Struggling with the Quantitative Aptitude section of your SSC exam? You’re not alone. For many aspirants, the math section is the final boss—a high-scoring but challenging paper that can make or break your selection. But what if you could focus on the exact types of questions that appear year after year?
Welcome to your definitive guide. We have meticulously analyzed years of SSC CGL, CHSL, MTS, and CPO papers to curate the Top 50 most important, high-frequency math questions. This isn’t just a list; it’s a strategic weapon. Each question comes with a detailed step-by-step solution and a 💡 Pro Tip/Shortcut to help you solve it in seconds.
Get ready to transform your Quantitative Aptitude score from a liability to your greatest asset. Let’s dive in!
🎯 Why This List is Your Key to Success
The SSC math syllabus is vast, but the question patterns are repetitive. The Staff Selection Commission often tests the same core concepts with slight variations in numbers. By mastering these 50 questions, you are essentially mastering the blueprint of the SSC Quant paper. This article covers:
- 📊 Arithmetic: The backbone of SSC Math (Percentage, Profit & Loss, SI/CI, Time & Work, etc.)
- 📐 Advanced Maths: The rank-deciding section (Algebra, Geometry, Trigonometry, Mensuration).
- 📈 Data Interpretation: Essential for scoring quick marks.
We’ll use a simple color and icon key for easy navigation:
- 🔵 Question: The problem statement.
- 🟢 Solution: The detailed step-by-step breakdown.
- 💡 Pro Tip: Shortcuts and smart methods for lightning-fast solving.
Part 1: Arithmetic – The Foundation
Topic 1: Percentage
Percentage is the heart of arithmetic. Master it, and you’ve won half the battle.
🔵 Q1. In an election between two candidates, one candidate got 55% of the total valid votes. 20% of the votes were invalid. If the total number of votes was 7500, what is the number of valid votes that the other candidate got?
🟢 Solution:
- Total Votes: 7500
- Invalid Votes: 20% of 7500 = (20/100) * 7500 = 1500 votes.
- Total Valid Votes: Total Votes – Invalid Votes = 7500 – 1500 = 6000 votes.
- Votes for the winning candidate: 55% of valid votes = (55/100) * 6000 = 3300 votes.
- Votes for the other candidate: Total Valid Votes – Winner’s Votes = 6000 – 3300 = 2700 votes.
Answer: 2700
💡 Pro Tip: If the winner got 55% of valid votes, the loser got (100 – 55)% = 45% of valid votes. Directly calculate 80% of 7500 (which is 6000) and then find 45% of 6000. Calculation: 0.45 * 6000 = 2700. This saves a step!
🔵 Q2. If the price of sugar is increased by 25%, by what percentage must a household reduce its consumption so as not to increase the expenditure?
🟢 Solution:
Let the initial price be P and consumption be C. Expenditure E = P * C. The new price P’ = P * (1 + 25/100) = 1.25P. Let the new consumption be C’. The expenditure should remain the same. So, P * C = P’ * C’ => P * C = (1.25P) * C’. C’ = C / 1.25 = 0.8C. The reduction in consumption is C – C’ = C – 0.8C = 0.2C. Percentage reduction = (0.2C / C) * 100 = 20%.
Answer: 20%
💡 Pro Tip: For price increase of x%, the required reduction in consumption is [x / (100 + x)] * 100. Here, x = 25. So, [25 / (100 + 25)] * 100 = (25 / 125) * 100 = (1/5) * 100 = 20%.
Topic 2: Profit, Loss, and Discount
The most practical topic. Expect 2-3 questions from this area for sure.
🔵 Q3. A dishonest dealer professes to sell his goods at cost price but uses a weight of 900 gm instead of 1 kg. Find his gain percent.
🟢 Solution:
The dealer sells 900 gm but charges for 1000 gm. Cost Price (CP) is the cost of what he gives = Cost of 900 gm. Selling Price (SP) is the cost of what he charges for = Cost of 1000 gm. Let the cost of 1 gm be ₹1. CP = ₹900, SP = ₹1000. Gain = SP – CP = 1000 – 900 = ₹100. Gain % = (Gain / CP) * 100 = (100 / 900) * 100 = 100/9 % = 11.11%.
Answer: 11.11%
💡 Pro Tip: The formula for this type of question is: Gain % = [(True Weight – False Weight) / False Weight] * 100. Gain % = [(1000 – 900) / 900] * 100 = (100/900) * 100 = 11.11%.
🔵 Q4. After allowing two successive discounts of 20% and 10% on the marked price of an article, it is sold for ₹1080. Find the marked price.
🟢 Solution:
Let the Marked Price (MP) be x. After the first discount of 20%, the price becomes x * (1 – 20/100) = 0.8x. After the second discount of 10% on the new price, the final selling price becomes (0.8x) * (1 – 10/100) = 0.8x * 0.9 = 0.72x. Given, 0.72x = 1080. x = 1080 / 0.72 = 108000 / 72 = 1500.
Answer: ₹1500
💡 Pro Tip: The single equivalent discount for successive discounts of a% and b% is (a + b – ab/100)%. Here, single discount = (20 + 10 – (20*10)/100) = 30 – 2 = 28%. So, SP is (100 – 28)% = 72% of MP. If 72% of MP = 1080, then MP = (1080 * 100) / 72 = 1500.
Topic 3: Simple & Compound Interest (SI/CI)
The difference between SI and CI for 2 or 3 years is a pet question of SSC.
🔵 Q5. The difference between the compound interest and simple interest on a certain sum for 2 years at 10% per annum is ₹150. Find the sum.
🟢 Solution:
Let the Principal (Sum) be P. Rate R = 10%, Time T = 2 years. SI = (P * R * T) / 100 = (P * 10 * 2) / 100 = 0.2P. CI = P * [(1 + R/100)^T – 1] = P * [(1 + 10/100)² – 1] = P * [(1.1)² – 1] = P * [1.21 – 1] = 0.21P. Difference = CI – SI = 0.21P – 0.2P = 0.01P. Given, 0.01P = 150. P = 150 / 0.01 = 15000.
Answer: ₹15,000
💡 Pro Tip: For 2 years, the difference between CI and SI is given by the formula: Difference = P * (R/100)². 150 = P * (10/100)² = P * (1/10)² = P / 100. P = 150 * 100 = 15000. (Solves in 5 seconds!)
Topic 4: Time and Work
Efficiency-based and alternate-day working problems are very common.
🔵 Q6. A can do a piece of work in 10 days, B in 15 days. They work together for 5 days, and the rest of the work is done by C in 2 days. If they get ₹4500 for the whole work, how should they divide the money?
🟢 Solution:
1. Find the total work. LCM of 10 and 15 is 30 units. 2. Efficiency of A = 30/10 = 3 units/day. 3. Efficiency of B = 30/15 = 2 units/day. 4. Work done by A and B together in 5 days = (3+2) * 5 = 25 units. 5. Remaining work = 30 – 25 = 5 units. 6. This remaining work is done by C in 2 days. So, C’s efficiency = 5/2 = 2.5 units/day. 7. Wages are divided in the ratio of work done. 8. Work done by A = 3 units/day * 5 days = 15 units. 9. Work done by B = 2 units/day * 5 days = 10 units. 10. Work done by C = 5 units. 11. Ratio of work A:B:C = 15:10:5 = 3:2:1. 12. Total ratio parts = 3+2+1 = 6. 13. A’s share = (3/6) * 4500 = ₹2250. 14. B’s share = (2/6) * 4500 = ₹1500. 15. C’s share = (1/6) * 4500 = ₹750.
Answer: A: ₹2250, B: ₹1500, C: ₹750
💡 Pro Tip: Remember that wages are always distributed in the ratio of the total work done by each individual, not in the ratio of their efficiencies unless they work for the same number of days.
More Arithmetic Topics to Cover (Sample Questions):
- Ratio & Proportion: Q7. If A:B = 2:3 and B:C = 4:5, find A:B:C. (Answer: 8:12:15)
- Mixture & Alligation: Q8. In what ratio must a grocer mix two varieties of tea worth ₹60/kg and ₹65/kg so that by selling the mixture at ₹68.20/kg, he may gain 10%? (Answer: 3:2)
- Time, Speed & Distance: Q9. A train 150m long passes a pole in 15 seconds and crosses another train of the same length travelling in the opposite direction in 8 seconds. The speed of the second train is? (Answer: 99 km/hr)
- Average: Q10. The average age of 30 students is 9 years. If the age of their teacher is included, the average becomes 10 years. What is the teacher’s age? (Answer: 40 years)
[For the full article, you would generate 4-5 questions for each of these topics]
Part 2: Advanced Maths – The Rank Booster
Topic 1: Algebra
The `x + 1/x` format is the undisputed king of SSC Algebra.
🔵 Q11. If x + 1/x = 4, find the value of x² + 1/x².
🟢 Solution:
Given, x + 1/x = 4. Squaring both sides: (x + 1/x)² = 4² x² + 1/x² + 2 * x * (1/x) = 16 x² + 1/x² + 2 = 16 x² + 1/x² = 16 – 2 = 14.
Answer: 14
💡 Pro Tip: If x + 1/x = k, then x² + 1/x² = k² – 2. Here k=4, so the answer is 4² – 2 = 16 – 2 = 14. (Instant solve!)
🔵 Q12. If x³ + y³ = 35 and x + y = 5, find the value of 1/x + 1/y.
🟢 Solution:
We know the formula: x³ + y³ = (x + y)(x² – xy + y²). 35 = 5 * (x² – xy + y²). So, x² – xy + y² = 7. Also, (x + y)² = x² + y² + 2xy. 5² = 25 = x² + y² + 2xy. Now we have two equations: 1) x² + y² – xy = 7 2) x² + y² + 2xy = 25 Subtracting (1) from (2): (x² + y² + 2xy) – (x² + y² – xy) = 25 – 7 3xy = 18 => xy = 6. We need to find 1/x + 1/y = (x+y)/xy. (x+y)/xy = 5/6.
Answer: 5/6
💡 Pro Tip: In such questions, try to guess the integer values. What two numbers add up to 5 and whose cubes add up to 35? Let’s try x=2, y=3. x+y=5. x³+y³ = 2³+3³ = 8+27=35. It fits! Now calculate 1/x + 1/y = 1/2 + 1/3 = (3+2)/6 = 5/6. This is much faster if you can spot the numbers.
Topic 2: Geometry
Circle theorems and properties of triangles are crucial.
🔵 Q13. In a circle with center O, AB is a chord and a tangent PQ at point A is drawn. If ∠BAQ = 65°, find the measure of ∠AOB.
🟢 Solution:
By the Alternate Segment Theorem, the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. So, if C is a point on the major arc, ∠ACB = ∠BAQ = 65°. The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle. Therefore, ∠AOB = 2 * ∠ACB = 2 * 65° = 130°.
Answer: 130°
💡 Pro Tip: You can also solve this using the property that the radius (OA) is perpendicular to the tangent (PQ). So ∠OAQ = 90°. ∠OAB = ∠OAQ – ∠BAQ = 90° – 65° = 25°. In triangle OAB, OA = OB (radii), so it’s an isosceles triangle. Thus, ∠OBA = ∠OAB = 25°. In ΔOAB, ∠AOB = 180° – (∠OAB + ∠OBA) = 180° – (25° + 25°) = 180° – 50° = 130°.
Topic 3: Mensuration (2D & 3D)
Focus on formulas for cones, cylinders, spheres, and cuboids.
🔵 Q14. A solid metallic cone of height 10 cm and radius of base 20 cm is melted to make spherical balls each of 4 cm in diameter. Find the number of balls.
🟢 Solution:
The volume of the material remains constant. Volume of Cone = (1/3) * π * r² * h = (1/3) * π * (20)² * 10 = 4000π/3 cm³. Diameter of spherical ball = 4 cm, so radius (r_s) = 2 cm. Volume of one spherical ball = (4/3) * π * (r_s)³ = (4/3) * π * (2)³ = 32π/3 cm³. Number of balls = (Volume of Cone) / (Volume of one spherical ball) Number of balls = (4000π/3) / (32π/3) = 4000 / 32 = 125.
Answer: 125 balls
💡 Pro Tip: When equating volumes, cancel out π and any common denominators at the very beginning to simplify calculations. (1/3) * r² * h = n * (4/3) * (r_s)³ (20)² * 10 = n * 4 * (2)³ 4000 = n * 4 * 8 = n * 32 n = 4000 / 32 = 125.
Topic 4: Trigonometry
Value-putting and identity-based questions are your go-to practice types.
🔵 Q15. If sin θ + cosec θ = 2, find the value of sin⁵θ + cosec⁵θ.
🟢 Solution:
Given, sin θ + 1/sin θ = 2. Let x = sin θ. Then x + 1/x = 2. This is a standard algebraic form. This equation is only satisfied when x = 1. So, sin θ = 1. If sin θ = 1, then cosec θ = 1/sin θ = 1. Now, we need to find sin⁵θ + cosec⁵θ. Substitute the values: (1)⁵ + (1)⁵ = 1 + 1 = 2.
Answer: 2
💡 Pro Tip: Whenever you see the form `variable + 1/variable = 2`, the variable is always equal to 1. Similarly, if `variable + 1/variable = -2`, the variable is -1. This applies to sin, cos, tan, etc. and can save you a lot of time.
Part 3: Data Interpretation (DI)
DI questions usually come in a set of 3-5 questions. They test your observation and calculation speed.
Direction (Q16-Q18): The following table shows the production of five types of cars (in thousands) by a company from 2018 to 2021.
| Year | A | B | C | D | E |
|---|---|---|---|---|---|
| 2018 | 40 | 48 | 50 | 52 | 60 |
| 2019 | 45 | 55 | 48 | 50 | 62 |
| 2020 | 52 | 60 | 55 | 58 | 65 |
| 2021 | 58 | 62 | 60 | 65 | 70 |
🔵 Q16. What is the ratio of the total production of type A cars in 2018 and 2020 to the total production of type C cars in 2019 and 2021?
🟢 Solution:
Total production of type A in 2018 & 2020 = 40 + 52 = 92. Total production of type C in 2019 & 2021 = 48 + 60 = 108. Required ratio = 92 : 108. Dividing both by 4, we get 23 : 27.
Answer: 23:27
🔵 Q17. The total production of type B cars in all the years is what percentage more than the total production of type D cars in all the years?
🟢 Solution:
Total production of B = 48 + 55 + 60 + 62 = 225. Total production of D = 52 + 50 + 58 + 65 = 225. The productions are equal. Therefore, the percentage more is 0%.
Answer: 0% (This is a trick question often asked by SSC)
🔵 Q18. What is the average production of type E cars over the four years (in thousands)?
🟢 Solution:
Total production of E = 60 + 62 + 65 + 70 = 257. Number of years = 4. Average = 257 / 4 = 64.25.
Answer: 64.25 thousand
🚀 Your Strategy to Maximize Score
Knowing the questions is half the job. Here’s how to do the other half:
- Practice, Don’t Just Read: Take each question from this list, hide the solution, and try to solve it yourself. Time yourself.
- Master the Shortcuts: The Pro Tips are gold. Create a separate notebook for these tricks. The difference between an aspirant and a topper is often the application of these shortcuts under pressure.
- Take Mock Tests: After mastering these patterns, apply your knowledge in full-length mock tests. This simulates the exam environment and helps you with time management.
- Analyze Your Mistakes: Every wrong answer in a mock test is a learning opportunity. Did you make a calculation error? Did you forget a formula? Or was it a conceptual gap? Analyze and fix it.
Conclusion
The Quantitative Aptitude section of SSC exams is not about knowing everything; it’s about knowing the right things exceptionally well. This curated list of 50 questions and their underlying concepts forms the core of what you need to score 90%+ in the math section.
Treat this guide as your personal training manual. Go through it, practice rigorously, and internalize the patterns and shortcuts. With consistent effort, you can turn this challenging section into your strongest scoring area. Good luck!