Numeracy Higher June 2025 Unit 1 Q10
10.
(a) In 2020, red squirrels were introduced into a woodland in Wales.
The following formula is used to calculate how many red squirrels there are in the woodland,
\[R = 200 \times 1{\cdot}1^{t}\]where:
- \(R\) is the number of red squirrels in the woodland
- \(t\) is the number of years after the red squirrels were introduced.
(i) How many red squirrels were introduced in 2020? [1]
(ii) Calculate the percentage increase in the number of red squirrels 2 years after they were introduced. [3]
(b) Workers are making a new area of woodland.
4 workers planted 15 000 trees in 3 days.
How many workers will be needed to plant another 17 500 trees in 2 days?
You can assume that all workers plant trees at the same rate. [3]
4 workers planted 15 000 trees in 3 days.
How many workers will be needed to plant another 17 500 trees in 2 days?
You can assume that all workers plant trees at the same rate. [3]
| Scheme | Marks | Comments |
|---|---|---|
| (i) 200 | B1 | |
| (ii) (Percentage increase =) \(\dfrac{200{\times}1.1{\times}1.1 - 200}{200}\) (× 100) OR \(\dfrac{200{\times}1.1{\times}1.1}{200}\) (× 100) − 1 (× 100) | M2 | Do not accept sight of 200×1.12 without evidence of understanding of meaning of 1.12 Accept a FT of ‘their 200’ from (a)(i) i.e. \(\dfrac{200{\times}1.1{\times}1.1 - \text{(a)(i)}}{\text{(a)(i)}}\) (× 100) M1 for appropriate sight of any one of the following:
|
| = 21 (%) | A1 |
Alternative method (a)(ii)
| Scheme | Marks | Comments |
|---|---|---|
| (1.1×1.1 =) 1.21 | B1 | Must be from a clear intent to only use 1.12 to arrive at 21% |
| (Percentage increase = ) 1.21(× 100) – 1(× 100) | M1 | FT ‘their 1.1×1.1’ |
| = 21 (%) | A1 |
| Scheme | Marks | Comments | ||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 4 × \(\dfrac{175(00)}{150(00)}\) or × \(\dfrac{7}{6}\) or × 1.166... or equivalent | M1 | A table method altering all 3 values in the same ratio at the same time is M0 M marks may be seen in either order e.g.
| ||||||||||||
| × \(\dfrac{3}{2}\) or × 1.5 or equivalent | M1 | FT from M0 previously awarded Must be from use of 4 e.g. if this calculation is performed first
| ||||||||||||
| = 7 (workers) | A1 | CAO |
Alternative method 1
| Scheme | Marks | Comments |
|---|---|---|
| (Number of trees planted per worker per day =) 15000 ÷ 3 ÷ 4 (= 1250) OR (Number of trees planted per worker in 2 days =) (15000 ÷ 3 ÷ 4) × 2 (=2500) | M1 | |
| (Number of workers needed =) 17500 ÷ ((15000 ÷ 3 ÷ 4) × 2) | M1 | |
| = 7 (workers) | A1 | CAO |
Alternative method 2
| Scheme | Marks | Comments | ||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| M1 | |||||||||||||
| 4 × \(\dfrac{17500}{10000}\) OR 4 × \(\dfrac{3.5}{2}\) or 4 × 1.75 or equivalent | M1 | FT from 2 (days), \(n\) trees, 4 workers for 4 × \(\dfrac{17500}{n}\) FT from \(n\) (days), 17500 trees, 4 workers for 4 × \(\dfrac{n}{2}\) | ||||||||||||
| = 7 (workers) | A1 | CAO |