Simple Interest deals with the following terms: principal, interest, rate of interest, time, amount. The term **Intrtest** is actually one of the most fundamental business terms, and without it, the financial trading of the world would come to standstill. Interest is defined as the â€œtime value of moneyâ€.

**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\(\frac{2}{3}\)% pen annum for 9 months.

**Solution**:

- P = Rs. 68000, R = \(\frac{50}{3}\)% p.a and T = \(\frac{9}{12}\) years = \(\frac{3}{4}\) years.

∴ S.I = (\(\frac{P \times R \times T}{100}\)) = Rs. (68000 x \(\frac{50}{3}\) x \(\frac{3}{4}\) x \(\frac{1}{100}\)) = Rs. 8500.

**Example 2**

Find the simple intrest on Rs. 3000 at 6\(\frac{1}{4}\)% per annum for the period from 4th Feb, 2005 to 18th April, 2005.

**Solution**:

- Time = (24 + 31 + 18) days = 73 days = \(\frac{73}{365}\) year = \(\frac{1}{5}\) year.

P = Rs. 3000 and R = 6\(\frac{1}{4}\)% p.a. = \(\frac{25}{4}\)% p.a.

∴ S.I = Rs. (3000 x \(\frac{25}{4}\) x \(\frac{1}{5}\) x \(\frac{1}{100}\)) = Rs. 37.50.

**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 = (\(\frac{100 \times 19.20}{64 \times 2}\))% = 15%.

Now, P = Rs. 86,. R = 15%, T = 4 years.

∴ S.I. = Rs. (\(\frac{86 \times 15 \times 4}{100}\)) = 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. (\(\frac{100 \times 5400}{12 \times 3}\)) = 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. (\(\frac{100 \times 4016.25}{9 \times 5}\)) = Rs. (\(\frac{401625}{45}\)) = Rs. 8925.

**Example 1**

A certain sum of money amounts to Rs. 1008 in 2 uears and to Rs. 1164 in 3\(\frac{1}{2}\) years. Find the sum and the rate of intrest.

**Solution**:

- S.I for 1\(\frac{1}{2}\) years = Rs. (1164 – 1008) = Rs. 156.

S.I for 2 years = Rs. (156 x \(\frac{2}{3}\) x 2) = Rs. 208.

∴ Principle = Rs. (1008 – 208) = Rs. 800.

Now, P = 800, T = 2 and S.I. = 208.

Rate = (\(\frac{100 \times 208}{800 \times 2}\)) % = 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 = (\(\frac{100 \times P}{P \times 16}\))% = 6\(\frac{1}{4}\)% p.a.

**Example 1**

In how many years, Rs. 150 will produce the same intrest @ 8% as Rs. 800 produce in 3 years @ 4\(\frac{1}{2}\)%?

**Solution**:

- P = Rs. 800, R = 4\(\frac{1}{2}\)% = \(\frac{9}{2}\)%, T = 3 years. Then,

S.I. = Rs. (800 x \(\frac{9}{2}\) x \(\frac{3}{100}\)) = Rs. 108.

Now, P = Rs. 150, S.I. = Rs. 108, R = 8%.

∴ Time = (\(\frac{100 \times 108}{150 \times 8}\)) years = 9 years.

**Example 2**

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

**Solution**:

- Let Sum = \(x\). Then, S.I = \(x\).

∴ Time = (\(\frac{100 \times S.I.}{P \times R}\)) = (\(\frac{100 \times x}{x \times 12}\)) years = 8\(\frac{1}{3}\) years = 8 years 4 months.

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**Example 1**

A sum at simple intrest at 13\(\frac{1}{2}\)% per annum amounts to Rs. 2502.50 after 4 years. Find the sum.

**Solution**:

- Let sum be Rs. \(x\). Then, S.I. = Rs. (\(x\) x \(\frac{27}{2}\) x 4 x \(\frac{1}{100}\)) = Rs. \(\frac{27x}{50}\).

∴ Amount = Rs. (\(x\) + \(\frac{27x}{50}\)) = Rs. \(\frac{77x}{50}\).

**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\(\frac{1}{2}\) years will grow to:

**Solution**:

- Let the sum be Rs. \(x\). Then, S.I. = Rs. (504 – \(x\)).

∴ (\(\frac{x \times 5 \times 4}{100}\)) = 504 – \(x\) ⇒ 20\(x\) = 50400 – 100\(x\) ⇒ 120\(x\) = 50400 ⇒ \(x\) = 420.

Now, P = Rs. 420, R = 10%, T = \(\frac{5}{2}\) years.

S.I. = Rs. (\(\frac{420 \times 10}{100}\) x \(\frac{5}{2}\)) = Rs. 105.

∴ Amount = Rs. (\(\frac{420 \times 10}{100}\) x \(\frac{5}{2}\)) = Rs. 105.

Amount = Rs. (420 + 105) = Rs. 525.

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**Solution**:

- Given,

Amount = Rs. 1120

Rate = 5 %

Time = \(\frac{12}{5}\) years

Now, Amount = P(1 + \( \frac{R * T}{100} \))

â‡’ 1120 = P(1 + \( \frac{5 * \frac{12}{5}}{100} \))

â‡’ P = 1000

Therefore, principal = Rs. 1000

**2. Find the simple interest on Rs. 625 at 6\(\frac{1}{2}\) % per annum for 2\(\frac{1}{2}\) years?**

**Solution**:

- Given,

Principal= Rs. 625

Rate = \(\frac{13}{2}\)

Time = \(\frac{5}{2}\)

Now, Simple interest(S.I.) = \(\frac{P * R * T }{100}\)

S.I. = \(\frac{625 * 13 * 5}{100}\)

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,

\(\frac{x * 21 * 1}{100}\) = \(\frac{(25000 – x) * 14 * 1}{100}\) = 3260

â‡’ 12\( x\) + 350000 – 14 \(x\) = 326000

â‡’ 2\(x\) = 24000

â‡’ \(x\) = 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 \(x\),

Then,

(\( x + \frac{x * 12 * 1}{100}\)) + (\( x + \frac{x * 12 * 2}{100}\)) + \(x\) = 1092

â‡’ \(\frac{28x}{25}\) + \(\frac{31x}{25}\) + \(x\)

â‡’ \(x\) = 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 = \(x\), original rate = R%

Then,

\(\frac{x * (R + 3) * 2}{100}\)

â‡’ \(\frac{x * R * 2}{100}\)

â‡’ 2 R\(\)x\(\) + 6\(\)x\(\) – 2R\(\)x\(\) = 300 x 100

â‡’ \(\)x\(\) = 5000

Therefore, sum = Rs. 5000