Population Growth MCQ Quiz - Objective Question with Answer for Population Growth - Download Free PDF

Given:

Initial Population = 110000

First Year Increase = 10%

Second Year Increase = 20%

Formula used:

Population after n years = Initial Population × (1 + Rate₁) × (1 + Rate₂)

Calculations:

Population after 2 years = 110000 × (1 + 10/100) × (1 + 20/100)

⇒ Population after 2 years = 110000 × 1.1 × 1.2

⇒ Population after 2 years = 110000 × 1.32

⇒ Population after 2 years = 145200

∴ The correct answer is option (4).

Given:

Total population = 360,000

Males = 225,000

Formula used:

Literate females = Total literate population - Literate males

Percentage of literate females = (Literate females ÷ Total females) × 100

Calculations:

Literate population = 35% of total = 0.35 × 360,000 = 126,000

Literate males = 17% of males = 0.17 × 225,000 = 38,250

Females = Total population - Males = 360,000 - 225,000 = 135,000

⇒ Literate females = 126,000 - 38,250 = 87,750

⇒ Percentage of literate females = (87,750 ÷ 135,000) × 100

⇒ Percentage of literate females = 0.65 × 100

⇒ Percentage of literate females = 65%

∴ The correct answer is option (1).

Given:

Current population = 31,110

Annual increase rate = 3.7%

Formula used:

Current Population = Last Year's Population × (1 + Rate of Increase / 100)

Calculations:

Let Last Year's Population be 'P'.

Current Population = P × (1 + \(\frac{3.7}{100}\))

⇒ 31,110 = P × (1 + 0.037)

⇒ 31,110 = P × 1.037

⇒ P = \(\frac{31110}{1.037}\)

⇒ P = 30,000

∴ Last year's population was 30,000.

Given:

Current population = 1,76,400

Annual increase rate = 5%

Formula Used:

Future Population = Present Population × (1 + (Rate)/(100))^(Time)

Calculation:

Present Population = 1,76,400

Rate = 5% = 0.05

Time = 2 years

Future Population = 1,76,400 × (1 + 0.05)²

Future Population = 1,76,400 × 1.1025

Future Population = 1,94,481

The population 2 years hence will be 1,94,481.

Given:

Present population of a town = 1,76,400

Rate of increase = 5% per annum

Formula Used:

Population after n years = Present Population × (1 + Rate/100)ⁿ

Calculation:

Population after 2 years = 1,76,400 × (1 + 5/100)²

⇒ Population after 2 years = 1,76,400 × (1.05)²

⇒ Population after 2 years = 1,76,400 × 1.1025

⇒ Population after 2 years = 1,94,481

The population after two years will be 1,94,481.

Given:

First year increase rate = 12%

Second year decrease rate = 8%

Calculations:

Let the initial population be x.

Then, According to question,

92% of 112% of x = 64,400

⇒ 0.92 × 1.12x = 64,400

⇒ x = 64,400/(0.92×1.12)

⇒ x = 64,400/1.0304

⇒ x = 62,500

Hence, The Required value is 62,500.

Given:

A district has 10,24,000 inhabitants.

The population increases at the rate of 2.5% per annum.

Concept used:

In case of the compound interest,

Amount = P(1 + R/100)^N

CI = P(1 + R/100)^N - P

where

P = Principal amount

R = Rate of interest per year

N = Time in years

Calculation:

Now, the number of inhabitants at the end of three years

⇒ 1024000(1 + 2.5%)³

⇒ 11,02,736

∴ The number of inhabitants at the end of three years is 11,02,736.

Given:

The population of country A decreased by p% and the population of country B decreased by q% from the year 2020 to 2021. Here 'p' is greater than ‘q’. Let 'x' be the ratio of the population of country A to the population of country B in the given year.

Calculation:

Let the initial population of Country A and B be J and K respectively.

Hence, x = J/K

Let the ratio of the population of country A to the population of country B for the next year be x₁.

​According to the question,

(J × (1 - p%)) : (K × (1 - q%)) = x₁

⇒ \(​\frac {J}{K} \times \frac {100 - p}{100 - q} = x_1\)

Now, % decrease

⇒ \(\frac {x - x_1}{x} \times 100\%\)

⇒ \(\frac {\frac {J}{K} - ​\frac {J}{K} \times \frac {100 - p}{100 - q}}{​\frac {J}{K} } \times 100\%\)

⇒ \(\frac {(p - q)}{(100 - q)} \times 100\%\)

⇒ \(\frac{100(p - q)}{(100 - q)} \%\)

∴ The percentage decrease in 'x' from 2020 to 2021 is \(\frac{100(p - q)}{(100 - q)} \%\).

Shortcut Trick

Assume,

The population of A in 2020 be 200 and B in 2020 be 100,

and p = 20% & q = 10% (as per the question)

hence according to the question,

So now,

x = 2 and x' = 16/9

now as per the question,

⇒ \({{2} - {16\over 9}}\over 2\) × 100 ⇒ (100/9)

Now substituting the values in the given options,

Only 1 option satisfies the value.

Given:

Population at the end of the third year = 9,47,625

Let the initial population of the village is 'x'

Concept used :

 5% of x =\(5x\over100\)

Calculations:

Initially, population is decreased by 5%.

 5% of x =\(5x\over100\)

 Now the population after decrement is:

\(x-{5x\over100}={95x\over100}\)

Population is further decreased by 5%:

 5% of \({95x\over100}={95x\over100}×{5\over100}={475x\over10000}\)

 Now the population after decrement is:

\({95x\over100}-{475x\over10000}={9025x\over10000}\)

Now the population is increased by 5%

 5% of \({9025x\over10000}={9025x\over10000}×{5\over100}={45125x\over1000000}\)

Now the population after the increment is:

\({9025x\over10000}+{45125x\over1000000}={947625x\over1000000}\)

From the given initial conditions:

\({947625x\over1000000}\) = 947625

 x = 1000000

Thus, the population at the beginning of the first year was 10,00,000.

Shortcut Trick

According to the question, 7581 unit → 947625

Then, 8000 unit → 947625/7581 × 8000 = 125 × 8000 = 1000000

Given:

20% of the inhabitants of the village died of malaria

30% of the inhabitants left the village

Formula used:

Inhabitants reduced = Population × (100 - Decreased %) 

Calculation:

According to the formula, 

The inhabitants initially, 

⇒ Reduced = P × (100 - 20%) × (100 - 30%)

⇒ 12000 = \(P×\frac{70}{100}×\frac{80}{100}\)

⇒ P = \(\frac{12000×100×100} {70×80} \)

⇒ P = 21428

⇒ Hence, The population initially was 21428

Given:

First year increase rate = 12%

Second year decrease rate = 8%

Calculations:

Let the initial population be x.

Then, According to question,

92% of 112% of x = 64,400

⇒ 0.92 × 1.12x = 64,400

⇒ x = 64,400/(0.92×1.12)

⇒ x = 64,400/1.0304

⇒ x = 62,500

Hence, The Required value is 62,500.

Given:

A district has 10,24,000 inhabitants.

The population increases at the rate of 2.5% per annum.

Concept used:

In case of the compound interest,

Amount = P(1 + R/100)^N

CI = P(1 + R/100)^N - P

where

P = Principal amount

R = Rate of interest per year

N = Time in years

Calculation:

Now, the number of inhabitants at the end of three years

⇒ 1024000(1 + 2.5%)³

⇒ 11,02,736

∴ The number of inhabitants at the end of three years is 11,02,736.

Given:

The population of country A decreased by p% and the population of country B decreased by q% from the year 2020 to 2021. Here 'p' is greater than ‘q’. Let 'x' be the ratio of the population of country A to the population of country B in the given year.

Calculation:

Let the initial population of Country A and B be J and K respectively.

Hence, x = J/K

Let the ratio of the population of country A to the population of country B for the next year be x₁.

​According to the question,

(J × (1 - p%)) : (K × (1 - q%)) = x₁

⇒ \(​\frac {J}{K} \times \frac {100 - p}{100 - q} = x_1\)

Now, % decrease

⇒ \(\frac {x - x_1}{x} \times 100\%\)

⇒ \(\frac {\frac {J}{K} - ​\frac {J}{K} \times \frac {100 - p}{100 - q}}{​\frac {J}{K} } \times 100\%\)

⇒ \(\frac {(p - q)}{(100 - q)} \times 100\%\)

⇒ \(\frac{100(p - q)}{(100 - q)} \%\)

∴ The percentage decrease in 'x' from 2020 to 2021 is \(\frac{100(p - q)}{(100 - q)} \%\).

Shortcut Trick

Assume,

The population of A in 2020 be 200 and B in 2020 be 100,

and p = 20% & q = 10% (as per the question)

hence according to the question,

So now,

x = 2 and x' = 16/9

now as per the question,

⇒ \({{2} - {16\over 9}}\over 2\) × 100 ⇒ (100/9)

Now substituting the values in the given options,

Only 1 option satisfies the value.