Simple Interest is a topic which is easy to understand and students can gain full marks. It is a very important topic for preliminary and mains exams. This topic covers 2 questions in SSC, Railway and Banking Preliminary exams and covers 4-5 questions in mains exam.   

Important Formulae of Simple Interest:- 

S.I = (P × r × t)/100 

Where, S.I = Simple interest, P = Principal, r = Rate of interest, t = Time 

Amount = Simple Interest + Principal 

Formula of Installment at Simple Interest:- 

Installment = (100A)/[100t + rt(t-1)/2]  

Where, A = debt, t = time, r = rate of interest 

Questions Based on Simple Interest:- 

Q1. How long will it take for a sum of money to grow from ₹2000 to ₹12000, invested at 12.5% p.a simple interest? 

a) 40 years 

b) 45 years 

c) 48 years 

d) can’t be determined 

Solution:- P = 2000, A = 12000 

S.I = 12000-2000 = 10000 

By using formula of simple interest, 

10000 =( 2000×12.5×t)/100 

t = 40 years 

Hence option a) is correct. 

Q2. If a sum of money at simple interest becomes 36 times in 100 years, in how many years sum of money becomes 29 times ? 

a) 90 

b) 85 

c) 80 

d) 75 

Solution:- Let sum of money:- x 

Amount:- 36x 

S.I = 36x – x = 35x 

35x = (x × r × 100)/100 

rate = 35% 

Now amount:- 29x 

S.I = 29x – x = 28x 

28x = (x × 35 × t)/100 

time = 80 years 

Hence option c) is correct. 

Q3. If the simple interest on a certain sum of money is ₹7480 at the rate of 18 1/3% p.a for 4 years 3 months. Find the principal. 

a) 8400 

b) 8800 

c) 9200 

d) 9600 

Solution:- Simple interest:- 7480 

Rate:- 55/3% , time:- 17/4 years 

7480 = (P × 55/3 × 17/4)/100 

P = 9600 

Hence option d) is correct. 

Q4. If a sum of money becomes ₹ 4000 in 2 years and ₹ 5500 in 4 years 6 months at the same rate of simple interest. Find the rate of interest. 

a) 21% 

b) 21 3/7% 

c) 21 5/7% 

d) 21 2/7% 

Solution:- P + 2.SI = 4000 ……(1)

P + 9/2. SI = 5500 ……..(2) 

Subtract equation (1) from equation (2) 

4.5SI – 2SI = 1500 

2.5SI = 1500 ⇒SI = 600 

P = 4000 – 1200 

P = 2800 

600 = (2800×r×1/100) 

r = 600/28 = 21 3/7% 

Hence option b) is correct.  

Q5. An amount of ₹ x was put at simple interest at a certain rate for 2 years. If it had been put at 3% higher rate, it fetched ₹ 300 more. Find the value of 5x. 

a) 17500 

b) 20500 

c) 23500

d) 25000 

Solution:- Principal amount:- x 

According to question, 

 [{x×(r+3)×2 – x×r×2}]/100 = 300 

[2rx + 6x – 2rx] = 300×100 

6x = 30000 

x = 5000 

5x = ₹ 25000 

Hence option d) is correct. 

Q6. Find the simple interest on ₹ 7300 from 11 may 1991 to 10 September 1991 (both days are included) at 5% p.a 

a) ₹ 123 

b) ₹ 315 

c) ₹ 73 

d) ₹ 369 

Solution:- rate of interest:- 5%, time:- 123 days = 123/365 year, principal:- 7300 

S.I = (7300×5×123)/(100×365) 

S.I = ₹ 123 

Hence option a) is correct. 

Q7. If the simple interest on ₹ 1600 is increased by 65 when the time increases by 6 1/2 years. Find the rate of interest p.a. 

a) 0.375% 

b) 0.875% 

c) 0.625% 

d) can’t be determined 

Solution:- principal = 1600, change in interest = 65, change in time = 13/2 years 

⇒ 65 = (1600×r×13)/200 

⇒ 65 = 8×13×r 

⇒ r = 5/8 = 0.625 

⇒ r = 0.625% 

Hence option c) is correct. 

Q8. The simple interest on ₹ 784 will be less than the interest on ₹ 889 at 2% simple interest by 15. Find the time. 

a) 7 years 

b) 6 years 

c) 6 1/7 years 

d) 7 1/7 years 

Solution:- P1 = 784, P2 = 889, r = 2%, difference in simple interest = 15 

15 = (889×2×t – 784×2×t)/100 

⇒ 1778t – 1568t = 1500 

⇒ 210t = 1500 

⇒ t = 150/21 = 50/7 

⇒ t = 7 1/7 years 

Hence option d) is correct. 

Q9. A man invested 1/3 of his capital at 7%, 1/4 of his capital at 8% and remaining at 12%. If his annual income is₹ 644, find the capital.

a) 6200

b) 6900 

c) 6500 

d) 7150 

Solution:- We can let total capital as 12 because LCM of 3 & 4 

Capital invested at 7% = 1/3 × 12 = 4 

Capital invested at 8% = 1/4 × 12 = 3 

Capital invested at 12% = 12–7 = 5 

Total income in units = 28% + 24% + 60% = 112% 

Capital:- 644×100/112 = 575

Required Capital = 575 × 12 = 6900

Hence option b) is correct. 

Q10. What annual installment will discharge a debt of ₹ 8400 due in 5 years at 10% simple interest? 

a) 1100 

b) 1200 

c) 1300 

d) 1400 

Solution:- Formula of Installment at simple interest:- (100A)/[100t + {rt(t-1)/2}] 

Debt A = 8400, time = 5 years, rate of interest = 10% 

Installment = (100×8400)/[100×5+{10×5(5-1)/2}] 

⇒ 840000/(500+100) = 840000/600 

⇒ 1400 

Hence option d) is correct. 

Q11. The difference between the interest received from two different persons on ₹1500 for 2 years is ₹4. What is the difference in the rate of interest? 

a) 0.133% 

b) 0.15% 

c) 0.67% 

d) 0.75% 

Solution:- rate of interest for two different banks:- r1 & r2 

Principal = 1500, difference in simple interest = 4, time = 2 years 

4 = (1500×2×r1 – 1500×2×r2)/100 

400 = 3000(r1 – r2) 

r1 – r2 = 4/30 = 2/15 

r1 – r2 = 0.133% 

Hence option a) is correct.