Numeracy Intermediate June 2023 Unit 1 Q10
10.
(a) A jar contains 300 g of chocolate spread.
In this spread:
- 58% of the mass is pure sugar,
- \(\dfrac{1}{8}\) of the mass is cocoa,
- the mass of the milk powder is \(\dfrac{4}{5}\) of the mass of cocoa,
- the remainder of the 300 g is palm oil.
Calculate the percentage of palm oil in the chocolate spread.
You must show all your working. [7]
(b) A different jar contains 840 g of chocolate spread.
The label on the jar says,
‘Offer: includes 20% extra chocolate spread for free.’
How many grams of chocolate spread did a jar contain before the offer started? [3]
| Scheme | Marks | Comments |
|---|---|---|
| (Sugar 0.58 × 300 =) 174 (g) | B1 | |
| (Cocoa 300 ÷ 8 =) 37.5 (g) | B1 | |
| (Milk powder) 4 × 37.5 ÷ 5 | M1 | FT ‘their derived mass of cocoa’ |
| (=) 30 (g) | A1 | |
| (Palm oil 300 – 174 – 37.5 – 30 =) 58.5 (g) OR (Ingredients other than palm oil 174 + 37.5 + 30 =) 241.5 (g) | B1 | May be implied in further working FT correctly evaluated sum of ‘their sugar, cocoa and milk’ provided at least 2 marks previously awarded and this sum is < 300 |
| (% of Palm oil) (100 ×) 58.5 300 or 58.5 ÷ 3 or 0.195 or (100 ×) 1 – 241.5 300 or 1 – 0.805 or equivalent | M1 | FT ‘their 58.5’ or ‘their 241.5’ as appropriate |
| 19.5 (%) | A1 | CAO, must be given as a percentage |
10(a) Alternative method 1:
| Scheme | Marks | Comments |
|---|---|---|
| (Cocoa) ⅛ × 100 or 1 ÷ 8 | M1 | |
| 12.5 % or 0.125 | A1 | |
| (Milk powder) ⅘ × ⅛ (× 100) | M1 | FT ‘their 12.5% or 0.125’ or ‘their % or decimal mass of cocoa’ |
| 10 % or 0.1 | A1 | |
| (Other than palm oil) (58 + 12.5 + 10 =) 80.5 % or (0.58 + 0.125 + 0.1 =) 0.805 | B1 | May be implied in further working FT 58 + ‘their 12.5 + 10’ or 0.58 + ‘their 0.125 + 0.1’ correctly evaluated provided at least 2 marks previously awarded and this sum is < 100 or < 1 respectively |
| (Palm oil) 100 – (58 + 12.5 + 10) or 1 – (0.58 + 0.125 + 0.1) or 0.195 | M1 | FT 58 + ‘their sum of %s cocoa and milk’ or equivalent working with decimals |
| 19.5 (%) | A1 | CAO, must be given as a percentage |
The alternative method is printed in italics in the original. In the original, the first fraction in the milk powder row is printed as an empty box (a missing character); the text layer of the PDF gives it as ⅘.
10(a) Alternative method 2:
| Scheme | Marks | Comments |
|---|---|---|
| (Proportion other than palm oil) \(\frac{58}{100}+\frac{1}{8}+\frac{4}{5}\times\frac{1}{8}\) or \(\frac{58}{100}+\frac{1}{8}+\frac{1}{10}\) | M3 | |
| 805 1000 or 161 200 | A1 | May be implied in further working |
| (Palm oil) 1 - \({}^{161}\!/_{200}\) or \({}^{39}\!/_{200}\) or \({}^{19.5}\!/_{100}\) | M2 | FT from M3 |
| 19.5 (%) | A1 | CAO, must be given as a percentage |
The alternative method is printed in italics in the original.
Corrected by Westie's Workshop
- The board's mark scheme printed “(Milk powder) □ × ⅛ (× 100)”. Changed to “(Milk powder) ⅘ × ⅛ (× 100)” because the first fraction is printed as an empty box (a missing character). The milk powder is 4/5 of the cocoa, as the main method (4 × 37.5 ÷ 5) shows.
Original: see the board's mark scheme PDF (link above).
| Scheme | Marks | Comments |
|---|---|---|
| Realising that 840g is 120% | B1 | |
| For appropriate use of 120% being 840g, e.g. 840 ÷ 1.2 or 8400 ÷ 12 or 100 × 840/120 | M1 | Also implies previous B1 |
| 700 (g) | A1 | Award all 3 marks for an answer of 700(g) provided not from incorrect working |