Compound Interest problems
- 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
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 %
- 2. 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 Discuss
Correct Answer: ? 7200
Explanation:
Let 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
- 3. 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
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
- 4. 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
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
- 5. 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
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
- 6. 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
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
- 7. 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
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
- 8. Sharon Stone deposits $2000 at the end of each year in an account earning 10% compounded annually. Determine how much money she has after 25 years. How much interest did she earn?
Options- A. 146694.12
- B. 13452
- C. 18232
- D. 15627 Discuss
Correct Answer: 146694.12
Explanation:
S=R[(1+i)^n-1]/i
- 9. Find compound interest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually
Options- A. 2109
- B. 3109
- C. 4109
- D. 6109 Discuss
Correct Answer: 3109
Explanation:
Time = 2 years 4 months = 2(4/12) years = 2(1/3) years.
Amount = Rs'. [8000 X (1+(15/100))^2 X (1+((1/3)*15)/100)]
=Rs. [8000 * (23/20) * (23/20) * (21/20)]
= Rs. 11109. .
:. C.I. = Rs. (11109 - 8000) = Rs. 3109.
- 10. A sum of money lent at compound interest for 2 years at 20% per annum would fetch Rs.482 more, if the interest was payable half yearly than if it was payable annually . The sum is
Options- A. 10000
- B. 20000
- C. 40000
- D. 50000 Discuss
Correct Answer: 20000
Explanation:
Let sum=Rs.x
C.I. when compounded half yearly =
C.I. when compounded annually =
=> x=20000
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