
Find the compound interest on \[{\text{Rs}}{\text{.24000}}\]at \[{\text{15% p}}{\text{.a}}\] for \[{\text{2}}\dfrac{{\text{1}}}{{\text{3}}}\]year.
A.\[{\text{Rs}}{\text{.9327}}\]
B.\[{\text{Rs}}{\text{.9723}}\]
C.\[{\text{Rs}}{\text{.9372}}\]
D.\[{\text{Rs}}{\text{.9237}}\]
Answer
593.1k+ views
Hint: Firstly among the data given in the question it is clear that the principle amount and time period are \[{\text{Rs}}{\text{.24000}}\] for \[{\text{2}}\dfrac{{\text{1}}}{{\text{3}}}\] year and rate of interest is of \[{\text{15% }}\]. Now using the compound interest formula \[{\text{A = P(1 + }}\dfrac{{\text{R}}}{{\text{n}}}{{\text{)}}^{{\text{nt}}}}\]. Substituting the values in this formula, we will reach our answer.
Complete step by step answer:
Given P\[{\text{ = Rs}}{\text{.24000}}\], Rate of interest\[{\text{ = 15% p}}{\text{.a}}\] and time as \[{\text{2}}\dfrac{{\text{1}}}{{\text{3}}}\]year
As , we know that the amount of interest can be easily calculated by \[{\text{A = P(1 + }}\dfrac{{\text{R}}}{{\text{n}}}{{\text{)}}^{{\text{nt}}}}\]
Where,
P = principle amount,
R=rate of interest
N= number of times interest applied per time periods
T= time periods elapsed
Substituting all the values, we get,
\[
\Rightarrow {\text{A = 24000(1 + }}\dfrac{{{\text{15}}}}{{{\text{100}}}}{{\text{)}}^{{\text{2}}\dfrac{1}{3}}} \\
\Rightarrow {\text{A = 24000(1 + }}\dfrac{{\text{3}}}{{{\text{20}}}}{{\text{)}}^{^{{\text{2}}\dfrac{1}{3}}}} \\
\Rightarrow {\text{24000(}}\dfrac{{{\text{23}}}}{{{\text{20}}}}{{\text{)}}^{{\text{2}}\dfrac{1}{3}}} \\
\Rightarrow {\text{24000(1}}{\text{.3855)}} \\
\Rightarrow {\text{Rs}}{\text{.33253}}{\text{.66}} \\
{\text{C}}{\text{.I = Rs}}{\text{.33253}}{\text{.66 - Rs}}{\text{.24000}} \\
{\text{C}}{\text{.I = Rs}}{\text{.9253}}{\text{.66}} \\
\]
Therefore , amount at the end will be \[{\text{Rs}}{\text{.9253}}{\text{.66}}\].
Note: Simple interest is calculated on the principal, or original, amount of a loan. Compound interest is calculated on the principal amount and also on the accumulated interest of previous periods, and can thus be regarded as "interest on interest."
Compound interest is calculated by multiplying the initial principal amount by one plus the annual interest rate raised to the number of compound periods minus one. The total initial amount of the loan is then subtracted from the resulting value.
Complete step by step answer:
Given P\[{\text{ = Rs}}{\text{.24000}}\], Rate of interest\[{\text{ = 15% p}}{\text{.a}}\] and time as \[{\text{2}}\dfrac{{\text{1}}}{{\text{3}}}\]year
As , we know that the amount of interest can be easily calculated by \[{\text{A = P(1 + }}\dfrac{{\text{R}}}{{\text{n}}}{{\text{)}}^{{\text{nt}}}}\]
Where,
P = principle amount,
R=rate of interest
N= number of times interest applied per time periods
T= time periods elapsed
Substituting all the values, we get,
\[
\Rightarrow {\text{A = 24000(1 + }}\dfrac{{{\text{15}}}}{{{\text{100}}}}{{\text{)}}^{{\text{2}}\dfrac{1}{3}}} \\
\Rightarrow {\text{A = 24000(1 + }}\dfrac{{\text{3}}}{{{\text{20}}}}{{\text{)}}^{^{{\text{2}}\dfrac{1}{3}}}} \\
\Rightarrow {\text{24000(}}\dfrac{{{\text{23}}}}{{{\text{20}}}}{{\text{)}}^{{\text{2}}\dfrac{1}{3}}} \\
\Rightarrow {\text{24000(1}}{\text{.3855)}} \\
\Rightarrow {\text{Rs}}{\text{.33253}}{\text{.66}} \\
{\text{C}}{\text{.I = Rs}}{\text{.33253}}{\text{.66 - Rs}}{\text{.24000}} \\
{\text{C}}{\text{.I = Rs}}{\text{.9253}}{\text{.66}} \\
\]
Therefore , amount at the end will be \[{\text{Rs}}{\text{.9253}}{\text{.66}}\].
Note: Simple interest is calculated on the principal, or original, amount of a loan. Compound interest is calculated on the principal amount and also on the accumulated interest of previous periods, and can thus be regarded as "interest on interest."
Compound interest is calculated by multiplying the initial principal amount by one plus the annual interest rate raised to the number of compound periods minus one. The total initial amount of the loan is then subtracted from the resulting value.
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