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Date November 2021 Marks available 2 Reference code 21N.2.SL.TZ0.4
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term Determine Question number 4 Adapted from N/A

Question

A large water reservoir is built in the form of part of an upside-down right pyramid with a horizontal square base of length 80 metres. The point C is the centre of the square base and point V is the vertex of the pyramid.

The bottom of the reservoir is a square of length 60 metres that is parallel to the base of the pyramid, such that the depth of the reservoir is 6 metres as shown in the diagram.

The second diagram shows a vertical cross section, MNOPC, of the reservoir.

Every day 80m3 of water from the reservoir is used for irrigation.

Joshua states that, if no other water enters or leaves the reservoir, then when it is full there is enough irrigation water for at least one year.

Find the angle of depression from M to N.

[2]
a.

Find CV.

[2]
b.i.

Hence or otherwise, show that the volume of the reservoir is 29600m3.

[3]
b.ii.

By finding an appropriate value, determine whether Joshua is correct.

[2]
c.

To avoid water leaking into the ground, the five interior sides of the reservoir have been painted with a watertight material.

Find the area that was painted.

[5]
d.

Markscheme

tanθ=610           (M1)

θ= 31.0°  30.9637°  OR  0.540  0.540419           A1

 

[2 marks]

a.

CV= 40tanθ  OR  CV= 4×6           (M1)


Note: Award (M1) for an attempt at trigonometry or similar triangles (e.g. ratios).


CV= 24m           A1

 

[2 marks]

b.i.

V=13802×24-13602×18           M1A1A1


Note:
Award M1 for finding the difference between the volumes of two pyramids, A1 for each correct volume expression. The final A1 is contingent on correct working leading to the given answer.
If the correct final answer is not seen, award at most M1A1A0. Award M0A0A0 for any height derived from V=29600, including 18.875 or 13.875.


V= 29600m3           AG

 

[3 marks]

b.ii.

METHOD 1

2960080= 370 (days)             A1

370>366  Joshua is correct             A1


Note: Award A0A0 for unsupported answer of “Joshua is correct”. Accept 1.01>1 for the first A1 mark.

 

METHOD 2

80×366=29280m3  OR  80×365=29200m3             A1

29280<29600  Joshua is correct             A1


Note: The second A1 can be awarded for an answer consistent with their result.

 

[2 marks]

c.

height of trapezium is 102+62 =11.6619            (M1)

area of trapezium is 80+602×102+62 =816.333            (M1)(A1)

SA= 4×80+602×102+62+602            (M1)


Note: Award M1 for adding 4 times their (MNOP) trapezium area to the area of the (60×60) base.


SA= 6870m2  6865.33m2            A1


Note: No marks are awarded if the correct shape is not identified.

 

[5 marks]

d.

Examiners report

Finding the angle of depression from M to N proved problematic for many. Some chose the angle inclined to the vertical or attempted a less efficient method such as solving triangle NMP. Most candidates attempted to find CV through trigonometry rather than similar triangles. Consistent with past examination sessions, it is clear that candidate do not always understand the demands of a 'show that' question. Working backwards to verify the given volume of 29600m3 accrued no marks. Various approaches were used, with most candidates earning at least one mark for one correct volume seen. Part (c) was accessible to all. It was surprising to see some candidates identify 360 as the number of days in a calendar year. Part (d) was a true high-order discriminator, proving to be challenging for even very capable candidates. Some candidates found the surface area of a shape/object that is not part of the reservoir, while the majority incorrectly identified the height of trapezoid MPON as 6.

a.

Finding the angle of depression from M to N proved problematic for many. Some chose the angle inclined to the vertical or attempted a less efficient method such as solving triangle NMP. Most candidates attempted to find CV through trigonometry rather than similar triangles. Consistent with past examination sessions, it is clear that candidate do not always understand the demands of a 'show that' question. Working backwards to verify the given volume of 29600m3 accrued no marks. Various approaches were used, with most candidates earning at least one mark for one correct volume seen. Part (c) was accessible to all. It was surprising to see some candidates identify 360 as the number of days in a calendar year. Part (d) was a true high-order discriminator, proving to be challenging for even very capable candidates. Some candidates found the surface area of a shape/object that is not part of the reservoir, while the majority incorrectly identified the height of trapezoid MPON as 6.

b.i.

Finding the angle of depression from M to N proved problematic for many. Some chose the angle inclined to the vertical or attempted a less efficient method such as solving triangle NMP. Most candidates attempted to find CV through trigonometry rather than similar triangles. Consistent with past examination sessions, it is clear that candidate do not always understand the demands of a 'show that' question. Working backwards to verify the given volume of 29600m3 accrued no marks. Various approaches were used, with most candidates earning at least one mark for one correct volume seen. Part (c) was accessible to all. It was surprising to see some candidates identify 360 as the number of days in a calendar year. Part (d) was a true high-order discriminator, proving to be challenging for even very capable candidates. Some candidates found the surface area of a shape/object that is not part of the reservoir, while the majority incorrectly identified the height of trapezoid MPON as 6.

b.ii.

Finding the angle of depression from M to N proved problematic for many. Some chose the angle inclined to the vertical or attempted a less efficient method such as solving triangle NMP. Most candidates attempted to find CV through trigonometry rather than similar triangles. Consistent with past examination sessions, it is clear that candidate do not always understand the demands of a 'show that' question. Working backwards to verify the given volume of 29600m3 accrued no marks. Various approaches were used, with most candidates earning at least one mark for one correct volume seen. Part (c) was accessible to all. It was surprising to see some candidates identify 360 as the number of days in a calendar year. Part (d) was a true high-order discriminator, proving to be challenging for even very capable candidates. Some candidates found the surface area of a shape/object that is not part of the reservoir, while the majority incorrectly identified the height of trapezoid MPON as 6.

c.

Finding the angle of depression from M to N proved problematic for many. Some chose the angle inclined to the vertical or attempted a less efficient method such as solving triangle NMP. Most candidates attempted to find CV through trigonometry rather than similar triangles. Consistent with past examination sessions, it is clear that candidate do not always understand the demands of a 'show that' question. Working backwards to verify the given volume of 29600m3 accrued no marks. Various approaches were used, with most candidates earning at least one mark for one correct volume seen. Part (c) was accessible to all. It was surprising to see some candidates identify 360 as the number of days in a calendar year. Part (d) was a true high-order discriminator, proving to be challenging for even very capable candidates. Some candidates found the surface area of a shape/object that is not part of the reservoir, while the majority incorrectly identified the height of trapezoid MPON as 6.

d.

Syllabus sections

Topic 3—Geometry and trigonometry » SL 3.1—3d space, volume, angles, midpoints
Show 90 related questions
Topic 3—Geometry and trigonometry » SL 3.2—2d and 3d trig
Topic 3—Geometry and trigonometry » SL 3.3—Angles of elevation and depression
Topic 3—Geometry and trigonometry

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