Question

sewer #1 contains 9 fewer gigantic rats than sewer #2. Sewer #2 contains 5 fewer gigantic rats than sewer #3. If there are 56 gigantic rats in all three sewers combined, how many gigantic rats are in sewer #1?

Answer 1

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:

1. \( x = y - 9 \) (sewer #1 contains 9 fewer rats than sewer #2)
2. \( y = z - 5 \) (sewer #2 contains 5 fewer rats than sewer #3)
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.