Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

The sum of Rs $1000$ is compounded continuously, the nominal rate of interest being four percent per annum. In how many years will the amount be twice the original principal ? $(\log_{e}$$2=0.6931)$

Can you answer this question?

1 Answer

0 votes
  • First order,first degree DE variable separatable : Variable of a DE are rearranged to separate there (i.e.) $f_1(x)g_2(y)dx+f_2(x)g_1(y)dy=0$
  • It can be written as $\large\frac{g_1(y)}{g_2(y)}$$dy=\large\frac{f_1(x)}{f_2(x)}$$dx$
  • The solution is therefore $\int \large\frac{g_1(y)}{g_2(y)}$$dy=-\int \large\frac{f_1(x)}{f_2(x)}$$dx+c$
Step 1:
The sum of Rs1000 is the principal or initial amount.($x_0)$
Step 2:
Let $x$ be the amount at time $t$ years,the amount being compounded continuously,at an annual rate of 4% per annum.
Step 3:
$\therefore$ rate of change of $x$ is 0.04
$\int \large\frac{dx}{x}$$=\int 0.04 dt+\log c$
$\log x=0.04t+\log c$
$\log\large\frac{x}{c}$$=0.04 t$
$\therefore x=ce^{0.04t}$
Step 4:
When $t=0,x_0=1000$
Step 5:
When the amount doubles ,$x=2000$
$\therefore 2000=1000e^{0.04t}$
$\Rightarrow e^{0.04t}=2$
$0.04t\log_e e=\log _e 2$
The time taken for the amount to double =17 years approximately.
answered Sep 6, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App