Mr. Tharu borrowed a sum of Rs 1,00,000 from an Agricultural Bank for 3 years to upgrade his poultry farming. The bank charged 5% interest compounded annually for the first year and the rate of interest was gradually increased by 1% every year. How much interest did he pay at the end of third year?

Answer: Rs 19091

Solution:

Given,

Principal (P) = Rs. 100000

Time (T) = 3 years

Interest in first year ($R_1%$) = 5%

According to the question, interest increased by 1% every year,

Interest in second year ($R_2%$) = 6%

Interest in third year ($R_3%$) = 7%

To find: Compound interest after 3 years (CI) = ?

We know,

$CI = P \left [ \left ( 1 + \dfrac{R_1}{100} \right ) \left ( 1 + \dfrac{R_2}{100} \right ) \left ( 1 + \dfrac{R_3}{100} \right ) -1 \right ]$

$CI = P \left [ \left ( 1 + \dfrac{5}{100} \right ) \left ( 1 + \dfrac{6}{100} \right ) \left ( 1 + \dfrac{7}{100} \right ) -1 \right ]$

$CI = P \left [ \left ( \dfrac{100+5}{100} \right ) \left ( 1 + \dfrac{100+6}{100} \right ) \left ( 1 + \dfrac{100+7}{100} \right ) -1 \right ]$

$CI = P \left [ \left ( \dfrac{105}{100} \right ) \left ( 1 + \dfrac{106}{100} \right ) \left ( 1 + \dfrac{107}{100} \right ) -1 \right ]$

$CI = P \left [ \dfrac{1190910}{1000000}  -1 \right ]$

$CI = P \left [ \dfrac{1190910 - 1000000}{1000000} \right ]$

$CI = P \left [ \dfrac{190910}{1000000} \right ]$

$CI = 100000 \left [ \dfrac{190910}{1000000} \right ]$

$CI = \dfrac{190910}{10}$

$\therefore CI = Rs 19091$

Hence, Mr. Tharu paid Rs. 19091 at the end of third year as the interest of the money burrowed.

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