
If $\$ 10,000$ is invested at $10$ percent annual interest, compounded semi-annually, what is the balance after $1$ year?
A. $\$ 10100.25$
B. $\$ 10200.25$
C. $\$ 11025$
D. $\$ 10100$
Answer
484.8k+ views
Hint: In this problem, given that the rate of interest is annual and the interest is compounded semi-annually (that is, six months). That means the interest paid at the end of every six months is one-half of the rate of interest per annum. So, the rate of annual interest is $\dfrac{R}{2}\% $ and the number of years is doubled (that is, $2T$). We will find the amount (balance) $A$ for $1$ year by using the formula $A = P{\left( {1 + \dfrac{R}{{100}}} \right)^T}$.
Complete step by step solution: Here given that the principal amount $P = $$\$ 10,000$, rate of interest $R = 10\% $ per annum and time $T = 1$ year. Also given that the interest is compounded semi-annually (that is, six months). So, the interest paid at the end of every six months is one-half of the rate of interest per annum. So, the rate of annual interest is $R = \dfrac{{10}}{2}\% = 5\% $ and the number of years is doubled. That is, $T = 2$ half years.
Now we are going to find the amount for $1$ year by using the formula $A = P{\left( {1 + \dfrac{R}{{100}}} \right)^T}$ where $P$ is principal amount, $R$ is rate of annual interest and $T$ is time in half years.
Now we are going to substitute the values of $P$, $R$ and $T$ in the formula of amount for $1$ year.
Therefore, $A = 10000{\left( {1 + \dfrac{5}{{100}}} \right)^2}$
$ \Rightarrow $ $A = 10000{\left( {\dfrac{{100 + 5}}{{100}}} \right)^2}$
$ \Rightarrow $$A = 10000{\left( {\dfrac{{105}}{{100}}} \right)^2} = 10000\left( {\dfrac{{105 \times 105}}{{100 \times 100}}} \right)$
$ \Rightarrow $ $A = 105 \times 105 = \$ 11025$
Therefore, the balance after $1$ year will be $\$ 11025$.
Therefore, option C is correct.
Note: Simple interest is calculated only on the principal amount but compound interest is calculated on principal amount as well as previous year’s interest. If interest is paid only for $T = 1$ year then there is no distinction between simple interest and compound interest. To find simple interest, we can use the formula $\dfrac{{PRT}}{{100}}$ where $P$ is principal amount, $R$ is rate of interest and $T$ is time in years.
Complete step by step solution: Here given that the principal amount $P = $$\$ 10,000$, rate of interest $R = 10\% $ per annum and time $T = 1$ year. Also given that the interest is compounded semi-annually (that is, six months). So, the interest paid at the end of every six months is one-half of the rate of interest per annum. So, the rate of annual interest is $R = \dfrac{{10}}{2}\% = 5\% $ and the number of years is doubled. That is, $T = 2$ half years.
Now we are going to find the amount for $1$ year by using the formula $A = P{\left( {1 + \dfrac{R}{{100}}} \right)^T}$ where $P$ is principal amount, $R$ is rate of annual interest and $T$ is time in half years.
Now we are going to substitute the values of $P$, $R$ and $T$ in the formula of amount for $1$ year.
Therefore, $A = 10000{\left( {1 + \dfrac{5}{{100}}} \right)^2}$
$ \Rightarrow $ $A = 10000{\left( {\dfrac{{100 + 5}}{{100}}} \right)^2}$
$ \Rightarrow $$A = 10000{\left( {\dfrac{{105}}{{100}}} \right)^2} = 10000\left( {\dfrac{{105 \times 105}}{{100 \times 100}}} \right)$
$ \Rightarrow $ $A = 105 \times 105 = \$ 11025$
Therefore, the balance after $1$ year will be $\$ 11025$.
Therefore, option C is correct.
Note: Simple interest is calculated only on the principal amount but compound interest is calculated on principal amount as well as previous year’s interest. If interest is paid only for $T = 1$ year then there is no distinction between simple interest and compound interest. To find simple interest, we can use the formula $\dfrac{{PRT}}{{100}}$ where $P$ is principal amount, $R$ is rate of interest and $T$ is time in years.
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