2.04 Applying rates

Applying rates

Now that we know how to make our rates, it's time to use them. Rates are very similar to ratios in that we can use them to calculate how much one measurement changes based on the change in another.

Returning to our sprinter, we found that they could run at a speed of $10$10 m/s. Assuming that they can maintain this speed, how far will the sprinter run in $15$15 seconds?

One way to solve this is to treat the rate like a ratio and multiply the top and bottom of the fraction by $15$15. This will give us:

Speed $=$= $10$10 m/s $\times$×$\frac{15}{15}$1515​ $=$= $150$150m/$15$15s

By turning the rate back into a fraction we can see that the sprinter will run $150$150 metres in $15$15 seconds.

Another way to solve this problem is to apply the rate directly to the question. We can do this by multiplying the time by the rate. This gives us:

Distance $=$= $15$15s $\times$× $10$10m/s $=$= $\left(15\times10\right)$(15×10) m$\times$×s/s $=$= $150$150m

Notice that the units for seconds from the time cancelled with the units for second in the compound unit leaving only metres as the unit for distance.

We can also ask the similar question, how long will it take for the sprinter to run $220$220 metres?

Again, one way to solve this is to treat the rate like a ratio and multiply the top and bottom of the fraction by $22$22. This gives us:

Speed $=$= $10$10 m/s $\times$×$\frac{22}{22}$2222​ $=$= $220$220m/$22$22s

By turning the rate back into a fraction we can see that the sprinter will take $22$22 seconds to run $220$220 metres.

Another way to solve this problem is to apply the rate directly to the question. We can do this by dividing the distance by the rate. This gives us:

Time $=$= $220$220m $\div$÷​ $10$10 m/s $=$= $\frac{220}{10}$22010​ s$\times$×m/m = $22$22s

This time we divided by the rate so that the compound unit would be flipped and the metres units would cancel out to leave only seconds as the unit for time.

 

 

Practice question

Question 1

 

Unit rates

Everyone loves a great deal or a sale when they're shopping! However, most products today come in different varieties or quantities and are sold at more than one store. To determine which item is better value, we need to find a common amount to compare. Often it's easiest to find the unit price of each item, where we find the cost per unit of measurement, for example per litre, per kilogram or per item. We can also compare the amount per dollar, where we work out how much of something we would get for one dollar.

 

Amount per unit

Are we better off paying $\$10.50$$10.50 for $3$3 kg of apples or $\$6.20$$6.20 for $2$2 kg of apples? An easy way to compare the two options is to find the price per kilogram for each option.

So we're better off paying $\$6.20$$6.20 for $2$2 kg of apples because it's a cheaper price per kilogram.

 

Amount per dollar

If David's deli sells $620$620 grams of salmon for $\$18$$18 and Fred's fish market sells $460$460 grams of salmon for $\$13$$13, and if the fish is of similar quality, which is better value? Let's see how much salmon we would get for one dollar at each shop.

David's:

Amount per dollar	$=$=	$\frac{620}{18}$62018​ grams per dollar	Expressing the rate as the amount of grams per dollar
	$=$=	$34.\overline{4}$34.4 grams per dollar	Expressing the rate as a decimal

So at David's, we'd get approximately $34.4$34.4 grams of salmon for a dollar.

Fred's:

Amount per dollar	$=$=	$\frac{460}{13}$46013​ grams per dollar	Expressing the rate as the amount of grams per dollar
	$\approx$≈	$35.4$35.4 grams per dollar	Expressing the rate as a decimal

We get approximately $35.4$35.4 grams of salmon at Fred's, which is slightly better value than at David's.

 

Worked example

example 1

At a growers' market, durians are sold at stand A for $80c$80c per kilogram. At stand B, each $4.5$4.5 kg durian is sold for $\$4.00$$4.00.

a) Calculate the price of $4.5$4.5 kg of durian from store A. Round your answer to two decimal places.

Think: How do we change the price per kilogram to work out the price for $4.5$4.5 kg?

Do: $4.5\times0.8=\$3.60$4.5×0.8=$3.60

b) Which store has the best price?

Think: In other words, at which store would you pay less for $4.5$4.5 kg of durian?

Do: Choose Store A, as you pay less for a $4.5$4.5 kg in this store.

c) What should the owner of store C charge for his $4.5$4.5 kg durians if he wants to beat the best price by $6%$6%? Round your answer to two decimal places.

Think: This means he wants to charge $6%$6% less that the best price or $94%$94% of the lowest price.

Do:

$0.94\times3.60$0.94×3.60	$=$=	$3.384$3.384	Finding $94%$94% of the price of $4.5$4.5 kg of durian
	$=$=	$\$3.38$$3.38 (to $2$2 d.p.)

 

So the owner of store C should charge $\$3.38$$3.38 for his $4.5$4.5 kg durians.

 

Practice questions

Question 2

question 3

question 4