CBSE Class 10 Mathematics Simple and Compound Interest MCQs Set 24

Answer: (d) 6%

Explanation:
1. SI = 3000
2. \( R = 3000 \times 100/(12500 \times 4) \) = 6%.

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) 8%

Explanation:
1. Rate × time stays the same for equal interest: \( 5 \times 8 = R \times 5 \)
2. \( R \) = 8%.

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) 18 years

Explanation:
1. SI = \( P \) in 6 years; for 4 times SI = \( 3P \), so time = 18 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: (a) Rs. 730

Explanation:
1. \( P(1 + 0.5) = 1100 \)
2. \( P = 733.33 \), approximately Rs. 730.

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) 10 years

Explanation:
1. SI = \( P \) in 5 years, so SI = \( 2P \) in 10 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: (d) Rs. 8400

Explanation:
1. \( T = 47/21 \) years
2. \( P = 658 \times 100/(3.5 \times 47/21) \) = Rs. 8400.

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) 3 years 10 months

Explanation:
1. \( T = 78.66 \times 100/(432 \times 4.75) = 3.8333 \) years = 3 years 10 months.

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. 137.60

Explanation:
1. SI = 19.20 in 2 years, so \( R = 19.20 \times 100/(64 \times 2) \) = 15%
2. Amount = \( 86 \times (1 + 0.15 \times 4) \) = Rs. 137.60.

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) Rs. 1750

Explanation:
1. CI on \( 4000 = 4000 \times 0.21 = 840 \)
2. SI = \( 420 = P \times 8 \times 3/100 \), so \( P \) = Rs. 1750.

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) Rs. 8920

Explanation:
1. Factor for 3 years = \( 20070/13380 = 1.5 \)
2. \( P = 13380/1.5 \) = Rs. 8920.

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

Answer: (c) 15 years

Explanation:
1. \( 8 = 2^{3} \), so the time = \( 3 \times 5 = 15 \) years.

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

Answer: (b) 6 years

Explanation:
1. \( 27 = 3^{3} \), so the time = \( 3 \times 2 = 6 \) years.

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

Answer: (c) 24

Explanation:
1. SI = \( 2P \) in 6 years, i.e. \( P/3 \) per year
2. For 9 times SI = \( 8P \), so time = 24 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: (d) 20%

Explanation:
1. Interest for 2 years = 500, i.e. Rs. 250 per year
2. \( P = 2000 - 750 = 1250 \)
3. \( R = 250/1250 \times 100 \) = 20%.

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) 40

Explanation:
1. \( 1029/375 = 2.744 = 1.4^{3} \)
2. The rate is 40%.

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

Answer: (c) Rs. 1600

Explanation:
1. Extra = \( P \times (R/100)^{2} = 160000 \times 0.01 \) = Rs. 1600.

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) Rs. 30000

Explanation:
1. \( P \times 0.11^{2} = 363 \)
2. \( P = 363/0.0121 \) = Rs. 30000.

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) 6

Explanation:
1. \( 9 = 3^{2} \), so the time = \( 2 \times 3 = 6 \) years.

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

Answer: (b) 2000

Explanation:
1. Interest on \( 2000 = 600 = 3000 \times R \times 2/100 \), so \( R \) = 10%
2. SI = \( 5000 \times 10 \times 4/100 \) = Rs. 2000.

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. 1536

Explanation:
1. Value after 2 years = \( 10000 \times 0.92^{2} = 8464 \)
2. Depreciation = 10000 − 8464 = Rs. 1536.

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: (a) Rs. 317772

Explanation:
1. After the loss = \( 300000 \times 0.97 = 291000 \)
2. After the second year = \( 291000 \times 1.04 = 302640 \)
3. After the third year = 302640 × 1.05 = Rs. 317772.

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) 20

Explanation:
1. SI per year = 1200
2. CI − SI = \( 240 = 1200 \times R/100 \), so \( R \) = 20%.

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 \( P(R/100)^{2}(3 + R/100) \).