Foundation January 2022 Paper 2 Q21
21 A bag contains only pink sweets, white sweets, green sweets and red sweets.
The table gives each of the probabilities that, when a sweet is taken at random from the bag, the sweet will be green or the sweet will be red.
| Sweet | pink | white | green | red |
|---|---|---|---|---|
| Probability | 0.2 | 0.35 |
The ratio
number of pink sweets : number of white sweets = 2 : 1
There are 28 red sweets in the bag.
Work out the number of white sweets in the bag.
(5)
| Scheme | Marks |
|---|---|
| 28 ÷ 0.35 (= 80) oe eg (28 ÷ 7) × 20 (= 80) | M1 |
| 1 – (0.2 + 0.35) (= 0.45) oe or (0.2 + 0.35) × “80” (= 44) or 28 + “16” (= 44) | M1 |
| “0.45” ÷ 3 (= 0.15) oe or “0.45” × “80” (= 36) or “80” – “44” (= 36) | M1 |
“80” × “0.15” or “80” × “0.3” (= 24) or “36” ÷ 3 or “36” ÷ \(\dfrac{3}{2}\) (= 24) | M1 |
| 12 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: indep for calculating total number of sweets
M1: or for a correct equation for missing values eg \(x + 2x + 0.2 + 0.35 = 1\) oe
(can be implied by 2 probabilities that total 0.45 in table if not contradicted in working space)
M1: (or 0.15 or 0.3 seen in table – either order)
M1: A correct calculation for the number of white sweets or the number of pink sweets
Alternative
| Scheme | Marks |
|---|---|
| 1 – (0.2 + 0.35) (= 0.45) or 100(%) – 20(%) – 35(%) = 45(%) | M1 |
| “0.45” ÷ 3 (= 0.15) 45(%) ÷ 3 (= 15(%)) | M1 |
| \(\dfrac{n}{28} = \dfrac{0.15}{0.35}\) or \(\dfrac{n}{0.15} = \dfrac{28}{0.35}\) oe or \(\dfrac{n}{28} = \dfrac{0.3}{0.35}\) or \(\dfrac{n}{0.3} = \dfrac{28}{0.35}\) or 35% = 28 so 5% = 4 | M1 |
| (\(n =\)) \(28 \times \dfrac{0.15}{0.35}\) or (\(n =\)) \(0.15 \times \dfrac{28}{0.35}\) or 15% = 3 × 4 or \(28 \times \dfrac{0.3}{0.35}\) or \(0.3 \times \dfrac{28}{0.35}\) or 30% = 6 × 4 (= 24) | M1 |
| 12 | A1 |
M1: or for a correct equation for missing values eg \(x + 2x + 0.2 + 0.35 = 1\) oe
M1: (or 0.15 or 0.3 seen in table – either order)
M1: for using proportion with an expression for \(n\) white sweets or
finding 5% oe to enable calculation to 15%
M1: a calculation using proportion that would lead to finding their \(n\) or \(2n\)