Question

The amount of Rs.10000 after 2 years, compounded annually with rate of interest being 10% per annum during the first year and 12% per annum during the second year, would be: -
(a) Rs.11320
(b) Rs.12000
(c) Rs.12320
(d) Rs.12500

Answer 1

Hint: Assume amount at the end of first year as \[{{A}_{1}}\] and at the end of second year as \[{{A}_{2}}\]. Apply the formula of compound interest given as: - \[{{A}_{1}}=P{{\left[ 1+\dfrac{{{r}_{1}}}{n} \right]}^{nt}}\] to calculate the amount \[{{A}_{1}}\] at the end of first year. Here, P = principal amount, \[{{r}_{1}}\] = rate for first year, t = time in years and n = number of times interest is paid or received in one year. Once \[{{A}_{1}}\] is calculated, assume it be the principal amount for second year and apply the formula: - \[{{A}_{2}}={{A}_{1}}{{\left[ 1+\dfrac{{{r}_{2}}}{n} \right]}^{nt}}\] to calculate the amount for the second year. Here, \[{{r}_{2}}\] = rate for second year.
Here, we have been provided with a principal amount of Rs.10000 and we have to calculate the amount after 2 years in which the rate for the first year is 10%, and for the second year, the rate is 12%.

Complete step-by-step solution
Now, let us assume the amount at the end of the first and second year as \[{{A}_{1}}\] and \[{{A}_{2}}\] respectively. Also, we are assuming the rate for the first and second year as \[{{r}_{1}}\] and \[{{r}_{2}}\] respectively. So, let us calculate the amounts one – by – one.
1. For first year: -
P = Principal amount = Rs.10000
t = 1 year
\[{{r}_{1}}\] = 10% = \[\dfrac{10}{100}\]
So, applying the formula of compound interest given as: -
\[{{A}_{1}}=P{{\left[ 1+\dfrac{{{r}_{1}}}{n} \right]}^{nt}}\], we get,
\[\Rightarrow {{A}_{1}}=10000{{\left[ 1+\dfrac{10}{100\times n} \right]}^{n\times 1}}\]
Here, it is given that interest is compounded annually, so the value of ‘n’ will be 1.
\[\Rightarrow {{A}_{1}}=10000\left[ 1+\dfrac{1}{10} \right]\]
\[\Rightarrow {{A}_{1}}=\] Rs.11000
Now, this amount \[{{A}_{1}}\] will be the principal amount for the second year. So, we have,
2. For second year: -
\[{{A}_{1}}\] = Principal amount = Rs.11000
t = 1 year
\[{{r}_{1}}\] = 12% = \[\dfrac{12}{100}\]
So, again applying the formula for compound interest, we get,
\[\begin{align}
  & \Rightarrow {{A}_{2}}={{A}_{1}}{{\left[ 1+\dfrac{{{r}_{2}}}{n} \right]}^{nt}} \\
 & \Rightarrow {{A}_{2}}=11000{{\left[ 1+\dfrac{12}{100\times n} \right]}^{n\times 1}} \\
\end{align}\]
Here, also it is given that the interest is compounded annually, so the value of ‘n’ will be 1.
\[\begin{align}
  & \Rightarrow {{A}_{2}}=11000\left[ 1+\dfrac{12}{100} \right] \\
 & \Rightarrow {{A}_{2}}=11000\times \dfrac{112}{100} \\
\end{align}\]
\[\Rightarrow {{A}_{2}}=\] Rs.12320
Hence, option (c) is the correct answer.

Note: One may note that we cannot find the amount for 2 years directly by considering t = 2 years in the formula. This is because the given rates for the first and second years are different. You may make a mistake by considering Rs.10000 as the principal amount for the second year. Here, we have to consider \[{{A}_{1}}\] as the principal amount for the second year. Note that here we have used the value of ‘n’ equal to 1 because it is given in the question that the interest is being compounded annually.