Question

A man borrowed 8000 from a money lender at 8% (p.a.) on 18th March, 2015. He
returns the money back with interest on 26th October 2015. Find the amount of interest
he had to pay.​

Answer 1

$\large{ \boxed{ \red{ \underline{ \purple{ \mid} \overline{ \sf \green{Question}} \purple{\mid} }} } }: -$

A man borrowed 8000 from a money lender at 8% (p.a.) on 18th March, 2015. He returns the money back with interest on 26th October 2015. Find the amount of interest he had to pay.

$\large{ \blue{\boxed{ \red{ \underline{ \purple{ \mid} \overline{ \sf \pink{Answer}} \purple{\mid} }} }}} : - \: \\ \sf He \: had \: to \: pay \: Rs. 389.260 \: \\ \large{ \purple{ \boxed{ \red{ \underline{ \purple{ \mid} \overline{ \sf \green{To \: Find}} \purple{\mid} }} } }}: - \: \\ \implies \sf amount \: of \: interest \: \\ \\ \large{ \red{ \boxed{ \red{ \underline{ \purple{ \mid} \overline{ \sf \red{Explanation}} \purple{\mid} }} } }}: - \: \\ \green{ \sf {Given}\begin{cases}\sf{ \pink{Principal (P) = }Rs. 8000 \: }\\\sf{ \blue{Rate (R) = }8\%}\\ \sf{Time (t) = \red{ {18}^{th} \: march \: to \: 26 \: october\: } }\\\sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sf \pink{ \frac{222}{365} }\: per \: year}\end{cases}}</p><p> \:$

$\large{\red{\boxed{ \red{ \underline{ \purple{ \mid} \overline{ \sf \purple{Solution}} \purple{\mid} }} } }}: - \:$

We know that

$\sf \mapsto \blue{ Simple }\: interest (SI) = \red{ \frac{ P \times R \times \: T}{100} }$

put the given values

$\mapsto \sf \green{ (SI) =} \frac{ \pink{8000 \times 8 } \red{ \times \: \frac{222}{365}} }{100} \: \: \\ \\ \mapsto \sf \red{ (SI) }= \green{ \frac{640 \times 222}{365} } \\ \\ \mapsto \sf \: (SI) \: = \pink{\frac{142080}{365}} \: \\ \\ \mapsto \boxed{ \red{\sf \: (SI) = 389.260}}$

Hence He had to pay Rs. 389.260

Answer 2

Answer:

${\boxed{\sf\red{Simple\:interest = Rs.390 [Approx.]}}}$

Question:-

A man borrowed Rs.8000 from a money lender at 8% per annum on 18th March,2015.He returns the money with interest on 26th October,2015.Find the amount of interest he had to pay.

At first,

Time(n) = 18th March to 26th October.

$\hookrightarrow\sf Time(n) = 222\: days.\\\\\hookrightarrow\sf\green{Time(n) = \dfrac{222}{365}\:years}$

$\frak\green{Given:-}\begin{cases}\tt\red{Principal(P) = Rs.8000}\\\tt\orange{Rate\: of \: interest(r) = 8\%}\\\tt\green{Time(n) =\dfrac{222}{365}\:years}\\\tt\pink{Simple\: interst(SI) = ?}\end{cases}$

We know that,

$\huge\boxed{\boxed{\sf\gray{SI = Prn}}}$

$\sf{*Substituting\: values:-}$

$\implies\sf SI = 8000 \times 8\% \times \dfrac{222}{365} \\\\\\\implies\sf SI = 80\cancel{00} \times \dfrac{8}{1\cancel{00}} \times \dfrac{222}{365} \\\\\\\implies\sf SI = 640 \times \dfrac{222}{365} \\\\\\\implies\sf SI = 640 \times 0.608 \\\\\\\implies\large{\boxed{\sf\pink{SI = Rs.390 \: \: [Approx.]}}}$

${\underline{\boxed{\therefore{\sf\blue{Amount\: of \: interest = Rs.390 \:[Approx.]}}}}}$