What will be the compound interest on rupees 40000 at 8% per annum when interest is compounded half yearly?

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10 Questions 10 Marks 9 Mins

Given:

The rate of interest = 20% per annum

The sum = Rs. 40,000

Formula used:

Amount = \({\ {P\ × [{1\ +\ ({R\over 100}})]^2}}\)

Calculation:

Let the compounded amount be X

The effective time = 2 × 2 = 4 Years

The effective rate of interest = \({20\over 2}\ =\ 10\%\) half yearly

The compounded amount = \(40000\ × {(1\ +\ {10\over 100})^4}\ =\ 40000\ × ({11\over 10})^4\) = 4 × 14641 = Rs. 58564

∴ The required result will be Rs. 58564.

Let's discuss the concepts related to Interest and Compound Interest. Explore more from Quantitative Aptitude here. Learn now!

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We will learn how to use the formula for calculating the compound interest when interest is compounded half-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 half-yearly (i.e., 6 months or, 2 times in a year) then the number of years (n) is doubled (i.e., made 2n) and the rate of annual interest (r) is halved (i.e., made \(\frac{r}{2}\)).  In such cases we use the following formula for compound interest when the interest is calculated half-yearly.

If the principal = P, rate of interest per unit time = \(\frac{r}{2}\)%, number of units of time = 2n, the amount = A and the compound interest = CI

Then

A = P(1 + \(\frac{\frac{r}{2}}{100}\))\(^{2n}\)

Here, the rate percent is divided by 2 and the number of years is multiplied by 2

Therefore,  CI = A - P = P{(1 + \(\frac{\frac{r}{2}}{100}\))\(^{2n}\) - 1}

Note:

A = P(1 + \(\frac{\frac{r}{2}}{100}\))\(^{2n}\) 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 + \(\frac{\frac{r}{2}}{100}\))\(^{2n}\) - 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 half-yearly:

1. Find the amount and the compound interest on $ 8,000 at 10 % per annum for 1\(\frac{1}{2}\) years if the interest is compounded half-yearly.

Solution:

Here, the interest is compounded half-yearly. So,

Principal (P) = $ 8,000

Number of years (n) = 1\(\frac{1}{2}\) × 2 = \(\frac{3}{2}\) × 2 = 3

Rate of interest compounded half-yearly (r) = \(\frac{10}{2}\)% = 5%

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

⟹ A = $ 8,000(1 + \(\frac{5}{100}\))\(^{3}\)

⟹ A = $ 8,000(1 + \(\frac{1}{20}\))\(^{3}\)

⟹ A = $ 8,000 × (\(\frac{21}{20}\))\(^{3}\)

⟹ A = $ 8,000 × \(\frac{9261}{8000}\)

⟹ A = $ 9,261 and

Compound interest = Amount - Principal

                          = $ 9,261 - $ 8,000

                          = $ 1,261

Therefore, the amount is $ 9,261 and the compound interest is $ 1,261

2. Find the amount and the compound interest on $ 4,000 is 1\(\frac{1}{2}\) years at 10 % per annum compounded half-yearly.

Solution:

Here, the interest is compounded half-yearly. So,

Principal (P) = $ 4,000

Number of years (n) = 1\(\frac{1}{2}\) × 2 = \(\frac{3}{2}\) × 2 = 3

Rate of interest compounded half-yearly (r) = \(\frac{10}{2}\)% = 5%

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

⟹ A = $ 4,000(1 + \(\frac{5}{100}\))\(^{3}\)

⟹ A = $ 4,000(1 + \(\frac{1}{20}\))\(^{3}\)

⟹ A = $ 4,000 × (\(\frac{21}{20}\))\(^{3}\)

⟹ A = $ 4,000 × \(\frac{9261}{8000}\)

⟹ A = $ 4,630.50 and

Compound interest = Amount - Principal

                          = $ 4,630.50 - $ 4,000

                          = $ 630.50

Therefore, the amount is $ 4,630.50 and the compound interest is $ 630.50

● Compound Interest

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Compound Interest when Interest is Compounded Yearly

Problems on Compound Interest

Variable Rate of Compound Interest

Practice Test on Compound Interest

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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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