Question

person deposited Rs.100,000 in a bank for three years. The bank paid interest at the rate of 8% per annum compounded half yearly during the first year and at the rate of 12% per annum compounded quarterly during the last two years. His balance after three years is: *​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that, A person deposited Rs.100,000 in a bank for three years. The bank paid interest at the rate of 8% per annum compounded half yearly during the first year and at the rate of 12% per annum compounded quarterly during the last two years.

Principal, P = Rs 100000

$\rm \: Rate\:of\:interest\:r_1 = 8\% \: per \: annum \: compounded \: half \: yarly$

$\rm \: Rate\:of\:interest\:r_2 = 12\% \: per \: annum \: compounded \: quarterly$

$\rm \: Time,\:t_1 \: = \: 1 \: year$

$\rm \: Time,\:t_2 \: = \: 2 \: years$

We know,

Amount received on a certain sum of money of Rs P invested at the rate of x % per annum compounded semi - annually for n years and y % per annum compounded quarterly for m years is given by

$\boxed{ \rm{ \:Amount \: = \: P \: {\bigg[1 + \dfrac{x}{200} \bigg]}^{2n} {\bigg[1 + \dfrac{y}{400} \bigg]}^{4m} \: }}$

So, on substituting the values, we get

$\rm \:Amount \: = \: 1000000 \: {\bigg[1 + \dfrac{8}{200} \bigg]}^{2} {\bigg[1 + \dfrac{12}{400} \bigg]}^{8} \:$

$\rm \:Amount \: = \: 1000000 \: {\bigg[1 + \dfrac{4}{100} \bigg]}^{2} {\bigg[1 + \dfrac{3}{100} \bigg]}^{8} \:$

$\rm \:Amount \: = \: 1000000 \: {\bigg[\dfrac{100 + 4}{100} \bigg]}^{2} {\bigg[\dfrac{100 + 3}{100} \bigg]}^{8} \:$

$\rm \:Amount \: = \: 1000000 \: {\bigg[\dfrac{104}{100} \bigg]}^{2} {\bigg[\dfrac{103}{100} \bigg]}^{8} \:$

$\rm \:Amount \: = \: 1000000 \: \times \: {(1.04)}^{2} \: \times \: {(1.03)}^{8}$

$\rm\implies \:Amount \: = \:Rs \: 137013.85$

So, his balance after 3 years = Rs 137013. 85

$\rule{190pt}{2pt}$

Note :-

Calculations with the help of logarithmics

$\rm \:Amount \: = \: 1000000 \: \times \: {(1.04)}^{2} \: \times \: {(1.03)}^{8}$

Let assume that

$\rm \:x \: = \: 1000000 \: \times \: {(1.04)}^{2} \: \times \: {(1.03)}^{8}$

Taking log on both sides, we get

$\rm \: logx = log100000 + 2log(1.04) + 8log(1.03)$

$\rm \: logx = 5 + 2(0.0170) + 8(0.0128)$

$\rm \: logx = 5 + 0.034 + 0.1024$

$\rm \: logx = 5.1364$

$\rm \: x = antilog(5.1364)$

$\rm \: x = {10}^{5} \times 1.369$

$\rm \: x = 136900 \: \{approx. \}$