High School

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There are only [tex]r[/tex] red counters and [tex]b[/tex] blue counters in a bag. A counter is taken at random from the bag. The probability that the counter is blue is [tex]\frac{4}{9}[/tex]. The counter is put back in the bag. Six more red counters and three more blue counters are put in the bag. A counter is taken at random from the bag. The probability that the counter is blue is [tex]\frac{5}{12}[/tex].

Find the number of red counters and the number of blue counters that were in the bag originally.

Answer :

Sure, let's solve this problem step-by-step.

Given:
- The probability that a counter taken from the bag is blue initially is [tex]\(\frac{4}{9}\)[/tex].
- After adding more counters, the probability that a counter taken from the bag is blue becomes [tex]\(\frac{5}{12}\)[/tex].

We need to find the original number of red and blue counters.

### Step 1: Set Up the Problem with Initial Conditions
Let:
- [tex]\( r \)[/tex] be the number of original red counters.
- [tex]\( b \)[/tex] be the number of original blue counters.

From the first piece of information, we get the equation:
[tex]\[
\frac{b}{r + b} = \frac{4}{9}
\][/tex]

Cross-multiplying to solve for [tex]\( b \)[/tex]:
[tex]\[
9b = 4(r + b) \implies 9b = 4r + 4b \implies 9b - 4b = 4r \implies 5b = 4r
\][/tex]
[tex]\[
b = \frac{4r}{5} \quad \text{(Equation 1)}
\][/tex]

### Step 2: Set Up the Problem After Adding Counters
After adding 6 red counters and 3 blue counters, we get:
- New number of red counters = [tex]\( r + 6 \)[/tex]
- New number of blue counters = [tex]\( b + 3 \)[/tex]
- The new total number of counters = [tex]\( (r + 6) + (b + 3) = r + b + 9 \)[/tex]

From the second piece of information, we get the equation:
[tex]\[
\frac{b + 3}{r + b + 9} = \frac{5}{12}
\][/tex]

Cross-multiplying to solve for [tex]\( b \)[/tex]:
[tex]\[
12(b + 3) = 5(r + b + 9) \implies 12b + 36 = 5r + 5b + 45 \implies 12b - 5b = 5r + 45 - 36
\][/tex]
[tex]\[
7b = 5r + 9 \quad \text{(Equation 2)}
\][/tex]

### Step 3: Solve the Two Equations Simultaneously
We now have the system of equations:
1. [tex]\( 5b = 4r \)[/tex]
2. [tex]\( 7b = 5r + 9 \)[/tex]

First, solve Equation 1 for [tex]\( r \)[/tex]:
[tex]\[
r = \frac{5b}{4}
\][/tex]

Substitute [tex]\( r \)[/tex] into Equation 2:
[tex]\[
7b = 5\left(\frac{5b}{4}\right) + 9 \implies 7b = \frac{25b}{4} + 9
\][/tex]

Multiply every term by 4 to clear the fraction:
[tex]\[
28b = 25b + 36
\][/tex]

Solving for [tex]\( b \)[/tex]:
[tex]\[
28b - 25b = 36 \implies 3b = 36 \implies b = 12
\][/tex]

Substitute [tex]\( b = 12 \)[/tex] back into Equation 1 to find [tex]\( r \)[/tex]:
[tex]\[
5 \times 12 = 4r \implies 60 = 4r \implies r = 15
\][/tex]

### Conclusion
The original number of red counters was 15, and the original number of blue counters was 12.

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