Compound interest is one of the most important and frequently tested topics in quantitative aptitude for competitive exams. It is asked in almost every major exam including CAT, SSC CGL, SSC CHSL, Bank PO, Bank Clerk, Railway RRB and CSAT. A strong understanding of compound interest concept, formulas and tricks is essential for scoring well in these exams. In this post we cover everything from the basic definition of compound interest, the CI formula with principal rate and time, successive method and tree method for fast calculation, ready reference CI table for rates 1 to 10 percent, difference between CI and SI for 2 and 3 years, periodic compounding formula for half yearly quarterly and monthly, equal installment problems at compound interest and doubling of money problems — all explained with multiple methods and solved examples.
Compound Interest: Compound interest is an interest is an interest calculated on the principal and existing interest together over a given period of time. The interest accumulated on a principal over a period of time is also added to the principal and becomes the new principal amount for the next period. Again, the interest for the next period is calculated on the accumulated principal value. Hence it is usually termed as “interest over the interest”.
Compound interest = interest on principal + interest over existing interest
⟶ principal = P
rate = r% p.a.
Amount = A
Time = t years
when the rate of interest for three years b r₁%, r₂ & r₃% then
Method of calculating compound Interest:-
● Successive ↑ is compound interest.
● If rate is fraction then (denominator)ᵗᶦᵐᵉ will be taken as principal for easier calculation.
rate = x% = \(\frac{1}{a}\)
time = n year
then take principal = (a)ⁿ
● If rate of compound interest for two years be x% & y%
then final rate of CI for two years = \(\left( {x + y + \frac{{xy}}{{100}}} \right)\)%
Final rate of (CI – SI) for two years = \(\left( {\frac{{xy}}{{100}}} \right)\)%
● Remember above tabe for time 3 year & rate of interest given from 1% to 10%. Final rate of CI & Final rate of (CI – SI) Given. Remember it to make calculation fast for those exams for which very high speed of solving is required in this cut throat competition.
● For two years
Final CI = \(\left( {a + b + \frac{{ab}}{{100}}} \right)\)%
Let r = 8%
∴ CI = 16/64 = 16.64%
r= 13%
● For three years
Final CI = \(\left( {a + b + c} \right) + \left( {\frac{{ab + bc + ca}}{{1000}}} \right) + \frac{{abc}}{{{{\left( {100} \right)}^2}}}\)
if rate is same then
● Remember above CI table to solve questions easily & fast for competitive exams.
● 2 year ⟹ (CI – SI) diff. = \(P{\left( {\frac{R}{{100}}} \right)^2}\)
● 3 year ⟹ (CI – SI) diff. = \(P{\left( {\frac{R}{{100}}} \right)^2} \times \left( {\frac{{300 + R}}{{100}}} \right)\)
● Effect of periodic compounding rate:
n = number of times the interest is compounding
∴ term = nt
& rate = \(\frac{r}{n}\) %
∴ new amount = \(P{\left( {1 + \frac{{\left( {\frac{r}{n}} \right)}}{{100}}} \right)^{nt}}\)
● Let Principal = P
Amount = A
rate = r% p.a. ⟶ rate is per annum i.e. rate is of 12 months or 365 days.
\(A = P{\left( {1 + \frac{r}{{100}}} \right)^t} = P{\left( {1 + r\% } \right)^t}\)
The above formula works when rate is compounded annually.
● Let if rate is compounded in n – months then \(r’ = \frac{{r \times n}}{{12}}\% \)
& \(t’ = \frac{{t \times 12}}{n}\) i.e. there will be total \(\frac{{12t}}{n}\) terms of n-n months in t years
\(A’ = P{(1 + r\% )^{t’}}\)
i.e. if we compound interest in n-n months then total \(\frac{{12t}}{n}\) terms will be of n – months in t years.
circle ① SI for first year
circle ② SI of second year
circle ③ combined SI of (i + ii)ⁿᵈ year
circle ④ CI of 2 year as well as CI of 2ⁿᵈ year
circle ⑤ CI + SI of 2 years
(P + circle ⑤) complete amount A
SI of first year = \(\frac{{\Pr }}{{100}}\)
SI of second year = \(\frac{{\Pr }}{{100}}\)
SI of third year = \(\frac{{\Pr }}{{100}}\)
CI of second year as well as CI of 2 years = \(P{\left( {\frac{r}{{100}}} \right)^2}\)
circle ① SI of first year
circle ② SI of second year
circle ③ SI of third year
circle ④ CI of second year as well
circle ⑤ CI of 3 years – CI of 2 years
circle ⑥ CI of 3 years
circle ⑦ total interest of second year
circle ⑧ total interest of third year
circle ⑨ CI – SI of 3 years
Time = 3 year
rate = r %
CI – SI = D
then, \(P{\left( {\frac{r}{{100}}} \right)^2}\left[ {\frac{{300 + r}}{{100}}} \right] = D\)
(Q). A invest a sum of 10000 Rs. at compound interest at the rate of 10 % per annum for a period of three years. What amount will he get after 3 years.
Sol:-
Method(1):
P = 10000
r = 10 %
t = 3
A = ?
\(A = P{\left( {1 + \frac{r}{{100}}} \right)^t}\)
= \(1000{\left( {1 + \frac{{10}}{{100}}} \right)^3} = 1000 \times \frac{{11 \times 11 \times 11}}{{10 \times 10 \times 10}}\)
= 13310 Rs. Answer
method(3): From table we remember that equivalent rate of 10% for 3 year in compounding case will be 33.1%
∴ A = \(10000\left( {1 + \frac{{33.1}}{{100}}} \right)\) = 13310 Answer
∴ A = 10000 + 1000×3 + 100×3 + 10
= 13310 Rs. Answer
(Q). A invested a sum of Rs. 64000 for 3 years at compound interest and received an amount 74088 Rs. on maturity. What is the rate of interest.
Sol:-
(Q). On what principal will the compound interest for 3 years at 5% p.a. amount to 63.05 ?
Sol: 5% = \(\frac{1}{{20}}\)
∴ P = 8000 × 0.05 = 400 Rs. Answer
Method(2): From table we remember that combined rate of 5% for 3 years is 15.7625%
∴ CI = \(\frac{{P \times 15.7625 \times 1}}{{100}}\)
= 400 Answer
(Q). Rs. 10000 is borrowed at CI at the rate of 1% for first year, 2% for second year and 3% for third year. Find the amount to be paid after 3 years.
Sol:-
Method(1):
∴ total amount after 3 years
= 100000 + (1000 + 2000 + 3000) + (60 + 30 + 20 + 0.6)
= 106110.6 Answer
Method(2):
∴ CI = 6.1106% of 100000
= 6110.6
∴ Amount = 100000 + 6110.6
= 106110.6 Rs. Answer
Method(3): 1% = \(\frac{1}{{100}}\) 2% = \(\frac{1}{{50}}\) 3% = \(\frac{3}{{100}}\)
(Q). Find the amount on Rs. 500000 in 1y8m at 24% p.a., compound interest being calculate in every 5 months.
Sol:-
P = 500000
r = 24% p.a.
t = 1y 8m = 20m
compounded in every 5 months
∴ r’ = \(\frac{{24}}{{12}} \times 5\) = 10%
& term = \(\frac{{20}}{5}\) = 4
Now apply tree method
∴ CI = 50000 × 4 + 5000 × 6 + 500 × 4 + 50
= 232050
∴ amount = 500000 + 232050
= 732050 Rs. Answer
Method(2): r’ = 10% = \(\frac{1}{{10}}\)
Method(3):
10% 10% 10% 10%
⟶ 10 + 10 + \(\frac{{10 \times 10}}{{100}}\) = 21%
21 + 10 + \(\frac{{21 \times 10}}{{100}}\)
= 32.1
⟶ 32.1 + 10 + \(\frac{{32.1 \times 10}}{{100}}\)
= 46.41%
∴ CI = 46.41% of 500000
= 232050
∴ A = 500000 + 232050 = 732050 Rs. Answer
(Q). If a certain sum of money becomes equal to 2 times in 6 years. In how much time it will be 128 times of itself.
Sol:-
(Q). A sum of money placed at compound interest thrice itself in 5 years. In how many years will it amount to 81 times itself and also find rate of interest ?
Sol:-
(Q). If a certain sum of money amounts to Rs. 3200 in 6 years and Rs. 4000 in 12 years. Find the principal.
Sol:-
⟹ \(P \times \frac{5}{4}\) ⟹ P = 2560 Rs. Answer
(Q). If a certain sum of money becomes Rs. 16000 in 3 years and Rs. 256000 in 4 years. Find the principal ?
Sol:
(Q). A man borrows Rs. 5000 at 10% compound rate of interest. At the end of each year he pays back Rs. 2000. How much amount should he pay at the end of third year to clear his all due ?
Sol:-
(Q). A man want to invest Rs. 16850 in bank account of his two sons whose ages are 12 years & 16 years in such a way so that they will get equal amount at the age of 120 years at the rate of \(33\frac{1}{3}\)% p.a. Find, the share of younger son. When the rate is compounded annually & when the rate is simple interest.
Sol:
r = \(33\frac{1}{3}\)% = \(\frac{1}{3}\)
A = \(P{\left( {1 + \frac{r}{{100}}} \right)^t}\)
∴ share of elder son E = 81 × 50 = 4050
& share of younger son Y = 256 × 50 = 12800 Answer
Short Trick
In these type of questions it does not matter that they get equal money in 120 years or 300 years or 2 lakh years, the thing which matters is what is the difference of their age. So we have to raise power equal to the difference in age. The younger one has more time to earn in comparison to elder one and also both have to get finally some amount. Hence younger one will get small part and elder one will get big part.
The above method works when rate is compounded annualy. For Simple interests this will not work.
⟹ So in above question
Age difference = 16 – 12 = 4
# When \(33\frac{1}{3}\% \) is simple interest.
Amount received by elder son = Amount received by younger son
⟹ \(\left( {100 + \frac{{100}}{3} \times \left( {120 – 16} \right)} \right)\% \times E = \left( {100 + \frac{{100}}{3} \times \left( {120 – 12} \right)} \right)\% \times Y\)
10700E = 11100Y
∴ E = \(111 \times \frac{{8425}}{{109}}\) = 8579.587 Answer
Y = \(107 \times \frac{{8425}}{{109}}\) = 8270.413 Rs. Answer
(Q). A certain sum amounts to Rs. 12884.08 in 5 years and to Rs. 17148.71408 in 8 years at compound interest p.a. What is the principal & rate of interest ?
Sol:-
now P×\(\frac{{11}}{{10}} \times \frac{{11}}{{10}} \times \frac{{11}}{{10}} \times \frac{{11}}{{10}} \times \frac{{11}}{{10}}\) = 1288408
P = 8000 Answer
Method(2):
Using Factor approach:
∴ P × 1.61051 = 12884.08
P = 8000 Answer
(Q) A sum of money becomes 64 times of itself in 3 years at compound interest. Find the rate of interest per annum ?
Sol:
(Q) Find the CI for Rs. 72000 at \(16\frac{2}{3}\% p.a. and time is 1y 73 days.
Sol: r = \(16\frac{2}{3}\% = \(\frac{1}{6}\)
(Q). A man purchases a motorbike for a certain price and promise to pay the price in 3 equal installments of Rs. 14040 at the rate of 20% per annum. Find the cost price of the motorbike.
Sol: 20% = \(\frac{1}{5}\)
(Q). A man borrowed a sum of Rs. 117425 from a bank and promises to pay the amount in 4 equal annual installment at the rate of 20% p.a. Find the value of each installment.
Sol: 20% = \(\frac{1}{5}\)
(Q). A sum of 3600 Rs. deposited at CI double after 5 years. How much it will be after 20 years ?
Sol:
7200 = 3600\({\left( {1 + \frac{r}{{100}}} \right)^5}\)
⟹ \({\left( {1 + \frac{r}{{100}}} \right)^5}\) = 2
Now A = 7200\({\left( {1 + \frac{r}{{100}}} \right)^{15}}\) = 7200\({\left( {{{\left( {1 + \frac{r}{{100}}} \right)}^5}} \right)^3}\)
= 7200\({\left( 2 \right)^3}\)
= 57600 Rs. Answer
alternate:
(Q). If the difference between CI & SI on a certain sum of money for 3 years at 5% p.a. is Rs. 183. What is the sum ?
Sol:
Method(1):
r% = \(\frac{1}{x}\)×100
5 = \(\frac{1}{x}\)×100 ⟹ x = 20
\(\frac{D}{P} = \frac{{3x + 1}}{{{x^3}}}\) ⟹ \(\frac{{183}}{P} = \frac{{61}}{{8000}}\)
⟹ P = 24000 Rs. Answer
Method(2):
SI for 3 years = 3 × 5% = 15%
CI for 3 years = 15.7625% (From table)
∴ (CI – SI) for 3 years = 0.7625%
∴ P × 0.7625% = 183
P = 24000 Rs. Answer
(Q). If the simple interest is 11.3% p.a. and compound interest is 10% annual. Find the difference between interests after 4 years on a sum of Rs. 800.
Sol:
SI for 4 years = 11.3 × 4 = 45.2%
CI for 4 years = 46.41% (From table)
∴ (CI – SI) for 4 years = 46.41 – 45.2 = 1.21%
∴ required interest = 800 × 1.21% = 9.68 Rs. Answer
(Q). Find the compound interest on Rs. 30000 for 1 year at the rate of 40% p.a. compounded quarterly.
Sol:
Compounding quarterly i.e. in every 3 months or 4 times in a year.
∴ t’ = 4 terms
& r = 40% p.a. ⟹ r’ = \(\frac{{40}}{4}\)% = \(\frac{1}{{10}}\)
Now apply Tree method