Question

▶Question:-

The population of a city was 20,000 in the year 1997. It increased at the rate of 5% p.a. Find the population at the end of the year 2000.

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Answer 1

Step-by-step explanation:

We have,

p=20,000

R=5%

T=3 year,

A=20,000(1+5/100)'3

=20,000×( 105/100)'3

=23,52.5

Hence this is the answer.

Answer 2

$\sf \: Given : -$

Population of city in 1997 = 20, 000

It Increases at the Rate of 5% Per Annum

Here, 5% is compounded Rate

$\sf \: Formula \: used:-$

$\sf \: A = P \bigg \lgroup\frac{ 1 + R}{100 } \bigg\rgroup^ n$

Here,

A = Population at end of 2000

P = Population in 1997 = 20, 000

R=5%

N = 2000 - 1997

= 3

Putting Values in Formula ,

$\sf \: Population \: at \: end \: of \: 2000 \: \\ \sf = 20000 \bigg \lgroup\frac{1 + 5}{100 \:} \bigg \rgroup ^ 3 \\ \\ \sf \: = 20000\bigg \lgroup1 + \frac{1}{20 \:} \bigg \rgroup ^ 3 \\ \\ \sf \: = 20000\bigg \lgroup\frac{20 + 1}{20} \bigg \rgroup ^ 3 \\ \\ \sf \: = 20000\bigg \lgroup\frac{21}{20 \:} \bigg \rgroup ^ 3 \\ \\ \sf \: = 20000 \times \frac{21 \times 21 \times 21}{20 \times 20 \times 20} \\ \\ \sf \: = \frac{5 \times 21 \times 21 \times 21}{2} \\ \\ \sf \: = \frac{441 \times 21 \times 5}{2} \\ \\ \sf \: = \frac{441 \times 105}{2} \\ \\ \sf \: = \frac{46305}{2} \\ \\ \sf \: =23152.5 \\ \\ \sf \: Since \: population \: cannot \: be \: in \: decimals \\ \sf\therefore Estimated \: population= 23, 153$