale or by coordinate pairs. Points need not be labelled A and B in the second diagram. Award a maximum of (A1)(A0)(A1)(ft)(A1)(ft) if coordinates are reversed. Do not penalise reversed coordinates if this has already been penalised in Q4(a)(iii).[4 marks]

c.

Question

A solid metal cylinder has a base radius of 4 cm and a height of 8 cm.

Find the area of the base of the cylinder.[2]

a.

Show that the volume of the metal used in the cylinder is 402 cm3, given correct to three significant figures.[2]

b.

Find the total surface area of the cylinder.[3]

c.

The cylinder was melted and recast into a solid cone, shown in the following diagram. The base radius OB is 6 cm.

Find the height, OC, of the cone.[3]

d.

The cylinder was melted and recast into a solid cone, shown in the following diagram. The base radius OB is 6 cm.

Find the size of angle BCO.

[2]
e.

The cylinder was melted and recast into a solid cone, shown in the following diagram. The base radius OB is 6 cm.

Find the slant height, CB.[2]

f.

The cylinder was melted and recast into a solid cone, shown in the following diagram. The base radius OB is 6 cm.

Find the total surface area of the cone.[4]

g.
Answer/Explanation

Markscheme

\( \pi \times 4^2\)     (M1)

= 50.3 (16\(\pi\)) cm(50.2654…)     (A1)(G2)

Note: Award (M1) for correct substitution in area formula. The answer is 50.3 cm2, the units are required.[2 marks]

a.

50.265…× 8     (M1)

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

= 402.123…     (A1)
= 402 (cm3)     (AG)

Note: Both the unrounded and the rounded answer must be seen for the (A1) to be awarded. The units are not required[2 marks]

b.

\(2 \times \pi \times 4 \times 8 + 2 \times \pi \times 4^2\)     (M1)(M1)

Note: Award (M1) for correct substitution in the curved surface area formula, (M1) for adding the area of their two bases.

= 302 cm2 (96π cm2) (301.592…)     (A1)(ft)(G2)

Notes: The answer is 302 cm2, the units are required. Do not penalise for missing or incorrect units if penalised in part (a). Follow through from their answer to part (a).[3 marks]

c.

\(\frac{1}{3} \pi \times 6^2 \times \text{OC} = 402\)     (M1)(M1)

Note: Award (M1) for correctly substituted volume formula, (M1) for equating to 402 (402.123…).

\({\text{OC}} = 10.7{\text{ (cm)}}\left( {{\text{10}}\frac{2}{3},{\text{ }}10.6666…} \right)\)     (A1)(G2)[3 marks]

d.

\(\tan \text{BCO} = \frac{6}{10.66…}\)     (M1)

Note: Award (M1) for use of correct tangent ratio.

\({\text{B}}{\operatorname{\hat C}}{\text{O}} = 29.4^\circ \) (29.3577…)     (A1)(ft)(G2)

Notes: Accept 29.3° (29.2814…) if 10.7 is used. An acceptable alternative method is to calculate CB first and then angle BCO. Allow follow through from parts (d) and (f). Answers range from 29.2° to 29.5°.[2 marks]

e.

\(\text{CB} = \sqrt{{6^2} + {(10.66…)^2}}\)     (M1)

OR

\(\sin 29.35…^\circ = \frac{6}{\text{CB}}\)     (M1)

OR

\(\cos 29.35…^\circ = \frac{10.66…}{\text{CB}}\)     (M1)

CB = 12.2 (cm) (12.2383…)     (A1)(ft)(G2)

Note: Accept 12.3 (12.2674…) if 10.7 (and/or 29.3) used. Follow through from part (d) or part (e) as appropriate.[2 marks]

f.

\(\pi \times 6 \times 12.2383… + \pi \times 6^2\)     (M1)(M1)(M1)

Note: Award (M1) for correct substitution in curved surface area formula, (M1) for correct substitution in area of circle formula, (M1) for addition of the two areas.

= 344 cm2 (343.785…)     (A1)(ft)(G3)

Note: The answer is 344 cm2, the units are required. Do not penalise for missing or incorrect units if already penalised in either part (a) or (c). Accept 345 cm2 if 12.3 is used and 343 cm2 if 12.2 is used. Follow through from their part (f).[4 marks]

g.

Question

The Great Pyramid of Cheops in Egypt is a square based pyramid. The base of the pyramid is a square of side length 230.4 m and the vertical height is 146.5 m. The Great Pyramid is represented in the diagram below as ABCDV . The vertex V is directly above the centre O of the base. M is the midpoint of BC.

(i) Write down the length of OM .

(ii) Find the length of VM .[3]

a.

Find the area of triangle VBC .[2]

b.

Calculate the volume of the pyramid.[2]

c.

Show that the angle between the line VM and the base of the pyramid is 52° correct to 2 significant figures.[2]

d.

Ahmed is at point P , a distance x metres from M on horizontal ground, as shown in the following diagram. The size of angle VPM is 27° . Q is a point on MP .

Write down the size of angle VMP .[1]

e.

Ahmed is at point P , a distance x metres from M on horizontal ground, as shown in the following diagram. The size of angle VPM is 27° . Q is a point on MP .

Using your value of VM from part (a)(ii), find the value of x.[4]

f.

Ahmed is at point P , a distance x metres from M on horizontal ground, as shown in the following diagram. The size of angle VPM is 27° . Q is a point on MP .

Ahmed walks 50 m from P to Q.

Find the length of QV, the distance from Ahmed to the vertex of the pyramid.[4]

g.
Answer/Explanation

Markscheme

(i) 115.2 (m)     (A1)

Note: Accept 115 (m)

(ii) \(\sqrt{(146.5^2 + 115.2^2)}\)     (M1)

Note: Award (M1) for correct substitution.

186 (m) (186.368…)     (A1)(ft)(G2)

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

a.

\(\frac{1}{2} \times 230.4 \times 186.368…\)     (M1)

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

21500 m2 (21469.6…m2)     (A1)(ft)(G2)

Notes: The final answer is 21500 m2; units are required. Accept 21400 m2 for use of 186 m and/or 115 m.[2 marks]

b.

\(\frac{1}{3} \times 230.4^2 \times 146.5\)     (M1)

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

2590000 m3 (2592276.48 m3)     (A1)(G2)

Note: The final answer is 2590000 m3; units are required but do not penalise missing or incorrect units if this has already been penalised in part (b). [2 marks]

c.

\(\tan^{-1}\left( {\frac{{146.5}}{{115.2}}} \right)\)     (M1)

Notes: Award (M1) for correct substituted trig ratio. Accept alternate correct trig ratios.

= 51.8203…= 52°     (A1)(AG)

Notes: Both the unrounded answer and the final answer must be seen for the (A1) to be awarded. Accept 51.96° = 52°, 51.9° = 52° or 51.7° = 52°

d.

128°     (A1)[1 mark]

e.

\(\frac{{186.368}}{{\sin27}} = \frac{{x}}{{\sin25}}\)     (A1)(M1)(A1)(ft)

Notes: Award (A1)(ft) for their angle MVP seen, follow through from their part (e). Award (M1) for substitution into sine formula, (A1) for correct substitutions. Follow through from their VM and their angle VMP.

x = 173 (m) (173.490…)     (A1)(ft)(G3)

Note: Accept 174 from use of 186.4.[4 marks]

f.

VQ2 = (186.368…)2 + (123.490…)2 − 2 × (186.368…) × (123.490…) × cos128     (A1)(ft)(M1)(A1)(ft)

Notes: Award (A1)(ft) for 123.490…(123) seen, follow through from their x (PM) in part (f), (M1) for substitution into cosine formula, (A1)(ft) for correct substitutions. Follow through from their VM and their angle VMP.

OR

173.490… − 50 = 123.490… (123)     (A1)(ft)

115.2 + 123.490… = 238.690…     (A1)(ft)

\(\text{VQ} = \sqrt{(146.5^2 + 238.690…^2)}\)     (M1)

VQ = 280 (m) (280.062…)     (A1)(ft)(G3)

Note: Accept 279 (m) from use of 3 significant figure answers.[4 marks]

g.

Question

A shipping container is to be made with six rectangular faces, as shown in the diagram.

The dimensions of the container are

length 2x
width x
height y.

All of the measurements are in metres. The total length of all twelve edges is 48 metres.

Show that y =12 − 3x .[3]

a.

Show that the volume V m3 of the container is given by

V = 24x2 − 6x3[2]

b.

Find \( \frac{{\text{d}V}}{{\text{d}x}}\).[2]

c.

Find the value of x for which V is a maximum.[3]

d.

Find the maximum volume of the container.[2]

e.

Find the length and height of the container for which the volume is a maximum.[3]

f.

The shipping container is to be painted. One litre of paint covers an area of 15 m2 . Paint comes in tins containing four litres.

Calculate the number of tins required to paint the shipping container.[4]

g.
Answer/Explanation

Markscheme

\(4(2x) + 4y + 4x = 48\)     (M1)

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

\(12x + 4y = 48\)     (M1)

Note: Award (M1) for simplifying (can be implied).

\(y = \frac{{48 – 12x}}{{4}}\)   OR   \(3x + y =12\)     (A1)

\(y =12 – 3x\)     (AG)

Note: The last line must be seen for the (A1) to be awarded.[3 marks]

a.

\(V = 2x \times x \times (12 – 3x)\)     (M1)(A1)

Note: Award (M1) for substitution into volume equation, (A1) for correct substitution.

\(= 24x^2 – 6x^3\)     (AG)

Note: The last line must be seen for the (A1) to be awarded.[2 marks]

b.

\(\frac{{\text{d}V}}{{\text{d}x}} = 48x – 18x^2\)     (A1)(A1)

Note: Award (A1) for each correct term.[2 marks]

c.

\(48x -18x^2 = 0\)     (M1)(M1)

Note: Award (M1) for using their derivative, (M1) for equating their answer to part (c) to 0.

OR

(M1) for sketch of \(V = 24x^2 – 6x^3\), (M1) for the maximum point indicated     (M1)(M1)

OR

(M1) for sketch of \(\frac{{\text{d}V}}{{\text{d}x}} = 48x – 18x^2\), (M1) for the positive root indicated     (M1)(M1)

\(2.67\left( {\frac{{24}}{9},{\text{ }}\frac{8}{3},{\text{ }}2.66666…} \right)\)     (A1)(ft)(G2)

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

d.

\(V = 24 \times {\left( {\frac{8}{3}} \right)^2} – 6 \times {\left( {\frac{8}{3}} \right)^3}\)     (M1)

Note: Award (M1) for substitution of their value from part (d) into volume equation.

\(56.9({{\text{m}}^3})\left( {\frac{{512}}{9},{\text{ }}56.8888…} \right)\)     (A1)(ft)(G2)

Note: Follow through from their answer to part (d).[2 marks]

e.

\(\text{length} = \frac{{16}}{{3}}\)     (A1)(ft)(G1)

Note: Follow through from their answer to part (d). Accept 5.34 from use of 2.67

\(\text{height} = 12 – 3 \times \left( {\frac{{8}}{{3}}} \right) = 4\)     (M1)(A1)(ft)(G2)

Notes: Award (M1) for substitution of their answer to part (d), (A1)(ft) for answer. Accept 3.99 from use of 2.67.[3 marks]

f.

\(\text{SA} = 2 \times \frac{{16}}{{3}} \times 4 + 2 \times \frac{{8}}{{3}} \times 4 + 2 \times \frac{{16}}{{3}} \times \frac{{8}}{{3}}\)     (M1)

OR

\(\text{SA} = 4 \left( {\frac{{8}}{{3}}}\right)^2 + 6 \times \frac{{8}}{{3}} \times 4\)     (M1)

Note: Award (M1) for substitution of their values from parts (d) and (f) into formula for surface area.

92.4 (m2) (92.4444…(m2))     (A1)

Note: Accept 92.5 (92.4622…) from use of 3 sf answers.

\(\text{Number of tins} = \frac{{92.4444…}}{{15 \times 4}}( = 1.54)\)     (M1)[4 marks] 

Note: Award (M1) for division of their surface area by 60.

2 tins required     (A1)(ft)

Note: Follow through from their answers to parts (d) and (f).

g.

Question

The graph of the function \(f(x) = \frac{{14}}{x} + x – 6\), for 1 ≤ x ≤ 7 is given below.

Calculate \(f (1)\).[2]

a.

Find \(f ′(x)\).[3]

b.

Use your answer to part (b) to show that the x-coordinate of the local minimum point of the graph of \(f\) is 3.7 correct to 2 significant figures.[3]

c.

Find the range of \(f\).[3]

d.

Points A and B lie on the graph of \(f\). The x-coordinates of A and B are 1 and 7 respectively.

Write down the y-coordinate of B.[1]

e.

Points A and B lie on the graph of f . The x-coordinates of A and B are 1 and 7 respectively.

Find the gradient of the straight line passing through A and B.[2]

f.

M is the midpoint of the line segment AB.

Write down the coordinates of M.[2]

g.

L is the tangent to the graph of the function \(y = f (x)\), at the point on the graph with the same x-coordinate as M.

Find the gradient of L.[2]

h.

Find the equation of L. Give your answer in the form \(y = mx + c\).[3]

i.
Answer/Explanation

Markscheme

\(\frac{{14}}{{(1)}} + (1) – 6\)     (M1)

Note: Award (M1) for substituting \(x = 1\) into \(f\).

\(= 9\)     (A1)(G2)

a.

\( – \frac{{14}}{{{x^2}}} + 1\)     (A3)

Note: Award (A1) for \(-14\), (A1) for \(\frac{{14}}{{{x^2}}}\) or for \(x^{-2}\), (A1) for \(1\).

   Award at most (A2) if any extra terms are present.

b.

\( – \frac{{14}}{{{x^2}}} + 1 = 0\) or \(f ‘ (x) = 0 \)     (M1)

Note: Award (M1) for equating their derivative in part (b) to 0.

\(\frac{{14}}{{{x^2}}} = 1\) or \({x^2} = 14\) or equivalent     (M1)

Note: Award (M1) for correct rearrangement of their equation.

\(x = 3.74165…(\sqrt {14})\)     (A1)

\(x = 3.7\)     (AG)

Notes: Both the unrounded and rounded answers must be seen to award the (A1). This is a “show that” question; appeals to their GDC are not accepted –award a maximum of (M1)(M0)(A0).

Specifically, \( – \frac{{14}}{{{x^2}}} + 1 = 0\) followed by \(x = 3.74165…, x = 3.7\) is awarded (M1)(M0)(A0).

c.

\(1.48 \leqslant y \leqslant 9\)     (A1)(A1)(ft)(A1)

Note: Accept alternative notations, for example [1.48,9]. (\(x = \sqrt{14}\) leads to answer 1.48331…)

Note: Award (A1) for 1.48331…seen, accept 1.48378… from using the given answer \(x = 3.7\), (A1)(ft) for their 9 from part (a) seen, (A1) for the correct notation for their interval (accept \( \leqslant y \leqslant \) or \( \leqslant f \leqslant \) ).

d.

3     (A1)

Note: Do not accept a coordinate pair.

e.

\(\frac{{3 – 9}}{{7 – 1}}\)     (M1)

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

\(= -1\)     (A1)(ft)(G2)

Note: Follow through from their answers to parts (a) and (e).

f.

(4, 6)     (A1)(ft)(A1)

Note: Accept \(x = 4\), \(y = 6\). Award at most (A1)(A0) if parentheses not seen.

If coordinates reversed award (A0)(A1)(ft).

Follow through from their answers to parts (a) and (e).

g.

\( – \frac{{14}}{{{4^2}}} + 1\)     (M1)

Note: Award (M1) for substitution into their gradient function.

Follow through from their answers to parts (b) and (g).

\( = \frac{1}{8}(0.125)\)     (A1)(ft)(G2)

h.

\(y – 1.5 = \frac{1}{8}(x – 4)\)     (M1)(ft)(M1)

Note: Award (M1) for substituting their (4, 1.5) in any straight line formula,

(M1) for substituting their gradient in any straight line formula.

\(y = \frac{x}{8} + 4\)     (A1)(ft)(G2)

Note: The form of the line has been specified in the question.

i.

Question

A greenhouse ABCDPQ is constructed on a rectangular concrete base ABCD and is made of glass. Its shape is a right prism, with cross section, ABQ, an isosceles triangle. The length of BC is 50 m, the length of AB is 10 m and the size of angle QBA is 35°.

Write down the size of angle AQB.[1]

a.

Calculate the length of AQ.[3]

b.

Calculate the length of AC.[2]

c.

Show that the length of CQ is 50.37 m, correct to 4 significant figures.[2]

d.

Find the size of the angle AQC.[3]

e.

Calculate the total area of the glass needed to construct

(i) the two rectangular faces of the greenhouse;

(ii) the two triangular faces of the greenhouse.[5]

f.

The cost of one square metre of glass used to construct the greenhouse is 4.80 USD.

Calculate the cost of glass to make the greenhouse. Give your answer correct to the nearest 100 USD.[3]

g.
Answer/Explanation

Markscheme

110°     (A1)

a.

\(\frac{{AQ}}{{\sin 35^\circ }} = \frac{{10}}{{\sin 110^\circ }}\)     (M1)(A1)

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

OR

\(AQ = \frac{5}{{\cos 35^\circ }}\)     (A1)(M1)

Note: Award (A1) for 5 seen, (M1) for correctly substituted trigonometric ratio.

\(AQ = 6.10\) (6.10387…)     (A1)(ft)(G2)

Notes: Follow through from their answer to part (a).

b.

\(AC^2 = 10^2 + 50^2\)     (M1)

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

\(AC = 51.0 (\sqrt{2600}, 50.9901…)\)     (A1)(G2)

c.

\(QC^2 = (6.10387…)^2 + (50)^2\)     (M1)

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

\(QC = 50.3711…\)     (A1)

\(= 50.37\)     (AG)

Note: Both the unrounded and rounded answers must be seen to award (A1).

   If 6.10 is used then 50.3707… is the unrounded answer.

   For an incorrect follow through from part (b) award a maximum of (M1)(A0) – the given answer must be reached to award the final (A1)(AG).

d.

\(\cos AQC = \frac{{{{(6.10387…)}^2} + {{(50.3711…)}^2} – {{(50.9901…)}^2}}}{{2(6.10387…)(50.3711…)}}\)     (M1)(A1)(ft)

Note: Award (M1) for substituted cosine rule formula, (A1)(ft) for their correct substitutions.

= 92.4°   (\({92.3753…^\circ }\))     (A1)(ft)(G2)


Notes: Follow through from their answers to parts (b), (c) and (d). Accept 92.2 if the 3 sf answers to parts (b), (c) and (d) are used.

     Accept 92.5° (\({92.4858…^\circ }\)) if the 3 sf answers to parts (b), (c) and 4 sf answers to part (d) used.

e.

(i) \(2(50 \times 6.10387…)\)     (M1)

Note: Award (M1) for their correctly substituted rectangular area formula, the area of one rectangle is not sufficient.

= 610 m2 (610.387…)     (A1)(ft)(G2)

Notes: Follow through from their answer to part (b).

    The answer is 610 m2. The units are required.

(ii) Area of triangular face \( = \frac{1}{2} \times 10 \times 6.10387… \times \sin 35^\circ \)     (M1)(A1)(ft)

OR

Area of triangular face \( = \frac{1}{2} \times 6.10387… \times 6.10387… \times \sin 110^\circ \)     (M1)(A1)(ft)

\(= 17.5051…\)

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

OR

(Height of triangle) \( = {(6.10387…)^2} – {5^2}\)

\(= 3.50103…\)

Area of triangular face \( = \frac{1}{2} \times 10 \times their{\text{ }} height\)

\(= 17.5051…\)

Note: Award (M1) for substituted triangle area formula, (A1)(ft) for correctly substituted area formula. If 6.1 is used, the height is 3.49428… and the area of both triangular faces 34.9 m2

Area of both triangular faces = 35.0 m2 (35.0103…)     (A1)(ft)(G2) 

Notes: The answer is 35.0 m2. The units are required. Do not penalize if already penalized in part (f)(i). Follow through from their part (b).

f.

(610.387… + 35.0103…) × 4.80     (M1)

= 3097.90…     (A1)(ft)

Notes: Follow through from their answers to parts (f)(i) and (f)(ii).

    Accept 3096 if the 3 sf answers to part (f) are used.

= 3100     (A1)(ft)(G2)

Notes: Follow through from their unrounded answer, irrespective of whether it is correct. Award (M1)(A2) if working is shown and 3100 seen without the unrounded answer being given.

g.

Question

Francesca is a chef in a restaurant. She cooks eight chickens and records their masses and cooking times. The mass m of each chicken, in kg, and its cooking time t, in minutes, are shown in the following table.

Draw a scatter diagram to show the relationship between the mass of a chicken and its cooking time. Use 2 cm to represent 0.5 kg on the horizontal axis and 1 cm to represent 10 minutes on the vertical axis.[4]

a.

Write down for this set of data

(i) the mean mass, \(\bar m\) ;

(ii) the mean cooking time, \(\bar t\) .[2]

b.

Label the point \({\text{M}}(\bar m,\bar t)\) on the scatter diagram.[1]

c.

Draw the line of best fit on the scatter diagram.[2]

d.

Using your line of best fit, estimate the cooking time, in minutes, for a 1.7 kg chicken.[2]

e.

Write down the Pearson’s product–moment correlation coefficient, r .[2]

f.

Using your value for r , comment on the correlation.[2]

g.

The cooking time of an additional 2.0 kg chicken is recorded. If the mass and cooking time of this chicken is included in the data, the correlation is weak.

(i) Explain how the cooking time of this additional chicken might differ from that of the other eight chickens.

(ii) Explain how a new line of best fit might differ from that drawn in part (d).[2]

h.
Answer/Explanation

Markscheme

(A1) for correct scales and labels (mass or m on the horizontals axis, time or t on the vertical axis)

(A3) for 7 or 8 correctly placed data points

(A2) for 5 or 6 correctly placed data points

(A1) for 3 or 4 correctly placed data points, (A0) otherwise.     (A4)

Note: If axes reversed award at most (A0)(A3)(ft). If graph paper not used, award at most (A1)(A0).

a.

(i) 1.91 (kg) (1.9125 kg)     (G1)

(ii) 83 (minutes)     (G1)

b.

Their mean point labelled.     (A1)(ft)

Note: Follow through from part (b). Accept any clear indication of the mean point. For example: circle around point, (m, t), M , etc.

c.

Line of best fit drawn on scatter diagram.     (A1)(ft)(A1)(ft)

Notes:Award (A1)(ft) for straight line through their mean point, (A1)(ft) for line of best fit with intercept 9(±2) . The second (A1)(ft) can be awarded even if the line does not reach the t-axis but, if extended, the t-intercept is correct.

d.

75     (M1)(A1)(ft)(G2)

Notes: Accept 74.77 from the regression line equation. Award (M1) for indication of the use of their graph to get an estimate OR for correct substitution of 1.7 in the correct regression line equation t = 38.5m + 9.32.

e.

0.960 (0.959614…)     (G2)

Note: Award (G0)(G1)(ft) for 0.95, 0.959

f.

Strong and positive     (A1)(ft)(A1)(ft)

Note: Follow through from their correlation coefficient in part (f).

g.

(i) Cooking time is much larger (or smaller) than the other eight     (A1)

(ii) The gradient of the new line of best fit will be larger (or smaller)     (A1)

Note: Some acceptable explanations may include but are not limited to:

The line of best fit may be further away from the plotted points
It may be steeper than the previous line (as the mean would change)
The t-intercept of the new line is smaller (larger)

Do not accept vague explanations, like:

The new line would vary
It would not go through all points
It would not fit the patterns
The line may be slightly tilted

h.

Question

A tent is in the shape of a triangular right prism as shown in the diagram below.

The tent has a rectangular base PQRS .

PTS and QVR are isosceles triangles such that PT = TS and QV = VR .

PS is 3.2 m , SR is 4.7 m and the angle TSP is 35°.

Show that the length of side ST is 1.95 m, correct to 3 significant figures.[3]

a.

Calculate the area of the triangle PTS.[3]

b.

Write down the area of the rectangle STVR.[1]

c.

Calculate the total surface area of the tent, including the base.[3]

d.

Calculate the volume of the tent.[2]

e.

A pole is placed from V to M, the midpoint of PS.

Find in metres,

(i) the height of the tent, TM;

(ii) the length of the pole, VM.[4]

f.

Calculate the angle between VM and the base of the tent.[2]

g.
Answer/Explanation

Markscheme

\({\text{ST}} = \frac{{1.6}}{{\cos 35^\circ }}\)     (M1)(A1)

Note: Award (M1) for correctly substituted trig equation, (A1) for 1.6 seen.

OR

\(\frac{{{\text{ST}}}}{{\sin 35^\circ }} = \frac{{3.2}}{{\sin 110^\circ }}\)     (M1)(A1)

Note: Award (M1) for substituted sine rule equation, (A1) for correct substitutions.

ST = 1.95323…     (A1)

= 1.95 (m)     (AG)

Notes: Both unrounded and rounded answer must be seen for final (A1) to be awarded.

a.

\(\frac{1}{2} \times 3.2 \times 1.95323… \times \sin 35^\circ \) or \(\frac{1}{2} \times 1.95323… \times 1.95323… \times \sin 110^\circ \)     (M1)(A1)

Note: Award (M1) for substituted area formula, (A1) for correct substitutions. Do not award follow through marks.

= 1.79 m2 (1.79253…m2)     (A1)(G2)

Notes: The answer is 1.79 m2, units are required. Accept 1.78955… from using 1.95.

OR

\(\frac{1}{2} \times 3.2 \times 1.12033…\)     (A1) (M1)

Note: Award (A1) for the correct value for TM (1.12033…) OR correct expression for TM (i.e. 1.6tan35°, \(\sqrt {{{(1.95323…)}^2} – {{1.6}^2}} \)), (M1) for correctly substituted formula for triangle area.

= 1.79 m2 (1.79253…m2)     (A1)(G2)

Notes: The answer is 1.79 m2, units are required. Accept 1.78 m2 from using 1.95.

b.

9.18 m2 (9.18022 m2)     (A1)(G1)

Notes: The answer is 9.18 m2, units are required. Do not penalize if lack of units was already penalized in (b). Do not award follow through marks here. Accept 9.17 m2 (9.165 m2) from using 1.95.

c.

\(2 \times 1.79253… + 2 \times 9.18022… + 4.7 \times 3.2\)     (M1)(A1)(ft)

Note: Award (M1) for addition of three products, (A1)(ft) for three correct products.

= 37.0 m2 (36.9855…m2)     (A1)(ft)(G2)

Notes: The answer is 37.0 m2, units are required. Accept 36.98 m2 from using 3sf answers. Follow through from their answers to (b) and (c). Do not penalize if lack of units was penalized earlier in the question.

d.

\(1.79253… \times 4.7\)     (M1)

Note: Award (M1) for their correctly substituted volume formula.

= 8.42 m3 (8.42489…m3)     (A1)(ft)(G2)

Notes: The answer is 8.42 m3, units are required. Accept 8.41 m3 from use of 1.79. An answer of 8.35, from use of TM = 1.11, will receive follow-through marks if working is shown. Follow through from their answer to part (b). Do not penalize if lack of units was penalized earlier in the question.

e.

(i) \({\text{TM}} = 1.6\tan {35^\circ }\)     (M1)

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

OR

\({\text{TM}} = \sqrt {{{(1.95323…)}^2} – {{1.6}^2}} \)     (M1)

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

OR

\(\frac{{3.2 \times {\text{TM}}}}{2} = 1.79253…\)     (M1)

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

= 1.12 (m) (1.12033…)     (A1)(ft)(G2)

Notes: Follow through from their answer to (b) if area of triangle is used. Accept 1.11 (1.11467) from use of ST = 1.95.

(ii) \({\text{VM}} = \sqrt {{{1.12033…}^2} + {{4.7}^2}} \)     (M1)

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

= 4.83 (m) (4.83168 )     (A1)(ft)(G2)

Notes: Follow through from (f)(i).

f.

\({\sin ^{ – 1}}\left( {\frac{{1.12033…}}{{4.83168…}}} \right)\)     (M1)

OR

\({\cos^{ – 1}}\left( {\frac{{4.7}}{{4.83168…}}} \right)\)     (M1)

OR

\({\tan^{ – 1}}\left( {\frac{{1.12033…}}{{4.7}}} \right)\)     (M1)

Note: Award (M1) for correctly substituted trig equation.

OR

\({\cos ^{ – 1}}\left( {\frac{{{{4.7}^2} + {{(4.83168…)}^2} – {{(1.12033…)}^2}}}{{2 \times 4.7 \times 4.83168…}}} \right)\)     (M1)

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

= 13.4° (13.4073…)     (A1)(ft)(G2)

Notes: Accept 13.3°. Follow through from part (f).

g.

Question

The diagram shows an aerial view of a bicycle track. The track can be modelled by the quadratic function

\(y = \frac{{ – {x^2}}}{{10}} + \frac{{27}}{2}x\), where \(x \geqslant 0,{\text{ }}y \geqslant 0\)

(x , y) are the coordinates of a point x metres east and y metres north of O , where O is the origin (0, 0) . B is a point on the bicycle track with coordinates (100, 350) .

The coordinates of point A are (75, 450). Determine whether point A is on the bicycle track. Give a reason for your answer.[3]

a.

Find the derivative of \(y = \frac{{ – {x^2}}}{{10}} + \frac{{27}}{2}x\).[2]

b.

Use the answer in part (b) to determine if A (75, 450) is the point furthest north on the track between O and B. Give a reason for your answer.[4]

c.

(i) Write down the midpoint of the line segment OB.

(ii) Find the gradient of the line segment OB.[3]

d.

Scott starts from a point C(0,150) . He hikes along a straight road towards the bicycle track, parallel to the line segment OB.

Find the equation of Scott’s road. Express your answer in the form \(ax + by = c\), where \(a, b {\text{ and }} c \in \mathbb{R}\).[3]

e.

Use your graphic display calculator to find the coordinates of the point where Scott first crosses the bicycle track.[2]

f.
Answer/Explanation

Markscheme

\(y =  – \frac{{{{75}^2}}}{{10}} + \frac{{27}}{2} \times 75\)     (M1)

Note: Award (M1) for substitution of 75 in the formula of the function.

= 450     (A1)

Yes, point A is on the bike track. (A1)

Note: Do not award the final (A1) if correct working is not seen.

a.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} =  – \frac{{2x}}{{10}} + \frac{{27}}{2}\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}} =  – 0.2x + 13.5} \right)\)     (A1)(A1)

Notes: Award (A1) for each correct term. If extra terms are seen award at most (A1)(A0). Accept equivalent forms.

b.

\( – \frac{{2x}}{{10}} + \frac{{27}}{2} = 0\)     (M1)

Note: Award (M1) for equating their derivative from part (b) to zero.

\(x = 67.5\)     (A1)(ft)

Note: Follow through from their derivative from part (b).

\( {\text{(Their) }} 67.5 \ne 75\)      (R1)

Note: Award (R1) for a comparison of their 67.5 with 75. Comparison may be implied (eg 67.5 is the x-coordinate of the furthest north point).

OR

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} =  – \frac{{2 \times (75)}}{{10}} + \frac{{27}}{2}\)     (M1)

Note: Award (M1) for substitution of 75 into their derivative from part (b).

\(= -1.5\)     (A1)(ft)


Note:
Follow through from their derivative from part (b)

\({\text{(Their)}} -1.5 \ne 0\)     (R1)


Note:
Award (R1) for a comparison of their –1.5 with 0. Comparison may be implied (eg The gradient of the parabola at the furthest north point (vertex) is 0).

Hence A is not the furthest north point.     (A1)(ft)   

Note: Do not award (R0)(A1)(ft). Follow through from their derivative from part (b).

c.

(i) M(50,175)     (A1)

Note: If parentheses are omitted award (A0). Accept x = 50, y = 175.

(ii) \(\frac{{350 – 0}}{{100 – 0}}\)     (M1)

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

\( = 3.5\left( {\frac{{350}}{{100}},\frac{7}{2}} \right)\)     (A1)(ft)(G2)

Note: Follow through from (d)(i) if midpoint is used to calculate gradient. Award (G1)(G0) for answer 3.5x without working.

d.

\(y = 3.5x + 150\)     (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for using their gradient from part (d), (A1)(ft) for correct equation of line.

\(3.5x – y = -150\) or \(7x – 2y = -300\) (or equivalent)     (A1)(ft)

Note: Award (A1)(ft) for expressing their equation in the form \(ax + by = c\).

e.

(18.4, 214) (18.3772…, 214.320…)     (A1)(ft)(A1)(ft)(G2)(ft)

Notes: Follow through from their equation in (e). Coordinates must be positive for follow through marks to be awarded. If parentheses are omitted and not already penalized in (d)(i) award at most (A0)(A1)(ft). If coordinates of the two intersection points are given award (A0)(A1)(ft). Accept x = 18.4, y = 214.

f.

Question

Tepees were traditionally used by nomadic tribes who lived on the Great Plains of North America. They are cone-shaped dwellings and can be modelled as a cone, with vertex O, shown below. The cone has radius, \(r\), height, \(h\), and slant height, \(l\).

A model tepee is displayed at a Great Plains exhibition. The curved surface area of this tepee is covered by a piece of canvas that is \(39.27{\text{ }}{{\text{m}}^2}\), and has the shape of a semicircle, as shown in the following diagram.

Show that the slant height, \(l\), is \(5\) m, correct to the nearest metre.[2]

a.

(i)     Find the circumference of the base of the cone.

(ii)     Find the radius, \(r\), of the base.

(iii)     Find the height, \(h\).[6]

b.

A company designs cone-shaped tents to resemble the traditional tepees.

These cone-shaped tents come in a range of sizes such that the sum of the diameter and the height is equal to 9.33 m.

Write down an expression for the height, \(h\), in terms of the radius, \(r\), of these cone-shaped tents.[1]

c.

A company designs cone-shaped tents to resemble the traditional tepees.

These cone-shaped tents come in a range of sizes such that the sum of the diameter and the height is equal to 9.33 m.

Show that the volume of the tent, \(V\), can be written as

\[V = 3.11\pi {r^2} – \frac{2}{3}\pi {r^3}.\][1]

d.

A company designs cone-shaped tents to resemble the traditional tepees.

These cone-shaped tents come in a range of sizes such that the sum of the diameter and the height is equal to 9.33 m.

Find \(\frac{{{\text{d}}V}}{{{\text{d}}r}}\).[2]

e.

A company designs cone-shaped tents to resemble the traditional tepees.

These cone-shaped tents come in a range of sizes such that the sum of the diameter and the height is equal to 9.33 m.

(i)     Determine the exact value of \(r\) for which the volume is a maximum.

(ii)     Find the maximum volume.[4]

f.
Answer/Explanation

Markscheme

\(\frac{{\pi {l^2}}}{2} = 39.27\)     (M1)(A1)

Note: Award (M1) for equating the formula for area of a semicircle to \(39.27\), award (A1) for correct substitution of \(l\) into the formula for area of a semicircle.

\(l = 5{\text{ (m)}}\)     (AG)

a.

(i)     \(5 \times \pi \)     (M1)

\( = 15.7\;\;\;(15.7079…,{\text{ }}5\pi )\;{\text{(m)}}\)     (A1)(G2)

(ii)     \(2\pi r = 15.7079…\;\;\;\)OR\(\;\;\;5\pi r = 39.27\)     (M1)

\((r = ){\text{ 2.5 (m)}}\)     (A1)(ft)(G2)

Note: Follow through from part (b)(i).

(iii)     \(({h^2} = ){\text{ }}{5^2} – {2.5^2}\)     (M1)

Notes: Award (M1) for correct substitution into Pythagoras’ theorem. Follow through from part (b)(ii).

\((h = ){\text{ 4.33 }}(4.33012 \ldots ){\text{ (m)}}\)     (A1)(ft)(G2)

b.

\(9.33 – 2 \times r\)     (A1)

c.

\(V = \frac{{\pi {r^2}}}{3} \times (9.33 – 2r)\)     (M1)

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

\(V = 3.11\pi {r^2} – \frac{2}{3}{\pi ^3}\)     (AG)

d.

\(6.22\pi r – 2\pi {r^2}\)     (A1)(A1)

Notes: Award (A1) for \(6.22\pi r\), (A1) for \( – 2\pi {r^2}\).

If extra terms present, award at most (A1)(A0).

e.

(i)     \(6.22\pi r – 2\pi {r^2} = 0\)     (M1)

Note: Award (M1) for setting their derivative from part (e) to 0.

\(r = 3.11{\text{ (m)}}\)     (A1)(ft)(G2)

Notes: Award (A1) for identifying 3.11 as the answer.

Follow through from their answer to part (e).

(ii)     \(\frac{1}{3}\pi {(3.11)^3}\;\;\;\)OR\(\;\;\;3.11\pi {(3.11)^2} – \frac{2}{3}\pi {(3.11)^3}\)     (M1)

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

\(31.5{\text{ (}}{{\text{m}}^3}{\text{)}}{\text{(31.4999}} \ldots {\text{)}}\)     (A1)(ft)(G2)

Note: Follow through from their answer to part (f)(i).

f.

Question

A surveyor has to calculate the area of a triangular piece of land, DCE.

The lengths of CE and DE cannot be directly measured because they go through a swamp.

AB, DE, BD and AE are straight paths. Paths AE and DB intersect at point C.

The length of AB is 15 km, BC is 10 km, AC is 12 km, and DC is 9 km.

The following diagram shows the surveyor’s information.

(i)     Find the size of angle \({\rm{ACB}}\).

(ii)     Show that the size of angle \({\rm{DCE}}\) is \(85.5^\circ\), correct to one decimal place.[4]

a.

The surveyor measures the size of angle \({\text{CDE}}\) to be twice that of angle \({\text{DEC}}\).

(i)     Using angle \({\text{DCE}} = 85.5^\circ \), find the size of angle \({\text{DEC}}\).

(ii)     Find the length of \({\text{DE}}\).[5]

b.

Calculate the area of triangle \({\text{DEC}}\).[4]

c.
Answer/Explanation

Markscheme

(i)     \(\cos {\rm{A\hat CB}} = \frac{{{{10}^2} + {{12}^2} – {{15}^2}}}{{2 \times 10 \times 12}}\)     (M1)(A1)

Note: Award (M1) for substituted cosine rule,

(A1) for correct substitution.

\({\rm{A\hat CB}} = 85.5^\circ \;\;\;({\text{85.4593}} \ldots {\text{)}}\)     (A1)(G2)

(ii)     \({\rm{D\hat CE}} = {\rm{A\hat CB}}\;\;\;{\text{and}}\;\;\;{\rm{A\hat CB}} = 85.5^\circ \;\;\;({\text{85.4593}} \ldots ^\circ {\text{)}}\)     (A1)

OR

\({\rm{B\hat CE}} = 180^\circ  – 85.5^\circ  = 94.5^\circ \;\;\;{\text{and}}\;\;\;{\rm{D\hat CE}} = 180^\circ  – 94.5^\circ  = 85.5^\circ \)     (A1)

Notes: Both reasons must be seen for the (A1) to be awarded.

\({\rm{D\hat CE}} = 85.5^\circ \)     (AG)

a.

(i)     \({\rm{D\hat EC}} = \frac{{180^\circ  – 85.5^\circ }}{3}\)     (M1)

\({\rm{D\hat EC}} = 31.5^\circ \)     (A1)(G2)

(ii)     \(\frac{{\sin (31.5^\circ )}}{9} = \frac{{\sin (85.5^\circ )}}{{{\text{DE}}}}\)     (M1)(A1)(ft)

Note: Award (M1) for substituted sine rule, (A1) for correct substitution.

\({\text{DE}} = 17.2{\text{ (km)}}(17.1718 \ldots )\).     (A1)(ft)(G2)

b.

\(0.5 \times 17.1718 \ldots  \times 9 \times \sin (63^\circ )\)     (A1)(ft)(M1)(A1)(ft)

Note: Award (A1)(ft) for \(63\) seen, (M1) for substituted triangle area formula, (A1)(ft) for \(0.5 \times 17.1718 \ldots  \times 9 \times \sin ({\text{their angle CDE}})\).

OR

\({\text{(triangle height}} = ){\text{ }}9 \times \sin (63^\circ )\)     (A1)(ft)(A1)(ft)

\({\text{0.5}} \times {\text{17.1718}} \ldots  \times {\text{9}} \times {\text{sin(their angle CDE)}}\)     (M1)

Note: Award (A1)(ft) for \(63\) seen, (A1)(ft) for correct triangle height with their angle \({\text{CDE}}\), (M1) for \({\text{0.5}} \times {\text{17.1718}} \ldots  \times {\text{9}} \times {\text{sin(their angle CDE)}}\).

\( = 68.9{\text{ k}}{{\text{m}}^2}\;\;\;(68.8509 \ldots )\)     (A1)(ft)(G3)

Notes: Units are required for the last (A1)(ft) mark to be awarded.

Follow through from parts (b)(i) and (b)(ii).

Follow through from their angle \({\text{CDE}}\) within this part of the question.

c.

Question

The following diagram shows two triangles, OBC and OBA, on a set of axes. Point C lies on the \(y\)-axis, and O is the origin.

The equation of the line BC is \(y = 4\).

Write down the coordinates of point C.[1]

a.

The \(x\)-coordinate of point B is \(a\).

(i)     Write down the coordinates of point B;

(ii)     Write down the gradient of the line OB.[2]

b.

Point A lies on the \(x\)-axis and the line AB is perpendicular to line OB.

(i)     Write down the gradient of line AB.

(ii)     Show that the equation of the line AB is \(4y + ax – {a^2} – 16 = 0\).[4]

c.

The area of triangle AOB is three times the area of triangle OBC.

Find an expression, in terms of a, for

(i)     the area of triangle OBC;

(ii)     the x-coordinate of point A.[3]

d.

Calculate the value of \(a\).[2]

e.
Answer/Explanation

Markscheme

\((0,{\text{ }}4)\)     (A1)

Notes: Accept \(x = 0,{\text{ }}y = 4\).

a.

(i)     \((a,{\text{ }}4)\)     (A1)(ft)

Notes: Follow through from part (a).

(ii)     \(\frac{4}{a}\)     (A1)(ft)

Note: Follow through from part (b)(i).

b.

(i)     \( – \frac{a}{4}\)     (A1)(ft)

Note: Follow through from part (b)(ii).

(ii)     \(y =  – \frac{a}{4}x + c\)     (M1)

Note: Award (M1) for substitution of their gradient from part (c)(i) in the equation.

\(4 =  – \frac{a}{4} \times a + c\)

\(c = \frac{1}{4} \times {a^2} + 4\)

\(y =  – \frac{a}{4}x + \frac{1}{4}{a^2} + 4\)     (A1)

OR

\(y – 4 =  – \frac{a}{4}(x – a)\)     (M1)

Note: Award (M1) for substitution of their gradient from part (c)(i) in the equation.

\(y =  – \frac{{ax}}{4} + \frac{{{a^2}}}{4} + 4\)     (A1)

\(4y =  – ax + {a^2} + 16\)

\(4y + ax – {a^2} – 16 = 0\)     (AG)

Note: Both the simplified and the not simplified equations must be seen for the final (A1) to be awarded.

c.

(i)     \(2a\)     (A1)

(ii)     \(\frac{{4x}}{2} = 3 \times 2a\)     (M1)

Note: Award (M1) for correct equation.

\(x = 3a\)     (A1)(ft)

Note: Follow through from part (d)(i).

OR

\(0 – 4 =  – \frac{a}{4}(x – a)\)     (M1)

Note: Award (M1) for correct substitution of their gradient and the coordinates of their point into the equation of a line.

\(\frac{{16}}{a} = x – a\)

\(x = a + \frac{{16}}{a}\)     (A1)(ft)

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

OR

\(4 \times 0 + ax – {a^2} – 16 = 0\)     (M1)

Note: Award (M1) for correct substitution of the coordinates of \({\text{A}}(x,{\text{ }}0)\) into the equation of line AB.

\(ax – {a^2} – 16 = 0\)

\(x = a + \frac{{16}}{a}\;\;\;\)OR\(\;\;\;x = \frac{{({a^2} + 16)}}{a}\)     (A1)(G1)

d.

\(4(0) + a(3a) – {a^2} – 16 = 0\)     (M1)

Note: Award (M1) for correct substitution of their \(3a\) from part (d)(ii) into the equation of line AB.

OR

\(\frac{1}{2}\left( {a + \frac{{16}}{a}} \right) \times 4 = 3\left( {\frac{{4a}}{2}} \right)\)     (M1)

Note: Award (M1) for area of triangle AOB (with their substituted \(a + \frac{{16}}{a}\) and 4) equated to three times their area of triangle AOB.

\(a = 2.83\;\;\;\left( {2.82842…,{\text{ }}2\sqrt 2 ,{\text{ }}\sqrt 8 } \right)\)     (A1)(ft)(G1)

Note: Follow through from parts (d)(i) and (d)(ii).

e.

Question

A boat race takes place around a triangular course, \({\text{ABC}}\), with \({\text{AB}} = 700{\text{ m}}\), \({\text{BC}} = 900{\text{ m}}\) and angle \({\text{ABC}} = 110^\circ \). The race starts and finishes at point \({\text{A}}\).

Calculate the total length of the course.[4]

a.

It is estimated that the fastest boat in the race can travel at an average speed of \(1.5\;{\text{m}}\,{{\text{s}}^{ – 1}}\).

Calculate an estimate of the winning time of the race. Give your answer to the nearest minute.[3]

b.

It is estimated that the fastest boat in the race can travel at an average speed of \(1.5\;{\text{m}}\,{{\text{s}}^{ – 1}}\).

Find the size of angle \({\text{ACB}}\).[3]

c.

To comply with safety regulations, the area inside the triangular course must be kept clear of other boats, and the shortest distance from \({\text{B}}\) to \({\text{AC}}\) must be greater than \(375\) metres.

Calculate the area that must be kept clear of boats.[3]

d.

To comply with safety regulations, the area inside the triangular course must be kept clear of other boats, and the shortest distance from \({\text{B}}\) to \({\text{AC}}\) must be greater than \(375\) metres.

Determine, giving a reason, whether the course complies with the safety regulations.[3]

e.

The race is filmed from a helicopter, \({\text{H}}\), which is flying vertically above point \({\text{A}}\).

The angle of elevation of \({\text{H}}\) from \({\text{B}}\) is \(15^\circ\).

Calculate the vertical height, \({\text{AH}}\), of the helicopter above \({\text{A}}\).[2]

f.

The race is filmed from a helicopter, \({\text{H}}\), which is flying vertically above point \({\text{A}}\).

The angle of elevation of \({\text{H}}\) from \({\text{B}}\) is \(15^\circ\).

Calculate the maximum possible distance from the helicopter to a boat on the course.[3]

g.
Answer/Explanation

Markscheme

\({\text{A}}{{\text{C}}^2} = {700^2} + {900^2} – 2 \times 700 \times 900 \times \cos 110^\circ \)     (M1)(A1)

\({\text{AC}} = 1315.65 \ldots \)     (A1)(G2)

length of course \( = 2920{\text{ (m)}}\;\;\;(2915.65 \ldots {\text{ m)}}\)     (A1)

Notes: Award (M1) for substitution into cosine rule formula, (A1) for correct substitution, (A1) for correct answer.

Award (G3) for \(2920\;\;\;(2915.65 \ldots )\) seen without working.

The final (A1) is awarded for adding \(900\) and \(700\) to their \({\text{AC}}\) irrespective of working seen.

a.

\(\frac{{2915.65}}{{1.5}}\)     (M1)

Note: Award (M1) for their length of course divided by \(1.5\).

Follow through from part (a).

\( = 1943.76 \ldots {\text{ (seconds)}}\)     (A1)(ft)

\( = 32{\text{ (minutes)}}\)     (A1)(ft)(G2)

Notes: Award the final (A1) for correct conversion of their answer in seconds to minutes, correct to the nearest minute.

Follow through from part (a).

b.

\(\frac{{700}}{{\sin {\text{ACB}}}} = \frac{{1315.65 \ldots }}{{\sin 110^\circ }}\)     (M1)(A1)(ft)

OR

\(\cos {\text{ACB}} = \frac{{{{900}^2} + 1315.65{ \ldots ^2} – {{700}^2}}}{{2 \times 900 \times 1315.65 \ldots }}\)     (M1)(A1)(ft)

\({\text{ACB}} = 30.0^\circ \;\;\;(29.9979 \ldots ^\circ )\)     (A1)(ft)(G2)

Notes: Award (M1) for substitution into sine rule or cosine rule formula, (A1) for their correct substitution, (A1) for correct answer.

Accept \(29.9^\circ\) for sine rule and \(29.8^\circ\) for cosine rule from use of correct three significant figure values. Follow through from their answer to (a).

c.

\(\frac{1}{2} \times 700 \times 900 \times \sin 110^\circ \)     (M1)(A1)

Note: Accept \(\frac{1}{2} \times {\text{their AC}} \times {\text{900}} \times {\text{sin(their ACB)}}\). Follow through from parts (a) and (c).

\( = 296000{\text{ }}{{\text{m}}^2}\;\;\;(296003{\text{ }}{{\text{m}}^2})\)     (A1)(G2)

Notes: Award (M1) for substitution into area of triangle formula, (A1) for correct substitution, (A1) for correct answer.

Award (G1) if \(296000\) is seen without units or working.

d.

\(\sin 29.9979 \ldots  = \frac{{{\text{distance}}}}{{900}}\)     (M1)

\({\text{(distance}} = ){\text{ }}450{\text{ (m)}}\;\;\;{\text{(449.971}} \ldots {\text{)}}\)     (A1)(ft)(G2)

Note: Follow through from part (c).

OR

\(\frac{1}{2} \times {\text{distance}} \times 1315.65 \ldots  = 296003\)     (M1)

\(({\text{distance}} = ){\text{ }}450{\text{ (m)}}\;\;\;{\text{(449.971}} \ldots {\text{)}}\)     (A1)(ft)(G2)

Note: Follow through from part (a) and part (d).

\(450\) is greater than \(375\), thus the course complies with the safety regulations     (R1)

Notes:  A comparison of their area from (d) and the area resulting from the use of \(375\) as the perpendicular distance is a valid approach and should be given full credit. Similarly a comparison of angle \({\text{ACB}}\) and \({\sin ^{ – 1}}\left( {\frac{{375}}{{900}}} \right)\) should be given full credit.

Award (R0) for correct answer without any working seen. Award (R1)(ft) for a justified reason consistent with their working.

Do not award (M0)(A0)(R1).

e.

\(\tan 15^\circ  = \frac{{{\text{AH}}}}{{700}}\)     (M1)

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

\({\text{AH}} = 188{\text{ (m)}}\;\;\;(187.564 \ldots )\)     (A1)(ft)(G2)

f.

\({\text{H}}{{\text{C}}^2} = 187.564{ \ldots ^2} + 1315.65{ \ldots ^2}\)     (M1)(A1)

Note: Award (M1) for substitution into Pythagoras, (A1) for their \(1315.65{ \ldots}\) and their \(187.564{ \ldots}\) correctly substituted in formula.

\({\text{HC}} = 1330 \ldots {\text{ (m)}}\;\;\;(1328.95 \ldots )\)     (A1)(ft)(G2)

Note: Follow through from their answer to parts (a) and (f).

g.

Question

The following diagram shows a perfume bottle made up of a cylinder and a cone.

The radius of both the cylinder and the base of the cone is 3 cm.

The height of the cylinder is 4.5 cm.

The slant height of the cone is 4 cm.

(i)     Show that the vertical height of the cone is \(2.65\) cm correct to three significant figures.

(ii)     Calculate the volume of the perfume bottle.[6]

a.

The bottle contains \({\text{125 c}}{{\text{m}}^{\text{3}}}\) of perfume. The bottle is not full and all of the perfume is in the cylinder part.

Find the height of the perfume in the bottle.[2]

b.

Temi makes some crafts with perfume bottles, like the one above, once they are empty. Temi wants to know the surface area of one perfume bottle.

Find the total surface area of the perfume bottle.[4]

c.

Temi covers the perfume bottles with a paint that costs 3 South African rand (ZAR) per millilitre. One millilitre of this paint covers an area of \({\text{7 c}}{{\text{m}}^{\text{2}}}\).

Calculate the cost, in ZAR, of painting the perfume bottle. Give your answer correct to two decimal places.[4]

d.

Temi sells her perfume bottles in a craft fair for 325 ZAR each. Dominique from France buys one and wants to know how much she has spent, in euros (EUR). The exchange rate is 1 EUR = 13.03 ZAR.

Find the price, in EUR, that Dominique paid for the perfume bottle. Give your answer correct to two decimal places.[2]

e.
Answer/Explanation

Markscheme

(i)     \({x^2} + {3^2} = {4^2}\)     (M1)

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

Accept correct alternative method using trigonometric ratios.

\(x = 2.64575 \ldots \)     (A1)

\(x = 2.65{\text{ }}({\text{cm}})\)     (AG)

Note: The unrounded and rounded answer must be seen for the (A1) to be awarded.

OR

\(\sqrt {{4^2} – {3^2}} \)     (M1)

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

\( = \sqrt 7 \)     (A1)

\( = 2.65{\text{ (cm)}}\)     (AG)

Note: The exact answer must be seen for the final (A1) to be awarded.

(ii)     \(\pi  \times {3^2} \times 4.5 + \frac{1}{3}\pi  \times {3^2} \times 2.65\)     (M1)(M1)(M1)

Note: Award (M1) for correct substitution into the volume of a cylinder formula, (M1) for correct substitution into the volume of a cone formula, (M1) for adding both of their volumes.

\( = 152{\text{ c}}{{\text{m}}^3}\;\;\;(152.210 \ldots {\text{ c}}{{\text{m}}^3},{\text{ }}48.45\pi {\text{ c}}{{\text{m}}^3})\)     (A1)(G3)

a.

\(\pi {3^2}h = 125\)     (M1)

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

Accept alternative methods. Accept \(4.43\) (\(4.42913 \ldots \)) from using rounded answers in \(h = \frac{{125 \times 4.5}}{{127}}\).

\(h = 4.42{\text{ (cm)}}\;\;\;\left( {4.42097 \ldots {\text{ (cm)}}} \right)\)     (A1)(G2)

b.

\(2\pi  \times 3 \times 4.5 + \pi  \times 3 \times 4 + \pi  \times {3^2}\)     (M1)(M1)(M1)

Note: Award (M1) for correct substitution into curved surface area of a cylinder formula, (M1) for correct substitution into the curved surface area of a cone formula, (M1) for adding the area of the base of the cylinder to the other two areas.

\( = 151{\text{ c}}{{\text{m}}^2}\;\;\;(150.796 \ldots {\text{ c}}{{\text{m}}^2},{\text{ }}48\pi {\text{ c}}{{\text{m}}^2})\)     (A1)(G3)

c.

\(\frac{{150.796 \ldots }}{7} \times 3\)     (M1)(M1)

Notes: Award (M1) for dividing their answer to (c) by \(7\), (M1) for multiplying by \(3\). Accept equivalent methods.

\( = 64.63{\text{ (ZAR)}}\)     (A1)(ft)(G2)

Notes: The (A1) is awarded for their correct answer, correctly rounded to 2 decimal places. Follow through from their answer to part (c). If rounded answer to part (c) is used the answer is \(64.71\) (ZAR).

d.

\(\frac{{325}}{{13.03}}\)     (M1)

Note: Award (M1) for dividing \(325\) by \(13.03\).

\( = 24.94{\text{ (EUR)}}\)     (A1)(G2)

Note: The (A1) is awarded for the correct answer rounded to 2 decimal places, unless already penalized in part (d).

e.

Question

An office block, ABCPQR, is built in the shape of a triangular prism with its “footprint”, ABC, on horizontal ground. \({\text{AB}} = 70{\text{ m}}\), \({\text{BC}} = 50{\text{ m}}\) and \({\text{AC}} = 30{\text{ m}}\). The vertical height of the office block is \(120{\text{ m}}\) .

Calculate the size of angle ACB.[3]

a.

Calculate the area of the building’s footprint, ABC.[3]

b.

Calculate the volume of the office block.[2]

c.

To stabilize the structure, a steel beam must be made that runs from point C to point Q.

Calculate the length of CQ.[2]

d.

Calculate the angle CQ makes with BC.[2]

e.
Answer/Explanation

Markscheme

\(\cos {\text{ACB}} = \frac{{{{30}^2} + {{50}^2} – {{70}^2}}}{{2 \times 30 \times 50}}\)     (M1)(A1)

Note: Award (M1) for substituted cosine rule formula, (A1) for correct substitution.

\({\text{ACB}} = {120^ \circ }\)     (A1)(G2)

a.

\({\text{Area of triangle ABC}} = \frac{{30(50)\sin {{120}^ \circ }}}{2}\)     (M1)(A1)(ft)

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

\( = 650{\text{ }}{{\text{m}}^2}\) \((649.519 \ldots {\text{ }}{{\text{m}}^2})\)     (A1)(ft)(G2)

Notes: The answer is \(650{\text{ }}{{\text{m}}^2}\) ; the units are required. Follow through from their answer in part (a).

b.

\({\text{Volume}} = 649.519 \ldots \times 120\)     (M1)
\( = 77900{\text{ }}{{\text{m}}^3}\) (\(77942.2 \ldots {\text{ }}{{\text{m}}^3}\))     (A1)(G2)

Note: The answer is \(77900{\text{ }}{{\text{m}}^3}\) ; the units are required. Do not penalise lack of units if already penalized in part (b). Accept \(78000{\text{ }}{{\text{m}}^3}\) from use of 3sf answer \(650{\text{ }}{{\text{m}}^2}\) from part (b).

c.

\({\text{C}}{{\text{Q}}^2} = {50^2} + {120^2}\)     (M1)
\({\text{CQ}} = 130{\text{ (m)}}\)     (A1)(G2)

Note: The units are not required.

d.

\(\tan {\text{QCB}} = \frac{{120}}{{50}}\)     (M1)

Note: Award (M1) for correct substituted trig formula.

\({\text{QCB}} = {67.4^ \circ }\) (\(67.3801 \ldots \))     (A1)(G2)

Note: Accept equivalent methods.

e.

Question

Nadia designs a wastepaper bin made in the shape of an open cylinder with a volume of \(8000{\text{ c}}{{\text{m}}^3}\).

Nadia decides to make the radius, \(r\) , of the bin \(5{\text{ cm}}\).

Merryn also designs a cylindrical wastepaper bin with a volume of \(8000{\text{ c}}{{\text{m}}^3}\). She decides to fix the radius of its base so that the total external surface area of the bin is minimized.

Let the radius of the base of Merryn’s wastepaper bin be \(r\) , and let its height be \(h\) .

Calculate
(i)     the area of the base of the wastepaper bin;
(ii)    the height, \(h\) , of Nadia’s wastepaper bin;
(iii)   the total external surface area of the wastepaper bin.[7]

a.

State whether Nadia’s design is practical. Give a reason.[2]

b.

Write down an equation in \(h\) and \(r\) , using the given volume of the bin.[1]

c.

Show that the total external surface area, \(A\) , of the bin is \(A = \pi {r^2} + \frac{{16000}}{r}\) .[2]

d.

Write down \(\frac{{{\text{d}}A}}{{{\text{d}}r}}\).[3]

e.

(i)     Find the value of \(r\) that minimizes the total external surface area of the wastepaper bin.
(ii)    Calculate the value of \(h\) corresponding to this value of \(r\) .[5]

f.

Determine whether Merryn’s design is an improvement upon Nadia’s. Give a reason.[2]

g.
Answer/Explanation

Markscheme

(i)     \({\text{Area}} = \pi {(5)^2}\)     (M1)
\( = 78.5{\text{ (c}}{{\text{m}}^2}{\text{)}}\) (\(78.5398 \ldots \))     (A1)(G2)

Note: Accept \(25\pi \) .

(ii)    \(8000 = 78.5398 \ldots  \times h\)     (M1)
\(h = 102{\text{ (cm)}}\) (\(101.859 \ldots \))     (A1)(ft)(G2)

Note: Follow through from their answer to part (a)(i).

(iii)   \({\text{Area}} = \pi {(5)^2} + 2\pi (5)(101.859 \ldots )\)     (M1)(M1)

Note: Award (M1) for their substitution in curved surface area formula, (M1) for addition of their two areas.

\( = 3280{\text{ (c}}{{\text{m}}^2}{\text{)}}\) (\(3278.53 \ldots \))     (A1)(ft)(G2)

Note: Follow through from their answers to parts (a)(i) and (ii).

a.

No, it is too tall/narrow.     (A1)(ft)(R1)

Note: Follow through from their value for \(h\).

b.

\(8000 = \pi {r^2}h\)     (A1)

c.

\(A = \pi {r^2} + 2\pi r\left( {\frac{{8000}}{{\pi {r^2}}}} \right)\)     (A1)(M1)

Note: Award (A1) for correct rearrangement of their part (c), (M1) for substitution of their rearrangement into area formula.

\( = \pi {r^2} + \frac{{16000}}{r}\)     (AG)

d.

\(\frac{{{\text{d}}A}}{{{\text{d}}r}} = 2\pi r – 16000{r^{ – 2}}\)     (A1)(A1)(A1)

Note: Award (A1) for \(2\pi r\) , (A1) for \( – 16000\) (A1) for \({r^{ – 2}}\) . If an extra term is present award at most (A1)(A1)(A0).

e.

(i)     \(\frac{{{\text{d}}A}}{{{\text{d}}r}} = 0\)     (M1)
\(2\pi {r^3} – 16000 = 0\)    (M1)
\(r = 13.7{\text{ cm}}\) (\(13.6556 \ldots \))     (A1)(ft)

Note: Follow through from their part (e).

(ii)    \(h = \frac{{8000}}{{\pi {{(13.65 \ldots )}^2}}}\)     (M1)
\( = 13.7{\text{ cm}}\) (\(13.6556 \ldots \))     (A1)(ft)

Note: Accept \(13.6\) if \(13.7\) used.

f.

Yes or No, accompanied by a consistent and sensible reason.     (A1)(R1)

Note: Award (A0)(R0) if no reason is given.

g.
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