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(B) Let the fixed amount be $Rs. x$ and the cost per unit be $Rs. y$.According to the problem:$540y + x = 1800$ $(i)$$620y + x = 2040$ (ii)Subtracting

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$1680$

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(B) Let the fixed amount be $Rs. x$ and the cost per unit be $Rs. y$.According to the problem:$540y + x = 1800$ $(i)$$620y + x = 2040$ (ii)Subtracting equation $(i)$ from equation (ii):$(620y - 540y) + (x - x) = 2040 - 1800$$80y = 240$$y = 3$Substituting $y = 3$ into equation $(i)$:$540(3) + x = 1800$$1620 + x = 1800$$x = 1800 - 1620 = 180$So,the fixed charge is $Rs. 180$ and the charge per unit is $Rs. 3$.For $500$ units,the total bill is:$500 	imes 3 + 180 = 1500 + 180 = 1680$Thus,the bill for $500$ units is $Rs. 1680$.

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