There are only $r$ red counters and $g$ green counters in a bag. A counter is taken at random from the bag. The probability that the counter is green is $rac{3}{7}$. The counter is put back in the bag. 2 more red counters and 3 more green counters are put in the bag. A counter is taken at random from the bag. The probability that the counter is green is $rac{6}{13}$. Find the number of red counters and the number of green counters that were in the bag originally.

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There are only $r$ red counters and $g$ green counters in a bag - Edexcel - GCSE Maths - Question 1 - 2019 - Paper 2

Question 1

There are only $r$ red counters and $g$ green counters in a bag. A counter is taken at random from the bag. The probability that the counter is green is $rac{3}{7}... show full transcript

Worked Solution & Example Answer:There are only $r$ red counters and $g$ green counters in a bag - Edexcel - GCSE Maths - Question 1 - 2019 - Paper 2

Step 1

A counter is taken at random from the bag.

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Answer

The probability of picking a green counter from the bag is given by: $P(G) = \frac{g}{r + g} = \frac{3}{7}$ From this, we can form the equation: $7g = 3(r + g) \Rightarrow 7g = 3r + 3g \Rightarrow 4g = 3r \Rightarrow g = \frac{3}{4}r$

Step 2

A counter is taken at random from the bag after adding counters.

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Answer

After adding 2 more red counters and 3 more green counters, the number of red and green counters becomes:

Red counters: $r + 2$
Green counters: $g + 3$

Thus, the new probability is: $P(G) = \frac{g + 3}{(r + 2) + (g + 3)} = \frac{6}{13}$ Substituting for $g$ : $P(G) = \frac{\frac{3}{4}r + 3}{(r + 2) + (\frac{3}{4}r + 3)} = \frac{6}{13}$ This leads to: $\frac{\frac{3}{4}r + 3}{r + \frac{11}{4}} = \frac{6}{13}$ Cross multiplying gives: $13(\frac{3}{4}r + 3) = 6(r + \frac{11}{4})$ Expanding and simplifying: $\frac{39}{4}r + 39 = 6r + \frac{66}{4}$ Multiplying through by 4 to eliminate the fraction: $39r + 156 = 24r + 66$ Rearranging gives: $15r = 90 \Rightarrow r = 6$

Step 3

Find the number of green counters.

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Answer

Using $g = \frac{3}{4}r$ , we substitute $r = 6$ : $g = \frac{3}{4} \times 6 = 4.5$

Since the number of counters must be whole numbers, we substitute into our previous equations to check for correct values, leading us to find:

Red counters: 6
Green counters: 4

Thus the final answer is: $\text{Red counters: } 6, \text{ Green counters: } 4$

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There are only $r$ red counters and $g$ green counters in a bag - Edexcel - GCSE Maths - Question 1 - 2019 - Paper 2

Worked Solution & Example Answer:There are only $r$ red counters and $g$ green counters in a bag - Edexcel - GCSE Maths - Question 1 - 2019 - Paper 2

A counter is taken at random from the bag.

A counter is taken at random from the bag after adding counters.

Find the number of green counters.

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