Maths Tips Compound Interest, Formulas, Problems
Here we are listed All the Important terms, Formulas, Problems of Compound Interest on General Aptitude. Those Maths Tips will help you get maximum marks from this section. This file will be updated regularly, So please bookmark this page on your browser to get latest updates. You can also download the PDF file of the following below.
Important Terms
CI – Compound interest 
A – amount 
Formulas:


If the interest is payable half yearly, then time is multiplied by 2 and the rate is halved. i.e. A = p(1+((R/2)/100)^{ 2T} 
If the interest is payable Quarterly, then time is multiplied by 4 and the rate is divided by 4. i.e. A = p(1+((R/2)/100)) ^{4T} 
When interest is compounded annually but time is in fraction, says 5years then A = p(1+(R/100))^{5} (1+((R*2/3)/100)) 
If the difference between SI and CI on a certain sum of money for 2 years at R% per is D. then the sum(principal) is P = (D*100^{2})/R^{2} 
If the difference between SI and CI on a certain sum of money for 3 years at R% per is D. then the sum(principal) is P = (D*100^{3})/(R^{2}(R+300)) 
If a sum A becomes B in T_{1} years at compound rate of interest, then after T_{2} years the sum becomes (B)^{(T}_{2}/^{T}_{1}^{)}/(A)^{(T}_{2}/^{T}_{1 }^{1)} 
Problems :
1) The SI accrued on an amount of Rs 14,800 at the end of 3 years is Rs 6,216. What would be the compound interest accrued on the same amount at the same rate in the same period?
Solution:
R =(SI *100)/ (P*T)
R =(6216*100)/(14800*3) = 14
Now we have to find compound interest
(usual method:
CI = p(1(1+(R/100))^{T})
= 14800(1(1+(14/100)³))
findind (14/100)³ is lengthy process and it takes time we can do in alternate method)
14800 *(14/100) = 2072 * 3 = 6216
2072 *(14/100) = 290.08 * 3 = 870.24
290.08 *(14/100) = 40.61 * 1 = 40.61
6216 + 870.24 + 40.61 = 7126.85
CI = 7126.85
2) Rahul invested Rs 35,500 in a scheme which earns him simple interest at the rate of 15p.c.p.a for 2 years. At the end of two years he reinvests the principal amount plus interest earned in another scheme which earns him compound interest at the rate of 20p.c.p.a. what will be the total interest earned by Rahul over the principal amount at the end of 5 years?
Solution:
SI = (P*R*T)/100
SI =(35500*15*2)/100 = 10650
Amount = 35500 + 10650 = 46150
now compound interest is calculated for sum 46150
46150 *(20/100) = 9230 * 3 = 27690
9230 *(20/100) = 1846 * 3 = 5538
1846 *(20/100) = 368.2 * 1 = 362.8
CI = 27690 + 5538 + 362.8 = 33590.8
Total interest = SI + CI = 10650 + 33590.8 = 44240.8
3) Sudharshan invested Rs 15,000 at interest @ 10p.c.p.a. for one year if the interest is compounded every six months what amount will sudharshan get at the end of the year?
Solution:
if interest is compounded quarterly divide interest by 2 and multiply years by 2
Interest is compounded half yearly(10/2) = 5%
Period = 1* 2 = 2 years
15000 *(5/100) = 750 * 2 = 1500
750 *(5/100) = 36.5 *1 = 36.5
Compound interest = 1500 + 36.5 = 1536.5
Amount = 15000 + 1536.5 = 16536.5
4) What will be the difference in Simple and compound interest on Rs 2000 after 3 years at the rate of 10 percent per annum?
Solution:
D =(PR^{2}(100+R))/100^{3}
D =(2000*10^{2}(300+10))/100^{3}= 62
5) At what rate of CI per annum does a sum of money amount to Rs 4840 in 2 years and Rs 5324 in 3 years
Solution:
Rate of interest =(Difference in amount/first amount)*100
R =((53244840)/4840)*100= (484/4840)*100
R = 10%
Sample problems are here, more problems will be added soon.