UK Secondary (7-11)
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Ratio Tables
Lesson

A ratio compares the relationship between two values. It compares how much there is of one thing compared to another.

For example, if a pie recipe calls for $2$2 tablespoons of brown sugar per $1$1 apple, we could write this as a ratio

$2:1$2:1

What if we wanted to make more than one pie?

Well, we'd need to keep the ingredients in equivalent ratios. For example, to make two pies, we'd need $4$4 tablespoons of sugar and $2$2 apples.

An easy way to display information about ratios is using a ratio table. Let's display our information from the above example:

Sugar $2$2 $4$4 $6$6 $8$8
Apples $1$1 $2$2 $3$3 $4$4

 

Ratio tables display the relationship between the two quantities and help us find the constant multiplicative factor between them. This constant multiplicative factor, which is also called the constant of proportionality, is a constant positive multiple between the two variables. As we've already learnt, ratios can also be represented as fractions, and by simplifying the fraction, we can work out the constant of proportionality.

In the table above, $\frac{4}{8}=\frac{3}{6}$48=36$=$=$\frac{2}{4}$24$=$=$\frac{1}{2}$12. In other words, the number of apples will always be half as much as the number of tablespoons of sugar. In other words, the constant multiplicative factor is $\frac{1}{2}$12. That means if I put $20$20 tablespoons of sugar in my recipe, I know I would need $10$10 apples because that is half of $20$20. Similarly, if I used $7$7 apples, I would need double the amount of sugar, which is $14$14 tablespoons.

Examples

Question 1

Fill in the ratio table below, then use it to answer the following questions.

a) 

Pens $10$10 $20$20 $30$30 $40$40 $50$50 $\editable{}$
Cost (dollars) $\editable{}$ $11.60$11.60 $\editable{}$ $\editable{}$ $29.00$29.00 $58.00$58.00

Think: What is the common multiplicative factor between the two quantities?

Do:

$11.60\div20=0.58$11.60÷​20=0.58

$29.00\div50=0.58$29.00÷​50=0.58

So our common multiplicative factor is $0.58$0.58. In other words, one pen costs $58$58c.

To work out the cost, we need to multiply the number of pens by $0.58$0.58.

For example, the cost of $10$10 pens would be $10\times0.58=\$5.80$10×0.58=$5.80.

Conversely, to work out the number of pens, we need to divide the cost by $0.58$0.58.

For example, $\$58$$58 would buy $58\div0.58=100$58÷​0.58=100 pens.  

Pens $10$10 $20$20 $30$30 $40$40 $50$50 $100$100
Cost (dollars) $5.80$5.80 $11.60$11.60 $17.40$17.40 $23.20$23.20 $29.00$29.00 $58.00$58.00

 

b) Calculate the cost of buying $90$90 pens.

Think: If $10$10 pens cost $\$5.80$$5.80, how much would $90$90 pens cost?

Do:

If we think, $90$90 pens is $9$9 times more than $10$10 pens. This means we need to increase the cost of $10$10 pens by $9$9 as well. $5.80\times9=\$52.20$5.80×9=$52.20.

We can also solve this algebraically. Let's let $x$x be the cost of $90$90 pens.

$10:5.80$10:5.80 $=$= $90:x$90:x
$\frac{10}{5.80}$105.80 $=$= $\frac{90}{x}$90x
$\frac{5.80}{10}$5.8010 $=$= $\frac{x}{90}$x90
$\frac{5.80\times90}{10}$5.80×9010 $=$= $x$x
$x$x $=$= $\$52.20$$52.20

 

c) How much would you expect to pay for $5$5 pens?

Think: The cost of $5$5 pens would be half the price of $10$10 pens.

Do: $5.80\div2=\$2.90$5.80÷​2=$2.90

 

Question 2

David and Justin both travel to school riding scooters. David covers $8$8 kilometres in $4$4 minutes. Justin travels $35$35 kilometres in $16$16 minutes.

a) Complete the table for the distance travelled by David in each time period. Assume that he travels at a constant speed.

 

b) Complete the table for the distance travelled by Justin in each time period. Assume that he travels at a constant speed.

c) Who travels faster?

 

Question 3

a) Fill in the ratio table.

 

b) How many pounds will you be able to buy with $\$10$$10 US?

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