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Date May 2007 Marks available 2 Reference code 07M.2.sl.TZ0.2
Level SL only Paper 2 Time zone TZ0
Command term Calculate Question number 2 Adapted from N/A

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.

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.

Examiners report

(i) Many candidates incurred the new one-off unit penalty here. Too many ignored the call for two decimal places and some extrapolated that instruction to later parts (which was clearly not intended). There was the predictable confusion of using radius instead of diameter. Another common error was to divide the cylinder volume by that of the ball, to decide how many would fit. Some follow-through was allowed later from this error, however, this led to zero or negligible air volume, which was clearly ridiculous.

Choice and use of the formulae for volumes was often competent but the conversion to cubic metres was very badly done. Almost no correct answers were seen at all.

i.a.

(i) Many candidates incurred the new one-off unit penalty here. Too many ignored the call for two decimal places and some extrapolated that instruction to later parts (which was clearly not intended). There was the predictable confusion of using radius instead of diameter. Another common error was to divide the cylinder volume by that of the ball, to decide how many would fit. Some follow-through was allowed later from this error, however, this led to zero or negligible air volume, which was clearly ridiculous.

Choice and use of the formulae for volumes was often competent but the conversion to cubic metres was very badly done. Almost no correct answers were seen at all.

i.b.

(i) Many candidates incurred the new one-off unit penalty here. Too many ignored the call for two decimal places and some extrapolated that instruction to later parts (which was clearly not intended). There was the predictable confusion of using radius instead of diameter. Another common error was to divide the cylinder volume by that of the ball, to decide how many would fit. Some follow-through was allowed later from this error, however, this led to zero or negligible air volume, which was clearly ridiculous.

Choice and use of the formulae for volumes was often competent but the conversion to cubic metres was very badly done. Almost no correct answers were seen at all.

i.c.

(ii) Candidates were often sloppy in reading the information. In particular, despite the statement BL = 120 clearly written, many took GL as 120. Triangle TBL was often taken as right-angled. Angle BTL presented few problems, though sometimes the method was very long-winded. Candidates often managed part (a) then went awry in later parts. Many unit penalties were applied, if not already used in questions 1 or 2.

ii.a.

(ii) Candidates were often sloppy in reading the information. In particular, despite the statement BL = 120 clearly written, many took GL as 120. Triangle TBL was often taken as right-angled. Angle BTL presented few problems, though sometimes the method was very long-winded. Candidates often managed part (a) then went awry in later parts. Many unit penalties were applied, if not already used in questions 1 or 2.

ii.b.

(ii) Candidates were often sloppy in reading the information. In particular, despite the statement BL = 120 clearly written, many took GL as 120. Triangle TBL was often taken as right-angled. Angle BTL presented few problems, though sometimes the method was very long-winded. Candidates often managed part (a) then went awry in later parts. Many unit penalties were applied, if not already used in questions 1 or 2.

ii.c.

Syllabus sections

Topic 5 - Geometry and trigonometry » 5.2 » Use of sine, cosine and tangent ratios to find the sides and angles of right-angled triangles.
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