Simple Interest problems
- 1. 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
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
- 2. 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
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%
- 3. 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
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
- 4. 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
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 %
- 5. 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
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
- 6. Simple interest for the sum of ? 1230 for 2 yr is ? 10 more than the simple interest for ? 1130 for the same duration. Find the rate of interest.
`
Options- A. 5%
- B. 6%
- C. 8%
- D. 2% Discuss
Correct Answer: 5%
Explanation:
According to the question,
[(1230 x 2 x R)/100] - [(1130 x 2 x R)/100] = 10
? (200 x R)/100 = 10
? R = 5%
- 7. For a certain sum, the simple interest in 2 yr at 8% per annum is ? 110 more than the simple interest in 1 yr at the ratio of 5% per annum for the same sum. Find the sum.
Options- A. ? 5000
- B. ? 1000
- C. ? 1050
- D. ? 950 Discuss
Correct Answer: ? 1000
Explanation:
Given, SI1 - SI2 = 110,
T1 = 2 Yr,
T2 = 1 yr, R1 = 8%, R2 = 5%
According to the question,
[(P x 8 x 2)/100] - [(P x 5 x 1)/100] = 110
? 11P/100 = 110
? P = ? 1000
- 8. A sum was invested for 4 yr at a certain rate of simple interest. If it had been invested at 2% more annual rate of interest, then ? 56 more would have been obtained. What is the sum?
Options- A. ? 680
- B. ? 700
- C. ? 720
- D. ? 820 Discuss
Correct Answer: ? 700
Explanation:
According to the question
[P x (R + 2) x 4]/100 - [P x R x 4]/100 = 56
? (4PR + 8P - 4PR)/100 = 56
? 8P/100 = 56
? P = (56 x 100)/8 = ? 700
- 9. A sum becomes 6 fold at 5% per annum. At what rate, the sum becomes 12 fold?
Options- A. 10%
- B. 12%
- C. 9%
- D. 11% Discuss
Correct Answer: 11%
Explanation:
SI at 5% = 6P - P = 5P
? 5P = (P x 5 x T) /100
? T = 100 yr
Now, for new rate (R),
11P = (P x R x 100)/100
? R =11%
- 10. A sum becomes two fold in 6 yr at a certain rate of interest. Find the time, in which the same mount will be 10 fold at the same rate of interest.
Options- A. 35 yr
- B. 49 yr
- C. 59 yr
- D. 54 yr Discuss
Correct Answer: 54 yr
Explanation:
Here, n = 2, m =10, T1 = 6 yr, T2 = ?
? T2 = [(m - 1) / (n - 1)] x T1
= [(10 - 1)/(2 - 1)] x 6
= 54 yr
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