Compound interest handwritten notes

 Compound interest 

Compound interest handwritten notes

Example 

Question

Sameerrao has taken a loan of ₹12500 at a rate of 12 p.c.p.a. for 3 years. If the interest is compounded annually then how many rupees should he pay to clear his loan?

ANSWER:

Here, P = ₹ 12500; R = 12 % ; N = 3 years

A=P(1+R100)N   =12500(1+12100)3   =12500(1+325)3   =12500(2825)3   =17561.60 Rupees$$\begin{gathered} \mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}} \\ =12500{\left(1+\frac{12}{100}\right)}^3 \\ =12500{\left(1+\frac{3}{25}\right)}^3 \\ =12500{\left(\frac{28}{25}\right)}^3 \\ =17561.60 \mathrm{Rupees} \end{gathered}$$

Hence, he should pay an amount of ₹ 17561.60 to clear his loan

Question 2:

A shepherd has 200 sheep with him. Find the number of sheeps with him after 3 years if the increase in number of sheeps is 8% every year.

ANSWER:

Here, P = Number of sheeps initially = 200
A = Number of sheeps after 3 years
R = Rate of increase of number of sheeps per year = 8 %
N = 3 years

A=P(1+R100)N   =200(1+8100)3   =200(1+225)3   =200(2725)3   =251.94   =252 (approx)$$\begin{gathered} \mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}} \\ =200{\left(1+\frac{8}{100}\right)}^3 \\ =200{\left(1+\frac{2}{25}\right)}^3 \\ =200{\left(\frac{27}{25}\right)}^3 \\ =251.94 \\ =252 \left(\mathrm{approx}\right) \end{gathered}$$

Hence, the number of sheeps after 3 years is 252.

Question 3:

In a forest there are 40,000 trees. Find the expected number of trees after 3 years if the objective is to increase the number at the rate 5% per year.

ANSWER:

Here, P = Number of trees initially = 40,000
A = Number of trees after 3 years
R = Rate of increase of number of trees per year = 5 %
N = 3 years

A=P(1+R100)N   =40000(1+5100)3   =40000(1+120)3   =40000(2120)3   =46305$$\begin{gathered} \mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}} \\ =40000{\left(1+\frac{5}{100}\right)}^3 \\ =40000{\left(1+\frac{1}{20}\right)}^3 \\ =40000{\left(\frac{21}{20}\right)}^3 \\ =46305 \end{gathered}$$

Hence, the expected number of trees after 3 years is 46,305.

Question 4:

The cost price of a machine is 2,50,000. If the rate of depreciation is 10% per year find the depreciation in price of the machine after two years.

ANSWER:

Here, P = Cost price of the machine = 2,50,000
A = Cost price after 2 years
I = Depreciation in price after 2 years
R = Rate of depreciation = 10 %
N = 2 years

A=P(1+R100)N   =250000(1+−10100)2   =250000(1−110)2   =250000(910)2   =202500$$\begin{gathered} \mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}} \\ =250000{\left(1+\frac{-10}{100}\right)}^2 \\ =250000{\left(1-\frac{1}{10}\right)}^2 \\ =250000{\left(\frac{9}{10}\right)}^2 \\ =202500 \end{gathered}$$

Also,
I = P − A
  = 250000 − 202500
  = 47500

Hence, the depreciation in price of the machine after two years is Rs 47,500.