Question

the population of a city was 20,000 in the year 1997. it incresed at the rate of 5%p.a find the population at the end of year 2000?

No spamming ​

Answer 1

Answer:

$\bold\green{Population\:will\:be}$ = ${\boxed{\mathfrak\purple{23,153}}}$

Step-by-step explanation:

$\frak{Given:-}\begin{cases}\tt{Population\:was(P) = 20,000}\\\tt{Rate\:of\:interest(r) = 5\%}\\\tt{Time(n) = (2000-1997) = 3\: years}\\\tt{Population\:in\:2000=?}\end{cases}$

We know that, population increases in a compounded way.So in 2000, the population will be equal to Compound amount after 3 years @ 5% per annum.

${\underline{\underline{\bold{According\:to\:question:-}}}}$

${\boxed{\bold\green{CA = P{(1+r)}^{n} }}}$

$\Rightarrow\tt CA = 20,000{(1+\dfrac{5}{100})}^{3} \\\\\\\Rightarrow\tt CA = 20,000 {(\dfrac{100+5}{100})}^{3} \\\\\\\Rightarrow\tt CA = 20,000 \times {(1.05)}^{3} \\\\\\\Rightarrow\tt CA = 20,000 \times 1.157625 \\\\\\\Rightarrow\huge{\boxed{\tt\green{CA = 23,152.5 }}}$

$\rule{100}{2}$

$\because\tt\red{Population\:can't\:be \: in\:fraction}$

$\therefore\bold\green{Population\:at\:the\:end\:of\:2000 = 23,153}$

${\boxed{\bold\purple{Population\:in\:2000 = 23153 }}}$

Answer 2

Answer:

Step-by-step explanation:

\begin{gathered}\frak{Given:-}\begin{cases}\tt{Population\:was(P) = 20,000}\\\tt{Rate\:of\:interest(r) = 5\%}\\\tt{Time(n) = (2000-1997) = 3\: years}\\\tt{Population\:in\:2000=?}\end{cases}\end{gathered}

Given:−

⎩

⎪

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎪

⎧

Populationwas(P)=20,000

Rateofinterest(r)=5%

Time(n)=(2000−1997)=3years

Populationin2000=?

We know that, population increases in a compounded way.So in 2000, the population will be equal to Compound amount after 3 years @ 5% per annum.

{\underline{\underline{\bold{According\:to\:question:-}}}}

Accordingtoquestion:−

{\boxed{\bold\green{CA = P{(1+r)}^{n} }}}

CA=P(1+r)

n

\begin{gathered}\Rightarrow\tt CA = 20,000{(1+\dfrac{5}{100})}^{3} \\\\\\\Rightarrow\tt CA = 20,000 {(\dfrac{100+5}{100})}^{3} \\\\\\\Rightarrow\tt CA = 20,000 \times {(1.05)}^{3} \\\\\\\Rightarrow\tt CA = 20,000 \times 1.157625 \\\\\\\Rightarrow\huge{\boxed{\tt\green{CA = 23,152.5 }}}\end{gathered}

⇒CA=20,000(1+

100

5

)

3

⇒CA=20,000(

100

100+5

)

3

⇒CA=20,000×(1.05)

3

⇒CA=20,000×1.157625

⇒

CA=23,152.5

\rule{100}{2}

\because\tt\red{Population\:can't\:be \: in\:fraction}∵Populationcan

′

tbeinfraction

\therefore\bold\green{Population\:at\:the\:end\:of\:2000 = 23,153}∴Populationattheendof2000=23,153

{\boxed{\bold\purple{Population\:in\:2000 = 23153 }}}

Populationin2000=23153