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Compound Interest Quiz - ICSE Class 9 Mathematics

Free practice quiz: Compound Interest (ICSE Class 9 Mathematics). 19 multiple-choice questions with answers and explanations. Play online, no sign-up needed.

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Sample questions

  1. Find the principal which yields ₹205 compound interest in 2 years at 5% per annum.

    1. ₹1000
    2. ₹2000
    3. ₹3000
    4. ₹4000
    Show answer

    Answer: B. ₹2000

    Let the principal be P. Rate (r) = 5%, Time (n) = 2 years. Compound Interest (CI) = P[(1 + r/100)^n - 1] 205 = P[(1 + 5/100)^2 - 1] 205 = P[(1.05)^2 - 1] 205 = P[1.1025 - 1] 205 = P[0.1025] P = 205 / 0.1025 P = 2000 So, the principal is ₹2000.

  2. If the compound interest on a sum for 2 years at 10% per annum is ₹210, find the principal.

    1. ₹1000
    2. ₹1500
    3. ₹2000
    4. ₹2100
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    Answer: A. ₹1000

    Let the principal be P. Rate (r) = 10%, Time (n) = 2 years. Compound Interest (CI) = P[(1 + r/100)^n - 1] 210 = P[(1 + 10/100)^2 - 1] 210 = P[(1.1)^2 - 1] 210 = P[1.21 - 1] 210 = P[0.21] P = 210 / 0.21 P = 1000 So, the principal is ₹1000.

  3. A sum of money doubles itself in 5 years at compound interest. In how many years will it become 4 times itself?

    1. 10 years
    2. 15 years
    3. 20 years
    4. 25 years
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    Answer: A. 10 years

    Let the principal be P. If the sum doubles in 5 years, then A = 2P after 5 years. 2P = P(1 + r/100)^5 2 = (1 + r/100)^5 We want to find the time (n) when the sum becomes 4 times itself (4P). 4P = P(1 + r/100)^n 4 = (1 + r/100)^n Since 4 = 2^2, we can substitute 2 = (1 + r/100)^5: [(1 + r/100)^5]^2 = (1 + r/100)^n (1 + r/100)^{5 imes 2} = (1 + r/100)^n (1 + r/100)^{10} = (1 + r/100)^n Therefore, n = 10 years. If a sum becomes k times itself in t years, it becomes k^m times itself in m*t years.

  4. A sum of money triples itself in 8 years at compound interest. How many years will it take for the sum to become 9 times itself?

    1. 16 years
    2. 20 years
    3. 24 years
    4. 32 years
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    Answer: A. 16 years

    Let the principal be P. If the sum triples in 8 years, then A = 3P after 8 years. 3P = P(1 + r/100)^8 3 = (1 + r/100)^8 We want to find the time (n) when the sum becomes 9 times itself (9P). 9P = P(1 + r/100)^n 9 = (1 + r/100)^n Since 9 = 3^2, we can substitute 3 = (1 + r/100)^8: [(1 + r/100)^8]^2 = (1 + r/100)^n (1 + r/100)^{8 imes 2} = (1 + r/100)^n (1 + r/100)^{16} = (1 + r/100)^n Therefore, n = 16 years. If a sum becomes k times itself in t years, it becomes k^m times itself in m*t years.

  5. Find the principal that amounts to ₹13310 in 3 years at 10% per annum compound interest.

    1. ₹10000
    2. ₹11000
    3. ₹12000
    4. ₹13000
    Show answer

    Answer: A. ₹10000

    Amount (A) = ₹13310 Rate (r) = 10% per annum Time (n) = 3 years Amount (A) = P(1 + r/100)^n 13310 = P(1 + 10/100)^3 13310 = P(1.1)^3 13310 = P * 1.331 P = 13310 / 1.331 P = 10000 So, the principal is ₹10000.

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