Question

Find the amount when Rs. 10,000 is invested for two and a half years at 10% interest compounded yearly.​

Answer 1

$\bf \underline{ \underline{\maltese\:Given} }$

$\sf \implies Principal = Rs. \: 10,000$

$\sf \implies Rate = 10 \: \%$

$\sf \implies Time = 2 \dfrac{1}{2} \: years$

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

$\sf \implies Amount = \: ?$

$\bf \underline{ \underline{\maltese\:Solution } }$

${\sf When\;time\;is\;in\;fraction\;\left(i.e.\;a\dfrac{b}{c}\right), {Then : }}$

$\sf{Amount=Principal\left(1+\dfrac{Rate}{100}\right)^a\times\left(1+\dfrac{\dfrac{b}{c}\times Rate}{100}\right)}$

$\sf \underline{Substituting \: the \: values},$

$\sf \implies{10000\left(1+\dfrac{10}{100}\right)^2\times\left(1+\dfrac{\dfrac{1}{2}\times10}{100}\right)}$

$\sf \implies{10000\left(1+\dfrac{1}{10}\right)^2\times\left(1+\dfrac{5}{100}\right)}$

$\sf \implies{10000\left(\dfrac{10 + 1}{10}\right)^2\times\left(1+\dfrac{1}{20}\right)}$

$\sf \implies{10000\left(\dfrac{11}{10}\right)^2\times\left(1+\dfrac{1}{20}\right)}$

$\sf \implies{10000\left(\dfrac{11}{10}\right)^2\times\left(\dfrac{20 + 1}{20}\right)}$

$\sf \implies{10000 \times \dfrac{121}{100} \times\dfrac{21}{20}}$

$\sf \implies{100\times \dfrac{121}{1} \times\dfrac{21}{20}}$

$\sf \implies \cancel{\dfrac{100\times 121 \times 21}{20} }$

$\sf \implies 12705$

$\underline{ \boxed{ \red{ \bf Hence, amount \: will \: be \: Rs. \: 12705}}}$

Answer 2

Answer:

Amt = 12,690.59

Step-by-step explanation:

Principle = 10,000

Rate of interest = 10%

Time = 2 yrs 6 months

Amt = p( 1 + r/100)ⁿ

Amt = 10,000( 1 + 0.10)⁵/₂

Amt = 10,000(1.10)⁵/₂

Amt = 10,000 * 1.269059

Amt = 12,690.59