There are only r red counters and g green counters in a bag - Edexcel - GCSE Maths - Question 5 - 2019 - Paper 1

After adding 2 red counters and 3 green counters, the new counts will be ( r + 2 ) and ( g + 3 ). The probability that the counter is green is now:
[ \frac{g + 3}{(r + 2) + (g + 3)} = \frac{6}{13} ]
This simplifies to:
[ \frac{g + 3}{r + g + 5} = \frac{6}{13} ]
Cross-multiplying gives us:
[ 13(g + 3) = 6(r + g + 5) ]
Expanding:
[ 13g + 39 = 6r + 6g + 30 ]
Rearranging terms leads to:
[ 7g - 6r = -9 ]

We now have the system of equations:

1. ( 4g = 3r )
2. ( 7g - 6r = -9 )
   We can express r in terms of g from the first equation:
   [ r = \frac{4}{3}g ]
   Substituting this into the second equation:
   [ 7g - 6\left(\frac{4}{3}g\right) = -9 ]
   This simplifies to:
   [ 7g - 8g = -9 ]
   [ -g = -9 ]
   Thus,
   [ g = 9 ]
   Substituting back to find r:
   [ r = \frac{4}{3} \times 9 = 12 ]
   Therefore, the original counts are:

• Red counters: 12
• Green counters: 9