
The compound interest on \[10000\] at \[10\% \] per annum for \[3\] years, compounded annually, is
A) \[1331\]
B) \[3310\]
C) \[3130\]
D) \[13310\]
Answer
522.3k+ views
Hint: First we have to know what compound interest is. Then use the compound interest formula to find the total interest for the number of years. If the interest is compounded half yearly then after every 6 months, interest will be added to the principal.
Complete step by step solution:
First understand some basic definitions and formulas.
Simple interest \[ = \dfrac{{P \times T \times R}}{{100}}\] where \[P\] is Principal, T is time and \[R\] is the rate of interest.
Sometimes it so happens that the borrower and the lender agree to fix up a certain unit of time, say yearly or half yearly or quarterly to settle the previous account. In such cases, the amount after a unit of time becomes the principal for the second unit, the amount after the second unit becomes the principal for the third unit and so on. After a specified period, the difference between the amount and the money borrowed is called the Compound Interest for that period. Compound interest is abbreviated as \[C.I\].
In case of compound interest, the interest will be added to the initial principal after every compounding period. Hence, compound interest keeps on increasing after every compounding period. Hence, Compound interest if given by
\[CI = P{\left[ {1 + \dfrac{R}{{100}}} \right]^N} - P\] Where, \[N\] is a number of years.
Given \[P = 10000\], \[N = 3\], \[R = 10\% \]
\[CI = 10000 \times {\left[ {1 + \dfrac{{10}}{{100}}} \right]^3} - 10000 = 13310 - 10000 = 3310\].
Therefore, the compound interest is ₹\[3310\]. So, Option B is correct.
Additional information:
If the interest is compounded half yearly, In this case, after every 6 months, interest will be added to the principal.
\[CI = P{\left[ {1 + \dfrac{{\left( {\dfrac{R}{2}} \right)}}{{100}}} \right]^{2N}} - P\]----(1), Where \[R\] is the rate of interest for one year.
Note:
Note that the unit of rate of interest and time should be the same. So, if the rate of interest is ‘per year’, then time should also be in ‘year’. Similarly, if the rate of interest is ‘per six months’, then time should also be in ‘per six-months’.
Complete step by step solution:
First understand some basic definitions and formulas.
Simple interest \[ = \dfrac{{P \times T \times R}}{{100}}\] where \[P\] is Principal, T is time and \[R\] is the rate of interest.
Sometimes it so happens that the borrower and the lender agree to fix up a certain unit of time, say yearly or half yearly or quarterly to settle the previous account. In such cases, the amount after a unit of time becomes the principal for the second unit, the amount after the second unit becomes the principal for the third unit and so on. After a specified period, the difference between the amount and the money borrowed is called the Compound Interest for that period. Compound interest is abbreviated as \[C.I\].
In case of compound interest, the interest will be added to the initial principal after every compounding period. Hence, compound interest keeps on increasing after every compounding period. Hence, Compound interest if given by
\[CI = P{\left[ {1 + \dfrac{R}{{100}}} \right]^N} - P\] Where, \[N\] is a number of years.
Given \[P = 10000\], \[N = 3\], \[R = 10\% \]
\[CI = 10000 \times {\left[ {1 + \dfrac{{10}}{{100}}} \right]^3} - 10000 = 13310 - 10000 = 3310\].
Therefore, the compound interest is ₹\[3310\]. So, Option B is correct.
Additional information:
If the interest is compounded half yearly, In this case, after every 6 months, interest will be added to the principal.
\[CI = P{\left[ {1 + \dfrac{{\left( {\dfrac{R}{2}} \right)}}{{100}}} \right]^{2N}} - P\]----(1), Where \[R\] is the rate of interest for one year.
Note:
Note that the unit of rate of interest and time should be the same. So, if the rate of interest is ‘per year’, then time should also be in ‘year’. Similarly, if the rate of interest is ‘per six months’, then time should also be in ‘per six-months’.
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