IB DP Mathematical Studies 1.3 Paper 2

Question

A manufacturer has a contract to make \(2600\) solid blocks of wood. Each block is in the shape of a right triangular prism, \({\text{ABCDEF}}\), as shown in the diagram.

\({\text{AB}} = 30{\text{ cm}},{\text{ BC}} = 24{\text{ cm}},{\text{ CD}} = 25{\text{ cm}}\) and angle \({\rm{A\hat BC}} = 35^\circ {\text{ }}\).

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

a.

Calculate the area of triangle \({\text{ABC}}\).[3]

b.

Assuming that no wood is wasted, show that the volume of wood required to make all \(2600\) blocks is \({\text{13}}\,{\text{400}}\,{\text{000 c}}{{\text{m}}^3}\), correct to three significant figures.[2]

c.

Write \({\text{13}}\,{\text{400}}\,{\text{000}}\) in the form \(a \times {10^k}\) where \(1 \leqslant a < 10\) and \(k \in \mathbb{Z}\).[2]

d.

Show that the total surface area of one block is \({\text{2190 c}}{{\text{m}}^2}\), correct to three significant figures.[3]

e.

The blocks are to be painted. One litre of paint will cover \(22{\text{ }}{{\text{m}}^2}\).

Calculate the number of litres required to paint all \(2600\) blocks.[3]

f.
Answer/Explanation

Markscheme

\({\text{A}}{{\text{C}}^2} = {30^2} + {24^2} – 2 \times 30 \times 24 \times \cos 35^\circ \)     (M1)(A1)

Note: Award (M1) for substituted cosine rule formula,

     (A1) for correct substitutions.

\({\text{AC}} = 17.2{\text{ cm}}\)   \((17.2168…)\)     (A1)(G2)

Notes: Use of radians gives \(52.7002…\) Award (M1)(A1)(A0).

     No marks awarded in this part of the question where candidates assume that angle \({\text{ACB}} = 90^\circ \).[3 marks]

a.

Units are required in part (b).

Area of triangle \({\text{ABC = }}\frac{1}{2} \times 24 \times 30 \times \sin 35^\circ \)     (M1)(A1)

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

     Special Case: Where a candidate has assumed that angle \({\text{ACB}} = 90^\circ \) in part (a), award (M1)(A1) for a correct alternative substituted formula for the area of the triangle \(\left( {ie{\text{ }}\frac{1}{2} \times {\text{base}} \times {\text{height}}} \right)\).

\( = 206{\text{ c}}{{\text{m}}^2}\)   \((206.487 \ldots {\text{c}}{{\text{m}}^2})\)     (A1)(G2)

Notes: Use of radians gives negative answer, \(–154.145…\) Award (M1)(A1)(A0).

     Special Case: Award (A1)(ft) where the candidate has arrived at an area which is correct to the standard rounding rules from their lengths (units required).[3 marks]

b.

\(206.487… \times 25 \times 2600\)     (M1)

Note: Award (M1) for multiplication of their answer to part (b) by \(25\) and \(2600\).

\({\text{13}}\,{\text{421}}\,{\text{688.61}}\)     (A1)

Note: Accept unrounded answer of \({\text{13}}\,{\text{390}}\,{\text{000}}\) for use of \(206\).

\({\text{13}}\,{\text{400}}\,{\text{000}}\)     (AG)

Note: The final (A1) cannot be awarded unless both the unrounded and rounded answers are seen.[2 marks]

c.

\(1.34 \times {10^7}\)     (A2)

Notes: Award (A2) for the correct answer.

     Award (A1)(A0) for \(1.34\) and an incorrect index value.

     Award (A0)(A0) for any other combination (including answers such as \(13.4 \times {10^6}\)).[2 marks]

d.

\(2 \times 206.487 \ldots  + 24 \times 25 + 30 \times 25 + 17.2168 \ldots  \times 25\)     (M1)(M1)

Note: Award (M1) for multiplication of their answer to part (b) by \(2\) for area of two triangular ends, (M1) for three correct rectangle areas using \(24\), \(30\) and their \(17.2\).

\(2193.26…\)     (A1)

Note: Accept \(2192\) for use of 3 sf answers.

\(2190\)     (AG)

Note: The final (A1) cannot be awarded unless both the unrounded and rounded answers are seen.[3 marks]

e.

\(\frac{{2190 \times 2600}}{{22 \times 10\,000}}\)     (M1)(M1)

Notes: Award (M1) for multiplication by \(2600\) and division by \(22\), (M1) for division by \({10\,000}\).

     The use of \(22\) may be implied ie division by \(2200\) would be acceptable.

\(25.9\) litres   \((25.8818…)\)     (A1)(G2)

Note: Accept \(26\).[3 marks]

f.

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

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.
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