Foundation June 2025 Paper 2R Q15
15 Joseph has some counters in a bag.
7 of the counters are red
5 of the counters are blue
The rest of the counters are green
Joseph is going to take at random a counter from the bag.
The probability that the counter is green is \(\dfrac{2}{5}\)
Work out the number of green counters in the bag.
(3)
| Scheme | Marks |
|---|---|
\(1 - \dfrac{2}{5}\) \(\left(= \dfrac{3}{5}\right)\) oe or (green : red + blue =) 2 : 3 OR eg \(\dfrac{x}{x + 12} = \dfrac{2}{5}\) or \(\dfrac{n - 12}{n} = \dfrac{2}{5}\) | M1 |
\(\dfrac{(7 + 5)}{\text{“}3\text{”}}\) oe or \(\dfrac{(7 + 5)}{\text{“}3\text{”}} \times 5\) oe (= 20) or \((7 + 5) \div \text{“}\dfrac{5}{3}\text{”}\) (= 20) OR \(x = \dfrac{12 \times 2}{5 - 2}\) or \(n = \dfrac{5 \times 12}{5 - 2}\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 8 | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for finding the proportion of counters that are not green (may be a percentage or decimal)
or for a correct ratio (allow ratio in any order)
Values in ratio do not need labels, but if labels are used, they must be correct
This is implied by eg 12 (counters) = 3 (parts)
OR for forming a correct equation in terms of the number of green counters or the total number of counters in the bag (allow use of any letter)
M1: for a method to find the value one-fifth of the counters or for a method to find the total number of counters in the bag
OR
for a correct method to solve a correct equation