Computing Electric Consumption
- To compute for water of electric consumption for a particular period of time , simply subtract the previous reading from the present reading.
Menwer made a record of their 3-month electric consumption . The initial reading is 973 kWh .
January : 1120 February : 1353 March : 1512
Assume that the basic charge for the first 30kWh is ₱ 120.00 and the succeeding kilowatt-hour is charged ₱ 6.00
a. How much will Menwer pay for each month ?
b. In what month did he pay the most? The least ?
c. What is the average monthly consumption
What is asked ?
- The amount that Menwer will pay for each month of electricity they consumed.
- The month with the highest and lowest bill.
- The average monthly consumption.
What are the given facts ?
- The initial reading is 973kWh.
- January , February and March readings are 1120 kWh , 1352kWh , and 1512kWh , respectively.
- The basic charge for the first 30 kWh is ₱ 120.00 and the next kilowatt-hour is charged ₱ 6.00.
Plan.
What is operations shall we use to solve the problem ? Select your own strategy . To get amount that Menwer will pay for each amount : subtract the initial reading from first month's reading. For the following month , subtract the previous month's reading from the current month's reading . Then , to get the amount Menwer will pay for , subtract 30kWh from from the monthly kWh consumption , then multiply the result by ₱ 6.00 . Afterwards , add the amount for the first 30kWh which is ₱120.00.
To get the average monthly consumption , get the total consumption for three months the divide it by 3.
Solution:
[tex]{\sf{a. To \: get \: every \: month's}} \\ { \sf \: {electric \: consumption:}}[/tex]
January: 1120-973=147kWh
February: 1352-1120=232kWh
March: 1512-1352=160kWh
Now , to get the amount for each monthly consumption with the conditions that first 30kWh is ₱120.00 and ₱6.00 per remaining kilowatt-hour.
That is ,
January: 147kWh⇒120+[(147-30)×6
120+[(147-30)×6] =120+(117×6)
= 120+702
= ₱822.00
Soo , for January he will pay ₱822.00.
February: 232kWh ⇒ 120 + [(232-30)×6]
120+[(232-30)×6] = 120 + (202×6)
= 120+1212
= ₱1,332.00
Soo , for February he will pay ₱ 1,332.00
March: 160kWh⇒120 + [(160-30)×6
120+[(160-30)×6] = 120 + (130×6
=120+780
= ₱900.00
Soo , for March he will pay ₱900.00.
b. See computation in (a) . The month with the highest (most) electricity consumption was on February with 232kWh consumption. They consumed the least amount of electricity of January with 147kWh.
c. The consumption below shows the average monthly consumption.
[tex]{\sf{Average \: Monthly \: Consumption}} \: {\sf \blue{ = { \frac{147 + 23 + 160}{3} = \frac{540}{3} = 179.67}}}[/tex]
Soo , the average monthly consumption is approximately 179.67 kWh.
Answer: The following are the answers to the question:
a. He will pay ₱822.00 , ₱1,332.00 , and ₱900.00 for the months of January , February and March , respectively.
b. The month with the highest (most) electricity consumption was on February.They consumed the least amount in January.
c. The average monthly electricity consumption is approximately 179.67kWh.
[tex]{\sf\blue{RyukiKawakami}}[/tex]
