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Arithmetic I

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Interest & Growth

Interest And Growth

MODULES

Basics & Simple Interest
Advanced Simple Interest
Basics of Compound Interest
Non-Annual Compounding
Present Value & EMI
Growth & CAGR
Common Types
Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Interest and growth 1
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Interest and growth 2
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PRACTICE

Interest & Growth : Level 1
Interest & Growth : Level 2
Interest & Growth : Level 3
ALL MODULES

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. 1,0001,0001,000 invested at the rate of 20%20 \%20% per annum for 222 years compounded annually, results in
A=1000×(1+20100)2=1,440\text{A} = 1000 \times \left(1 + \dfrac{20}{100} \right) ^{2} = 1,440A=1000×(1+10020​)2=1,440

If the above investment is compounded semi-annually (every 666 months), then the rate becomes 10%10 \%10% per semi-annum and the time period is for 444 semi-annums.
A=1000×(1+10100)4=1,464.1\text{A} = 1000 \times \left(1 + \dfrac{10}{100} \right)^{4} = 1,464.1A=1000×(1+10010​)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 \%8% per annum compounded quarterly for 333 years,
n=3×4=12n = 3 \times 4 = 12n=3×4=12 quarters
r=84%=2%r = \dfrac{8}{4} \% = 2\%r=48​%=2% per quarter

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

Example 9

What is the amount due on Rs. 10,00010,00010,000 loaned for 222 months at interest rate of 24%24 \%24% per annum compounded monthly?

Solution

p=Rs.10,000p = \text{Rs}. 10,000p=Rs.10,000
n=2n = 2n=2 months
r=2412%=2%r = \dfrac{24}{12} \% = 2\%r=1224​%=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} =A=p(1+100r​)n=10000×1.022= Rs. 10,40410,40410,404

Answer: Rs. 10,40410,40410,404

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}}p=(1+100r​)nA​

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 \%25% per annum for 101010 consecutive years, ending with a value of Rs. 6,2506,2506,250 at the end of the year 202020202020. What was the value of the artwork at the start of the year 201820182018?

Solution

Value of the artwork at the start of 201820182018 will be the same at the end of 201720172017. From end of 201720172017 to end of 202020202020, the number of years is 333.

Let the value at the start of 201820182018 be ppp.

p×(1+25100)3=6250p \times \left(1 + \dfrac{25}{100} \right)^{3} = 6250p×(1+10025​)3=6250

⇒ p×(54)3=6250 p \times \left(\dfrac{5}{4} \right)^{3} = 6250p×(45​)3=6250

⇒ p=6250×64125=Rs.3200 p = 6250 \times \dfrac{64}{125} = \text{Rs.} 3200p=6250×12564​=Rs.3200

Answer: Rs. 3,2003,2003,200

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