Question

In what time will Rs.64000 amount to Rs.68921 at 5% interest per annum, interest being compounded half annually?
\[\begin{align}
  & A.1\dfrac{1}{2}\text{years} \\
 & B.\dfrac{2}{3}\text{years} \\
 & C.2\text{years} \\
 & D.\text{None of these} \\
\end{align}\]

Answer 1

Hint: In this question, we are given the principal amount, compounded amount, and the rate of compound interest per annum. We need to find the time period if interest is compounded half annually. For this, we will use the formula of the compound amount given as $A=P{{\left( 1+\dfrac{r}{100} \right)}^{T}}$ where A is compounded amount, P is the principal amount, r is the rate of interest per annum, T is the time in years. Since the rate of interest is compounded half annually so will use time 2T in place of T and $\dfrac{r}{2}$ in place of r.

Complete step-by-step solution
Here we are given the principal amount as Rs.64000 and the compound amount as Rs.68921. We are also given the rate of interest per annum as 5%. Now the rate of interest is compounded half annually. So we will have to decrease the rate of interest and increase the time period i.e. rate of interest will become $\dfrac{5}{2}\%$ per half annum or we can say $r=\dfrac{5}{2}\%$ per half annum.
Also, our time period will become twice. So, we can say t = 2T, where T is in years.
We know the formula of the compound amount is given by $A=P{{\left( 1+\dfrac{r}{100} \right)}^{T}}$ where A is the compound amount, P is the principal amount, r is the rate of interest and t is the time period.
Putting all the found values we get:
\[\begin{align}
  & 68921=64000{{\left( 1+\dfrac{\dfrac{5}{2}}{100} \right)}^{2T}} \\
 & \Rightarrow \dfrac{68921}{64000}={{\left( 1+\dfrac{5}{200} \right)}^{2T}} \\
\end{align}\]
Taking LCM of 200 on the left side of the equation we get:
\[\begin{align}
  & \Rightarrow \dfrac{68921}{64000}={{\left( \dfrac{200+5}{200} \right)}^{2T}} \\
 & \Rightarrow \dfrac{68921}{64000}={{\left( \dfrac{205}{200} \right)}^{2T}} \\
 & \Rightarrow \dfrac{68921}{64000}={{\left( \dfrac{41}{40} \right)}^{2T}} \\
\end{align}\]
Now, we know that 68921 can be written as $41\times 41\times 41$ and 64000 can be written as $40\times 40\times 40$ so we get:
\[\Rightarrow {{\left( \dfrac{41}{40} \right)}^{3}}={{\left( \dfrac{41}{40} \right)}^{2T}}\]
By law of indices, if base are equal then power are equal, therefore we can say that,
\[\begin{align}
  & \Rightarrow 3=2T \\
 & \Rightarrow T=\dfrac{3}{2} \\
\end{align}\]
Or we can write $\dfrac{3}{2}$ as $1\dfrac{1}{2}$ years.
Hence the required time period is $1\dfrac{1}{2}$ years.
Hence option A is the correct answer.

Note: Students can forget to decrease the rate of interest also. The rate of interest is given per annum but we want it per half annum so we have to decrease it. Make sure the base values are the same, before putting exponential values equally. Take care while dealing with the fractional part in the formula.