The population of foxes in a town increased by 20% from Year 1 to Year 2. The population of foxes in Year 2 is 36.

Question

We’ll use x to represent the population in Year 1. Which tape diagram shows the population in Year 2?

x
Year 1

Year 2
36

x
Year 1

Year 2
36
Excellent!
What do we know about the population in Year 2?
x
Year 1

Year 2
36

20% more than Year 1 population
= Year 2 population
You got it!
Fill in the equation to show that the population in Year 2 is 20% more than the population in Year 1. What percent of Year 1’s population is Year 2’s population?
x
Year 1

Year 2
36
20% more than Year 1 population = Year 2 population
% of Year 1 population = Year 2 population
Excellent!
Now write an equation that shows that the population in Year 2 is a 20% increase of the population in Year 1.
x
Year 1

Year 2
36
120 % of Year 1 population = Year 2 population
·
=

Answer 1

To summarize the information:

• Let \( x \) be the population of foxes in Year 1.
• The population in Year 2 is given as 36.

When the population increases by 20%, the population in Year 2 can be represented mathematically as:

\[ \text{Year 2 Population} = \text{Year 1 Population} + 20% \text{ of Year 1 Population} \]

This can be rewritten as:

\[ \text{Year 2 Population} = x + 0.2x = 1.2x \]

Given that the population in Year 2 is 36, we can set up the equation:

\[ 1.2x = 36 \]

To fill in the equation you provided:

• \( 120% \) of Year 1 population equals Year 2 population, so:

\[ \frac{120}{100} \cdot x = 36 \]

Or:

\[ 1.2x = 36 \]

To find \( x \), we can solve for it:

\[ x = \frac{36}{1.2} = 30 \]

Thus, the missing equations can be filled in as follows:

• \( 120% \text{ of Year 1 population } = \text{ Year 2 population} \)
• Solving this gives \( 1.2x = 36 \).

So, to conclude:

• Population in Year 1: \( 30 \)
• Population in Year 2: \( 36 \)

You can express it in the final format:

\[ 120% \text{ of Year 1 population } = Year 2 population \]

\[ 1.2x = 36 \]