We can set up equations based on the information provided. Let:
- \( x \) be the number of gigantic rats in sewer #1.
- \( y \) be the number of gigantic rats in sewer #2.
- \( z \) be the number of gigantic rats in sewer #3.
From the problem, we have the following relationships:
- \( x = y - 9 \) (sewer #1 contains 9 fewer rats than sewer #2)
- \( y = z - 5 \) (sewer #2 contains 5 fewer rats than sewer #3)
- \( x + y + z = 56 \) (the total number of rats in all three sewers is 56)
Substituting \( y \) from the second equation into the first equation:
From \( y = z - 5 \):
\[ x = (z - 5) - 9 \] \[ x = z - 14 \]
Now we substitute \( x \) and \( y \) in terms of \( z \) into the total equation:
Using \( y = z - 5 \) and \( x = z - 14 \):
\[ (z - 14) + (z - 5) + z = 56 \]
Combining terms:
\[ 3z - 19 = 56 \]
Adding 19 to both sides:
\[ 3z = 75 \]
Dividing by 3:
\[ z = 25 \]
Now we can find \( y \) and \( x \):
Using \( y = z - 5 \):
\[ y = 25 - 5 = 20 \]
Using \( x = y - 9 \):
\[ x = 20 - 9 = 11 \]
Thus, the number of gigantic rats in sewer #1 is:
\[ \boxed{11} \]
For verification:
- sewer #1 \( x = 11 \)
- sewer #2 \( y = 20 \)
- sewer #3 \( z = 25 \)
Calculating the total:
\[ x + y + z = 11 + 20 + 25 = 56 \]
The numbers are consistent with the given conditions. Therefore, sewer #1 contains 11 gigantic rats.
1 answer