Higher June 2025 Paper 2 Q11
11 Khalid has a box of counters.
17 of the counters are red
28 of the counters are blue
the rest of the counters are orange
Khalid is going to take at random a counter from the box.
The probability that Khalid will take an orange counter is \(\dfrac{4}{9}\)
Work out the number of orange counters that are in the box.
(3)
| Scheme | Marks |
|---|---|
(17 + 28) ÷ (9 – 4) or 45 ÷ 5 or \(1 - \dfrac{4}{9}\left(= \dfrac{5}{9}\right)\) or \((17 + 28 =)\dfrac{5}{9}\) oe or \(\dfrac{p}{p + 28 + 17} = \dfrac{4}{9}\) oe or \(\dfrac{p}{p + 45} = \dfrac{4}{9}\) or \(\dfrac{m - 45}{m} = \dfrac{4}{9}\) | M1 |
(17 + 28) ÷ 5 × 4 or 45 ÷ 5 × 4 or \((17 + 28) \div \text{``}{\dfrac{5}{9}}\text{''}\) (= 81) or \(45 \div \text{``}{\dfrac{5}{9}}\text{''}\) (= 81) or \((17 + 28) \times \text{``}{\dfrac{9}{5}}\text{''}\) (= 81) or \(45 \times \text{``}{\dfrac{9}{5}}\text{''}\) (= 81) or (17 + 28) ÷ 5 × 4 or 45 ÷ 5 × 4 or \(\text{``}{\dfrac{5}{9}}\text{''} = \dfrac{17 + 28}{n}\) oe or \(n = 81\) or \(9p = 4(p + 28 + 17)\) or \(9p - 4p = 180\) oe or \(5p = 180\) oe or \(9(m - 45) = 4m\) or \(9m - 4m = 405\) oe or \(5m = 405\) oe or \(m = 81\) | M1 |
| Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: 36 | A1 |
| (3) | |
| (3 marks) |
Notes
M1: Allow 0.55(555…) or 55(.555…)% truncated or rounded
M1: for the correct calculation for the total number of counters
or
for the correct calculation for the number of orange counters
or
for the correct equation for the total number of counters (removing the denominators)
or
for the correct equation for the number of orange counters (removing the denominators)
A1: cao