## Definitions

**Principal:** The money borrowed (lent or invested) is called **principal**.

**Interest:** The additional money paid by the borrower to the moneylender in lieu of the money used by him is called **interest**.

**Amount:** The total money paid by the borrower to the moneylender is called **amount**.

Thus, **amount** = **principal** + **interest **

**Rate:** It is the interest paid on Rs 100 for specified period.

For example:

(i) Rate of 6% per annum means that the interest paid on Rs 100 for one year is Rs 6

(ii) Rate of 1.25% per month means that the interest paid on Rs 100 for one month is Rs 1.25

(iii) Rate of 2.5% per quarterly means that the interest paid on Rs 100 for 3 months is Rs 2.5

However, if the time period for the interest rate is not given, then we shall take the time period as one year.

Time: It is the time for which the money is borrowed (or invested).

**Simple interest**: It is the interest calculated on the original money (principal) at given rate of interest for any given time.

**Following example explains the above terms :**

**Vijay borrowed Rs 10,000 from a credit society for 1 year. At the end of 1 year, he was required to return Rs 11,200 instead of Rs 10,000 which he borrowed.**

Why did Vijay not return the same amount of money i.e., Rs 10,000 which he borrowed?

Why was he required to pay Rs 1200 extra to the credit society?

Vijay was required to pay Rs 1200 extra to the credit society because he used the credit society’s money (Rs 10,000) for 1 year. The extra money is the charge for using the credit society’s money which Vijay borrowed. It is called the interest charged by the credit society.

• The money that Vijay borrowed (Rs 10,000) is called the **Principal. **

• The duration (1 year) for which he borrowed the money is called the **Period.**

• The charges (Rs 1200) for using the money is called the **Interest.**

• The total money which Vijay returned at the end of 1 year

(Rs 10,000 + Rsl200 = Rs 11,200) is called the **Amount**.

## Formulas

If **P** denotes the principal, **R **the rate of interest, **T** the time for which the money is borrowed (or invested), **I** (or S.I.) the simple interest and** A** the amount, then

I = \(\frac{(P * R * T)}{100}\)

P = \(\frac{I * 100}{R * T}\), R = \(\frac{I * 100}{P * T}\), T = \(\frac{I * 100}{P * R}\)

A = P + I = P + \(\frac{P * R * T}{100}\) = (1 + \(\frac{R * T}{100}\)) P

**Points To Remember:**

-> For counting the time between two given dates, only one of the two dates counted (either first or last). Usually, we exclude the date of start and include the date of return.

-> For converting the time in days into years, always divide by 365, whether it is leap year or not.

-> The time must be taken in accordance with the interest rate percent. Thus, if interest rate is per month then time must be taken in months.

## Illustrative Examples

**Example1:** Find the simple Interest on Rs 7200 at 5% per annum for 8 months. Also, find the amount.

**Solution.** Principal = Rs 7200, rate = 5% p.a. and time = \(\frac{8}{12}\) year = \(\frac{2}{3}\) year.

Therefore, P = Rs 7200, R = 5% p.a. and T = \(\frac{2}{3}\) year.

=> SI = \(\frac{P * T * R}{100}\)

= Rs (7200 x 5 x \(\frac{2}{3}\) x \(\frac{1}{100}\)) = Rs 240.

=> amount = (principal + SI)

= Rs (7200 + 240) = Rs 7440.

Therefore, SI = Rs 240 and amount = Rs 7440.

**Example 2:** Find the simple interest on Rs 4500 at 8% per annum for 73 days. Also, find the amount.

**Solution.** Principal = Rs 4500, rate = 8% p.a. and time = \(\frac{73}{365}\) year = \(\frac{1}{5}\) year.

Therefore, P = Rs 4500, R = 8% p.a. and T = \(\frac{1}{5}\) year.

=> SI = \(\frac{P * T * R}{100}\)

= Rs (4500 x 8 x \(\frac{1}{5}\) x \(\frac{1}{100}\)) = Rs 72.

=> amount = (principal + SI)

= Rs (4500 + 72) = Rs 4572.

Therefore, SI = Rs 72 and amount = Rs 4572.

**Example 3:** How long will it take for Rs 5660 invested at 10% per annum simple interest to amount to Rs 7641?

**Solution.** Here, P = Rs 5660, A = Rs 7641, R = 10% p.a.

I = A — P = Rs 7641 — Rs 5660 = Rs 1981

Let T years be the required time.

Using T = \(\frac{I * 100}{P * R}\),

we get T = \(\frac{1981 * 100}{5660 * 10}\)

= \(\frac{1981}{566}\)

= \(\frac{7}{2}\) = \(3\frac{1}{2}\)

Hence the required time = \(3\frac{1}{2}\) years = 3 years 6 months.

**Example 4:** In what time will the simple interest on a certain sum of money at 6% per annum be of itself?

**Solution.** Let the sum of money (principal) be Rs P, then

Interest = \(\frac{3}{8}\) of Rs P = Rs \(\frac{3}{8}\)P

Rate of simple interest = \(6\frac{1}{4}\) p.a. = \(\frac{25}{4}\) % p.a.

Let T years be the required time

Using, T = \(\frac{I * 100}{P * R}\)

we get T = \( \frac { \frac { 3 }{ 8 } \times P\times 100 }{ P\times \frac { 25 }{ 4 } } \)

= \(\frac{3}{8}\) x 100 x \(\frac{4}{25}\)

= 6

Hence the required time = 6 years.

**Example 5:** If the interest charged for 9 months be 0.09 times the money borrowed, find the rate of simple interest per annum.

**Solution.** Let the money borrowed (principal) be Rs P, then

I (simple interest) = 0.09 of Rs P = Rs \(\frac{9}{100}\) P

Time = 9 months = \(\frac{9}{12}\) years = \(\frac{3}{4}\) years

Let R be the rate percent of simple interest per annum.

Using R = \(\frac{I \times 100}{P \times T}\),

we get R = \( \frac { \frac { 9 }{ 100 } \times P\times 100 }{ P\times \frac { 3 }{ 4 } } \)

= 9 x \(\frac{4}{3}\)

= 12

Hence the rate of simple interest per annum 12%.

**Example 6:** At what rate percent simple interest will a sum of money will amount to \(\frac{5}{3}\) of itself in 6 years 8 months?

**Solution.** Let the money borrowed (principal) be Rs.P, then

amount = \(\frac{5}{3}\) P = Rs \(\frac{5}{3}\) P

I (simple interest) = Amount – Principal = Rs \(\frac{5}{3}\) P – Rs P

= ( \(\frac{5}{3}\) P – P)

= Rs ( \(\frac{5}{3}\) – 1) P

= Rs \(\frac{2}{3}\) P

Time = 6 years 9 months = \( 6\frac{8}{12}\) years = \(6\frac{2}{3}\) years = \(\frac{20}{3}\) years

Let R be the rate percent of simple interest per annum.

Using R = \(\frac{I \times 100}{P \times T}\),

we get R = \( \frac { \frac { 2 }{ 3 } \times P\times 100 }{ P\times \frac { 20 }{ 3 } } \)

= \(\frac{200}{3}\) x \(\frac{3}{20}\)

= 10

Hence the rate of simple interest per annum 10%.

**Example 7:** Sudhir borrowed Rs 3,00,000 at 12% per annum from a money-lender. At the end of 3 years, he cleared the account by paying Rs 2,60,000 and a gold necklace. Find the cost of the necklace.

**Solution.** SI = \(\frac{P * T * R}{100}\)

Here P = Rs 3,00,000, R = 12%, T= 3 years

Therefore, S.I. = Rs \(\frac{300000 \times 12 \times 3}{100}\) = Rs 108000

So, amount to be returned by Sudhir

= Principal + Interest = Rs300000 + Rs 108000 = Rs 408000

Amount returned by Sudhir = Rs 2,60,000

So, cost of the necklace = Rs 4,08,000— Rs2,60,000 = Rs 1,48,000.

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