Question

Find the compound interest on $3000 at the rate of 6% per annum for 1 *1/2 years compounded half yearly

Answer 1

• If the compound interest is half yearly.

Then

   Amount = p [ 1 + r/200 ] ² ⁿ

                  = $ 3000 [ 1+6/200 ] ³

                   = $ 3000  x 103/100 x 103/100 x 103/100

                    = $ 3000 x 1092727/1000000

                     = $3278 . 181

CI = A - P = $ 3278.181 - 3000 = 278.181

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Answer 2

Answer:

Given :

➤ Principle = Rs.3000

➤ Rate = 6% per annum

➤ Time = 1½ year compound half yearly

$\begin{gathered}\end{gathered}$

To Find :

➤ Amount

➤ Compound Interest

$\begin{gathered}\end{gathered}$

Using Formulas :

$\small\longrightarrow{\underline{\boxed{\pmb{\sf{A= P\bigg(1 + \dfrac{ \frac{R}{2} }{100} \bigg)^{2n}}}}}}$

$\small\longrightarrow{\underline{\boxed{\pmb{\sf{{C.I=A- P}}}}}}$

☼ Where :-

➢ A = Amount

➢ P = Principle

➢ R = Rate

➢ n = Time

➢ C.I = Compound Interest

$\begin{gathered}\end{gathered}$

Solution :

☼ Firstly, finding the amount by substituting the given values in the formula :-

$\longrightarrow{\sf{A= P\bigg(1 + \dfrac{ \frac{R}{2} }{100} \bigg)^{2n}}}$

$\longrightarrow{\sf{A= P\bigg(1 + \dfrac{R}{100 \times 2} \bigg)^{2n}}}$

$\longrightarrow{\sf{A= P\bigg(1 + \dfrac{R}{200} \bigg)^{2n}}}$

$\longrightarrow{\sf{A= 3000\bigg(1 + \dfrac{6}{200} \bigg)^{2 \times \frac{3}{2} }}}$

$\longrightarrow{\sf{A= 3000\bigg(1 + \dfrac{6}{200} \bigg)^{\frac{6}{2} }}}$

$\longrightarrow{\sf{A= 3000\bigg(1 + \dfrac{6}{200} \bigg)^{\cancel{\frac{6}{2}}}}}$

$\longrightarrow{\sf{A= 3000\bigg(1 + \dfrac{6}{200} \bigg)^{3}}}$

${\longrightarrow{\sf{A= 3000\bigg(\dfrac{(1 \times 200) + (6 \times 1)}{200} \bigg)^{3}}}}$

${\longrightarrow{\sf{A= 3000\bigg(\dfrac{200 + 6}{200} \bigg)^{3}}}}$

${\longrightarrow{\sf{A= 3000\bigg(\dfrac{206}{200} \bigg)^{3}}}}$

${\longrightarrow{\sf{A= 3000\bigg(\cancel{\dfrac{206}{200}} \bigg)^{3}}}}$

${\longrightarrow{\sf{A= 3000\bigg({\dfrac{103}{100}} \bigg)^{3}}}}$

${\longrightarrow{\sf{A= 3000\bigg({\dfrac{103}{100}} \times \dfrac{103}{100} \times \dfrac{103}{100}\bigg)}}}$

${\longrightarrow{\sf{A= 3000\bigg({\dfrac{1092727}{1000000}}\bigg)}}}$

${\longrightarrow{\sf{A= 3000 \times \dfrac{1092727}{1000000}}}}$

${\longrightarrow{\sf{A= 3\cancel{000} \times \dfrac{1092727}{1000\cancel{000}}}}}$

${\longrightarrow{\sf{A= 3\times \dfrac{1092727}{1000}}}}$

${\longrightarrow{\sf{A= \dfrac{3 \times 1092727}{1000}}}}$

${\longrightarrow{\sf{A= \dfrac{3278181}{1000}}}}$

${\longrightarrow{\underline{\underline{\sf{A= Rs.3278.181}}}}}$

${\bigstar{\underline{\boxed{\sf{\pink{Amount= Rs.3278.181}}}}}}$

∴ The amonut is Rs.3278.181.

$\begin{gathered}\end{gathered}$

☼ Now, finding the compound interest by substituting the given values in the formula :-

$\longrightarrow{\sf{C.I=A- P}}$

$\longrightarrow{\sf{C.I=3278.181 - 3000}}$

$\longrightarrow{\underline{\underline{\sf{C.I=Rs.278.181}}}}$

${\bigstar{\underline{\boxed{\sf{\pink{Compound \: Interest=Rs.278.181}}}}}}$

∴ The compound interest is Rs.278.181.

$\begin{gathered}\end{gathered}$

Learn More :

$\small\circ{\underline{\boxed{\pmb{\sf{ Simple \: Interest = \dfrac{P \times R \times T}{100}}}}}}$

$\small\circ{\underline{\boxed{\pmb{\sf{Amount={P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}}}$

$\small\circ{\underline{\boxed{\pmb{\sf{Amount = Principle + Interest}}}}}$

$\small\circ{\underline{\boxed{\pmb{\sf{ Principle=Amount - Interest }}}}}$

$\small\circ{\underline{\boxed{\pmb{\sf{Principle = \dfrac{Amount\times 100 }{100 + (Time \times Rate)}}}}}}$

$\small\circ{\underline{\boxed{\pmb{\sf{Principle = \dfrac{Interest \times 100 }{Time \times Rate}}}}}}$

$\small\circ{\underline{\boxed{\pmb{\sf{Rate = \dfrac{Simple \: Interest \times 100}{Principle \times Time}}}}}}$

$\small\circ{\underline{\boxed{\pmb{\sf{Time = \dfrac{Simple \: Interest \times 100}{Principle \times Rate}}}}}}$

$\rule{300}{1.5}$