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Interest & Growth
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CAT 2025 Lesson : Interest & Growth - Non-Annual Compounding

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3.1 Non Annual Compounding

Annual compounding of interest means that interest is computed at the end of every year. This means interest earned during a year is idle. This rate of compounding when increased, results in higher interest earned as the idle period for the interest is reduced.

For instance, Rs. 1,0001,000 invested at the rate of 20%20 \% per annum for 22 years compounded annually, results in
A=1000×(1+20100)2=1,440\text{A} = 1000 \times \left(1 + \dfrac{20}{100} \right) ^{2} = 1,440

If the above investment is compounded semi-annually (every 66 months), then the rate becomes 10%10 \% per semi-annum and the time period is for 44 semi-annums.
A=1000×(1+10100)4=1,464.1\text{A} = 1000 \times \left(1 + \dfrac{10}{100} \right)^{4} = 1,464.1

In cases where interest is provided as a percentage per annum and the compounding period is not per annum, r and n need to be converted.

In a question with interest of 8%8 \% per annum compounded quarterly for 33 years,
n=3×4=12n = 3 \times 4 = 12 quarters
r=84%=2%r = \dfrac{8}{4} \% = 2\% per quarter

These values can then be substituted into the Compound Interest formula.

Example 9

What is the amount due on Rs. 10,00010,000 loaned for 22 months at interest rate of 24%24 \% per annum compounded monthly?

Solution

p=Rs.10,000p = \text{Rs}. 10,000
n=2n = 2 months
r=2412%=2%r = \dfrac{24}{12} \% = 2\% per month

A=p(1+r100)n=10000×1.022=\text{A} = p \left(1 + \dfrac{r}{100} \right)^{n} = 10000 \times 1.02 ^{2} = Rs. 10,40410,404

Answer: Rs. 10,40410,404

3.2 Present Value of CI

Present Value of CI is nothing but the value of the Principal when the amount, rate of interest and time period are provided.

p=A(1+r100)np = \dfrac{A}{\left(1 + \dfrac{r}{100} \right)^{n}}

This is deduced from the Compound Interest formula and need not be separately memorised.

Example 10

The value of an artwork grew at the rate of 25%25 \% per annum for 1010 consecutive years, ending with a value of Rs. 6,2506,250 at the end of the year 20202020. What was the value of the artwork at the start of the year 20182018?

Solution

Value of the artwork at the start of 20182018 will be the same at the end of 20172017. From end of 20172017 to end of 20202020, the number of years is 33.

Let the value at the start of 20182018 be pp.

p×(1+25100)3=6250p \times \left(1 + \dfrac{25}{100} \right)^{3} = 6250

⇒ p×(54)3=6250 p \times \left(\dfrac{5}{4} \right)^{3} = 6250

⇒ p=6250×64125=Rs.3200 p = 6250 \times \dfrac{64}{125} = \text{Rs.} 3200

Answer: Rs. 3,2003,200

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