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Maths Quiz for Class 8 Compound Interest |
In this post, we are providing 20 online maths quiz
questions for class 8 compound interest. Online maths quiz will take around 20
minutes to complete it.
Question 1: If P = ₹ 5000, R = 8% and T = 3 years, calculate the simple interest.
A) ₹ 1600
B) ₹ 1400
C) ₹ 1200
D) ₹ 120
Explanation: Simple interest = (P × R × T)/100 = (₹ 5000 × 8 × 3)/100 = ₹ 50 × 8 × 3 = ₹ 1200
Question 2: If P = ₹ 1500, R = 5% and simple interest earned is ₹ 300, then find the time duration.
A) 4 years
B) 3 years
C) 2 years
D) 1 year
Explanation: T = (SI × 100)/(P × R) = (300 × 100)/(1500 × 5) = 30000/7500 = 4 years
Question 3: If P = ₹ 12,000, T = 3 years and simple interest earned is ₹ 2160, then find the rate of interest.
A) 6%
B) 5%
C) 4%
D) 8%
Explanation: R = (SI × 100)/(P × T) = (2160 × 100)/(12,000 × 3) = 2,16,000/36,000 = 6%
Question 4: If R = 8%, T = 2 years and simple interest earned is ₹ 400, then find the principal.
A) ₹ 2000
B) ₹ 2500
C) ₹ 1500
D) ₹ 3500
Explanation: P = (SI × 100)/(R × T) = (400 × 100)/(8 × 2) = 40,000/16 = ₹ 2500
Question 5: Find the amount on ₹ 20,000 for 2 years at 8% p.a. compounded annually.
A) ₹ 23,320
B) ₹ 23,328
C) ₹ 23,300
D) ₹ 23,350
Explanation: Amount = P(1 + R/100)T = 20,000(1 + 8/100)2 = 20,000(1.08)2 = 20,000 × 1.08 × 1.08 = ₹ 23,328
Question 6: Find the compound interest on ₹ 12,000 for 3 years at 10% p.a. compounded annually.
A) ₹ 3900
B) ₹ 3970
C) ₹ 3972
D) ₹ 3872
Explanation: Amount = P(1 + R/100)T = 12,000(1 + 10/100)3 = 12,000(1.1)3 = 12,000 × 1.1 × 1.1 × 1.1 = ₹ 15,972. CI = A – P = ₹ 15,972 – ₹ 12,000 = ₹ 3972
Question 7: Find the compound interest on ₹ 15,000 for 2 years at 6% p.a. compounded annually.
A) ₹ 1850
B) ₹ 1854
C) ₹ 1800
D) ₹ 1954
Explanation: Amount = P(1 + R/100)T = 15,000(1 + 6/100)2 = 15,000(1.06)2 = 15,000 × 1.06 × 1.06 = ₹ 16,854. CI = A – P = ₹ 16,854 – ₹ 15,000 = ₹ 1854
Question 8: Find the amount on ₹ 8000 for 3 years at 5% p.a. compounded annually.
A) ₹ 9271
B) ₹ 9200
C) ₹ 9260
D) ₹ 9261
Explanation: Amount = P(1 + R/100)T = 8000(1 + 5/100)3 = 8000(1.05)3 = 8000 × 1.05 × 1.05 × 1.05 = ₹ 9261
Question 9: Find the difference between the compound interest and simple interest on ₹ 16,000 for 3 years at 15% p.a. compounded annually.
A) ₹ 1134
B) ₹ 1130
C) ₹ 1124
D) ₹ 1234
Explanation: Amount = P(1 + R/100)T = 16,000(1 + 15/100)3 = 16,000(1.15)3 = 16,000 × 1.15 × 1.15 × 1.15 = ₹ 24,334. CI = A – P = ₹ 24,334 – ₹ 16,000 = ₹ 8334. Simple interest = (P × R × T)/100 = (₹ 16,000 × 15 × 3)/100 = ₹ 160 × 15 × 3 = ₹ 7200. Difference between CI and SI = ₹ 8334 – ₹ 7200 = ₹ 1134
Question 10: Find the number of conversion periods in 1 year when the interest is compounded half yearly.
A) 4
B) 1
C) 2
D) 3
Explanation: When the interest is compounded half yearly, the number of conversion periods = 2, that is 2 half years.
Question 11: If interest is compounded quarterly, then how many conversion periods are there in 2 years.
A) 4
B) 8
C) 6
D) 2
Explanation: When interest is compounded quarterly, the number of conversion periods in 2 years = 8, that is 4 quarters in one year.
Question 12: If the interest is compounded quarterly, then the rate of interest 8% p.a. is converted to __________ quarterly.
A) 1%
B) 2%
C) 4%
D) 8%
Explanation: When the rate of interest 8% p.a. converted to quarterly, we get 8/4 = 2%
Question 13: Find the compound interest on ₹ 8000 for 1 ½ years at 10% p.a., interest being compounded half yearly.
A) ₹ 1261
B) ₹ 1260
C) ₹ 1200
D) ₹ 1270
Explanation: Here, R = 10% p.a. = 10/2 = 5% half yearly and T = 1 ½ years = 3 half years. Amount = P(1 + R/100)T = 8000(1 + 5/100)3 = 8000(1.05)3 = 8000 × 1.05 × 1.05 × 1.05 = ₹ 9261. CI = A – P = ₹ 9261 – ₹ 8000 = ₹ 1261
Question 14: Find the compound interest on ₹ 10,000 for 1 year at 8% p.a., interest being compounded quarterly.
A) ₹ 804.32
B) ₹ 800.32
C) ₹ 820.32
D) ₹ 824.32
Explanation: Here, R = 8% p.a. = 8/4 = 2% quarterly and T = 1 year = 4 quarters. Amount = P(1 + R/100)T = 10,000(1 + 2/100)4 = 10,000(1.02)4 = 10,000 × 1.02 × 1.02 × 1.02 × 1.02 = ₹ 10,824.32. CI = A – P = ₹ 10,824.32 – ₹ 10,000 = ₹ 824.32
Question 15: Find the compound interest on ₹ 25,000 for 9 months at 16% p.a., interest being compounded quarterly.
A) ₹ 3221.60
B) ₹ 3125.60
C) ₹ 3121.60
D) ₹ 3122.60
Explanation: Here, R = 16% p.a. = 16/4 = 4% quarterly and T = 9 months = 3 quarters. Amount = P(1 + R/100)T = 25,000(1 + 4/100)3 = 25,000(1.04)3 = 25,000 × 1.04 × 1.04 × 1.04 = ₹ 28,121.60. CI = A – P = ₹ 28,121.60 – ₹ 25,000 = ₹ 3121.60
Question 16: The population of a town is 2,00,000. If the rate of growth is 5% p.a., find the population after 2 years.
A) 2,20,000
B) 2,20,500
C) 2,00,500
D) 2,22,500
Explanation: In case of growth, A = P(1 + R/100)T = 2,00,000(1 + 5/100)2 = 2,00,000(1.05)2 = 2,00,000 × 1.05 × 1.05 = 2,20,500
Question 17: The population of a city is 40,00,000. If the rate of growth is 8% p.a., find the population after 3 years.
A) 50,38,848
B) 50,38,840
C) 50,38,800
D) 50,38,040
Explanation: In case of growth, A = P(1 + R/100)T = 40,00,000(1 + 8/100)3 = 40,00,000(1.08)3 = 40,00,000 × 1.08 × 1.08 × 1.08 = 50,38,848
Question 18: Rahul purchased a bike for ₹ 95,000. Its value is depreciating at the rate of 8% p.a. Find its value after 2 years.
A) 80,500
B) 80,000
C) 80,408
D) 80,400
Explanation: In case of depreciation, A = P(1 – R/100)T = 95,000(1 – 8/100)2 = 95,000(0.92)2 = 95,000 × 0.92 × 0.92 = 80,408
Question 19: The count of bacteria in a culture was initially 6,25,000. If it is increasing at the rate of 4% per hour, find the count of bacteria at the end of 3 hours.
A) 7,03,000
B) 7,03,040
C) 7,00,040
D) 7,03,140
Explanation: In case of growth, A = P(1 + R/100)T = 6,25,000(1 + 4/100)3 = 6,25,000(1.04)3 = 6,25,000 × 1.04 × 1.04 × 1.04 = 7,03,040
Question 20: A car was purchased at ₹ 12,00,000. If its value is depreciating at 10% p.a., what will be its value after 2 years?
A) 9,60,000
B) 9,74,000
C) 9,72,000
D) 9,72,200
Explanation: In case of depreciation, A = P(1 – R/100)T = 12,00,000(1 – 10/100)2 = 12,00,000(0.9)2 = 12,00,000 × 0.9 × 0.9 = 9,72,000
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