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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
Compound Interest
Compound Interest with Growing Principal
Compound Interest with Periodic Deductions
Compound Interest by Using Formula
Compound Interest when Interest is Compounded 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
Worksheet on Compound Interest with Periodic Deductions
8th Grade Math Practice
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