Foundation November 2018 Paper 2 Q18
18 There are 240 cows on a farm.
(a) On the farm,
number of bulls : number of cows = 1 : 30
How many bulls are there? [1 mark]
(b) Assume
the 240 cows produce milk for 10 months each year
each cow produces an average of 25 litres of milk per day.
Estimate the total milk production, in litres, of the 240 cows in one year.
You must show your working. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| 8 | B1 |
Additional guidance
| Ignore mention of bulls or cows eg condone 8 cows | B1 |
| Condone an answer of 8 : 240 | B1 |
| 8 : 240 followed by 1 : 30 | B0 |
| 8 : 30 | B0 |
| Do not accept 8 from an incorrect method eg \(240 \div 31 = 7.7\ldots\) and answer 8 | B0 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \([28, 31] \times 10\) or [280, 310] | M1 | appropriate days in 10-month year |
| their \([280, 310] \times 25\) or [7000, 7750] or their \([280, 310] \times 240\) or [67 200, 74 400] | M1dep | litres per year per cow milkings per year for 240 cows |
| their \([7000, 7750] \times 240\) or their \([67\,200, 74\,400] \times 25\) | M1dep | |
| [1 680 000, 1 860 000] with correct working | A1 | accept to 1 or 2 sf with correct working SC2 answer of [2 016 000, 2 232 000] with the only error using 12 months and working shown |
| Alternative method 2 | ||
| \(25 \times 240\) or 6000 | M1 | litres per day for 240 cows may be seen embedded in a product eg \(25 \times 10 \times 240\) |
| their \(6000 \times [28, 31]\) or [168 000, 186 000] or \(25 \times 240\) or 6000 and \([28, 31] \times 10\) or [280, 310] | M1dep | litres per month for 240 cows litres per day for 240 cows and appropriate days in 10-month year |
| their \([168\,000, 186\,000] \times 10\) or \(25 \times 240 \times [28, 31] \times 10\) or their \(6000 \times\) their [280, 310] | M1dep | |
| [1 680 000, 1 860 000] with correct working | A1 | accept to 1 or 2 sf with correct working SC2 answer of [2 016 000, 2 232 000] with the only error using 12 months and working shown |
| Alternative method 3 | ||
| \([28, 31] \times 25\) or [700, 775] | M1 | litres per month per cow |
| their \([700, 775] \times 10\) or [7000, 7750] or their \([700, 775] \times 240\) or [168 000, 186 000] | M1dep | litres per year per cow litres per month for 240 cows |
| their \([7000, 7750] \times 240\) or their \([168\,000, 186\,000] \times 10\) | M1dep | |
| [1 680 000, 1 860 000] with correct working | A1 | accept to 1 or 2 sf with correct working SC2 answer of [2 016 000, 2 232 000] with the only error using 12 months and working shown |
| Alternative method 4 | ||
| \([28, 31] \times 240\) or [6720, 7440] | M1 | milkings per month for 240 cows |
| their \([6720, 7440] \times 10\) or [67 200, 74 400] or their \([6720, 7440] \times 25\) or [168 000, 186 000] | M1dep | milkings per year for 240 cows litres per month for 240 cows |
| their \([67\,200, 74\,400] \times 25\) or their \([168\,000, 186\,000] \times 10\) | M1dep | |
| [1 680 000, 1 860 000] with correct working | A1 | accept to 1 or 2 sf with correct working SC2 answer of [2 016 000, 2 232 000] with the only error using 12 months and working shown |
Additional guidance
| Use the scheme that awards the most marks and ignore choice | |
| A value in the range [280, 310] may come from subtracting two months from a year eg uses 303 (may come from 365 – 31 – 31) | M1 |
| The special case allows 2 marks for those using 12 months or using [336, 372] days | |
| Allow consistent use of approximations to 1 sf throughout (this leads to an answer in the given range) ie \(30 \times 10 \times 30 \times 200 = 1\,800\,000\) | M3A1 |
| Mark inconsistent use of approximations to 1sf as the scheme | |
| Their final answer must be in range and correct for their product but may be given to 1 or 2 sf eg 280 days: \(28 \times 10 \times 25 \times 240 = 1\,680\,000\) 300 days: \(30 \times 10 \times 25 \times 240 = 1\,800\,000\) 310 days: \(31 \times 10 \times 25 \times 240 = 1\,860\,000\) 303 days: \(303 \times 25 \times 240 = 1\,818\,000\) 304 days: \(304 \times 25 \times 240 = 1\,824\,000\) 305 days: \(305 \times 25 \times 240 = 1\,830\,000\) | M3A1 |
| eg 12 months of 28 days: \(28 \times 12 \times 25 \times 240 = 2\,016\,000\) 12 months of 30 days: \(30 \times 12 \times 25 \times 240 = 2\,160\,000\) 12 months of 31 days: \(31 \times 12 \times 25 \times 240 = 2\,232\,000\) 365 days: \(365 \times 25 \times 240 = 2\,190\,000\) 366 days: \(366 \times 25 \times 240 = 2\,196\,000\) | SC2 |