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- Question
- A farmer borrowed ? 4200 at 10% per annum. At the end of 5 yr he cleared his account by paying ? 5000 and a cow. The cost of the cow is?
Options- A. ? 1300
- B. ? 1240
- C. ? 1200
- D. ? 1340
- Correct Answer
- ? 1300
ExplanationAmount after 5 yr, 4200 + (4200 x 10 x 5)/100
= (4200 + 2100) = Rs 6300
? Cost of the cow = (6300 - 5000) = ? 1300 - 1. When a bank reduce the rate of interest from 6 1/ 2% per annum to 5 1/ 2% per annum, depositor withdrew Rs. 600 and thereby his interest reduced by ? 73. Find the initial deposit.?
Options- A. ? 4000
- B. ? 6000
- C. ? 7000
- D. ? 9000 Discuss
- 2. A man invests ? 3000 at the rate of 5% per annum. How much more should he invest at the rate of 8%, so that he can earn a total of 6% per annum?
Options- A. ? 1200
- B. ? 1300
- C. ? 1500
- D. ? 2000 Discuss
- 3. A sum of ? 4000 is lent out in two parts, one at 8% simple interest and other at 10% simple interest. If the annual interest is ? 352, the sum lent at 8% is
Options- A. ? 1600
- B. ? 2400
- C. ? 1800
- D. ? 2800 Discuss
- 4. If a certain sum of money becomes double at simple interest in 12 yr. What would be the rate of interest per annum?
Options- A. 81/3%
- B. 10%
- C. 12%
- D. 14% Discuss
- 5. Simple interest on ? 1680 for 4 yr at 7 1/ 2 per annum is equal to the simple interest on ? 1200 at 7% per annum for a certain period of time. Time period of time is?
Options- A. 7 yr
- B. 6 yr
- C. 51/3 yr
- D. 71/4 yr Discuss
- 6. The simple interest on a sum of money will be ? 600 after 10 yr. It the principle is trebled after 5 yr, what will be the total interest at the end of tenth year?
Options- A. ? 600
- B. ? 900
- C. ? 1200
- D. ? 1500 Discuss
- 7. Two equal sums of money were lent at simple interest at 10% per annum for 4 yr and 5 yr, respectively. If the difference in interest for two period was ? 220, then each sum is?
Options- A. ? 880
- B. ? 1100
- C. ? 2200
- D. ? 1640 Discuss
- 8. A man borrowed ? 12000 for 4 yr at 7 3/ 4 per annum and a year later he again borrowed another ? 12000 for 3 yr at the same rate. How much should he pay at the end to settle the loans?
Options- A. ? 30510
- B. ? 28610
- C. ? 31450
- D. ? 29590 Discuss
- 9. A man took a loan at simple interest at 7% in the first year with an increase of 0.5% in each subsequent year. He pays ? 3690 as interest after 5 yr . How much loan did he take?
Options- A. ? 9250
- B. ? 9440
- C. ? 9225
- D. ? 9360 Discuss
- 10. What will be the difference in the compound interest on ? 50000 at 12% for one year, when the interest is paid yearly and half- yearly?
Options- A. ?. 500
- B. ? 600
- C. ? 180
- D. ? 360 Discuss
Simple Interest problems
Search Results
Correct Answer: ? 4000
Explanation:
Initially ,
let P = ? A, R = 13/2% per annum and T = 1yr
SI = PRT/100 = (A x 1 x 13/2)/100 = ? 13A/200
Now, new deposit = ? (A - 600), R = 11/2 % per annum and T = 1yr
SI = PTR/100 = [(A - 600) x 1 x 11/2)/100] = ? 11(A - 600)/200
By the given condition ,
13A/200 - 11(A - 600)/200 = 73
? 13A - 11(A - 600 ) = 200 x 73
? 2A = 200 x 73 - 600 x 11
? 2A = 14600 - 6600
? 2A = 8000
? A = 4000
Hence, the initial investment is ? 4000.
Correct Answer: ? 1500
Explanation:
Let the extra money invested = ? P
According to the question .
(3000 x 5 x 1)/100 + (P x 8 x 1)/100 = [(3000 + P) x 6 x 1)]/100
? 15000 + 8P = 18000 + 6P
? 2P = 18000 - 15000
? P = 3000/2 = ? 1500
Correct Answer: ? 2400
Explanation:
Let the money interest at 8% interest be ? P .
Then, the money interest at 10% interest = ?(4000 - P)
According to the question,
(P x 8 x 1)/100 + [(4000 - P) x 10 x 1]/100 = 352
? 8P + 40000 - 10P = 35200
? 40000 - 35200 = 2P
? P = 4800/2 = ? 2400
Correct Answer: 81/3%
Explanation:
Let the amount be P, then amount after 12 yr = 2 P
SI = 2P - P = P
SI = (P x R x T)/ 100
P = (P x R x 12)/ 100
? R = 100/12 = 81/3%
Correct Answer: 6 yr
Explanation:
Let the required period of time be T yr.
Then, (1680 x 4 x 15) /100 = (1200 x T x 7)/100
T = (1680 x 2 x 12)/(1200 x 7)
= 6 yr
Hence, the required time period is 6 yr.
Correct Answer: ? 1200
Explanation:
Let the principle and rate of interest be P and R%
Then, (P x R x 10)/100 = 600
? PR = 6000
Total interest at the end of year = 300 + (3P x R x 5)/100 = 300 + (15 x 6000)/100 = ? 1200
Correct Answer: ? 2200
Explanation:
Let each sum be ? P then,
(P x 10 x 5)/100 - (P x 10 x 4)/100 = 220
? P/10 = 220
? P = 2200
Hence, each sum is ? 2200.
Correct Answer: ? 30510
Explanation:
In the first case
P = Rs. 12000, T = 4 yr and R = 31/4 % per annum
SI = PTR/100 = (12000 x 4 x 31/4)/100 = ? 3720
A = (P + SI) = (12000 + 3720) = ? 15720
In the second case
P = ? 12000, T = 3 yr and R = 31/4 % per annum
SI = PTR/100 = (12000 x 3 x 31/4)/100 = ? 2790
A = (P + SI ) = (12000 + 2790) = ? 14790
? Total memory to be paid at the end = (15720 + 14790) = ? 30510
Correct Answer: ? 9225
Explanation:
Let the loan taken be ? P Then,
[(P x 7 x 1)/100] + [(P x 7.5 x 1)/100] + [(P x 8 x 1)/100] + [(P x 8.5 x 1)/100] + [(P x 9 x 1)/100] = ? 3690
? P x (7 + 7.5 + 8 + 8.5 + 9)/100 = 3690
? (P x 40)/100 = 3690
? P = (3690 x 100)/40 = ? 9225
Hence, the loan taken was ? 9225.
Correct Answer: ? 180
Explanation:
CI on ? 50000 at the rate of 12% for one year, when the interest is paid half-yearly
= 50000(1 + 6/100)2 - 50000
= [50000 x (53/50) x (53/50)] - 50000 = ? 6180
CI when the interest is paid yearly
= 50000(1 + 12/100)1 - 50000
= 50000(28/25 - 1) = 50000(3/25) = ? 6000
? Required difference = CI - SI = 6180 - 6000 = ? 180
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