1.
A = P [ 1 + [ R/(100*n) ^{n*t}] ] |
P -----> Principle
R -----> Rate % per annum
n -----> No.of convertions per year
T -----> Time in years
2.C.I=A-P i.e.,
[ P [ 1+(R/(100*n)^{n*t}] - 1 ] |
3.When interest is calculated anually n=1,
4.When time is in fraction, t = x * (1/y) year:
A = P[1+(R/100)^{x}] + [1 + (1/y)*R/100 ] |
5.When rate od interest is R_{1}% R_{2}% R_{3}% for I^{st} year,II^{nd}year, III^{rd} year respectively then amount,
A = P [ 1 + (R_{1}100) * [1+(R_{2}/100)] [1+(R_{3}/100)] |
6.When difference between C.I and S.I on certain sum at rate% on Rs.x,
[ C.I - S.I = sum * (r/100)^{2} ] |
i.e., [ D = P * (r/100)^{2} ] |
Note: Applicable only for two years. 7.
D =[ (P*R^{2})(300+R)/100^{3} ] |
Note:Applicable only for 3 years.
8.
[ C.I/(200+R) = S.I/200 ] |
Note: Applicable only for 2 years.
9.
R = [ (2*difference of C.I and S.I)/S.I ] * 100 |
10.R% amounts after 2 successive years we given:-
R = [ (A_{n+1}-A_{n})/A_{n} ] * 100 |
A_{n+1} -----> Amount after (n+1) years.
A_{n} -----> Amount after (n+1) years.
11.
P =[ A_{3}^{2}/A_{6} ] = [ A_{2}^{2}/A_{4} ] = [ A_{1}^{2}/A_{2} ] = [A_{4}^{2}/A_{8} ] |
Note: Double the years.
12.
P =[ A_{2}^{3}/A_{3}^{2} ] = [ A_{3}^{4}/A_{4}^{3} ] = [ A_{4}^{5}/A_{5}^{4} ] |
Note: Consecutive years.
P =[ ÖA_{2}^{3}/A_{6} ] = [ ÖA_{1}^{3}/A_{3} ] = [ ÖA_{3}^{3}/A_{9} ] |
13.
R =[( A_{6}/A_{3})^{1/3} - 1 ] = [ ( A_{4}/A_{2} )^{1/2} - 1 ] = [ ( A_{5}/A_{2} )^{1/3} - 1 ]
R =[( A_{7}/A_{2})^{1/3} - 1 ] = [ ( A_{10}/A_{2} )^{1/8} - 1 ] = [ ( A_{10}/A_{7} )^{1/3} - 1 ] |
14.Installment problems:
a [ 100/(100+R) + 100/(100+R)^{2} + 100/(100+r)^{3} + ....... ] = B |
a -----> Annual installment
B -----> Borrowed amount.
15. R = [ (A/P)^{1/T} - 1 ] * 100 |
16. P = [ A_{2} * [100/(100+R)]^{2} ] | |
1 comment:
HI.
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http://infibee.com/general-aptitude/compound-interest/
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