Year 12: Measurement

NSWMaths Standard 2Syllabus dot point

How are rates used to solve practical problems involving fuel consumption, energy use and dosage, and how do we convert between units?

Use rates and unit conversions to solve practical problems including fuel consumption, dosage, power consumption and energy efficiency

A focused answer to the HSC Maths Standard 2 dot point on rates and unit conversions. Definition of a rate, the unitary method, converting between SI units, fuel consumption (L per 100 km), energy use (kWh), and dosage calculations with worked Australian examples.

Generated by Claude OpusReviewed by Better Tuition Academy8 min answer

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What this dot point is asking

NESA wants you to compute rates, use the unitary method to solve worded rate problems, convert between metric units, and apply rates to everyday Australian contexts including fuel consumption, energy use and medication dosages.

The answer

What a rate is

A rate compares two quantities with different units, expressed per single unit of the denominator.

  • IMATH_5 km/h: a speed.
  • IMATH_6 L per 100100 km: fuel consumption.
  • IMATH_8 1.92$/L: unit cost.
  • IMATH_9 mg per kg per day: a dosage rate.

A rate is a fraction with units on both top and bottom.

The unitary method

To solve "if aa produces bb, how much does cc produce", scale by the ratio ca\frac{c}{a}:

answer=b×ca.\text{answer} = b \times \frac{c}{a}.

Equivalently, find the value for 11 unit, then multiply by cc.

Example: if 55 kg of meat costs \84,then, then 1kgcosts kg costs \\frac{84}{5} = \16.80,and, and 8kgcosts kg costs \16.80 \times 8 = \134.40$.

Metric conversions

From To Multiply by
km m IMATH_24
m cm IMATH_25
cm mm IMATH_26
kg g IMATH_27
L mL IMATH_28
h min IMATH_29
min s IMATH_30

Divide to go the other way. For area, the factor is squared (1 m2^2 = 1000010000 cm2^2); for volume it is cubed (1 m3^3 = 10001000 L).

Fuel consumption

Standard Australian unit: litres per 100100 km. Smaller is better.

To find fuel used for a trip of dd km at consumption cc L per 100100 km:

fuel=c100×d.\text{fuel} = \frac{c}{100} \times d.

Then cost is fuel times price per litre.

Energy use (kWh)

Household electricity is billed in kilowatt-hours (kWh). One kWh is the energy used by a 10001000 watt appliance running for 11 hour.

For an appliance rated PP watts running tt hours:

E=P×t1000 kWh.E = \frac{P \times t}{1000} \text{ kWh}.

Cost is energy times price per kWh.

Dosage calculations

For weight-based dosing: total dose = (mg per kg) ×\times patient weight (kg).

Number of tablets or doses: total dose divided by strength per tablet.

Always sanity-check: if a patient ends up taking 100100 tablets a day, you have made a unit error.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2023 HSC Q143 marksA car uses fuel at a rate of 8.58.5 L per 100100 km. Petrol costs \1.92perlitre.Findthefuelcostforatripof per litre. Find the fuel cost for a trip of 620$ km.
Show worked answer →

Fuel used: 8.5100×620=0.085×620=52.7\frac{8.5}{100} \times 620 = 0.085 \times 620 = 52.7 L.

Cost: 52.7 \times 1.92 = \101.18$ (round to cents).

Markers reward the fuel-used calculation, the cost calculation, and the answer rounded to cents with the dollar sign.

2021 HSC Q183 marksA patient is prescribed 400400 mg of medication per kg of body weight per day. The patient weighs 6565 kg. The medication comes in 250250 mg tablets. How many tablets per day?
Show worked answer →

Daily dose: 400 mg/kg×65 kg=26000400 \text{ mg/kg} \times 65 \text{ kg} = 26000 mg/day.

Convert to tablets: 26000250=104\frac{26000}{250} = 104 tablets per day.

That is implausibly many, so the question likely intended 44 mg per kg, giving 260260 mg per day and roughly 11 tablet per day. In the exam, work the maths as stated and add a brief flag if the result is implausible. Markers reward correct unit-by-unit computation and an answer to the right precision.

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