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

Given, 30000 + 4347

           = 34347

Formula is p(1+r/100)^n

               = 30000(1+7/100)^n  = 34347

               = (107/100)^n = 34347/30000

               = 11449/10000

               = (107/100)^2

So, the time period = 2 years.


Hope this helps!

Answer 2

${\mathbf {\blue{S}{\underline{\underline{olution:-}}}}}$

$\mathfrak{\underline{Answer:-}}$

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

$\mathfrak{\underline{Given:-}}$

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

$\mathfrak{\underline{Need\:To\: Find:-}}$

• The time period if the interest is compound annually = ?

${\mathbf {\blue{E}{\underline{\underline{xplanation:-}}}}}$

In Case 1:

$\:\:\:\:\dag\bf{\underline \green{Formula\:used\:here:-}}$

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

$\:\:\:\:\dag\bf{\underline \blue{Putting\:the\:values:-}}$

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

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

$\rule{200}{2}$

In Case 2:

Let the time be N Years.

$\:\:\:\:\dag\bf{\underline \green{Formula\:used\:here:-}}$

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

$\:\:\:\:\dag\bf{\underline \blue{Putting\:the\:values:-}}$

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

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

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

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

$\longrightarrow</p><p>\large\boxed{\sf{\purple{N\:=\: 2 \: years}}}$

$\:\:\:\:\dag\bf{\underline{\underline \blue{Hence:-}}}$

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

$\rule{200}{2}$