Answer :
[tex]\huge\mathrm{\underline{\underline{Answer}}}[/tex]
★ Given :-
- Difference between the CI and SI = ₹363
- Years = 2 years
- Rate of interest = 11% pa
★ Notations :-
- p is principal
- r is the rate of interest
- t is the time period
★ Formula :-
- Difference between the compound interest and simple interest =
- p(1+(r/100))^t - p - (p×r×t)/100 = 363
★ To find :-
- Let the sum be :- p
★ Solution :-
- p(1+(r/100))^t - p - (p×r×t)/100 = 363
- p( (1+11/100)^2 - 1 - 11×2/100 ) = 363
- p( (111/100)^2 - 1 - 22/100 ) = 363
- p ( 12321 / 10000 - 1 - 22 / 100 ) = 363
- p ( (1.2321 - 1.22 ) = 363
- p ( 0.0121 ) = 363
- p = 363 / 0.0121
- p = 3630000 / 121
- p = ₹ 30,000
Ans:-The sum is ₹30,000