Question

A sum of money becomes rs. 13380 after 3 years and rs. 20070 after 6 years on compound interest. the sum is​

Answer 1

Answer of the question is 8920

Answer 2

Given

• Amount For 3 Years=Rs 13380
• Amount For 6 years=Rs 20070

To Find

• The principal/Sum

Solution

$\mathfrak{ \: as \: we \: know \: that}$

$~~~~~~~~~~\boxed{ \pink{ \sf \: a = p\Large \left( 1 + \dfrac{R}{100} \right)} }$

Sum For 3 Years

$\sf \: 13380 = p( 1 + \dfrac{R}{100} ) ^{3}$

$\sf \: p ( 1 + \dfrac{R}{100} ) ^{3} = 13380 \: \: \: \: - - - - (1)$

Sum For 6 years

$\sf \: 20040 = p ( 1 + \dfrac{R}{100} ) ^{6}$

$\sf \: p ( 1 + \dfrac{R}{100} ) ^{6} = 20070 \: \: \: \: - - - (2)$

On Dividing Equation 2 by 1 we get,

$~~~~ \sf \dfrac{\sf \gray{\: p ( 1 + \dfrac{R}{100} ) ^{6} = 20070} }{\sf \: \: \: \: \: \: p ( 1 + \dfrac{R}{100} ) ^{3} = 13380~~~~ }$

${\implies\sf \: ( 1 + \dfrac{R}{100} ) ^{3} = \dfrac{2}{3} \: \: \: \: - - - (3)}$

On Putting 3 in Equation 1 we get,

${\implies \sf \: p\times \dfrac{3}{2} = 13380}$

$\implies\sf \: p = 13380 \times \dfrac{2}{3}$

$\implies \sf \: p = 4460 \times 2$

$~~~~~~~~~\underline{\boxed{ \bf \: p = 8920}}$

Therefore The Sum is Rs.8920