1. Simple interest(S.I): If the interest on a sum borrowed for a certain period is calculated uniformly, then it is called simple interest.
2. Principal(P): The money borrowed or lent out for a certain period is called principal.
3. Interest(I): Paying of extra money for using other's money is called interest.
4. Rate(R): A rate of interest is the rate at which the interest is calculated on the principal amount and it is a special ratio in which the two terms are in different units.
5. Time(T): The specific period for which the money is borrowed is called as time/ time period.
6. Amount(A): The money to be assured or matured or paid finally is called amount which is equal to the principal+simple interest.

Example 1 Find the simple intrest on Rs. 68,000 at 16[latex]\frac{2}{3}[/latex]% pen annum for 9 months.
Solution:

P = Rs. 68000, R = [latex]\frac{50}{3}[/latex]% p.a and T = [latex]\frac{9}{12}[/latex] years = [latex]\frac{3}{4}[/latex] years.
∴ S.I = ([latex]\frac{P \times R \times T}{100}[/latex]) = Rs. (68000 x [latex]\frac{50}{3}[/latex] x [latex]\frac{3}{4}[/latex] x [latex]\frac{1}{100}[/latex]) = Rs. 8500.

Example 2 Find the simple intrest on Rs. 3000 at 6[latex]\frac{1}{4}[/latex]% per annum for the period from 4th Feb, 2005 to 18th April, 2005.
Solution:

Time = (24 + 31 + 18) days = 73 days = [latex]\frac{73}{365}[/latex] year = [latex]\frac{1}{5}[/latex] year.
P = Rs. 3000 and R = 6[latex]\frac{1}{4}[/latex]% p.a. = [latex]\frac{25}{4}[/latex]% p.a.
∴ S.I = Rs. (3000 x [latex]\frac{25}{4}[/latex] x [latex]\frac{1}{5}[/latex] x [latex]\frac{1}{100}[/latex]) = Rs. 37.50.
Remark: The day on which money is deposited is not counted while the day on which money is withdrawn is counted.

Example 3 If Rs. 64 amounts to Rs. 83.20 in 2 years, what will Rs. 86 amount to in 4 yewars at the same rate percent per annum?
Solution:

P = Rs. 64, S.I. = Rs. (83.20 - 64) = Rs. 19.20, T = 2 years.
So, rate = ([latex]\frac{100 \times 19.20}{64 \times 2}[/latex])% = 15%.
Now, P = Rs. 86,. R = 15%, T = 4 years.
∴ S.I. = Rs. ([latex]\frac{86 \times 15 \times 4}{100}[/latex]) = Rs. 51.60

Example 1 A man took loan from a bank at the rate of 12% p.a. simple intrest. After 3 years he had to pay Rs.5400 intrest only for the period. The principal amount borrowed by him was:
Solution:

Principal = Rs. ([latex]\frac{100 \times 5400}{12 \times 3}[/latex]) = Rs. 15000.

Example 2 A sum fetched a total simple intrest of Rs. 4016.25 at the rate of 9 p.c.p.a in 5 years. What is the sum?
Solution:

Principal = Rs. ([latex]\frac{100 \times 4016.25}{9 \times 5}[/latex]) = Rs. ([latex]\frac{401625}{45}[/latex]) = Rs. 8925.

Example 1 A certain sum of money amounts to Rs. 1008 in 2 uears and to Rs. 1164 in 3[latex]\frac{1}{2}[/latex] years. Find the sum and the rate of intrest.
Solution:

S.I for 1[latex]\frac{1}{2}[/latex] years = Rs. (1164 - 1008) = Rs. 156.
S.I for 2 years = Rs. (156 x [latex]\frac{2}{3}[/latex] x 2) = Rs. 208.
∴ Principle = Rs. (1008 - 208) = Rs. 800.
Now, P = 800, T = 2 and S.I. = 208.
Rate = ([latex]\frac{100 \times 208}{800 \times 2}[/latex]) % = 13%.

Example 2 At what rate percent per annum will a sum of money double in 15 years?
Solution:

Let principal = P. Then, S.I. = P and T = 16 yrs.
∴ Rate = ([latex]\frac{100 \times P}{P \times 16}[/latex])% = 6[latex]\frac{1}{4}[/latex]% p.a.

Example 1 In how many years, Rs. 150 will produce the same intrest @ 8% as Rs. 800 produce in 3 years @ 4[latex]\frac{1}{2}[/latex]%?
Solution:

P = Rs. 800, R = 4[latex]\frac{1}{2}[/latex]% = [latex]\frac{9}{2}[/latex]%, T = 3 years. Then,
S.I. = Rs. (800 x [latex]\frac{9}{2}[/latex] x [latex]\frac{3}{100}[/latex]) = Rs. 108.
Now, P = Rs. 150, S.I. = Rs. 108, R = 8%.
∴ Time = ([latex]\frac{100 \times 108}{150 \times 8}[/latex]) years = 9 years.

Example 2 In how many years will a sum of money double itself at 12% per annum?
Solution:

Let Sum = [latex]x[/latex]. Then, S.I = [latex]x[/latex].
∴ Time = ([latex]\frac{100 \times S.I.}{P \times R}[/latex]) = ([latex]\frac{100 \times x}{x \times 12}[/latex]) years = 8[latex]\frac{1}{3}[/latex] years = 8 years 4 months.

Example 1 A sum at simple intrest at 13[latex]\frac{1}{2}[/latex]% per annum amounts to Rs. 2502.50 after 4 years. Find the sum.
Solution:

Let sum be Rs. [latex]x[/latex]. Then, S.I. = Rs. ([latex]x[/latex] x [latex]\frac{27}{2}[/latex] x 4 x [latex]\frac{1}{100}[/latex]) = Rs. [latex]\frac{27x}{50}[/latex].
∴ Amount = Rs. ([latex]x[/latex] + [latex]\frac{27x}{50}[/latex]) = Rs. [latex]\frac{77x}{50}[/latex].

Example 2 A sum invested at 5% simple intrest per annum grows to Rs. 504 in 4 years. The same amount at 10% simple intrest per annum in 2[latex]\frac{1}{2}[/latex] years will grow to:
Solution:

Let the sum be Rs. [latex]x[/latex]. Then, S.I. = Rs. (504 - [latex]x[/latex]).
∴ ([latex]\frac{x \times 5 \times 4}{100}[/latex]) = 504 - [latex]x[/latex] ⇒ 20[latex]x[/latex] = 50400 - 100[latex]x[/latex] ⇒ 120[latex]x[/latex] = 50400 ⇒ [latex]x[/latex] = 420.
Now, P = Rs. 420, R = 10%, T = [latex]\frac{5}{2}[/latex] years.
S.I. = Rs. ([latex]\frac{420 \times 10}{100}[/latex] x [latex]\frac{5}{2}[/latex]) = Rs. 105.
∴ Amount = Rs. ([latex]\frac{420 \times 10}{100}[/latex] x [latex]\frac{5}{2}[/latex]) = Rs. 105.
Amount = Rs. (420 + 105) = Rs. 525.

1. Simple interest(S.I) = [latex]\frac{principal(P) \times rate(R) \times time(T)}{100}[/latex]
2. Principal(P) = [latex]\frac{100 \times Simple interest}{Rate \times time}[/latex]
3. Rate = [latex]\frac{100 \times Simple interest}{principal \times time}[/latex]
4. Time = [latex]\frac{100 \times Simple interest}{principal \times rate}[/latex]
5. Amount = Principal+ simple interest (or) Amount = P(1 + [latex]\frac{R \times T}{100} [/latex])

1. Find the principal on a certain sum of money at 5% per annum for 2 [latex]\frac{2}{5}[/latex] years if the amount being Rs. 1120?
Solution:

Given,
Amount = Rs. 1120
Rate = 5 %
Time = [latex]\frac{12}{5}[/latex] years
Now, Amount = P(1 + \( \frac{R * T}{100} \))
⇒ 1120 = P(1 + \( \frac{5 * \frac{12}{5}}{100} \))
⇒ P = 1000
Therefore, principal = Rs. 1000

2. Find the simple interest on Rs. 625 at 6[latex]\frac{1}{2}[/latex] % per annum for 2[latex]\frac{1}{2}[/latex] years?
Solution:

Given,
Principal= Rs. 625
Rate = [latex]\frac{13}{2}[/latex]
Time = [latex]\frac{5}{2}[/latex]
Now, Simple interest(S.I.) = [latex]\frac{P * R * T }{100}[/latex]
S.I. = [latex]\frac{625 * 13 * 5}{100}[/latex]
S.I. = Rs. 101.56

3. A shopkeeper borrowed Rs. 25000 from the two money-lenders. For one loan paid 12 % per annum and for other 14 % per annum. The total interest paid for one year was Rs. 3260. How much did shopkeeper borrow at each rate?
Solution:

Suppose money borrowed at 12 % = Rs.\( x \)
Then, the money borrowed at 14 % = Rs. (25000 - \( x \))
Therefore,
[latex]\frac{x * 21 * 1}{100}[/latex] = [latex]\frac{(25000 - x) * 14 * 1}{100}[/latex] = 3260
⇒ 12[latex] x[/latex] + 350000 - 14 [latex]x[/latex] = 326000
⇒ 2[latex]x[/latex] = 24000
⇒ [latex]x[/latex] = 12000
Money borrowed at 12 % = Rs. 12000
Money borrowed at 14 % = Rs. 13000

4. What installment will discharge a debt of Rs. 1092 due in 3 years at 12 % S.I.?
Solution:

Let each installment be [latex]x[/latex],
Then,
([latex] x + \frac{x * 12 * 1}{100}[/latex]) + ([latex] x + \frac{x * 12 * 2}{100}[/latex]) + [latex]x[/latex] = 1092
⇒ [latex]\frac{28x}{25}[/latex] + [latex]\frac{31x}{25}[/latex] + [latex]x[/latex]
⇒ [latex]x[/latex] = 325
Therefore, each installment = Rs. 325

5. A sum was put at simple interest at a certain rate for 2 years. Had it been put at 3 % higher rate, it would have fetched Rs. 300 more. Find the sum?
Solution:

Let,
sum = [latex]x[/latex], original rate = R%
Then,
[latex]\frac{x * (R + 3) * 2}{100}[/latex]
⇒ [latex]\frac{x * R * 2}{100}[/latex]
⇒ 2 R[latex]x[latex] + 6[latex]x[latex] - 2R[latex]x[latex] = 300 x 100
⇒ [latex]x[latex] = 5000
Therefore, sum = Rs. 5000