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

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

1. Given

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

   To Find

   • The principal/Sum

   Solution

   [tex] \mathfrak{ \: as \: we \: know \: that}[/tex]

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

   Sum For 3 Years

   [tex]\sf \: 13380 = p( 1 + \dfrac{R}{100} ) ^{3} [/tex]

   [tex]\sf \: p ( 1 + \dfrac{R}{100} ) ^{3} = 13380 \: \: \: \: – – – – (1)[/tex]

   Sum For 6 years

   [tex]\sf \: 20040 = p ( 1 + \dfrac{R}{100} ) ^{6} [/tex]

   [tex]\sf \: p ( 1 + \dfrac{R}{100} ) ^{6} = 20070 \: \: \: \: – – – (2)[/tex]

   On Dividing Equation 2 by 1 we get,

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

   [tex]{\implies\sf \: ( 1 + \dfrac{R}{100} ) ^{3} = \dfrac{2}{3} \: \: \: \: – – – (3)}[/tex]

   On Putting 3 in Equation 1 we get,

   [tex]{\implies \sf \: p\times \dfrac{3}{2} = 13380} [/tex]

   [tex]\implies\sf \: p = 13380 \times \dfrac{2}{3}[/tex]

   [tex]\implies \sf \: p = 4460 \times 2[/tex]

   [tex]~~~~~~~~~\underline{\boxed{ \bf \: p = 8920}}\\\\[/tex]

   Therefore The Sum is Rs.8920

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2. Answer of the question is 8920

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