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- Question
- A sum of ? 11000 was taken as loan. This is to be paid in two equal annual installments. If the rate of interest be 20% compounded annually, then the value of each installment is
Options- A. ? 7500
- B. ? 7000
- C. ? 7100
- D. ? 7200
- Correct Answer
- ? 7200
ExplanationLet value of each installment be ? N .
Then, N/(1 + 20/100) + N/(1 + 20/100)2 = 11000
? N(5/6 + 25/36) = 11000
? N(56/36) = 11000
? N = ? 7200 - 1. An amount is invested in a bank at compound rate of interest . The total amount, including interest, after first and third year is ? 1200 and ? 1587, respectively . What is the rate of interest?
Options- A. 10 %
- B. 3.9 %
- C. 12 %
- D. 15 % Discuss
- 2. A man borrows ? 5100 to be paid back with compound interest at the rate of 4 % pa by the end of 2 yr in two equal yearly installments. How much will each installment be?
Options- A. ? 2704
- B. ? 2800
- C. ? 3000
- D. ? 2500 Discuss
- 3. A sum invested for 3 yr compounded at 5%, 10% and 20% respectively. In 3 yr, if the sum amounts to ? 16632, then find the sum.
Options- A. ? 11000
- B. ? 12000
- C. ? 9000
- D. ? 15000 Discuss
- 4. The simple interest for certain sum in 2 yr at 4% pa is ? 80. What will be the compound interest for the same sum, if conditions of rate and time period are same?
Options- A. ? 91.60
- B. ? 81.60
- C. ? 71.60
- D. ? 80 Discuss
- 5. Income of Shantanu was ? 4000. In the first 2 yr. his income decreased by 10% and 5% respectively but in the third year, the income increased by 15%. What was his income at the end of third year?
Options- A. ? 3933
- B. ? 4000
- C. ? 3500
- D. ? 3540 Discuss
- 6. A sum of ? 8448 is to be divided between A and B who are respectively 18 and 19 yr old, in such a way that if their shares be invested at 6.25 % per annum at compound interest, they will receive equal amounts on attaining the age of 21 yr. The present share of A is
Options- A. ? 4225
- B. ? 4352
- C. ? 4096
- D. ? 4000 Discuss
- 7. The simple interest on a certain sum of money for 3 yr at 8 % pa is half the compound interest on ? 8000 for 2 yr at 10 % pa. Find the sum on which simple interest is calculated
Options- A. ? 3500
- B. ? 3800
- C. ? 4000
- D. ? 3600 Discuss
- 8. During the first year, the population of a village is increased by 5% and the second year, it is diminished by 5%. At the end of the second year, its population was 47880. What was the population at the beginning of the first year?
Options- A. 45500
- B. 48000
- C. 43500
- D. 53000 Discuss
- 9. What amount will be received on a sum of ? 1750 in 2 1/ 2 yr, if the interest is compounded at the rate of 8% pa?
Options- A. ? 2125
- B. ? 2122.85
- C. ? 2100
- D. ? 2200 Discuss
- 10. Find what is that first years in which a sum of money will become more than double in amount if put out at compound interest at the rate of 10% per annum?
Options- A. 6th years
- B. 7th years
- C. 8th years
- D. Data inadequate Discuss
Compound Interest problems
Search Results
Correct Answer: 15 %
Explanation:
Let amount be ? P and rate of interest be R % annually .
According to the question,
Amount after 1st yr = ? 1200
? P(1 + R/100) = 1200 ...(i)
Amount after 3rd yr = 1587
? P(1 + R/100)3 = 1587 ...(ii)
On dividing Eq. (ii) from Eq. (i), we get
(1 + R/100)2 = 1587/1200 = 529/400
? 1 + R/100 = 23/20
? R/100 = 3/20
? R = 15 %
Correct Answer: ? 2704
Explanation:
Let each installment be ? N .
Then , according to the question ,
N/(1 + 4/100 ) + N/(1 + 4/100 )2 = 5100
? 25N/26 + 625N/676 = 5100
? 1275N = 5100 x 676
? N = (5100 x 676)/1275 = ? 2704
Correct Answer: ? 12000
Explanation:
Let the required sum be P Then,
P (1 + 5/100) (1 + 10/100) (1 + 20/100) = 16632
? P x (105/100) x (110/100) x (120/100) = 16632
? P = 16632 x [(100 x 100 x 100) / (105 x 110 x 120)]
? P = ? 12000
Correct Answer: ? 81.60
Explanation:
Given, n = 2 yr, R = 4% and SI = 80
According to the formula,
CI = SI(1 + 4/200) = 80 x 51/50 = ? 81.60
Correct Answer: ? 3933
Explanation:
Given, P = ? 4000
R1= 10% (decreased), R2 = 5% (decreases) and R3 = 15% (growth)
? According to the formula,
Income at the end of third year = P(1 - R1/100)(1 - R2/100)(1 + R3/100)
= 4000(1- 10/100) (1 - 5/100) (1 + 15/100)
= 4000 x (9/10) x (19/20) x (23/20)
= 9 x 19 x 23
= ? 3933
Correct Answer: ? 4096
Explanation:
Let shares of A and B be ? N and ? (8448 - N), respectively,
Amount got by A after 3 yr = Amount got by B after 2 yr
N(1 + 6.25/100)3 = (8448 - N) x (1 + 6.25/100)2
? 1 + 6.25/100 = (8448 - N)/N
? 1 + 1/16 = (8448 - N)/N
? 17/16 = (8448 - N)/N
? 17N = 135168 - 16N
? N = 4096
Correct Answer: ? 3500
Explanation:
CI = 8000 x (1 + 10/100)2 - 8000
= 8000 x (11/10) x (11/10) - 8000
= ? 9680 - 8000
= ? 1680
? Sum = (840 x 100)/(3 x 8) = ? 3500
[ ? SI is half of CI ]
? SI = 1680/2 = ? 840
Correct Answer: 48000
Explanation:
Let the population at the beginning of the first year be N .
Then, according to the question,
N x [1+ 5/100] x [1 - 5/100] = 47880
? N x (105/100) x (95/100) = 47880
? N = 47880 x (100/105) x (100/95) = 48000
Correct Answer: ? 2122.85
Explanation:
Given, P = ? 1750, R = 8%,
n = 2 and a/b = 1/2
According to the formula,
Amount = P(1 + R/100)n x [1 + {(a/b) x R}/100]
=1750 (1 + 8/100 )2 [1 + {(1/2) x 8} /100]
=1750 (27/25)2 x 26/25
= 1750 x (27/25) x (27/25) x (26/25)
=? 2122.848
= ? 2122.85
Correct Answer: 8th years
Explanation:
Here, P(1 + 10/100)t > 2P
? (11/10)t > 2
When t = 8 ? (11/10)8 = 2.14358
t =7 ? (11/10)7 = 1.9487
By trial, [11 x 11 x 11 x 11 x 11 x 11 x 11 x 11] / [10 x 10 x 10 x 10 x 10 x 10 x 10 x 10] >2
Hence, the first years in which sum of money will become more than double in amount is 8th year
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