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Let's solve this problem step by step.

### Problem Breakdown:
- The price of the article increases by \(x\%\) and then decreases by \(x\%\), leading to a drop in price by Rs. 180.
- After one more iteration (increase by \(x\%\) and then decrease by \(x\%\)), the price falls further by Rs. 160.
- We need to find the original price of the article.

### Key Concept:
When the price is increased by \(x\%\) and then decreased by \(x\%\), the net effect on the price is a decrease, as the percentage decrease applies to the increased price. The formula for the new price after an increase and then decrease of \(x\%\) is:
\[
\text{New Price} = P \times \left(1 - \frac{x^2}{100^2}\right)
\]
where \(P\) is the price before the operation.

Let the original price be \(P_0\), and the change factor be:
\[
k = \left(1 - \frac{x^2}{100^2}\right)
\]
After the first iteration:
\[
P_1 = P_0 \times k
\]
The difference after the first iteration is:
\[
P_0 - P_1 = 180 \quad \Rightarrow \quad P_0 - P_0 \times k = 180 \quad \Rightarrow \quad P_0 (1 - k) = 180
\]

After the second iteration:
\[
P_2 = P_1 \times k = P_0 \times k^2
\]
The difference after the second iteration is:
\[
P_1 - P_2 = 160 \quad \Rightarrow \quad P_0 \times k - P_0 \times k^2 = 160 \quad \Rightarrow \quad P_0 k (1 - k) = 160
\]

### System of Equations:
We now have two equations:
1. \(P_0 (1 - k) = 180\)
2. \(P_0 k (1 - k) = 160\)

We can solve these two equations simultaneously to find \(P_0\) (the original price) and \(k\). Let's solve this now.The solution to the system of equations gives us the original price of the article, \(P_0 = 1620\) rupees, and the change factor \(k = \frac{8}{9}\).

Thus, the original price of the article is **Rs. 1620**.

Would you like more details or have any further questions? Here are 5 related questions for practice:

1. How does the percentage increase and decrease lead to an overall price drop?
2. What is the impact on the final price if the increase and decrease percentages are different?
3. If the owner repeated this process three times, what would be the price reduction?
4. Can you find the percentage \(x\%\) that caused the drop in price?
5. What is the total percentage drop in price after two iterations?

**Tip**: When prices are increased and then decreased by the same percentage, the overall effect is always a decrease. The percentage loss is quadratic, not linear.

The owner of a shop increases the price of an article by x% and then decreases it by x%. As a result, the price of the article falls by Rs. 180. After one more such iteration, the price falls by a further Rs. 160. Find the original price (in rupees) of the article.

Learn how to solve a real-world problem involving percentage increases and decreases, leading to a drop in price. This problem includes key algebraic concepts such as simultaneous equations and percentage changes, suitable for students in Grades 10-12.

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