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

##### 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

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