Question

The simple interest on a certain sum of money for 2 years at 10% per annum is $800. What would be the
corresponding compound interest?​

Answer 1

Given :

• Simple interest = Rs.800
• Time = 2 years
• Rate of interst = 10%

To Find :

• The corresponding compound interest

Solution :

We have ,

• Simple interst (SI) = Rs.800
• Time (T) = 2 years
• Rate of interst (R) = 10%

Simple interest is calculated by ,

$\\ \star \: \boxed{\purple{\sf{SI = \dfrac{PTR}{100}}}}$

Where ,

• P is principal

By substituting the values we have ,

$\\ : \implies \sf \: 800 = \frac{P \times 2 \times 10}{100}$

$\\ : \implies \sf \: 800 \times 100 = P \times 20$

$\\ : \implies \sf \:P = \frac{800 \times 100}{20}$

$\\ : \implies \boxed{\pink{\sf{P = Rs.4000 }}}$

Now , Comppund interst is calculated by ;

$\\ \star\boxed{\purple{\sf{CI = P\bigg(1+\dfrac{R}{100}\bigg)^{n} - P}}}$

Where ,

• P is principal
• R is Rate of interest

$\\ : \implies \sf \: CI = 4000 \bigg(1 + \frac{10}{100} \bigg)^{2} - 4000$

$\\ : \implies \sf \:CI = 4000 \bigg( \frac{100 + 10}{100} \bigg)^{2} - 4000$

$\\ : \implies \sf \:CI = 4000 \bigg( \frac{110}{100} \bigg)^{2} - 4000$

$\\ : \implies \sf \:CI = 4000 \bigg( \frac{110 \times 110}{100 \times 100} \bigg) - 4000$

$\\ : \implies \sf \: CI = 400 \bigg( \frac{12100}{10000}\bigg) - 4000$

$\\ : \implies \sf \:CI = 4000 \bigg( \frac{121}{100}\bigg) - 4000$

$\\ : \implies \sf \:CI = 4840 - 4000$

$\\ : \implies \boxed{\pink {\sf{ \: CI =Rs.840 }}} \: \bigstar$

Hence , The corresponding compound Interest is Rs.840

Answer 2

Answer:

Given :

Simple interest = Rs.800

Time = 2 years

Rate of interst = 10%

To Find :

The corresponding compound interest

Solution :

We have ,

Simple interst (SI) = Rs.800

Time (T) = 2 years

Rate of interst (R) = 10%

Simple interest is calculated by ,

\begin{gathered} \\ \star \: \boxed{\purple{\sf{SI = \dfrac{PTR}{100}}}}\end{gathered}

⋆

SI=

100

PTR

Where ,

P is principal

By substituting the values we have ,

\begin{gathered} \\ : \implies \sf \: 800 = \frac{P \times 2 \times 10}{100} \\ \\ \end{gathered}

:⟹800=

100

P×2×10

\begin{gathered} \\ : \implies \sf \: 800 \times 100 = P \times 20 \\ \\ \end{gathered}

:⟹800×100=P×20

\begin{gathered} \\ : \implies \sf \:P = \frac{800 \times 100}{20} \\ \\ \end{gathered}

:⟹P=

20

800×100

\begin{gathered} \\ : \implies \boxed{\pink{\sf{P = Rs.4000 }}} \\ \\ \end{gathered}

:⟹

P=Rs.4000

Now , Comppund interst is calculated by ;

\begin{gathered} \\ \star\boxed{\purple{\sf{CI = P\bigg(1+\dfrac{R}{100}\bigg)^{n} - P}}}\end{gathered}

⋆

CI=P(1+

100

R

)

n

−P

Where ,

P is principal

R is Rate of interest

\begin{gathered} \\ : \implies \sf \: CI = 4000 \bigg(1 + \frac{10}{100} \bigg)^{2} - 4000 \\ \\ \end{gathered}

:⟹CI=4000(1+

100

10

)

2

−4000

\begin{gathered} \\ : \implies \sf \:CI = 4000 \bigg( \frac{100 + 10}{100} \bigg)^{2} - 4000\\ \\ \end{gathered}

:⟹CI=4000(

100

100+10

)

2

−4000

\begin{gathered} \\ : \implies \sf \:CI = 4000 \bigg( \frac{110}{100} \bigg)^{2} - 4000 \\ \\ \end{gathered}

:⟹CI=4000(

100

110

)

2

−4000

\begin{gathered} \\ : \implies \sf \:CI = 4000 \bigg( \frac{110 \times 110}{100 \times 100} \bigg) - 4000 \\ \\ \end{gathered}

:⟹CI=4000(

100×100

110×110

)−4000

\begin{gathered} \\ : \implies \sf \: CI = 400 \bigg( \frac{12100}{10000}\bigg) - 4000\\ \\ \end{gathered}

:⟹CI=400(

10000

12100

)−4000

\begin{gathered} \\ : \implies \sf \:CI = 4000 \bigg( \frac{121}{100}\bigg) - 4000\\ \\ \end{gathered}

:⟹CI=4000(

100

121

)−4000

\begin{gathered} \\ : \implies \sf \:CI = 4840 - 4000\\ \\\end{gathered}

:⟹CI=4840−4000

\begin{gathered} \\ : \implies \boxed{\pink {\sf{ \: CI =Rs.840 }}} \: \bigstar \\ \\ \end{gathered}

:⟹

CI=Rs.840

★

Hence , The corresponding compound Interest is Rs.840