Let the original price per dozen pens be \(x\) rupees. ∴ After a reduction of ₹6, the new price per dozen becomes \((x - 6)\) rupees. Previously, the number of pens that could be bought for ₹30 = \(\frac{30}{x}\) dozens = \(\frac{30 \times 12}{x}\) pens [since 1 dozen = 12 pens] Now, the number of pens that can be bought for ₹30 = \(\frac{30}{x - 6}\) dozens = \(\frac{30 \times 12}{x - 6}\) pens According to the condition: \(\frac{360}{x - 6} - \frac{360}{x} = 3\) i.e., \(\frac{360(x - x + 6)}{x(x - 6)} = 3\) i.e., \(\frac{360 \times 6}{x^2 - 6x} = 3\) i.e., \(\frac{2160}{x^2 - 6x} = 3\) i.e., \(x^2 - 6x = 720\) i.e., \(x^2 - 6x - 720 = 0\) Solving the quadratic: \(x^2 - 30x + 24x - 720 = 0\) \(x(x - 30) + 24(x - 30) = 0\) \((x - 30)(x + 24) = 0\) So, either \(x = 30\) or \(x = -24\) But price cannot be negative, so \(x = 30\) Therefore, the original price per dozen pens was ₹30.