Question

A district contains 64000 inhabitants. if the population increases at the rate of 2/2/1% per annum find the number of inhabitants at the end of 3 year​

Answer 1

Question:-

A district contains 64000 inhabitants. if the population increases at the rate of. 2 2/1% per annum find the number of inhabitants at the end of 3 year

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Solution:-

Given:-

• A district contains 64000 inhabitants
• Rate:- 2 2/1

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

To Find:-

• number of inhabitants at the end of 3 year

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Here, the initial population, P = 64000

Annual rate of interest,$\sf \: r = 2 \: \: \: \frac{1}{2}\%=\frac{5}{2}$

• And, time, t = 3 years

Hence, the population after 3 years is,

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\sf \: A = P \bigg( \: \: 1+\frac{r}{100} \: \: \bigg)^t$

$\sf \:\longrightarrow 64000 \bigg( \: \: 1+\frac{ \dfrac{5}{2}}{100} \: \: \bigg)\\ \\ \sf =64000\bigg(1+\frac{5}{200}\bigg) \\ \\ \sf \:\longrightarrow 64000\times \frac{205}{200} \\ \\ \qquad \sf \longrightarrow\frac{13120000}{200} \\ \\ \qquad\longrightarrow \underline{ \boxed{ \sf \purple{68921} }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

• Hence, the number of inhabitants at the end of 3 years is 68921.

Answer 2

Answer:

A district contains 64000 inhabitants

Rate:- 2 2/1

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

To Find:-

number of inhabitants at the end of 3 year

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Here, the initial population, P = 64000

Annual rate of interest,\begin{gathered} \sf \: r = 2 \: \: \: \frac{1}{2}\%=\frac{5}{2} \\ \end{gathered}

r=2

2

1

%=

2

5

And, time, t = 3 years

Hence, the population after 3 years is,

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

\begin{gathered} \sf \: A = P \bigg( \: \: 1+\frac{r}{100} \: \: \bigg)^t \\ \end{gathered}

A=P(1+

100

r

)

t

\begin{gathered} \sf \:\longrightarrow 64000 \bigg( \: \: 1+\frac{ \dfrac{5}{2}}{100} \: \: \bigg)\\ \\ \sf =64000\bigg(1+\frac{5}{200}\bigg) \\ \\ \sf \:\longrightarrow 64000\times \frac{205}{200} \\ \\ \qquad \sf \longrightarrow\frac{13120000}{200} \\ \\ \qquad\longrightarrow \underline{ \boxed{ \sf \purple{68921} }}\end{gathered}

⟶64000(1+

100

2

5

)

=64000(1+

200

5

)

⟶64000×

200

205

⟶

200

13120000

⟶

68921

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Hence, the number of inhabitants at the end of 3 years is 68921.