Quantitative Aptitude
Simple and Compound Interest
3,394 Questions
Simple and Compound Interest Questions
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Rs./रू. 4200
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Rs./रू. 4480
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Rs./रू. 4840
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Rs./रू. 4680
C
Correct answer
Explanation
For a loan repaid in equal annual installments under compound interest, use the formula: P = x[1 - (1 + r)⁻ⁿ]/r where x is the installment. Here: 8400 = x[1 - (1.1)⁻²]/0.1 = x[1 - 1/1.21]/0.1 = x[0.21/1.21]/0.1 = x(21/121). Solving: x = 8400 × 121/21 = 400 × 121 = 4840. Verification: After year 1, balance = 8400(1.1) - 4840 = 9240 - 4840 = 4400; after year 2: 4400(1.1) - 4840 = 4840 - 4840 = 0. Key distractor: 4680 (uses simple interest approximation).
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Rs. 5000
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Rs. 6000
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Rs. 7500
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Rs. 6250
D
Correct answer
Explanation
For 3 years at r%: CI - SI = P[(1 + r)³ - 1 - 3r] = P[(1.04)³ - 1 - 0.12] = P[1.124864 - 1.12] = P(0.004864) = 30.4. Therefore P = 30.4/0.004864 = 6250. Verification: CI = 6250(1.04)³ - 6250 = 780.4; SI = 6250 × 0.04 × 3 = 750; difference = 30.4 ✓. Key distractor: 6000 (uses simple approximation).
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Rs./रू.180
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Rs./रू.1260
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Rs./रू.960
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Rs./रू.1125
D
Correct answer
Explanation
Let principal be P and rate be R%. Simple Interest for 2 years = P×R×2/100 = 450, so PR = 22500. Compound Interest for 2 years = P[(1+R/100)² - 1] = 495. Expanding: P[2R/100 + R²/10000] = 495. Using PR = 22500: 2×22500/100 + PR²/10000 = 495, so 450 + PR²/10000 = 495, PR²/10000 = 45. Since PR = 22500: R = 45×10000/22500 = 20%. Then P = 22500/20 = 1125.
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Rs. 6000
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Rs. 6200
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Rs. 6300
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Rs. 6400
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None of these
D
Correct answer
Explanation
The difference between compound and simple interest for 3 years at 15% is P[(1.15)³ - 1 - 0.45]. Solving: P[1.520875 - 1 - 0.45] = P[0.070875] = 453.6. Therefore P = 453.6/0.070875 = 6400. For verification: SI = 6400 × 0.15 × 3 = 2880. CI = 6400 × 0.520875 = 3333.6. Difference = 453.6.
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5.4%
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11.23%
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9.45%
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12.03%
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7.45%
A
Correct answer
Explanation
Principal = Rs. 18000, Rate = 20% p.a., Time = 2 years. Annual compounding: Amount = 18000 × (1.2)² = 18000 × 1.44 = 25920. Interest = 25920 - 18000 = 7920. Half-yearly compounding: Rate per half-year = 10%, Periods = 4. Amount = 18000 × (1.1)⁴ = 18000 × 1.4641 = 26353.8. Interest = 26353.8 - 18000 = 8353.8. Difference = 8353.8 - 7920 = 433.8. Percentage more = (433.8/7920) × 100 ≈ 5.4%. Option A is correct.
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25 years/वर्ष
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15 years/वर्ष
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40 years/वर्ष
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50 years/वर्ष
D
Correct answer
Explanation
Let t years be when debts are equal. Paritosh's debt after t years: 8000 + 8000 × 0.04 × t = 8000 + 320t. Arunesh's debt after t years: 6000 + 6000 × 0.06 × t = 6000 + 360t. Equating: 8000 + 320t = 6000 + 360t. Solving: 2000 = 40t, giving t = 50 years. After 50 years, Paritosh owes 8000 + 320(50) = 8000 + 16000 = 24000, and Arunesh owes 6000 + 360(50) = 6000 + 18000 = 24000, making option D correct.
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Rs./रू. 1000
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Rs./रू. 620
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Rs./रू. 625
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Rs./रू. 640
D
Correct answer
Explanation
SI for 3 years = 240, so SI for 1 year = 80. CI for 2 years = 240 - 70 = 170. Let principal = P, rate = r%. SI for 2 years = 2Pr/100 = 160 (since SI for 3 years = 240). So Pr = 8000. CI for 2 years = P(1 + r/100)² - P = 170. Using approximation: CI ≈ SI for 2 years + interest on first year's interest. First year interest = 80, second year interest on 80 = 80r/100. CI = 160 + 80r/100 = 170, so 80r/100 = 10, r = 12.5%. Then Pr = 8000 means P × 12.5 = 8000, P = 640. The claimed answer D is correct.
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Rs. 18000
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Rs. 15000
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Rs. 20000
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Rs. 12000
D
Correct answer
Explanation
Compound Interest formula: CI = P(1 + r/100)³ - P = 6250.5. At 15% for 3 years: CI = P(1.15)³ - P = P(1.520875) = 0.520875P. If P = 12000: CI = 12000 × 0.520875 = 6250.5 ✓. Option A (18000) would give CI = 9375.75. Option B (15000) would give CI = 7813.13. Option C (20000) would give CI = 10417.5. Note: Question states CI as 6250.5 in one place and 6252.56 in another - assuming 6250.5 is correct based on calculation.
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Rs. 5572
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Rs. 5472
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Rs. 6652
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Rs. 4572
B
Correct answer
Explanation
Let each installment = x. After 1 year: amount = 8360 × 1.2 = 10032. After paying x: remaining = 10032 - x. After 2nd year: (10032 - x) × 1.2 = x. Solving: 12038.4 - 1.2x = x, 2.2x = 12038.4, x = 5472. Use the compound interest formula recursively, equating final amount to the second installment.
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Rs. /रु.7500
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Rs. /रु.9000
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Rs. /रु.10000
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Rs. /रु.12000
D
Correct answer
Explanation
In compound interest, if amount doubles every 3 years (3000 to 6000), the pattern continues: after 9 years = 12000. This follows the rule that in compound interest, when amount doubles over equal intervals, each interval doubles the amount. The 3-6 year doubling means 6-9 years also doubles.
D
Correct answer
Explanation
Simple Interest = Principal × Rate × Time / 100. When principal doubles in 5 years, SI equals the principal P. So P = P × R × 5 / 100, which gives 100 = 5R, so R = 20% per annum. Option D is correct.
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Rs. 3246
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Rs.3136
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Rs. 4210
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Rs. 3926
B
Correct answer
Explanation
Simple Interest = 24500 × 20 × 3 / 100 = Rs. 14700. Compound Interest after 3 years = 24500 × (1.2)^3 - 24500 = 24500 × 1.728 - 24500 = Rs. 17836. The difference is 17836 - 14700 = Rs. 3136. Option B is correct.
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Rs./रू.6063.75
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Rs./रू.6381.25
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Rs./रू.6588.75
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Rs./रू.6863.75
A
Correct answer
Explanation
Year 1: Loan grows to 15000 × 1.05 = 15750. After payment: 15750 - 5250 = 10500. Year 2: Grows to 10500 × 1.05 = 11025. After payment: 11025 - 5250 = 5775. Year 3: Grows to 5775 × 1.05 = 6063.75, which is the final payment needed.
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Rs./रू.1990
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Rs./रू.2500
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Rs./रू. 2000
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Rs./रू.2625
C
Correct answer
Explanation
Let principal be P and rate be R%. Given: 4-year simple interest = P/4, so 4PR/100 = P/4. Solving: R = 100/16 = 6.25%. For 2 years at same rate: Amount = P + (2 × 6.25/100)P = 1.125P. Setting 1.125P = 2250 gives P = 2000. Check: 2-year SI on Rs.2000 at 6.25% is Rs.250, so amount = Rs.2250.
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Rs. /रू.10800
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Rs. /रू.15000
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Rs. /रू.18000
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Rs. /रू.12000
D
Correct answer
Explanation
For 2 years at R = 30%: CI = P(1.3)² - P = 0.69P, SI = 2PR/100 = 0.6P. Difference = CI - SI = 0.69P - 0.6P = 0.09P. Given 0.09P = 1080, so P = 12000. Standard 2-year formula: CI - SI = PR²/10000. Here: P × 900/10000 = 1080, giving P = 12000. This difference is the 'interest on interest' for year 2.