16000 invested at 10% p.a. compounded semi-annually amounts to 18522. Find the
time period of the investment.​

1 thought on “16000 invested at 10% p.a. compounded semi-annually amounts to 18522. Find the<br />time period of the investment.​”

1. [tex]\bf \underline{ \underline{\maltese\:Given} }[/tex]

   [tex] \sf \implies Principal \: (P) = Rs. \: 16000[/tex]

   [tex] \sf \implies Rate \: of \: interest \: (R) = 10 \: \% \: \: \: [Compounded \: semi- annually][/tex]

   [tex] \sf \implies Rate \: of \: interest \: (R) = 10 \: \% = \dfrac{10}{2} = 5 \: \%[/tex]

   [tex] \sf \implies Amount = Rs. \: 18522[/tex]

   [tex]\bf \underline{\underline{\maltese\: To \: find }}[/tex]

   [tex] \sf \implies Time \: period \: of \: the \: investment = \: ? [/tex]

   [tex]\bf \underline{\underline{\maltese\: Solution }}[/tex]

   [tex] \underline{ \boxed{\sf Amount = Principal \bigg(1+ \dfrac{Rate }{100} \bigg ) ^{Time }}}[/tex]

   [tex] \sf Substituting \: the \: values, [/tex]

   [tex]\sf \implies 18522 = 16000 \bigg(1+ \dfrac{5 }{100} \bigg ) ^{Time }[/tex]

   [tex]\sf \implies 18522 = 16000 \bigg(1+ \dfrac{1}{20} \bigg ) ^{Time }[/tex]

   [tex]\sf \implies 18522 = 16000 \bigg( \dfrac{20 + 1}{20} \bigg ) ^{Time }[/tex]

   [tex] \sf \implies18522 = 16000 \bigg( \dfrac{21}{20} \bigg ) ^{Time }[/tex]

   [tex] \sf \implies \cancel{ \dfrac{18522}{16000}} = \bigg( \dfrac{21}{20} \bigg ) ^{Time }[/tex]

   [tex] \sf \implies \dfrac{9261}{8000} = \bigg( \dfrac{21}{20} \bigg ) ^{Time }[/tex]

   [tex] \sf \implies \bigg( \dfrac{21}{20} \bigg ) ^{3}= \bigg( \dfrac{21}{20} \bigg ) ^{Time }[/tex]

   [tex] \sf If \: the \: \bf bases \sf \: are \: \bf{ same } \sf \: then, \bf{index } \sf\: will \: also \: be \: equal. [/tex]

   [tex] \sf \implies Time = 3 \: half \:years [/tex]

   [tex] \bf \underline{ Therefore},[/tex]

   [tex] \underline{ \boxed{ \red{\bf Time \: period \: of \: interest \: is \: 3 \: half \: years = \dfrac{3}{2} = 1.5 \: Years }}}[/tex]

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