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# Calculation of Interest - Solved Examples

### Simple Interest

Example 1. Find the simple interest on ₹ 25,000 for a period of 5 years at 10% per annum.
Solution The formula to calculate simple interest is \begin{align} SI & = \frac{P × N × R}{100} \end{align} where
P is the Principal (Amount borrowed or lent)
N is the number of years for which the sum is borrowed or lent (N = 5 in this example)
R is the annual rate of interest

\begin{align} SI &= \frac{25000 × 5 × 10}{100} \\ &= \frac{25000}{2} \\ &= 12500 \end{align}

Example 2. Find the simple interest on ₹ 60,000 for a period of 6 months at 5% per annum.
Solution The formula to calculate simple interest is \begin{align} SI & = \frac{P × N × R}{100} \end{align} where
P is the Principal (Amount borrowed or lent)
N is the number of years for which the sum is borrowed or lent (N = $\frac{6}{12}$ = 0.5 in this example)
R is the annual rate of interest

\begin{align} SI &= \frac{60000 × 0.5 × 5}{100} \\ &= \frac{60000 × 2.5}{100} \\ &= 1500 \end{align}

Example 3. The simple interest on a certain sum of money for 2 years at 6% per annum is Rs. 900 less than the simple interest on the same sum for 3 years at 7% per annum. Find the sum.
Solution Let the sum be $x$.
(Simple interest on $x$ for 3 years at 7% per annum) - (Simple interest on $x$ for 2 years at 6% per annum) = 900 \begin{align} ∴ \frac{x × 3 × 7}{100} - \frac{x × 2 × 6}{100} &= 900\\ \frac{21x}{100} - \frac{12x}{100} &= 900 \\ \frac{9x}{100} &= 900 \\ x &= 10000 \end{align}

Example 4. A person invests a certain amount in three different Fixed deposits FD1, FD2 and FD3 with the rate of interest 5% p.a., 6% p.a. and 7% p.a. respectively. If the total interest that accumulates in one year was Rs. 10400 and the amount he invested in FD2 was 100% more of the amount he invested in FD1 and the amount he invested in scheme FD3 was 400% more of the amount he investeds in FD1, what was the amount he invested in scheme FD1?
Solution Let the sum invested in FD1 be $x$.
∴ the sum invested in FD2 = $x + \frac{100}{100} × x = 2x$
and the sum invested in FD3 = $x + \frac{400}{100} × x = 5x$

(Simple interest on FD1) + (Simple interest on FD2) + (Simple interest on FD3) = 10400
\begin{align} \frac{x × 1 × 5}{100} + \frac{2x × 1 × 6}{100} + \frac{5x × 1 × 7}{100} &= 10400 \\ \frac{5x}{100} + \frac{12x}{100} + \frac{35x}{100} &= 10400 \\ \frac{52x}{100} &= 10400 \\ x &= 20000 \end{align}

Example 5. A person invests ₹ 50000 in two different shares. At the end of 1 year, the first share provides a yield of 11% p.a. and the second share gives yield of 14% p.a. If the total dividend (interest) at the end of the year amounts to 12.8%, calculate the amount invested in each share.
Solution Let the sum invested in the first share be $x$.
∴ the sum invested in second share = ₹ 50000 - $x$

(Simple interest on share 1) + (Simple interest on share 2) = (Simple interest on ₹ 50000)
\begin{align} \frac{x × 1 × 11}{100} + \frac{(50000 - x) × 1 × 14}{100} &= \frac{50000 × 1 × 12.8}{100} \\ \frac{11x}{100} + \frac{(50000 - x) × 14}{100} &= 6400 \\ \frac{11x}{100} + \frac{(700000 - 14x)}{100} &= 6400 \\ \frac{11x - 14x + 700000}{100} &= 6400 \\ 11x - 14x &= 640000 - 700000 \\ -3x &= -60000 \\ x &= 20000 \end{align} ∴ The sum invested in the first share is ₹ 20000 and the sum invested in second share = ₹ (50000 - 20000) = ₹ 30000

### Compound Interest

Example 1. Find the compound interest on ₹ 25,000 for a period of 5 years at 10% per annum.
Solution The formula to calculate compound interest is \begin{align} CI & = P × (1 + \frac{r}{n})^{nt} - P \end{align} where
P is the Principal (Amount borrowed or lent)
n is the number of times interest is compounded per year (n = 1 in this example)
t is the number of years the money is borrowed or lent (t = 5 in this example)
r is the annual rate of interest

\begin{align} CI &= 25000 × (1 + \frac{10}{100})^{1 × 5} - 25000 \\ &= 25000 × (1 + 0.1)^{1 × 5} - 25000 \\ &= 25000 × 1.61051 - 25000 \\ &= 15262.75 \end{align}

Example 2. Manoj deposited a sum of ₹ 64,000 in a bank for a period of 3 years compounded annually at 7$\frac{1}{2}$% per annum. What amount will he get on maturity?
Solution The formula to calculate amount at maturity is \begin{align} A & = P × (1 + \frac{r}{n})^{nt} \end{align} where
A is the amount at the end of period t
P is the Principal (Amount borrowed or lent)
n is the number of times interest is compounded per year (n = 1 in this example)
t is the number of years the money is borrowed or lent (t = 3 in this example)
r is the annual rate of interest

\begin{align} A &= 64000 × (1 + \frac{7 \frac{1}{2}}{100})^{1 × 3} \\ &= 64000 × (1 + 0.075)^{1 × 3}\\ &= 64000 × 1.2423 \\ &= 79507 \end{align}

Example 3. Find the compound interest on ₹ 30,000 in a bank for a period of 3$\frac{3}{4}$ years compounded annually at 8% per annum.
Solution The formula to calculate compound interest when the time is a fraction e.g. 3$\frac{3}{4}$ years in this case is \begin{align} CI & = P × (1 + \frac{r}{n})^{nt} × (1 + \frac{3}{4} r) \end{align} where
P is the Principal (Amount borrowed or lent)
n is the number of times interest is compounded per year (n = 1 in this example)
t is the number of complete years (not partial) the money is borrowed or lent (t = 3 in this example)
r is the annual rate of interest

\begin{align} CI & = 30000 × (1 + \frac{8}{100})^{3} × (1 + \frac{\frac{3}{4} × 8}{100}) \\ &= 30000 × (\frac{108}{100})^{3} × (1 + \frac{6}{100}) \\ &= 30000 × (\frac{108}{100})^{3} × (\frac{106}{100}) \\ &= 30000 × (\frac{108}{100}) × (\frac{108}{100}) × (\frac{108}{100}) × (\frac{106}{100}) \\ &= 40058.84 \end{align}

Example 4. Find the amount of ₹ 50,000 after 2 years, compounded annually; the rate of interest being 8% per annum during the first year and 9% per annum during the second year.
Solution The formula to calculate the amount after annual compounding is \begin{align} A & = P × (1 + \frac{p}{100}) × (1 + \frac{q}{100}) \end{align} where
P is the Principal (Amount borrowed or lent)
p is the annual rate of interest during the first year (p = 8% in this example)
q is the annual rate of interest during the second year (q = 9% in this example)

\begin{align} A &= 50000 × (1 + \frac{8}{100}) × (1 + \frac{9}{100}) \\ &= 25000 × (\frac{108}{100}) × (\frac{109}{100}) \\ &= 58860 \end{align}

Example 5. In how many years will ₹ 6250 amount to ₹ 7290 at 8% per annum, compounded annually?
Solution The formula to calculate amount after a certain period compunded annually is \begin{align} A & = P × (1 + \frac{r}{n})^{nt} \end{align} where
A is the amount at the end of period t
P is the Principal (Amount borrowed or lent)
n is the number of times interest is compounded per year (n = 1 in this example)
t is the number of years the money is borrowed or lent (t = 3 in this example)
r is the annual rate of interest

\begin{align} 7290 &= 6250 × (1 + \frac{8}{100})^{1 × t} \\ &= 6250 × (\frac{108}{100})^{1 × t}\\ \frac{7290}{6250} &= (\frac{108}{100})^{1 × t}\\ (\frac{27}{25})^{2} &= (\frac{27}{25})^{t}\\ ∴ t &= 2 \end{align}

Example 6. The population of a town is 125000. It is increasing at the rate of 2% per annum. What will be its population at the end of 3 years?
Solution The formula to calculate the population at the end of 3 years is \begin{align} A & = P × (1 + \frac{r}{n})^{nt} \end{align} where
A is the amount at the end of period t
P is the Principal (Amount borrowed or lent)
t is the number of years the money is borrowed or lent (t = 3 in this example)
r is the annual rate of interest
Please note that n is treated as 1 here i.e. it is assumed that compounding is happening on an annual basis

\begin{align} A &= 125000 × (1 + \frac{2}{100})^{3} \\ &= 125000 × (\frac{102}{100})^{3}\\ &= 125000 × (1.02)^{3}\\ &= 125000 × 1.061208\\ &= 132651 \end{align}

Example 7. The population of a town was 120000 in 2011. During next year it increased by 6% but due to an epidemic it decreased by 5% in the following year. What would have been its population in year 2013?
Solution The formula to calculate the population with annual compunding is \begin{align} A & = P × (1 + \frac{p}{100}) × (1 - \frac{q}{100}) \end{align} where
A is the population at the end of 2 years period
P is the initial population (in year 2011, in this example, 120000)
p is the annual rate of increase during the first year (p = 6% in this example)
q is the annual rate of decrease during the second year (q = 5% in this example)

\begin{align} A &= 120000 × (1 + \frac{6}{100}) × (1 - \frac{5}{100}) \\ &= 120000 × (\frac{106}{100}) × (\frac{95}{100}) \\ &= 120000 × 1.06 × 0.95 \\ &= 120840 \end{align}

Example 8. Ajay took a loan of ₹ 390,625 from a finance institution. If the institution charges interest at 16% per annum, compounded quarterly, what amount will discharge his debt after 1 year?
Solution The formula to calculate the amount after quarterly compounding is \begin{align} A & = P × (1 + \frac{r}{n})^{nt} \end{align} where
A is the amount at the end of period t
P is the Principal (Amount borrowed or lent)
n is the number of times interest is compounded per year (n = 4 in this example)
t is the number of years the money is borrowed or lent (t = 1 in this example)
r is the annual rate of interest

\begin{align} A &= 390625 × (1 + \frac{16}{4 × 100})^{4 × 1} \\ &= 390625 × (1 + \frac{4}{100})^{4}\\ &= 390625 × (1.04)^{4}\\ &= 390625 × 1.16986\\ &= 456976.5 \end{align}