Basic Trigonometry: WJEC GCSE Numeracy Intermediate November 2023 Unit 2 Q8

Numeracy Intermediate November 2023 Unit 2 Q8

WJECIntermediateCurrent spec15 marksBasic Trigonometry Units & Time Volume & Surface Area

(a)

(i) A single tree can absorb 48 pounds of carbon dioxide per year.
Calculate the carbon dioxide absorbed per year by a forest of 440 of these trees.
Give your answer in kilograms. [2]

Carbon dioxide absorbed per year is …………… kg

(ii) A forest of trees absorbs \(2{\cdot}3 \times 10^{11}\) grams of carbon dioxide per year.
Which of the following is \(2{\cdot}3 \times 10^{11}\)?
Circle your answer. [1]

230 000 000 000 23 000 000 000 2 300 000 000 000

0·000 000 000 002 3 0·000 000 000 023

(b)

Remember: 10 000 m² ≈ 2·47 acres

A report states that a fire in a forest has a high risk of spreading when there are more than 60 trees per acre.

There are 615 trees in Grancwm Forest.
The forest covers an area of 40 000 m².

Would a fire in Grancwm Forest have a high risk of spreading?

Yes ☐ No ☐

You must show all your working to support your answer. [4]

(c)

A vertical pine tree stands on horizontal ground.
From a point on the ground 21 metres from its base, the angle of elevation of the top of the pine tree is 39°.

A vertical pine tree on horizontal ground. A line from a point on the ground 21 metres from the base of the tree to the top of the tree makes an angle of 39° with the ground

Diagram not drawn to scale

(i) Show that the pine tree has a vertical height of 17 metres. [3]

(ii) A cylindrical log is cut from this pine tree.
The circumference of the cross-section of the log is 1·75 m.
The length of the log is half the height of the tree.

Calculate the volume of the log.
Give your answer in m³.
You must show all your working. [5]

Volume of the log is …………… m³

Mark scheme (a)

Scheme	Marks	Comments
(i) 440 × 48 ÷ 2.2	M1	May be seen in stages
9600 (kg)	A1	Mark final answer Allow answers in the inclusive range 9588 to 9601 from premature approximation Answer space takes precedence
(ii) 230 000 000 000	B1

Mark scheme (b)

Scheme	Marks	Comments
		Throughout, if 4 marks are awarded, penalise -1 if conclusion ‘Yes’ is not indicated On FT the conclusion may be different to ‘Yes’
(Area) 2.47 × 40000 ÷ 10000 or equivalent	M1
9.88 (acres)	A1	May be implied in further working Allow 9.8 (acres), 9.9 (acres) or 10 (acres)
(Density of trees) 615 ÷ 9.88	m1
62(.2...trees per acre) (>60)	A1	Depends on M1 m1 previously awarded

Alternative

8(b) Alternative method 1

Scheme	Marks	Comments
(Area) 2.47 × 40 000 ÷ 10 000 or equivalent	M1
9.88 (acres)	A1	May be implied in further working Allow 9.8 (acres), 9.9 (acres) or 10 (acres)
(Maximum number of trees) 9.88 × 60	m1
592(.8) (trees) or 593 (trees) (< 615)	A1	Depends on M1 m1 previously awarded Allow suitable rounding, e.g. 590 or 600

Alternative

8(b) Alternative method 2

Scheme	Marks	Comments
(Area) 2.47 × 40 000 ÷ 10 000 or equivalent	M1
9.88 (acres)	A1	May be implied in further working Allow 9.8 (acres), 9.9 (acres) or 10 (acres)
(Minimum area) 615 ÷ 60	M1	Do not allow embedded in further working
10.25 (acres) (> 9.88)	A1	Allow rounded to 10 (acres) provided ‘their area’ (9.88 acres) has not been rounded to 10

Alternative

8(b) Alternative method 3

Scheme	Marks	Comments
(Minimum area) 615 ÷ 60	M1
10.25 (acres)	A1	May be implied in further working Allow 10 (acres)
(Convert to m²) 10 000 × 10.25 ÷ 2.47	m1
41 497(.97 m²) or 41 498(m²) (>40 000)	A1	Depends on M1 m1 previously awarded Accept suitable rounding, e.g. 41 000 or 41 500

Alternative

8(b) Alternative method 4

Scheme	Marks	Comments
(Trees in 2.47 acres) 615 ÷ (40 000 ÷ 10 000) or equivalent	M1
153.75 (trees)	A1	May be implied in further working Allow 153, 153.8 or 154 (trees)
(Density of trees) 153.75 ÷ 2.47	m1
62(.2...trees per acre) (> 60)	A1	Depends on M1 m1 previously awarded

Alternative

8(b) Alternative method 5

Scheme	Marks	Comments
(Forest area per tree) 40 000 ÷ 615	M1	Do not allow embedded in further working
65(.0406.. m²)	A1
(Fire risk, area per tree) 10 000 ÷ (60 × 2.47)	M1
67(.476…m²) (> 65)	A1

Corrected by Westie's Workshop

The board's mark scheme printed “‘their area’ (9.88m²)”. Changed to “(9.88 acres)” because the area was worked out in acres (9.88 acres), not square metres.

Original: see the board's mark scheme PDF (link above).

Mark scheme (c)

Scheme	Marks	Comments
(i) (Height of the tree =) 21 × tan 39	M2	OR alternative full method M1 for tan 39 = \(\dfrac{\text{height of tree}}{21}\)
17.(…. m)	A1	CAO
(ii) diameter = \(\dfrac{1.75}{\pi}\) or (radius =) \(\dfrac{1.75}{2 \times \pi}\)	M2	M1 for any one of the following: 1.75 = π × diameter 1.75 = 2 × π × radius (Note: radius = \(\dfrac{7}{8\pi}\) m, radius ≈ 0.28m)
(Area of cross section =) π × (1.75 ÷ 2π)²	M1	FT for ‘their derived radius’ provided it is from a calculation involving the use of π (Note: area of cross section = \(\dfrac{49}{64\pi}\) m² area of cross section ≈ 0.24 m²)
× 17 ÷ 2	m1	FT provided previous M1awarded
(Volume) answer in the range 2.07 (m³) to 2.15 (m³)	A1	CAO, accept an answer of 2 (m³) from correct working without sight of premature approximation leading to an answer outside the range

Alternative

8(c)(i) Alternative method 1

Scheme	Marks	Comments
Hypotenuse = 21/cos39 (= 27.02…) AND Height = √(27(.02)² – 21²)	M2	M1 for Hypotenuse = 21/cos39 (= 27.02…) AND Height² = 27(.02)² – 21²
16.9(7…m) to 17.(0..m)	A1	CAO

Alternative

8(c)(i) Alternative method 2

Scheme	Marks	Comments
(Angle of elevation) tan^-1\(\dfrac{17}{21}\)	M2	M1 tan (elevation) = \(\dfrac{17}{21}\)
38.9(9...°) or 39(°)	A1	CAO

Alternative

8(c)(i) Alternative method 3

Scheme	Marks	Comments
(Horizontal distance) \(\dfrac{17}{\tan 39}\)	M2	M1 for tan 39 = \(\dfrac{17}{\text{distance}}\)
20.9(98…m) or 21m	A1	CAO