Question:
Mel uses a 900-watt air conditioner for 8 hours a day. This means that each hour the AC uses 900 watts of energy. If he reduces the time he uses the air conditioner by 5 hours a day, how many kilowatts of electric energy will he save in 30 days?
Answer:
An air conditioner uses 900 x 8 = <<9008=7200>>7200 watts for 8 hours a day.\nAn air conditioner uses 900 x 5 = <<9005=4500>>4500 watts for 5 hours a day.\nSo, Mel saves 7200 - 4500 = <<7200-4500=2700>>2700 watts per day.\nThat is 2700/1000 = <<2700/1000=2.7>>2.7 kilowatts per day since 1 kilowatt is equal to 1000 watts.\nHence, in 30 days he will have 2.7 x 30 = <<2.7*30=81>>81 kilowatts of electric energy saved.\n#### 81
The answer is incorrectly returning the energy he is USING after reducing by 5 hours per day (i.e. 3 hours), instead of the energy he is SAVING.
0.9 kW per hour 3 hours = 2.7 kWh 30 days = 81 kWh
The correct answer should be the 5 hours a day he is saving.
0.9 kW per hours 5 hours = 4.5 kWh 30 days = 135 kWh
Question: Mel uses a 900-watt air conditioner for 8 hours a day. This means that each hour the AC uses 900 watts of energy. If he reduces the time he uses the air conditioner by 5 hours a day, how many kilowatts of electric energy will he save in 30 days?
Answer: An air conditioner uses 900 x 8 = <<9008=7200>>7200 watts for 8 hours a day.\nAn air conditioner uses 900 x 5 = <<9005=4500>>4500 watts for 5 hours a day.\nSo, Mel saves 7200 - 4500 = <<7200-4500=2700>>2700 watts per day.\nThat is 2700/1000 = <<2700/1000=2.7>>2.7 kilowatts per day since 1 kilowatt is equal to 1000 watts.\nHence, in 30 days he will have 2.7 x 30 = <<2.7*30=81>>81 kilowatts of electric energy saved.\n#### 81
The answer is incorrectly returning the energy he is USING after reducing by 5 hours per day (i.e. 3 hours), instead of the energy he is SAVING. 0.9 kW per hour 3 hours = 2.7 kWh 30 days = 81 kWh
The correct answer should be the 5 hours a day he is saving. 0.9 kW per hours 5 hours = 4.5 kWh 30 days = 135 kWh