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Measurement, STD2 M7 2025 HSC 21

A house has a reverse-cycle air conditioner which uses 2.5 kW of power for cooling and 3.2 kW of power for heating. The cost of electricity is 29 cents per kWh .

  1. Find the cost, in dollars and cents, of cooling the house for 6 hours.   (1 mark)

  2. The cost of operating the air conditioner to heat the house during winter last year was $640. There are 92 days in winter.
  3. Find the number of hours, to 1 decimal place, that the air conditioner was used on average per day.   (2 marks)

Show Answers Only

a.    \(\text{Cooling cost (6 hours )}\ = 6 \times 2.5 \times 0.29 = \$4.35\)

b.    \(7.5\ \text{hours}\)

Show Worked Solution

a.    \(\text{Cooling cost (6 hours )}\ = 6 \times 2.5 \times 0.29 = \$4.35\)
 

b.    \(\text{Let \(h\) = hours used per day}\)

\(\text{Cost per day}\ = h \times 3.2 \times 0.29\)

\(\text{Cost (92 days )}\ = 92 \times h \times 3.2 \times 0.29\)

\(\text{Find \(h\) when cost = \$640:}\)

\(640\) \(=92 \times h \times 3.2 \times 0.29\)  
\(h\) \(=\dfrac{640}{92 \times 3.2 \times 0.29}=7.5\ \text{hours (1 d.p.)}\)  

v1 Measurement, STD2 M1 2015 HSC 30a

A school projector left in idle mode consumes 95 watts of power. The cost of electricity is 28 cents per kWh.

There are 12 projectors in the school, each idle for 6 hours a day, 5 days a week.

How much does it cost the school to leave all the projectors idle for a 10-week school term? (2 marks)

Show Answers Only

`$95.76`

Show Worked Solution

`text(95 watts per projector)`

`text(Hours per term:) \ 6 xx 5 xx 10 = 300`

`text(Total watt-hours)` `=95 xx 12 xx 300`
  `= 342\ 000`
  `= 342\ text(kWh)`

 

`:.\ \text{Cost}` `=342 xx $0.28`
  `= \ $95.76`

v1 Measurement, STD2 M1 2013 HSC 26d

A section of Jim’s electricity bill is shown.

  1. What is the value of `X`?    (1 mark)
  2. How much will Jim save if he uses 523.5 kWh of energy at the Off-peak rate rather than at the Shoulder rate?   (2 marks)
Show Answers Only

a.    `531.2`

b.    `$51.30`

Show Worked Solution
a.     `X` `= \text{Last Reading} + \text{Energy Used}`
    `= 274.8+256.4`
    `= 531.2`

 

b.     `text(Cost)_{text(Shoulder)}` `= 523.5 × 19.4 = 10155.9 \text{ cents}`
    `= \$101.56`
  `text(Cost)_{text(Offpeak)}` `= 523.5 × 9.6 = 5025.6 \text{ cents}`
    `= \$50.26`
  `text(Saving)` `= 101.56-50.26 = \$51.30`

v1 Measurement, STD2 M7 2021 HSC 27

The price and the power consumption of two different models of air purifiers are shown.

\[ \begin{array}{|l|l|} \hline \text{Air Purifier X} & \text{Air Purifier Y} \\ \hline \text{Price: \$480} & \text{Price: \$462.40} \\ \hline \text{Power: 95 W} & \text{Power: 88 W} \\ \hline \end{array} \]

The average cost for electricity is 30c/kWh. A household runs an air purifier for an average of 10 hours a day.

  1. The annual cost of electricity for Air Purifier X for this household is \$104.03.
  2. For this household, what is the difference in the annual cost of electricity between Air Purifier X and Air Purifier Y? (2 marks)
  3. For this household, how many years will it take for the total cost of buying and using Air Purifier X to be equal to the cost of buying and using Air Purifier Y? (2 marks)
Show Answers Only
  1. `$7.66`
  2. `2.3\ \text{years}`
Show Worked Solution
a. `text{Annual power usage (Y)}` `= 88 \times 10 \times 365 = 321200\ \text{Wh}`
    `= 321.2\ \text{kWh}`
  `text{Annual cost (Y)}` `= 321.2 \times 0.30 = \$96.36`
  `text{Annual cost (X)}` `= 95 \times 10 \times 365 / 1000 \times 0.30 = \$104.03`
  `text{Difference}` `= 104.03-96.36 = \$7.67`
b. `text{Price difference}` `= 480-462.40 = \$17.60`
  `text{Years to equal total cost}` `= 17.60 / 7.67 ≈ 2.3\ \text{years}`

v1 Measurement, STD2 M7 2018 HSC 28c

A 900-watt air purifier runs for 2 hours per day at 60% power. Electricity costs $0.30 per kWh.

What is the total cost of running the air purifier for 150 days? (3 marks)

Show Answers Only

`$48.60`

Show Worked Solution
`text(Daily usage)` `= 900 xx 2 xx 60text(%)`
  `= 1080\ \text{Wh}`

 

`text(Total usage over 150 days)` `= 150 xx 1080`
  `=162 \ 000\ \text{Wh}`
  `= 162\ \text{kWh}`

 

`∴ \  text(Cost)` `= 162 xx 0.30`
  `= $48.60`

v1 Measurement, STD2 M1 2014 HSC 20 MC

In a household of 5, each member uses an average of 8 minutes of hot water per day.

The household uses a 7.5 kW hot water unit.

Electricity is charged at 30.2 c/kWh when the hot water unit is being used.

What is the electricity cost for the hot water used by this household in one week?

  1. $7.92
  2. $9.84
  3. $10.57
  4. $11.32
Show Answers Only

`C`

Show Worked Solution

`text(Usage per day) = 5 xx 8 = 40\ text(mins)`

`text(Usage per week) = 7 xx 40 = 280\ text(mins)`

`text(Convert minutes to hours) = 280/60 = 4.67\ \text{hours}`

`text(Energy used in kWh) = 4.67 xx 7.5 = 35.0\ \text{kWh}`

`text(Cost in cents) = 35.0 xx 30.2 = 1057`

`= 1057\text(¢)`

`= $10.57`

`=> C`

Measurement, STD2 M7 2024 HSC 17

The cost of electricity is 30.13 cents per kWh .

Calculate the cost of using a 650 W air conditioner for 6 hours.   (2 marks)

Show Answers Only

\(\text{Cost} =\$ 1.18\)

Show Worked Solution

\(\text{Usage}=6 \times 650=3900\, \text{Wh}=3.9\, \text{kWh}\)

\(\text{Cost} =3.9 \times 30.13=117.507 \,\text{c}=\$ 1.18 \, \text {(nearest cent)}\)

Measurement, STD2 M7 2021 HSC 27

The price and the power consumption of two different brands of television are shown.

The average cost for electricity is 25c/kWh. A particular family watches an average of 3 hours of television per day.

  1. The annual cost of electricity for Television A for this family is $48.18.
  2. For this family, what is the difference in the annual cost of electricity between Television A and Television B?   (2 marks)

  3. For this family, how many years will it take for the total cost of buying and using Television A to be equal to the cost of buying and using Television B?   (2 marks)

Show Answers Only

a.    `$4.38`

b.    `5\ text(years)`

Show Worked Solution
a.     `text{Annual power usage (B)}` `= 160 xx 3 xx 365`
    `=175\ 200`
    `=175.2\ text(kWh)`

♦ Mean mark part (a) 49%.

`text{Annual cost (B)}= 175.2 xx 0.25=$43.80`

`text{Difference in cost}= 48.18-43.80=$4.38`
  

♦♦ Mean mark part (b) 23%.

b.    `text{Difference in price}= 921.90-900=$21.90`

`text(Years to even out cost)=21.90/4.38=5\ text{years}`

Measurement, STD2 M7 2018 HSC 28c

Every day, a 1200-watt microwave oven is used for 45 minutes at 40% power. Electricity is charged at $0.25 per kWh.

What is the cost of running this microwave oven for 180 days?   (3 marks)

Show Answers Only

`$16.20`

Show Worked Solution

`text(Daily usage)= 1200 xx 45/60 xx 40text(%)= 360\ text(watts)`

 

`text(180 day usage)` `= 180 xx 360`
  `= 64\ 800\ text(watt hours)= 64.8\ text(kWh)`

  
`:.\ text(C)text(ost over 180 days)= 64.8 xx 0.25= $16.20`

Measurement, STD2 M7 EQ-Bank 32

Bikram runs a hot yoga studio.

If it costs 34 cents for 1-kilowatt (1000 watts) for 1 hour, how much does it cost him to run three 3200-watt heaters from 9:00 am to 12:30 pm on a single day? (Give your answer to the nearest cent)   (2 marks)

Show Answers Only

`$11.42`

Show Worked Solution

`text(Total energy usage)`

`= 3 xx 3200 xx 3.5\ text(hours)= 33\ 600\ text(Wh)=33.6\ text(kWh)`

  `:.\ text(C)text(ost)= (33.6 xx 0.34= 11.424= $11.42\ \ text{(nearest cent)}`

Measurement, STD2 M1 SM-Bank 3

A 250-watt television is turned on for an average of 4 hours per day during off-peak periods for a week.

If the television is not running at any other time and electricity is charged at $0.36/kWh during off-peak, how much does it cost to run the television for a week?  (2 marks)

Show Answers Only

`$2.52`

Show Worked Solution

`text(Total electricity used)`

`= 7\ text(days × 4 hours × 250)`

`= 7000\ text(watt hours)`

`= 7.0\ text(kWh)`
 

`:.\  text(Running Cost)` `= 7.0 xx 0.36`
  `= $2.52`

Measurement, STD2 M1 2016 HSC 28b

The cost of buying a new heater is $990. It has an energy consumption of 505 kWh per year.

Energy is charged at the rate of $0.35 kWh.

How much will it cost in total to purchase and then run this heater for five years?   (2 marks)

Show Answers Only

`$1873.75`

Show Worked Solution

`text(C)text(ost to run heater for 5 years)`

`= 5 xx 505 xx 0.35= $883.75`
 

`:.\ text(Total purchase and running cost)`

`= 883.75 + 990= $1873.75`

Measurement, STD2 M1 2015 HSC 30a

The energy consumption of a computer in standby mode is 21 watts. The cost of electricity is 31 cents per kWh.

A school computer room has 20 computers.

How much will the school save by switching off all 20 computers during 11 weeks of school holidays?   (2 marks)

Show Answers Only

`$240.61`

Show Worked Solution

`text(21 watts usage per computer per hour.)`

♦ Mean mark 49%.

`text(Watts used by 20 computers in 11 weeks)`

`= 21 xx 20 xx 24 xx 7 xx 11`

`= 776\ 160\ text(Wh)= 776.16\ text(kWh)` 
  

`:.\ text(C)text(ost of energy)`

`= 776.16 xx $0.31`

`= $240.6096= $240.61\ \ text{(nearest cent)}`
  

`:.\ text(The school will save $240.61 by switching)`

`text(off all the computers.)`

Measurement, STD2 M1 2014 HSC 20 MC

In a household of 4, each member uses an average of 13 minutes of hot water per day.

The household uses a 9 kW hot water unit.

Electricity is charged at 11.97 c/kWh when the hot water unit is being used.

What is the electricity cost for the hot water used by this household in one week?

  1. $1.63
  2. $6.54
  3. $392.14
  4. $653.56
Show Answers Only

`B`

Show Worked Solution

`text(Usage per day) = 4 xx 13 = 52\ text(mins)`

♦ Mean mark 39%.

`text(Usage per week) = 7 xx 52 = 364\ text(mins)`

`text(Converting to kWh)`

`= text{(hours of usage)} xx 9\ text(kW)`

`= 364/60 xx 9 = 54.6\ text(kWh)`
 

`:.\ text(C)text(ost)` `= 54.6 xx 11.97 text(c)`
  `~~ 654 text(c)=$6.54`

`=>  B`

Measurement, STD2 M1 2013 HSC 26d

A section of Jim’s electricity bill is shown.

2013 26d

  1. What is the value of `A`?   (1 mark)

  2. How much will Jim save if he uses 154 kWh of energy at the Off-peak rate rather than at the Peak rate?   (2 marks)

Show Answers Only

a.    `1084.4`

b.    `$ 58.78\ \ \ (text(nearest cent) )`

Show Worked Solution
a.     `A` `=\ text(Last reading + Energy used)`
    `= 560.9 + 523.5= 1084.4`
♦ Mean mark
part (a) 43%
part (b) 46%.

 

b.     `text(C)text(ost at off-peak)` `= 154 xx 9.6`
    `= 1478.4\ text(cents)`

 

`text(C)text(ost at peak)` `=154 xx 47.77`
  `= 7356.58\ text(cents)`

 

`:.\ text(Saving)` `=7356.58-1478.4`
  `=5878.18= $58.78\ text{(nearest cent)}`