AFCAT Numerical Ability · Percentage

Population Growth and Depreciation

A population or a value that changes by a percentage each year is multiplied by the same factor every year: go forward, go back, change the rate year by year, or find the rate.

All Percentage notes

Why this matters

About a quarter of this page is HARD, more than any page before it, because the numbers are large and the factor is raised to a power. Two things decide the marks: dividing (not subtracting) when going back in time, and spotting that 1.44 or 1.728 is a familiar square or cube.

Concept 1 of 5: Growth and depreciation year after year

A town that grows 12.5% a year is multiplied by 1.125 every year, and each year's growth is on the new, larger figure. Three years multiply by

1.125^3

Definition

After $n$ years at $r\%$ growth: $P_n = P \left(1 + \tfrac{r}{100}\right)^n$ .
Depreciation (a machine, a car, a falling population) uses $1 - \tfrac{r}{100}$ .
Births and deaths, or arrivals and departures, in the same year: use the net rate, births minus deaths.
The period need not be a year. 'Rises 20% every 5 years' over 10 years is two periods.

Compound growth

P_n = P\left(1 + \dfrac{r}{100}\right)^n

Worked example

A town of 64,000 grows by 12.5% a year. Find its population after 3 years.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionEasy

Example 1 · Percentage · Population and Growth

The population of a city is

50,000

. If it increases at the rate of

8

% per annum, what will be the population after

2

years?

Compound, not simple

12.5% a year for 3 years is not 37.5% in all. Each year's growth is on the new figure, so 64,000 becomes 91,125, not 88,000.

Drill 8 more on growth and depreciation year after year

Concept 2 of 5: Going back in time

To find a population two years ago, undo two years of growth: divide by the factor twice. Subtracting the percentage from today's figure gives the wrong base.

Definition

$n$ years ago: $P = \dfrac{P_n}{\left(1 + \tfrac{r}{100}\right)^n}$ . For depreciation, divide by $\left(1 - \tfrac{r}{100}\right)^n$ .
Friendly divisors: $0.9^2 = 0.81$ , $0.8^2 = 0.64$ , $0.85^2 = 0.7225$ , $1.2^2 = 1.44$ , $1.15^2 = 1.3225$ .
If an option is a round number, check it forward: multiply the option by the factors and see if you reach the given figure.

The starting value

P = \dfrac{P_n}{\left(1 \pm \tfrac{r}{100}\right)^n}

Worked example

After 2 years of 10% growth, a population is 60,500. What was it 2 years ago?

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionHard

Example 2 · Percentage · Population and Growth

A village loses

5

% of its population each year. If after

3

years the count is

85,737

, what was the starting population?

Divide, do not subtract

For 60,500 after two years of 10% growth, taking 10% off twice gives 49,005. The answer is 50,000: the growth was on the smaller, earlier figure.

Drill 11 more on going back in time

Concept 3 of 5: A different rate each year

When the rate changes from year to year, multiply by each year's own factor. A rise, a fall and a rise is three different factors in a row, the successive-change idea again.

Definition

$P_{\text{final}} = P \times \left(1 + \tfrac{r_1}{100}\right)\left(1 + \tfrac{r_2}{100}\right)\cdots$ , with a minus sign for each fall.
Losses stack the same way: 10% die, then 20% of the survivors leave, keeps $0.9 \times 0.8 = 0.72$ .
Up $x\%$ and then down $x\%$ leaves $1 - \tfrac{x^2}{10000}$ , so a final drop of 9% means $x^2 = 900$ .
Given the final figure, divide by the product of the factors.

Changing rates

P_{\text{final}} = P\left(1 + \tfrac{r_1}{100}\right)\left(1 + \tfrac{r_2}{100}\right)\left(1 + \tfrac{r_3}{100}\right)

Worked example

A population of 50,000 grows by 10%, then falls by 20%, then grows by 25%. Find the final population.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Percentage · Population and Growth

The population of a district was

8,00,000

2020

. It rose

4

% in 2021,

6

% in 2022 and

5

% in 2023. Find the 2023 population.

The second loss is on the survivors

10% die and then 20% of the rest leave is a loss of 28%, not 30%. The 20% is of the 90% left.

Drill 13 more on a different rate each year

Concept 4 of 5: Finding the rate of growth

If a town grows from 40,000 to 57,600 in two years, the two-year factor is 1.44. The yearly factor is its square root, 1.2, so the rate is 20%.

Definition

Factor over $n$ years $=$ final $\div$ start. The yearly factor is its $n$ th root.
Recognise the common powers: $1.2^2 = 1.44$ , $1.25^2 = 1.5625$ , $1.06^2 = 1.1236$ , $1.2^3 = 1.728$ , $1.05^3 = 1.157625$ .
When the factor is not a neat power, test the options forward: raise each $1 + \tfrac{r}{100}$ to the power and compare.

Rate from two figures

1 + \dfrac{r}{100} = \left(\dfrac{P_n}{P}\right)^{1/n}

Worked example

A town grows from 40,000 to 57,600 in 2 years. Find the yearly rate of growth.

Practice this conceptself-check · 2 quick reps

The same idea in a practice question:

Practice questionHard

Example 4 · Percentage · Population and Growth

A city's population rises from

2,00,000

2,42,000

2

years. Assuming compounded growth, what is the annual rate?

Do not divide the total growth by the years

44% growth in two years is 20% a year, not 22%. The yearly rate is a root, not an average.

Drill 3 more on finding the rate of growth

Concept 5 of 5: The make-up of a population

Men, women and children are shares of one total. Any one share given as a number fixes the total, and every other share follows from it.

Definition

Shares add to 100%: if men are 37.5% and women 40%, children are 22.5%.
A difference of two shares is a share too: men 55% and women 45% differ by 10% of the total.
When parts change by different percentages (urban up 10%, rural down 10%), weight each change by its share: $0.7 \times 1.1 + 0.3 \times 0.9 = 1.04$ .
A sex ratio of 900 per 1,000 males means females are 90% of males.

A share of the whole

\text{total} = \dfrac{\text{size of a part} \times 100}{\text{its percentage}}

Worked example

A town has only men and women. Men are 55% of it and outnumber women by 4,500. Find the population.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 5 · Percentage · Population and Growth

In a town, males exceed females by

2,600

and constitute

52

% of the population. Find the total population.

Weight each part's change

Urban up 10% and rural down 10% is not 'no change' when the parts are 70% and 30%. The larger part pulls the total its way, here to a 4% rise.

Drill 16 more on the make-up of a population

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Growth and depreciation year after year
Compound growth
$P_n = P\left(1 + \dfrac{r}{100}\right)^n$
Going back in time
The starting value
$P = \dfrac{P_n}{\left(1 \pm \tfrac{r}{100}\right)^n}$
A different rate each year
Changing rates
$P_{\text{final}} = P\left(1 + \tfrac{r_1}{100}\right)\left(1 + \tfrac{r_2}{100}\right)\left(1 + \tfrac{r_3}{100}\right)$
Finding the rate of growth
Rate from two figures
$1 + \dfrac{r}{100} = \left(\dfrac{P_n}{P}\right)^{1/n}$
The make-up of a population
A share of the whole
$\text{total} = \dfrac{\text{size of a part} \times 100}{\text{its percentage}}$

Watch out for (5)

Compound, not simple
→ Growth and depreciation year after year
Divide, do not subtract
→ Going back in time
The second loss is on the survivors
→ A different rate each year
Do not divide the total growth by the years
→ Finding the rate of growth
Weight each part's change
→ The make-up of a population

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