Question

The compound interest on rupees 30,000 at 7% per annum for a certain time is rupees 4347. Find the time period if the interest is compound annually.​

Answer 1

ATQ

$30000 {( 1 + \frac{7}{100} )}^{t} - 30000 = 4347$

${( \frac{107}{100} )}^{t} = \frac{34347}{30000}$

${( \frac{107}{100} )}^{t} = \frac{11449}{10000}$

${( \frac{107}{100} )}^{t} = {\frac{107}{100}}^{2}$

Therefore time = 2 years

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Answer 2

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

• The compound interest on rupees 30,000 at 7% per annum for a certain time is rupees 4347.

$\;\;\underline{\textbf{\textsf{ To Find :-}}}$

•What's the time period if the interest is compound annually ?

$\;\;\underline{\textbf{\textsf{ Formula Used :-}}}$

$\bigstar{\underline{\boxed{\sf\green{Amount = Principal + Interest}}}}$

$\bigstar{\underline{\boxed{\sf\green{Amount = P\bigg(1 + \dfrac{R}{100}\bigg)^X}}}}$

$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

Let the time be X Years.

We know that,

$\bigstar{\underline{\boxed{\sf\green{Amount = Principal + Interest}}}}$

$\longrightarrow \sf {Amount = Rs.(30000 + 4347) }$

$\longrightarrow \sf {Amount = Rs. 34347 }$

_______________________________________________________

Again, we know

$\bigstar{\underline{\boxed{\sf\green{Amount = P\bigg(1 + \dfrac{R}{100}\bigg)^X}}}}$

$\longrightarrow \sf {34347 = 30000\bigg(1 + \dfrac{7}{100}\bigg)^X }$

$\longrightarrow \sf {\dfrac{34347}{30000} = \bigg(\dfrac{107}{100}\bigg)^X}$

$\longrightarrow \sf {\dfrac{11449}{10000} = \bigg(\dfrac{107}{100}\bigg)^X}$

$\longrightarrow \sf {\bigg(\dfrac{107}{100}\bigg)^2 = \bigg(\dfrac{107}{100}\bigg)^X}$

$\\ \\ \tt{ \green{ \longrightarrow \: X = 2 \: years}}$

$\small{\underline{\sf{\green{Hence -}}}}$

The time period if the interest is compound annually is 2 years.