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

Options

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

Answer

To solve the problem step by step, let's denote the original price of the article as \( X \) rupees.

### Step 1: Increase the original price by \( r\% \)
When the price is increased by \( r\% \), the new price becomes:
\[ \text{New Price} = X + \frac{r}{100} \times X = X \left(1 + \frac{r}{100}\right) \]

### Step 2: Decrease the new price by \( r\% \)
Now, we decrease the new price by \( r\% \). The amount of decrease is:
\[ \text{Decrease} = \frac{r}{100} \times \text{New Price} = \frac{r}{100} \times X \left(1 + \frac{r}{100}\right) \]
Thus, the latest price after the decrease becomes:
\[ \text{Latest Price} = \text{New Price} - \text{Decrease} = X \left(1 + \frac{r}{100}\right) - \frac{r}{100} \times X \left(1 + \frac{r}{100}\right) \]

### Step 3: Simplify the latest price expression
Factoring out \( X \left(1 + \frac{r}{100}\right) \):
\[ \text{Latest Price} = X \left(1 + \frac{r}{100}\right) \left(1 - \frac{r}{100}\right) \]
Using the identity \( (a + b)(a - b) = a^2 - b^2 \):
\[ \text{Latest Price} = X \left(1 - \left(\frac{r}{100}\right)^2\right) = X \left(\frac{10000 - r^2}{10000}\right) \]

### Step 4: Set the latest price equal to Rs. 1
According to the problem, the latest price is Rs. 1:
\[ X \left(\frac{10000 - r^2}{10000}\right) = 1 \]

### Step 5: Solve for the original price \( X \)
Rearranging the equation gives:
\[ X = \frac{10000}{10000 - r^2} \]

### Conclusion
Thus, the original price of the article is:
\[ \boxed{\frac{10000}{10000 - r^2}} \]