IB DP Mathematical Studies 5.5 Paper 2

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

A closed rectangular box has a height \(y{\text{ cm}}\) and width \(x{\text{ cm}}\). Its length is twice its width. It has a fixed outer surface area of \(300{\text{ c}}{{\text{m}}^2}\) .

Factorise \(3{x^2} + 13x – 10\).[2]

i.a.

Solve the equation \(3{x^2} + 13x – 10 = 0\).[2]

i.b.

Consider a function \(f(x) = 3{x^2} + 13x – 10\) .

Find the equation of the axis of symmetry on the graph of this function.[2]

i.c.

Consider a function \(f(x) = 3{x^2} + 13x – 10\) .

Calculate the minimum value of this function.[2]

i.d.

Show that \(4{x^2} + 6xy = 300\).

[2]
ii.a.

Find an expression for \(y\) in terms of \(x\).[2]

ii.b.

Hence show that the volume \(V\) of the box is given by \(V = 100x – \frac{4}{3}{x^3}\).[2]

ii.c.

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

ii.d.

(i)     Hence find the value of \(x\) and of \(y\) required to make the volume of the box a maximum.

(ii)    Calculate the maximum volume.[5]

ii.e.
Answer/Explanation

Markscheme

\((3x – 2)(x + 5)\)     (A1)(A1)[2 marks]

i.a.

\((3x – 2)(x + 5) = 0\)

\(x = \frac{2}{3}\) or \(x = – 5\)     (A1)(ft)(A1)(ft)(G2)[2 marks]

i.b.

\(x = \frac{{ – 13}}{6}{\text{ }}( – 2.17)\)     (A1)(A1)(ft)(G2)

Note: (A1) is for \(x = \), (A1) for value. (ft) if value is half way between roots in (b).[2 marks]

i.c.

Minimum \(y = 3{\left( {\frac{{ – 13}}{6}} \right)^2} + 13\left( {\frac{{ – 13}}{6}} \right) – 10\)     (M1)

Note: (M1) for substituting their value of \(x\) from (c) into \(f(x)\) .

\( = – 24.1\)     (A1)(ft)(G2)[2 marks]

i.d.

\({\text{Area}} = 2(2x)x + 2xy + 2(2x)y\)     (M1)(A1)

Note: (M1) for using the correct surface area formula (which can be implied if numbers in the correct place). (A1) for using correct numbers.

\(300 = 4{x^2} + 6xy\)     (AG)


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

ii.a.

\(6xy = 300 – 4{x^2}\)     (M1)

\(y = \frac{{300 – 4{x^2}}}{{6x}}\) or \(\frac{{150 – 2{x^2}}}{{3x}}\)     (A1)[2 marks]

ii.b.

\({\text{Volume}} = x(2x)y\)     (M1)

\(V = 2{x^2}\left( {\frac{{300 – 4{x^2}}}{{6x}}} \right)\)     (A1)(ft)

\( = 100x – \frac{4}{3}{x^3}\)     (AG)

Note: Final line must be seen or previous (A1) mark is lost.[2 marks]

ii.c.

\(\frac{{{\text{d}}V}}{{{\text{d}}x}} = 100 – \frac{{12{x^2}}}{3}\)  or  \(100 – 4{x^2}\)     (A1)(A1)

 Note: (A1) for each term.[2 marks]

ii.d.

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

(i)     For maximum \(\frac{{{\text{d}}V}}{{{\text{d}}x}} = 0\)  or  \(100 – 4{x^2} = 0\)     (M1)

\(x = 5\)     (A1)(ft)

\(y = \frac{{300 – 4{{(5)}^2}}}{{6(5)}}\)  or  \(\left( {\frac{{150 – 2{{(5)}^2}}}{{3(5)}}} \right)\)     (M1)

\( = \frac{{20}}{3}\)     (A1)(ft)

(UP)     (ii)    \(333\frac{1}{3}{\text{ c}}{{\text{m}}^3}{\text{ }}(333{\text{ c}}{{\text{m}}^3})\)


Note: (ft)
from their (e)(i) if working for volume is seen.
[5 marks]

ii.e.

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

A contractor is building a house. He first marks out three points A , B and C on the ground such that AB = 5 m , AC = 7 m and angle BAC = 112°.

Find the length of BC.[3]

a.

He next marks a fourth point, D, on the ground at a distance of 6 m from B , such that angle BDC is 40° .

Find the size of angle DBC .[4]

b.

He next marks a fourth point, D, on the ground at a distance of 6 m from B , such that angle BDC is 40° .

 

Find the area of the quadrilateral ABDC.[4]

c.

He next marks a fourth point, D, on the ground at a distance of 6 m from B , such that angle BDC is 40° .

The contractor digs up and removes the soil under the quadrilateral ABDC to a depth of 50 cm for the foundation of the house.

Find the volume of the soil removed. Give your answer in m3 .[3]

d.

He next marks a fourth point, D, on the ground at a distance of 6 m from B , such that angle BDC is 40° .

The contractor digs up and removes the soil under the quadrilateral ABDC to a depth of 50 cm for the foundation of the house.

To transport the soil removed, the contractor uses cylindrical drums with a diameter of 30 cm and a height of 40 cm. 

(i) Find the volume of a drum. Give your answer in m3 .

(ii) Find the minimum number of drums required to transport the soil removed.[5]

e.
Answer/Explanation

Markscheme

Units are required in part (c) only.

BC2 = 52 + 72 − 2(5)(7)cos112°     (M1)(A1)

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

BC = 10.0 (m) (10.0111…)     (A1)(G2)

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

a.

Units are required in part (c) only.


\(\frac{{\sin 40^\circ }}{{10.0111…}} = \frac{{\sin {\text{D}}{\operatorname{\hat C}}{\text{B}}}}{6}\)     (M1)(A1)(ft)

Notes: Award (M1) for substitution in sine formula, (A1)(ft) for their correct substitutions. Follow through from their part (a).

\({\text{D}}{\operatorname {\hat C}}{\text{B}}\) = 22.7° (22.6589…)     (A1)(ft) 

Notes: Award (A2) for 22.7° seen without working. Use of radians results in unrealistic answer. Award a maximum of (M1)(A1)(ft)(A0)(ft). Follow through from their part (a).

\({\text{D}}{\operatorname {\hat C}}{\text{B}}\) = 117° (117.341…)     (A1)(ft)(G3) 

Notes: Do not penalize if use of radians was already penalized in part (a). Follow through from their answer to part (a).

OR

From use of cosine formula

DC = 13.8(m)   (13.8346…)     (A1)(ft)

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

\(\frac{{\sin \alpha }}{{13.8346…}} = \frac{{\sin 40^\circ }}{{10.0111…}}\)     (M1)

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

α = 62.7°  (62.6589)     (A1)(ft)

Note: Accept 62.5° from use of 3sf.

\({\text{D}}{\operatorname {\hat B}}{\text{C}}\) = 117(117.341…)     (A1)(ft) 

Note: Follow through from their part (a). Use of radians results in unrealistic answer, award a maximum of (A1)(M1)(A0)(A0).[4 marks]

b.

Units are required in part (c) only.

\({\text{ABDC}} = \frac{1}{2}(5)(7)\sin 112^\circ  + \frac{1}{2}(6)(10.0111…)\sin 117.341…^\circ \)     (M1)(A1)(ft)(M1)N

Note: Award (M1) for substitution in both triangle area formulae, (A1)(ft) for their correct substitutions, (M1) for seen or implied addition of their two triangle areas. Follow through from their answer to part (a) and (b).

= 42.9 m2 (42.9039…)     (A1)(ft)(G3)

Notes: Answer is 42.9 m2 i.e. the units are required for the final (A1)(ft) to be awarded. Accept 43.0 m2 from using 3sf answers to parts (a) and (b). Do not penalize if use of radians was previously penalized.[4 marks]

c.

Units are required in part (c) only.

42.9039… × 0.5     (M1)(M1)

Note: Award (M1) for 0.5 seen (or equivalent), (M1) for multiplication of their answer in part (c) with their value for depth.

= 21.5 (m3) (21.4519…)     (A1)(ft)(G3)

Note: Follow through from their part (c) only if working is seen. Do not penalize if use of radians was previously penalized. Award at most (A0)(M1)(A0)(ft) for multiplying by 50.[3 marks]

d.

Units are required in part (c) only.

(i) π(0.15)2(0.4)     (M1)(A1)

OR

π × 152 × 40  (28274.3…)     (M1)(A1)

Notes: Award (M1) for substitution in the correct volume formula. (A1) for correct substitutions.

= 0.0283 (m3) (0.0282743…, 0.09π)

(ii) \(\frac{{21.4519…}}{{0.0282743…}}\)     (M1)

Note: Award (M1) for correct division of their volumes.

= 759     (A1)(ft)(G2)

Notes: Follow through from their parts (d) and (e)(i). Accept 760 from use of 3sf answers. Answer must be a positive integer for the final (A1)(ft) mark to be awarded.[5 marks]

e.

Question

A parcel is in the shape of a rectangular prism, as shown in the diagram. It has a length \(l\) cm, width \(w\) cm and height of \(20\) cm.

The total volume of the parcel is \(3000{\text{ c}}{{\text{m}}^3}\).

Express the volume of the parcel in terms of \(l\) and \(w\).[1]

a.

Show that \(l = \frac{{150}}{w}\).[2]

b.

The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Show that the length of string, \(S\) cm, required to tie up the parcel can be written as

\[S = 40 + 4w + \frac{{300}}{w},{\text{ }}0 < w \leqslant 20.\][2]

c.

The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Draw the graph of \(S\) for \(0 < w \leqslant 20\) and \(0 < S \leqslant 500\), clearly showing the local minimum point. Use a scale of \(2\) cm to represent \(5\) units on the horizontal axis \(w\) (cm), and a scale of \(2\) cm to represent \(100\) units on the vertical axis \(S\) (cm).[2]

d.

The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Find \(\frac{{{\text{d}}S}}{{{\text{d}}w}}\).[3]

e.

The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Find the value of \(w\) for which \(S\) is a minimum.[2]

f.

The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Write down the value, \(l\), of the parcel for which the length of string is a minimum.[1]

g.

The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Find the minimum length of string required to tie up the parcel.[2]

h.
Answer/Explanation

Markscheme

\(20lw\)   OR   \(V = 20lw\)     (A1)[1 mark]

a.

\(3000 = 20lw\)     (M1)

Note: Award (M1) for equating their answer to part (a) to \(3000\).

\(l = \frac{{3000}}{{20w}}\)     (M1)

Note: Award (M1) for rearranging equation to make \(l\) subject of the formula. The above equation must be seen to award (M1).

OR

\(150 = lw\)     (M1)

Note: Award (M1) for division by \(20\) on both sides. The above equation must be seen to award (M1).

\(l = \frac{{150}}{w}\)     (AG)[2 marks]

b.

\(S = 2l + 4w + 2(20)\)     (M1)

Note: Award (M1) for setting up a correct expression for \(S\).

\(2\left( {\frac{{150}}{w}} \right) + 4w + 2(20)\)     (M1)

Notes: Award (M1) for correct substitution into the expression for \(S\). The above expression must be seen to award (M1).

\( = 40 + 4w + \frac{{300}}{w}\)     (AG)[2 marks]

c.


    
 (A1)(A1)(A1)(A1)

 

Note: Award (A1) for correct scales, window and labels on axes, (A1) for approximately correct shape, (A1) for minimum point in approximately correct position, (A1) for asymptotic behaviour at \(w = 0\).

     Axes must be drawn with a ruler and labeled \(w\) and \(S\).

     For a smooth curve (with approximately correct shape) there should be one continuous thin line, no part of which is straight and no (one-to-many) mappings of \(w\).

     The \(S\)-axis must be an asymptote. The curve must not touch the \(S\)-axis nor must the curve approach the asymptote then deviate away later.[4 marks]

d.

\(4 – \frac{{300}}{{{w^2}}}\)     (A1)(A1)(A1)

Notes: Award (A1) for \(4\), (A1) for \(-300\), (A1) for \(\frac{1}{{{w^2}}}\) or \({w^{ – 2}}\). If extra terms present, award at most (A1)(A1)(A0).[3 marks]

e.

\(4 – \frac{{300}}{{{w^2}}} = 0\)   OR   \(\frac{{300}}{{{w^2}}} = 4\)   OR   \(\frac{{{\text{d}}S}}{{{\text{d}}w}} = 0\)     (M1)

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

\(w = 8.66{\text{ }}\left( {\sqrt {75} ,{\text{ 8.66025}} \ldots } \right)\)     (A1)(ft)(G2)

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

f.

\(17.3 \left( {\frac{{150}}{{\sqrt {75} }},{\text{ 17.3205}} \ldots } \right)\)     (A1)(ft)

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

g.

\(40 + 4\sqrt {75}  + \frac{{300}}{{\sqrt {75} }}\)     (M1)

Note: Award (M1) for substitution of their answer to part (f) into the expression for \(S\).

\( = 110{\text{ (cm) }}\left( {40 + 40\sqrt 3 ,{\text{ 109.282}} \ldots } \right)\)     (A1)(ft)(G2)

Note: Do not accept \(109\).

     Follow through from their answers to parts (f) and (g).[2 marks]

h.

Question

The front view of the edge of a water tank is drawn on a set of axes shown below.

The edge is modelled by \(y = a{x^2} + c\).



Point \({\text{P}}\) has coordinates \((-3, 1.8)\), point \({\text{O}}\) has coordinates \((0, 0)\) and point \({\text{Q}}\) has coordinates \((3, 1.8)\).

Write down the value of \(c\).[1]

a.

Find the value of \(a\).[2]

b.

Hence write down the equation of the quadratic function which models the edge of the water tank.[1]

c.

The water tank is shown below. It is partially filled with water.

Calculate the value of y when \(x = 2.4{\text{ m}}\).[2]

d.

The water tank is shown below. It is partially filled with water.

State what the value of \(x\) and the value of \(y\) represent for this water tank.[2]

e.

The water tank is shown below. It is partially filled with water.

Find the value of \(x\) when the height of water in the tank is \(0.9\) m.[2]

f.

The water tank is shown below. It is partially filled with water.

The water tank has a length of 5 m.

When the water tank is filled to a height of \(0.9\) m, the front cross-sectional area of the water is \({\text{2.55 }}{{\text{m}}^2}\).

(i)     Calculate the volume of water in the tank.

The total volume of the tank is \({\text{36 }}{{\text{m}}^3}\).

(ii)     Calculate the percentage of water in the tank.[2]

g.
Answer/Explanation

Markscheme

\(0\)     (A1)(G1)[1 mark]

a.

\(1.8 = a{(3)^2} + 0\)     (M1)

OR

\(1.8 = a{( – 3)^2} + 0\)     (M1)

Note: Award (M1) for substitution of \(y = 1.8\) or \(x = 3\) and their value of \(c\) into equation. \(0\) may be implied.

\(a = 0.2\)   \(\left( {\frac{1}{5}} \right)\)     (A1)(ft)(G1)

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

     Award (G1) for a correct answer only.[2 marks]

b.

\(y = 0.2{x^2}\)     (A1)(ft)

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

     Answer must be an equation.[1 mark]

c.

\(0.2 \times {(2.4)^2}\)     (M1)

\( = 1.15{\text{ (m)}}\)   \((1.152)\)     (A1)(ft)(G1)

Notes: Award (M1) for correctly substituted formula, (A1) for correct answer. Follow through from their answer to part (c).

     Award (G1) for a correct answer only.[2 marks]

d.

\(y\) is the height     (A1)

positive value of \(x\) is half the width (or equivalent)     (A1)[2 marks]

e.

\(0.9 = 0.2{x^2}\)     (M1)

Note: Award (M1) for setting their equation equal to \(0.9\).

\(x =  \pm 2.12{\text{ (m)}}\)   \(\left( { \pm \frac{3}{2}\sqrt 2 ,{\text{ }} \pm \sqrt {4.5} ,{\text{ }} \pm {\text{2.12132}} \ldots } \right)\)     (A1)(ft)(G1)

Note: Accept \(2.12\). Award (G1) for a correct answer only.[2 marks]

f.

(i)     \(2.55 \times 5\)     (M1)

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

\( = 12.8{\text{ (}}{{\text{m}}^3}{\text{)}}\)   \(\left( {{\text{12.75 (}}{{\text{m}}^3}{\text{)}}} \right)\)     (A1)(G2)[2 marks]

(ii)     \(\frac{{12.75}}{{36}} \times 100\)     (M1)

Note: Award (M1) for correct quotient multiplied by \(100\).

\( = 35.4 (\%)\)  \((35.4166 \ldots )\)     (A1)(ft)(G2)

Note: Award (G2) for \(35.6 (\%) (35.5555… (\%))\).

     Follow through from their answer to part (g)(i).[2 marks]

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

The Great Pyramid of Giza in Egypt is a right pyramid with a square base. The pyramid is made of solid stone. The sides of the base are \(230\,{\text{m}}\) long. The diagram below represents this pyramid, labelled \({\text{VABCD}}\).

\({\text{V}}\) is the vertex of the pyramid. \({\text{O}}\) is the centre of the base, \({\text{ABCD}}\) . \({\text{M}}\) is the midpoint of \({\text{AB}}\). Angle \({\text{ABV}} = 58.3^\circ \) .

Show that the length of \({\text{VM}}\) is \(186\) metres, correct to three significant figures.[3]

a.

Calculate the height of the pyramid, \({\text{VO}}\) .[2]

b.

Find the volume of the pyramid.[2]

c.

Write down your answer to part (c) in the form \(a \times {10^k}\)  where \(1 \leqslant a < 10\) and \(k \in \mathbb{Z}\) .[2]

d.

Ahmad is a tour guide at the Great Pyramid of Giza. He claims that the amount of stone used to build the pyramid could build a wall \(5\) metres high and \(1\) metre wide stretching from Paris to Amsterdam, which are \(430\,{\text{km}}\) apart.

Determine whether Ahmad’s claim is correct. Give a reason.[4]

e.

Ahmad and his friends like to sit in the pyramid’s shadow, \({\text{ABW}}\), to cool down.
At mid-afternoon, \({\text{BW}} = 160\,{\text{m}}\)  and angle \({\text{ABW}} = 15^\circ .\)

i)     Calculate the length of \({\text{AW}}\) at mid-afternoon.

ii)    Calculate the area of the shadow, \({\text{ABW}}\), at mid-afternoon.[6]

f.
Answer/Explanation

Markscheme

\(\tan \,(58.3) = \frac{{{\text{VM}}}}{{115}}\)   OR \(115 \times \tan \,(58.3^\circ )\)              (A1)(M1)

Note: Award (A1) for \(115\,\,\left( {ie\,\frac{{230}}{2}} \right)\)   seen, (M1) for correct substitution into trig formula.

\(\left( {{\text{VM}} = } \right)\,\,186.200\,({\text{m}})\)        (A1)

\(\left( {{\text{VM}} = } \right)\,\,186\,({\text{m}})\)           (AG)

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

a.

\({\text{V}}{{\text{O}}^2} + {115^2} = {186^2}\)  OR \(\sqrt {{{186}^2} – {{115}^2}} \)       (M1)

Note: Award (M1) for correct substitution into Pythagoras formula. Accept alternative methods.

\({\text{(VO}} = )\,\,146\,({\text{m}})\,\,(146.188…)\)       (A1)(G2)

Note: Use of full calculator display for \({\text{VM}}\) gives \(146.443…\,{\text{(m)}}\).

b.

Units are required in part (c)

\(\frac{1}{3}({230^2} \times 146.188…)\)       (M1)

Note: Award (M1) for correct substitution in volume formula. Follow through from part (b).

\( = 2\,580\,000\,{{\text{m}}^3}\,\,(2\,577\,785…\,{{\text{m}}^3})\)       (A1)(ft)(G2)

Note: The answer is \(2\,580\,000\,{{\text{m}}^3}\) , the units are required. Use of \({\text{OV}} = 146.442\) gives  \(2582271…\,{{\text{m}}^3}\)

Use of \({\text{OV}} = 146\) gives  \(2574466…\,{{\text{m}}^3}.\)

c.

\(2.58 \times {10^6}\,({{\text{m}}^3})\)       (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for \(2.58\) and (A1)(ft) for \( \times {10^6}.\,\)

Award (A0)(A0) for answers of the type: \(2.58 \times {10^5}\,({{\text{m}}^3}).\)

Follow through from part (c).

d.

the volume of a wall would be \(430\,000 \times 5 \times 1\)       (M1)

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

\(2150000\,({{\text{m}}^3})\)       (A1)(G2)

which is less than the volume of the pyramid       (R1)(ft)

Ahmad is correct.       (A1)(ft)

OR

the length of the wall would be \(\frac{{{\text{their part (c)}}}}{{5 \times 1 \times 1000}}\)       (M1)

Note: Award (M1) for dividing their part (c) by \(5000.\)

\(516\,({\text{km}})\)          (A1)(ft)(G2)

which is more than the distance from Paris to Amsterdam       (R1)(ft)

Ahmad is correct.       (A1)(ft)

Note: Do not award final (A1) without an explicit comparison. Follow through from part (c) or part (d). Award (R1) for reasoning that is consistent with their working in part (e); comparing two volumes, or comparing two lengths.

e.

Units are required in part (f)(ii).

i)     \({\text{A}}{{\text{W}}^2} = {160^2} + {230^2} – 2 \times 160 \times 230 \times \cos \,(15^\circ )\)       (M1)(A1)

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

\({\text{AW}} = 86.1\,({\text{m}})\,\,\,(86.0689…)\)       (A1)(G2)

Note: Award (M0)(A0)(A0) if \({\text{BAW}}\) or \({\text{AWB}}\)  is considered to be a right angled triangle.

ii)    \({\text{area}} = \frac{1}{2} \times 230 \times 160 \times \sin \,(15^\circ )\)         (M1)(A1)

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

\( = 4760\,{{\text{m}}^2}\,\,\,(4762.27…\,{{\text{m}}^2})\)       (A1)(G2)

Note: The answer is \(4760\,{{\text{m}}^2}\) , the units are required.

f.

Question

A playground, when viewed from above, is shaped like a quadrilateral, \({\text{ABCD}}\), where \({\text{AB}} = 21.8\,{\text{m}}\) and \({\text{CD}} = 11\,{\text{m}}\) . Three of the internal angles have been measured and angle \({\text{ABC}} = 47^\circ \) , angle \({\text{ACB}} = 63^\circ \) and angle \({\text{CAD}} = 30^\circ \) . This information is represented in the following diagram.

Calculate the distance \({\text{AC}}\).[3]

a.

Calculate angle \({\text{ADC}}\).[3]

b.

There is a tree at \({\text{C}}\), perpendicular to the ground. The angle of elevation to the top of the tree from \({\text{D}}\) is \(35^\circ \).

Calculate the height of the tree.[2]

c.

Chavi estimates that the height of the tree is \(6\,{\text{m}}\).

Calculate the percentage error in Chavi’s estimate.[2]

d.

Chavi is celebrating her birthday with her friends on the playground. Her mother brings a \(2\,\,{\text{litre}}\) bottle of orange juice to share among them. She also brings cone-shaped paper cups.

Each cup has a vertical height of \(10\,{\text{cm}}\) and the top of the cup has a diameter of \(6\,{\text{cm}}\).

Calculate the volume of one paper cup.[3]

e.

Calculate the maximum number of cups that can be completely filled with the \(2\,\,{\text{litre}}\) bottle of orange juice.[3]

f.
Answer/Explanation

Markscheme

\(\frac{{21.8}}{{\sin 63^\circ }} = \frac{{{\text{AC}}}}{{\sin 47^\circ }}\)          (M1)(A1)

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

\(({\text{AC}} = )\,\,17.9\,({\text{m}})\,\,\,(17.8938…\,({\text{m}}))\)           (A1)(G2)

a.

\(\frac{{11}}{{\sin 30}} = \frac{{17.8938…}}{{\sin {\text{ADC}}}}\)          (M1)(A1)(ft)

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

\(\left( {{\text{Angle ADC}} = } \right)\,\,\,54.4^\circ \,\,\,(54.4250…^\circ )\)          (A1)(ft)(G2)

Note: Accept \(54.5\,\,(54.4527…)\) or \(126\,\,(125.547…)\) from using their \(3\) sf answer.
Follow through from part (a). Accept \(125.575…\)

b.

\(11 \times \tan 35^\circ \)  (or equivalent)          (M1)

Note: Award (M1) for correct substitution into trigonometric ratio.

\(7.70\,({\text{m}})\,\,\,(7.70228…\,({\text{m}}))\)          (A1)(G2)

c.

\(\left| {\frac{{6 – 7.70228…}}{{7.70228…}}} \right| \times 100\,\% \)          (M1)

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

OR

\(100 – \left| {\frac{{6 \times 100}}{{7.70228…}}} \right|\)          (M1)

Note: Award (M1) for the alternative method.

\(22.1\,(\% )\,\,\,(22.1009…\,(\% ))\)          (A1)(ft)(G2)

Note: Award at most (M1)(A0) for a final answer that is negative. Follow through from part (c).

d.

\(\frac{1}{3}\pi  \times {3^2} \times 10\)          (A1)(M1)

Note: Award (A1) for \(3\) seen, (M1) for their correct substitution into volume of a cone formula.

\(94.2\,{\text{c}}{{\text{m}}^3}\,\,\,(30\pi \,{\text{c}}{{\text{m}}^3},\,\,94.2477…\,{\text{c}}{{\text{m}}^3})\)          (A1)(G3)

Note: The answer is \(94.2\,{\text{c}}{{\text{m}}^3}\), units are required. Award at most (A0)(M1)(A0) if an incorrect value for \(r\) is used.

e.

\(\frac{{2000}}{{94.2477…}}\) OR \(\frac{2}{{0.0942477…}}\)          (M1)(M1)

Note: Award (M1) for correct conversion (litres to \({\text{c}}{{\text{m}}^3}\) or \({\text{c}}{{\text{m}}^3}\) to litres), (M1) for dividing by their part (e) (or their converted part (e)).

\(21\)          (A1)(ft)(G2)

Note: The final (A1) is not awarded if the final answer is not an integer. Follow through from part (e), but only if the answer is rounded down.

f.

Question

A water container is made in the shape of a cylinder with internal height \(h\) cm and internal base radius \(r\) cm.

N16/5/MATSD/SP2/ENG/TZ0/06

The water container has no top. The inner surfaces of the container are to be coated with a water-resistant material.

The volume of the water container is \(0.5{\text{ }}{{\text{m}}^3}\).

The water container is designed so that the area to be coated is minimized.

One can of water-resistant material coats a surface area of \(2000{\text{ c}}{{\text{m}}^2}\).

Write down a formula for \(A\), the surface area to be coated.[2]

a.

Express this volume in \({\text{c}}{{\text{m}}^3}\).[1]

b.

Write down, in terms of \(r\) and \(h\), an equation for the volume of this water container.[1]

c.

Show that \(A = \pi {r^2}\frac{{1\,000\,000}}{r}\).[2]

d.

Show that \(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\).[2]

d.

Find \(\frac{{{\text{d}}A}}{{{\text{d}}r}}\).[3]

e.

Using your answer to part (e), find the value of \(r\) which minimizes \(A\).[3]

f.

Find the value of this minimum area.[2]

g.

Find the least number of cans of water-resistant material that will coat the area in part (g).[3]

h.
Answer/Explanation

Markscheme

\((A = ){\text{ }}\pi {r^2} + 2\pi rh\)    (A1)(A1)

Note:     Award (A1) for either \(\pi {r^2}\) OR \(2\pi rh\) seen. Award (A1) for two correct terms added together.[2 marks]

a.

\(500\,000\)    (A1)

Notes:     Units not required.[1 mark]

b.

\(500\,000 = \pi {r^2}h\)    (A1)(ft)

Notes:     Award (A1)(ft) for \(\pi {r^2}h\) equating to their part (b).

Do not accept unless \(V = \pi {r^2}h\) is explicitly defined as their part (b).[1 mark]

c.

\(A = \pi {r^2} + 2\pi r\left( {\frac{{500\,000}}{{\pi {r^2}}}} \right)\)    (A1)(ft)(M1)

Note:     Award (A1)(ft) for their \({\frac{{500\,000}}{{\pi {r^2}}}}\) seen.

Award (M1) for correctly substituting only \({\frac{{500\,000}}{{\pi {r^2}}}}\) into a correct part (a).

Award (A1)(ft)(M1) for rearranging part (c) to \(\pi rh = \frac{{500\,000}}{r}\) and substituting for \(\pi rh\) in expression for \(A\).

\(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\)    (AG)

Notes:     The conclusion, \(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\), must be consistent with their working seen for the (A1) to be awarded.

Accept \({10^6}\) as equivalent to \({1\,000\,000}\).[2 marks]

d.

\(A = \pi {r^2} + 2\pi r\left( {\frac{{500\,000}}{{\pi {r^2}}}} \right)\)    (A1)(ft)(M1)

Note:     Award (A1)(ft) for their \({\frac{{500\,000}}{{\pi {r^2}}}}\) seen.

Award (M1) for correctly substituting only \({\frac{{500\,000}}{{\pi {r^2}}}}\) into a correct part (a).

Award (A1)(ft)(M1) for rearranging part (c) to \(\pi rh = \frac{{500\,000}}{r}\) and substituting for \(\pi rh\) in expression for \(A\).

\(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\)    (AG)

Notes:     The conclusion, \(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\), must be consistent with their working seen for the (A1) to be awarded.

Accept \({10^6}\) as equivalent to \({1\,000\,000}\).[2 marks]

d.

\(2\pi r – \frac{{{\text{1}}\,{\text{000}}\,{\text{000}}}}{{{r^2}}}\)    (A1)(A1)(A1)

Note:     Award (A1) for \(2\pi r\), (A1) for \(\frac{1}{{{r^2}}}\) or \({r^{ – 2}}\), (A1) for \( – {\text{1}}\,{\text{000}}\,{\text{000}}\).[3 marks]

e.

\(2\pi r – \frac{{1\,000\,000}}{{{r^2}}} = 0\)    (M1)

Note:     Award (M1) for equating their part (e) to zero.

\({r^3} = \frac{{1\,000\,000}}{{2\pi }}\) OR \(r = \sqrt[3]{{\frac{{1\,000\,000}}{{2\pi }}}}\)     (M1)

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

OR

sketch of derivative function     (M1)

with its zero indicated     (M1)

\((r = ){\text{ }}54.2{\text{ }}({\text{cm}}){\text{ }}(54.1926 \ldots )\)    (A1)(ft)(G2)[3 marks]

f.

\(\pi {(54.1926 \ldots )^2} + \frac{{1\,000\,000}}{{(54.1926 \ldots )}}\)    (M1)

Note:     Award (M1) for correct substitution of their part (f) into the given equation.

\( = 27\,700{\text{ }}({\text{c}}{{\text{m}}^2}){\text{ }}(27\,679.0 \ldots )\)    (A1)(ft)(G2)[2 marks]

g.

\(\frac{{27\,679.0 \ldots }}{{2000}}\)    (M1)

Note:     Award (M1) for dividing their part (g) by 2000.

\( = 13.8395 \ldots \)    (A1)(ft)

Notes:     Follow through from part (g).

14 (cans)     (A1)(ft)(G3)

Notes:     Final (A1) awarded for rounding up their \(13.8395 \ldots \) to the next integer.[3 marks]

h.

Question

A pan, in which to cook a pizza, is in the shape of a cylinder. The pan has a diameter of 35 cm and a height of 0.5 cm.

M17/5/MATSD/SP2/ENG/TZ1/04

A chef had enough pizza dough to exactly fill the pan. The dough was in the shape of a sphere.

The pizza was cooked in a hot oven. Once taken out of the oven, the pizza was placed in a dining room.

The temperature, \(P\), of the pizza, in degrees Celsius, °C, can be modelled by

\[P(t) = a{(2.06)^{ – t}} + 19,{\text{ }}t \geqslant 0\]

where \(a\) is a constant and \(t\) is the time, in minutes, since the pizza was taken out of the oven.

When the pizza was taken out of the oven its temperature was 230 °C.

The pizza can be eaten once its temperature drops to 45 °C.

Calculate the volume of this pan.[3]

a.

Find the radius of the sphere in cm, correct to one decimal place.[4]

b.

Find the value of \(a\).[2]

c.

Find the temperature that the pizza will be 5 minutes after it is taken out of the oven.[2]

d.

Calculate, to the nearest second, the time since the pizza was taken out of the oven until it can be eaten.[3]

e.

In the context of this model, state what the value of 19 represents.[1]

f.
Answer/Explanation

Markscheme

\((V = ){\text{ }}\pi  \times {{\text{(17.5)}}^2} \times 0.5\)     (A1)(M1)

Notes:     Award (A1) for 17.5 (or equivalent) seen.

Award (M1) for correct substitutions into volume of a cylinder formula.

\( = 481{\text{ c}}{{\text{m}}^3}{\text{ }}(481.056 \ldots {\text{ c}}{{\text{m}}^3},{\text{ }}153.125\pi {\text{ c}}{{\text{m}}^3})\)     (A1)(G2)[3 marks]

a.

\(\frac{4}{3} \times \pi  \times {r^3} = 481.056 \ldots \)     (M1)

Note:     Award (M1) for equating their answer to part (a) to the volume of sphere.

\({r^3} = \frac{{3 \times 481.056 \ldots }}{{4\pi }}{\text{ }}( = 114.843 \ldots )\)     (M1)

Note:     Award (M1) for correctly rearranging so \({r^3}\) is the subject.

\(r = 4.86074 \ldots {\text{ (cm)}}\)     (A1)(ft)(G2)

Note:     Award (A1) for correct unrounded answer seen. Follow through from part (a).

\( = 4.9{\text{ (cm)}}\)     (A1)(ft)(G3)

Note:     The final (A1)(ft) is awarded for rounding their unrounded answer to one decimal place.[4 marks]

b.

\(230 = a{(2.06)^0} + 19\)     (M1)

Note:     Award (M1) for correct substitution.

\(a = 211\)     (A1)(G2)[2 marks]

c.

\((P = ){\text{ }}211 \times {(2.06)^{ – 5}} + 19\)      (M1)

Note:     Award (M1) for correct substitution into the function, \(P(t)\). Follow through from part (c). The negative sign in the exponent is required for correct substitution.

\( = 24.7\) (°C) \((24.6878 \ldots \) (°C))     (A1)(ft)(G2)[2 marks]

d.

\(45 = 211 \times {(2.06)^{ – t}} + 19\)     (M1)

Note:     Award (M1) for equating 45 to the exponential equation and for correct substitution (follow through for their \(a\) in part (c)).

\((t = ){\text{ }}2.89711 \ldots \)     (A1)(ft)(G1)

\(174{\text{ (seconds) }}\left( {173.826 \ldots {\text{ (seconds)}}} \right)\)     (A1)(ft)(G2)

Note:     Award final (A1)(ft) for converting their \({\text{2.89711}} \ldots \) minutes into seconds.[3 marks]

e.

the temperature of the (dining) room     (A1)

OR

the lowest final temperature to which the pizza will cool     (A1)[1 mark]

f.

Question

A restaurant serves desserts in glasses in the shape of a cone and in the shape of a hemisphere. The diameter of a cone shaped glass is 7.2 cm and the height of the cone is 11.8 cm as shown.

N17/5/MATSD/SP2/ENG/TZ0/06

The volume of a hemisphere shaped glass is \(225{\text{ c}}{{\text{m}}^3}\).

The restaurant offers two types of dessert.

The regular dessert is a hemisphere shaped glass completely filled with chocolate mousse. The cost, to the restaurant, of the chocolate mousse for one regular dessert is $1.89.

The special dessert is a cone shaped glass filled with two ingredients. It is first filled with orange paste to half of its height and then with chocolate mousse for the remaining volume.

N17/5/MATSD/SP2/ENG/TZ0/06.d.e.f

The cost, to the restaurant, of \(100{\text{ c}}{{\text{m}}^3}\) of orange paste is $7.42.

A chef at the restaurant prepares 50 desserts; \(x\) regular desserts and \(y\) special desserts. The cost of the ingredients for the 50 desserts is $111.44.

Show that the volume of a cone shaped glass is \(160{\text{ c}}{{\text{m}}^3}\), correct to 3 significant figures.[2]

a.

Calculate the radius, \(r\), of a hemisphere shaped glass.[3]

b.

Find the cost of \(100{\text{ c}}{{\text{m}}^3}\) of chocolate mousse.[2]

c.

Show that there is \(20{\text{ c}}{{\text{m}}^3}\) of orange paste in each special dessert.[2]

d.

Find the total cost of the ingredients of one special dessert.[2]

e.

Find the value of \(x\).[3]

f.
Answer/Explanation

Markscheme

\((V = ){\text{ }}\frac{1}{3}\pi {(3.6)^2} \times 11.8\)     (M1)

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

\( = 160.145 \ldots {\text{ }}({\text{c}}{{\text{m}}^3})\)     (A1)

\( = 160{\text{ }}({\text{c}}{{\text{m}}^3})\)     (AG)

Note:     Both rounded and unrounded answers must be seen for the final (A1) to be awarded.[2 marks]

a.

\(\frac{1}{2} \times \frac{4}{3}\pi {r^3} = 225\)     (M1)(A1)

Notes:     Award (M1) for multiplying volume of sphere formula by \(\frac{1}{2}\) (or equivalent).

Award (A1) for equating the volume of hemisphere formula to 225.

OR

\(\frac{4}{3}\pi {r^3} = 450\)     (A1)(M1)

Notes:     Award (A1) for 450 seen, (M1) for equating the volume of sphere formula to 450.

\((r = ){\text{ }}4.75{\text{ }}({\text{cm}}){\text{ }}(4.75380 \ldots )\)     (A1)(G2)[3 marks]

b.

\(\frac{{1.89 \times 100}}{{225}}\)     (M1)

Note:     Award (M1) for dividing 1.89 by 2.25, or equivalent.

\( = 0.84\)     (A1)(G2)

Note: Accept 84 cents if the units are explicit.[2 marks]

c.

\({r_2} = 1.8\)     (A1)

\({V_2} = \frac{1}{3}\pi {(1.8)^2} \times 5.9\)     (M1)

Note:     Award (M1) for correct substitution into volume of a cone formula, but only if the result rounds to 20.

\( = 20{\text{ c}}{{\text{m}}^3}\)     (AG)

OR

\({r_2} = \frac{1}{2}r\)     (A1)

\({V_2} = {\left( {\frac{1}{2}} \right)^3}160\)     (M1)

Notes:     Award (M1) for multiplying 160 by \({\left( {\frac{1}{2}} \right)^3}\). Award (A0)(M1) for \(\frac{1}{8} \times 160\) if \(\frac{1}{2}\) is not seen.

\( = 20{\text{ }}({\text{c}}{{\text{m}}^3})\)     (AG)

Notes:     Do not award any marks if the response substitutes in the known value \((V = 20)\) to find the radius of the cone.[2 marks]

d.

\(\frac{{20}}{{100}} \times 7.42 + \frac{{140}}{{100}} \times 0.84\)     (M1)

Note:     Award (M1) for the sum of two correct products.

$ 2.66     (A1)(ft)(G2)

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

e.

\(x + y = 50\)     (M1)

Note:     Award (M1) for correct equation.

\(1.89x + 2.66y = 111.44\)     (M1)

Note:     Award (M1) for setting up correct equation, including their 2.66 from part (e).

\((x = ){\text{ }}28\)     (A1)(ft)(G3)

Note:     Follow through from part (e), but only if their answer for \(x\) is rounded to the nearest positive integer, where \(0 < x < 50\).

Award at most (M1)(M1)(A0) for a final answer of “28, 22”, where the \(x\)-value is not clearly defined.[3 marks]

f.

Question

Farmer Brown has built a new barn, on horizontal ground, on his farm. The barn has a cuboid base and a triangular prism roof, as shown in the diagram.

The cuboid has a width of 10 m, a length of 16 m and a height of 5 m.
The roof has two sloping faces and two vertical and identical sides, ADE and GLF.
The face DEFL slopes at an angle of 15° to the horizontal and ED = 7 m .

The roof was built using metal supports. Each support is made from five lengths of metal AE, ED, AD, EM and MN, and the design is shown in the following diagram.

ED = 7 m , AD = 10 m and angle ADE = 15° .
M is the midpoint of AD.
N is the point on ED such that MN is at right angles to ED.

Farmer Brown believes that N is the midpoint of ED.

Calculate the area of triangle EAD.[3]

a.

Calculate the total volume of the barn.[3]

b.

Calculate the length of MN.[2]

c.

Calculate the length of AE.[3]

d.

Show that Farmer Brown is incorrect.[3]

e.

Calculate the total length of metal required for one support.[4]

f.
Answer/Explanation

Markscheme

(Area of EAD =) \(\frac{1}{2} \times 10 \times 7 \times {\text{sin}}15\)    (M1)(A1)

Note: Award (M1) for substitution into area of a triangle formula, (A1) for correct substitution. Award (M0)(A0)(A0) if EAD or AED is considered to be a right-angled triangle.

= 9.06 m2  (9.05866… m2)     (A1)   (G3)[3 marks]

a.

(10 × 5 × 16) + (9.05866… × 16)     (M1)(M1)

Note: Award (M1) for correct substitution into volume of a cuboid, (M1) for adding the correctly substituted volume of their triangular prism.

= 945 m3  (944.938… m3)     (A1)(ft)  (G3)

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

b.

\(\frac{{{\text{MN}}}}{5} = \,\,\,{\text{sin}}15\)     (M1)

Note: Award (M1) for correct substitution into trigonometric equation.

(MN =) 1.29(m) (1.29409… (m))     (A1) (G2)[2 marks]

c.

(AE2 =) 102 + 72 − 2 × 10 × 7 × cos 15     (M1)(A1)

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

(AE =) 3.71(m)  (3.71084… (m))     (A1) (G2)[3 marks]

d.

ND2 = 52 − (1.29409…)2     (M1)

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

(ND =) 4.83  (4.82962…)     (A1)(ft)

Note: Follow through from part (c).

OR

\(\frac{{1.29409 \ldots }}{{{\text{ND}}}} = {\text{tan}}\,15^\circ \)     (M1)

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

(ND =) 4.83  (4.82962…)     (A1)(ft)

Note: Follow through from part (c).

OR

\(\frac{{{\text{ND}}}}{5} = {\text{cos }}15^\circ \)     (M1)

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

(ND =) 4.83  (4.82962…)     (A1)(ft)

Note: Follow through from part (c).

OR

ND2 = 1.29409…2 + 52 − 2 × 1.29409… × 5 × cos 75°     (M1)

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

(ND =) 4.83  (4.82962…)     (A1)(ft)

Note: Follow through from part (c).

4.82962… ≠ 3.5   (ND ≠ 3.5)     (R1)(ft)

OR

4.82962… ≠ 2.17038…   (ND ≠ NE)     (R1)(ft)

(hence Farmer Brown is incorrect)

Note: Do not award (M0)(A0)(R1)(ft). Award (M0)(A0)(R0) for a correct conclusion without any working seen.[3 marks]

e.

(EM2 =) 1.29409…2 + (7 − 4.82962…)2     (M1)

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

OR

(EM2 =) 52 + 72 − 2 × 5 × 7 × cos 15     (M1)

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

(EM =) 2.53(m) (2.52689…(m))     (A1)(ft) (G2)(ft)

Note: Follow through from parts (c), (d) and (e).

(Total length =) 2.52689… + 3.71084… + 1.29409… +10 + 7     (M1)

Note: Award (M1) for adding their EM, their parts (c) and (d), and 10 and 7.

= 24.5 (m)    (24.5318… (m))     (A1)(ft) (G4)

Note: Follow through from parts (c) and (d).[4 marks]

f.

Question

A manufacturer makes trash cans in the form of a cylinder with a hemispherical top. The trash can has a height of 70 cm. The base radius of both the cylinder and the hemispherical top is 20 cm.

A designer is asked to produce a new trash can.

The new trash can will also be in the form of a cylinder with a hemispherical top.

This trash can will have a height of H cm and a base radius of r cm.

There is a design constraint such that H + 2r = 110 cm.

The designer has to maximize the volume of the trash can.

Write down the height of the cylinder.[1]

a.

Find the total volume of the trash can.[4]

b.

Find the height of the cylinder, h , of the new trash can, in terms of r.[2]

c.

Show that the volume, V cm3 , of the new trash can is given by

\(V = 110\pi {r^3}\).[3]

d.

Using your graphic display calculator, find the value of r which maximizes the value of V.[2]

e.

The designer claims that the new trash can has a capacity that is at least 40% greater than the capacity of the original trash can.

State whether the designer’s claim is correct. Justify your answer.[4]

f.
Answer/Explanation

Markscheme

50 (cm)      (A1)[1 mark]

a.

\(\pi  \times 50 \times {20^2} + \frac{1}{2} \times \frac{4}{3} \times \pi  \times {20^3}\)     (M1)(M1)(M1)

Note: Award (M1) for their correctly substituted volume of cylinder, (M1) for correctly substituted volume of sphere formula, (M1) for halving the substituted volume of sphere formula. Award at most (M1)(M1)(M0) if there is no addition of the volumes.

\( = 79600\,\,\left( {{\text{c}}{{\text{m}}^3}} \right)\,\,\left( {79587.0 \ldots \left( {{\text{c}}{{\text{m}}^3}} \right)\,,\,\,\frac{{76000}}{3}\pi } \right)\)     (A1)(ft) (G3)

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

b.

h = H − r (or equivalent) OR H = 110 − 2r     (M1)

Note: Award (M1) for writing h in terms of H and r or for writing H in terms of r.

(h =) 110 3r     (A1) (G2)[2 marks]

c.

\(\left( {V = } \right)\,\,\,\,\frac{2}{3}\pi {r^3} + \pi {r^2} \times \left( {110 – 3r} \right)\)    (M1)(M1)(M1)

Note: Award (M1) for volume of hemisphere, (M1) for correct substitution of their h into the volume of a cylinder, (M1) for addition of two correctly substituted volumes leading to the given answer. Award at most (M1)(M1)(M0) for subsequent working that does not lead to the given answer. Award at most (M1)(M1)(M0) for substituting H = 110 − 2r as their h.

\(V = 110\pi {r^2} – \frac{7}{3}\pi {r^3}\)    (AG)[3 marks]

d.

(r =) 31.4 (cm)  (31.4285… (cm))     (G2)

OR

\(\left( \pi  \right)\left( {220r – 7{r^2}} \right) = 0\)      (M1)

Note: Award (M1) for setting the correct derivative equal to zero.

(r =) 31.4 (cm)  (31.4285… (cm))     (A1)[2 marks]

e.

\(\left( {V = } \right)\,\,\,\,110\pi {\left( {31.4285 \ldots } \right)^3} – \frac{7}{3}\pi {\left( {31.4285 \ldots } \right)^3}\)     (M1)

Note: Award (M1) for correct substitution of their 31.4285… into the given equation.

= 114000 (113781…)     (A1)(ft)

Note: Follow through from part (e).

(increase in capacity =) \(\frac{{113.781 \ldots  – 79587.0 \ldots }}{{79587.0 \ldots }} \times 100 = 43.0\,\,\left( {\text{% }} \right)\)     (R1)(ft)

Note: Award (R1)(ft) for finding the correct percentage increase from their two volumes.

OR

1.4 × 79587.0… = 111421.81…     (R1)(ft)

Note: Award (R1)(ft) for finding the capacity of a trash can 40% larger than the original.

Claim is correct (A1)(ft)

Note: Follow through from parts (b), (e) and within part (f). The final (R1)(A1)(ft) can be awarded for their correct reason and conclusion. Do not award (R0)(A1)(ft).[4 marks]

f.

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.
Scroll to Top