4.06 Applications of Ratios

Ratios tell us about the relative sizes of two or more values. They are often used in everyday life, whether it's for dividing up money, betting odds, cooking, or mixing cement. So knowing how to apply knowledge about ratios can be handy. Almost all ratio problems can be solved using what's known as the unitary method.

The unitary method is named for the key step in which we find one part (one unit) of the whole amount. From there we can find the value of any number of parts.

Worked examples

Example 1

Amir and Keira shared $\$720$$720 in the ratio $4:5$4:5. How much did each person get?

Think: There are $4+5=9$4+5=9 parts in total, so we can find one part by dividing $\$720$$720 by $9$9 parts to get $\$80$$80. We can now use the knowledge that Amir gets $4$4 parts and Keira gets $5$5 parts to find each share of the money.

Do:

Amir's share	$=$=	$4\times\$80$4×$80
	$=$=	$\$320$$320
Keira's share	$=$=	$5\times\$80$5×$80
	$=$=	$\$400$$400

Check: The total of Amir's share and Keira's share should sum to the total amount:

$\$320+\$400=\$720$$320+$400=$720

Reflect: We can also solve problems like this using a variation of the unitary method known as the fraction method. Since we know there are $9$9 parts in total, and Amir gets $4$4 parts and Keira gets $5$5 parts, then Amir will get $\frac{4}{9}$49​ of the total and Keira will get $\frac{5}{9}$59​ of the total.

Amir's share	$=$=	$\frac{4}{9}\times\$720$49​×$720
	$=$=	$\$320$$320
Keira's share	$=$=	$\frac{5}{9}\times\$720$59​×$720
	$=$=	$\$400$$400

Notice that multiplying $720$720 by $\frac{5}{9}$59​ is effectively the same as dividing it by $9$9 (the total number of parts) and then multiplying it by $5$5 (the number of parts we want to find).

Example 2

Divide $60$60 cm in the ratio $1:2:3$1:2:3.

Think: First we find the length of one part. There are $1+2+3=6$1+2+3=6 parts in total, and $60\div6$60÷​6 is $10$10. This means that one part is $10$10 cm in length.

Do: $60$60 cm divided in the ratio $1:2:3$1:2:3 would give one length of $10$10 cm, another length of $20$20 cm, and a final length of $30$30 cm. The total of these three lengths is $10+20+30=60$10+20+30=60 cm, as expected.

 

An application: population estimation with observational study

A commonly used method of estimating the population of a species is capture-recapture. As the name suggests, this method involves capturing individuals, releasing them, then capturing them again. Simple, right? In practice, it is slightly more complicated, and is based on ratios.

The idea is to capture some individuals, mark them in some way so you know you have already captured them, then release them back into the environment. After a little time has passed, we do the same thing again but this time we compare the number of marked individuals to the number captured. With this information we can estimate the total population!

Seems a bit of a stretch, doesn't it? Actually it boils down to a single ratio: the number of individuals we capture in one go compared to the total population. Let's introduce some notation: let $n$n be the number in the first capture, $N$N be the total population, $k$k be the number of marked individuals in the second capture and $K$K the total number in the second capture. Let's say that during the first capture we mark a certain proportion of the total population, this proportion would be $n:N$n:N. During the second capture, given random conditions and no bias in the selection of individuals, we would expect the ratio $k:K$k:K of marked individuals to the number in the capture to be the same as $n:N$n:N. That's it! with this assumption we can form an equation and solve for $N$N.

$n:N$n:N	$=$=	$k:K$k:K
$\frac{N}{n}$Nn​	$=$=	$\frac{K}{k}$Kk​
$N$N	$=$=	$\frac{nK}{k}$nKk​

 

Although this method is quite straight-forward, in practice there are some drawbacks:

• It depends on the mobility of individuals. For example, it doesn't work with plants or other stationary organisms. Also, If your sampling area is too small, marked individuals may range outside of it and affect your result
• For it to be effective you must do the second capture shortly after the first, to avoid the possibility of individuals dying or seasonal change etc, but this also reduces the probability that the population has adequately mixed back together.
• It actually estimates the population in the sampling area, which can then be extrapolated to a larger area. This is often reasonable but one area does not always reflect the population density of a more general area.

That said, there are ways around all these issues and ecologists still use this method every day. One way to get more accurate and reliable results is to do three or more captures, comparing the ratio of marked individuals multiple times. Can you think of any other solutions to the drawbacks above?

 

Worked example

Example 3

Idat wants to estimate the total population of skinks in his backyard. He performs a capture-recapture, marking 12 individuals in the first capture and finding 3 marked out of the 9 individuals in the second capture. What is the total population?

Think: We have three of the variables used in the capture-recapture formula, the number in the first capture $n$n, the number of marked individuals in the second capture $k$k, and the number in the second capture $K$K.

Do: Assign the variables correctly then, using the formula, substitute and evaluate $N$N.

$N$N	$=$=	$\frac{nK}{k}$nKk​
	$=$=	$\frac{12\times9}{3}$12×93​
	$=$=	$36$36

 

Reflect: Would this number be normal of any backyard? What sort of factors could affect the population of skinks in a particular backyard?

Practice questions

Question 1

Question 2

Question 3