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- Question
- Subbarao was approached by two neighbour for loan. He had ? 2540, a part of which he lent to one person at 12% interest per annum and the other part was lent to the second person at 12.5%. At the end of a year, Subbarao received ? 311.60 as interest on the total loan . Calculate the amount of money of money lent by him at 12% interest .?
Options- A. ? 1360
- B. ? 1340
- C. ? 1240
- D. ? 1180
- Correct Answer
- ? 1180
ExplanationLet the amount of money lent at 12% interest be ? P.
? Amount of money lent at 12.5% interest = ? (2540 - P)
According to the question;
(P x 12 x 1)/100 + [(2540 - P) x 12.5 x 1]/100 = 311.60
? -0.5P/100 + 31750/100 = 311.60
? 0.5P = 31750 - 31160 = 590
? P = ? 1180 - 1. Vikram borrowed ?. 6450 at 5% simple interest repayable in 4 equal instalments. What will be the annual instalment payable by him?
Options- A. ? 1710
- B. ? 1810
- C. ? 1910
- D. ? 1860 Discuss
- 2. 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
- 3. 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
- 4. 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
- 5. 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
- 6. At the rate of 8 1/ 2% per annum simple interest, a sum of ? 5800 will earn how much interest in 2 yr 3 months?
Options- A. ? 1300
- B. ? 1109
- C. ? 1509
- D. ? 1725 Discuss
- 7. A certain sum becomes 3 fold at 4% annual rate of interest. At what rate, it will become 6 fold?
Options- A. 10%
- B. 12%
- C. 8%
- D. 9% Discuss
- 8. Kriya deposits an amount of ? 65800 to obtain a simple interest at the rate of 14% per annum for 4 yr. What total amount will Kriya get at the end of 4 yr?
Options- A. ? 102648
- B. ? 115246
- C. ? 125578
- D. ? 110324 Discuss
- 9. In a certain time, a sum becomes 4 times at the rate of 5% per annum. At what rate of simple interest, the same sum becomes 8 times in the same duration?
Options- A. 122/3 %
- B. 113/5 %
- C. 112/3 %
- D. 123/5 % Discuss
- 10. In what time does a sum of money become four fold at the simple interest ratio of 10% per annum?
Options- A. 40 yr
- B. 30 yr
- C. 15 yr
- D. 25 yr Discuss
Simple Interest problems
Search Results
Correct Answer: ? 1810
Explanation:
Let the annual instalment be Rs. P.
Amount of ? 100 after 4 yr = 100 + (100 x 5 x 4)/100 = ? 120
? Present Value (PV) of ? 120 due after 4 yr = ? 100
Present Value (PV) of ? due after 4 yr = 100P/120 = 5P/6
Similary, PV of ? P due after 3 yr = 20P/23
PV of ? P due after 2 yr = 10P/11
PV of ? P due after 1 yr = 20P/21
? 5P/6 + 20P/23 + 10P/11 + 20P/11 = 6450
? P = ? 1810
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
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: ? 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: ? 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: ? 1109
Explanation:
Time = 2 yr 3 months = 21/4 yr = 9/4 yr
? SI = [5800 x (17/2) x (9/4)] / 100
= 8874/8
= ? 1109.25
= ? 1109
Correct Answer: 10%
Explanation:
Here, n = 3, m = 6, R1 = 4%
? R2 = [(m - 1)/(n - 1)] x R1
= [(6 - 1)/(3 - 1)] x 4
=(5/2) x 4
= 10%
Correct Answer: ? 102648
Explanation:
Here, P = ? 65800, R = 14% T = 4 yr
Hence, SI = (65800 x 14 x 4) / 100 =? 36848
? Required amount = P + SI
= 65800 + 36848
= ? 102648
Correct Answer: 112/3 %
Explanation:
Let the sum be p
Then, for R = 5%
SI = (4p - p) = 3p
? 3p = (P x 5 x T)/100 = PT/20
? T = 60 yr
Again, for another rate (R)
SI = (8P - P) = 7P
? 7P = (P x R x 60)/100
? R = (7 x 100)/60 = 35/3%
= 112/3 %
Correct Answer: 30 yr
Explanation:
Here, n = 4, R = 10, T = ?
According to the formula,
R = 100 (n - 1)/T
? 10 = [100(4 - 1)]/T = (100 x 3)/T
? T = 30 yr
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