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Answer :
We start by letting the number of black counters be [tex]$2k$[/tex] and the number of white counters be [tex]$7k$[/tex], based on the given ratio of black to white counters, which is [tex]$2:7$[/tex].
Similarly, if we let the number of white counters be [tex]$3m$[/tex] and the number of red counters be [tex]$8m$[/tex], according to the ratio of white to red counters being [tex]$3:8$[/tex], then we have two expressions for the white counters. Since the number of white counters is the same in both cases, we set them equal:
[tex]$$
7k = 3m.
$$[/tex]
Solving for [tex]$m$[/tex], we get
[tex]$$
m = \frac{7k}{3}.
$$[/tex]
The number of red counters can now be expressed in terms of [tex]$k$[/tex]:
[tex]$$
\text{red counters} = 8m = 8 \times \frac{7k}{3} = \frac{56k}{3}.
$$[/tex]
The total number of counters is the sum of the black, white, and red counters:
[tex]$$
2k + 7k + \frac{56k}{3}.
$$[/tex]
To add these, we write [tex]$2k+7k$[/tex] as [tex]$\frac{27k}{3}$[/tex] and then get
[tex]$$
\frac{27k}{3} + \frac{56k}{3} = \frac{83k}{3}.
$$[/tex]
Since the total number of counters is [tex]$1079$[/tex], we set up the equation:
[tex]$$
\frac{83k}{3} = 1079.
$$[/tex]
Multiplying both sides by [tex]$3$[/tex] gives:
[tex]$$
83k = 1079 \times 3.
$$[/tex]
Dividing both sides by [tex]$83$[/tex], we find
[tex]$$
k = \frac{1079 \times 3}{83}.
$$[/tex]
This calculation yields [tex]$k = 39$[/tex].
Now, substituting [tex]$k = 39$[/tex] back into the expression for the red counters, we have:
[tex]$$
\text{red counters} = \frac{56k}{3} = \frac{56 \times 39}{3}.
$$[/tex]
Since [tex]$39 \div 3 = 13$[/tex], this simplifies to:
[tex]$$
\text{red counters} = 56 \times 13 = 728.
$$[/tex]
Thus, the box contains [tex]$\boxed{728}$[/tex] red counters.
Similarly, if we let the number of white counters be [tex]$3m$[/tex] and the number of red counters be [tex]$8m$[/tex], according to the ratio of white to red counters being [tex]$3:8$[/tex], then we have two expressions for the white counters. Since the number of white counters is the same in both cases, we set them equal:
[tex]$$
7k = 3m.
$$[/tex]
Solving for [tex]$m$[/tex], we get
[tex]$$
m = \frac{7k}{3}.
$$[/tex]
The number of red counters can now be expressed in terms of [tex]$k$[/tex]:
[tex]$$
\text{red counters} = 8m = 8 \times \frac{7k}{3} = \frac{56k}{3}.
$$[/tex]
The total number of counters is the sum of the black, white, and red counters:
[tex]$$
2k + 7k + \frac{56k}{3}.
$$[/tex]
To add these, we write [tex]$2k+7k$[/tex] as [tex]$\frac{27k}{3}$[/tex] and then get
[tex]$$
\frac{27k}{3} + \frac{56k}{3} = \frac{83k}{3}.
$$[/tex]
Since the total number of counters is [tex]$1079$[/tex], we set up the equation:
[tex]$$
\frac{83k}{3} = 1079.
$$[/tex]
Multiplying both sides by [tex]$3$[/tex] gives:
[tex]$$
83k = 1079 \times 3.
$$[/tex]
Dividing both sides by [tex]$83$[/tex], we find
[tex]$$
k = \frac{1079 \times 3}{83}.
$$[/tex]
This calculation yields [tex]$k = 39$[/tex].
Now, substituting [tex]$k = 39$[/tex] back into the expression for the red counters, we have:
[tex]$$
\text{red counters} = \frac{56k}{3} = \frac{56 \times 39}{3}.
$$[/tex]
Since [tex]$39 \div 3 = 13$[/tex], this simplifies to:
[tex]$$
\text{red counters} = 56 \times 13 = 728.
$$[/tex]
Thus, the box contains [tex]$\boxed{728}$[/tex] red counters.
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