Rate Problems
Rate problems use a known rate to find an amount, a time or a cost, so a hose that gives \(12\) L/min fills \(180\) L in \(15\) minutes. In NSW Year 8 maths this is where rates meet everyday money and household decisions (NESA outcome MA4‑RAT‑C‑01).
You will choose the right way round for the rule, match hours and minutes to the rate, and find unit prices per kilogram or per \(100\) g to pick the best buy, including items on special. You will work out pay from clock times, add the rates of two pumps or pipes working together, use a bill with two rates in its formula, and decide how many years an efficient fridge takes to pay back its extra cost. It is part of Year 8 Rates and Ratios.
Theory
In a rate problem, write the rule first and then substitute: amount \(=\) rate \(\times\) quantity. An aircraft at \(800\) km/h flies \(800 \times \dfrac{6}{60} = 80\) km in \(6\) minutes. The units tell you whether the rule is the right way round. This Year 8 maths page shows how.
Rule first. Litres per minute \(\times\) minutes gives litres, and litres \(\div\) litres per minute gives minutes. If the units do not work out, the rule is upside down.
Match the time to the rate. A rate per hour needs hours: \(1\) h \(15\) min \(= 1.25\) h \(= 75\) min. Change a decimal back at the end: \(0.25\) h \(= 0.25 \times 60 = 15\) min. A time that is not a tidy decimal of an hour, such as \(20\) min, stays a fraction: \(\dfrac{1}{3}\) h.
No rate given? Work out the amount first (an area \(l \times w\), a circumference \(2\pi r\), a mass in kg), then divide by the time.
Unit price and best buy. The unit price is the price for one unit, or for a fixed amount such as \(100\) g. A \(250\) g pack is \(2.5\) lots of \(100\) g, so its price per \(100\) g is the price \(\div 2.5\). Use the same unit for every item, take off any special first, and round to the nearest cent. The lowest unit price is the best buy.
Working together, rates add. Two pumps at \(35\) L/min and \(20\) L/min remove \(55\) L/min, so \(13\,200\) L takes \(13\,200 \div 55 = 240\) min. “Fills it alone in \(8\) hours” means the rate is the amount \(\div 8\).
Two rates in one formula. Substitute both and add: a bill \(C = ad + bk\) has a charge for each day and a charge for each unit used.
Decisions. Compare unit prices or hourly rates, not totals. A pay-back time is the extra cost \(\div\) the saving each year, rounded up to a whole year.
Which rule?
| You want | Rule |
|---|---|
| the amount | rate \(\times\) quantity |
| the quantity | amount \(\div\) rate |
| the rate | amount \(\div\) quantity |
| the unit price | price \(\div\) quantity |
| the total cost | quantity \(\times\) unit price |
Solving a rate problem
- Write the rule with units, such as distance \(=\) speed \(\times\) time.
- Match units: change the time (or mass, or volume) to suit the rate.
- Substitute and calculate.
- Answer with units, changing decimal hours back to hours and minutes if needed.
Finding the best buy
- Change every size to the same unit, and take off any special.
- Divide price by quantity (or by the number of \(100\) g lots).
- Choose the lowest unit price.
The rate is per hour, so change the time to hours: \(6\) min \(= \dfrac{6}{60} = 0.1\) h.
| \(800 \times 0.1\) | \(=\) | \(80\) |
\(80\) km
| \(\text{Loss needed}\) | \(=\) | \(50\%\) |
| \(50 \div 2\) | \(=\) | \(25\) |
After \(25\) weeks.
(i) It runs from 5:50 pm to 10:10 pm. How many kilograms does it burn?
(ii) How long does a \(15\) kg bag last?
(i) \(4\) h \(20\) min \(= 4\dfrac{1}{3} = \dfrac{13}{3}\) h.
| \(2.4 \times 13 \div 3\) | \(=\) | \(10.4\) |
| \(15 \div 2.4\) | \(=\) | \(6.25\) |
| \(0.25 \times 60\) | \(=\) | \(15\) |
(i) \(10.4\) kg
(ii) \(6.25\) h \(= 6\) h \(15\) min
| \(3.60 \div 0.9\) | \(=\) | \(4.00\) |
| \(6.45 \div 1.5\) | \(=\) | \(4.30\) |
| \(4.30 - 4.00\) | \(=\) | \(0.30\) |
The box (\(\$4.00\)/kg against \(\$4.30\)/kg), by \(\$0.30\) per kg, even though the bag is bigger.
| \(4.40 \div 2.5\) | \(=\) | \(1.76\) |
| \(6.80 \div 4\) | \(=\) | \(1.70\) |
\(\$1.76\) and \(\$1.70\) per \(100\) g: the \(400\) g pack is the better buy.
(i) Find each rate in kg/min.
(ii) How long do they take together?
\(6\) t \(= 6000\) kg.
| \(6000 \div 40\) | \(=\) | \(150\) |
| \(6000 \div 24\) | \(=\) | \(250\) |
| \(6000 \div (150 + 250)\) | \(=\) | \(15\) |
(i) \(150\) and \(250\) kg/min
(ii) \(15\) min, less than either alone.
\(37\,400\) L \(= 37.4\) kL.
| \(0.95 \times 91\) | \(=\) | \(86.45\) |
| \(2.86 \times 37.4\) | \(=\) | \(106.964\) |
| \(86.45 + 106.964\) | \(=\) | \(193.414\) |
\(\$193.41\) (to the nearest cent).
(ii) A Ferris wheel of radius \(20\) m turns once every \(4\) minutes. How far does a seat travel each minute?
| \(240 \div 55\) | \(=\) | \(4.36\ldots\) |
| \(2 \times \pi \times 20 \div 4\) | \(=\) | \(31.41\ldots\) |
(i) \(5\) years (round up: after \(4\) years it has not quite paid back).
(ii) About \(31.4\) m/min (one circumference every \(4\) minutes).
Common pitfalls
Frequently asked questions
How do I know whether to multiply or divide by a rate?
Use the units: L/min × min gives L, and L ÷ L/min gives min.
How do I change minutes to a decimal of an hour?
Divide by 60. 45 minutes is 45 ÷ 60 = 0.75 hours.
How do I find the best buy?
Work out the price for the same amount of each item, such as per kilogram or per 100 g. The lowest unit price is the best buy.
How long do two pumps take together?
Add their rates, then divide the total amount by the combined rate.
What is a pay-back time?
The extra cost divided by the saving each year, rounded up to a whole number of years.