Compound Interest, Population Growth,and Depriciation

Compound interest is interest on interest. It is the result of reinvesting interest, rather than paying it out, so that interest in the next period is then earned on the principal sum plus previously-accumulated interest. Compound interest is standard in finance and economics.

Compound interest may be contrasted with simple interest, where interest is not added to the principal, so there is no compounding. The simple annual interest rate is the interest amount per period, multiplied by the number of periods per year.

Compound Amount (C. A) = P (1+ R100$\frac{\mathrm{R}}{100}$) T

C.I= C.A – P

= P (1+ R100$\frac{\mathrm{R}}{100}$) T - P

= P [(1+ R100$\frac{\mathrm{R}}{100}$) T-1]

Interested compounded half yearly

When the compound interest is paid half yearly, then the rate R% per year will be R/2% and time would be doubled (2T).

(C. A)halfyearly= P (1+ R/2100$\frac{\mathrm{R}/2}{100}$) 2T

(C. A) = P (1+ R200$\frac{\mathrm{R}}{200}$) 2T

Population Growth and Depreciation

Population of the certain place increases every year with the certain rate. The increase in population is same as the compound interest.

If PTbe the population after T years, P be the present population the then formula reduces to

PT=P (1+R100)T${\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{T}}$

Compound depreciation

The value of the machines and other goods decreases every year. The reduced value of goods is known as depreciation. If the reduced value of the goods is compounded for fixed time then it is called compound depreciation. It is opposite to the compound interest.

If VTbe the value of the goods after T years and VPbe the present values .Then,

 VT=VP(1-R100$\frac{\mathrm{R}}{100}$)T

Where R and T are the Rate and the time respectively.

Example 1

The compound amount of a certain sum of money in 2 years and 3 years become Rs 8820 and Rs 926 respectively. Find the sum and the rate of interest.

Compound amount = P(1+R100)T$\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{T}}$

Or, 8820 = x(1+R100)2$\mathrm{x}{\left(1+\frac{\mathrm{R}}{100}\right)}^2$…..(i)

Or, 9261 = x(1+R100)3${\left(1+\frac{\mathrm{R}}{100}\right)}^3$….(ii)

Dividing (ii) by (i),

Or, 1.05 = (1+R100)$\left(1+\frac{\mathrm{R}}{100}\right)$

Or, 1.05 = (100+R100)$\left(\frac{100+\mathrm{R}}{100}\right)$

Or, 105 = 100 + R

So, R = 5%

Or, 9261 = x(1+R100)3${\left(1+\frac{\mathrm{R}}{100}\right)}^3$

Or, 9261 = x(1+5100)3${\left(1+\frac{5}{100}\right)}^3$

Or, 9261 = x(1+0.05)3${\left(1+0.05\right)}^3$

Or, x = 92611.16$\frac{9261}{1.16}$

So, x = Rs. 8,000

Examples 2

Find the difference between compound interest compounded semi – annually and simple interest on Rs 8000 at 10% per annum in 112$\frac{1}{2}$years.$.$

Soln

For semi annually,

Principle (P) = Rs. 8,000

Rate (R) = 10%

Time (T) = 32$\frac{3}{2}$ years

Now,

I = PTR100$\frac{\mathrm{P}\mathrm{T}\mathrm{R}}{100}$

= 8000∗10∗32100$\frac{8000\ast 10\ast \frac{3}{2}}{100}$

= 1200

Now, For semi-annually,

= P[(1+R2∗100)2T−1]$\left[{\left(1+\frac{\mathrm{R}}{2\ast 100}\right)}^{2\mathrm{T}}-1\right]$

= 8000 [(1+10200)2∗32−1]$\left[{\left(1+\frac{10}{200}\right)}^{2\ast \frac{3}{2}}-1\right]$

= 8000[(1+120)3−1]$8000\left[{\left(1+\frac{1}{20}\right)}^3-1\right]$

= 8,000 * [1.1576 – 1]

= 8000 * 0.1576

= 1260.8

Now, difference = 1260.8 – 12600

= 60.8

Example 3

At a certain rate of yearly compound interest, a sum of money amounts to Rs 66550 in 3 years and Rs 73205 in 4 years. Find the interest and the sum.

Soln

For 1st case,

Amount (A) = Rs. 66,550

Time (T) = 3years

A = P(1+R100)T$\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{T}}$

Or, 66550 = P (1+R100)3${\left(1+\frac{\mathrm{R}}{100}\right)}^3$…..(i)

For 2nd case,

Amount (A) = 73205

Time (T) = 4yrs

Or, A = P(1+R100)T$\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{T}}$

Or, 73205 = P(1+R100)4$\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^4$…(i)

Dividing (ii) by (i)

7320566550$\frac{73205}{66550}$ = (1+R100)4−3${\left(1+\frac{\mathrm{R}}{100}\right)}^{4-3}$

Or, 1.1 = (1+R100)$\left(1+\frac{\mathrm{R}}{100}\right)$

Or, 1.1 * 100 = (100+R)$\left(100+\mathrm{R}\right)$

Or, 110 = 100 + R

So, R = 10%

or, 73205 = P(1+R100)4$\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^4$

or, 73205 = P (1+10100)4${\left(1+\frac{10}{100}\right)}^4$

or, 73205 = P (1.1)4${\left(1.1\right)}^4$

or, P = 73205(1.1)4$\frac{73205}{{\left(1.1\right)}^4}$

So, P = Rs. 50,000

 Examples 4

The value of the article depreciated From Rs 18000 to Rs 14580 in 2 years. Find the yearly rate depreciation.

Depreciation amount = 18,000- 14580 = Rs 16520

T = 2yr

DT= Di(1−R100)T${\left(1-\frac{\mathrm{R}}{100}\right)}^{\mathrm{T}}$

14850= 18000(1−R100)2${\left(1-\frac{\mathrm{R}}{100}\right)}^2$

 R = 10%