Question

Questions of the Day..
(1). A car was purchased for 150,000 in the year 2017. If it's value decresed by 16 % every year. Find the value after 2 years.

(2). The population of a town is increasing at the rate of 10 % per annum. What will be the population of the town after three years if the present population is 15000. ​

Answer 1

Solution - 1

$\sf \: Value \: after \: 2 \: Years = Present \: value \: \bigg( 1 - \dfrac{R }{100} \bigg)^{n}$

$\sf \: Value \: after \: 2 \: Years = P \: \bigg( 1 - \dfrac{R }{100} \bigg)^{n}$

$\sf \longrightarrow \: 1,50,000 \: \bigg( \: 1 - \dfrac{16}{100} \: \bigg)^{2}$

$\sf \longrightarrow \: 1,50,000 \: \bigg( \dfrac{84}{100} \: \bigg)^{2}$

$\sf \longrightarrow \: 1,05,840$

$\underline{ \boxed{\pink{ \sf \: Therefore \: value \: of \: car \: after \: 2 \: Years = Rs. \: 1,05,840 }}}$

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Solution - 2

$\bf \: \underline{Given} :$ ↴

$\sf \: Present \: population \: (P) = 15000$

$\sf \: Rate \: of \: growth \: or \: Increase \: = 10 \: \% \: per \: annum$

$\sf \: Time \: (n) = 3 \: Years$

$\sf \: Population \: after \: 3 \: Years = P \: \bigg( 1 + \dfrac{R }{100} \bigg)^{n}$

$\sf \longrightarrow \: 15000\: \bigg( 1 + \dfrac{1 }{100} \bigg)^{3}$

$\sf \longrightarrow \: 15000\: \bigg( \dfrac{11 }{10} \bigg)^{3}$

$\sf \longrightarrow \: 15000\: \bigg( \dfrac{11 }{10} \times \dfrac{11 }{10} \times \dfrac{11 }{10} \bigg)$

$\sf \longrightarrow \: 19,965$

$\underline{ \boxed{ \pink{ \sf \: Hence \: Population \: after \: three \: years = 19,965}}}$