Numeracy Foundation June 2024 Unit 2 Q9
9. During the 30 days in the month of September, Andras used 138 m3 of gas.
The cost of this gas was 12p per kWh.
| The formula used to calculate the number of kWh for the gas used is as follows: Number of kWh = number of m3 of gas × 39·5 × 1·02264 ÷ 3·6 |
The standing charge was 32p per day.
VAT at 5% was payable on the sum of the cost of the gas used and the standing charge.
Calculate Andras’s gas bill for September.
You must show all your working. [7]
| Scheme | Mark | Comments |
|---|---|---|
| Penalise incorrect units -1 only on first occurrence, by withholding A or B mark, not M marks | ||
| (Number of kWh =) 138 × 39·5 × 1.02264 ÷ 3.6 (Cost of gas = Number of kWh) × 0(.)12 | M1 m1 | (= 1548.4474 kWh) |
| (£)185.76 to (£)185.82 or 18576(p) to 18582(p) | A1 | (1548.4474 × 0.12 = £185.813688) |
| (Standing charge 30 × (0.)32 =) (£)9.6(0) or 960(p) | B1 | CAO |
| (Total of gas and standing charge) *(£)195.36 to (£)195.42 or 19536(p) to 19542(p) | B1 | FT ‘their derived cost of gas’ + ‘their 30 × (0.)32’ correctly evaluated, provided ‘their derived cost of gas’ ≠ ‘their 138 × 39·5 × 1.02264 ÷ 3.6’ or 1548.4474 kWh May be implied in later working |
| (Total including VAT =) 1.05 × 195(.)36 to 1.05 × 195(.)42 | M1 | FT ‘their derived total cost of gas + ‘their standing charge’ |
| *(£)205.12 to (£)205.19(1) or 20512(p) to 20519(.1p) | A1 | ISW further rounding, e.g. (£)205, (£)205.20 If final B0 B0 M0 A0, award SC1 for correctly evaluated final answer of 1.05 × ‘derived cost of gas’ having omitted the standing charge, provided ‘their derived cost of gas’ ≠ ‘their 138 × 39·5 × 1.02264 ÷ 3.6’ or 1548.4474 kWh *Allow answers in these ranges that may include unseen rounding or truncation from a previously written value |
Alternative
Alternative method: Gas per day (printed in italics in the mark scheme)
| Scheme | Mark | Comments |
|---|---|---|
| Penalise incorrect units -1 only on first occurrence, by withholding A or B mark, not M marks | ||
| (Number of kWh =) 138 × 39·5 × 1.02264 ÷ 3.6 (Number of kWh per day) ÷ 30 (Cost of gas per day) × 0(.)12 | M1 m1 m1 | (= 1548.4474 kWh) (= 51.6149133…kWh) |
| (Cost of gas per day =) (£)6.19(…) or 619(...p) | A1 | CAO |
| (Total of gas and standing charge) 6.51(…) or 651.(…p) | B1 | FT ‘their derived cost of gas per day’ + (0.)32 correctly evaluated May be implied in later working |
| (Total including VAT =) 1.05 × 6.51(…) × 30 | M1 | (= 6.83(94…) × 30 or 6.84 × 30) FT ‘their derived cost of gas per day + (0.)32, provided ‘their derived cost of gas’ ≠ ‘their 138 × 39·5 × 1.02264 ÷ 3.6’ or 1548.4474 kWh |
| *(£)205.12 to (£)205.19(1) or 20512(p) to 20519(.1p) | A1 | ISW further rounding, e.g. (£)205, (£)205.20 If final B0 M0 A0, award SC1 for correctly evaluated final answer of 1.05 × ‘derived cost of gas per day’ × 30 having omitted the standing charge, provided ‘their cost of gas’ ≠ ‘their 138 × 39·5 × 1.02264 ÷ 3.6’ or 1548.4474 kWh *Allow answers in these ranges that may include unseen rounding or truncation from a previously written value |