Answer
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Hint We first find out the units consumed by all the appliances per day and then calculate the bill by multiplying per day power consumption with \[30\,days\]. To get the per day power consumption, we individually compute the units consumed by each appliance and then add them. Using this power consumption, we compare the quantity if it is bigger than \[50\,units\], in which case the units will be charged at \[Rs\,3.50\]. The rest of the units (if any remaining), will be charged at \[Rs\,4.50\].
Complete step by step answer:
We will calculate the total amount of units consumed in the entire month.
Each appliance will consume the product of its wattage and number of hours it is in use.
For that we start with
(A) \[1\] tube light of \[40\,W\]used for \[8\,hours\]consumes \[40\,W \times 8\,hours = 320\,Whr\]of power
(B) \[\therefore \]\[3\]tube lights of \[40\,W\]used for \[8\,hours\]consumes \[320\,Whr \times 3 = 960\,Whr\]of power
(C) \[1\] refrigerator of \[300\,W\]used for \[24\,hours\]consumes \[300\,W \times 24\,hours = 7200\,Whr\]of power
(D) \[1\] mixer of \[750\,W\]used for \[1\,hour\]consumes \[750\,W \times 1\,hour = 750\,Whr\]of power
(E) \[1\] TV of \[100\,W\]used for \[6\,hours\]consumes \[100\,W \times 6\,hours = 600\,Whr\]of energy.
Adding the power consumed in all the four sub-parts we get the total units of power consumed as
\[960\,Whr + 7200\,Whr + 750\,Whr + 600\,Whr = 9510\,Whr = 9.51\,kWhr\]per day.
Now, the amount of power consumed in the entire month is
\[9.51\,kWhr \times 30 = 285.3\,kWhr = 285.3\,units\].
It is given that the first \[50\,units\]will be charged at \[Rs\,3.50\]
\[ \Rightarrow 50\,units \times Rs\,3.50\,per\,unit = Rs\,175\]
And the next \[(285.3 - 50)\,units = 235.3\,units\]will be charged at \[Rs\,4.50\]
\[ \Rightarrow 235.3\,units \times Rs\,4.50\,per\,unit = Rs\,1058.85\]
Adding these, we get the required electricity bill as \[Rs\,1058.85 + Rs\,175 = Rs\,1233.85\]
Note It needs to be remembered that \[1\,unit\] of electricity means \[1\,kWhr\] of electricity consumption which is the unit of electrical power. The term power consumption & units consumed is interchangeable (if the power consumption is given in \[kWhr\]).
Complete step by step answer:
We will calculate the total amount of units consumed in the entire month.
Each appliance will consume the product of its wattage and number of hours it is in use.
For that we start with
(A) \[1\] tube light of \[40\,W\]used for \[8\,hours\]consumes \[40\,W \times 8\,hours = 320\,Whr\]of power
(B) \[\therefore \]\[3\]tube lights of \[40\,W\]used for \[8\,hours\]consumes \[320\,Whr \times 3 = 960\,Whr\]of power
(C) \[1\] refrigerator of \[300\,W\]used for \[24\,hours\]consumes \[300\,W \times 24\,hours = 7200\,Whr\]of power
(D) \[1\] mixer of \[750\,W\]used for \[1\,hour\]consumes \[750\,W \times 1\,hour = 750\,Whr\]of power
(E) \[1\] TV of \[100\,W\]used for \[6\,hours\]consumes \[100\,W \times 6\,hours = 600\,Whr\]of energy.
Adding the power consumed in all the four sub-parts we get the total units of power consumed as
\[960\,Whr + 7200\,Whr + 750\,Whr + 600\,Whr = 9510\,Whr = 9.51\,kWhr\]per day.
Now, the amount of power consumed in the entire month is
\[9.51\,kWhr \times 30 = 285.3\,kWhr = 285.3\,units\].
It is given that the first \[50\,units\]will be charged at \[Rs\,3.50\]
\[ \Rightarrow 50\,units \times Rs\,3.50\,per\,unit = Rs\,175\]
And the next \[(285.3 - 50)\,units = 235.3\,units\]will be charged at \[Rs\,4.50\]
\[ \Rightarrow 235.3\,units \times Rs\,4.50\,per\,unit = Rs\,1058.85\]
Adding these, we get the required electricity bill as \[Rs\,1058.85 + Rs\,175 = Rs\,1233.85\]
Note It needs to be remembered that \[1\,unit\] of electricity means \[1\,kWhr\] of electricity consumption which is the unit of electrical power. The term power consumption & units consumed is interchangeable (if the power consumption is given in \[kWhr\]).
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