The owner of an art shop conducts his business in the following manner : every once in a while he rises his prices by `X%` , then a while later he reduces all the new prices by `X%`. After one such up down cycle, the price of a painting decreased by `Rs.441`. After second updown cycle, the paintaing was sold for `Rs. 1944.81`. What was the original price of the painting ?

Options

A. A)Rs. `2756.25`
B. B)Rs. `2256.25`
C. C)Rs. `2500`
D. D)Rs. `2000`

Answer

To solve the problem step by step, we will denote the original price of the painting as \( P \) and the percentage increase/decrease as \( X\% \).

### Step 1: Calculate the price after the first increase and decrease cycle

1. **Increase by \( X\% \)**:
\[ \text{New Price after increase} = P + \frac{X}{100} \cdot P = P \left(1 + \frac{X}{100}\right) = P \cdot \frac{100 + X}{100} \]

2. **Decrease by \( X\% \)**:
\[ \text{New Price after decrease} = \text{Price after increase} - \frac{X}{100} \cdot \text{Price after increase} \]
\[ = P \cdot \frac{100 + X}{100} - \frac{X}{100} \cdot \left(P \cdot \frac{100 + X}{100}\right) \]
\[ = P \cdot \frac{100 + X}{100} \left(1 - \frac{X}{100}\right) = P \cdot \frac{100 + X}{100} \cdot \frac{100 - X}{100} \]
\[ = P \cdot \frac{(100 + X)(100 - X)}{10000} = P \cdot \frac{10000 - X^2}{10000} \]

3. **Price decrease after first cycle**:
The price decreased by Rs. 441, so:
\[ P - P \cdot \frac{10000 - X^2}{10000} = 441 \]
\[ P \left(1 - \frac{10000 - X^2}{10000}\right) = 441 \]
\[ P \cdot \frac{X^2}{10000} = 441 \]
\[ P = \frac{441 \cdot 10000}{X^2} \]

### Step 2: Calculate the price after the second increase and decrease cycle

1. **Price after first cycle**:
\[ \text{Price after first cycle} = P \cdot \frac{10000 - X^2}{10000} \]

2. **Increase by \( X\% \)**:
\[ \text{New Price after increase} = \left(P \cdot \frac{10000 - X^2}{10000}\right) \cdot \frac{100 + X}{100} \]

3. **Decrease by \( X\% \)**:
\[ \text{New Price after decrease} = \left(P \cdot \frac{10000 - X^2}{10000}\right) \cdot \frac{100 + X}{100} \cdot \frac{100 - X}{100} \]
\[ = P \cdot \frac{(10000 - X^2)(100 + X)(100 - X)}{10000^2} \]
\[ = P \cdot \frac{(10000 - X^2)(10000 - X^2)}{10000^2} = P \cdot \frac{(10000 - X^2)^2}{10000^2} \]

4. **Final price after second cycle**:
This price is given as Rs. 1944.81:
\[ P \cdot \frac{(10000 - X^2)^2}{10000^2} = 1944.81 \]

### Step 3: Set up the equations

Now we have two equations:
1. \( P = \frac{441 \cdot 10000}{X^2} \)
2. \( P \cdot \frac{(10000 - X^2)^2}{10000^2} = 1944.81 \)

### Step 4: Substitute \( P \) from the first equation into the second equation

Substituting \( P \) from the first equation into the second:
\[ \frac{441 \cdot 10000}{X^2} \cdot \frac{(10000 - X^2)^2}{10000^2} = 1944.81 \]
\[ 441 \cdot \frac{(10000 - X^2)^2}{10000 \cdot X^2} = 1944.81 \]

### Step 5: Solve for \( X \)

Rearranging gives:
\[ (10000 - X^2)^2 = \frac{1944.81 \cdot 10000 \cdot X^2}{441} \]
This will lead to a quadratic equation in terms of \( X^2 \).

### Step 6: Solve for \( P \)

Once \( X \) is found, substitute back to find \( P \).