Answer :
After adding 20 blue counters to the initial 15 red counters, the total number of counters in the bag is 61 counters.
Given:
Initial ratio of red, green, and blue counters = 3 : 4 : 5
Total number of red counters initially = 15
Total number of red counters after addition = 15
Let's denote the number of blue counters added as 'b'.
After the addition, the ratio becomes:
7 : 6 : 8
Now, let's find the ratio of red counters before and after the addition:
Initial red counters : Final red counters
3 : 7
Since the ratio of red counters remains the same, we can write the equation:
[tex]\( \frac{15}{3} = \frac{15 + b}{7} \)[/tex]
Cross multiply to solve for 'b':
[tex]\( 15 \times 7 = 3 \times (15 + b) \)[/tex]
[tex]\( 105 = 45 + 3b \)[/tex]
[tex]\( 3b = 105 - 45 \)[/tex]
[tex]\( 3b = 60 \)[/tex]
[tex]\( b = \frac{60}{3} \)[/tex]
[tex]\( b = 20 \)[/tex]
So, 20 blue counters were added.
Now, let's find the total number of counters after the addition:
Total red counters = 15 + 15 = 30
Total blue counters = 20
Total green counters = (6/7) * Total red counters = [tex]\( \frac{6}{7} \)[/tex] * 30 = 25.71 (approximated to nearest integer)
However, since the number of counters must be a whole number, let's adjust the green counters:
Let's consider the ratio:
[tex]\( \frac{7}{6} = \frac{30 + g}{25.71} \)[/tex]
Solving for 'g':
[tex]\( 7 \times 25.71 = 6 \times (30 + g) \)[/tex]
[tex]\( 179.97 = 180 + 6g \)[/tex]
[tex]\( 6g = 179.97 - 180 \)[/tex]
[tex]\( 6g = -0.03 \)[/tex]
[tex]\( g = \frac{-0.03}{6} \)[/tex]
[tex]\( g = -0.005 \)[/tex]
Since 'g' is negative, it implies there are fewer green counters than in the initial ratio. So, let's adjust the green counters to 26.
Now, the total number of counters:
Total red counters = 30
Total green counters = 26
Total blue counters = 20
Total counters in the bag = 30 (red) + 26 (green) + 20 (blue) = 76 counters
So, there are 76 counters in the bag after the red and blue counters were added.
