Quantitative Aptitude Fundamentals
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Introduction: Why Quantitative Aptitude Matters in Placement
When you walk into your campus placement drive, the first filter you encounter is the aptitude test. Companies like TCS, Infosys, Wipro, and Amazon use these tests to screen thousands of candidates quickly. Among the three main sections—Quantitative Aptitude, Logical Reasoning, and Verbal Ability—the quant section often determines whether you move to the next round. The good news? Quantitative aptitude is the most learnable and improvable section if you understand the fundamentals properly.
Many students struggle not because they lack intelligence, but because they approach quant like they’re memorizing historical dates. They memorize formulas, attempt random problems, and hope for the best. This approach fails 9 times out of 10. In reality, quantitative aptitude is about building conceptual muscle memory—understanding why a formula works, not just how to apply it.
What Makes Quantitative Aptitude Different from School Math
Here’s what catches most students off guard: placement aptitude questions aren’t about solving complex mathematical proofs. They’re about solving practical real-world problems under time pressure. A typical quant question tests whether you can identify patterns, apply formulas creatively, and calculate accurately—all in 2-3 minutes.
Think of it this way: in school, your math teacher might spend 40 minutes teaching about compound interest with 15 worked examples. In a placement test, you get a 3-minute window to solve a problem you’ve never seen before using concepts you learned months ago. The difference isn’t in the math—it’s in the speed, confidence, and problem-solving intuition.
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Section 1: Number Systems - The Foundation
What Are Number Systems and Why They Matter
Every quantitative problem, no matter how complex, ultimately breaks down into working with numbers. Before diving into percentages, ratios, or profit-loss calculations, you need rock-solid clarity on number systems.
A number system is simply a way of representing quantities. But for placement purposes, understanding number systems means knowing:
- How numbers are classified (whole, natural, integers, rational, irrational)
- How divisibility work
- How to find HCF (Highest Common Factor) and LCM (Least Common Multiple)
- How remainders behave
Let me explain with a practical example.
Real-World Example: Understanding Divisibility
Imagine you’re managing inventory for a retail company. You have 144 notebooks and need to pack them into boxes. Each box must contain the same number of notebooks, with zero wastage. How many different box sizes can you create?
This is a divisibility problem. 144 is divisible by 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, and 144. In placement tests, divisibility questions appear as:
“A number when divided by 12 leaves remainder 7. When divided by 16, it leaves remainder 3. Find the number.”
Understanding HCF and LCM Through Real Scenarios
HCF (Highest Common Factor) and LCM (Least Common Multiple) aren’t just abstract concepts—they solve real problems.
Scenario 1: HCF Application
You’re organizing a workshop with 48 software engineers and 36 data analysts. You need to form teams where each team has equal numbers of engineers and analysts, with no one left out. What’s the maximum team size?
Answer: HCF of 48 and 36 = 12
So you can form 12 teams (each with 4 engineers and 3 analysts).
Scenario 2: LCM Application
Buses from your office leave every 15 minutes. Trains leave every 20 minutes. Both depart at 9 AM. When’s the next time they depart together?
Answer: LCM of 15 and 20 = 60 minutes
They depart together again at 10 AM.
Key Formulas to Remember:
For two numbers A and B:
A×B=HCF(A,B)×LCM(A,B)A \times B = \text{HCF}(A, B) \times \text{LCM}(A, B)A×B=HCF(A,B)×LCM(A,B)
Why This Matters: Many students memorize this formula but don’t understand it. Knowing this relationship helps you verify whether your HCF and LCM calculations are correct—a lifesaver during timed exams.
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Section 2: Simplifications and Approximations - Speed is Key
Quantitative aptitude isn’t about getting correct answers—it’s about getting correct answers fast. Simplifications teach you shortcuts that save crucial minutes.
The BODMAS Rule (Order of Operations)
Every calculation follows a sequence. Mess this up, and your entire answer is wrong.
BODMAS=Brackets→Orders→Division→Multiplication→Addition→Subtraction\text{BODMAS} = \text{Brackets} \rightarrow \text{Orders} \rightarrow \text{Division} \rightarrow \text{Multiplication} \rightarrow \text{Addition} \rightarrow \text{Subtraction}BODMAS=Brackets→Orders→Division→Multiplication→Addition→Subtraction
Example Problem:
12+18÷3×2−412 + 18 \div 3 \times 2 – 412+18÷3×2−4
Wrong Approach: Most students work left to right and get 35.
Correct Approach (Using BODMAS):
- Division first: 18 ÷ 3 = 6
- Multiplication: 6 × 2 = 12
- Addition: 12 + 12 = 24
- Subtraction: 24 – 4 = 20
Approximation Technique for Speed
Sometimes, you don’t need the exact answer. Many placement tests offer options with significant differences. Using approximation, you can eliminate options and calculate faster.
Example: What is 47.8% of 1248?
Standard Method: 0.478 × 1248 = 596.544
Approximation Method:
- 47.8% ≈ 50%
- 1248 ≈ 1250
- 50% of 1250 = 625
Your approximated answer (625) is very close to the actual answer (596.54). If options are like 400, 550, 625, 750, you can confidently choose 625 without doing exact calculations.
Section 3: Ratios, Proportions & Averages
Understanding Ratios
A ratio compares two quantities. Think of it as a recipe—if a cake recipe calls for flour and sugar in the ratio 3:2, it means for every 3 parts flour, you use 2 parts sugar.
Practical Example:
In a college, the ratio of boys to girls is 5:3. If there are 120 more boys than girls, how many total students are there?
Solution:
- Boys = 5x, Girls = 3x
- Difference = 5x – 3x = 2x = 120
- Therefore x = 60
- Boys = 300, Girls = 180
- Total = 480 students
Why This Matters: This problem type appears frequently in placement tests. The key insight is that ratios are variable quantities—you multiply the ratio by a common factor.
Averages – More Than Just Adding and Dividing
Average = Sum of all values ÷ Number of values
But placement tests test averages in tricky ways.
Example: A student scores 45 in Math, 60 in English, and 50 in Science. The average is (45+60+50)/3 = 55. But what if you need to find the missing score?
“A student scores 45 in Math and 60 in English. To get an average of 55 across three subjects, what must be the Science score?”
Solution:
- Average = 55 across 3 subjects
- Total needed = 55 × 3 = 165
- Current total = 45 + 60 = 105
- Science score = 165 – 105 = 60
Section 4: Percentage - The Most Tested Concept
Percentages appear in nearly every placement test section. Yet most students struggle because they view percentages as abstract “percent signs” rather than one part per hundred.
Percentage Fundamentals
50% simply means 50 out of 100, or 1 out of 2, or 0.5 as a decimal.
Example: If a phone costs ₹20,000 and there’s a 20% discount, the discount amount is:
20% of 20,000=20100×20,000=₹4,00020\% \text{ of } 20,000 = \frac{20}{100} \times 20,000 = ₹4,00020% of 20,000=10020×20,000=₹4,000
Final price = ₹20,000 – ₹4,000 = ₹16,000
Percentage Increase and Decrease
This is where students make mistakes. The formula isn’t complicated, but the application often confuses.
Formula:
Percentage Change=New Value−Old ValueOld Value×100\text{Percentage Change} = \frac{\text{New Value} – \text{Old Value}}{\text{Old Value}} \times 100Percentage Change=Old ValueNew Value−Old Value×100
Scenario: A company’s revenue was ₹50 lakhs last year and ₹60 lakhs this year. What’s the percentage increase?
Solution:
60−5050×100=1050×100=20%\frac{60 – 50}{50} \times 100 = \frac{10}{50} \times 100 = 20\%5060−50×100=5010×100=20%
Common Mistake: Students sometimes calculate: (60-50)/60 × 100 = 16.67%. Remember: always divide by the original value, not the new value.
Section 5: Profit and Loss - Understanding Business Mathematics
Why Profit-Loss Appears in Every Placement Test
Here’s something interesting: placement tests love P&L questions because companies want employees who understand business impact. When you move to a corporate role, decisions about pricing, discounts, and cost management directly affect company profits. That’s why companies test this concept.
Fundamental Definitions
Let’s start with basic terminology that sometimes confuses students:
- Cost Price (CP): What a seller pays to buy the item
- Selling Price (SP): What the seller charges the customer
- Profit: When SP > CP, the difference is profit
- Loss: When SP < CP, the difference is loss
Core Formulas:
Profit=SP−CP\text{Profit} = \text{SP} – \text{CP}Profit=SP−CP
Profit%=SP−CPCP×100\text{Profit\%} = \frac{\text{SP} – \text{CP}}{\text{CP}} \times 100Profit%=CPSP−CP×100
Loss%=CP−SPCP×100\text{Loss\%} = \frac{\text{CP} – \text{SP}}{\text{CP}} \times 100Loss%=CPCP−SP×100
Real Example: Understanding Through a Store Scenario
Imagine you run a small shop. You buy notebooks from a supplier for ₹5 each (CP = ₹5). You sell them for ₹8 each (SP = ₹8).
Profit per notebook = ₹8 – ₹5 = ₹3
Profit percentage = (3/5) × 100 = 60% profit per notebook
But here’s where many students make mistakes. Let me show you a common pitfall:
Mistake: A student calculates profit% as (3/8) × 100 = 37.5%
Why it’s wrong: You always divide the profit by the original cost, not the selling price. Think about it logically: if you invested ₹5 and made ₹3, that’s a 60% return on your investment, not 37.5%.
Advanced P&L Concept: Markup vs. Profit Margin
These two terms confuse even intermediate learners.
Markup = Profit as a percentage of Cost Price
Profit Margin = Profit as a percentage of Selling Price
Example to Clarify:
CP = ₹100, SP = ₹150
Markup = (150-100)/100 × 100 = 50% markup
Profit Margin = (150-100)/150 × 100 = 33.33% profit margin
See the difference? Same profit amount, but different percentages because we’re dividing by different bases.
Placement Test Problem Type 1: Finding Selling Price with Profit%
“A shopkeeper buys shirts for ₹400 and wants to make 25% profit. At what price should he sell them?”
Solution:
Profit% = 25%
SP=CP×(1+Profit%100)\text{SP} = \text{CP} \times \left(1 + \frac{\text{Profit\%}}{100}\right)SP=CP×(1+100Profit%)
SP=400×(1+25100)=400×1.25=₹500\text{SP} = 400 \times \left(1 + \frac{25}{100}\right) = 400 \times 1.25 = ₹500SP=400×(1+10025)=400×1.25=₹500
Why This Formula Works: When you add 1 to the percentage fraction, you’re essentially calculating 100% of CP (the original cost) plus the profit percentage. It’s a shortcut that saves you from calculating separately and then adding.
Placement Test Problem Type 2: Multiple Transactions
“A trader buys 100 items for ₹10 each. He sells 60 items at ₹12 each and 40 items at ₹8 each. What’s his overall profit or loss?”
Solution:
Total CP = 100 × ₹10 = ₹1000
Revenue from first 60 items = 60 × ₹12 = ₹720
Revenue from remaining 40 items = 40 × ₹8 = ₹320
Total SP = ₹720 + ₹320 = ₹1040
Overall Profit = ₹1040 – ₹1000 = ₹40
Profit% = (40/1000) × 100 = 4% profit
Key Learning: In multi-transaction problems, calculate total CP and total SP separately, then find overall profit/loss. Don’t calculate profit% for each transaction individually and average them out—that’s a common mistake.
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Section 6: Simple Interest and Compound Interest
Why Interest Calculations Matter in Placements
Interest calculations test whether you understand exponential growth. Companies analyze investment returns, loan calculations, and financial projections using these concepts. For a management trainee role, understanding interest calculations could mean the difference between securing millions in loans or losing company money.
Simple Interest – The Linear Growth Model
Simple Interest assumes that interest is calculated only on the principal amount, not on accumulated interest.
Formula:
SI=P×R×T100\text{SI} = \frac{P \times R \times T}{100}SI=100P×R×T
Where:
- P = Principal (initial amount borrowed/invested)
- R = Rate of interest (per annum, in %)
- T = Time (in years)
Amount after Simple Interest:
A=P+SI=P+P×R×T100=P(1+R×T100)A = P + \text{SI} = P + \frac{P \times R \times T}{100} = P \left(1 + \frac{R \times T}{100}\right)A=P+SI=P+100P×R×T=P(1+100R×T)
Practical Example: Bank Savings
You deposit ₹10,000 in a bank account that offers 5% simple interest per annum. How much money do you have after 3 years?
Solution:
SI=10,000×5×3100=150,000100=₹1,500\text{SI} = \frac{10,000 \times 5 \times 3}{100} = \frac{150,000}{100} = ₹1,500SI=10010,000×5×3=100150,000=₹1,500
Total Amount = ₹10,000 + ₹1,500 = ₹11,500
Understanding the Timeline:
- Year 1: Interest earned = ₹500 (5% of ₹10,000)
- Year 2: Interest earned = ₹500 (5% of ₹10,000, not on the new total)
- Year 3: Interest earned = ₹500 (5% of ₹10,000)
Notice how the interest remains constant each year? That’s simple interest—it doesn’t compound.
Compound Interest – The Exponential Growth Model
Compound Interest is where interest is calculated on the principal plus the accumulated interest from previous periods.
Formula:
A=P(1+R100)TA = P \left(1 + \frac{R}{100}\right)^TA=P(1+100R)T
Where A is the final amount after T years.
Compound Interest Amount:
CI=A−P=P[(1+R100)T−1]\text{CI} = A – P = P \left[\left(1 + \frac{R}{100}\right)^T – 1\right]CI=A−P=P[(1+100R)T−1]
Same Example with Compound Interest:
₹10,000 invested at 5% compound interest per annum for 3 years.
A=10,000(1+5100)3=10,000×(1.05)3A = 10,000 \left(1 + \frac{5}{100}\right)^3 = 10,000 \times (1.05)^3A=10,000(1+1005)3=10,000×(1.05)3
A=10,000×1.157625=₹11,576.25A = 10,000 \times 1.157625 = ₹11,576.25A=10,000×1.157625=₹11,576.25
Compound Interest = ₹11,576.25 – ₹10,000 = ₹1,576.25
The Key Difference:
- Simple Interest: ₹1,500
- Compound Interest: ₹1,576.25
- Difference: ₹76.25 (extra earnings because interest earns interest)
This might seem small, but over longer periods, the difference becomes massive. This is why Warren Buffett emphasizes compound interest as “the eighth wonder of the world.”
Critical Placement Test Variation: Half-Yearly and Quarterly Compounding
Most students memorize the annual compounding formula and miss questions about half-yearly or quarterly compounding.
If interest compounds semi-annually (twice per year):
A=P(1+R200)2TA = P \left(1 + \frac{R}{200}\right)^{2T}A=P(1+200R)2T
Notice: R is divided by 200 (not 100), and T is multiplied by 2.
Why? When compounding happens twice a year, the annual rate R is split into two parts (R/2 each), and the number of compounding periods doubles.
Example: ₹5,000 at 10% per annum, compounded half-yearly, for 2 years.
A=5,000(1+10200)4=5,000×(1.05)4A = 5,000 \left(1 + \frac{10}{200}\right)^{4} = 5,000 \times (1.05)^4A=5,000(1+20010)4=5,000×(1.05)4
A=5,000×1.21550625=₹6,077.53A = 5,000 \times 1.21550625 = ₹6,077.53A=5,000×1.21550625=₹6,077.53
Common Student Mistakes to Avoid:
- Using simple interest formula for compound interest – Always check the problem statement
- Forgetting to adjust R and T for compounding frequency – If quarterly, use R/400 and 4T
- Calculating total amount instead of just the interest – Some questions ask for SI/CI amount alone, not the total
Section 7: Time, Speed, and Distance - The Practical Dynamics
Why TSD Questions Appear Everywhere
Time, Speed, and Distance questions test real-world problem-solving. Whether calculating delivery times, travel expenses, or resource allocation, companies need employees who think spatially and temporally.
Fundamental Relationships
Distance=Speed×Time\text{Distance} = \text{Speed} \times \text{Time}Distance=Speed×Time
Speed=DistanceTime\text{Speed} = \frac{\text{Distance}}{\text{Time}}Speed=TimeDistance
Time=DistanceSpeed\text{Time} = \frac{\text{Distance}}{\text{Speed}}Time=SpeedDistance
Unit Conversion – The Hidden Pitfall
Many students solve TSD problems correctly but get marks deducted because of unit mismatches. Here’s the conversion you must memorize:
1 km/h=518 m/s1 \text{ km/h} = \frac{5}{18} \text{ m/s}1 km/h=185 m/s
1 m/s=185 km/h1 \text{ m/s} = \frac{18}{5} \text{ km/h}1 m/s=518 km/h
Why This Ratio?
- 1 km = 1000 m
- 1 hour = 3600 seconds
- So, 1 km/h = 1000/3600 m/s = 5/18 m/s
Example: A car travels at 72 km/h. What’s the speed in m/s?
72×518=36018=20 m/s72 \times \frac{5}{18} = \frac{360}{18} = 20 \text{ m/s}72×185=18360=20 m/s
Basic TSD Problem: Straightforward Calculation
“A cyclist travels 150 km in 6 hours. What’s his average speed?”
Solution:
Speed=150 km6 hours=25 km/h\text{Speed} = \frac{150 \text{ km}}{6 \text{ hours}} = 25 \text{ km/h}Speed=6 hours150 km=25 km/h
Intermediate Problem: Variable Speed
“A man drives from City A to City B (120 km) at 40 km/h. On the return journey, he drives at 60 km/h. What’s his average speed for the entire trip?”
Wrong Approach: Average = (40 + 60)/2 = 50 km/h
Why it’s wrong: Average speed is NOT the average of speeds. It’s total distance divided by total time.
Correct Solution:
Distance from A to B = 120 km
Time taken (A to B) = 120/40 = 3 hours
Distance from B to A = 120 km
Time taken (B to A) = 120/60 = 2 hours
Total Distance = 240 km
Total Time = 5 hours
Average Speed=2405=48 km/h\text{Average Speed} = \frac{240}{5} = 48 \text{ km/h}Average Speed=5240=48 km/h
This is slightly less than 50 km/h because the man spends more time driving at the slower speed.
Advanced Concept: Relative Speed
This appears frequently in placement tests because it mimics real-world chase and meeting problems.
Relative Speed When Moving in Same Direction:
Relative Speed=∣V1−V2∣\text{Relative Speed} = |V_1 – V_2|Relative Speed=∣V1−V2∣
Relative Speed When Moving Towards Each Other:
Relative Speed=V1+V2\text{Relative Speed} = V_1 + V_2Relative Speed=V1+V2
Practical Example: Two Trains Meeting
Train A travels from Station X towards Station Y at 60 km/h. Train B travels from Station Y towards Station X at 40 km/h. The distance between stations is 500 km. When will they meet?
Solution:
Since trains move towards each other:
Relative Speed = 60 + 40 = 100 km/h
Time to Meet=500100=5 hours\text{Time to Meet} = \frac{500}{100} = 5 \text{ hours}Time to Meet=100500=5 hours
Verification:
In 5 hours, Train A covers: 60 × 5 = 300 km
In 5 hours, Train B covers: 40 × 5 = 200 km
Total distance covered: 300 + 200 = 500 km ✓
Section 8: Work and Time - Collaboration Mathematics
Why Work-Time Problems Are Placement Favorites
Companies love work-time problems because they test whether you understand productivity, efficiency, and team collaboration—skills critical for any organization.
Core Concept: Work as Fractions
If a person completes a job in 10 days, their work rate (productivity) is 1/10 of the job per day.
Formula:
Work Rate=1Time Taken to Complete\text{Work Rate} = \frac{1}{\text{Time Taken to Complete}}Work Rate=Time Taken to Complete1
Example: If Person A completes a task in 12 days, their work rate = 1/12
Combined Work Problem
“Person A completes a project in 12 days. Person B completes the same project in 18 days. If they work together, how many days will it take?”
Solution:
Person A’s work rate = 1/12 per day
Person B’s work rate = 1/18 per day
Combined work rate = 1/12 + 1/18
To add these fractions, find LCM of 12 and 18 = 36
112+118=336+236=536\frac{1}{12} + \frac{1}{18} = \frac{3}{36} + \frac{2}{36} = \frac{5}{36}121+181=363+362=365
If they complete 5/36 of work per day together, time to complete entire work:
Time=1 (entire work)536=365=7.2 days\text{Time} = \frac{1 \text{ (entire work)}}{\frac{5}{36}} = \frac{36}{5} = 7.2 \text{ days}Time=3651 (entire work)=536=7.2 days
Advanced Scenario: Work with Different Efficiency
“A does half the work done by B in the same time.” This statement means if B completes 1 unit of work, A completes 0.5 units.
Example Problem:
“A does half the work done by B in the same time. If A completes the project in 20 days, in how many days do they complete the project working together?”
Solution:
If A completes 1 unit, B completes 2 units in the same time.
So B is twice as efficient as A.
Time for A to complete project = 20 days
A’s work rate = 1/20 per day
B’s work rate = 2 × (1/20) = 2/20 = 1/10 per day
(Meaning B completes in 10 days alone)
Combined rate = 1/20 + 1/10 = 1/20 + 2/20 = 3/20
Time to complete together = 1 ÷ (3/20) = 20/3 ≈ 6.67 days