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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Find the rate when principal = ₹ 8250, SI = ₹ 1100 and time = 2 years. |
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Answer» Given: – P = ₹ 8250, SI = ₹ 1100, t = 2 years. We know that, R = (100 × SI) / (P × T) = (100 × 1100)/ (8250 × 2) = (50 × 1100) / (8250 × 1) = (55000/ 8250) = 6.67 = 6(2/3) % |
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| 2. |
Find the rate when principal = ₹ 3560, amount = ₹ 4521.20 and time = 3 years. |
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Answer» Given: – P = ₹ 3560, amount = ₹ 4521.20, t = 3 years. We know that, SI = A – P = 4521.20 – 3560 = ₹ 961.2 R = (100 × SI) / (P × T) = (100 × 961.20)/ (3560 × 3) = (100 × 96120) / (8250 × 3 ×100) = (32040 /3560) = 9% p.a. |
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| 3. |
Nalini borrowed Rs 550 from her friend at 8% per annum. She returned the amount after 6 months. How much did she pay? |
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Answer» Given Principal amount P = Rs 550 We know that simple interest = (P × T × R)/100 On substituting these values in above equation we get SI = (550 × ½ × 8)/100 = Rs 22 Total amount paid after ½ year = Principal amount + Interest = Rs 550 + Rs 22 |
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| 4. |
Shanta borrowed ₹ 6000 from the State Bank of India for 3 years 8 months at 12% per annum. What amount will clear off her debt? |
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Answer» From the question, Shanta borrowed ₹ 6000 from the State Bank of India (Principal) Time = 3 years 8 months We know that, 1 year = 12 months ∴ 3 years 8 months = (44/12) = (11/3) SI = 12 % .p.a. First we have to find Simple Interest, SI = (P × R × T)/100 = (6000 × 12 × (11/3))/ 100 = 6000 × 12 × (11/3) × (1/100) = (6000 × 12 × 11 × 1)/ (3 × 100) = (60 × 4 × 11 × 1)/ (1 × 1) = (60 × 4 × 11 × 1) = ₹ 2640 Amount = (principal + SI) = (6000 + 2640) = ₹ 8640 ∴The amount will clear off her debt is ₹ 8640 |
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| 5. |
A man borrowed Rs 8000 from a bank at 8% per annum. Find the amount he has to pay after 4 ½ years. |
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Answer» Given Principal amount P = Rs 8000 We know that simple interest = (P × T × R)/100 On substituting these values in above equation we get SI = (8000 × (9/2) × 8)/100 = Rs 2880 = Rs 8000 + 2880 = Rs 10880 |
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| 6. |
Find the simple interest and the amount when principal = ₹ 2650, rate = 8% p.a. and time = 2 ½ years. |
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Answer» Given: – P = ₹ 2650, R = 8% p.a. and time = 2 ½ years = (5/2) If interest is calculated uniformly on the original principal throughout the loan period, it is called simple interest. SI = (P × R × T)/100 = (2650 × 8 × (5/2))/ 100 = 2650 × 8 × (5/2) × (1/100) = (2650 × 8 × 5 × 1)/ (2 × 100) = (2650 × 4 × 1 × 1)/ (1 × 20) = (2650 × 1 × 1 × 1)/ (1 × 5) = (2650 / 5) = ₹ 530 Amount = (principal + SI) = (2650 + 530) = ₹ 3180 |
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| 7. |
Find the simple interest and the amount when principal = ₹ 9600, rate = 7 ½ % p.a. and time = 5 months. |
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Answer» Given: – P = ₹ 9600, R = 7 ½ % p.a. = (15/2) and time = 5 months = (5/12) years If interest is calculated uniformly on the original principal throughout the loan period, it is called simple interest. SI = (P × R × T)/100 = (9600 × (15/2) × (5/12))/ 100 = 9600 × (15/2) × (5/12) × (1/100) = (9600 × 15 × 5 × 1)/ (2 × 12 × 100) = (96 × 15 × 5 × 1)/ (2 × 12) = (7200)/ (24) = ₹ 300 Amount = (principal + SI) = (9600 + 300) = ₹ 9900 |
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| 8. |
At simple interest a sum becomes (6/5) of itself in 2½ years. The rate of interest per annum is(a) 6% (b) 7½% (c) 8% (d) 9% |
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Answer» (c) 8% Because, Let the required sum be ₹ x Rate of interest = r % Time = 2½ years = 5/2 years Amount = (6/5) × sum Amount = principal + SI (6/5) × x = x + [(P × R × T)/100] = (6/5) x = x + [(x × r × 5)/ (100 × 2)] = (6/5) = (1 + (r/40)) = r = (40 × (1/5)) = r = 8% |
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| 9. |
At what rate percent per annum simple interest will a sum double itself in 10 years?(a) 8% (b) 10% (c) 12% (d) 12 ½ % |
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Answer» (b) 10% Because, Let the sum be ₹ x and the rate be r% Then, Amount = 2x = P + SI = 2x = P + [(P × R × T)/100] = 2x = x (1 + ((r × 10)/ 100)) = 2x = (100 + (10 × r))/100 = 2 = 10 × r = 200 – 100 = r = 100/10 = r = 10 % |
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| 10. |
A owes B Rs 1573 payable \(1\frac12\) years hence. Also B owes A Rs 1444.50 payable 6 months hence. If they want to settle the account forth with, keeping 14 % as the rate of interest, then who should pay whom and how much ? (a) A to B, Rs 28.50 (b) B to A, Rs 37.50 (c) A to B, Rs 50 (d) B to A, Rs 50 |
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Answer» (d) B to A, Rs 50 Let the present value of what A owes B be Rs x. Then, x + \(\frac{x\times14\times3}{2\times100} = 1573\) ⇒ x + \(\frac{21x}{100}\) 1573 ⇒ \(\frac{21x}{100}\) = 1573 ⇒ x = \(\frac{1573\times100}{121}=1300\) Let y be the present value of what B owes A. Then, y + y x \(\frac12\times\frac{14}{100}\) = 1444.50 ⇒ y + \(\frac{7y}{100}\) = 1444.50 ⇒ 107 y = 1444.50 x 100 ⇒ y = \(\frac{144450}{107}\) = Rs 1350 Hence B must pay Rs 50 to A. |
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| 11. |
David invested certain amount in three different schemes A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs 3200 and the amount invested in Scheme C was 150% of the amount invested in Scheme A and 240% of the amount invested in Scheme B, what was the amount invested in scheme B ? (a) Rs 5000 (b) Rs 6500 (c) Rs 6000 (d) Rs 8000 |
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Answer» (a) Rs 5000. Let x, y and z be the amounts invested in schemes A, B and C respectively. Then, \(\bigg(\frac{x\times10\times1}{100}\bigg)\) + \(\bigg(\frac{y\times12\times1}{100}\bigg)\) + \(\bigg(\frac{z\times15\times1}{100}\bigg)\) = 3200 ⇒ 10x + 12y + 15z = 320000 ......(i) Given, Z = 240% of y = \(\frac{240}{100} \) x y = \(\frac{12}{5} \)y .......(ii) and z = 150% of x = \(\frac{150}{100} \) x \(x = \frac{3}{2}x \) ⇒ x = \(\frac{2}{3} \)z = \(\frac{2}{3} \) x \(\frac{12}{5} \)y = \(\frac{8}{5} \)y .....(iii) ∴ From (i), (ii) and (iii) we have 10 x \(\frac{8}{5} \)y + 12y + 15 x \(\frac{12}{5} \)y = 320000 ⇒ 16y + 12y +36y = 320000 ⇒ 64y = 320000 ⇒ y = \(\frac{320000}{64} \) = 5000. ∴ Sum invested in scheme B = Rs 5000. |
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| 12. |
Find the simple interest, when Principal = Rs 1000, Rate of Interest = 10% per annum and Time = 73 days. |
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Answer» Given Principal = Rs 1000, Rate of Interest = 10% per annum and Time = 73 days = (73/365) days We know that simple interest = (P × T × R)/100 On substituting these values in above equation we get SI = (1000 × (73/365) × 10)/100 SI = (1000 × 73 × 10)/100 × 365 = Rs 20 |
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| 13. |
Divide Rs 1586 in three parts in such a way that their amounts at the end of 2, 3 and 4 years at 5% per annum simple interest be equal.(a) Rs 552, Rs 528, Rs 506 (b) Rs 560, Rs 520, Rs 506 (c) Rs 556, Rs 524, Rs 506 (d) Rs 548, Rs 528, Rs 510 |
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Answer» (a) Rs 552, Rs 528, Rs 506 |
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| 14. |
Anand deposited Rs 6000 on simple interest. He with- draw Rs 4000 and its interest from that amount after 2 years. After next 3 years, he withdraw the rest of the amount and its interest accrued till that time. In all he obtained Rs 900 as interest. The rate of interest per annum was (a) 3% (b) 4% (c) 5% (d) 6% |
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Answer» (c) 5%. Let the rate of interest p.a. be x % Then, interest on Rs 6000 after 2 years = \(\frac{2\times6000\times{x}}{100}\) = 120x = I1 After withdrawing Rs 4000, Rs 2000 remained in the account. ∴ Interest on Rs 2000 for 3 years = \(\frac{3\times2000\times{x}}{100}\) = 60x = I2 . Given, I1 + I2 = 900 ⇒ 120x + 60x = 900 ⇒ 180x = 900 ⇒ x = 5%. |
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| 15. |
Formula of Simple interest : |
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Answer» Answer is: = \(\frac{P\times{R}\times{T}}{100}\) P = \(\frac{S.I\times100}{R\times{T}}\) R = \(\frac{S.I\times100}{P\times{T}}\) T = \(\frac{S.I\times100}{P\times{R}}\) |
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| 16. |
The difference between the simple interest received from two different sources on Rs 1500 for 3 years is Rs 13.50. The difference between their rates of interests is (a) 0.1% (b) 0.2% (c) 0.3% (d) 0.4% |
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Answer» (c) 0.3% Let the two rates of simple interests p.a. be x % and y %. Then, \(\frac{1500\times{x}\times3}{100}-\frac{1500\times{y}\times3}{100}=13.50\) ⇒ 45x - 45y = 13.50 ⇒ 45(x-y) = 13.50 ⇒ x-y = \(\frac{13.50}{45}\) = 0.3% |
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| 17. |
Formula of Amount (A) : |
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Answer» Answer is: = P + S.I. |
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| 18. |
Find the interest on Rs 500 for a period of 4 years at the rate of 8% per annum. Also, find the amount to be paid at the end of the period. |
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Answer» Given Principal amount P = Rs 500 We know that simple interest = (P × T × R)/100 On substituting these values in above equation we get SI = (500 × 4 × 8)/100 = Rs 160 Amount = Principal amount + Interest = Rs 500 + 160 = Rs 660 |
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| 19. |
With a given rate of simple interest, the ratio of principal and amount for a certain period of time is 4 : 5. After 3 years, with the same rate of interest, the ratio of the principal and amount becomes 5 : 7. The rate of interest per annum is (a) 4% (b) 5% (c) 6% (d) 7% |
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Answer» (b) 5% p.a. Suppose after t years, P = Rs 4x and Amount = Rs 5x ⇒ P + S.I. after t years = Rs 5x … (i) Also, given P : [P + S.I. after (t + 3) years] = 5 : 7 = 1 : \(\frac75\) = 4x : \(\frac75\) x 4x = 4x : \(\frac{28x}{5}\) ⇒ P + S.I. after (t + 3) years = Rs \(\frac{28x}{5}\) ....(ii) ∴ Eq (ii) – Eq (i) gives S.I. after 3 years = Rs \(\bigg(\frac{28x}{5}-5x\bigg)\) = Rs \(\frac{3x}{5}\) ∴ Rate of interest = \(\frac{\frac{3x}{5}\times100}{4x\times3}\) = 5% p.a. |
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| 20. |
A sum of Rs 400 is lent at the rate of 5% per annum. Find the interest at the end of 2 years. |
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Answer» Given Principal amount P = Rs 400 We know that simple interest = (P × T × R)/100 On substituting these values in above equation we get SI = (400 × 2 × 5)/100 = Rs 40 |
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| 21. |
Rs 1000 invested at 5% p.a. simple interest. If the interest is added to the principal after every 10 years, the amount will become Rs 2000 after (a) 15 years (b) \(6\frac23\) years (c) 18 years (d) 20 years |
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Answer» (b) \(6\frac23\) years. S.I. for 10 years = Rs \(\bigg(1000\times\frac{5}{100}\times10\bigg)\) = Rs 500 Principal after 10 years becomes = Rs (1000 + 500) = Rs 1500 Amount on that principal after t years = Rs 2000 ∴ S.I. on it = Rs (2000 – 1500) = Rs 500 ∴ t = \(\bigg(\frac{500\times100}{1500\times5}\bigg)\) years = \(6\frac23\) years ∴ Total time = \(\bigg(10+6\frac{2}3{}\bigg)\)years = \(6\frac23\) years. |
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| 22. |
If the simple interest for 6 years be equal to 30% of the principal, it will be equal to the principal after (a) 10 years (b) 20 years (c) 22 years (d) 30 years |
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Answer» (b) 20 years. Let the principal = Rs x and rate of interest = R% p.a. Then, S.I. = 30% of Rs x = \(\frac{30}{100}\times x\) ∴ \(\frac{x\times{R}\times6}{100} = \) \(\frac{30}{100}\times x\) ⇒ R = \(\frac{30}{6}=\) 5 % p.a. Let the time in which the principal is equal to simple interest be t years, then \(\frac{x\times{5}\times{t}}{100} = x\) ⇒ t = \(\frac{100}{5}years\) = 20 years. |
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| 23. |
Find the simple interest, when Principal = Rs 2000, Rate of Interest = 5% per annum and Time = 5 years. |
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Answer» Given Principal = Rs 2000, Rate of Interest = 5% per annum and Time = 5 years. We know that simple interest = (P × T × R)/100 On substituting these values in above equation we get SI = (2000 × 5 × 5)/100 = Rs 500 |
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| 24. |
Kruti took a loan at simple interest at 6% in the first year with an increase of 0.5% in each subsequent year. She paid Rs 3375 as interest after 4 years. How much loan did she take ? (a) Rs 12500 (b) Rs 15800 (c) Rs 33250 (d) Rs 30,000 |
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Answer» (a) Rs 12500. Let the loan taken be Rs x. Then, \(\frac{x\times{6}\times1}{100}+\frac{x\times6.5\times1}{100}+\frac{x\times7\times1}{100}+\frac{x\times7.5\times1}{100}= 3375\) ⇒ \(\frac{27x}{100} = 3375 \) ⇒ x = \(\frac{3375\times100}{27}\) = Rs 12500. |
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| 25. |
Shikha deposited Rs 2000 in a bank which pays 6% simple interest. She withdrew Rs 700 at the end of first year. What will be her balance after 3 years? |
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Answer» Given Principal amount P = Rs 2000 We know that simple interest = (P × T × R)/100 On substituting these values in above equation we get SI = (2000 × 1 × 6)/100 = Rs 120 So amount after 1 year = Principal amount + Interest = 2000 + 120 = Rs 2120 We know that simple interest = (P × T × R)/100 On substituting these values in above equation we get SI = (1420 × 2 × 6)/100 Interest after two years = Rs 170.40 Total amount after 3 years = Rs 1420 + Rs 170.40 = Rs 1590.40 |
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| 26. |
Find the simple interest, when Principal = Rs 500, Rate of Interest = 12.5% per annum and Time = 4 years. |
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Answer» Given Principal = Rs 500, Rate of Interest = 12.5% per annum and Time = 4 years. We know that simple interest = (P × T × R)/100 On substituting these values in above equation we get SI = (500 × 4 × 12.5)/100 = Rs 250 |
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| 27. |
Reema took a loan of Rs 8000 from a money lender, who charged interest at the rate of 18% per annum. After 2 years, Reema paid him Rs 10400 and wrist watch to clear the debt. What is the price of the watch? |
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Answer» Given Principal amount P = Rs 8000 We know that simple interest = (P × T × R)/100 On substituting these values in above equation we get SI = (8000 × 2 × 18)/100 = Rs 2880 Total amount payable by Reema after 2 years = Rs 8,000 + Rs 2,880 = Rs 10,880 |
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| 28. |
Find the simple interest, when Principal = Rs 4500, Rate of Interest = 4% per annum and Time = 6 months. |
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Answer» Given Principal = Rs 4500, Rate of Interest = 4% per annum and Time = 6 months = ½ years We know that simple interest = (P × T × R)/100 On substituting these values in above equation we get SI = (4500 × ½ × 12.5)/100 SI = (4500 × 1 × 12.5)/100 × 2 = Rs 90 |
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| 29. |
Find the simple interest, when Principal = Rs 12000, Rate of Interest = 18% per annum and Time = 4 months. |
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Answer» Given Principal = Rs 12000, Rate of Interest = 18% per annum and Time = 4 months = (4/12) = (1/3) years We know that simple interest = (P × T × R)/100 On substituting these values in above equation we get SI = (12000 × (1/3) × 18)/100 SI = (12000 × 1 × 18)/100 × 3 = Rs 720 |
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