Simple interest is one of the most fundamental 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 simple interest concept, formulas and tricks is essential for scoring well in these exams. In this post we cover everything from the basic definition of simple interest, the SI formula with principal rate and time, amount formula, annual installment formula for repayment of debt in equal installments, problems on doubling and tripling of money, difference in simple interest formula and multiple approach methods for all types of simple interest problems — all explained with clear formulas and solved examples.
Simple Interest:
If the interest on a certain sum borrowed for a certain period is reckoned uniformly then it is called simple interest denoted by SI.
P = principal
R = rate % p.a.
T = time in years
⟹ Simple interest SI = Amount(A) – Principal (P)
⟹ \(\frac{{{A_1} – {P_1}}}{{{A_2} – {P_2}}} = \frac{{{P_1} \times {R_1} \times {T_1}}}{{{P_2} \times {R_2} \times {T_2}}}\)
⟹
SI ∝ P
SI ∝ R
SI ∝ T
If P = 100%
then SI = rt%
A = (P + rt)% = (P + SI)%
Example: SI 5600 r = 7% t = 8 year A = ?
● Annual installment = \(\frac{{duedebt \times 100}}{{100t + \frac{{r \times t \times (t – 1)}}{2}}}\)
P< installment total amount < P + I
Repayment of debt in equal installment:
r = rate of interest p.a.
y = number of installments per annum
So when installment is paid yearly then y = 1
when installment is paid half yearly then y = 2
when installment is paid quarterly then y = 4
when installment is paid monthly then y = 12
(Q). A sum of Rs. 300 amounts to Rs. 900 in 4 years. what will it amount in 15 years if the rate of interest is increased by 2% p.a. ?
Sol:
Method(1):
SI = 900 – 300 = 600
t = 4%
P = 300
600 = \(\frac{{300 \times r \times 4}}{{100}}\)
r = 50%
now required amount A = 300 + \(\frac{{300 \times 52 \times 15}}{{100}}\)
= 2640 Rs.
Method(2):
(Q). What annual installment will discharge a debt of Rs. 2625 in 6 years at 10 % Simple Interest.
Sol:
debt = \(na + \frac{{ra}}{{100y}} \times \frac{{n(n – 1)}}{2}\)
2625 = \(6a + \frac{{10a}}{{100y}} \times \frac{{6 \times 5}}{2}\)
a = 350 Rs. Answer
Method(2):
∴ annual installment = 100 × 3.5 = 350 Rs. Answer
(Q). A sum of money doubles itself in 5 years at simple interest. In how many years it will become 4 times of itself.
Sol:
(Q). The simple interest on Rs. 1450 will be less than the simple interest on Rs. 1800 at 4% SI by Rs. 35. Find the time ?
Sol:
35 = \(\frac{{\left( {1800 – 1450} \right)}}{{100}} \times 4 \times t\)
t = 2.5 years Answer
(Q). If the simple interest for 4 years be equal to 40 % of the principal. After how many years it will be equal to the principal ?
Sol:
Method(1):
\(\frac{{40}}{{100}}P = \frac{{P \times 4 \times r}}{{100}}\)
r = 10%
now P = \(\frac{{P \times 10 \times t}}{{100}}\)
t = 10 Answer
Method(2):
i.e. for SI 2, time required = 4 years
∴ for SI 5, time required = \(\frac{5}{2} \times 4\) = 10 years Answer
(Q). Rs. 700 becomes Rs. 840 in 7 years at a certain rate of simple interest. If the rate of interest is increased by 5%, what amount will Rs. 700 become in 7 years ?
Sol:
if rate is increased by 5% then extra ↑ in SI = 700 × 7.5% = 245
∴ new amount = 840 + 245
= 1085 Rs. Answer
(Q). A sum of money at simple interest amounts to Rs. 1060 in \(3\frac{1}{2}\) years and Rs. 1100 in 5 years. Find the rate of interest p.a.
Sol:
Method(1):
1060 = P(1 + r% × \(\frac{7}{2}\)) ……………… (i)
1100 = P(1 + r% × 5) …………………. (ii)
divide (i) by (ii)
\(\frac{{1060}}{{1100}} = \frac{{1 + \frac{7}{2}r\% }}{{1 + 5r\% }}\)
\(\frac{{53}}{{55}} = \frac{{2 + 7r\% }}{{2 + 10r\% }}\)
\(\frac{{53}}{2} = \frac{{2 + 7r\% }}{{3r\% }}\)
159r% = 4 + 14r%
145r% = 4
r = \(2\frac{{22}}{{29}}\)% Answer
Method(2):
P + SI for 3.5 years = 1060 …………..(i)
P + SI for 5 years = 1100 ……………..(ii)
subtract (i) & (ii)
SI for 1.5 year = 40
SI for 3.5 year = \(\frac{{40}}{{1.5}}\)×3.5 = \(\frac{{280}}{3}\)
put this value in equation (i)
P + \(\frac{{280}}{3}\) = 1060
P = \(\frac{{2900}}{3}\)
now SI for 3.5 year = \(\frac{{P \times R \times T}}{{100}}\)
\(\frac{{280}}{3} = \frac{{2900}}{3} \times R \times \frac{{3.5}}{{100}}\)
r = \(2\frac{{22}}{{29}}\% \) Answer