Q1): Find the simple interest on ₹12,000 at 8% per annum for 2 years 6 months.
A) ₹2,200
B) ₹2,300
C) ₹2,400
D) ₹2,500
Answer: C) ₹2,400
Explanation:
- Step 1: Convert time: 2 years 6 months = 2.5 years
- Step 2: Formula: SI = (P × R × T) / 100
- Step 3: Symbols: P = 12000, R = 8, T = 2.5
- Step 4: SI = (12000 × 8 × 2.5) / 100 = (12000 × 20) / 100
- Final: SI = 2400, so option C is correct
Q2): The simple interest on ₹11,000 at some rate for 3 years is ₹1,650. The rate is:
A) 4%
B) 5%
C) 6%
D) 7%
Answer: B) 5%
Explanation:
- Step 1: Formula: SI = (P × R × T) / 100
- Step 2: Rearrange: R = (SI × 100) / (P × T)
- Step 3: Substitute: R = (1650 × 100) / (11000 × 3)
- Step 4: R = 165000 / 33000 = 5
- Final: Rate = 5% p.a., so option B is correct
Q3): A sum amounts to ₹15,000 in 3 years at 10% per annum simple interest. The principal is (approx.):
A) ₹11,000
B) ₹11,250
C) ₹11,538.46
D) ₹12,000
Answer: C) ₹11,538.46
Explanation:
- Step 1: For SI, Amount: A = P + SI
- Step 2: Formula: SI = (P × R × T) / 100
- Step 3: SI = (P × 10 × 3) / 100 = 0.30P
- Step 4: A = P + 0.30P = 1.30P ⇒ P = 15000 / 1.30
- Final: P ≈ ₹11,538.46, so option C is correct
Q4): Find the compound interest on ₹8,000 at 12% per annum for 2 years (compounded annually).
A) ₹1,920
B) ₹2,000
C) ₹2,035.20
D) ₹2,120
Answer: C) ₹2,035.20
Explanation:
- Step 1: Formula: A = P(1 + R/100)^T
- Step 2: Symbols: P = 8000, R = 12, T = 2
- Step 3: A = 8000(1.12)^2 = 8000 × 1.2544 = 10035.20
- Step 4: CI = A − P = 10035.20 − 8000
- Final: CI = ₹2,035.20, so option C is correct
Q5): Find the compound interest on ₹2,500 at 8% per annum for 1 year, compounded half-yearly.
A) ₹200
B) ₹204
C) ₹208
D) ₹210
Answer: B) ₹204
Explanation:
- Step 1: Half-yearly ⇒ rate per half-year = 8%/2 = 4%, periods = 2
- Step 2: Formula: A = P(1 + r/100)^n
- Step 3: A = 2500(1.04)^2 = 2500 × 1.0816 = 2704
- Step 4: CI = A − P = 2704 − 2500
- Final: CI = ₹204, so option B is correct
Q6): Find the compound interest on ₹5,000 at 8% per annum for 1 year, compounded quarterly (to 2 decimals).
A) ₹400.00
B) ₹408.00
C) ₹412.16
D) ₹416.00
Answer: C) ₹412.16
Explanation:
- Step 1: Quarterly ⇒ rate per quarter = 8%/4 = 2%, periods = 4
- Step 2: Formula: A = P(1 + r/100)^n
- Step 3: A = 5000(1.02)^4 = 5000 × 1.08243216 = 5412.1608
- Step 4: CI = A − P = 5412.1608 − 5000 = 412.1608
- Final: CI ≈ ₹412.16, so option C is correct
Q7): In 2 years at 6% per annum, compound interest exceeds simple interest by ₹72. The principal is:
A) ₹15,000
B) ₹18,000
C) ₹20,000
D) ₹24,000
Answer: C) ₹20,000
Explanation:
- Step 1: For 2 years, CI − SI = P × (R/100)^2
- Step 2: Symbols: P = principal, R = 6
- Step 3: 72 = P × (6/100)^2 = P × (0.06)^2 = P × 0.0036
- Step 4: P = 72 / 0.0036 = 20000
- Final: Principal = ₹20,000, so option C is correct
Q8): Find (CI − SI) on ₹5,000 at 10% per annum for 3 years.
A) ₹145
B) ₹150
C) ₹155
D) ₹160
Answer: C) ₹155
Explanation:
- Step 1: SI: SI = (P × R × T) / 100 = (5000 × 10 × 3)/100 = 1500
- Step 2: CI amount: A = 5000(1.10)^3 = 5000 × 1.331 = 6655
- Step 3: CI = A − P = 6655 − 5000 = 1655
- Step 4: Difference = CI − SI = 1655 − 1500
- Final: CI − SI = ₹155, so option C is correct
Q9): ₹20,000 is increased by 10% in the first year and then decreased by 10% in the second year. Net result is:
A) Gain ₹200
B) Loss ₹200
C) No gain, no loss
D) Loss ₹400
Answer: B) Loss ₹200
Explanation:
- Step 1: Successive change factor = (1 + 10/100)(1 − 10/100)
- Step 2: Factor = 1.10 × 0.90 = 0.99
- Step 3: Final amount = 20000 × 0.99 = 19800
- Step 4: Change = 19800 − 20000 = −200
- Final: Loss = ₹200, so option B is correct
Q10): For 2 years at 10% per annum, by how much does CI exceed SI on ₹12,000?
A) ₹100
B) ₹110
C) ₹120
D) ₹130
Answer: C) ₹120
Explanation:
- Step 1: For 2 years, CI − SI = P × (R/100)^2
- Step 2: Symbols: P = 12000, R = 10
- Step 3: CI − SI = 12000 × (10/100)^2 = 12000 × (0.1)^2
- Step 4: 12000 × 0.01 = 120
- Final: Difference = ₹120, so option C is correct
Q11): A nominal rate is 10% per annum compounded half-yearly. The effective annual rate is:
A) 10.00%
B) 10.25%
C) 10.50%
D) 10.75%
Answer: B) 10.25%
Explanation:
- Step 1: Half-yearly ⇒ rate per half-year = 10%/2 = 5%
- Step 2: Effective annual factor = (1 + 0.05)^2
- Step 3: (1.05)^2 = 1.1025
- Step 4: Effective rate = 1.1025 − 1 = 0.1025 = 10.25%
- Final: Effective rate = 10.25%, so option B is correct
Q12): Find the compound interest on ₹10,000 at 10% per annum for 2 years, compounded half-yearly (approx.).
A) ₹2,000
B) ₹2,100
C) ₹2,155.06
D) ₹2,200
Answer: C) ₹2,155.06
Explanation:
- Step 1: Half-yearly ⇒ rate per half-year = 10%/2 = 5%, periods = 4
- Step 2: Formula: A = P(1 + r/100)^n
- Step 3: A = 10000(1.05)^4 = 10000 × 1.21550625 = 12155.0625
- Step 4: CI = A − P = 12155.0625 − 10000
- Final: CI ≈ ₹2,155.06, so option C is correct
Q13): What principal will amount to ₹8,820 in 2 years at 5% per annum compound interest (annual)?
A) ₹7,800
B) ₹8,000
C) ₹8,200
D) ₹8,400
Answer: B) ₹8,000
Explanation:
- Step 1: Formula: A = P(1 + R/100)^T
- Step 2: Here, (1 + 5/100)^2 = (1.05)^2 = 1.1025
- Step 3: P = A / 1.1025 = 8820 / 1.1025
- Step 4: 1.1025 × 8000 = 8820
- Final: P = ₹8,000, so option B is correct
Q14): ₹10,000 is invested at CI with rates 10% (1st year), 12% (2nd year), and 8% (3rd year). The amount after 3 years is:
A) ₹13,200
B) ₹13,305.60
C) ₹13,350.60
D) ₹13,500
Answer: B) ₹13,305.60
Explanation:
- Step 1: With different yearly rates: A = P(1+r1)(1+r2)(1+r3)
- Step 2: A = 10000 × 1.10 × 1.12 × 1.08
- Step 3: 1.10 × 1.12 = 1.232; then 1.232 × 1.08 = 1.33056
- Step 4: A = 10000 × 1.33056 = 13305.60
- Final: Amount = ₹13,305.60, so option B is correct
Q15): The compound interest on a sum at 10% per annum for 2 years is ₹2,100. The principal is:
A) ₹8,000
B) ₹9,000
C) ₹10,000
D) ₹12,000
Answer: C) ₹10,000
Explanation:
- Step 1: For 2 years: CI = P[(1 + R/100)^2 − 1]
- Step 2: Here, (1.10)^2 = 1.21 ⇒ bracket = 1.21 − 1 = 0.21
- Step 3: So, 2100 = P × 0.21
- Step 4: P = 2100 / 0.21 = 10000
- Final: Principal = ₹10,000, so option C is correct
Q16): A sum becomes ₹11,025 in 2 years at 5% compound interest (annual). The principal is:
A) ₹9,500
B) ₹10,000
C) ₹10,500
D) ₹11,000
Answer: B) ₹10,000
Explanation:
- Step 1: Formula: A = P(1.05)^2
- Step 2: (1.05)^2 = 1.1025
- Step 3: P = A / 1.1025 = 11025 / 1.1025
- Step 4: 1.1025 × 10000 = 11025
- Final: Principal = ₹10,000, so option B is correct
Q17): ₹10,000 is invested for 1 year at 10% p.a. Find the extra amount earned by half-yearly compounding over annual compounding.
A) ₹20
B) ₹25
C) ₹30
D) ₹40
Answer: B) ₹25
Explanation:
- Step 1: Annual compounding (1 year): A₁ = 10000 × 1.10 = 11000
- Step 2: Half-yearly ⇒ 5% each half-year: A₂ = 10000(1.05)^2 = 10000 × 1.1025 = 11025
- Step 3: Extra amount = A₂ − A₁ = 11025 − 11000
- Step 4: Extra amount = 25
- Final: Extra earned = ₹25, so option B is correct
Q18): Find the compound interest on ₹8,000 at 10% per annum for 1.5 years (annual compounding).
A) ₹1,200
B) ₹1,240
C) ₹1,280
D) ₹1,320
Answer: B) ₹1,240
Explanation:
- Step 1: For fractional years: compound for full years, then SI on the amount for remaining time
- Step 2: After 1 year: Amount = 8000 × 1.10 = 8800
- Step 3: Remaining time = 0.5 year ⇒ Interest = (8800 × 10 × 0.5)/100 = 440
- Step 4: Final amount = 8800 + 440 = 9240
- Final: CI = 9240 − 8000 = ₹1,240, so option B is correct
Q19): Find the amount on ₹10,000 at 12% per annum compound interest for 2 years 3 months (annual compounding, approx.).
A) ₹12,544.00
B) ₹12,880.32
C) ₹12,920.32
D) ₹13,054.40
Answer: C) ₹12,920.32
Explanation:
- Step 1: Compound for 2 full years: A₂ = 10000(1.12)^2
- Step 2: (1.12)^2 = 1.2544 ⇒ A₂ = 10000 × 1.2544 = 12544
- Step 3: Remaining 3 months = 0.25 year ⇒ SI on 12544 = (12544 × 12 × 0.25)/100
- Step 4: Extra interest = 376.32 ⇒ Final amount = 12544 + 376.32 = 12920.32
- Final: Amount ≈ ₹12,920.32, so option C is correct
Q20): A man borrows ₹15,000 at 12% per annum simple interest and repays ₹18,900. The time period is:
A) 2 years
B) 2 years 2 months
C) 2 years 6 months
D) 3 years
Answer: B) 2 years 2 months
Explanation:
- Step 1: Interest paid = Amount − Principal = 18900 − 15000 = 3900
- Step 2: Formula: SI = (P × R × T) / 100
- Step 3: T = (SI × 100) / (P × R) = (3900 × 100) / (15000 × 12)
- Step 4: T = 390000 / 180000 = 2.1666… years = 2 years + (0.1666… × 12) months
- Final: 0.1666… × 12 ≈ 2 months ⇒ time = 2 years 2 months, so option B is correct