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. The simple interest on a sum of Rs 3500 for 6 years is Rs 840. What would the simple interest be on a sum of Rs 6500, at the same rate and for the same period?
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Answer: C
Let Principal P1 = Rs 3500, S.I.1 = Rs 840, Time T = 6 years. First find the rate: R = (100 x S.I.)/(P x T) = (100 x 840)/(3500 x 6) = 4%. The second sum P2 = Rs 6500 earns interest at the same rate 4% for the same 6 years. S.I.2 = (P2 x R x T)/100 = (6500 x 4 x 6)/100 = Rs 1560. Check: (100 x 1560)/(6500 x 6) = 4%, confirming the same rate.
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Q2. The simple interest on a sum of Rs 19000 for 4 years is Rs 7600. What would the simple interest be on a sum of Rs 8500, at the same rate and for the same period?
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Answer: D
Let Principal P1 = Rs 19000, S.I.1 = Rs 7600, Time T = 4 years. First find the rate: R = (100 x S.I.)/(P x T) = (100 x 7600)/(19000 x 4) = 10%. The second sum P2 = Rs 8500 earns interest at the same rate 10% for the same 4 years. S.I.2 = (P2 x R x T)/100 = (8500 x 10 x 4)/100 = Rs 3400. Check: (100 x 3400)/(8500 x 4) = 10%, confirming the same rate.
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Q3. The simple interest on a sum of Rs 16000 for 2 years is Rs 4800. What would the simple interest be on a sum of Rs 6000, at the same rate and for the same period?
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Answer: B
Let Principal P1 = Rs 16000, S.I.1 = Rs 4800, Time T = 2 years. First find the rate: R = (100 x S.I.)/(P x T) = (100 x 4800)/(16000 x 2) = 15%. The second sum P2 = Rs 6000 earns interest at the same rate 15% for the same 2 years. S.I.2 = (P2 x R x T)/100 = (6000 x 15 x 2)/100 = Rs 1800. Check: (100 x 1800)/(6000 x 2) = 15%, confirming the same rate.
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Q4. The simple interest on a sum of Rs 17500 for 2 years is Rs 3150. What would the simple interest be on a sum of Rs 16500, at the same rate and for the same period?
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Answer: A
Let Principal P1 = Rs 17500, S.I.1 = Rs 3150, Time T = 2 years. First find the rate: R = (100 x S.I.)/(P x T) = (100 x 3150)/(17500 x 2) = 9%. The second sum P2 = Rs 16500 earns interest at the same rate 9% for the same 2 years. S.I.2 = (P2 x R x T)/100 = (16500 x 9 x 2)/100 = Rs 2970. Check: (100 x 2970)/(16500 x 2) = 9%, confirming the same rate.
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Q5. The simple interest on a sum of Rs 3000 for 1 year is Rs 360. What would the simple interest be on a sum of Rs 20000, at the same rate and for the same period?
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Answer: C
Let Principal P1 = Rs 3000, S.I.1 = Rs 360, Time T = 1 year. First find the rate: R = (100 x S.I.)/(P x T) = (100 x 360)/(3000 x 1) = 12%. The second sum P2 = Rs 20000 earns interest at the same rate 12% for the same 1 year. S.I.2 = (P2 x R x T)/100 = (20000 x 12 x 1)/100 = Rs 2400. Check: (100 x 2400)/(20000 x 1) = 12%, confirming the same rate.
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Q6. The simple interest on a sum of Rs 8000 for 4 years is Rs 2240. What would the simple interest be on a sum of Rs 18500, at the same rate and for the same period?
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Answer: A
Let Principal P1 = Rs 8000, S.I.1 = Rs 2240, Time T = 4 years. First find the rate: R = (100 x S.I.)/(P x T) = (100 x 2240)/(8000 x 4) = 7%. The second sum P2 = Rs 18500 earns interest at the same rate 7% for the same 4 years. S.I.2 = (P2 x R x T)/100 = (18500 x 7 x 4)/100 = Rs 5180. Check: (100 x 5180)/(18500 x 4) = 7%, confirming the same rate.
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Q7. The simple interest on a sum of Rs 7500 for 4 years is Rs 3000. What would the simple interest be on a sum of Rs 19500, at the same rate and for the same period?
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Answer: B
Let Principal P1 = Rs 7500, S.I.1 = Rs 3000, Time T = 4 years. First find the rate: R = (100 x S.I.)/(P x T) = (100 x 3000)/(7500 x 4) = 10%. The second sum P2 = Rs 19500 earns interest at the same rate 10% for the same 4 years. S.I.2 = (P2 x R x T)/100 = (19500 x 10 x 4)/100 = Rs 7800. Check: (100 x 7800)/(19500 x 4) = 10%, confirming the same rate.
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Q8. The simple interest on a sum of Rs 3500 for 3 years is Rs 1260. What would the simple interest be on a sum of Rs 8000, at the same rate and for the same period?
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Answer: D
Let Principal P1 = Rs 3500, S.I.1 = Rs 1260, Time T = 3 years. First find the rate: R = (100 x S.I.)/(P x T) = (100 x 1260)/(3500 x 3) = 12%. The second sum P2 = Rs 8000 earns interest at the same rate 12% for the same 3 years. S.I.2 = (P2 x R x T)/100 = (8000 x 12 x 3)/100 = Rs 2880. Check: (100 x 2880)/(8000 x 3) = 12%, confirming the same rate.
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Q9. The simple interest on a sum of Rs 4000 for 4 years is Rs 1920. What would the simple interest be on a sum of Rs 15000, at the same rate and for the same period?
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Answer: D
Let Principal P1 = Rs 4000, S.I.1 = Rs 1920, Time T = 4 years. First find the rate: R = (100 x S.I.)/(P x T) = (100 x 1920)/(4000 x 4) = 12%. The second sum P2 = Rs 15000 earns interest at the same rate 12% for the same 4 years. S.I.2 = (P2 x R x T)/100 = (15000 x 12 x 4)/100 = Rs 7200. Check: (100 x 7200)/(15000 x 4) = 12%, confirming the same rate.
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Q10. The simple interest on a sum of Rs 3500 for 1 year is Rs 140. What would the simple interest be on a sum of Rs 15000, at the same rate and for the same period?
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Answer: B
Let Principal P1 = Rs 3500, S.I.1 = Rs 140, Time T = 1 year. First find the rate: R = (100 x S.I.)/(P x T) = (100 x 140)/(3500 x 1) = 4%. The second sum P2 = Rs 15000 earns interest at the same rate 4% for the same 1 year. S.I.2 = (P2 x R x T)/100 = (15000 x 4 x 1)/100 = Rs 600. Check: (100 x 600)/(15000 x 1) = 4%, confirming the same rate.
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Q11. The simple interest on a sum of Rs 14000 for 3 years is Rs 4200. What would the simple interest be on a sum of Rs 8000, at the same rate and for the same period?
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Answer: A
Let Principal P1 = Rs 14000, S.I.1 = Rs 4200, Time T = 3 years. First find the rate: R = (100 x S.I.)/(P x T) = (100 x 4200)/(14000 x 3) = 10%. The second sum P2 = Rs 8000 earns interest at the same rate 10% for the same 3 years. S.I.2 = (P2 x R x T)/100 = (8000 x 10 x 3)/100 = Rs 2400. Check: (100 x 2400)/(8000 x 3) = 10%, confirming the same rate.
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Q12. The simple interest on a sum of Rs 8500 for 5 years is Rs 1700. What would the simple interest be on a sum of Rs 8000, at the same rate and for the same period?
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Answer: C
Let Principal P1 = Rs 8500, S.I.1 = Rs 1700, Time T = 5 years. First find the rate: R = (100 x S.I.)/(P x T) = (100 x 1700)/(8500 x 5) = 4%. The second sum P2 = Rs 8000 earns interest at the same rate 4% for the same 5 years. S.I.2 = (P2 x R x T)/100 = (8000 x 4 x 5)/100 = Rs 1600. Check: (100 x 1600)/(8000 x 5) = 4%, confirming the same rate.
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Q13. At what rate of simple interest per annum will a sum of money become 5 times itself in 4 years?
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Answer: A
Let Principal = P and Time T = 4 years. Becoming 5 times itself means Amount = 5P. So S.I. = Amount – P = 5P – P = 4P. Formula: Rate R = (100 x S.I.)/(P x T). R = (100 x 4P)/(P x 4). The P cancels: R = 400/4 = 100%. Check: R x T/100 = (100 x 4)/100 = 4, matching (multiple-1).
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Q14. At 4% simple interest per annum, in how many years will a sum of money become 3 times itself?
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Answer: B
Let Principal = P and Rate R = 4% per annum. Becoming 3 times itself means Amount = 3P. So S.I. = Amount – P = 3P – P = 2P. Formula: Time T = (100 x S.I.)/(P x R). T = (100 x 2P)/(P x 4). The P cancels: T = 200/4 = 50 years. Check: R x T/100 = (4 x 50)/100 = 2, matching (multiple-1).
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Q15. At 4% simple interest per annum, in how many years will a sum of money become 2 times itself?
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Answer: D
Let Principal = P and Rate R = 4% per annum. Becoming 2 times itself means Amount = 2P. So S.I. = Amount – P = 2P – P = 1P. Formula: Time T = (100 x S.I.)/(P x R). T = (100 x 1P)/(P x 4). The P cancels: T = 100/4 = 25 years. Check: R x T/100 = (4 x 25)/100 = 1, matching (multiple-1).
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Q16. At 8% simple interest per annum, in how many years will a sum of money become 3 times itself?
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Answer: C
Let Principal = P and Rate R = 8% per annum. Becoming 3 times itself means Amount = 3P. So S.I. = Amount – P = 3P – P = 2P. Formula: Time T = (100 x S.I.)/(P x R). T = (100 x 2P)/(P x 8). The P cancels: T = 200/8 = 25 years. Check: R x T/100 = (8 x 25)/100 = 2, matching (multiple-1).
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Q17. At what rate of simple interest per annum will a sum of money become 2 times itself in 5 years?
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Answer: B
Let Principal = P and Time T = 5 years. Becoming 2 times itself means Amount = 2P. So S.I. = Amount – P = 2P – P = 1P. Formula: Rate R = (100 x S.I.)/(P x T). R = (100 x 1P)/(P x 5). The P cancels: R = 100/5 = 20%. Check: R x T/100 = (20 x 5)/100 = 1, matching (multiple-1).
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Q18. At 8% simple interest per annum, in how many years will a sum of money become 5 times itself?
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Answer: A
Let Principal = P and Rate R = 8% per annum. Becoming 5 times itself means Amount = 5P. So S.I. = Amount – P = 5P – P = 4P. Formula: Time T = (100 x S.I.)/(P x R). T = (100 x 4P)/(P x 8). The P cancels: T = 400/8 = 50 years. Check: R x T/100 = (8 x 50)/100 = 4, matching (multiple-1).
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Q19. At what rate of simple interest per annum will a sum of money become 4 times itself in 20 years?
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Answer: D
Let Principal = P and Time T = 20 years. Becoming 4 times itself means Amount = 4P. So S.I. = Amount – P = 4P – P = 3P. Formula: Rate R = (100 x S.I.)/(P x T). R = (100 x 3P)/(P x 20). The P cancels: R = 300/20 = 15%. Check: R x T/100 = (15 x 20)/100 = 3, matching (multiple-1).
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Q20. At 10% simple interest per annum, in how many years will a sum of money become 3 times itself?
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Answer: C
Let Principal = P and Rate R = 10% per annum. Becoming 3 times itself means Amount = 3P. So S.I. = Amount – P = 3P – P = 2P. Formula: Time T = (100 x S.I.)/(P x R). T = (100 x 2P)/(P x 10). The P cancels: T = 200/10 = 20 years. Check: R x T/100 = (10 x 20)/100 = 2, matching (multiple-1).
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Q21. At 5% simple interest per annum, in how many years will a sum of money become 2 times itself?
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Answer: D
Let Principal = P and Rate R = 5% per annum. Becoming 2 times itself means Amount = 2P. So S.I. = Amount – P = 2P – P = 1P. Formula: Time T = (100 x S.I.)/(P x R). T = (100 x 1P)/(P x 5). The P cancels: T = 100/5 = 20 years. Check: R x T/100 = (5 x 20)/100 = 1, matching (multiple-1).
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Q22. At 12% simple interest per annum, in how many years will a sum of money become 4 times itself?
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Answer: B
Let Principal = P and Rate R = 12% per annum. Becoming 4 times itself means Amount = 4P. So S.I. = Amount – P = 4P – P = 3P. Formula: Time T = (100 x S.I.)/(P x R). T = (100 x 3P)/(P x 12). The P cancels: T = 300/12 = 25 years. Check: R x T/100 = (12 x 25)/100 = 3, matching (multiple-1).
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Q23. At 10% simple interest per annum, in how many years will a sum of money become 4 times itself?
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Answer: C
Let Principal = P and Rate R = 10% per annum. Becoming 4 times itself means Amount = 4P. So S.I. = Amount – P = 4P – P = 3P. Formula: Time T = (100 x S.I.)/(P x R). T = (100 x 3P)/(P x 10). The P cancels: T = 300/10 = 30 years. Check: R x T/100 = (10 x 30)/100 = 3, matching (multiple-1).
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Q24. At 4% simple interest per annum, in how many years will a sum of money become 4 times itself?
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Answer: A
Let Principal = P and Rate R = 4% per annum. Becoming 4 times itself means Amount = 4P. So S.I. = Amount – P = 4P – P = 3P. Formula: Time T = (100 x S.I.)/(P x R). T = (100 x 3P)/(P x 4). The P cancels: T = 300/4 = 75 years. Check: R x T/100 = (4 x 75)/100 = 3, matching (multiple-1).
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Q25. At what rate of simple interest per annum will a sum of money become 3 times itself in 4 years?
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Answer: A
Let Principal = P and Time T = 4 years. Becoming 3 times itself means Amount = 3P. So S.I. = Amount – P = 3P – P = 2P. Formula: Rate R = (100 x S.I.)/(P x T). R = (100 x 2P)/(P x 4). The P cancels: R = 200/4 = 50%. Check: R x T/100 = (50 x 4)/100 = 2, matching (multiple-1).
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