The price of an article was in creased by r%. Later the new price was decreased by r%.. If the latest price was 1, then the original price was:

Options

A. 1
B. `(1-r)^(2)/(100)`
C. `sqrt(1-r^(2))/(100)`
D. `(10000)/(1000-r^(2))`

Answer

To solve the problem step by step, we will denote the original price of the article as \( P \).

### Step 1: Increase the Price
The price of the article is increased by \( r\% \). Therefore, the new price after the increase can be calculated as:
\[ \text{New Price} = P + \left( \frac{r}{100} \times P \right) = P \left( 1 + \frac{r}{100} \right) \]

### Step 2: Decrease the Price
Next, the new price is decreased by \( r\% \). The price after the decrease can be calculated as:
\[ \text{Latest Price} = \text{New Price} - \left( \frac{r}{100} \times \text{New Price} \right) = \text{New Price} \left( 1 - \frac{r}{100} \right) \]
Substituting the expression for the new price from Step 1:
\[ \text{Latest Price} = P \left( 1 + \frac{r}{100} \right) \left( 1 - \frac{r}{100} \right) \]

### Step 3: Simplify the Expression
Using the identity \( (a + b)(a - b) = a^2 - b^2 \), we can simplify:
\[ \text{Latest Price} = P \left( 1 - \left( \frac{r}{100} \right)^2 \right) \]
This means:
\[ \text{Latest Price} = P \left( 1 - \frac{r^2}{10000} \right) \]

### Step 4: Set the Latest Price to 1
According to the problem, the latest price is given as 1:
\[ P \left( 1 - \frac{r^2}{10000} \right) = 1 \]

### Step 5: Solve for the Original Price \( P \)
Now, we can rearrange the equation to find \( P \):
\[ P = \frac{1}{1 - \frac{r^2}{10000}} \]

### Step 6: Final Expression for Original Price
Thus, the original price \( P \) can be expressed as:
\[ P = \frac{10000}{10000 - r^2} \]

### Summary
The original price of the article, given that the latest price is 1 after an increase and a subsequent decrease of \( r\% \), is:
\[ P = \frac{10000}{10000 - r^2} \]