A bag contains only red counters, blue counters and green counters. The ratio of the number of red counters to the number of blue counters is 3 : 4 The ratio of the number of blue counters to the number of green counters is 6 : 5 There are less than 100 counters in the bag. Work out the greatest possible number of green counters in the bag.

Mark scheme

Marking points from the official mark scheme. The mark or mark type is shown in brackets.

1. R : B = 3 : 4 = 9 : 12
2. B : G = 6 : 5 = 12 : 10
3. R : B : G = 9 : 12 : 10 M1 for R:B:G in the form a:b:c (e.g. 3:4:12/6*5 or 3*6:4*6:5*4 or 18:24:20)
4. M1 for finding a common value for B (e.g. 12 or 24 or 36, etc.) and writing 3 part ratio e.g. 9:12:10 or 18:24:20
5. 9 + 12 + 10 = 31 M1 for finding the total parts in their ratio e.g. 9 + 12 + 10 = 31
6. 31 × 1 = 31, 31 × 2 = 62, 31 × 3 = 93 M1 for scaling their total sum to be less than 100 (e.g. 31k < 100)
7. 10 × 3 = 30 A1 30