Please give me a formula. A water tank has a capacity of 500 litres. Starting from day 1, 50 litres is added to the tank. Amount of water added increases by 2 litres on each odd day. On even days, water is remove. 50 litres is removed on day 2, and on subsequesnt consecutive even days the amount of water removed reduces by 10% from the previous day. When will the tank overflow.

To model this problem, we'll set up a formula that accounts for both the water added on odd days and the water removed on even days. The key variables are:

- **Day \( n \)**: Refers to the day number.
- **Odd days**: Water added increases by 2 liters starting from 50 liters on day 1.
- **Even days**: Water removed starts at 50 liters on day 2, reducing by 10% on each consecutive even day.

Let's define the following:

- **\( A(n) \)**: Amount of water added on the \( n \)-th odd day.
- **\( R(n) \)**: Amount of water removed on the \( n \)-th even day.
- **\( W_n \)**: Total amount of water in the tank after day \( n \).

### 1. Water added on odd days:

- Day 1: \( A(1) = 50 \) liters.
- Day 3: \( A(3) = 50 + 2 = 52 \) liters.
- Day 5: \( A(5) = 50 + 2 \times 2 = 54 \) liters.
- General formula for water added on the \( k \)-th odd day:  
  \[
  A(n_{\text{odd}}) = 50 + 2 \times (k - 1)
  \]
  where \( k \) is the index of the odd day (1st odd day, 2nd odd day, etc.).

### 2. Water removed on even days:

- Day 2: \( R(2) = 50 \) liters.
- Day 4: \( R(4) = 50 \times (1 - 0.1) = 50 \times 0.9 = 45 \) liters.
- Day 6: \( R(6) = 45 \times (1 - 0.1) = 40.5 \) liters.
- General formula for water removed on the \( k \)-th even day:  
  \[
  R(n_{\text{even}}) = 50 \times (0.9)^{k - 1}
  \]
  where \( k \) is the index of the even day (2nd day, 4th day, etc.).

### 3. Total amount of water in the tank:

Starting from day 1 with 0 liters in the tank, the water added or removed on each day modifies the total amount. The total water in the tank after day \( n \) is:

\[
W_n = W_{n-1} + A(n) \quad \text{if \( n \) is odd}
\]
\[
W_n = W_{n-1} - R(n) \quad \text{if \( n \) is even}
\]

Where \( W_0 = 0 \) liters.

### 4. Overflow condition:

The tank has a capacity of 500 liters, so the tank will overflow when \( W_n > 500 \).

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To find when the tank overflows, you will need to iteratively apply the water added and removed formulas to compute the total water for each day, continuing until \( W_n \) exceeds 500 liters.

Would you like me to compute the exact day when the tank overflows?

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### 5 Follow-Up Questions:
1. How does the amount of water added change over time?
2. How does the reduction in the removal rate on even days affect the tank's filling rate?
3. Can the tank overflow only due to water added, or do the removal rates slow this process down significantly?
4. How would the overflow day change if the initial amount of water added or removed was different?
5. How would the situation change if the tank’s capacity were smaller or larger?

**Tip**: When solving problems involving gradual change, always keep an eye on patterns in the increase or decrease of quantities, like the 2-liter increase or 10% reduction in this problem.

A water tank has a capacity of 500 litres. Starting from day 1, 50 litres is added to the tank. The amount of water added increases by 2 litres on each odd day. On even days, water is removed. 50 litres is removed on day 2, and on subsequent consecutive even days, the amount of water removed reduces by 10% from the previous day. When will the tank overflow?

This problem models the water level in a 500-litre tank where water is added progressively on odd days and removed with a decreasing rate on even days. Key concepts include arithmetic progression for water addition, exponential decay for water removal, and iterative calculations to determine when the tank will overflow. Suitable for students in Grades 9-11.

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