Higher January 2020 Paper 2 Q12
12 Astrid wants to buy some oil.
She can buy the oil from either Dane Oil or Arctic Oil.
Here is information about the price that each company will charge Astrid.
| Dane Oil | Arctic Oil |
|---|---|
| \((4.2 \times 10^5)\) litres for 2 500 000 Krone | \((8.6 \times 10^5)\) litres for 770 000 Dollars |
Astrid wants to get the better value for money for the oil.
1 Dollar = 6.57 Krone
From which company should she buy her oil, Dane Oil or Arctic Oil?
You must show your working.
(4)
Litres per amount of money and then conversion
| Scheme | Marks |
|---|---|
| \(\dfrac{8.6 \times 10^5}{770\,000}\) (= 1.1168) l/$ | M1 |
| \(\dfrac{4.2 \times 10^5}{2\,500\,000}\) (= 0.168) l/k | M1 |
| A: l/$ to l/k ‘1.1168’ ÷ 6.57 (= 0.1699..) or D: l/k to l/$ ‘0.168’ × 6.57 (= 1.103..) | M1 |
| Working required Answer: Arctic Oil and relevant figures | A1 |
| (4) | |
| (4 marks) |
Notes
M1: Number of litres per $ for A
M1: Number of litres per Krone for D
M1: l/$ to l/k for A or l/k to l/$ for D
A1: for Arctic Oil with 1.1168… and 1.10376… or 0.168 and 0.1699..
(corrected from the printed mark scheme: in this first method the notes for the first two marks print “for D” and “for A” the wrong way round; 770 000 Dollars is the Arctic Oil (A) price and 2 500 000 Krone is the Dane Oil (D) price)
Conversion then litres per amount of money
| Scheme | Marks |
|---|---|
| \(\dfrac{2\,500\,000}{6.57}\) (= 380517.5..) or 770 000 × 6.57 (= 5 058 900) | M1 |
| \(\dfrac{4.2 \times 10^5}{2\,500\,000}\) (= 0.168) or \(\dfrac{4.2 \times 10^5}{\text{‘}380517.5\text{’}}\) (= 1.103..) | M1 |
| \(\dfrac{8.6 \times 10^5}{770\,000}\) (= 1.1168) or \(\dfrac{8.6 \times 10^5}{\text{‘}5058900\text{’}}\) (= 0.1699..) | M1 |
| Working required Answer: Arctic Oil and relevant figures | A1 |
Notes
M1: Changing Krone to $ or $ to Krone
M1: Litres per Krone or litres per $ for D
M1: Litres per Krone or litres per $ for A
A1: for Arctic Oil with 1.1168… and 1.10376… or 0.168 and 0.1699..
Cost per litre then conversion
| Scheme | Marks |
|---|---|
| \(\dfrac{2\,500\,000}{4.2 \times 10^5}\) (= 5.952..) | M1 |
| \(\dfrac{770\,000}{8.6 \times 10^5}\) (0.895..) | M1 |
| ‘5.952’ ÷ 6.57 (= 0.9059..) or ‘0.895’ × 6.57 (= 5.882..) | M1 |
| Working required Answer: Arctic Oil and relevant figures | A1 |
Notes
M1: Price per litre in Krone for D
M1: Price per litre in $ for A
M1: Conversion of Krone to $ or $ to Krone
A1: For Arctic Oil with 5.952 and 5.882 or 0.895 and 0.9059
Conversion then cost per litre
| Scheme | Marks |
|---|---|
| \(\dfrac{2\,500\,000}{6.57}\) (= 380517.5..) or 770 000 × 6.57 (= 5 058 900) | M1 |
| \(\dfrac{2\,500\,000}{4.2 \times 10^5}\) (= 5.952) or \(\dfrac{\text{‘}380517.5..\text{’}}{4.2 \times 10^5}\) (= 0.9059..) | M1 |
| \(\dfrac{770\,000}{8.6 \times 10^5}\) (= 0.895) or \(\dfrac{\text{‘}5058900\text{’}}{8.6 \times 10^5}\) (= 5.882..) | M1 |
| Working required Answer: Arctic Oil and relevant figures | A1 |
Notes
M1: Changing Krone to $ or $ to Krone
M1: Cost per litre in Krone or cost per litre in $ for D
M1: Cost per litre in $ or cost per litre in Krone for A
A1: For Arctic Oil with 5.952 and 5.882 or 0.895 and 0.9059
Comparing equal amounts
| Scheme | Marks |
|---|---|
\(\dfrac{8.6 \times 10^5}{4.2 \times 10^5} \left(= \dfrac{43}{21} = 2.047..\right)\) or \(\dfrac{4.2 \times 10^5}{8.6 \times 10^5} \left(= \dfrac{21}{43} = 0.488..\right)\) | M1 |
| ‘2.047..’ × 2 500 000 K (= 5119047.619..) K or ‘0.488..’ × 770 000 $ (= 376046.511..) $ | M1 |
| ‘5119047.619’ ÷ 6.57 = 779154.88… $ or 770 000 × 6.57 = 5058900 K or ‘376046.511..’ × 6.57 = 2470625.58.. K or 2 500 000 ÷ 6.57 = 380517.. $ | M1 |
| Working required Answer: Arctic Oil and relevant figures | A1 |
Notes
M1: Multiplier for same amount of D as A or same amount of A as D
M1: Cost of equal amount of D as A or A as D
M1: Converts so can compare costs – either K to $ or original A to K or $ to K or original D to $
A1: Arctic Oil and 779154.. or with 2470625.. (figures may be rounded)
Or
Arctic Oil with 5119047… and 5058900 or with 376046.. and 380517
(corrected from the printed mark scheme: the second column prints ‘2.047..’ × 770 000 $; the multiplier there is ‘0.488..’, which gives 376046.511..)
Students may compare other equal amounts – please use the scheme that best fits their method and award marks appropriately.