The problem states that in a pond, $45\%$ of the fish are perch, $\frac{3}{10}$ of the fish are carp, and the rest are trout. We need to find the ratio of perch to carp to trout in its simplest form.

1. Problem Description

The problem states that in a pond, $45\%$ of the fish are perch, $\frac{3}{10}$ of the fish are carp, and the rest are trout. We need to find the ratio of perch to carp to trout in its simplest form.

2. Solution Steps

First, convert the percentage of perch to a fraction. $45\% = \frac{45}{100} = \frac{9}{20}$.

The fraction of carp is given as $\frac{3}{10}$.

To find the fraction of trout, we subtract the fractions of perch and carp from 1 (representing the whole pond):

Fraction of trout $= 1 - \frac{9}{20} - \frac{3}{10}$.

To subtract the fractions, we need a common denominator, which is

2

0. So we have:

Fraction of trout $= 1 - \frac{9}{20} - \frac{6}{20} = \frac{20}{20} - \frac{9}{20} - \frac{6}{20} = \frac{20-9-6}{20} = \frac{5}{20} = \frac{1}{4}$.

Now we have the fractions of perch, carp, and trout as $\frac{9}{20}$, $\frac{3}{10}$, and $\frac{1}{4}$ respectively.

To find the ratio, we can write them as $\frac{9}{20} : \frac{3}{10} : \frac{1}{4}$.

To simplify the ratio, we can multiply each fraction by the least common multiple (LCM) of the denominators, which is

2

0. So, we have:

$(\frac{9}{20} \times 20) : (\frac{3}{10} \times 20) : (\frac{1}{4} \times 20) = 9 : 6 : 5$.

Since 9, 6, and 5 do not have any common factors other than 1, the ratio is already in its simplest form.

3. Final Answer

$9:6:5$