SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

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
  1. `531.2`
  2. `$51.30`
Show Worked Solution
i.     `X` `= \text{Last Reading} + \text{Energy Used}`
    `= 274.8+256.4`
    `= 531.2`

 

ii.     `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 = \$7.66`
b. `text{Price difference}` `= 480-462.40 = \$17.60`
  `text{Years to equal total cost}` `= 17.60 / 7.66 ≈ 2.3\ \text{years}`

v1 Measurement, STD2 M1 SM-Bank 7

Ben and Lily select a meal each from the table below.

 
Ben chooses BBQ Chicken and Lily selects Butter Chicken with Steamed Rice.

After eating, Ben cycles and expends energy at 22 kJ/minute.

Lily does yoga and expends 10 kJ/minute.

Who will burn off their meal faster, and by how many minutes? (3 marks)

Show Answers Only

`text(Ben will burn off his meal)`

`text(in 141 minutes less than Lily.)`

Show Worked Solution

`text(Ben’s energy intake = 3120)`

`text(Time to work off)` `= 3120/22`
  `≈ 142\ text(minutes)`

 

`text(Lily’s energy intake = 2827)`

`text(Time to work off)` `= 2827/10`
  `= 283\ text(minutes)`

`283-142 = 141\ text(minutes)`

`:.\ text(Ben will burn off his meal)`

`text(in 141 minutes less than Lily.)`

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{watts}`

 

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

 

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

v1 Measurement, STD2 M1 SM-Bank 8

A cannonball is made out of lead and has a diameter of 18 cm.

  1. Find the volume of the sphere in cubic centimetres (correct to 1 decimal place). (2 marks)
  2. It is known that the mass of lead is 11.3 tonnes/m³. Use this information to find the mass of the cannonball to the nearest gram. (2 marks)
Show Answers Only
  1. `3053.6\ \text{cm³  (1 d.p.)}`
  2. `34\ 506\ \text{grams}`
Show Worked Solution

i.     `text(Radius) = 18 / 2 = 9\ \text{cm}`

`text(Volume)` `= \frac{4}{3} \pi r^3`
  `= \frac{4}{3} \pi (9)^3`
  `= \frac{4}{3} \pi (729) = 3053.628…`
  `= 3053.6\ \text{cm³ (1 d.p.)}`

 

ii.     `text(1 m³ = 1,000,000 cm³)`

`11.3\ \text{tonnes} = 11,300\ \text{kg} = 11,300,000\ \text{grams}`

`text(Mass of cannon ball)`

`= 3053.6 \times ({11,300,000} / {1,000,000})`

`= 3053.6 \times 11.3 = 34,505.68…`

`= 34,506\ \text{grams}`

v1 Measurement, STD2 M1 2019 HSC 24

Ben uses 45 kilocalories of energy per kilometre when cycling.

He drinks a smoothie that contains 1260 kilojoules of energy. How many kilometres will he need to cycle to use up all the energy from the smoothie? Give your answer correct to one decimal place. (1 kilocalorie = 4.184 kilojoules)  (2 marks)

Show Answers Only

`6.7\ \text{km (1 d.p.)}`

Show Worked Solution
`text(Kilocalories in smoothie)` `= 1260 / 4.184`
  `= 301.252…`

 
`:.\ \text(Kilometres required to cycle)`

`= 301.252 / 45`

`= 6.694…`

`= 6.7\ \text{km (1 d.p.)}`

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`

v1 Measurement, STD2 M1 SM-Bank 13 MC

The table shows the average energy used, in kilojoules per kilogram of body mass, by a person running on a treadmill for 20 minutes at different speeds.

\begin{array} {|c|c|} \hline \rule{0pt}{2.5ex} \text{Running speed} \rule[-1ex]{0pt}{0pt} & \text{Energy used in 20 minutes} \\ \hline \rule{0pt}{2.5ex} \text{6 km/h} \rule[-1ex]{0pt}{0pt} & \text{6.2 kJ/kg} \\ \hline \rule{0pt}{2.5ex} \text{9 km/h} \rule[-1ex]{0pt}{0pt} & \text{9.8 kJ/kg} \\ \hline \end{array}

Jess weighs 72 kg and eats a muffin containing 305 kilocalories.

For approximately how long must Jess run at 9 km/hr to burn off the energy from the muffin? (1 kilocalorie = 4.184 kilojoules)

  1. 29 minutes
  2. 32 minutes
  3. 36 minutes
  4. 45 minutes
Show Answers Only

`text(C)`

Show Worked Solution

`text{Kilojoules in muffin}`

`= 305 xx 4.184`

`= 1276.72 \text{ kJ}`

`text{Energy used by Jess in 20 minutes at 9 km/h}`

`= 72 xx 9.8`

`= 705.6 \text{ kJ}`

`:.\ text(Time of the walk)`

`= 1276.72/(705.6) xx 20`

`= 36.1…\ text(minutes)`