Question

1
Find the amount and the compound interest on 64000 for 1 years at 15%
per annum
compounded half-yearly.​

Answer 1

Answer:

a=p(1+r/100)^t1

Step-by-step explanation:

Compound interest= A-P hope it helped☺️☺️

Answer 2

${\large{\underline{\underline{\bf{\green{Given : -}}}}}}$

⟶ Principle = Rs.64000

⟶ Time = 1½ years

⟶ Rate = 15% per annum componded half yearly.

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

${\large{\underline{\underline{\bf{\green{To \: Find : -}}}}}}$

⟶ Amount

⟶ Compound Interest

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

${\large{\underline{\underline{\bf{\green{ Using \: Formulae: -}}}}}}$

⟶ A = P(1 + R/100)ᵀ

⟶ C.I = A - P

$\red\bigstar$ Where

⟶ A = Amount

⟶ P = Principle

⟶ R = Rate

⟶ T = Time

⟶ C.I = Compound Interest

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

${\large{\underline{\underline{\bf{\green{Solution : -}}}}}}$

$\red\bigstar$ Here, the interest is compounded half yearly and time period is 1½, then :-

${\dashrightarrow{\sf{Time = 1\dfrac{1}{2}}}}$

${\dashrightarrow{\sf{Time = \dfrac{(1 \times 2) + 1}{2}}}}$

${\dashrightarrow{\sf{Time = \dfrac{ 2 + 1}{2}}}}$

${\dashrightarrow{\sf{Time = \dfrac{3}{2}}}}$

${\dashrightarrow{\sf{Time =3 \: Half \: Years}}}$

$\bigstar{\underline{\boxed{\bf{\blue{Time =3 \: Half \: Years}}}}}$

∴ The time is 3 half years.

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

$\red\bigstar$ Let us find out the Amount by Substituting the values in the formula :-

${\dashrightarrow{\sf{Amount = P\bigg(1 + \dfrac{R}{100} \bigg)^{T}}}}$

Substituting the values

${\dashrightarrow{\sf{Amount = 64000 \bigg(1 + \dfrac{15}{2 \times 100} \bigg)^{3}}}}$

${\dashrightarrow{\sf{Amount = 64000\bigg(1 + \dfrac{15}{200} \bigg)^{3}}}}$

${\dashrightarrow{\sf{Amount = 64000\bigg( \dfrac{(1 \times 200) + 15}{200} \bigg)^{3}}}}$

${\dashrightarrow{\sf{Amount = 64000\bigg( \dfrac{200+ 15}{200} \bigg)^{3}}}}$

${\dashrightarrow{\sf{Amount = 64000\bigg( \dfrac{215}{200} \bigg)^{3}}}}$

${\dashrightarrow{\sf{Amount = 64000\bigg( \cancel{\dfrac{215}{200}} \bigg)^{3}}}}$

${\dashrightarrow{\sf{Amount = 64000\bigg({\dfrac{43}{40}} \bigg)^{3}}}}$

${\dashrightarrow{\sf{Amount = 64000\bigg({\dfrac{43}{40}} \times \dfrac{43}{40} \times \dfrac{43}{40} \bigg)}}}$

${\dashrightarrow{\sf{Amount = 64000\bigg( \dfrac{79507}{64000} \bigg)}}}$

${\dashrightarrow{\sf{Amount = 64000 \times \dfrac{79507}{64000}}}}$

${\dashrightarrow{\sf{Amount = \cancel{64000} \times \dfrac{79507}{\cancel{64000}}}}}$

${\dashrightarrow{\sf{Amount =Rs.79507}}}$

$\bigstar{\underline{\boxed{\bf{\purple{Amount =Rs.79507}}}}}$

∴ The Amount is Rs.79507.

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

$\red\bigstar$ Let us find out the compound interest by substituting the values in the formula :-

${\dashrightarrow{\sf{Compound \: Interest=Amount - Principal}}}$

Substituting the values

${\dashrightarrow{\sf{Compound \: Interest=79507 - 64000}}}$

${\dashrightarrow{\sf{Compound \: Interest=Rs.15507}}}$

$\bigstar{\underline{\boxed{\bf{\pink{Compound \: Interest=Rs.15507}}}}}$

∴ The compound interest is Rs.15507.

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

${\large{\underline{\underline{\bf{\green{Learn \: More : -}}}}}}$

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

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

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

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

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

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

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

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

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