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Questions to simple interest to calculate the time
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Here are practice questions (with solutions) all focused on finding the time in simple interest problems. Useful formula: Simple interest SI = P * R * T / 100, so
T = (SI * 100) / (P * R).
If you’re given the final amount A instead of SI, first compute SI = A − P.
1) Problem
Principal = 1000, Rate = 5% p.a., Interest earned = 150. Find time.
Solution
T = (150 * 100) / (1000 * 5) = 15000 / 5000 = 3 years.
2) Problem
P = 6000, Rate = 4% p.a., Amount after T years = 6720. Find T.
Solution
SI = 6720 − 6000 = 720.
T = (720 * 100) / (6000 * 4) = 72000 / 24000 = 3 years.
3) Problem
P = 2400, Rate = 6% p.a., Interest = 360. Find T (also give months).
Solution
T = (360 * 100) / (2400 * 6) = 36000 / 14400 = 2.5 years = 2 years 6 months = 30 months.
4) Problem
Borrowed 5000 at 8% p.a.; repaid 6200. Find time.
Solution
SI = 6200 − 5000 = 1200.
T = (1200 * 100) / (5000 * 8) = 120000 / 40000 = 3 years.
5) Problem
At 10% p.a. simple interest, how long for a sum to increase by 25%?
Solution
An increase of 25% means SI = 25% of P.
T = 25 / 10 = 2.5 years = 2 years 6 months.
6) Problem
Interest earned equals half the principal at 12% p.a. Find time.
Solution
SI = 0.5P, so T = (50) / 12 = 25/6 years ≈ 4.1667 years = 4 years 2 months.
7) Problem
P = 1500, Rate = 9% p.a., Interest earned = 405. Find T.
Solution
T = (405 * 100) / (1500 * 9) = 40500 / 13500 = 3 years.
8) Problem
P = 8000, Rate = 7.5% p.a., Interest = 1800. Find T.
Solution
T = (1800 * 100) / (8000 * 7.5) = 180000 / 60000 = 3 years.
9) Problem
A loan of 2000 at 12% p.a. simple interest produces interest of 200. How long in months?
Solution
T = (200 * 100) / (2000 * 12) = 20000 / 24000 = 5/6 year = (5/6)*12 = 10 months.
10) Problem
A sum becomes 1.4 times itself (i.e., increases by 40%) at simple interest of 8% p.a. Find the time.
Solution
Increase = 40% so T = 40 / 8 = 5 years.
If you want more problems, problems with fractional years, or a worksheet without solutions for practice, tell me how many and the difficulty level.
T = (SI * 100) / (P * R).
If you’re given the final amount A instead of SI, first compute SI = A − P.
1) Problem
Principal = 1000, Rate = 5% p.a., Interest earned = 150. Find time.
Solution
T = (150 * 100) / (1000 * 5) = 15000 / 5000 = 3 years.
2) Problem
P = 6000, Rate = 4% p.a., Amount after T years = 6720. Find T.
Solution
SI = 6720 − 6000 = 720.
T = (720 * 100) / (6000 * 4) = 72000 / 24000 = 3 years.
3) Problem
P = 2400, Rate = 6% p.a., Interest = 360. Find T (also give months).
Solution
T = (360 * 100) / (2400 * 6) = 36000 / 14400 = 2.5 years = 2 years 6 months = 30 months.
4) Problem
Borrowed 5000 at 8% p.a.; repaid 6200. Find time.
Solution
SI = 6200 − 5000 = 1200.
T = (1200 * 100) / (5000 * 8) = 120000 / 40000 = 3 years.
5) Problem
At 10% p.a. simple interest, how long for a sum to increase by 25%?
Solution
An increase of 25% means SI = 25% of P.
T = 25 / 10 = 2.5 years = 2 years 6 months.
6) Problem
Interest earned equals half the principal at 12% p.a. Find time.
Solution
SI = 0.5P, so T = (50) / 12 = 25/6 years ≈ 4.1667 years = 4 years 2 months.
7) Problem
P = 1500, Rate = 9% p.a., Interest earned = 405. Find T.
Solution
T = (405 * 100) / (1500 * 9) = 40500 / 13500 = 3 years.
8) Problem
P = 8000, Rate = 7.5% p.a., Interest = 1800. Find T.
Solution
T = (1800 * 100) / (8000 * 7.5) = 180000 / 60000 = 3 years.
9) Problem
A loan of 2000 at 12% p.a. simple interest produces interest of 200. How long in months?
Solution
T = (200 * 100) / (2000 * 12) = 20000 / 24000 = 5/6 year = (5/6)*12 = 10 months.
10) Problem
A sum becomes 1.4 times itself (i.e., increases by 40%) at simple interest of 8% p.a. Find the time.
Solution
Increase = 40% so T = 40 / 8 = 5 years.
If you want more problems, problems with fractional years, or a worksheet without solutions for practice, tell me how many and the difficulty level.
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