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 with Growing Principal**

**Compound Interest with Periodic Deductions**

**Compound Interest by Using Formula**

**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**

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