Compound Interest when Interest is Compounded Quarterly

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

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 quarterly (i.e., 3 months or, 4 times in a year) then the number of years (n) is 4 times (i.e., made 4n) and the rate of annual interest (r) is one-fourth (i.e., made $\frac{r}{4}$).  In such cases we use the following formula for compound interest when the interest is calculated quarterly.

If the principal = P, rate of interest per unit time = $\frac{r}{4}$%, number of units of time = 4n, the amount = A and the compound interest = CI

Then

A = P(1 + $\frac{\frac{r}{4}}{100}$)$^{4n}$

Here, the rate percent is divided by 4 and the number of years is multiplied by 4.

Therefore, CI = A - P = P{(1 + $\frac{\frac{r}{4}}{100}$)$^{4n}$ - 1}

Note:

A = P(1 + $\frac{\frac{r}{4}}{100}$)$^{4n}$ 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 + $\frac{\frac{r}{4}}{100}$)$^{4n}$ - 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 quarterly:

1. Find the compound interest when $1,25,000 is invested for 9 months at 8% per annum, compounded quarterly.

Solution:

Here, P = principal amount (the initial amount) = $ 1,25,000

Rate of interest (r) = 8 % per annum

Number of years the amount is deposited or borrowed for (n) = $\frac{9}{12}$ year = $\frac{3}{4}$ year.

Therefore,

The amount of money accumulated after n years (A) = P(1 + $\frac{\frac{r}{4}}{100}$)$^{4n}$

                                                                       = $ 1,25,000 (1 + $\frac{\frac{8}{4}}{100}$)$^{4 ∙ \frac{3}{4}}$

                                                                       = $ 1,25,000 (1 + $\frac{2}{100}$)$^{3}$

                                                                       = $ 1,25,000 (1 + $\frac{1}{50}$)$^{3}$

                                                                       = $ 1,25,000 × ($\frac{51}{50}$)$^{3}$

                                                                       = $ 1,25,000 × $\frac{51}{50}$ × $\frac{51}{50}$ × $\frac{51}{50}$

                                                                       = $ 1,32,651

Therefore, compound interest $ (1,32,651 - 1,25,000) = $ 7,651.

2. Find the compound interest on $10,000 if Ron took loan from a bank for 1 year at 8 % per annum, compounded quarterly.

Solution:

Here, P = principal amount (the initial amount) = $ 10,000

Rate of interest (r) = 8 % per annum

Number of years the amount is deposited or borrowed for (n) = 1 year

Using the compound interest when interest is compounded quarterly formula, we have that

A = P(1 + $\frac{\frac{r}{4}}{100}$)$^{4n}$

   = $ 10,000 (1 + $\frac{\frac{8}{4}}{100}$)$^{4 ∙ 1}$

   = $ 10,000 (1 + $\frac{2}{100}$)$^{4}$

   = $ 10,000 (1 + $\frac{1}{50}$)$^{4}$

   = $ 10,000 × ($\frac{51}{50}$)$^{4}$

   = $ 10,000 × $\frac{51}{50}$ × $\frac{51}{50}$ × $\frac{51}{50}$ × $\frac{51}{50}$

   = $ 10824.3216

   = $ 10824.32 (Approx.)

Therefore, compound interest $ (10824.32 - $ 10,000) = $ 824.32

3. Find the amount and the compound interest on $ 1,00,000 compounded quarterly for 9 months at the rate of 4% per annum.

Solution:

Here, P = principal amount (the initial amount) = $ 1,00,000

Rate of interest (r) = 4 % per annum

Number of years the amount is deposited or borrowed for (n) = $\frac{9}{12}$ year = $\frac{3}{4}$ year.

Therefore,

The amount of money accumulated after n years (A) = P(1 + $\frac{\frac{r}{4}}{100}$)$^{4n}$

                                                                       = $ 1,00,000 (1 + $\frac{\frac{4}{4}}{100}$)$^{4 ∙ \frac{3}{4}}$

                                                                       = $ 1,00,000 (1 + $\frac{1}{100}$)$^{3}$

                                                                       = $ 1,00,000 × ($\frac{101}{100}$)$^{3}$

                                                                       = $ 1,00,000 × $\frac{101}{100}$ × $\frac{101}{100}$ × $\frac{101}{100}$

                                                                       = $ 103030.10

Therefore, the required amount = $ 103030.10 and compound interest $ ($ 103030.10 - $ 1,00,000) = $ 3030.10

4. If $1,500.00 is invested at a compound interest rate 4.3% per annum compounded quarterly for 72 months, find the compound interest.

Solution:

Here, P = principal amount (the initial amount) = $1,500.00

Rate of interest (r) = 4.3 % per annum

Number of years the amount is deposited or borrowed for (n) = $\frac{72}{12}$ years = 6 years.

A = amount of money accumulated after n years

Using the compound interest when interest is compounded quarterly formula, we have that

A = P(1 + $\frac{\frac{r}{4}}{100}$)$^{4n}$

   = $1,500.00 (1 + $\frac{\frac{4.3}{4}}{100}$)$^{4 ∙ 6}$

   = $1,500.00 (1 + $\frac{1.075}{100}$)$^{24}$

   = $1,500.00 × (1 + 0.01075)$^{24}$

   = $1,500.00 × (1.01075)$^{24}$

   = $ 1938.83682213

   = $ 1938.84 (Approx.)

Therefore, the compound interest after 6 years is approximately $ (1,938.84 - 1,500.00) = $ 438.84.

● Compound Interest

Compound Interest

Compound Interest with Growing Principal

Compound Interest with Periodic Deductions

Compound Interest by Using Formula

Compound Interest when Interest is Compounded Yearly

Compound Interest when Interest is Compounded Half-Yearly

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

8th Grade Math Practice 

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