Question

What amount would become Rs.18,522 at an annual rate of interest of 10% after
one and a half years if the interest is calculated every six months?​

Answer 1

${ \huge{ \tt{ \green{ \underline{ \underline{ QUESTION- }}}}}}$




▪ What amount of money would become Rs. 18,522 at an annual rate of interest of 10% after one and a half years, if the interest is calculated every six months??




${ \huge{ \tt{ \green{ \underline{ \underline{SOLUTION- }}}}}}$





${ \blue{ \bold{GIVEN- }}}$




${ \sf{ \pink{amount = Rs . \: 18522}}} \\ \\ { \sf{ \pink{rate = 10 \: percent}}} \\ \\{ \sf{ \pink{time = 1 \times \frac{1}{2} years = \frac{3}{2} years}}}$




${ \bold{ \blue{TO \: FIND- }}}$



${ \rightarrow{ \sf{principal(P)}}}$




Since, the money is compounded half- yearly..




${ \star{ \blue{ \sf{ \: \: rate \: will \: be \: halved}}}} \\ \\ { \blue{ \sf{R = \frac{10}{2} = 5 \: percent}}}$




${ \star{ \blue{ \sf{ \: \: time \: will \: get \: doubled}}}} \\ \\ { \sf{ \blue{T = \frac{3}{2} \times 2 \: years = 3 \: years}}}$




${ \boxed{ \boxed{ \red{ \sf{ \: \: amount = P{(1 + \frac{R}{100}) }^{T} \: \: }}}}}$




putting the above given values in the formula...




${ \sf{18522 = P {(1 + \frac{5}{100}) }^{3}}}$





${ \implies{ \sf{18522 = P {(1 + \frac{1}{20} )}^{3}}}}$




${ \implies{ \sf{18522 = P{( \frac{21}{20}) }^{3}}}}$





${ \implies{ \sf{18522 = P \times \frac{9261}{8000}}}}$




${ \implies{ \sf{P = 18522 \times \frac{8000}{9261}}}}$




${ \implies{ \sf{P = 2 \times 8000}}}$




${ \implies{ \boxed{ \boxed{ \pink{ \sf{P = Rs. \: 16000}}}}}}$

Answer 2

Answer:

the avove answer is right