You keep Rs 10,000 in a bank at 15% a year. How much extra money do you earn in 2 years? The answer is Rs 3,000 . That extra money is called simple interest .
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Q1. A sum of money becomes double itself at simple interest after a certain time. What is true about the interest earned in that time?
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Answer: A
Let the original Principal be P, so the Amount doubles to 2P. Interest earned = Amount – Principal = 2P – P = P. So the interest earned exactly equals the original principal. Check: adding this interest back, P + P = 2P, confirming the amount doubled.
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Q2. What is the money borrowed or lent out for a certain period called?
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Answer: D
The principal, also called the sum, is the original money borrowed or invested. Interest is the extra amount earned or paid on top of it. The rate is the percentage used to calculate that interest. The amount is principal plus interest combined, at the end of the period.
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Q3. If a sum P earns simple interest S.I. over time T at rate R, what is the correct formula for the amount?
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Answer: D
The amount is what the borrower owes, or the lender receives, at the end. It equals the original principal plus the interest earned. Formula: Amount = P + S.I. For example, if P = Rs 500 and S.I. = Rs 50, Amount = 500 + 50 = Rs 550. Check: Amount minus S.I. always gives back the original principal P.
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Q4. What is the correct formula for simple interest, where P is principal, R is the rate percent per annum, and T is time in years?
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Answer: A
Simple interest grows in direct proportion to Principal P, Rate R, and Time T. The formula is S.I. = (P x R x T)/100. For example, Rs 1000 at 10% for 2 years gives S.I. = (1000 x 10 x 2)/100 = Rs 200. Check: doubling the time to 4 years doubles the interest to Rs 400, since S.I. is linear in T.
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Q5. A sum of Rs 2800 is lent out in two parts, one at 5% and the other at 9% simple interest per annum. If the total annual interest earned is Rs 198, find the amount lent at 5%.
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Answer: B
Let the amount lent at 5% be x, so (2800 – x) is lent at 9%. Total interest equation: x x 5/100 + (2800 – x) x 9/100 = 198. Multiply through by 100: 5x + 9(2800 – x) = 19800. Expand: 5x + 25200 – 9x = 19800. Combine x terms: (5-9)x = 19800 – 25200 = -5400. x = -5400/-4 = Rs 1350. Check: amount at 9% = 2800 – 1350 = Rs 1450. Verify: (1350 x 5 + 1450 x 9)/100 = Rs 198.
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Q6. A sum of Rs 1350 is lent out in two parts, one at 4% and the other at 6% simple interest per annum. If the total annual interest earned is Rs 60, find the amount lent at 4%.
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Answer: C
Let the amount lent at 4% be x, so (1350 – x) is lent at 6%. Total interest equation: x x 4/100 + (1350 – x) x 6/100 = 60. Multiply through by 100: 4x + 6(1350 – x) = 6000. Expand: 4x + 8100 – 6x = 6000. Combine x terms: (4-6)x = 6000 – 8100 = -2100. x = -2100/-2 = Rs 1050. Check: amount at 6% = 1350 – 1050 = Rs 300. Verify: (1050 x 4 + 300 x 6)/100 = Rs 60.
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Q7. A sum of Rs 2400 is lent out in two parts, one at 8% and the other at 12% simple interest per annum. If the total annual interest earned is Rs 242, find the amount lent at 8%.
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Answer: C
Let the amount lent at 8% be x, so (2400 – x) is lent at 12%. Total interest equation: x x 8/100 + (2400 – x) x 12/100 = 242. Multiply through by 100: 8x + 12(2400 – x) = 24200. Expand: 8x + 28800 – 12x = 24200. Combine x terms: (8-12)x = 24200 – 28800 = -4600. x = -4600/-4 = Rs 1150. Check: amount at 12% = 2400 – 1150 = Rs 1250. Verify: (1150 x 8 + 1250 x 12)/100 = Rs 242.
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Q8. A sum of Rs 1200 is lent out in two parts, one at 8% and the other at 6% simple interest per annum. If the total annual interest earned is Rs 84, find the amount lent at 8%.
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Answer: B
Let the amount lent at 8% be x, so (1200 – x) is lent at 6%. Total interest equation: x x 8/100 + (1200 – x) x 6/100 = 84. Multiply through by 100: 8x + 6(1200 – x) = 8400. Expand: 8x + 7200 – 6x = 8400. Combine x terms: (8-6)x = 8400 – 7200 = 1200. x = 1200/2 = Rs 600. Check: amount at 6% = 1200 – 600 = Rs 600. Verify: (600 x 8 + 600 x 6)/100 = Rs 84.
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Q9. A sum of Rs 1950 is lent out in two parts, one at 6% and the other at 4% simple interest per annum. If the total annual interest earned is Rs 85, find the amount lent at 6%.
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Answer: A
Let the amount lent at 6% be x, so (1950 – x) is lent at 4%. Total interest equation: x x 6/100 + (1950 – x) x 4/100 = 85. Multiply through by 100: 6x + 4(1950 – x) = 8500. Expand: 6x + 7800 – 4x = 8500. Combine x terms: (6-4)x = 8500 – 7800 = 700. x = 700/2 = Rs 350. Check: amount at 4% = 1950 – 350 = Rs 1600. Verify: (350 x 6 + 1600 x 4)/100 = Rs 85.
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Q10. A sum of Rs 2250 is lent out in two parts, one at 12% and the other at 5% simple interest per annum. If the total annual interest earned is Rs 249, find the amount lent at 12%.
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Answer: D
Let the amount lent at 12% be x, so (2250 – x) is lent at 5%. Total interest equation: x x 12/100 + (2250 – x) x 5/100 = 249. Multiply through by 100: 12x + 5(2250 – x) = 24900. Expand: 12x + 11250 – 5x = 24900. Combine x terms: (12-5)x = 24900 – 11250 = 13650. x = 13650/7 = Rs 1950. Check: amount at 5% = 2250 – 1950 = Rs 300. Verify: (1950 x 12 + 300 x 5)/100 = Rs 249.
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Q11. A sum of Rs 2750 is lent out in two parts, one at 8% and the other at 5% simple interest per annum. If the total annual interest earned is Rs 178, find the amount lent at 8%.
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Answer: B
Let the amount lent at 8% be x, so (2750 – x) is lent at 5%. Total interest equation: x x 8/100 + (2750 – x) x 5/100 = 178. Multiply through by 100: 8x + 5(2750 – x) = 17800. Expand: 8x + 13750 – 5x = 17800. Combine x terms: (8-5)x = 17800 – 13750 = 4050. x = 4050/3 = Rs 1350. Check: amount at 5% = 2750 – 1350 = Rs 1400. Verify: (1350 x 8 + 1400 x 5)/100 = Rs 178.
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Q12. A sum of Rs 1900 is lent out in two parts, one at 4% and the other at 10% simple interest per annum. If the total annual interest earned is Rs 178, find the amount lent at 4%.
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Answer: A
Let the amount lent at 4% be x, so (1900 – x) is lent at 10%. Total interest equation: x x 4/100 + (1900 – x) x 10/100 = 178. Multiply through by 100: 4x + 10(1900 – x) = 17800. Expand: 4x + 19000 – 10x = 17800. Combine x terms: (4-10)x = 17800 – 19000 = -1200. x = -1200/-6 = Rs 200. Check: amount at 10% = 1900 – 200 = Rs 1700. Verify: (200 x 4 + 1700 x 10)/100 = Rs 178.
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Q13. A sum of Rs 1350 is lent out in two parts, one at 10% and the other at 7% simple interest per annum. If the total annual interest earned is Rs 105, find the amount lent at 10%.
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Answer: C
Let the amount lent at 10% be x, so (1350 – x) is lent at 7%. Total interest equation: x x 10/100 + (1350 – x) x 7/100 = 105. Multiply through by 100: 10x + 7(1350 – x) = 10500. Expand: 10x + 9450 – 7x = 10500. Combine x terms: (10-7)x = 10500 – 9450 = 1050. x = 1050/3 = Rs 350. Check: amount at 7% = 1350 – 350 = Rs 1000. Verify: (350 x 10 + 1000 x 7)/100 = Rs 105.
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Q14. A sum of Rs 650 is lent out in two parts, one at 8% and the other at 9% simple interest per annum. If the total annual interest earned is Rs 54, find the amount lent at 8%.
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Answer: D
Let the amount lent at 8% be x, so (650 – x) is lent at 9%. Total interest equation: x x 8/100 + (650 – x) x 9/100 = 54. Multiply through by 100: 8x + 9(650 – x) = 5400. Expand: 8x + 5850 – 9x = 5400. Combine x terms: (8-9)x = 5400 – 5850 = -450. x = -450/-1 = Rs 450. Check: amount at 9% = 650 – 450 = Rs 200. Verify: (450 x 8 + 200 x 9)/100 = Rs 54.
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Q15. A sum of Rs 2900 is lent out in two parts, one at 5% and the other at 7% simple interest per annum. If the total annual interest earned is Rs 164, find the amount lent at 5%.
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Answer: C
Let the amount lent at 5% be x, so (2900 – x) is lent at 7%. Total interest equation: x x 5/100 + (2900 – x) x 7/100 = 164. Multiply through by 100: 5x + 7(2900 – x) = 16400. Expand: 5x + 20300 – 7x = 16400. Combine x terms: (5-7)x = 16400 – 20300 = -3900. x = -3900/-2 = Rs 1950. Check: amount at 7% = 2900 – 1950 = Rs 950. Verify: (1950 x 5 + 950 x 7)/100 = Rs 164.
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Q16. A sum of Rs 2300 is lent out in two parts, one at 4% and the other at 10% simple interest per annum. If the total annual interest earned is Rs 185, find the amount lent at 4%.
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Answer: B
Let the amount lent at 4% be x, so (2300 – x) is lent at 10%. Total interest equation: x x 4/100 + (2300 – x) x 10/100 = 185. Multiply through by 100: 4x + 10(2300 – x) = 18500. Expand: 4x + 23000 – 10x = 18500. Combine x terms: (4-10)x = 18500 – 23000 = -4500. x = -4500/-6 = Rs 750. Check: amount at 10% = 2300 – 750 = Rs 1550. Verify: (750 x 4 + 1550 x 10)/100 = Rs 185.
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Q17. What equal annual installment will discharge a debt of Rs 477, due in 2 years, at 12% simple interest?
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Answer: D
Let each installment = x, number of installments n = 2, Rate R = 12% per annum. Debt D = Rs 477 is discharged by n equal yearly installments, each earning interest until due. Formula: D = x x [n + n(n-1)R/200]. n(n-1) = 2 x 1 = 2. n(n-1)R/200 = 2 x 12/200 = 0.12. Bracket = n + that = 2 + 0.12 = 2.12. x = D/bracket = 477/2.12 = Rs 225. Check: x x bracket = 225 x 2.12 = Rs 477, matching the debt.
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Q18. What equal annual installment will discharge a debt of Rs 1417, due in 4 years, at 6% simple interest?
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Answer: A
Let each installment = x, number of installments n = 4, Rate R = 6% per annum. Debt D = Rs 1417 is discharged by n equal yearly installments, each earning interest until due. Formula: D = x x [n + n(n-1)R/200]. n(n-1) = 4 x 3 = 12. n(n-1)R/200 = 12 x 6/200 = 0.36. Bracket = n + that = 4 + 0.36 = 4.36. x = D/bracket = 1417/4.36 = Rs 325. Check: x x bracket = 325 x 4.36 = Rs 1417, matching the debt.
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Q19. What equal annual installment will discharge a debt of Rs 714, due in 2 years, at 4% simple interest?
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Answer: D
Let each installment = x, number of installments n = 2, Rate R = 4% per annum. Debt D = Rs 714 is discharged by n equal yearly installments, each earning interest until due. Formula: D = x x [n + n(n-1)R/200]. n(n-1) = 2 x 1 = 2. n(n-1)R/200 = 2 x 4/200 = 0.04. Bracket = n + that = 2 + 0.04 = 2.04. x = D/bracket = 714/2.04 = Rs 350. Check: x x bracket = 350 x 2.04 = Rs 714, matching the debt.
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Q20. What equal annual installment will discharge a debt of Rs 3245, due in 5 years, at 9% simple interest?
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Answer: B
Let each installment = x, number of installments n = 5, Rate R = 9% per annum. Debt D = Rs 3245 is discharged by n equal yearly installments, each earning interest until due. Formula: D = x x [n + n(n-1)R/200]. n(n-1) = 5 x 4 = 20. n(n-1)R/200 = 20 x 9/200 = 0.9. Bracket = n + that = 5 + 0.9 = 5.9. x = D/bracket = 3245/5.9 = Rs 550. Check: x x bracket = 550 x 5.9 = Rs 3245, matching the debt.
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Q21. What equal annual installment will discharge a debt of Rs 645, due in 4 years, at 5% simple interest?
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Answer: C
Let each installment = x, number of installments n = 4, Rate R = 5% per annum. Debt D = Rs 645 is discharged by n equal yearly installments, each earning interest until due. Formula: D = x x [n + n(n-1)R/200]. n(n-1) = 4 x 3 = 12. n(n-1)R/200 = 12 x 5/200 = 0.3. Bracket = n + that = 4 + 0.3 = 4.3. x = D/bracket = 645/4.3 = Rs 150. Check: x x bracket = 150 x 4.3 = Rs 645, matching the debt.
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Q22. What equal annual installment will discharge a debt of Rs 1180, due in 5 years, at 9% simple interest?
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Answer: A
Let each installment = x, number of installments n = 5, Rate R = 9% per annum. Debt D = Rs 1180 is discharged by n equal yearly installments, each earning interest until due. Formula: D = x x [n + n(n-1)R/200]. n(n-1) = 5 x 4 = 20. n(n-1)R/200 = 20 x 9/200 = 0.9. Bracket = n + that = 5 + 0.9 = 5.9. x = D/bracket = 1180/5.9 = Rs 200. Check: x x bracket = 200 x 5.9 = Rs 1180, matching the debt.
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Q23. What equal annual installment will discharge a debt of Rs 1530, due in 2 years, at 4% simple interest?
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Answer: D
Let each installment = x, number of installments n = 2, Rate R = 4% per annum. Debt D = Rs 1530 is discharged by n equal yearly installments, each earning interest until due. Formula: D = x x [n + n(n-1)R/200]. n(n-1) = 2 x 1 = 2. n(n-1)R/200 = 2 x 4/200 = 0.04. Bracket = n + that = 2 + 0.04 = 2.04. x = D/bracket = 1530/2.04 = Rs 750. Check: x x bracket = 750 x 2.04 = Rs 1530, matching the debt.
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Q24. What equal annual installment will discharge a debt of Rs 3720, due in 5 years, at 12% simple interest?
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Answer: A
Let each installment = x, number of installments n = 5, Rate R = 12% per annum. Debt D = Rs 3720 is discharged by n equal yearly installments, each earning interest until due. Formula: D = x x [n + n(n-1)R/200]. n(n-1) = 5 x 4 = 20. n(n-1)R/200 = 20 x 12/200 = 1.2. Bracket = n + that = 5 + 1.2 = 6.2. x = D/bracket = 3720/6.2 = Rs 600. Check: x x bracket = 600 x 6.2 = Rs 3720, matching the debt.
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Q25. What equal annual installment will discharge a debt of Rs 1248, due in 3 years, at 4% simple interest?
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Answer: B
Let each installment = x, number of installments n = 3, Rate R = 4% per annum. Debt D = Rs 1248 is discharged by n equal yearly installments, each earning interest until due. Formula: D = x x [n + n(n-1)R/200]. n(n-1) = 3 x 2 = 6. n(n-1)R/200 = 6 x 4/200 = 0.12. Bracket = n + that = 3 + 0.12 = 3.12. x = D/bracket = 1248/3.12 = Rs 400. Check: x x bracket = 400 x 3.12 = Rs 1248, matching the debt.
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