Q. 55.0( 1 Vote )

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

[Given e^{0.6} = 1.822]

Answer :

Let the principal, rate and time be Rs P, r and t years.

Also, let the initial principal be P_{o}.

⇒

Integrating both sides, we have

⇒ ∫∫dt

⇒ log|P| = t + c ……(1)

Now, at t = 0, P = P_{o}

log| P_{o} | = 0 + c

⇒ c = log| P_{o} |……(2)

Putting the value of c in equation (1) we have,

log|P| = t + log|P_{o}|

⇒ log|P| – log|P_{o}| = t

⇒ (log |P| – log|P_{o}|) = t []

⇒ log ( = t ……(3)

Now, P_{o} = 1000, t = 10years, r = 6

∴ log ( = ×10

⇒ log ( = 0.6

⇒

⇒ P = ×1000

⇒ P = 1.822×1000 (Given: = 1.822)

⇒ P = 1822

Rs 1000 will be Rs 1822 after 10 years at 6% rate.

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