
At what rate percentage a sum of \[Rs.640\] be compounded to \[Rs.774.40\] in two years?
Answer
507.9k+ views
Hint: In this question, we have to find the rate percentage at which it was compounded that we reached the given amount in two years. We do this by using the formula of Amount in terms of Principle, Time and Rate percentage. On solving the formula \[A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}\] we will get the desired result.
Formula used: If \[P\] is the principal money at the starting, which is compounded for \[n\] years at the rate of \[r\] percentage, the total amount we will reach is given by,
\[A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}\].
Complete step-by-step answer:
In this question, we are given principal rupees which get compounded to amount rupees given above in two years. We have to find the rate percentage at which it was compounded that we reached the given amount in two years. So,
Principle money, \[P = Rs.640\]
Amount, \[A = Rs.774.40\]
Time compounded in years, \[n = 2\]
Let rate percentage \[ = r\]
We know that Amount in terms of Principle, Time and Rate percentage is given by,
\[A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}\]
On putting the values from above,
\[
\Rightarrow 774.40 = 640{\left( {1 + \dfrac{r}{{100}}} \right)^2} \\
\Rightarrow \dfrac{{774.40}}{{640}} = {\left( {1 + \dfrac{r}{{100}}} \right)^2} \\
\]
\[\dfrac{{774.40}}{{640}}\] can also be written as \[{\left( {\dfrac{{11}}{{10}}} \right)^2}\]. Using this in the above step,
\[
\Rightarrow {\left( {\dfrac{{11}}{{10}}} \right)^2} = {\left( {1 + \dfrac{r}{{100}}} \right)^2} \\
\Rightarrow \dfrac{{11}}{{10}} = 1 + \dfrac{r}{{100}} \\
\]
On further solving the above step,
\[
\Rightarrow \dfrac{{11}}{{10}} - 1 = \dfrac{r}{{100}} \\
\Rightarrow \dfrac{1}{{10}} = \dfrac{r}{{100}} \\
\Rightarrow r = \dfrac{{100}}{{10}} \\
\Rightarrow r = 10 \\
\]
Hence, \[r = 10\% \], we compounded \[Rs.640\] to \[Rs.774.40\] in two years.
Note: In above question it is to note if the principle is compounded annually for two years or the principle is compounded in two yearly basis. In this case it is the former one. In compound interest, we get interest on the in interest as well, Amount in compound interest is more than amount in simple interest form.
Formula used: If \[P\] is the principal money at the starting, which is compounded for \[n\] years at the rate of \[r\] percentage, the total amount we will reach is given by,
\[A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}\].
Complete step-by-step answer:
In this question, we are given principal rupees which get compounded to amount rupees given above in two years. We have to find the rate percentage at which it was compounded that we reached the given amount in two years. So,
Principle money, \[P = Rs.640\]
Amount, \[A = Rs.774.40\]
Time compounded in years, \[n = 2\]
Let rate percentage \[ = r\]
We know that Amount in terms of Principle, Time and Rate percentage is given by,
\[A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}\]
On putting the values from above,
\[
\Rightarrow 774.40 = 640{\left( {1 + \dfrac{r}{{100}}} \right)^2} \\
\Rightarrow \dfrac{{774.40}}{{640}} = {\left( {1 + \dfrac{r}{{100}}} \right)^2} \\
\]
\[\dfrac{{774.40}}{{640}}\] can also be written as \[{\left( {\dfrac{{11}}{{10}}} \right)^2}\]. Using this in the above step,
\[
\Rightarrow {\left( {\dfrac{{11}}{{10}}} \right)^2} = {\left( {1 + \dfrac{r}{{100}}} \right)^2} \\
\Rightarrow \dfrac{{11}}{{10}} = 1 + \dfrac{r}{{100}} \\
\]
On further solving the above step,
\[
\Rightarrow \dfrac{{11}}{{10}} - 1 = \dfrac{r}{{100}} \\
\Rightarrow \dfrac{1}{{10}} = \dfrac{r}{{100}} \\
\Rightarrow r = \dfrac{{100}}{{10}} \\
\Rightarrow r = 10 \\
\]
Hence, \[r = 10\% \], we compounded \[Rs.640\] to \[Rs.774.40\] in two years.
Note: In above question it is to note if the principle is compounded annually for two years or the principle is compounded in two yearly basis. In this case it is the former one. In compound interest, we get interest on the in interest as well, Amount in compound interest is more than amount in simple interest form.
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