Answer: C
Let the man have bought \(x\) kilograms of sugar. ∴ The price per kilogram of sugar was \(\cfrac{80}{x}\) rupees. If he had received 4 kilograms more, he would have gotten \((x + 4)\) kilograms of sugar. Then, the price per kilogram would have been \(\cfrac{80}{x + 4}\) rupees. ∴ According to the question: \[ \cfrac{80}{x} - \cfrac{80}{x + 4} = 1 \] \[ \cfrac{80(x + 4) - 80x}{x(x + 4)} = 1 \] \[ \cfrac{80x + 320 - 80x}{x(x + 4)} = 1 \] \[ \cfrac{320}{x(x + 4)} = 1 \] \[ x(x + 4) = 320 \] \[ x^2 + 4x - 320 = 0 \] \[ x^2 + 20x - 16x - 320 = 0 \] \[ x(x + 20) - 16(x + 20) = 0 \] \[ (x + 20)(x - 16) = 0 \] ∴ Either \((x + 20) = 0\) or \((x - 16) = 0\) When \((x + 20) = 0\), then \(x = -20\) [But the quantity of sugar cannot be negative] When \((x - 16) = 0\), then \(x = 16\) ∴ He bought 16 kilograms of sugar and the price per kilogram was \(\cfrac{80}{16} = 5\) rupees.
Let the man have bought \(x\) kilograms of sugar. ∴ The price per kilogram of sugar was \(\cfrac{80}{x}\) rupees. If he had received 4 kilograms more, he would have gotten \((x + 4)\) kilograms of sugar. Then, the price per kilogram would have been \(\cfrac{80}{x + 4}\) rupees. ∴ According to the question: \[ \cfrac{80}{x} - \cfrac{80}{x + 4} = 1 \] \[ \cfrac{80(x + 4) - 80x}{x(x + 4)} = 1 \] \[ \cfrac{80x + 320 - 80x}{x(x + 4)} = 1 \] \[ \cfrac{320}{x(x + 4)} = 1 \] \[ x(x + 4) = 320 \] \[ x^2 + 4x - 320 = 0 \] \[ x^2 + 20x - 16x - 320 = 0 \] \[ x(x + 20) - 16(x + 20) = 0 \] \[ (x + 20)(x - 16) = 0 \] ∴ Either \((x + 20) = 0\) or \((x - 16) = 0\) When \((x + 20) = 0\), then \(x = -20\) [But the quantity of sugar cannot be negative] When \((x - 16) = 0\), then \(x = 16\) ∴ He bought 16 kilograms of sugar and the price per kilogram was \(\cfrac{80}{16} = 5\) rupees.