Mathematics Questions and Answers – Compound Interest

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Compound Interest”.

1. Calculate the simple interest if the principal amount is 50000 and the rate is 2% for 4 years.
a) 4000
b) 400
c) 40000
d) 40
View Answer

Answer: a
Explanation: S.I. = PNR/100
⇒ S.I. = 50000 × 2 × 4/100 = 4000.

2. Find the Compound Interest on Rs. 1000 for two years at 2% per annum.
a) 20
b) 20.5
c) 20.4
d) 20.6
View Answer

Answer: c
Explanation: Principal for the first year = Rs. 1000
Interest for the first year = 1000 × 2 × 1/100 = Rs. 20
Amount at the end of one year = 1000 + 20 = Rs. 1020
Interest for second year = 1020 × 2 × 1/100 = Rs. 20.4
Principal for the second year = Rs. 1020
Amount at the end of one year = 1000 + 20. 4 = Rs. 1040.4
C.I. = Amount – Principal = 1040.4 – 1020 = Rs. 20.4.
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3. Evaluate the compound interest on Rs. 10101 for 3 years at the rate of 9% per annum compounded annually.
a) 2980
b) 30000
c) 10101
d) 33333
View Answer

Answer: a
Explanation: A = P (1 + \(\frac{R}{100})\)n
⇒ A = 10101 (1 + \(\frac{9}{100}\))3 = 10101 (\(\frac{109}{100}\))3 = Rs. 13081.08
C.I. = A – P = 13081.08 – 10101 = Rs. 2980.08.

4. Vidhya lent Rs. 5000 to Kavya for 3 years at the rate of 5% per annum compound interest. Calculate the amount that Vidhya will get after 3 years.
a) 5789
b) 5788.12
c) 5788.13
d) 5788
View Answer

Answer: c
Explanation: A = P (1 + \(\frac{R}{100})\)n
Amount for 3 years = 5000 (1 + \(\frac{5}{100}\))3 = 5000 (\(\frac{21}{20}\))3 = 5788.13.
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5. A farmer gets a loan of Rs. 100000 against his fixed deposits. If the rate of interest is 1.5 paise per rupee per annum, calculate the compound interest payable after 2 years.
a) 22250
b) 42250
c) 52250
d) 32250
View Answer

Answer: d
Explanation: R = 1.5 paise per rupee per annum = 1.5 × 100 paise per hundred rupee per annum
= 1.5 × \(\frac{100}{100}\) rupee per hundred rupee per annum = 1.5%
Amount = 100000 (1 + \(\frac{1.5}{100}\))2 = 100000 (\(\frac{23}{20}\))2 = 132250
C.I. = 132250 – 100000 = Rs. 32250.

6. Calculate the compound interest at the rate of 6% per annum for 2 years on the principle which in 2 years at the rate of 2% per annum gives Rs. 8000 as simple interest.
a) 50000
b) 49440
c) 59440
d) 49000
View Answer

Answer: b
Explanation: P = 8000 × \(\frac{100}{2}\) × 2 = 400000
A = 400000(1 + \(\frac{6}{100}\))2 = 400000 (\(\frac{53}{50}\))2 = Rs. 449440
C.I. = 449440 – 400000 = Rs. 49440.

7. Compute the compound interest on Rs. 16000 for 2 years 10% per annum when compounded half yearly.
a) 18600
b) 17640
c) 18640
d) 17600
View Answer

Answer: b
Explanation: A = P (1 + \(\frac{R}{200}\))2n
⇒ A = 16000 (1 + \(\frac{10}{200}\))2 = 16000 (\(\frac{21}{20}\))2 = Rs. 17640.
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8. Find the amount on Rs. 5000 at the rate of 20% per annum for 18 months when interest is compounded half yearly.
a) 6644
b) 6666
c) 6000
d) 6655
View Answer

Answer: d
Explanation: n = 18/12 = 3/2
A = P (1 + \(\frac{R}{200}\))2n = 5000 (1 + \(\frac{20}{200}\))2n = Rs. 6655.

9. If the amount is Rs. 400 and Principal is Rs. 100 which is compounded half yearly for 1 year, calculate the rate of interest.
a) 10
b) 200
c) 2
d) 20
View Answer

Answer: b
Explanation: A = P (1 + \(\frac{R}{200}\))2n
⇒ 400 = 100 (1 + \(\frac{R}{200}\))2
⇒ 2 = 1 + \(\frac{R}{200}\)
R = 200.
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10. Calculate the compound interest on Rs. 4000 for 2 years at 20% per annum when compounded annually.
a) 1856.4
b) 1756.4
c) 1846.4
d) 1746.4
View Answer

Answer: a
Explanation: A = P (1 + \(\frac{R}{200}\))2n = 4000(\(\frac{11}{10}\))4 = Rs. 5856.4
C.I. = 5856.4 – 4000 = Rs. 1856.4.

Sanfoundry Global Education & Learning Series – Mathematics – Class 8.

To practice all areas of Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter