Question

21. In how many years will 6250 amount to 7290 at 8% per annum, compounded annually?
22. The population of a town is 125000. It is increasing at the rate of 2% per annum. What will
be its population after 3 years?
23. Three years ago, the population of a town was 50000. If the annual increase during
three successive years be at the rate of 5%, 4% and 3% respectively, what is its present
population?
24. The population of a city was 120000 in the year 2013. During next year it increased by 6%
but due to an epidemic it decreased by 5% in the following year. What was its population in
the year 2015?​

Answer 1

$\bf{\underline{\underline{Solutions}}}$

21. $\bf{\underline{\underline{Given}}}$

• Principal = Rs. 6250
• Amount = Rs. 7290
• Rate = 8% p.a.

$\bf{\underline{\underline{To\:Find}}}$

The time aftter which Rs.6250 will amount to Rs.7290.

$\bf{\underline{\underline{Solution}}}$

$\bf{\underline{We\:know}}$

$\bf{A = P\bigg(1+\dfrac{r}{100}\bigg)^t}$

= $\bf{7290 = 6250\bigg(1+\dfrac{8}{100}\bigg)^t}$

= $\bf{\dfrac{7290}{6250} = \bigg(1+\dfrac{2}{25}\bigg)^t}$

= $\bf{\dfrac{729}{625} = \bigg(\dfrac{25+2}{25}\bigg)^t}$

= $\bf{\bigg(\dfrac{27}{25}\bigg)^2 = \bigg(\dfrac{27}{25}\bigg)^t}$

$\bf{As \:the \:bases \:are\: same,}$

t = 2

$\bf{\therefore t = 2\:years}$

$\bf{After\:2\:years\:Rs.6258\:will\:amount\:to\:Rs.7290}$

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22. $\bf{\underline{\underline{Given}}}$

• Population of the town (P) = 125000
• Rate (R) = 2%
• Time (T) = 3 years

$\bf{\underline{\underline{To\:Find}}}$

Population after 3 years.

$\bf{\underline{\underline{Assumption}}}$

Let the population of town after 3 years be A

$\bf{\underline{\underline{Solution}}}$

The population of town after three years,

$\bf{A = P\bigg(1+\dfrac{r}{100}\bigg)^t}$

= $\bf{A = 125000\bigg(1+\dfrac{2}{100}\bigg)^3}$

= $\bf{A = 125000\bigg(1+\dfrac{1}{50}\bigg)^3}$

= $\bf{A = 125000\bigg(\dfrac{50+1}{50}\bigg)^3}$

= $\bf{A = 125000\times\dfrac{51}{50}\times\dfrac{51}{50}\times\dfrac{51}{50}}$

= $\bf{A = 132651}$

Therefore The population after 3 years will be 132651.

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23. $\bf{\underline{\underline{Given}}}$

• $\bf{Population\: of \:a \:town\: 3 \:years\: ago\: (P) = 50000}$
• $\bf{Rate \:of \:increase\: for \:1st \:year\: (r_1)= 5\%}$
• $\bf{Rate \:of \:increase\: for \:2nd \:year\: (r_2)= 4\%}$
• $\bf{Rate \:of \:increase\:for \:3rd \:year \:(r_3)= 3\%}$
• $\bf{Time \:(t) = 3 \:years }$

$\bf{\underline{\underline{To\:find}}}$

The present population of the town.

$\bf{\underline{\underline{Assumption}}}$

Let the present population be A

$\bf{\underline{\underline{Solution}}}$

$\bf{Present\:population}$

$\bf{A = 50000\bigg(1+\dfrac{r_1}{100}\bigg)\bigg(1+\dfrac{r_2}{100}\bigg)\bigg(1+\dfrac{r_3}{100}\bigg)}$

= $\bf{A = 50000\bigg(1+\dfrac{5}{100}\bigg)\bigg(1+\dfrac{4}{100}\bigg)\bigg(1+\dfrac{3}{100}\bigg)}$

= $\bf{A = 50000\bigg(\dfrac{100+5}{100}\bigg)\bigg(\dfrac{100+4}{100}\bigg)\bigg(\dfrac{100+3}{100}\bigg)}$

= $\bf{A = 50000\times\dfrac{105}{100}\times\dfrac{104}{100}\times\dfrac{103}{100}}$

= $\bf{A = 56238}$

Therefore the present population is 56328.

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24. $\bf{\underline{\underline{Given}}}$

• $\bf{Population\: of\: a\: city\: in \:2013 = 120000}$
• $\bf{Rate\: of\: 1st \:year\: (r_1) = 6\%}$
• $\bf{Rate\:of\:2nd\:year\:(r_2) = 5\%}$
• $\bf{Time = 2015-2013 = 2\:years}$

$\bf{\underline{\underline{To\:find}}}$

Population in the year 2015 i.e., after 2 years.

$\bf{\underline{\underline{Assumption}}}$

Let the Population in the year 2015 be A

$\bf{\underline{\underline{Solution}}}$

$\bf{Population\:in\:the\:year\:2015}$

$\bf{A = P\bigg(1+\dfrac{r_1}{100}\bigg)\bigg(1-\dfrac{r}{100}\bigg)}$

= $\bf{A = 120000\bigg(1+\dfrac{6}{100}\bigg)\bigg(1-\dfrac{5}{100}\bigg)}$

= $\bf{A = 120000\bigg(\dfrac{100+6}{100}\bigg)\bigg(\dfrac{100-5}{100}\bigg)}$

= $\bf{A = 120000\times\dfrac{106}{100}\times\dfrac{95}{100}}$

= $\bf{A = 120840}$

Therefore the population in the year 2015 is 120840.

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