Compound Interest when Interest is Compounded Yearly

We will learn how to use the formula for calculating the compound interest when interest is compounded yearly.

Computation of compound interest by using growing principal becomes lengthy and complicated when the period is long. If the rate of interest is annual and the interest is compounded annually then in such cases we use the following formula for compound interest.

If the principal = P, rate of interest per unit time = r %, number of units of time = n, the amount = A and the compound interest = CI

Then

A = P(1 + \(\frac{r}{100}\))\(^{n}\) and CI = A - P = P{(1 + \(\frac{r}{100}\))\(^{n}\) - 1}

Note:

A = P(1 + \(\frac{r}{100}\))\(^{n}\) is the relation among the four quantities P, r, n and A.

Given any three of these, the fourth can be found from this formula.

CI = A - P = P{(1 + \(\frac{r}{100}\))\(^{n}\) - 1} is the relation among the four quantities P, r, n and CI.

Given any three of these, the fourth can be found from this formula.


Word problems on compound interest when interest is compounded yearly:

1. Find the amount and the compound interest on $ 7,500 in 2 years and at 6% compounded yearly.

Solution:

Here,

 Principal (P) = $ 7,500

Number of years (n) = 2

Rate of interest compounded yearly (r) = 6%

A = P(1 + \(\frac{r}{100}\))\(^{n}\)

   = $ 7,500(1 + \(\frac{6}{100}\))\(^{2}\)

   = $ 7,500 × (\(\frac{106}{100}\))\(^{2}\)

   = $ 7,500 × \(\frac{11236}{10000}\)

   = $ 8,427

Therefore, the required amount = $ 8,427 and

Compound interest = Amount - Principal

                          = $ 8,427 - $ 7,500

                          = $ 927

2. In how many years will a sum of $ 1,00,000 amount to $ 1,33,100 at the compound interest rate of 10% per annum?

Solution:

Let the number of years = n

Here,

Principal (P) = $ 1,00,000

Amount (A) = $ 1,33,100

Rate of interest compounded yearly (r) = 10

Therefore,

A = P(1 + \(\frac{r}{100}\))\(^{n}\)

133100 = 100000(1 + \(\frac{10}{100}\))\(^{n}\)

\(\frac{133100}{100000}\) = (1 + \(\frac{1}{10}\))\(^{n}\)

\(\frac{1331}{1000}\)= (\(\frac{11}{10}\))\(^{n}\)

(\(\frac{11}{10}\))\(^{3}\) = (\(\frac{11}{10}\))\(^{n}\)

n = 3

Therefore, at the rate of compound interest 10% per annum, Rs. 100000 will amount to $ 133100 in 3 years.

3. A sum of money becomes $ 2,704 in 2 years at a compound interest rate 4% per annum. Find

(i) the sum of money at the beginning

(ii) the interest generated.

Solution:

Let the sum of money at the beginning = $ P

Here,

Amount (A) = $ 2,704

Rate of interest compounded yearly (r) = 4

Number of years (n) = 2

(i) A = P(1 + \(\frac{r}{100}\))\(^{n}\)

⟹ 2,704 = P(1 + \(\frac{4}{100}\))\(^{2}\)

⟹ 2,704 = P(1 + \(\frac{1}{25}\))\(^{2}\)

⟹ 2,704 = P(\(\frac{26}{25}\))\(^{2}\)

⟹ 2,704 = P × \(\frac{676}{625}\)

⟹ P = 2,704 × \(\frac{625}{676}\)

 P = 2,500

Therefore, the sum of money at the beginning was $ 2,500

(ii) The interest generated = Amount – Principal

                                    = $2,704 - $2,500

                                    = $ 204


4. Find the rate of compound interest for $ 10,000 amounts to $ 11,000 in two years.

Solution:

Let the rate of compound interest be r% per annum.

Principal (P) = $ 10,000

Amount (A) = $ 11,000

Number of years (n) = 2

Therefore,

A = P(1 + \(\frac{r}{100}\))\(^{n}\)

 10000(1 + \(\frac{r}{100}\))\(^{2}\) = 11664

 (1 + \(\frac{r}{100}\))\(^{2}\) = \(\frac{11664}{10000}\)

 (1 + \(\frac{r}{100}\))\(^{2}\) = \(\frac{729}{625}\)

 (1 + \(\frac{r}{100}\))\(^{2}\) = (\(\frac{27}{25}\))

⟹ 1 + \(\frac{r}{100}\) = \(\frac{27}{25}\)

⟹ \(\frac{r}{100}\) = \(\frac{27}{25}\) - 1

 \(\frac{r}{100}\) = \(\frac{2}{25}\)

⟹ 25r = 200

 r = 8

Therefore, the required rate of compound interest is 8 % per annum.

Compound Interest

Compound Interest

Compound Interest with Growing Principal

Compound Interest with Periodic Deductions

Compound Interest by Using Formula

Problems on Compound Interest

Variable Rate of Compound Interest

Practice Test on Compound Interest


Compound Interest - Worksheet

Worksheet on Compound Interest

Worksheet on Compound Interest with Growing Principal

Worksheet on Compound Interest with Periodic Deductions




8th Grade Math Practice 

From Compound Interest when Interest is Compounded Yearly to HOME PAGE


New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.



Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



Share this page: What’s this?

Recent Articles

  1. Properties of Multiplication | Multiplicative Identity | Whole Numbers

    Mar 29, 24 09:02 AM

    Properties of Multiplication of Whole Numbers
    There are six properties of multiplication of whole numbers that will help to solve the problems easily. The six properties of multiplication are Closure Property, Commutative Property, Zero Property…

    Read More

  2. Multiplication of a Number by a 3-Digit Number |3-Digit Multiplication

    Mar 28, 24 06:33 PM

    Multiplying by 3-Digit Number
    In multiplication of a number by a 3-digit number are explained here step by step. Consider the following examples on multiplication of a number by a 3-digit number: 1. Find the product of 36 × 137

    Read More

  3. Multiply a Number by a 2-Digit Number | Multiplying 2-Digit by 2-Digit

    Mar 27, 24 05:21 PM

    Multiply 2-Digit Numbers by a 2-Digit Numbers
    How to multiply a number by a 2-digit number? We shall revise here to multiply 2-digit and 3-digit numbers by a 2-digit number (multiplier) as well as learn another procedure for the multiplication of…

    Read More

  4. Multiplication by 1-digit Number | Multiplying 1-Digit by 4-Digit

    Mar 26, 24 04:14 PM

    Multiplication by 1-digit Number
    How to Multiply by a 1-Digit Number We will learn how to multiply any number by a one-digit number. Multiply 2154 and 4. Solution: Step I: Arrange the numbers vertically. Step II: First multiply the d…

    Read More

  5. Multiplying 3-Digit Number by 1-Digit Number | Three-Digit Multiplicat

    Mar 25, 24 05:36 PM

    Multiplying 3-Digit Number by 1-Digit Number
    Here we will learn multiplying 3-digit number by 1-digit number. In two different ways we will learn to multiply a two-digit number by a one-digit number. 1. Multiply 201 by 3 Step I: Arrange the numb…

    Read More