Numeracy Higher November 2023 Unit 2 Q4
4.
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
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
| Remember: 10 000 m\(^2\) ≈ 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 m\(^2\).
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]
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°.
Diagram not drawn to scale
Show that the pine tree has a vertical height of 17 metres. [3]
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 m\(^3\).
You must show all your working. [5]
Volume of the log is …………… m\(^3\)
(i)
| Scheme | Marks | Comments |
|---|---|---|
| 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)
| Scheme | Marks | Comments |
|---|---|---|
| 230 000 000 000 | B1 |
| 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 9.88 (acres) | M1 A1 | May be implied in further working Allow 9.8 (acres), 9.9 (acres) or 10 (acres) |
| (Density of trees) 615 ÷ 9.88 62(.2...trees per acre) (>60) | m1 A1 | Depends on M1 m1 previously awarded |
Alternative
4(b) Alternative method 1
| Scheme | Marks | Comments |
|---|---|---|
| (Area) 2.47 × 40 000 ÷ 10 000 or equivalent 9.88 (acres) | M1 A1 | May be implied in further working Allow 9.8 (acres), 9.9 (acres) or 10 (acres) |
| (Maximum number of trees) 9.88 × 60 592(.8) (trees) or 593 (trees) (< 615) | m1 A1 | Depends on M1 m1 previously awarded Allow suitable rounding, e.g. 590 or 600 |
Alternative
4(b) Alternative method 2
| Scheme | Marks | Comments |
|---|---|---|
| (Area) 2.47 × 40 000 ÷ 10 000 or equivalent 9.88 (acres) | M1 A1 | May be implied in further working Allow 9.8 (acres), 9.9 (acres) or 10 (acres) |
| (Minimum area) 615 ÷ 60 10.25 (acres) (> 9.88) | M1 A1 | Do not allow embedded in further working Allow rounded to 10 (acres) provided ‘their area’ (9.88 acres) has not been rounded to 10 |
Alternative
4(b) Alternative method 3
| Scheme | Marks | Comments |
|---|---|---|
| (Minimum area) 615 ÷ 60 10.25 (acres) | M1 A1 | May be implied in further working Allow 10 (acres) |
| (Convert to m\(^2\)) 10 000 × 10.25 ÷ 2.47 41 497(.97 m\(^2\)) or 41 498(m\(^2\)) (>40 000) | m1 A1 | Depends on M1 m1 previously awarded Accept suitable rounding, e.g. 41 000 or 41 500 |
Alternative
4(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 62(.2...trees per acre) (> 60) | m1 A1 | Depends on M1 m1 previously awarded |
Alternative
4(b) Alternative method 5
| Scheme | Marks | Comments |
|---|---|---|
| (Forest area per tree) 40 000 ÷ 615 65(.0406.. m\(^2\)) | M1 A1 | Do not allow embedded in further working |
| (Fire risk, area per tree) 10 000 ÷ (60 × 2.47) 67(.476…m\(^2\)) (> 65) | M1 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).
(i)
| Scheme | Marks | Comments |
|---|---|---|
| (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 |
Alternative
4(c)(i) Alternative method 1
| Scheme | Marks | Comments |
|---|---|---|
| Hypotenuse = 21/cos39 (= 27.02…) AND Height = \(\sqrt{(27(.02)^2 - 21^2)}\) | M2 | M1 for Hypotenuse = 21/cos39 (= 27.02…) AND Height\(^2\) = 27(.02)\(^2\) – 21\(^2\) |
| 16.9(7…) (m) to 17.(0..m) | A1 | CAO |
Alternative
4(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
4(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 |
(ii)
| Scheme | Marks | Comments |
|---|---|---|
| diameter = \(\dfrac{1.75}{\pi}\) or (radius =) \(\dfrac{1.75}{2 \times \pi}\) | M2 | M1 for any one of the following:
|
| (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 = \(\dfrac{49}{64\pi}\) m\(^2\) area of cross section ≈ 0.24 m\(^2\)) |
| × 17 ÷ 2 | m1 | FT provided previous M1 awarded |
| (Volume) answer in the range 2.07 (m\(^3\)) to 2.15 (m\(^3\)) | A1 | CAO, accept an answer of 2 (m\(^3\)) from correct working without sight of premature approximation leading to an answer outside the range |