5.02 Equivalent ratios and dividing quantities

We previously learnt how to write ratios that match the information given to us. Writing a ratio can let us compare things mathematically but has only limited use in solving problems. We can build upon this with the use of equivalent ratios and simplified ratios.

 

Exploration

Consider a cake recipe that uses $1$1 cup of milk and $4$4 cups of flour. What is the ratio of milk to flour used in the cake?

Letting the unit be "the number of cups", we can express the information given as the ratio $1:4$1:4.

What if we want to make two cakes? We would need to double the amount of milk and flour we use. This means we will need $2$2 cups of milk and $8$8 cups of flour. Now the ratio of milk to flour is $2:8$2:8.

But how do we get two different ratios from the same recipe? The secret is that the two ratios actually represent the same proportion of milk to flour. We say that $1:4$1:4 and $2:8$2:8 are equivalent ratios.

Now consider if we wanted to make enough cakes to use up $4$4 cups of milk. How many cakes would this make, and how much flour would we need?

 

Equivalent ratios

Equivalent ratios are useful for when we want to change the value of one quantity but also keep it in the same proportion to another quantity. After calculating how much the value of the first quantity has increased, we can increase the value of the second quantity by the same multiple to preserve the ratio.

We saw in the cake example that increasing both the amount of milk and the amount of flour by the same multiple preserved the ratio. That's because this is the same as having multiple sets of the same ratio.

 

And since this is an equivalence relation, we can also say the same for the reverse:

 

 

Worked examples

Example 1

The ratio of tables to chairs is $1:2$1:2. If there are $14$14 chairs, how many tables are there?

Think: The ratio $1:2$1:2 says that each table has two chairs. We want to increase both sides of this ratio by some multiple to get the equivalent ratio that looks like $\editable{?}:14$?:14. The missing number in this ratio will be the number of tables needed for $14$14 chairs.

Do: The first thing we can figure out is by what multiple the chairs have been increased. We can find this by dividing $14$14 by $2$2. This tells us how many sets of two chairs there are:

Number of pairs of chairs	$=$=	$14\div2$14÷​2
	$=$=	$7$7

If there are $7$7 sets of two chairs, and each table has two chairs, then we will need $7\times1=7$7×1=7 tables.

 

Example 2

The ratio of players to teams is $60:10$60:10. If there are only $12$12 students present, how many teams can be made?

Think: We only have $12$12 out of $60$60 students. What fraction of the students are present? To preserve the ratio, we want to take the same fraction of the usual $10$10 teams.

Do: The fraction of students present is $12$12 out of $60$60 which we can write as:

Fraction of students present	$=$=	$12$12 out of $60$60
	$=$=	$\frac{12}{60}$1260​
	$=$=	$\frac{1}{5}$15​

If only one fifth of the students are present, they can only make one fifth of the usual number of teams. So we can make $10\times\frac{1}{5}=2$10×15​=2 teams with $12$12 students.

In both of these worked examples we were able to preserve the ratio by either increasing or decreasing the initial ratio by some multiple. It is important to note that whenever we performed an operation we applied it to both sides of the ratio.

Tables	to	Chairs				Students	to	Teams
$1$1	:	$2$2				$60$60	:	$10$10
$\times7$×7		$\times7$×7				$\div5$÷​5		$\div5$÷​5
$7$7	:	$14$14				$12$12	:	$2$2

 

Simplified Ratios

A ratio is a simplified ratio if there is no equivalent ratio with smaller integer values. This is the same as saying that the two integers in the ratio have a greatest common factor of $1$1.

Since the simplified ratio is the smallest integer valued ratio, this also means that all the ratios equivalent to it are multiples of it. This makes the simplified ratio very useful for solving equivalent ratio questions that don't have very nice numbers.

 

 

Worked example

EXAMPLE 3

The ratio of raisins to nuts in a bag of trail mix is always $32:56$32:56. If there are $12$12 raisins left, how many nuts are left?

Think: Since $12$12 is not a factor of $32$32, we can first turn $32:56$32:56 into a simplified ratio, then find the equivalent ratio that looks like $12:\editable{?}$12:?.

Do: We can simplify the ratio $32:56$32:56 by decreasing both sides of the ratio until the greatest common factor of the two numbers is $1$1. Use the fact that both $32$32 and $56$56 are divisible by $8$8.

$32:56$32:56	$=$=	$\frac{32}{8}:\frac{56}{8}$328​:568​
	$=$=	$4:7$4:7

Since $4$4 and $7$7 have no common factors except for $1$1, this is a simplified ratio. So we now know that for every $4$4 raisins there are $7$7 nuts.

To get from $4$4 raisins to $12$12 raisins we multiply by $3$3. But we know the ratio of raisins and nuts is always $4:7$4:7, which means there must be $7\times3=21$7×3=21 nuts left.

Reflect: We started with the ratio $32:56$32:56, then found the simplified ratio $4:7$4:7, and used this to find the equivalent ratio $12:21$12:21 that was relevant to our problem. Notice that the three ratios are all equivalent, but only $4:7$4:7 is a simplified ratio.

 

 

The application of equivalent and simplified ratios is useful for when we want to keep things in proper proportion while changing their size, or when we want to measure large objects by considering their ratio with smaller objects.

 

 

Practice questions

Question 1

Question 2

Question 3

We previously learned how to write equivalent and simplified ratios. We can apply these to finding part or whole measures of quantities using ratios. This is useful for when we want to find an unknown value that is in a ratio with a known ratio.

 

Exploration

Consider a bag containing $56$56 red and blue marbles where the ratio of red marbles to blue marbles is $3:5$3:5.
How many red marbles are there? How many blue marbles are there?

To find the number of red and blue marbles we would normally use equivalent ratios by increasing or decreasing the ratio by some multiple. However, in this case, we only know the total number of marbles so this method won't work. So what do we do?

To solve our problem we can make the total number of marbles a component in our ratio. To do this we need to know how many ratio parts correspond to the total number of marbles. Since there are $3$3 parts red marbles and $5$5 parts blue marbles, the total number of marbles is simply $3+5=8$3+5=8 parts.

Taking this information, our ratio of red to blue to the total number of marbles is written $3:5:8$3:5:8.

Now we can use equivalent ratios to solve our problem.

Since we get $56$56 from $8$8 by increasing by a multiple of $7$7, we can find the number of red and blue marbles by increasing their ratio parts by the same multiple.

 

Red

to

Blue

to

Total

Ratio parts

$3$3

:

$5$5

:

$8$8

$\times7$×7

$\times7$×7

$\times7$×7

No. marbles

$21$21

:

$35$35

:

$56$56

 

Using the equivalent ratio we find that there are $21$21 red marbles and $35$35 blue marbles.

 

Part to whole ratios

In the exploration we used the ratio $3:5:8$3:5:8 to represent the ratio of red to blue to the total number of marbles. This is an example of a part to whole ratio. A part to whole ratio is a ratio that shows the ratio of one component compared to the whole and is useful when considering what fraction of the total represents that component.

For example, we can write the part to whole ratio of blue marbles to the total as $5:8$5:8 which is equivalent to saying that $\frac{5}{8}$58​ of the marbles are blue. This can then be used to find the number of blue marbles from the total with the calculation

Number of blue marbles$=$=$56\times\frac{5}{8}$56×58​$=$=$35$35

This is equivalent to finding the number of marbles in one part, that is we divide the total, $56$56, by $8$8, and then multiplying by $5$5 to find the number of marbles in $5$5 parts. This is known as the unitary method.

Number of blue marbles$=$=$\frac{56}{8}\times5$568​×5$=$=$35$35

Notice that the only difference is which number we divide by $8$8,

We can use the ratio $3:8$3:8 to perform a similar calculation to find the number of red marbles.

 

 

Can we use these types of ratios to find the total quantities when we only know the quantity of the component?

Yes, we can do this by multiplying the quantity of the component by the reciprocal of the fraction equivalent to the ratio.

For example, if we know that there are $21$21 red marbles and the ratio of red to the total number of marbles is $3:8$3:8 we can find the total number of marbles with the calculation

Total number of marbles $=$=$21\times\frac{8}{3}$21×83​$=$=$56$56

which returns the expected answer.

We can also solve this using the unitary method.

If $21$21 red marbles is $3$3 parts, we can divide $21$21 by $3$3 to find the size of one part, and then multiply this amount by $8$8 to find the total number of marbles,

Total number of marbles $=$=$\frac{21}{3}\times8$213​×8$=$=$56$56

 

Let's have a look at how we can use our part to whole ratios to split total lengths into parts.

 

Worked example

EXAMPLE 4

A plank of wood is cut so that the length of the two pieces are in the ratio $3:11$3:11.
If the plank of wood was $280$280 cm long, what is the length of the shorter piece?

Think: Since we are given the total length and want to find the length of the short piece, we want to find the length of one part first, and then we can find the length of $3$3 parts, that is, the length of the shorter piece.

Do: We can find how many ratio parts are in the total component by adding the ratio components of the two pieces together. This gives us:

Ratio parts for the total $=$= $3+11$3+11 $=$= $14$14

So we can find the length of one part by dividing the total length by $14$14.

We can then multiply this by $3$3, to find the total length of the shorter piece.

Length of the short piece $=$= $\frac{280}{14}\times3$28014​×3 $=$= $20\times3$20×3 $=$= $60$60

So the length of the short piece is $60$60 cm.

Reflect: We found the total component for the ratio so that we could write the appropriate part to whole ratio. Then we used the unitary method to find the length of one part, and used this to find the length of three parts.

It should be noted that we could have instead constructed a parts to whole ratio and used equivalent ratios. Using this method we could write "short piece to long piece to total" as $3:11:14$3:11:14 and multiply each component by $20$20 to get $60:220:280$60:220:280. This also tells us that the shorter piece has a length of $60$60 cm.

Practice questions

Question 4

Question 5

Question 6