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- Question
- 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
- Correct Answer
- ? 3933
ExplanationGiven, 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 - 1. The compound interest on a sum of ? 4000 becomes ? 630.50 in 9 months. Find the rate of interest, if interest is compounded quarterly.
Options- A. 20%
- B. 23%
- C. 19%
- D. 21% Discuss
- 2. A sum of money lent at compound interest for 2 yr at 20 % pa would fetch ? 964 more, if the interest was payable half-yearly than if it was payable annually. What is the sum?
Options- A. ? 40000
- B. ? 60000
- C. ? 90000
- D. ? 500000 Discuss
- 3. SBI lent ? 1331 lakh to the TATA group at compound interest and got ?1728 lakh after 3 yr. What is the rate of interest charged, if compounded annually?
Options- A. 11%
- B. 9.09%
- C. 12%
- D. 8.33% Discuss
- 4. Divide ? 2602 between M and N, so that the amount of M after 7 yr is equal to the amount of N after 9 yr, the interest being compounded at 4% pa.
Options- A. ? 1352, ? 1250
- B. ? 1400, ? 1350
- C. ?1415, ?1300
- D. ?1500, ?1450 Discuss
- 5. What is the difference between the compound interest and simple interest calculated on an amount of ? 16200 at the end of 3 yr at 25% pa? (Rounded off to two digits after decimal)
Options- A. ? 3213.44
- B. ? 3302.42
- C. ? 3495.28
- D. ? 3290.63 Discuss
- 6. 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
- 7. 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
- 8. 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
- 9. 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
- 10. 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
Compound Interest problems
Search Results
Correct Answer: 20%
Explanation:
Given, P =? 4000,
n = 9 months = 3/4 yr and
CI = ? 630.50
Amount = P + CI = 4000 + 630.50 = ? 4630.50
According to the formula,
? Amount = P[(1+R/(100 x 4)]4n
? 4630.50 = 4000(1+R/400)4 x 3/4
? 4630.50 = 4000[(400 + R)/400]3
? 4630.50/4000 = [(400 + R)/400]3
? 9261/8000 = [(400 + R)/400]3
? (21/20)3 = [(400 + R)/400]3
? [(400 + R)/400]/400 = 21/20
? 400 + R = 21 x 20 = 420
? R = 420 - 400 = 20%
Correct Answer: ? 40000
Explanation:
Let the sum be ? P .
Then, CI when compounded half - yearly = [P x (1 + 10/100)4 - P] = 4641P/10000
CI when compounded annually = [ P x (1 + 20/100)2 - P] = 11P/25
According to the question, 4641P/10000 - 11P/25 = 964
? [(4641 - 4400)/10000] x P = 964
? P = (964 x 10000)/241
= ? 40000
Correct Answer: 9.09%
Explanation:
According to the question,
1728 = 1331(1 + R/100)3
1728/1331 = (1 + R/100)3
? (12/11)3 = (1 + R/100)3
? 1 + R/100 = 12/11
? R/100 = (12/11) - 1 = 1/11
? R = 100/11= 9.09%
Correct Answer: ? 1352, ? 1250
Explanation:
Let the first part be ? M .
Then , second part = (2602 - M)
According to the question,
M(1 + 4/100)4 = (2602 - M) (1+4/100)9
? M/(2602 - M) = [(1 + 4/100)9]/[(1 + 4/100)7]
? M/(2602 - M) = (1 + 4/100)2 = (26/25) x (26/25) = 676/625
? 625M = 676 x 2602 - 676M
? 1301M = 676 x 2602
? M = (676 x 2602)/1301 = 676 x 2 = 1352
? The two parts are ? 1352 and ? (2602 - 1352) = ? 1250
Correct Answer: ? 3290.63
Explanation:
Let required difference be ? D.
By formula,
D = P x (R/100)2 x [(3 + R)/100]
= 16200 x (25/100)2 x [(3 + 25)/100]
= 16200 x 625/10000 x 13/4
= 162 x 625 x 13/4 x 100 = ? 3290.63
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: ? 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: ? 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: 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: ? 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
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