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Exercise 5 · Q2

Q.Gaurav deposited ₹5000 in an account paying 3% interest compounded continuously for 5 years.
i. Find the total amount at the end of 5 years.
ii. How long will it take for the money to double?

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✓ Free question

With A=PertA=Pe^{rt}, P=₹5000P=₹5000, r=0.03r=0.03: (i) A=5000e0.15≈₹5809.17A=5000e^{0.15}\approx ₹5809.17;

(ii) money doubles when 0.03t=ln⁡20.03t=\ln2, i.e. t≈23.10t\approx23.10 years.

Continuous compound interest: A=PertA=Pe^{rt}, where PP = principal, rr = annual rate (decimal), tt = time in years, AA = amount.

Given: P=₹5000P=₹5000, r=3%=0.03r=3\%=0.03, t=5t=5 years.

Part (i): amount after 5 years

  1. Substitute into A=PertA=Pe^{rt}:

A=5000 e0.03×5=5000 e0.15.A=5000\,e^{0.03\times5}=5000\,e^{0.15}.

  1. Evaluate e0.15=1.161834e^{0.15}=1.161834:

A=5000×1.161834=5809.17.A=5000\times1.161834=5809.17.

Part (ii): doubling time

  1. Money doubles: A=2PA=2P, so

2P=Pe0.03t  ⇒  e0.03t=2.2P=Pe^{0.03t}\;\Rightarrow\;e^{0.03t}=2.

  1. Take natural logs:

0.03t=log⁡2=0.6931  ⇒  t=0.69310.03=23.10 years.0.03t=\log 2=0.6931\;\Rightarrow\;t=\frac{0.6931}{0.03}=23.10\text{ years}.

✓Final answer

  1. Amount after 5 years ≈₹5809.17\approx ₹5809.17.
  2. The money doubles in t≈23.10t\approx23.10 years.

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