Simple Interest & Compound Interest Practice
Take sectional tests or a full 40-question Simple Interest & Compound Interest mock for CUET UG with timers, answer review, and detailed explanations.
Take sectional tests or a full 40-question Simple Interest & Compound Interest mock for CUET UG with timers, answer review, and detailed explanations.
Reserved ad placement in the same competitive-exams practice layout.
Use the sectional sessions to isolate SI fundamentals, annual CI, periodic compounding plus comparison, and application-based growth questions. Then switch to the full mixed mock to test speed and model recognition under pressure.
One question at a time, 60 seconds per question, with score, accuracy, and subtopic insight at the end.
1. What is the simple interest on Rs 14,000 at 15% for 3 years?
Explanation: $SI=\frac{PRT}{100}=\frac{14000\times15\times3}{100}=6300$.
2. A sum earns Rs 1,920 as SI in 4 years at 8%. Find the principal.
Explanation: $P=\frac{SI\times100}{RT}=\frac{1920\times100}{8\times4}=6000$.
3. At what rate will Rs 8,000 earn Rs 2,400 in 5 years under SI?
Explanation: $R=\frac{SI\times100}{PT}=\frac{2400\times100}{8000\times5}=6\%$.
4. In how many years will Rs 12,500 amount to Rs 16,250 at 10% SI?
Explanation: Interest = 16250 - 12500 = 3750. $T=\frac{3750\times100}{12500\times10}=3$ years.
5. A sum doubles under SI in 12 years. It will become 1.5 times in:
Explanation: If a sum doubles, interest = P in 12 years. For 1.5 times, interest needed = 0.5P, so the time is half of 12 years.
6. A loan of Rs 30,000 is taken for 9 months at 16% SI. Find the interest.
Explanation: Time = 9/12 = 0.75 year. $SI=30000\times16\times0.75/100=3600$.
7. A sum becomes Rs 28,600 in 2 years at 5% SI. Find the principal.
Explanation: $A=P(1+RT/100)=1.1P$. So $P=28600/1.1=26000$.
8. The SI on Rs 18,000 at 7.5% for 2 years is:
Explanation: $SI=\frac{18000\times7.5\times2}{100}=2700$.
9. A sum triples in 16 years at SI. What is the rate?
Explanation: Tripling means interest = 2P. So $RT=200$. With $T=16$, $R=12.5\%$.
10. The amount on Rs 10,000 at 8% SI for 2 years is:
Explanation: Interest = 10000 x 8 x 2 / 100 = 1600. Amount = 10000 + 1600 = 11600.
11. Find the amount on Rs 9,000 for 2 years at 10% compounded annually.
Explanation: $A=9000(1.1)^2=10890$.
12. The CI on Rs 12,000 for 3 years at 5% annually is:
Explanation: $A=12000(1.05)^3=13891.50$. So $CI=1891.50$.
13. A sum becomes Rs 13,225 in 2 years at 15% annually. Find the principal.
Explanation: $P=13225/(1.15)^2=10000$.
14. At what rate will Rs 25,000 amount to Rs 30,250 in 2 years at annual CI?
Explanation: $(1+r)^2=30250/25000=1.21$. So $r=10\%$.
15. How many years will Rs 16,000 become Rs 19,360 at 10% annual CI?
Explanation: $16000(1.1)^2=19360$, so the time is 2 years.
16. CI on Rs 20,000 at 12% for 1 year is:
Explanation: For one year, CI = SI. $20000\times12/100=2400$.
17. The amount after 3 years on Rs 5,000 at 20% annual CI is:
Explanation: $A=5000(1.2)^3=8640$.
18. A sum earns Rs 1,680 as CI in 2 years at 20%. Find the principal.
Explanation: For 2 years at 20%, CI fraction = $(1.2)^2-1=0.44$. So $0.44P=1680$, giving $P=4000$.
19. The amount on Rs 30,000 for 2 years at 6% CI is:
Explanation: $A=30000(1.06)^2=33708$.
20. A principal becomes Rs 46,656 in 3 years at 8% CI. Find the nearest principal.
Explanation: $P=46656/(1.08)^3\approx37050.75$. The nearest option is Rs 37,050.
21. Find the amount on Rs 16,000 at 10% p.a. compounded half-yearly for 1 year.
Explanation: Half-yearly rate = 5% and periods = 2. $A=16000(1.05)^2=17640$.
22. CI on Rs 25,000 for 6 months at 8% p.a. compounded half-yearly is:
Explanation: One half-year period at 4% gives CI = 25000 x 4 / 100 = 1000.
23. The difference between CI and SI on Rs 18,000 for 2 years at 10% is:
Explanation: For 2 years, difference = $P(R/100)^2$. = 18000 x 0.01 = 180.
24. On what sum will the difference between CI and SI for 2 years at 5% be Rs 25?
Explanation: At 5%, difference for 2 years = $P\times25/10000=P/400$. So $P/400=25$ gives $P=10000$.
25. A sum is invested at 10%, 10%, and 20% in three successive years under CI. If principal is Rs 5,000, the amount is:
Explanation: $A=5000\times1.1\times1.1\times1.2=7260$.
26. What is the effective annual rate of 12% p.a. compounded half-yearly?
Explanation: Effective rate = $(1.06)^2 - 1 = 0.1236 = 12.36\%$.
27. The amount on Rs 40,000 for 9 months at 12% p.a. compounded quarterly is:
Explanation: Quarterly rate = 3% and periods = 3. $A=40000(1.03)^3=43708.36$, approximately Rs 43,708.
28. A sum at 8% compounded annually becomes Rs 11,664 in 2 years. Find SI on the same principal, same rate, same time.
Explanation: First find the principal: $P=11664/(1.08)^2=10000$. Then SI = $10000\times8\times2/100 = 1600$.
29. The difference between half-yearly and annual amount on Rs 20,000 for 1 year at 10% is:
Explanation: Half-yearly amount = 20000(1.05)^2 = 22050. Annual amount = 22000. Difference = 50.
30. At 10% annual CI, in how many years will a sum become 1.331 times?
Explanation: Since $1.1^3=1.331$, the time is 3 years.
31. The population of a village is 80,000 and grows by 5% every year. Find the population after 2 years.
Explanation: Population after 2 years = $80000(1.05)^2=88200$.
32. A machine depreciates by 20% every year. If current value is Rs 25,600 after 2 years, original price was:
Explanation: $P(0.8)^2=25600$. So $P=25600/0.64=40000$.
33. A tree grows by 10% every year. If height is 121 cm after 2 years, present height is:
Explanation: Present height = $121/(1.1)^2 = 100$ cm.
34. A debt of Rs 12,000 at 10% annual CI is cleared after 2 years. What amount is paid?
Explanation: Amount = $12000(1.1)^2=14520$.
35. If Rs 5,000 is borrowed at 12% SI and repaid after 8 months, interest is:
Explanation: Time = 8/12 year. $SI=5000\times12\times(8/12)/100=400$.
36. Present worth of Rs 12,100 due after 2 years at 10% CI is:
Explanation: Present worth = $12100/(1.1)^2 = 10000$.
37. A student deposits Rs 25,000 at 8% annual CI. What will be the amount after 3 years?
Explanation: $A=25000(1.08)^3=31492.8$, approximately Rs 31,493.
38. An article depreciates from Rs 50,000 to Rs 40,500 in 2 years. The annual depreciation rate is:
Explanation: $(1-r)^2=40500/50000=0.81$. So $1-r=0.9$ and $r=10\%$.
39. At SI, a sum becomes four times in 15 years. In how many years will it become seven times?
Explanation: Four times means interest = 3P in 15 years. Seven times means interest = 6P, which needs double the time, i.e. 30 years.
40. If Rs 10,000 under CI grows to Rs 12,100 in 2 years, what is the annual rate?
Explanation: $(1+r)^2=12100/10000=1.21$. So $r=10\%$.
41. A sum gives Rs 2,160 simple interest in 3 years at 12% p.a. Find the principal.
Explanation: $SI=PRT/100$. $2160=P\times12\times3/100$, so $P=6000$.
42. The difference between compound interest and simple interest on a sum for 2 years at 10% is Rs 80. Find the principal.
Explanation: For 2 years, CI - SI $=P\times(r/100)^2$. $80=P\times(10/100)^2=P/100$, so $P=8000$.
43. Find the amount on Rs 16,000 for 1 year at 10% p.a. compounded half-yearly.
Explanation: Half-yearly rate = 5%, periods = 2. Amount $=16000(1.05)^2=17640$.
44. A laptop worth Rs 72,000 depreciates by 20% in the first year and 10% in the second year. Its value after 2 years is:
Explanation: Value factor $=0.8\times0.9=0.72$. Value $=72000\times0.72=51840$.
45. A sum becomes Rs 13,310 in 3 years at annual compound interest. If the principal is Rs 10,000, the annual rate is:
Explanation: $13310/10000=1.331$. Since $1.1^3=1.331$, the annual rate is 10%.