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Simple Interest Questions Placement

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Simple Interest Questions for Placement Exams

Comprehensive Topic Guide with 30 practice questions, shortcuts, and detailed solutions for TCS, Infosys, Wipro, Banking, SSC, and all major placement exams.

Last Updated: March 2026

Introduction to Simple Interest

Simple Interest (SI) is one of the most scoring topics in quantitative aptitude sections of placement exams and government recruitment tests. Unlike compound interest, simple interest follows a linear growth pattern, making calculations straightforward and quick.

Why is Simple Interest Important?

  • Easy to Master: Only one main formula to remember
  • High Accuracy Potential: Minimal chance of calculation errors
  • Foundation for Banking: Essential for banking and financial sector jobs
  • Quick Solves: Most questions can be solved in 30-60 seconds with practice
  • Direct Applications: Loans, savings accounts, fixed deposits in real life

Exam Distribution

Exam TypeQuestionsDifficulty Level
TCS NQT2-3Easy to Medium
Infosys2-3Easy
Wipro2-3Easy to Medium
Banking PO3-5Easy to Medium
SSC CGL2-3Medium
Railway Exams2-3Easy

Complete Formula Sheet & Shortcuts

Basic Formulas

FormulaExpressionDescription
Simple InterestSI = (P × R × T)/100Principal × Rate × Time
AmountA = P + SIPrincipal + Interest
AmountA = P(1 + RT/100)Direct formula
PrincipalP = (100 × SI)/(R × T)From SI formula
RateR = (100 × SI)/(P × T)Annual interest rate
TimeT = (100 × SI)/(P × R)In years

Variable Definitions

  • P (Principal): The initial sum borrowed or invested
  • R (Rate): Annual interest rate (in percentage)
  • T (Time): Duration in years
  • SI (Simple Interest): Interest earned/paid
  • A (Amount): Total value after interest

🚀 SHORTCUT TRICKS

Trick 1: Quick SI Calculation

For P = ₹100, SI for T years at R% = R × T

Example: SI on ₹100 for 3 years at 8% = 8 × 3 = ₹24

Trick 2: Amount Doubling Time

If Amount = 2P (doubles), then RT = 100

Example: At 10% rate, time to double = 100/10 = 10 years

Trick 3: Amount Tripling Time

If Amount = 3P (triples), then RT = 200

Trick 4: Finding Rate from Doubling Time

If money doubles in T years, R = 100/T %

Example: If money doubles in 8 years, R = 100/8 = 12.5%

Trick 5: Installment Formula

For calculating equal installments:

  • Each installment = (100 × A)/(100 × n + R × n(n-1)/2)
  • Where n = number of installments

Trick 6: Different Rates for Different Years

If rates are R₁%, R₂%, R₃% for consecutive years: SI = P(R₁ + R₂ + R₃)/100 (for 3 years)

Trick 7: When Time is in Months

T (in years) = months/12 Example: 8 months = 8/12 = 2/3 years

Trick 8: When Time is in Days

T (in years) = days/365 (or 360 for banking convention)

Practice Questions with Solutions

EASY LEVEL (Questions 1-10)

Question 1 [Easy]

Find the simple interest on ₹5,000 for 3 years at 8% per annum.

Solution: Given: P = ₹5,000, R = 8%, T = 3 years

SI = (P × R × T)/100 SI = (5000 × 8 × 3)/100 SI = 120000/100 SI = ₹1,200

Question 2 [Easy]

What is the amount when ₹8,000 is invested at 6% simple interest for 4 years?

Solution: P = ₹8,000, R = 6%, T = 4 years

SI = (8000 × 6 × 4)/100 = ₹1,920 Amount = P + SI = 8000 + 1920 = ₹9,920

Shortcut: A = P(1 + RT/100) = 8000(1 + 24/100) = 8000 × 1.24 = ₹9,920 ✓

Question 3 [Easy]

In how many years will ₹6,000 amount to ₹8,400 at 8% simple interest?

Solution: P = ₹6,000, A = ₹8,400 SI = A - P = 8400 - 6000 = ₹2,400

SI = (P × R × T)/100 2400 = (6000 × 8 × T)/100 2400 = 480T T = 2400/480 = 5 years

Question 4 [Easy]

At what rate percent per annum will ₹4,000 yield an interest of ₹1,200 in 3 years?

Solution: P = ₹4,000, SI = ₹1,200, T = 3 years

R = (100 × SI)/(P × T) R = (100 × 1200)/(4000 × 3) R = 120000/12000 R = 10%

Question 5 [Easy]

A sum of money amounts to ₹7,200 in 2 years and to ₹8,400 in 4 years at simple interest. Find the principal.

Solution: Amount in 4 years - Amount in 2 years = SI for 2 years 8400 - 7200 = ₹1,200 (SI for 2 years)

SI for 2 years = ₹1,200 SI for 1 year = ₹600 SI for 2 years (to reach ₹7,200) = ₹1,200

Principal = 7200 - 1200 = ₹6,000

Question 6 [Easy]

What sum will amount to ₹8,800 in 4 years at 12% simple interest?

Solution: A = ₹8,800, R = 12%, T = 4 years

A = P(1 + RT/100) 8800 = P(1 + 48/100) 8800 = P × 1.48 P = 8800/1.48 = ₹5,945.95 ≈ ₹5,946

Question 7 [Easy]

Find the simple interest on ₹15,000 from March 15, 2023 to June 15, 2023 at 9% per annum.

Solution: Time period: March 15 to June 15

  • March: 16 days (excluding March 15)
  • April: 30 days
  • May: 31 days
  • June: 15 days Total = 16 + 30 + 31 + 15 = 92 days

T = 92/365 years

SI = (15000 × 9 × 92)/(100 × 365) SI = (15000 × 9 × 92)/36500 SI = 12420000/36500 SI = ₹340.27

Question 8 [Easy]

At what rate percent per annum will a sum of money double itself in 8 years?

Solution: If money doubles, A = 2P, so SI = P

Using SI = (P × R × T)/100 P = (P × R × 8)/100 1 = 8R/100 R = 100/8 = 12.5%

Shortcut: For doubling, R = 100/T = 100/8 = 12.5% ✓

Question 9 [Easy]

A man borrowed ₹12,000 at 10% per annum simple interest. He lent it to another person at 15% per annum. What is his gain in 3 years?

Solution: P = ₹12,000, T = 3 years

Interest paid = (12000 × 10 × 3)/100 = ₹3,600 Interest earned = (12000 × 15 × 3)/100 = ₹5,400

Gain = 5400 - 3600 = ₹1,800

Question 10 [Easy]

Find the simple interest on ₹8,500 for 6 months at 16% per annum.

Solution: T = 6 months = 6/12 = 0.5 years

SI = (8500 × 16 × 0.5)/100 SI = (8500 × 8)/100 SI = 68000/100 SI = ₹680

MEDIUM LEVEL (Questions 11-20)

Question 11 [Medium]

A sum of money becomes 7/6 of itself in 3 years at simple interest. Find the rate percent.

Solution: Let Principal = P Amount = 7P/6 SI = 7P/6 - P = P/6

SI = (P × R × T)/100 P/6 = (P × R × 3)/100

1/6 = 3R/100 100 = 18R R = 100/18 = 50/9 = 5.56%

Question 12 [Medium]

₹5,000 was invested at 12% per annum simple interest and a certain sum at 10% per annum. If total interest after 4 years is ₹4,000, find the second sum.

Solution: Let second sum = ₹x

Interest from first sum = (5000 × 12 × 4)/100 = ₹2,400 Interest from second sum = (x × 10 × 4)/100 = 0.4x

Total interest = 2400 + 0.4x = 4000 0.4x = 1600 x = 1600/0.4 = ₹4,000

Question 13 [Medium]

The simple interest on a sum for 3 years at 8% is ₹600 less than the simple interest on the same sum for 5 years at 6%. Find the sum.

Solution: Let the sum be ₹P

SI at 8% for 3 years = (P × 8 × 3)/100 = 24P/100 SI at 6% for 5 years = (P × 6 × 5)/100 = 30P/100

According to question: 30P/100 - 24P/100 = 600 6P/100 = 600 P = 600 × 100/6 = ₹10,000

Question 14 [Medium]

A sum of ₹8,000 amounts to ₹11,600 in 5 years at simple interest. If the rate is increased by 3%, what will be the new amount?

Solution: First, find original rate: SI = 11600 - 8000 = ₹3,600

3600 = (8000 × R × 5)/100 3600 = 400R R = 9%

New rate = 9% + 3% = 12% New SI = (8000 × 12 × 5)/100 = ₹4,800

New Amount = 8000 + 4800 = ₹12,800

Question 15 [Medium]

Divide ₹8,400 into two parts such that the simple interest on the first part at 10% for 4 years equals the simple interest on the second part at 8% for 6 years.

Solution: Let first part = ₹x, second part = ₹(8400 - x)

SI on first part = (x × 10 × 4)/100 = 40x/100 SI on second part = ((8400-x) × 8 × 6)/100 = 48(8400-x)/100

Equating: 40x = 48(8400 - x) 40x = 403200 - 48x 88x = 403200 x = ₹4,582 (approximately)

Second part = 8400 - 4582 = ₹3,818

Question 16 [Medium]

A certain sum of money amounts to ₹6,720 in 3 years and to ₹8,400 in 6 years at simple interest. Find the rate percent.

Solution: SI for 3 years = 8400 - 6720 = ₹1,680 SI for 6 years = 2 × 1680 = ₹3,360

P = 8400 - 3360 = ₹5,040

Rate calculation: 3360 = (5040 × R × 6)/100 3360 = 302.4R R = 3360/302.4 = 11.11% ≈ 100/9 %

Question 17 [Medium]

A person invests ₹20,000 in two schemes A and B. Scheme A offers 12% per annum and scheme B offers 15% per annum. If total interest earned in one year is ₹2,640, how much is invested in scheme A?

Solution: Let investment in A = ₹x, in B = ₹(20000 - x)

Interest from A = 12x/100 Interest from B = 15(20000-x)/100

Total: 12x/100 + 15(20000-x)/100 = 2640 12x + 300000 - 15x = 264000 -3x = -36000 x = ₹12,000

Question 18 [Medium]

The simple interest on a sum for 8 years is ₹2,400. What will be the compound interest on the same sum at the same rate for 2 years if the interest is compounded annually?

Solution: First, note that we cannot determine CI from SI without knowing P and R separately.

Wait - let me check if there's enough information: SI for 8 years = ₹2,400 SI for 1 year = ₹300

So P × R/100 = 300 (i.e., PR = 30000)

For CI for 2 years: CI = P[(1 + R/100)² - 1]

We need both P and R. Since we only know PR = 30000, we cannot find unique values.

Data Insufficient - Cannot be determined uniquely.

(Note: This is a trick question to test conceptual understanding)

Question 19 [Medium]

A sum was put at simple interest at a certain rate for 4 years. Had it been put at 2% higher rate, it would have fetched ₹400 more. Find the sum.

Solution: Let sum = ₹P, original rate = R%

SI at original rate = (P × R × 4)/100 SI at (R+2)% = (P × (R+2) × 4)/100

Difference: (P × 4 × (R+2))/100 - (P × 4 × R)/100 = 400 (4P/100) × 2 = 400 8P/100 = 400 P = 400 × 100/8 = ₹5,000

Question 20 [Medium]

A sum of money at simple interest amounts to ₹8,160 in 3 years and to ₹10,080 in 6 years. Find the sum and the rate of interest.

Solution: SI for 3 years = 10080 - 8160 = ₹2,240 SI for 6 years = 2 × 2240 = ₹4,480

Principal = 10080 - 4480 = ₹5,600

Rate calculation: 2240 = (5600 × R × 3)/100 2240 = 168R R = 2240/168 = 13.33% ≈ 40/3 %

HARD LEVEL (Questions 21-30)

Question 21 [Hard]

A man borrowed ₹25,000 at 12% per annum simple interest. He immediately lent it to another person at 15% per annum. After 3 years, he reinvested the entire amount at 18% for 2 more years. What is his total profit?

Solution: First 3 years: Interest paid = (25000 × 12 × 3)/100 = ₹9,000 Interest earned = (25000 × 15 × 3)/100 = ₹11,250 Profit = 11250 - 9000 = ₹2,250

Amount after 3 years = 25000 + 11250 = ₹36,250

Next 2 years: Interest earned = (36250 × 18 × 2)/100 = ₹13,050 Interest paid (on original 25000) = (25000 × 12 × 2)/100 = ₹6,000 Profit = 13050 - 6000 = ₹7,050

Total profit = 2250 + 7050 = ₹9,300

Question 22 [Hard]

A part of ₹15,000 is lent at 10% per annum and the rest at 18% per annum. If the average rate of interest is 14%, find the amount lent at 18%.

Solution: Using alligation method:

10% 4 (18-14) \ / 14 /
18% 4 (14-10)

Ratio = 4:4 = 1:1

Amount at 18% = (1/2) × 15000 = ₹7,500

Question 23 [Hard]

A sum of money lent at simple interest amounts to ₹9,920 in 2 years and to ₹11,200 in 4 years. Find the sum and rate of interest.

Solution: SI for 2 years = 11200 - 9920 = ₹1,280 SI for 4 years = 2 × 1280 = ₹2,560

Principal = 11200 - 2560 = ₹8,640

Rate: 1280 = (8640 × R × 2)/100 1280 = 172.8R R = 1280/172.8 = 7.41% ≈ 200/27 %

Question 24 [Hard]

₹10,000 is divided into three parts and lent at 5%, 8%, and 10% per annum simple interest. The interest from the first two parts after 4 years is equal to the interest from all three parts after 2 years. If the third part is ₹3,000, find the first part.

Solution: Let first part = ₹x, second part = ₹(10000 - 3000 - x) = ₹(7000 - x) Third part = ₹3,000

Interest from first two parts (4 years): = (x × 5 × 4)/100 + ((7000-x) × 8 × 4)/100 = 20x/100 + 32(7000-x)/100

Interest from all three parts (2 years): = (x × 5 × 2)/100 + ((7000-x) × 8 × 2)/100 + (3000 × 10 × 2)/100 = 10x/100 + 16(7000-x)/100 + 600

Equating: 20x + 224000 - 32x = 10x + 112000 - 16x + 60000 -12x + 224000 = -6x + 172000 52000 = 6x x = ₹8,666.67

Question 25 [Hard]

A person invests a certain sum at 15% per annum for 2 years, then reinvests the amount at 20% for 3 years. If the final amount is ₹38,500, what was the initial investment?

Solution: Let initial investment = ₹P

After 2 years at 15%: Amount = P(1 + 15×2/100) = P × 1.30

After 3 more years at 20%: Final amount = 1.30P × (1 + 20×3/100) 38500 = 1.30P × 1.60 38500 = 2.08P P = 38500/2.08 = ₹18,509.62

Question 26 [Hard]

A sum of ₹18,000 is lent out in two parts, one at 12% and the other at 18%. If the total annual income is ₹2,520, find the amount lent at 18%.

Solution: Let amount at 12% = ₹x, at 18% = ₹(18000 - x)

Total annual interest: (x × 12)/100 + ((18000-x) × 18)/100 = 2520 12x + 324000 - 18x = 252000 -6x = -72000 x = 12,000

Amount at 18% = 18000 - 12000 = ₹6,000

Question 27 [Hard]

The difference between the simple interest received from two different sources on ₹5,000 for 3 years is ₹450. The difference between their rates of interest is:

Solution: Let rates be R₁% and R₂%

SI₁ = (5000 × R₁ × 3)/100 = 150R₁ SI₂ = (5000 × R₂ × 3)/100 = 150R₂

Difference: 150(R₁ - R₂) = 450 R₁ - R₂ = 450/150 = 3%

Question 28 [Hard]

A sum of money at simple interest amounts to ₹10,240 in 4 years and to ₹11,200 in 6 years. If invested at compound interest compounded annually at the same rate, what would be the amount after 2 years?

Solution: SI for 2 years = 11200 - 10240 = ₹960 SI for 4 years = ₹1,920

Principal = 10240 - 1920 = ₹8,320

Finding rate: 960 = (8320 × R × 2)/100 R = (960 × 100)/(8320 × 2) = 5.77% ≈ 75/13 %

For compound interest (2 years): A = P(1 + R/100)² A = 8320(1 + 5.77/100)² A = 8320 × 1.0577² A ≈ ₹9,310 (approximate due to rounding)

Question 29 [Hard]

A man borrowed ₹50,000 at two different rates of simple interest. The interest on one part at 10% for 5 years is equal to the interest on another part at 12.5% for 4 years. Find the amount borrowed at 10%.

Solution: Let amount at 10% = ₹x, at 12.5% = ₹(50000 - x)

Equating interests: (x × 10 × 5)/100 = ((50000-x) × 12.5 × 4)/100 50x = 50(50000 - x) x = 50000 - x 2x = 50000 x = ₹25,000

Question 30 [Hard]

A person deposits ₹40,000 in a bank which offers simple interest. At the end of 3 years, he withdraws the amount and deposits it in another bank offering 2% higher rate. If he gets ₹4,800 more interest in the next 2 years, find the original rate of interest.

Solution: Let original rate = R%

Interest for first 3 years: SI₁ = (40000 × R × 3)/100 = 1200R

Amount after 3 years = 40000 + 1200R

New rate = (R + 2)% Interest for next 2 years: SI₂ = ((40000 + 1200R) × (R+2) × 2)/100

According to question, extra interest = ₹4,800 SI₂ - (interest at original rate for 2 years) = 4800

Actually, the question says he gets ₹4,800 more in next 2 years, meaning: SI at new rate - SI would have been at old rate = 4800

SI at new rate (on new amount): = ((40000 + 1200R) × (R+2) × 2)/100

This is complex. Let me simplify: Interest in next 2 years at (R+2)% = (40000+1200R)(R+2)×2/100 Interest would have been at R% = (40000+1200R)×R×2/100

Difference: (40000+1200R)×2/100 × [(R+2) - R] = 4800 (40000+1200R)×2/100 × 2 = 4800 (40000+1200R) × 0.04 = 4800 1600 + 48R = 4800 48R = 3200 R = 6.67% ≈ 20/3 %

Companies and Exams That Frequently Ask Simple Interest

🏢 IT Companies

  • TCS: 2-3 questions (mostly formula-based)
  • Infosys: 2-3 questions (Easy level)
  • Wipro: 2-3 questions (Mixed difficulty)
  • Accenture: 2-3 questions (Medium level)
  • Cognizant: 2-3 questions
  • Capgemini: 1-2 questions
  • IBM: 1-2 questions

🏛️ Government Exams

  • SBI PO: 3-5 questions (High weightage)
  • IBPS PO/Clerk: 3-4 questions
  • RBI Grade B: 3-4 questions (Higher difficulty)
  • SSC CGL: 2-3 questions
  • SSC CHSL: 2-3 questions
  • Railway Exams: 2-3 questions

📊 Other Exams

  • CAT: 0-1 questions (as part of Arithmetic)
  • MAT/XAT: 1-2 questions
  • AMCAT: 1-2 questions
  • eLitmus: 1-2 questions

Preparation Tips for Simple Interest

🎯 Before the Exam

  1. Memorize the Main Formula: SI = (P × R × T)/100 is the foundation. Everything else is derived from this.

  2. Practice Unit Conversions: Be comfortable converting months and days to years instantly.

    • 6 months = 0.5 year
    • 9 months = 0.75 year
    • 3 months = 0.25 year

  3. Master the Shortcut: If amount doubles in T years, R = 100/T. This saves 30 seconds per question.

  4. Practice Comparison Problems: Questions comparing two investments are very common.

  5. Work on Speed: Aim to solve easy questions in 30 seconds and medium in 60 seconds.

📝 During the Exam

  1. Check Units First: Ensure all time periods are in years before plugging into the formula.

  2. Use Approximation: For complex divisions, use approximation to eliminate wrong options quickly.

  3. Verify Your Answer: Always do a quick sanity check - SI should be less than amount, rate is usually 5-20% in real-world scenarios.

📚 Study Resources

  • RS Aggarwal Quantitative Aptitude: Chapter 11 (Simple Interest)
  • Arun Sharma CAT Quant: Interest section
  • Previous Year Bank PO Papers: Essential for banking aspirants

Frequently Asked Questions

Q1: What's the difference between Simple Interest and Compound Interest?

Q2: Can Simple Interest exceed the Principal?

Q3: How do I handle time given in months or days?

  • Months: divide by 12 (e.g., 8 months = 8/12 = 2/3 year)
  • Days: divide by 365 (or 360 for banking exams)

Q4: What if the rate changes every year?

Q5: How important is Simple Interest for banking exams?

Quick Reference Card

SIMPLE INTEREST - KEY POINTS

→ SI = (P × R × T)/100
→ A = P + SI = P(1 + RT/100)
→ P = 100×SI/(R×T)
→ R = 100×SI/(P×T)
→ T = 100×SI/(P×R)

SHORTCUTS:
• Amount doubles when RT = 100
• Amount triples when RT = 200
• If doubles in T years, R = 100/T %
• For months: T = months/12
• For days: T = days/365

COMMON RATES:
• TCS: 2-3 questions (Easy)
• Banking: 3-5 questions (Mixed)
• SSC: 2-3 questions (Medium)

Best of luck with your placement preparation! 🎯

Master Simple Interest thoroughly—it's one of the easiest topics to score full marks in!

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