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 

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