Advance Level of Simple Interest
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.
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