Rate

A rate is a ratio of two quantities measured in different units. The word "per" signals a rate.

Rate	Units
Speed	km per hour, m per second
Wage	dollars per hour, dollars per day
Flow	litres per minute, m³ per second
Population growth	people per year
Heart rate	beats per minute
Fuel efficiency	km per litre, miles per gallon

Computing a rate from two known quantities: $\text{rate} = \dfrac{\text{quantity}}{\text{time (or base unit)}}.$

Worked. "A car uses $42\,\text{L}$ of petrol to travel $588\,\text{km}$. Find the fuel efficiency in km per litre."

• $\dfrac{588}{42} = 14\,\text{km/L}$.

Using a rate to scale up or down: $\text{quantity} = \text{rate} \times \text{time}.$

Worked. "A printer prints at $24$ pages per minute. How many pages in $7$ minutes?"

• $24 \times 7 = 168$ pages.

Converting between rate units. Same trick as speed: convert each part separately.

• $20\,\text{L/min} = 20 \times 60 = 1200\,\text{L/h}$.
• $15 per hour: $15 \div 60 = 0.25$, so $0.25 per minute.

When two products come in different sizes/prices, compute the unit rate for each — usually "price per unit quantity" — and pick the smallest.

Worked. Pack A: $4.50 for $750\,\text{g}$. Pack B: $6.20 for $1\,\text{kg}$. Which is better value?

• A: $4.50 / 750 = 0.006$, so $0.006 per gram = $6.00 per kg.
• B: $6.20 / 1000 = 0.0062$, so $0.0062 per gram = $6.20 per kg.

A is cheaper per kg ($6.00 vs $6.20).

State the answer in context. Examiners want a sentence that names the cheaper pack AND quotes the unit price for both.

Reverse direction. Sometimes you compare "quantity per dollar" instead — the LARGER value wins. For Pack A: $750/4.50 = 166.7$ g per $. Pack B: $1000/6.20 = 161.3$ g per $. A again. The choice of direction doesn't matter as long as you stay consistent.

Flow problems give you a flow rate (e.g. $20\,\text{L/min}$) and ask either how much fills in a given time, or how long to fill a given volume.

Volume from rate and time: $V = R \times t$. Time from volume and rate: $t = V / R$.

Worked. A tap fills a $240\,\text{L}$ tank at $20\,\text{L/min}$. How long?

• $t = 240 / 20 = 12\,\text{min}$.

Two taps. When two taps fill the same tank, the combined rate is the SUM of the individual rates. (Same idea as workers in inverse-proportion problems.)

Worked. Tap A fills at $30\,\text{L/min}$, tap B at $20\,\text{L/min}$. Both running, how long to fill $300\,\text{L}$?

• Combined rate: $30 + 20 = 50\,\text{L/min}$.
• Time: $300 / 50 = 6\,\text{min}$.

Filling AND draining. If a drain operates at $10\,\text{L/min}$ alongside the two taps, net rate is $50 - 10 = 40\,\text{L/min}$. Time = $300 / 40 = 7.5\,\text{min}$.

Both ratios and rates use division — and it's easy to slip from one into the other. The distinction:

• Ratio — same units. e.g. "$50 to $30" simplifies to $5 : 3$. The unit cancels.
• Rate — different units. e.g. "$50 per $30\,\text{kg}$" is $1.67 per kg, NOT a ratio of $50 to 30.

A useful rule of thumb: if you can drop the unit safely, it's a ratio. If you'd lose meaning by dropping it, it's a rate.