Question 1209500: Hi
Containers A B and C contained water. 1/5 was poured from A to B. 1/6 of the water from B was then poured into C. Finally 1/6 was transferred from C to A. In the end there were 8.5 litres of water in each container. How much water was there in B at first.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let the initial amounts of water in containers \( A \), \( B \), and \( C \) be \( a \), \( b \), and \( c \) liters, respectively.
### Step 1: Transfer water from \( A \) to \( B \)
\[
\text{Amount poured from } A \text{ to } B = \frac{1}{5}a
\]
After this transfer:
\[
A = a - \frac{1}{5}a = \frac{4}{5}a, \quad B = b + \frac{1}{5}a
\]
### Step 2: Transfer water from \( B \) to \( C \)
\[
\text{Amount poured from } B \text{ to } C = \frac{1}{6}(b + \frac{1}{5}a)
\]
After this transfer:
\[
B = b + \frac{1}{5}a - \frac{1}{6}(b + \frac{1}{5}a) = b + \frac{1}{5}a - \frac{1}{6}b - \frac{1}{30}a
\]
\[
C = c + \frac{1}{6}(b + \frac{1}{5}a)
\]
### Step 3: Transfer water from \( C \) to \( A \)
\[
\text{Amount poured from } C \text{ to } A = \frac{1}{6}[c + \frac{1}{6}(b + \frac{1}{5}a)]
\]
After this transfer:
\[
C = c + \frac{1}{6}(b + \frac{1}{5}a) - \frac{1}{6}[c + \frac{1}{6}(b + \frac{1}{5}a)]
\]
\[
A = \frac{4}{5}a + \frac{1}{6}[c + \frac{1}{6}(b + \frac{1}{5}a)]
\]
### Step 4: Final state
At the end, all containers have the same amount of water: \( 8.5 \) liters.
#### Equations:
1. \( A = 8.5 \):
\[
\frac{4}{5}a + \frac{1}{6}[c + \frac{1}{6}(b + \frac{1}{5}a)] = 8.5
\]
2. \( B = 8.5 \):
\[
b + \frac{1}{5}a - \frac{1}{6}(b + \frac{1}{5}a) = 8.5
\]
3. \( C = 8.5 \):
\[
c + \frac{1}{6}(b + \frac{1}{5}a) - \frac{1}{6}[c + \frac{1}{6}(b + \frac{1}{5}a)] = 8.5
\]
We will solve the system of equations to find \( b \), the initial amount of water in container \( B \).
The initial amount of water in container \( B \) was \( \mathbf{8.5 \, \text{liters}} \). It turns out that all three containers started with the same amount of water, \( 8.5 \) liters.
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