Simple and Compound Interest
CSAT · Quantitative Aptitude
Introduction
Interest questions in CSAT are short and formula-driven. Knowing a few relationships – especially the 2-year difference between compound and simple interest – turns them into mental arithmetic.
Key concepts
Simple interest (SI)
Interest is earned only on the principal: SI = PRT/100 and amount A = P + SI. The interest is the same every year.
Compound interest (CI)
Interest is added to the principal each period: A = P(1 + R/100)ᵀ and CI = A − P. Half-yearly compounding: use R/2 and 2T; quarterly: R/4 and 4T.
CI versus SI
For 2 years: CI − SI = P(R/100)². For 3 years: CI − SI = P(R/100)²(3 + R/100).
Year-by-year view
At 10% on ₹1,000: SI gives 100 + 100 = 200 in two years; CI gives 100 + 110 = 210. The extra ₹10 is interest on the first year's interest.
Doubling and multiplying
Under SI, a sum becomes k times itself when RT = 100(k − 1). Under CI, if a sum doubles in n years it becomes 4 times in 2n years and 8 times in 3n years.
Rule of 72 (approximation)
Under compound interest, money roughly doubles in 72/R years – handy for elimination.
- SI = PRT/100
- A = P(1 + R/100)ᵀ
- CI − SI (2 years) = P(R/100)²
- SI: k-fold when RT = 100(k − 1)
Solved examples
Example 1
What is the simple interest on ₹8,000 at 8% per annum for 2 years?
(a) ₹1280 (b) ₹1920 (c) ₹2560 (d) ₹640
- SI = P × R × T / 100 = 8000 × 8 × 2 / 100 = ₹1280.
Answer: (a) ₹1280
Example 2
In how many years will a sum become three times itself at 5% per annum simple interest?
(a) 60 years (b) 40 years (c) 42 years (d) 20 years
- Interest needed = 2 × principal = 200% of P.
- Years = 200 ÷ 5 = 40.
Answer: (b) 40 years
Example 3
What is the compound interest on ₹16,000 at 10% per annum for 3 years, compounded annually?
(a) ₹21296 (b) ₹4800 (c) ₹5326 (d) ₹5296
- A = P(1 + R/100)^T = 16000 × (1 1/10)^3 = ₹21296.
- CI = A − P = ₹5296.
Answer: (d) ₹5296
Example 4
What is the compound interest on ₹5,000 at 10% per annum for 2 years, compounded annually?
(a) ₹6050 (b) ₹1000 (c) ₹1050 (d) ₹1070
- A = P(1 + R/100)^T = 5000 × (1 1/10)^2 = ₹6050.
- CI = A − P = ₹1050.
Answer: (c) ₹1050
Example 5
What is the compound interest on ₹8,000 at 5% per annum for 2 years, compounded annually?
(a) ₹820 (b) ₹800 (c) ₹8820 (d) ₹840
- A = P(1 + R/100)^T = 8000 × (1 1/20)^2 = ₹8820.
- CI = A − P = ₹820.
Answer: (a) ₹820
Example 6
In how many years will a sum become three times itself at 8% per annum simple interest?
(a) 37 1/2 years (b) 27 years (c) 25 years (d) 12 1/2 years
- Interest needed = 2 × principal = 200% of P.
- Years = 200 ÷ 8 = 25.
Answer: (c) 25 years
Example 7
What is the simple interest on ₹6,000 at 5% per annum for 2 years?
(a) ₹600 (b) ₹1200 (c) ₹900 (d) ₹300
- SI = P × R × T / 100 = 6000 × 5 × 2 / 100 = ₹600.
Answer: (a) ₹600
Example 8
In how many years will a sum become four times itself at 10% per annum simple interest?
(a) 32 years (b) 40 years (c) 15 years (d) 30 years
- Interest needed = 3 × principal = 300% of P.
- Years = 300 ÷ 10 = 30.
Answer: (d) 30 years
Example 9
The difference between compound and simple interest on a sum for 2 years at 10% per annum is ₹50. What is the sum?
(a) ₹500 (b) ₹5000 (c) ₹10000 (d) ₹2500
- For 2 years, CI − SI = P(R/100)².
- 50 = P × (10/100)² → P = 50 × 100 = ₹5,000.
Answer: (b) ₹5000
Example 10
What is the simple interest on ₹6,000 at 7.5% per annum for 5 years?
(a) ₹4500 (b) ₹2250 (c) ₹2700 (d) ₹450
- SI = P × R × T / 100 = 6000 × 7.5 × 5 / 100 = ₹2250.
Answer: (b) ₹2250
- For CI at 10% or 20%, multiply step by step (×1.1, ×1.2) rather than expanding.
- Successive-change formula: two years at r% CI = 2r + r²/100 % in total.
- When the options are close, compute the 2-year difference formula first.
- Using total years instead of compounding periods for half-yearly interest.
- Reporting the amount when the question asks for the interest.
- Treating 'doubles in n years' under SI the same as under CI.
Practice set
Attempt all 12 questions (about 90 seconds each), then check the worked solutions below.
Q1. The difference between compound and simple interest on a sum for 2 years at 5% per annum is ₹25. What is the sum?
(a) ₹500 (b) ₹10000 (c) ₹20000 (d) ₹5000
Q2. In how many years will a sum become three times itself at 10% per annum simple interest?
(a) 20 years (b) 22 years (c) 10 years (d) 30 years
Q3. The difference between compound and simple interest on a sum for 2 years at 10% per annum is ₹20. What is the sum?
(a) ₹200 (b) ₹2000 (c) ₹1000 (d) ₹4000
Q4. In how many years will a sum become three times itself at 20% per annum simple interest?
(a) 10 years (b) 5 years (c) 15 years (d) 12 years
Q5. What is the simple interest on ₹15,000 at 6% per annum for 2 years?
(a) ₹2700 (b) ₹900 (c) ₹1800 (d) ₹3600
Q6. What is the simple interest on ₹8,000 at 5% per annum for 3 years?
(a) ₹400 (b) ₹1600 (c) ₹1200 (d) ₹2400
Q7. What is the simple interest on ₹5,000 at 12% per annum for 3 years?
(a) ₹2400 (b) ₹1800 (c) ₹600 (d) ₹3600
Q8. In how many years will a sum become double itself at 6.25% per annum simple interest?
(a) 16 years (b) 8 years (c) 18 years (d) 32 years
Q9. The difference between compound and simple interest on a sum for 2 years at 5% per annum is ₹5. What is the sum?
(a) ₹100 (b) ₹4000 (c) ₹1000 (d) ₹2000
Q10. The difference between compound and simple interest on a sum for 2 years at 4% per annum is ₹20. What is the sum?
(a) ₹6250 (b) ₹500 (c) ₹25000 (d) ₹12500
Q11. In how many years will a sum become three times itself at 6.25% per annum simple interest?
(a) 34 years (b) 48 years (c) 32 years (d) 16 years
Q12. What is the compound interest on ₹8,000 at 20% per annum for 3 years, compounded annually?
(a) ₹13824 (b) ₹5854 (c) ₹4800 (d) ₹5824
Q1 – (b) ₹10000. For 2 years, CI − SI = P(R/100)². 25 = P × (5/100)² → P = 25 × 400 = ₹10,000.
Q2 – (a) 20 years. Interest needed = 2 × principal = 200% of P. Years = 200 ÷ 10 = 20.
Q3 – (b) ₹2000. For 2 years, CI − SI = P(R/100)². 20 = P × (10/100)² → P = 20 × 100 = ₹2,000.
Q4 – (a) 10 years. Interest needed = 2 × principal = 200% of P. Years = 200 ÷ 20 = 10.
Q5 – (c) ₹1800. SI = P × R × T / 100 = 15000 × 6 × 2 / 100 = ₹1800.
Q6 – (c) ₹1200. SI = P × R × T / 100 = 8000 × 5 × 3 / 100 = ₹1200.
Q7 – (b) ₹1800. SI = P × R × T / 100 = 5000 × 12 × 3 / 100 = ₹1800.
Q8 – (a) 16 years. Interest needed = 1 × principal = 100% of P. Years = 100 ÷ 6.25 = 16.
Q9 – (d) ₹2000. For 2 years, CI − SI = P(R/100)². 5 = P × (5/100)² → P = 5 × 400 = ₹2,000.
Q10 – (d) ₹12500. For 2 years, CI − SI = P(R/100)². 20 = P × (4/100)² → P = 20 × 625 = ₹12,500.
Q11 – (c) 32 years. Interest needed = 2 × principal = 200% of P. Years = 200 ÷ 6.25 = 32.
Q12 – (d) ₹5824. A = P(1 + R/100)^T = 8000 × (1 1/5)^3 = ₹13824. CI = A − P = ₹5824.
- SI = PRT/100; CI: A = P(1+R/100)^T
- CI − SI (2 yrs) = P(R/100)²
- Half-yearly: R/2, 2T
- SI doubling: RT = 100
