Question 44 of 96intermediate🔧 ApplyLong Answer4 marks

In a bank, principal increases continuously at the rate of 5%5\% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0.5=1.648)(e^{0.5} = 1.648).

Correct Answer

Rs 1648

Exercise: EXERCISE 9.3 | Q: 21 | (Page 13)
For More Understanding

Explanation

This question is from Exercise 9.3, Question 21 of NCERT Class 12 Mathematics. It applies the concept of forming and solving differential equations for continuous growth problems.

The context from Example 9 provides the exact methodology: the rate of change of principal is proportional to the principal itself, leading to an exponential growth formula P=P0×ert/100P = P_0 \times e^{rt/100}.

The student must correctly set up the differential equation, solve it using variable separation, apply the initial condition to find the constant, and then substitute t=10t = 10 years to find the final amount.

Solution Steps

  1. Step 1: Let PP be the principal at time tt. Since principal increases at 5%5\% per year, the differential equation is:

    dpdt=5100×P=P20\frac{dp}{dt} = \frac{5}{100} \times P = \frac{P}{20}

  2. Step 2: Separating the variables:

    dpP=dt20\frac{dp}{P} = \frac{dt}{20}

  3. Step 3: Integrating both sides:

    log⁡P=t20+C1\log P = \frac{t}{20} + C_1

    or

    P=C×et/20P = C \times e^{t/20}

    (where C=eC1C = e^{C_1})

  4. Step 4: Using initial condition: When t=0t = 0, P=1000P = 1000. Substituting:

    1000=C×e0=C1000 = C \times e^0 = C

    so C=1000C = 1000

  5. Step 5: Therefore,

    P=1000×et/20P = 1000 \times e^{t/20}

  6. Step 6: For t=10t = 10 years:

    P=1000×e10/20=1000×e0.5P = 1000 \times e^{10/20} = 1000 \times e^{0.5}

  7. Step 7: Given e0.5=1.648e^{0.5} = 1.648, so

    P=1000×1.648=Rs 1648P = 1000 \times 1.648 = \textbf{Rs 1648}
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