10 important Compound Interest (C.I.) questions with solutions

Here are 10 important Compound Interest (C.I.) questions with solutions, useful for Coal India exams and other competitive exams:

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Formula for Compound Interest:

For annual compounding,

A = P \left(1 + \frac{R}{100}\right)^T

C.I. = A - P ] Where:

A = Final Amount

P = Principal Amount

R = Rate of Interest per annum

T = Time in years

C.I. = Compound Interest

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1. Finding Compound Interest

Q: Find the Compound Interest on ₹5,000 at 10% per annum for 2 years.

Solution:

A = 5000 \left(1 + \frac{10}{100}\right)^2

= 5000 \times \left(\frac{110}{100} \times \frac{110}{100}\right) ]

= 5000 \times \frac{121}{100} = 6050

C.I. = A - P = 6050 - 5000 = 1050 ] So, C.I. = ₹1,050.

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2. Finding Amount after 3 Years

Q: A sum of ₹10,000 is invested at 8% per annum for 3 years. Find the amount.

Solution:

A = 10000 \left(1 + \frac{8}{100}\right)^3

= 10000 \times \left(\frac{108}{100} \times \frac{108}{100} \times \frac{108}{100}\right) ]

= 10000 \times \frac{125.97}{100} = 12597

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3. Finding Principal

Q: The Compound Interest for 2 years at 5% per annum is ₹205 on a principal. Find the principal amount.

Solution:

A = P \left(1 + \frac{5}{100}\right)^2

P = \frac{A}{\left(1 + \frac{5}{100}\right)^2} = \frac{P + 205}{1.1025} ] Using approximation, P = ₹1,800.

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4. Rate of Interest Calculation

Q: ₹6,400 becomes ₹7,744 in 2 years under Compound Interest. Find the rate of interest.

Solution:

A = P \left(1 + \frac{R}{100}\right)^T

7744 = 6400 \left(1 + \frac{R}{100}\right)^2 ]

\frac{7744}{6400} = \left(1 + \frac{R}{100}\right)^2

1.21 = \left(1 + \frac{R}{100}\right)^2 ]

1.1 = 1 + \frac{R}{100}

R = 10% ] So, Rate of Interest = 10%.

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5. Time Calculation

Q: At 6% per annum, ₹5,000 amounts to ₹5,950. Find the time period.

Solution:

A = P \left(1 + \frac{6}{100}\right)^T

5950 = 5000 \times \left(\frac{106}{100}\right)^T ] Using approximation, T = 3 years.

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6. C.I. for Half-Yearly Compounding

Q: Find the C.I. on ₹4,000 for 1 year at 10% per annum, compounded half-yearly.

Solution:

For half-yearly compounding,

R = \frac{10}{2} = 5\%, \quad T = 2 \text{ half-years}

A = 4000 \left(1 + \frac{5}{100}\right)^2 ]

= 4000 \times \left(\frac{105}{100} \times \frac{105}{100}\right)

= 4000 \times \frac{110.25}{100} = 4410 ]

C.I. = 4410 - 4000 = 410

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7. C.I. for Quarterly Compounding

Q: Find the amount on ₹8,000 at 12% per annum, compounded quarterly for 1 year.

Solution:

For quarterly compounding,

R = \frac{12}{4} = 3\%, \quad T = 4 \text{ quarters}

A = 8000 \left(1 + \frac{3}{100}\right)^4 ]

= 8000 \times \left(\frac{103}{100} \times \frac{103}{100} \times \frac{103}{100} \times \frac{103}{100}\right)

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8. Investment Doubling Time

Q: In how many years will ₹10,000 double at 10% per annum?

Solution:

A = 2P \Rightarrow 2P = P \left(1 + \frac{10}{100}\right)^T

2 = \left(\frac{110}{100}\right)^T ] Using logarithm, T ≈ 7.27 years.

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9. Population Growth Model

Q: The population of a town is 50,000 and increases by 4% annually. Find the population after 2 years.

Solution:

A = 50000 \left(1 + \frac{4}{100}\right)^2

= 50000 \times \frac{108.16}{100} = 54,080 ] So, Population after 2 years = 54,080.

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10. Loan Repayment Calculation

Q: A person takes a loan of ₹25,000 at 7% per annum for 3 years, compounded annually. Find the total amount to be repaid.

Solution:

A = 25000 \left(1 + \frac{7}{100}\right)^3

= 25000 \times \left(\frac{107}{100} \times \frac{107}{100} \times \frac{107}{100}\right) ] Using calculation, A ≈ ₹30,612.75.

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These Compound Interest questions cover basic, half-yearly, quarterly, and real-life applications, essential for Coal India exams. Let me know if you need more!