Question

At what rate percent per annum will rupee 6000 amount s to
6615 in 2 years when interest is compounded annually

Answer 1

Answer:

${\sf{\red{\underline{\underline{\huge{Answer:}}}}}}$

Given:

• ⇒ Principal amount (P) = Rs 6000
• ⇒ Total amount (A) = Rs 6615
• ⇒ Time (n) = 2 years

To Find:

• ⇒ We need to find the rate of interest.

Solution:

As it is given that the principle amount is Rs 6000, time is 2 years and Total amount is Rs 6615.

We can find the rate of interest by this formula:

A = P[1 + r/100]^n

Substituting the given values, we have

6615 = 6000[1 + r/100]^2

➤ 6615/6000 = (1 + r/100)^2

➤ 1.1025 = (1 + r/100)^2

➤ √1.1025 = (1 + r/100)

➤ 1.05 = (1 + r/100)

➤ 0.05 = r/100

➤ r = 5%

• Therefore, the rate of interest at which Rs 6000 will amount to Rs 6615 in 2 years is 5%.

Answer 2

Answer:

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}\end{gathered}$

• ➽ Principle = Rs.6000
• ➽ Amount = Rs.6615
• ➽ Time = 2 years

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

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{To Find :}}}}}}\end{gathered}$

• ➽ Rate of Interest

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

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Using Formula :}}}}}}\end{gathered}$

$\bigstar{\underline{\boxed{\rm{\purple{A = P\left(1 + \dfrac{R}{100} \right)^{n}}}}}}$

Where

• ➽ A = Amount
• ➽ P = Principle
• ➽ R = Rate of Interest
• ➽ N = Time

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

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}\end{gathered}$

$\red\bigstar$ Here

• ➽ Amount = Rs.6615
• ➽ Principle = Rs.6000
• ➽ Time = 2 years
• ➽ Rate of Interest = ?

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

$\red\bigstar$ Calculating the Rate of Interest,

$\quad{:\implies{\rm{A = P\left(1 + \dfrac{R}{100} \right)^{n}}}}$

• Substuting the values

$\quad{:\implies{\rm{6615 = 6000\left(1 + \dfrac{R}{100} \right)^{2}}}}$

$\quad{:\implies{\rm{\dfrac{6615}{6000} =\left(1 + \dfrac{R}{100} \right)^{2}}}}$

$\quad{:\implies{\rm{\cancel{\dfrac{6615}{6000}} =\left(1 + \dfrac{R}{100} \right)^{2}}}}$

$\quad{:\implies{\rm{\dfrac{441}{400} =\left(1 + \dfrac{R}{100} \right)^{2}}}}$

$\quad{:\implies{\rm{\cancel{\dfrac{441}{400}} =\left( \dfrac{(1 \times 100) + R}{100} \right)^{2}}}}$

$\quad{:\implies{\rm{1.1025=\left( \dfrac{ 100+ R}{100} \right)^{2}}}}$

$\quad{:\implies{\rm{ \sqrt{1.1025} =\left( \dfrac{ 100+ R}{100} \right)}}}$

$\quad{:\implies{\rm{1.05 =\left( \dfrac{ 100+ R}{100} \right)}}}$

$\quad{:\implies{\rm{1.05 \times 100 =({ 100+ R})}}}$

$\quad{:\implies{\rm{105=({ 100+ R})}}}$

$\quad{:\implies{\rm{105 - 100 = R}}}$

$\quad{:\implies{\rm{5 \% = R}}}$

$\quad\bigstar{\underline{\boxed{\rm{\purple{Rate = 5 \%}}}}}$

∴ The Rate of Interest is 5%

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

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Learn More :}}}}}}\end{gathered}$

$\quad\bigstar{\underline{\boxed{\rm{\red{A ={P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}}}$

$\quad\bigstar{\underline{\boxed{\rm{\red{Amount = Principle + Interest}}}}}$

$\quad\bigstar{\underline{\boxed{\rm{\red{ P=Amount - Interest }}}}}$

$\quad\bigstar{\underline{\boxed{\rm{\red{ S.I = \dfrac{P \times R \times T}{100}}}}}}$

$\quad\bigstar{\underline{\boxed{\rm{\red{P = \dfrac{Amount\times 100 }{100 + (Time \times Rate)}}}}}}$

$\quad\bigstar{\underline{\boxed{\rm{\red{P = \dfrac{Interest \times 100 }{Time \times Rate}}}}}}$

$\quad\bigstar{\underline{\boxed{\rm{\red{Rate = \dfrac{Simple \: Interest \times 100}{Principle \times Time}}}}}}$

$\quad\bigstar{\underline{\boxed{\rm{\red{ Time= \dfrac{Simple \: Interest \times 100}{Principle \times Rate}}}}}}$

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