Question

A sum of money is borrowed at compound interest payable annually. the interest for second and third year rupees 330 and rupees 335 respectively. find the sum.​

Answer 1

Given :

A sum of money is borrowed at compound interest payable annually

The interest for second year = $I_2$ = Rs 330

The interest for third year = $I_3$ = Rs 335

To Find :

The sum of money borrowed

Solution :

Let The sum of money borrowed = p

From compound Interest

Amount = Principal × $(1+\dfrac{Rate}{ 100})^{Time}$

∵  Interest = Amount - Principal

i.e  Interest = Principal × $(1+\dfrac{Rate}{ 100})^{Time}$ - Principal

For second year, t = 2

  $I_2$ = p × $(1+\dfrac{Rate}{ 100})^{2}$ - p  × $(1+\dfrac{Rate}{ 100})^{}$

i.e Rs 330 = p  × $(1+\dfrac{Rate}{ 100})^{}$  [  $(1+\dfrac{Rate}{ 100})^{}$ - 1 ]                          

Or,  Rs 330 = p  × $(1+\dfrac{Rate}{ 100})^{}$  × $\dfrac{Rate}{100}$                                    .....................1

For Third year, t = 3    

  $I_3$ = p × $(1+\dfrac{Rate}{ 100})^{3}$ - p× $(1+\dfrac{Rate}{ 100})^{2}$

i.e Rs 335 = p × $(1+\dfrac{Rate}{ 100})^{2}$ [  $(1+\dfrac{Rate}{ 100})^{}$ - 1 ]                        

Or, Rs 335 = p × $(1+\dfrac{Rate}{ 100})^{2}$ × $\dfrac{Rate}{100}$                ....................2

From eq 1 and eq 2 , we get

  $\dfrac{Rs 335}{Rs 330}$ = $\dfrac{p[(1+\dfrac{Rate}{ 100})^{2}\dfrac{Rate}{100} ]}{p[(1+\dfrac{Rate}{ 100})^{}\dfrac{Rate}{100} ]}$

or, $\dfrac{Rs 335}{Rs 330}$  =    1 + $\dfrac{Rate}{100}$

i.e   $\dfrac{67}{66}$  = 1 + $\dfrac{Rate}{100}$

or,   $\dfrac{67}{66}$  - 1 = $\dfrac{Rate}{100}$

Or,    $\dfrac{1}{66}$  =  $\dfrac{Rate}{100}$

Put the value of $\dfrac{Rate}{100}$ into eq 1

Rs 330 = p  × $(1+\dfrac{1}{ 66})^{}$  × $\dfrac{1}{66}$

or, Rs 330 = p  × $(\dfrac{67}{ 66})^{}$  × $\dfrac{1}{66}$

∴  p = $\dfrac{1437480}{67}$

i.e Principal = Rs 21454.92

Hence, The Sum of principal borrowed is  Rs 21454.92  Answer