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)
subtact (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