Foundation November 2021 Paper 2 Q14
14. Omar used the following recipe to make a drink to sell at a school fete.
For each glass of drink:
|
He squeezed:
- 68 ml of juice from each orange,
- 232 ml of juice from each grapefruit.
Omar sold 72 glasses of his drink.
The bottles of lemonade Omar used each contained 2 litres.
(a)
How many bottles of lemonade did Omar need to open?
You must show all your working. [5]
……………… bottles
(b)
All the lemonade was donated, so cost Omar nothing.
He only paid for the fruit that he used.
The oranges cost £1.08 for a bag of 6.
The grapefruit cost 56p each.
Omar made 60% profit on each glass he sold.
What was the selling price of each glass of Omar’s drink? [4]
| Scheme | Marks | Comments |
|---|---|---|
| 68 + 232 ÷ 8 (= 68 + 29 = 97) | M1 | Allow equivalent working in litres |
| (300 – 97) × 72 (= 14 616) | M2 | FT ‘their 97’ provided it is not 68 or 232 M1 for 300 – 97 (= 203) |
| 14 616 ÷ 2000 (= 7.308) or 7 bottles = 14 000 | M1 | FT ‘their 14 616’; provided at least M2 previously awarded; implied by sight of 7.3(...) |
| 8 | A1 | CAO with no incorrect working seen An answer of 8 does not imply full marks but allow full marks if the first 3 marks have been awarded and an answer of 8 stated |
Alternative
| Scheme | Marks | Comments |
|---|---|---|
| 68 × 72 + 232 ÷ 8 × 72 (= 4896 + 2088 = 6984) | M2 | M1 for 232 ÷ 8 × 72 |
| 72 × 300 – (68 × 72 + 232 ÷ 8 × 72) (= 14 616) | M1 | FT ‘their 68 × 72 + 232 ÷ 8 × 72’ |
| 14 616 ÷ 2000 (= 7.308) or 7 bottles = 14 000 | M1 | FT ‘their 14 616’; provided at least M2 previously awarded; implied by sight of 7.3(...) |
| 8 | A1 | CAO with no incorrect working seen An answer of 8 does not imply full marks but allow full marks if the first 3 marks have been awarded and an answer of 8 stated |
| Scheme | Marks | Comments |
|---|---|---|
| (cost of fruit for one glass = ) 108 ÷ 6 + 56 ÷ 8 (= 25p or £0.25) | M1 | may be in pounds or pence but units must be consistent |
| \(25 + \dfrac{60}{100} \times 25\) oe | M2 | For M2 or M1, FT ‘their derived 25’ provided obtained using 108(p) and 56 (p) or equivalent; M1 for \(\dfrac{60}{100} \times\) ‘their 25’ oe |
| 40(p) or (£)0.40 | A1 | CAO If units are given they must be correct. Allow £0.40p |
| (9) |
Alternative
| Scheme | Marks | Comments |
|---|---|---|
| (cost of fruit for 72 glasses = ) 72(1.08 ÷ 6 + 0.56 ÷ 8) (= £18 or 1800p) or 72 ÷ 6 × 1.08 + 72 ÷ 8 × 0.56 or 12 × 1.08 + 9 × 0.56 | M1 | may be in pounds or pence but units must be consistent; |
| \(\left(18 + \dfrac{60}{100} \times 18\right) \div 72\) (= 28.80 ÷ 72) oe | M2 | For M2 or M1, FT ‘their derived 18’ provided obtained using 108(p) and 56 (p) oe; M1 for \(\dfrac{60}{100} \times\) ‘their 18’ oe |
| 40(p) or (£)0.40 | A1 | CAO If units are given they must be correct. Allow £0.40p |