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- Question
- Raja invested ? 15000 at the rate of 10 % pa for 1 yr. If the interest is compound half-yearly, then find the amount received by Raja at the end of the year.
Options- A. ? 16537.50
- B. ? 18000
- C. ? 19000.50
- D. ? 20000
- Correct Answer
- ? 16537.50
ExplanationGiven P = ? 15000 , R = 10 %
and n = 1 yr
Accounting to the formula,
Amount = P(1 + R/(2 x 100))2n
= 15000 x (1 + 5/100 )2
= 15000 x (105/100 )2
= 15000 x (21/20) x (21/20)
= ? 16537.50 - 1. Vandana invested an amount of ? 8000 in a fixed deposit scheme for 2 yr at the rate of 5% pa compound interest. How much amount will Vandana get on maturity of the fixed deposit?
Options- A. ? 8000
- B. ? 8890
- C. ? 8888
- D. ? 8820 Discuss
- 2. A loan was repaid in two annual instalment of ? 1089 each. If the rate of interest be 10% per annum compounded annually, the sum borrowed was?
Options- A. ? 1840
- B. ? 1890
- C. ? 1850
- D. ? 1860 Discuss
- 3. A sum of Rs. 12000 deposited at compound interest becomes doubles after 5 years. After 20 years it will become?
Options- A. 120000
- B. 192000
- C. 124000
- D. 96000 Discuss
- 4. A loan was repaid in two annual instalments of Rs. 121 each. if the rate of interest be 10% per annum. compounded annually, the sum borrowed was?
Options- A. Rs. 200
- B. Rs. 210
- C. Rs. 217.80
- D. Rs. 280 Discuss
- 5. A tree increases annually by 1/8 th of its height. By how much will it increase after 2 years, if it stands today 64 cm high?
Options- A. 72 cm
- B. 74 cm
- C. 75 cm
- D. 81 cm Discuss
- 6. What amount will be received on a sum of Rs 15000 in 1 1/ 4 yr at 12% pa, if interest is compounded quarterly?
Options- A. Rs 16596.88
- B. Rs 16789.08
- C. Rs 17630.77
- D. Rs 17389.12 Discuss
- 7. The compound interest on ? 30000 at 7% pa for a certain time is ?4347. The time is
Options- A. 2 yr
- B. 2.5 yr
- C. 3 yr
- D. 4 yr Discuss
- 8. What is the difference between compound interest and simple interest for 2 yr on the sum of ? 1250 at 4% pa?
Options- A. ? 3
- B. ? 4
- C. ? 2
- D. ? 8 Discuss
- 9. The difference between compound and simple rates of interest on ? 10000 for 3 yr at 5% per annum is
Options- A. ? 76.25
- B. ? 76.75
- C. ? 76.50
- D. ? 76 Discuss
- 10. A borrowed sum was paid in the two annual installments of ? 121 each. If the rate of compound interest is 10% pa, what sum was borrowed?
Options- A. ? 217.80
- B. ? 210
- C. ? 220
- D. ? 200 Discuss
Compound Interest problems
Search Results
Correct Answer: ? 8820
Explanation:
Given, P = ? 8000, R = 5% and n = 2 Yr
According to the formula,
Amount = p ( 1 + R / 100 )n
= 8000 ( 1 + 5/100 )2
= 8000 x (105/100)2
=8000 x (21/20) x (21/20)
= 20 x 21 x 21
= ? 8820
Correct Answer: ? 1890
Explanation:
Principle = ? 1089
? P/{(1 + 10/100) + (1 + 10/100)2 } = 1089
? P = {(1089 x 10/11) + (1089 x 100/121)} = (990 + 900) = ? 1890
Correct Answer: 192000
Explanation:
? 2P = P(1 +r/100)5
? (1 +r/100)5 = 2
? (1 + r/100)20 = 24 = 16
Thus, P(1 +r/100)20 = 16P
= Rs. (12000 x 16)
= Rs. 192000
Correct Answer: Rs. 210
Explanation:
Principal = (P.W. of Rs. 121 due 1 year later) + (P.W. of Rs. 121 due 2 years later)
= Rs. [ 121 /(1 + 10/100) + (121 / (1 + 10/100)2]
= Rs. 210
Correct Answer: 81 cm
Explanation:
Increase% = (1/8) x 100% = 12.5%
Height after 2 years =64 x {1 +25/(2 x 100)}2
= 64 x 9/8 x 9/8
= 81 cm
Correct Answer: Rs 17389.12
Explanation:
Given, P = Rs 15000, R = 12 %
and n = 11/4 = 5/4 Yr
According to the formula,
Amount = p [1 + R/(100 x 4)]4n
= 15000 x [1 + 12/(100 x 4)]4 x 5/4
= 15000(412/400)5
= 15000 (103/100)5
= 15000 x (103/100) x (103/100) x (103/100) x (103/100) x (103/100)
= (15 x 103 x 103 x 103 x 103 x 103) / 10000000
= Rs 17389.111
= Rs 17389.12 (approx )
Correct Answer: 2 yr
Explanation:
Given, CI = ? 4347, P = ? 30000 and R = 7%
By formula, CI = P[(1 + R/100)n - 1 ]
? 4347 = 30000 [(1+7/100)n - 1]
? (107/100)n = (4347/30000) + 1
? (107/100)n = 34347/30000 = 11449/10000
? (107/100)n = (107/100)2
? n = 2
Correct Answer: ? 2
Explanation:
Given, P = ? 1250, n = 2 yr and R = 4%
According to the formula,
Difference between compound interest and simple interest = PR2/1002
? Required difference = (1250 x 4 x 4)/(100 x 100) = ? 2
Correct Answer: ? 76.25
Explanation:
Required difference = P(R/100)2 x [(300 + R)/100]
= 10000(5/100)2 (305/100)
= 76.25
Correct Answer: ? 210
Explanation:
According to the question,
Borrowed sum = [121/{1/(1 + 10/100)}] + 121/(1 + 10/100)2
= 121/(11/10) + 121/{(11/10) x (11/10)]
=110 + 100
= ? 210
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