IB DP Mathematical Studies 5.5 Paper 1

 

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Question

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

Find the volume of one tennis ball.[2]

a.

Calculate the volume of the empty space in the tube when four tennis balls have been placed in it.[4]

b.
Answer/Explanation

Markscheme

Unit penalty (UP) applies

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

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

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

a.

Unit penalty (UP) applies

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

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

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

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

b.

Question

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

The surface area of a sphere is 500 cm2 .

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

a.

Write down your answer to (a) correct to the nearest integer.[1]

b.

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

c.
Answer/Explanation

Markscheme

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

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

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

a.

1051     (A1)(ft)[1 mark]

b.

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

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

c.

Question

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

Find the volume of the box.[2]

a.

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

Find Joaquin’s estimated number of cubes in the box.[2]

b.

The actual number of cubes in the box is 350.

Find the percentage error in Joaquin’s estimated number of cubes in the box.[2]

c.
Answer/Explanation

Markscheme

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

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

a.

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

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

b.

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

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

Award (A1) for answer, follow through from part (b).

Accept –14.3 %.

% sign not necessary.[2 marks]

c.

Question

The base of a prism is a regular hexagon. The centre of the hexagon is O and the length of OA is 15 cm.

Write down the size of angle AOB.[1]

a.

Find the area of the triangle AOB.[3]

b.

The height of the prism is 20 cm.

Find the volume of the prism.[2]

c.
Answer/Explanation

Markscheme

60°     (A1)     (C1)[1 mark]

a.

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

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

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

OR

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

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

Follow through from their answer to part (a).

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

b.

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

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

c.

Question

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

Calculate the total volume of all \(75\) metal cannon balls excavated.[3]

a.

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

Calculate the height of the cone, assuming that no metal is wasted.[3]

b.
Answer/Explanation

Markscheme

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

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

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

a.

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

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

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

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

b.

Question

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

Calculate the volume of the observatory.[4]

a.

The hemispherical roof is to be painted.

Calculate the area that is to be painted.[2]

b.
Answer/Explanation

Markscheme

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

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

\( = 15550.8 \ldots \)

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

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

a.

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

\( = 1413.71 \ldots \)

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

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

b.

Question

The area of a circle is equal to 8 cm2.

Find the radius of the circle.[2]

a.

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

Write down the volume of the solid cylinder.[1]

b.

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

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

c.
Answer/Explanation

Markscheme

πr2 = 8     (M1)

Note: Award (M1) for correct area formula.

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

a.

200 cm3     (A1)(ft)     (C1)

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

b.

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

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

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

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

c.

Question

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

Calculate the exact value of the volume of the cuboid, in cm3.[2]

a.

Write your answer to part (a) correct to

(i) one decimal place;

(ii) three significant figures.[2]

b.

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

c.
Answer/Explanation

Markscheme

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

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

\(=165.648\)     (A1)     (C2)

a.

(i) 165.6     (A1)(ft)

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

(ii) 166     (A1)(ft)     (C2)

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

b.

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

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

c.

Question

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

Angle OVB is \({27.9^ \circ }\).

Diagram not to scale.

Calculate OV, the height of the cone.[2]

a.

Calculate the volume of wood used to make the toy.[4]

b.
Answer/Explanation

Markscheme

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

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

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

a.

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

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

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

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

b.

Question

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

Each chocolate has a radius of 1 cm.

Find the volume of 1 chocolate.[2]

a.

Write down the volume of 20 chocolates.[1]

b.

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

Calculate the volume of the box.[2]

c.

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

Calculate the volume of empty space in the box.[1]

d.
Answer/Explanation

Markscheme

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

\(\frac{4}{3}\pi {(1)^3}\)     (M1)

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

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

a.

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

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

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

b.

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

\(10 \times 8 \times 2\)     (M1)

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

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

c.

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

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

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

d.

Question

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

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

a.

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

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

Give your answer correct to the nearest integer.[3]

b.
Answer/Explanation

Markscheme

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

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

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

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

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

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

a.

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

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

\( = 49.3308 \ldots \)     (A1)(ft)

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

\( = 49\)     (A1)     (C3)

Notes: Follow through from part (a).

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

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

b.

Question

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

Calculate \({\text{VM}}\).[4]

a.

Calculate the volume of the pyramid.[2]

b.
Answer/Explanation

Markscheme

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

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

\({\text{AC}} = 10\)     (A1)

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

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

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

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

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

a.

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

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

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

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

b.

Question

A cuboid has a rectangular base of width \(x\) cm and length 2\(x\) cm . The height of the cuboid is \(h\) cm . The total length of the edges of the cuboid is \(72\) cm.

The volume, \(V\), of the cuboid can be expressed as \(V = a{x^2} – 6{x^3}\).

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

a.

Find the value of \(x\) that makes the volume a maximum.[3]

b.
Answer/Explanation

Markscheme

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

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

\(V = 2{x^2}(18 – 3x)\)     (A1)

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

a.

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

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

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

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

Note: Follow through from part (a).

OR

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

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

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

Notes: Follow through from part (a).

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

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

b.

Question

FreshWave brand tuna is sold in cans that are in the shape of a cuboid with length \(8\,{\text{cm}}\), width \({\text{5}}\,{\text{cm}}\) and height \({\text{3}}{\text{.5}}\,{\text{cm}}\). HappyFin brand tuna is sold in cans that are cylindrical with diameter \({\text{7}}\,{\text{cm}}\) and height \({\text{4}}\,{\text{cm}}\).

Find the volume, in \({\text{c}}{{\text{m}}^3}\), of a can of

i)    FreshWave tuna;

ii)   HappyFin tuna.[4]

a.

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

Calculate the price, in cents, of a can of HappyFin tuna.[2]

b.
Answer/Explanation

Markscheme

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

\( = 140\)        (A1)

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

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

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

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

a.

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

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

\( = 99\)       (A1)(ft) (C2)

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

b.

Question

Assume that the Earth is a sphere with a radius, \(r\) , of \(6.38 \times {10^3}\,{\text{km}}\) .

i)     Calculate the surface area of the Earth in \({\text{k}}{{\text{m}}^2}\).

ii)    Write down your answer to part (a)(i) in the form \(a \times {10^k}\) , where \(1 \leqslant a < 10\) and \(k \in \mathbb{Z}\) .[4]

a.

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

Calculate the percentage of the surface area of the Earth that is covered by water.[2]

b.
Answer/Explanation

Markscheme

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

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

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

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

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

Note: Award (A1) for \(5.12\) and (A1) for \( \times {10^8}\).
Award (A0)(A0) for answers of the type: \(5.12 \times {10^7}\).
Follow through from part (a)(i).

a.

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

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

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

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

b.

Question

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

Write down the radius, in \({\text{cm}}\), of the base of the container.[1]

a.

Calculate the area of the base of the container.[2]

b.

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

Calculate the area to be painted.[3]

c.
Answer/Explanation

Markscheme

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

a.

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

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

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

b.

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

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

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

c.

Question

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

The volume of the balloon is increased by 40%.

Calculate the volume of the balloon.[2]

a.

Calculate the radius of the balloon following this increase.[4]

b.
Answer/Explanation

Markscheme

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

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

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

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

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

a.

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

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

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

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

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

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

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

b.

Question

A type of candy is packaged in a right circular cone that has volume \({\text{100 c}}{{\text{m}}^{\text{3}}}\) and vertical height 8 cm.

M17/5/MATSD/SP1/ENG/TZ1/09

Find the radius, \(r\), of the circular base of the cone.[2]

a.

Find the slant height, \(l\), of the cone.[2]

b.

Find the curved surface area of the cone.[2]

c.
Answer/Explanation

Markscheme

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

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

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

a.

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

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

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

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

b.

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

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

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

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

c.

Question

A cylindrical container with a radius of 8 cm is placed on a flat surface. The container is filled with water to a height of 12 cm, as shown in the following diagram.

M17/5/MATSD/SP1/ENG/TZ2/12

A heavy ball with a radius of 2.9 cm is dropped into the container. As a result, the height of the water increases to \(h\) cm, as shown in the following diagram.

M17/5/MATSD/SP1/ENG/TZ2/12.b

Find the volume of water in the container.[2]

a.

Find the value of \(h\).[4]

b.
Answer/Explanation

Markscheme

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

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

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

a.

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

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

OR

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

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

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

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

b.

Question

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

Calculate the radius of the base of the cone which has been removed.[2]

a.

Calculate the curved surface area of the cone which has been removed.[2]

b.

Calculate the curved surface area of the remaining solid.[2]

c.
Answer/Explanation

Markscheme

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

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

OR

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

Note: Award (M1) for a correct equation.

= 9 (cm)     (A1) (C2)[2 marks]

a.

\(\pi  \times 9 \times 15\)      (M1)

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

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

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

b.

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

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

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

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

c.

Question

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

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

The cone has a base radius of r cm and a height of 10 cm.

Find an expression for the slant height of the cone in terms of r.[2]

a.

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

Using your graphic display calculator, calculate the value of r.[4]

b.
Answer/Explanation

Markscheme

(slant height2 =) 102 + r 2     (M1)

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

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

a.

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

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

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

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

b.

Question

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

Calculate the length, \(l\), of the slant height of the cone.[2]

a.

Calculate the area that is painted red.[4]

b.
Answer/Explanation

Markscheme

\(\sqrt {{5^2} + {{12}^2}} \)     (M1)

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

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

a.

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

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

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

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

b.
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