To solve this problem, let's first define the variables we'll use:
Let \( b \) be the number of blue fish and \( r \) be the number of red fish.
According to the problem:
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There is 1 fewer red fish than blue fish, which can be expressed as: \[ r = b - 1 \]
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We also know that 60% of the total fish are blue. The total number of fish can be defined as: \[ \text{Total fish} = b + r \]
Now we can substitute the expression for \( r \) into the equation for the total number of fish: \[ \text{Total fish} = b + (b - 1) = 2b - 1 \]
Since 60% of the total fish are blue, we can set up the following equation: \[ b = 0.6 \times (2b - 1) \]
Now we can simplify and solve this equation. First, distribute 0.6: \[ b = 1.2b - 0.6 \]
Next, we'll move all terms involving \( b \) to one side: \[ b - 1.2b = -0.6 \] \[ -0.2b = -0.6 \]
Now, divide both sides by -0.2: \[ b = \frac{-0.6}{-0.2} = 3 \]
Thus, the number of blue fish in the aquarium is: \[ \boxed{3} \]
To verify the solution, we can find the number of red fish: \[ r = b - 1 = 3 - 1 = 2 \]
Now, let's calculate the total number of fish: \[ \text{Total fish} = b + r = 3 + 2 = 5 \]
Finally, we check if 60% of the total fish are indeed blue: \[ 0.6 \times 5 = 3 \]
Since 3 is the number of blue fish, our solution is confirmed to be correct, and the number of blue fish in the aquarium is \( \boxed{3} \).