Skip to main content
ExamExplained

Rates, unit conversions, fuel consumption, energy and heart rate: HSC Maths Standard 1 Year 12

Syllabus dot point

“Use rates to solve practical problems, including converting between units for rates (such as km/h to m/s), comparing rates, fuel consumption, energy use and costs, and heart rate and target heart rate”

HSCMaths Standard 1Year 12: Measurement7 min read

Quick answer

A rate compares quantities with different units. Convert each unit (km/h ÷ 3.6 = m/s), compare using a common unit, calculate fuel in L/100 km, energy in kWh (kW × hours) and its cost, and heart rate in bpm, using 220−age220 - \text{age} or the given formula for maximum and target heart rates.

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Practice questions

What this dot point is asking

Rates compare two quantities with different units. You need to convert rates between units, compare them to make decisions, and apply them in practical contexts NESA highlights: fuel consumption, energy use, and heart rate.

The answer

Converting rates

Convert each unit separately:

72 km/h=72×1000 m3600 s=20 m/s72 \text{ km/h} = \frac{72 \times 1000 \text{ m}}{3600 \text{ s}} = 20 \text{ m/s}

Speed conversions
  • km/h to m/s: divide by 3.6.
  • m/s to km/h: multiply by 3.6.
  • Distance = speed × time; time = distance ÷ speed.

Comparing rates (best buys)

Convert to the same unit (price per 100 g, cost per litre, cost per minute) before comparing.

Fuel consumption

Measured in litres per 100 km (L/100 km):

fuel consumption=litres useddistance (km)×100\text{fuel consumption} = \frac{\text{litres used}}{\text{distance (km)}} \times 100

Fuel needed =distance100×rate= \frac{\text{distance}}{100} \times \text{rate}.

Energy and power

  • Power is measured in watts (W) or kilowatts (kW); 1 kW = 1000 W.
  • Energy used (kWh) = power (kW) × time (h).
  • Cost = kWh × price per kWh.

Heart rate

  • Heart rate is measured in beats per minute (bpm): count beats for 15 seconds and multiply by 4, or 30 seconds and multiply by 2.
  • A common estimate of maximum heart rate is 220−age220 - \text{age}; a target heart rate range for exercise is a percentage of the maximum (for example 50% to 85%). Use the formula given in the question.
Worked example

A 17-year-old wants to exercise at 60% to 80% of maximum heart rate (using 220−age220 - \text{age}).

  1. Maximum =220−17=203= 220 - 17 = 203 bpm.
  2. Lower =0.6×203=121.8= 0.6 \times 203 = 121.8, about 122 bpm.
  3. Upper =0.8×203=162.4= 0.8 \times 203 = 162.4, about 162 bpm.
  4. Target range: about 122 to 162 bpm.
Common traps

Converting only one unit of a rate (for example km to m but not h to s).

Mixing watts and kilowatts. Convert to kW before finding kWh.

Comparing prices without a common unit.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marks
A cyclist travels 18 km in 45 minutes. Find the average speed in km/h and in m/s.
Show worked solution →

45 minutes = 0.75 h, so speed =180.75=24= \frac{18}{0.75} = 24 km/h.

In m/s: 24×1000÷3600=6.6724 \times 1000 \div 3600 = 6.67 m/s.

Marking guide: 1 mark for converting time, 1 mark for km/h, 1 mark for m/s.

core4 marks
A family car uses fuel at 8.5 L/100 km. They drive 640 km on holiday and fuel costs 1.95 dollars per litre. Find the fuel used and its cost.
Show worked solution →

Fuel used =640100×8.5=54.4= \frac{640}{100} \times 8.5 = 54.4 L.

Cost =54.4×$1.95=$106.08= 54.4 \times \$1.95 = \$106.08.

Marking guide: 2 marks for fuel used, 2 marks for cost.

exam5 marks
A household's reverse-cycle air conditioner is rated at 2.4 kW and runs 6 hours a day for 90 days of summer. Electricity costs 32 cents per kWh. (a) Find the energy used and the cost. (b) A newer model rated at 1.8 kW costs 1100 dollars. How many summers would its savings take to pay for it?
Show worked solution →

(a) Energy =2.4×6×90=1296= 2.4 \times 6 \times 90 = 1296 kWh. Cost =1296×$0.32=$414.72= 1296 \times \$0.32 = \$414.72.

(b) New model energy =1.8×6×90=972= 1.8 \times 6 \times 90 = 972 kWh, costing 972×0.32=$311.04972 \times 0.32 = \$311.04.

Saving per summer =414.72−311.04=$103.68= 414.72 - 311.04 = \$103.68.

Payback =1100103.68=10.6= \frac{1100}{103.68} = 10.6, so about 11 summers (assuming the same usage and price).

Marking guide: 2 marks for (a), 1 mark for the new cost, 1 mark for the saving, 1 mark for the payback with an assumption.

Practise this

Sources & how we know this

Based on the official NESA syllabus. Not yet individually reviewed by a person; always check NESA and your school's materials for current rules.

ExamExplained