If the interest is compounded continuously at 6% per annum, how much worth Rs. 1000 will be after 10 years? How long will it take to double Rs. 1000?

Asked by Sakshi | 1 year ago |  138

1 Answer

Solution :-

From the question it is given that,

The interest is compounded continuously at 6% per annum

So, let us assume P be the principal,

\( (\dfrac{dP}{dt}) = \dfrac{Pr}{100}\)

Then,

\( \dfrac{dP}{dt}=(\dfrac{r}{100})dt\)

Integrating on both side we get,

\( ∫ \dfrac{dP}{P} = ∫ \dfrac{r}{100} dt\)

Log P = (\( \dfrac{rt}{100}\)) + c … [equation (i)]

Let us assume Po be the initial principal at t = 0

log (Po) = 0 + C

C = log (Po)

Now, substitute the value of C in equation (i)

log (P) = \( \dfrac{rt}{100}\) + log (Po)

 \(log (\dfrac{P}{P_0}) = \dfrac{rt}{100}\)

Now in case 1:

P= 1000, t = 10 years and r = 6

\( log (\dfrac{P}{1000}) = \dfrac{(6 × 10)}{100}\)

log P – log 1000 = 0.6

log P = loge0.6 + log 1000

Taking log common in both terms,

log P = log (e0.6 + 1000)

log P = log (1.822 + 1000)

log P = log 1822

Then,

P = ₹ 1822

₹ 1000 will be ₹ 1822 after 10 years,

Now in case 2:

Let us assume t1 ne the time to double ₹ 1000,

P = 2000, Po = 1000, r = 6%

\( log (\dfrac{P}{P_0}) = \dfrac{rt}{100}\)

\( log\dfrac{2000}{1000}=\dfrac{6t_1}{100}\)

\( \dfrac{100log_2}{6}=t_1\)

\( \dfrac{(100 × 0.6931)}{6}=t_1\)

t1 = 11.55 years

Therefore, it will take 12 years approximately to double

Answered by Aaryan | 1 year ago

Related Questions

Find the particular solution satisfying the given condition \( (\dfrac{dy}{dx}) = (x – 1)\dfrac{dy}{dx} = 2x^3y\)

Class 12 Maths Differential Equations View Answer

In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Class 12 Maths Differential Equations View Answer

The rate of growth of population is proportional to the number present. If the population doubled in the last 25 years and the present population is 1 lac, when will the city have population 5,00,000 ?

Class 12 Maths Differential Equations View Answer

A population grows at the rate of 5% per year. If x=x(t) denotes the number of individuals in the population after t years, then the rate of change of x is equal to 5% of x. Form the desired differential equation . Find the time period

Class 12 Maths Differential Equations View Answer

The surface area of a spherical balloon,being inflated changes at a rate proportional to time the if initially its radius is 1 units and after 3 seconds it is 2 units, find the radius after t seconds.

Class 12 Maths Differential Equations View Answer