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Sagot :
Given the initial ratio of cats to dogs is 2:5 and there are 10 cats, we can find the number of dogs:
\[
\text{Let } C \text{ be the number of cats and } D \text{ be the number of dogs.}
\]
\[
C = 10 \text{ cats}
\]
Since the ratio of cats to dogs is 2:5, we can set up the following proportion:
\[
\frac{C}{D} = \frac{2}{5}
\]
Substituting \( C = 10 \):
\[
\frac{10}{D} = \frac{2}{5}
\]
Cross-multiplying to solve for \( D \):
\[
10 \times 5 = 2 \times D \implies 50 = 2D \implies D = \frac{50}{2} = 25
\]
So, initially, there are 10 cats and 25 dogs.
After a group of animals arrives, the new ratio of cats to dogs becomes 5:3. Let the number of new cats be \( x \) and the number of new dogs be \( y \). The new number of cats and dogs are:
\[
10 + x \text{ cats and } 25 + y \text{ dogs}
\]
According to the new ratio:
\[
\frac{10 + x}{25 + y} = \frac{5}{3}
\]
Cross-multiplying to solve for \( x \) and \( y \):
\[
3(10 + x) = 5
\[
\text{Let } C \text{ be the number of cats and } D \text{ be the number of dogs.}
\]
\[
C = 10 \text{ cats}
\]
Since the ratio of cats to dogs is 2:5, we can set up the following proportion:
\[
\frac{C}{D} = \frac{2}{5}
\]
Substituting \( C = 10 \):
\[
\frac{10}{D} = \frac{2}{5}
\]
Cross-multiplying to solve for \( D \):
\[
10 \times 5 = 2 \times D \implies 50 = 2D \implies D = \frac{50}{2} = 25
\]
So, initially, there are 10 cats and 25 dogs.
After a group of animals arrives, the new ratio of cats to dogs becomes 5:3. Let the number of new cats be \( x \) and the number of new dogs be \( y \). The new number of cats and dogs are:
\[
10 + x \text{ cats and } 25 + y \text{ dogs}
\]
According to the new ratio:
\[
\frac{10 + x}{25 + y} = \frac{5}{3}
\]
Cross-multiplying to solve for \( x \) and \( y \):
\[
3(10 + x) = 5
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