The population of a town in 2020 was 100000 . The population decreased by y%
from the year 2020 to 2021 , and increased by x%
from the year 2021 to 2022, where x
and y
are two natural numbers. If population in 2022 was greater than the population in 2020 and the difference between x
and y
is 10 , then the lowest possible population of the town in 2021 was

To solve this problem, let's break it down step by step.

### Step 1: Establish the relationship between the populations and percentage changes
- Let the population in 2021 be \( P_{2021} \).
- The population decreased by \( y\% \) from 2020 to 2021, so:
  \[
  P_{2021} = 100000 \times \left(1 - \frac{y}{100}\right)
  \]
  
- Let the population in 2022 be \( P_{2022} \).
- The population increased by \( x\% \) from 2021 to 2022, so:
  \[
  P_{2022} = P_{2021} \times \left(1 + \frac{x}{100}\right)
  \]

### Step 2: Express the population in 2022 in terms of the population in 2020
Substitute the expression for \( P_{2021} \) into the equation for \( P_{2022} \):
\[
P_{2022} = 100000 \times \left(1 - \frac{y}{100}\right) \times \left(1 + \frac{x}{100}\right)
\]

### Step 3: Apply the condition that the population in 2022 is greater than in 2020
Since \( P_{2022} > 100000 \), we have:
\[
100000 \times \left(1 - \frac{y}{100}\right) \times \left(1 + \frac{x}{100}\right) > 100000
\]
Dividing both sides by 100000:
\[
\left(1 - \frac{y}{100}\right) \times \left(1 + \frac{x}{100}\right) > 1
\]

### Step 4: Simplify the inequality
Expand the left side:
\[
1 - \frac{y}{100} + \frac{x}{100} - \frac{xy}{10000} > 1
\]
Subtract 1 from both sides:
\[
-\frac{y}{100} + \frac{x}{100} - \frac{xy}{10000} > 0
\]
Multiply by 100 to clear the fractions:
\[
-x + y + \frac{xy}{100} > 0
\]
Rearrange the inequality:
\[
y - x > \frac{xy}{100}
\]

### Step 5: Utilize the given condition \( x - y = 10 \)
Given \( x - y = 10 \), we substitute \( y = x - 10 \) into the inequality:
\[
x - (x - 10) > \frac{x(x - 10)}{100}
\]
Simplifying:
\[
10 > \frac{x(x - 10)}{100}
\]
Multiply by 100:
\[
1000 > x(x - 10)
\]

### Step 6: Solve for the minimum \( P_{2021} \)
We need to find the lowest \( P_{2021} \) that satisfies this inequality. Let's test natural numbers for \( x \) and calculate \( y = x - 10 \).

- If \( x = 20 \) and \( y = 10 \):
  \[
  20 \times 10 = 200 \quad (\text{which is less than } 1000)
  \]
  Calculate \( P_{2021} \):
  \[
  P_{2021} = 100000 \times \left(1 - \frac{10}{100}\right) = 100000 \times 0.9 = 90000
  \]

- If \( x = 15 \) and \( y = 5 \):
  \[
  15 \times 5 = 75 \quad (\text{which is also less than } 1000)
  \]
  Calculate \( P_{2021} \):
  \[
  P_{2021} = 100000 \times \left(1 - \frac{5}{100}\right) = 100000 \times 0.95 = 95000
  \]

Here, the lowest possible population in 2021 is \( 90000 \) when \( x = 20 \) and \( y = 10 \).

### Conclusion:
The lowest possible population of the town in 2021 was **90000**.

Would you like further details on any step or have any questions? Here are some related questions:

1. What would the population in 2022 be for different values of \( x \) and \( y \)?
2. How would the answer change if the difference between \( x \) and \( y \) was more than 10?
3. Can you explain why we took specific values of \( x \) and \( y \) to minimize \( P_{2021} \)?
4. What happens if \( P_{2022} \) is less than \( P_{2020} \)?
5. How can we generalize this approach for any given percentage difference?

**Tip**: When dealing with percentage changes, it's often useful to translate them into multipliers for easier manipulation.

Learn how to calculate the lowest possible population in 2021 given percentage changes in population from 2020 to 2022. This problem involves understanding percentage changes, inequalities, and population dynamics. Suitable for high school students.

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