A town’s population increased by in 2019, decreased by in 2020, and then increased by in 2021. At the end of 2021, the population was . What was the town’s population at the beginning of 2019?

Solution available with step-by-step explanation

SAT Percentages Question 18 | Free SAT Prep | Aniko

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Question 18·Hard·Percentages

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A town’s population increased by $15\%$ in 2019, decreased by $12\%$ in 2020, and then increased by $10\%$ in 2021. At the end of 2021, the population was $22,264$ . What was the town’s population at the beginning of 2019?

Your answer

(Express the answer as an integer)

For multi-year percent change problems, immediately convert each percent to a multiplier (like $1.15$ , $0.88$ , $1.10$ ) and recognize that the final amount equals the initial amount times the product of these factors. When you’re given the final amount and asked for the original, it is usually fastest and most accurate to work backward one step at a time—divide by each factor in reverse order of the years—using simple fractions when possible to keep the arithmetic exact and avoid rounding errors.

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Translate percent changes into multipliers

Think about what number you multiply the population by for each year: what do you multiply by for a $15\%$ increase, a $12\%$ decrease, and a $10\%$ increase?

Relate the beginning and ending populations

If $P$ is the population at the beginning of 2019, write an expression for the population after all three yearly changes using your multipliers, and set it equal to $22{,}264$ .

Work backward instead of forward

To find the original population from the final one, think about undoing each year’s percent change in reverse order (2021, then 2020, then 2019). What operation undoes multiplying by $1.10$ ? By $0.88$ ? By $1.15$ ?

Use fractions to keep the arithmetic clean

It may help to rewrite $1.10$ , $0.88$ , and $1.15$ as simple fractions (like $11/10$ , $22/25$ , $23/20$ ) to make the divisions easier to do exactly.

Desmos Guide

Compute the original population directly

In the Desmos expression line, type 22264/(1.15*0.88*1.10) and press Enter. The value Desmos outputs is the town’s population at the beginning of 2019.

Step-by-step Explanation

Express each percent change as a multiplication factor

A percent increase of $p\%$ means you multiply by $1+p/100$ , and a percent decrease of $p\%$ means you multiply by $1-p/100$ .

2019: increased by $15\%$ $\Rightarrow$ multiply by $1.15$ .
2020: decreased by $12\%$ $\Rightarrow$ multiply by $0.88$ .
2021: increased by $10\%$ $\Rightarrow$ multiply by $1.10$ .

If $P$ is the population at the beginning of 2019, then after these three years, the population is:

P \times 1.15 \times 0.88 \times 1.10.

We know this equals $22{,}264$ .

Set up the equation relating initial and final population

Using the expression from Step 1, write the equation:

P \times 1.15 \times 0.88 \times 1.10 = 22{,}264.

To find $P$ , we need to divide the final population by the product of the three factors:

P = \frac{22{,}264}{1.15 \times 0.88 \times 1.10}.

You can compute this either directly, or by working backward one year at a time (which keeps the numbers nicer).

Work backward one year at a time

Start from the end of 2021 and undo each year’s change.

Undo the 2021 increase of $10\%$ (factor $1.10$ ):
- End of 2021 population: $22{,}264$ .
- Start of 2021 population = $\dfrac{22{,}264}{1.10}$ .
Since $1.10 = 11/10$ , dividing by $1.10$ is multiplying by $10/11$ :

\frac{22{,}264}{1.10} = 22{,}264 \times \frac{10}{11} = 20{,}240.

So, population at the start of 2021 is $20{,}240$ .

Undo the 2020 decrease of $12\%$ (factor $0.88$ ):

Start of 2020 population = $\dfrac{20{,}240}{0.88}$ .

Since $0.88 = 22/25$ , dividing by $0.88$ is multiplying by $25/22$ :

\frac{20{,}240}{0.88} = 20{,}240 \times \frac{25}{22} = 23{,}000.

So, population at the end of 2019 is $23{,}000$ .

Undo the 2019 increase and state the initial population

Finally, undo the 2019 increase of $15\%$ (factor $1.15$ ):

Beginning of 2019 population $P = \dfrac{23{,}000}{1.15}$ .

Since $1.15 = 23/20$ , dividing by $1.15$ means multiplying by $20/23$ :

P = 23{,}000 \times \frac{20}{23} = 1{,}000 \times 20 = 20{,}000.

So the town’s population at the beginning of 2019 was $20{,}000$ .

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation