Question

find the compound interest on 3000 at the rate of 6% for one by 1/1/2 year compound half yearly​

Answer 1

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 :

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

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

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

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

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

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

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

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

$\rule{300}{1.5}$

Answer 2

Solution :-

First, we should find the amount of the given values. We have an formula to find the same. It's mentioned below,

Amount :-

$\sf\implies Amount = Principle \bigg(1 + \dfrac{Rate}{100} \bigg)^{time(2)}$

$\sf \implies \:3000 \bigg(1 + \dfrac{6}{100} \bigg)^{1 \dfrac{1}{2} (2)}$

$\sf \implies 3000 \bigg(1 + \dfrac{6}{100} \bigg)^{\dfrac{3}{2} (2)}$

$\sf \implies 3000 \bigg(1 + \dfrac{6}{100} \bigg)^{\dfrac{6}{2}}$

$\sf \implies 3000 \bigg(1 + \dfrac{6}{100} \bigg)^{3}$

$\sf \implies 3000 \bigg(1 + \dfrac{3}{50} \bigg)^{3}$

$\sf \implies 3000 \bigg( \dfrac{50 + 3}{50} \bigg)^{3}$

$\sf \implies 3000 \bigg( \dfrac{53}{50} \bigg)^{3}$

$\sf \implies 3000 \bigg( \dfrac{53^3}{50^3} \bigg)$

$\sf \implies3000 \bigg( \dfrac{148877}{125000} \bigg)$

$\sf \implies 3 \bigg( \dfrac{148877}{125} \bigg)$

$\sf \implies \dfrac{3 \times 148877}{125} = \dfrac{446631}{125}$

$\sf \implies \cancel \dfrac{446631}{125} = 3573.048$

Now, we can find the compound interest.

Compound interest :-

$\sf \leadsto CI = Amount - Principle$

$\sf \leadsto 3573.048 - 3000$

$\sf \leadsto Rs \: . \: 573.048$

Therefore, the compound interest is ₹573.048.