Question

In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 doubles itself in 10 years (log­_e 2 = 0.6931).

Answer 1

Let P be the principal at any time t.

Now, `(dP)/dt = r/100 *P`                  ....(1)

 Integrating (1) both sides, we get

`int (dP)/P = int r/100 dt`

⇒ `log P = r/100 t + C_1`

⇒ `P = e^((rt)/100) . eC_1`

⇒ `P = Ce ^((rt)/100)` (Where e^C₁ = C)              .....(2)

Now, P = 100, when t = 0

Substituting the values of P and t in (2), we get

100 = Ce⁰

⇒ C = 100

∴ Equation (2) becomes, `P = 100 e^((rt)/100)`              .....(3)

When P = 200, t = 10

Substituting the values of P and t in (3), we get

⇒ `200 = e^((10r)/100) xx 100`

⇒ `2 = e^(r/10)`

⇒ `log 2 r/10`

⇒ r = 10 log 2

⇒ r = 10 × 0.6931

⇒ r = 6.931

Hence, r = 6.93% per annum