CBSE Class 10 Mathematics Simple and Compound Interest MCQs Set 30

Answer: (c) Rs. 4998

Explanation:
1. \( T = 12.25 \) years
2. SI = \( 4800 \times 8.5 \times 12.25/100 \) = Rs. 4998.

Teacher's Note:
1. Simple interest = \( P \times R \times T/100 \) and amount = \( P \) + SI.
2. Keep time in years and the rate per annum.

Answer: (d) Rs. 8925

Explanation:
1. \( P = 4016.25 \times 100/(9 \times 5) \) = Rs. 8925.

Teacher's Note:
1. Simple interest = \( P \times R \times T/100 \) and amount = \( P \) + SI.
2. Keep time in years and the rate per annum.

Answer: (b) Rs. (100/x)

Explanation:
1. \( x = P \times x \times x/100 \)
2. \( P = 100/x \).

Teacher's Note:
1. Simple interest = \( P \times R \times T/100 \) and amount = \( P \) + SI.
2. Keep time in years and the rate per annum.

Answer: (c) 9

Explanation:
1. Interest on \( 800 = 800 \times 4.5 \times 3/100 \) = Rs. 108
2. \( 150 \times 8 \times t/100 = 108 \), so \( t = 9 \) years.

Teacher's Note:
1. Simple interest = \( P \times R \times T/100 \) and amount = \( P \) + SI.
2. Keep time in years and the rate per annum.

Answer: (b) Rs. 22.50

Explanation:
1. Deposit = 202.50/0.045 = Rs. 4500
2. Extra interest = \( 4500 \times 0.5/100 \) = Rs. 22.50.

Teacher's Note:
1. Simple interest = \( P \times R \times T/100 \) and amount = \( P \) + SI.
2. Keep time in years and the rate per annum.

Answer: (a) 1 1/4 years

Explanation:
1. \( 0.125 = 10 \times t/100 \), so \( t = 1.25 \) years = \( 1 1/4 \) years.

Teacher's Note:
1. Simple interest = \( P \times R \times T/100 \) and amount = \( P \) + SI.
2. Keep time in years and the rate per annum.

Answer: (b) Rs. 3

Explanation:
1. CI = \( 1200 \times (1.05^{2} - 1) = 123 \)
2. SI = 120, so the difference = Rs. 3.

Teacher's Note:
1. The difference between CI and SI for 2 years is \( P(R/100)^{2} \).
2. The difference for 3 years is \( P(R/100)^{2}(3 + R/100) \).

Answer: (a) 625

Explanation:
1. \( P \times (4/100)^{2} = 1 \), so \( P = 1/0.0016 \) = Rs. 625.

Teacher's Note:
1. The difference between CI and SI for 2 years is \( P(R/100)^{2} \).
2. The difference for 3 years is \( P(R/100)^{2}(3 + R/100) \).

Answer: (d) Rs. 1,92,000

Explanation:
1. Doubles every 5 years, so after 20 years it becomes \( 2^{4} = 16 \) times
2. Amount = 12000 × 16 = Rs. 192000.

Teacher's Note:
1. Interest rate questions are solved from the ratio of amounts.
2. Spot perfect squares or cubes in the ratio.

Answer: (d) Rs. 14,320

Explanation:
1. Year 1: \( 12500 \times 1.2 - 2000 = 13000 \)
2. Year 2: \( 13000 \times 1.2 - 2000 = 13600 \)
3. Year 3: 13600 × 1.2 − 2000 = Rs. 14320.

Teacher's Note:
1. Read which sum is the principal and which is the amount.
2. Break the problem into simple one-step calculations.

Answer: (a) Rs. 5200

Explanation:
1. \( P \times 0.15^{2} = 117 \), so \( P = 117/0.0225 \) = Rs. 5200.

Teacher's Note:
1. The difference between CI and SI for 2 years is \( P(R/100)^{2} \).
2. The difference for 3 years is \( P(R/100)^{2}(3 + R/100) \).

Answer: (c) 20/3%

Explanation:
1. Let \( R = T \): \( P \times R^{2}/100 = 4P/9 \)
2. \( R^{2} = 400/9 \), so \( R \) = 20/3%.

Teacher's Note:
1. Simple interest = \( P \times R \times T/100 \) and amount = \( P \) + SI.
2. Keep time in years and the rate per annum.

Answer: (a) Rs. 4000

Explanation:
1. Interest for 3 years = 1800, i.e. Rs. 600 per year
2. \( P = 6400 - 4 \times 600 \) = Rs. 4000.

Teacher's Note:
1. Simple interest = \( P \times R \times T/100 \) and amount = \( P \) + SI.
2. Keep time in years and the rate per annum.

Answer: (d) Rs. 4058

Explanation:
1. Amount = \( 15625 \times 1.08^{3} = 19683 \)
2. CI = 19683 − 15625 = Rs. 4058.

Teacher's Note:
1. Compound interest uses amount = \( P(1 + R/100)^n \) and CI = amount \( - P \).
2. Compounding at a different frequency changes both rate and number of periods.

Answer: (c) 625

Explanation:
1. SI = \( 0.6P = P - 250 \)
2. \( 0.4P = 250 \), so \( P \) = Rs. 625.

Teacher's Note:
1. Simple interest = \( P \times R \times T/100 \) and amount = \( P \) + SI.
2. Keep time in years and the rate per annum.

Answer: (c) 5901

Explanation:
1. Population = \( 5620 \times 1.05 = 5901 \).

Teacher's Note:
1. Compound interest uses amount = \( P(1 + R/100)^n \) and CI = amount \( - P \).
2. Compounding at a different frequency changes both rate and number of periods.

Answer: (b) 5%

Explanation:
1. \( 22050/20000 = 1.1025 = 1.05^{2} \)
2. The rate is 5%.

Teacher's Note:
1. Interest rate questions are solved from the ratio of amounts.
2. Spot perfect squares or cubes in the ratio.

Answer: (a) 1200

Explanation:
1. \( P = 300 \times 100/(5 \times 5) \) = Rs. 1200.

Teacher's Note:
1. Simple interest = \( P \times R \times T/100 \) and amount = \( P \) + SI.
2. Keep time in years and the rate per annum.

Answer: (a) Rs. 15300

Explanation:
1. Rate = 1/6, so the factor = \( (7/6)^{2} = 49/36 \)
2. \( P = 20825 \times 36/49 \) = Rs. 15300.

Teacher's Note:
1. Compound interest uses amount = \( P(1 + R/100)^n \) and CI = amount \( - P \).
2. Compounding at a different frequency changes both rate and number of periods.

Answer: (b) Half year

Explanation:
1. Rate per quarter = 12.5% = 1/8
2. \( 4050/3200 = 1.265625 = (9/8)^{2} \), so 2 quarters = half a year.

Teacher's Note:
1. Compound interest uses amount = \( P(1 + R/100)^n \) and CI = amount \( - P \).
2. Compounding at a different frequency changes both rate and number of periods.

Answer: (c) Rs. 1160.25

Explanation:
1. Rate 10% per half year for 4 half years
2. Amount = \( 2500 \times 1.1^{4} = 3660.25 \)
3. CI = Rs. 1160.25.

Teacher's Note:
1. Compound interest uses amount = \( P(1 + R/100)^n \) and CI = amount \( - P \).
2. Compounding at a different frequency changes both rate and number of periods.

Answer: (c) 3124 years

Explanation:
1. 4% = 20% of 20%, so two periods of 1562 years
2. Age = \( 2 \times 1562 = 3124 \) years.

Teacher's Note:
1. Interest rate questions are solved from the ratio of amounts.
2. Spot perfect squares or cubes in the ratio.