SSC GD 2026 Maths Top 501 Most Important Questions

Preparing for SSC GD 2026 requires a strong command of Mathematics, as it plays a key role in boosting your overall score. The SSC GD 2026 Maths Top 501 Most Important Questions are carefully selected to cover all high-weightage topics like Arithmetic, Algebra, Geometry, and Mensuration. 

Practising these questions helps candidates improve speed, accuracy, and confidence while understanding the exam pattern and frequently asked concepts.

SSC GD 2026 Maths Top 501 Most Important Questions

To solve the Top 501 Most Important Questions, first understand the approach and shortcuts from the video. For additional practice and clarity on key concepts, refer to the questions and explanations provided below.

SSC GD 2026 Maths Section

The Mathematics section of the SSC GD exam features 20 questions, with each question carrying 2 marks. There is negative marking. Mathematics is often a deciding factor for selection, as many applicants struggle here. A strong score in this section can determine rank and help secure preferred jobs. Due to negative marking, avoid random guesses.

High-Weightage Topics 

• Average: At least one, often two questions.

• Profit & Loss and Discount: Combined, these account for three questions.

• Time & Work

• Time, Speed & Distance

• Mensuration

• Compound Interest (CI) & Simple Interest (SI)

• Percentage

• Ratio & Proportions

Low-Weightage Topics (Yielding "Free Marks")

These topics, though lower in weightage, often provide easy, guaranteed marks.

• Number System: One question, consistently repeated.

• HCF & LCM: One question, consistently asked.

• Simplification: If BODMAS rule is known, one question is guaranteed.

• Power & Indices: Sometimes appears if Simplification is not present.

• Pipe & Cistern

• Boat & Stream

• Statistics & Data Interpretation (DI): (Mean, Mode, Median) Occasionally asked.

Candidates can refer to the most important topics with few examples provided below for your reference. 

SSC GD 2026 Maths Topic: Discount

Discount is a small topic that guarantees one free mark.

1. What Is Discount?

Discount is calculated on the Marked Price (MP), also known as the Maximum Retail Price (MRP).

• Definition: Discount is the difference between the Marked Price and the Selling Price (SP).

• Formula: Discount = Marked Price – Selling Price

2. How To Calculate Discount Percentage

• Formula: Discount % = (Discount / Marked Price) * 100

• Key Principle: Discount percentage is always calculated on the Marked Price.

Example Problem 1: Basic Discount Percentage Calculation

Problem: An item has a Marked Price of ₹650 and is sold for ₹572. What is the discount percentage?

Solution:

1. Discount Amount = ₹650 – ₹572 = ₹78

2. Discount % = (₹78 / ₹650) * 100 = 12%

3. Relation Between Selling Price, Marked Price, And Discount Percentage

This is a very important relationship.

• Formula: Selling Price (SP) = Marked Price (MP) * [(100 - Discount %) / 100]

• Explanation: If an article has a 20% discount, its selling price is 80% of its marked price (100-20 = 80).

Example Problem 2: Finding Marked Price From Selling Price And Discount %

Problem: An item is sold with a 14% discount for ₹387. Find its Marked Price.

Solution:

1. SP = MP * [(100 - Discount %) / 100]

2. 387 = MP * [(100 - 14) / 100] => 387 = MP * (86 / 100)

3. MP = (387 * 100) / 86 = ₹450

• (Memory Tip: When faced with calculations, consider using unit digits or digital sums for option elimination if applicable, especially when time is limited.)

Example Problem 3: Finding Selling Price With Discount And Approximation

Problem: Find the selling price of an item with a Marked Price of ₹8980 after a 19% discount, to the nearest rupee.

Solution:

1. SP = MP * [(100 - Discount %) / 100]

2. SP = 8980 * [(100 - 19) / 100] = 8980 * (81 / 100)

3. SP = 89.80 * 81 = 7273.8

4. To the nearest rupee, SP = ₹7274.

Example Problem 4: Conditional Discount Scheme

Problem: A retailer offers a 10% discount on purchases between ₹1000 and ₹5000, and a 20% discount on purchases above ₹5000. If a customer buys goods worth ₹6000, what is their saving compared to the original price?

Solution:

1. The purchase of ₹6000 falls under the "above ₹5000" category, so a 20% discount applies.

2. Savings = 20% of ₹6000 = (20 / 100) * 6000 = ₹1200.

• (Memory Tip: For such problems in an exam, quickly identify the applicable condition to save time. The length of the problem description does not always imply complexity.)

Recap Of Concepts

• Concept 1 (Discount Amount): Discount = Marked Price – Selling Price

• Concept 2 (Discount Percentage): Discount % = (Discount / Marked Price) * 100

• Concept 3 (SP, MP, Discount % Relation): SP = MP * [(100 - Discount %) / 100]

SSC GD 2026 Maths Topic: Successive Discounts

1. Formula For Two Successive Discounts

If two successive discounts are x% and y%, the effective or net discount is given by:

• Net Discount % = x + y - (xy / 100)

• (Memory Tip: This formula and concept are highly likely to appear in the exam.)

2. Ratio Method For Successive Discounts

The ratio method is useful for three or more successive discounts.

1. Convert Percentage to Fraction: P% = P/100. (e.g., 20% = 1/5 means 1 unit discount on 5 units MP).

2. Build Ratios (MP : SP): For a 1/5 discount, MP : SP = 5 : (5-1) = 5 : 4.

3. Multiply Ratios: For multiple discounts, multiply the MP values together and the SP values together. Cross-cancellation can simplify.

Example Problem 5: Effective Discount For Two Successive Discounts

Problem: What is the effective discount for two successive discounts of 20% and 25%?

Solution:

Method 1: Formula Method

1. Net Discount % = 20 + 25 - (20 * 25 / 100) = 45 - 5 = 40%.
   Method 2: Ratio Method

2. 20% = 1/5 (MP:SP = 5:4)

3. 25% = 1/4 (MP:SP = 4:3)

4. Multiply ratios: (5:4) * (4:3) = 5:3 (after cancelling 4).

5. Net Discount = (5-3)/5 * 100 = (2/5) * 100 = 40%.

Successive Discounts For Three Or More Discounts

For three or more successive discounts, the ratio method is generally more efficient.

• (Memory Tip: Choose the ratio method when dealing with three or more successive discounts or when the fractions are straightforward. Choose the successive formula when dealing with two discounts or when fractions are complicated.)

Discount On "Buy A, Get B Free" Offers

This applies to promotions like "Buy 2, Get 1 Free".

• Formula: Discount % = (Number of Free Articles / Total Number of Articles) * 100

• Example: "Buy 3 shirts, Get 2 shirts Free"

• Discount % = (2 / (3 + 2)) * 100 = (2 / 5) * 100 = 40%.

Relation Between Market Price, Cost Price, Profit/Loss, And Discount

This concept establishes a direct relationship between Market Price (MP) and Cost Price (CP).

• Formula: MP / CP = (100 + Profit %) / (100 - Discount %)

• (Use -Loss % if there is a loss instead of profit).

• Example Problem: An item is marked such that after a 20% discount, there is a 25% profit. How much above the cost price was it marked?
  Solution:

1. MP / CP = (100 + 25) / (100 - 20) = 125 / 80 = 25 / 16.

2. If CP is 16 units, MP is 25 units. Increase = 25 - 16 = 9 units.

3. Percentage increase = (9 / 16) * 100 = 56.25%.

SSC GD 2026 Maths Topic: Least Common Multiple (LCM) And Highest Common Factor (HCF)

Introduction To LCM

LCM stands for Least Common Multiple. It is the smallest positive integer that is a multiple of two or more given integers.

• Example: LCM of 12 and 36 is 36. (36 is the smallest number divisible by both 12 and 36).

Basic LCM Calculation & Property Review

• LCM Calculation (Prime Factorization Method): Divide numbers by prime factors until all are 1. Product of factors is LCM.

• Fundamental LCM Property: If K is the LCM of numbers a, b, and c, then K will always be divisible by a, b, and c.

Strategy: Option Elimination Using Divisibility (For LCM)

• Rule 1: Last Digit Divisibility: If any number has a last digit of zero, then their LCM must also end in zero.

• Rule 2: Digital Sum Divisibility: If the LCM must be divisible by a number like 9 (e.g., from 36 or 45), its digital sum must be divisible by 9.

Relationship Between Two Numbers, LCM, And HCF

• Fundamental Formula: For any two positive integers (First Number and Second Number), their product is always equal to the product of their LCM and HCF.

• First Number × Second Number = LCM × HCF

Example Problem: Finding Second Number

Problem: Given: LCM = 144, HCF = 8, First Number = 16. Find Second Number.

Solution: 16 × Second Number = 144 × 8 => Second Number = (144 × 8) / 16 = 72.

Example Problem: Sum Of Reciprocals

Problem: Sum of two numbers (a + b) = 15. HCF = 3, LCM = 18. Find the sum of their reciprocals (1/a + 1/b).

Solution:

1. Sum of reciprocals = (a + b) / (ab)

2. We know ab = LCM × HCF.

3. So, Sum of reciprocals = (a + b) / (LCM × HCF) = 15 / (18 × 3) = 15 / 54 = **5 / 18**.

HCF (Highest Common Factor)

• Definition: HCF is the largest number that divides two or more numbers exactly.

• Method: Prime Factorization: Take the lowest power of each common prime factor.

• Example: HCF of 42 (2×3×7), 60 (2²×3×5), 72 (2³×3²) is 2¹×3¹ = 6.

Fundamental HCF Property (And Option Elimination Strategy)

• If K is the HCF of numbers a, b, and c, then K will always divide a, b, and c.

• Option Elimination: Check which options divide all given numbers, then choose the largest.

HCF Of Numbers With Powers

For numbers expressed as products of prime powers (e.g., 2^a * 3^b * 5^c), for each common prime factor, take the lowest power.

• Example: HCF of (2² × 3⁵ × 5³) and (2⁴ × 3² × 5¹) is 2² × 3² × 5¹.

Example Problem: Ratio Of LCM To HCF

Problem: Find the ratio of LCM to HCF for 42 and 49.

Solution:

1. HCF(42, 49) = 7

2. LCM(42, 49) = 294

3. Ratio = LCM : HCF = 294 : 7 = 42 : 1.

Relation Between HCF, LCM, And Ratio Of Numbers

This is a very important topic.

• If two numbers are in the ratio A : B (coprime), and their HCF is H, then:

• LCM = H × A × B

• First Number = A × H

• Second Number = B × H

• (Memory Tip: Any number derived from or related to a given HCF will always be a multiple of that HCF. This is a powerful tool for option elimination.)

Example Problem: Finding Larger Number

Problem: Two numbers are in the ratio 13 : 9. Their HCF = 13. Find the larger number.

Solution:

1. First Number = 13 × 13 = 169

2. Second Number = 9 × 13 = 117

3. The larger number is 169.

SSC GD 2026 Maths Topic: Calculating HCF And Numbers From Ratio And LCM

The relationship: LCM = HCF × a × b (where a:b is the ratio).

Example Problem: Finding HCF And Numbers

Problem: Ratio of two numbers = 3:5; LCM = 240. Find HCF and the two numbers.

Solution:

1. Calculate HCF: 240 = HCF × 3 × 5 => HCF = 240 / 15 = 16.

2. Numbers: 16 × 3 = 48 and 16 × 5 = 80.

SSC GD 2026 Maths Topic: Prime Numbers And Their Application In LCM/HCF

• Definition: A prime number is a natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself (e.g., 2, 3, 5, 7).

• Important Concept: If two numbers are prime, their LCM is their product.

Example Problem: LCM Of Prime Numbers

Problem: Two prime numbers x and y (where x < y). Their LCM = 141. Find y - 5x.

Solution:

1. Since x and y are prime, their product is 141. Prime factorization of 141 = 3 × 47.

2. So, x = 3, y = 47.

3. y - 5x = 47 - (5 × 3) = 47 - 15 = 32.

SSC GD 2026 Maths Topic: HCF And Sum Of Numbers With Ratios

If numbers are in ratio a:b:c… and their HCF is K, then numbers are a×K, b×K, c×K.

Example Problem: Sum Of Numbers

Problem: Three numbers in the ratio 4:3:7; HCF = 6. Find the sum of the three numbers.

Solution:

Method 1: Individual Numbers

1. Numbers: 4×6 = 24, 3×6 = 18, 7×6 = 42.

2. Sum: 24 + 18 + 42 = 84.
   Method 2: Sum of Ratio Parts

3. Sum of ratio parts = 4 + 3 + 7 = 14.

4. Total sum = 14 × HCF = 14 × 6 = 84.

SSC GD 2026 Maths Topic: HCF Of Powers

Example Problem: HCF Of 16³ And 48²

Problem: Find the HCF of 16³ and 48².

Solution:

1. 16³ = 16 × 16 × 16

2. 48² = (16 × 3) × (16 × 3) = 16 × 16 × 3 × 3

3. Highest common factors are 16 × 16.

4. HCF = 16 × 16 = 256.

SSC GD 2026 Maths Topic: Finding Smallest Numbers With A Remainder

To find the smallest number that leaves a specific remainder (R) when divided by several numbers (N1, N2, N3…):

• Required number = LCM (N1, N2, N3…) + R

Example Problem: Smallest 4-Digit Number With Remainder 9

Problem: Find the smallest 4-digit number that leaves a remainder of 9 when divided by 10, 12, 15, 24, 30.

Solution:

1. LCM (10, 12, 15, 24, 30) = 120.

2. Smallest 4-digit multiple of 120: 1000 / 120 ≈ 8.33, so 9 × 120 = 1080.

3. Required number = 1080 + 9 = 1089.

SSC GD 2026 Maths Topic: Applications Of HCF (Finding Maximum Quantities)

• (Memory Tip: When a problem asks for the maximum, greatest, or highest common quantity, calculate the HCF. When it asks for the minimum, least, or smallest, calculate the LCM.)

Example Problem: Distribution Of Cakes And Chocolates

Problem: 16 cake pieces and 40 chocolates are to be distributed among children such that each child receives an equal number of both. Find the maximum number of children.

Solution: This is a "maximum distribution" problem, so calculate the HCF of 16 and 40.

HCF(16, 40) = 8. Maximum children = 8.

SSC GD 2026 Maths Topic: LCM And HCF Of Fractions And Decimals

A fraction is a numerical representation a/b.

Rules for LCM and HCF of Fractions:

1. LCM of Fractions: LCM = (LCM of Numerators) / (HCF of Denominators)

2. HCF of Fractions: HCF = (HCF of Numerators) / (LCM of Denominators)

Example Problem: LCM And HCF Of Reciprocals

Problem: Find the LCM and HCF of the reciprocals of 18 and 24.

Solution: Reciprocals are 1/18 and 1/24.

1. LCM (1/18, 1/24) = LCM(1, 1) / HCF(18, 24) = 1 / 6 = 1/6.

2. HCF (1/18, 1/24) = HCF(1, 1) / LCM(18, 24) = 1 / 72 = 1/72.

Product Of LCM And HCF Of Fractions

Example Problem: Product Of LCM And HCF

Problem: Find the product of the LCM and HCF of the fractions: 4/5, 9/15, 3/25, 2/10, 2/3.

Solution:

1. LCM of Fractions: Numerators (4, 9, 3, 2, 2) => LCM = 36. Denominators (5, 15, 25, 10, 3) => HCF = 1.
   LCM = 36/1 = 36.

2. HCF of Fractions: Numerators (4, 9, 3, 2, 2) => HCF = 1. Denominators (5, 15, 25, 10, 3) => LCM = 150.
   HCF = 1/150.

3. Product (LCM × HCF) = 36 × (1/150) = 36/150 = 6/25.

• (Note: Original lecture stated LCM as 252 for numerators (4,9,3,2,2) and simplified it from there. Recalculation shows LCM(4,9,3,2,2) = 36. Following the correct calculation: 36 * (1/150) = 6/25)

LCM And HCF Of Decimals

To calculate the LCM or HCF of decimal numbers, convert them into fractions with a common denominator (power of 10).

Example Problem: HCF Of Decimals

Problem: Find the HCF of 0.05, 0.5, 0.25, 1.25.

Solution:

1. Convert to fractions with denominator 100: 5/100, 50/100, 25/100, 125/100.

2. HCF = (HCF of Numerators (5, 50, 25, 125)) / (LCM of Denominators (100, 100, 100, 100))

3. HCF(5, 50, 25, 125) = 5. LCM(100, 100, 100, 100) = 100.

4. HCF = 5/100 = 0.05.

Distinguishing LCM And HCF For Minimum/Maximum Scenarios

Example Problem: Minimum Number Of Soldiers

Problem: What is the minimum number of soldiers required to arrange them in rows of 12, 15, 16, and 18?

Solution: This requires finding the LCM of 12, 15, 16, 18.

LCM(12, 15, 16, 18) = 720.

Application Of LCM: Total Interval

When problems involve events occurring at different intervals and ask when they will occur together again, find the LCM of the individual intervals.

Example Problem: Bells Ringing Together

Problem: Four bells ring at intervals of 6, 8, 9, and 10 seconds. After how many minutes will they ring together again?

Solution:

1. LCM(6, 8, 9, 10) = 360 seconds.

2. Convert to minutes: 360 / 60 = 6 minutes.

Relationship Between LCM And HCF (Theoretical Question)

The LCM of two or more numbers must always be a multiple of their HCF. Any number that is not a multiple of the HCF cannot be the LCM.

• (Memory Tip: The LCM can never be a number that is not a multiple of the HCF.)

LCM And HCF With Ratios

Example Problem: Product Of Two Numbers From Ratio And LCM

Problem: Two numbers are in the ratio 2:7. Their LCM is 42. Find the product of the two numbers.

Solution:

1. Let numbers be 2x and 7x. HCF = x. LCM = 14x.

2. Given LCM = 42, so 14x = 42 => x = 3.

3. Numbers are 2*3 = 6 and 7*3 = 21.

4. Product = 6 × 21 = 126.

• Alternatively, using Product = LCM × HCF: 42 × 3 = 126.

Introduction To Simple Interest (SI)

Simple Interest is a type of interest calculation where the interest amount remains the same every year on the principal amount.

Key Terms In Simple Interest

• Principal (P): Initial amount invested/borrowed.

• Rate of Interest (R): Percentage per annum.

• Time (T): Duration, usually in years.

• Simple Interest (SI): Interest earned/paid.

• Amount (A): Total sum (Principal + SI).

Simple Interest Formula

• SI = (P × R × T) / 100

Shortcut Method For Simple Interest Calculation

• (Memory Tip: To quickly calculate Simple Interest: Calculate the Simple Interest Percentage = R × T. Then, Simple Interest = Principal × (Simple Interest Percentage / 100).)

Example Problem: Basic Simple Interest Calculation

Problem: Find the simple interest on ₹2,00,000 at 7% for 2 years.

Solution:

1. SI Percentage = 7% × 2 = 14%.

2. SI = 14% of ₹2,00,000 = (14/100) * 2,00,000 = ₹28,000.

3. Amount = ₹2,00,000 + ₹28,000 = ₹2,28,000.

Calculating Simple Interest Percentage

1. Multiply the rate (R) by the time (T).

Calculating Simple Interest For A Principal

1. Determine the simple interest percentage (R*T).

2. Multiply the principal amount by this percentage.

Converting Days To Years For Simple Interest

When time is in days, convert to years by dividing by 365.

• (Memory Tip: 73 days = 1/5 year.)

Example Problem: Simple Interest With Days Converted To Years

Problem: Calculate SI for P = ₹48,750, R = 16%, T = 73 days.

Solution:

1. Time (T) = 73 / 365 = 1/5 year.

2. SI = (48750 * 16 * (1/5)) / 100 = (48750 * 16) / 500 = ₹1560.

Calculating Duration Between Two Dates For Simple Interest

To calculate days between two dates, count days in each month carefully.

• Example: Feb 7, 2022 to April 20, 2022 (2022 is not a leap year).

• Feb: (28 - 7) + 1 = 22 days.

• Mar: 31 days.

• Apr: 20 days.

• Total = 22 + 31 + 20 = 73 days (1/5 year).

Example Problem: Detailed Simple Interest Calculation (Date Range)

Problem: Calculate SI for P = ₹2,000, R = 8.25%, for Feb 7, 2022 to April 20, 2022.

Solution:

1. Time (T) = 73 days = 1/5 year.

2. SI = (2000 * 8.25 * (1/5)) / 100 = (2000 * 8.25) / 500 = ₹33.

Simple Interest Application Problem (Down Payment Scenario)

Problem: A bike costs ₹60,000. Down payment = ₹10,000. Remaining amount at 15% SI for 2 years. Calculate SI to be paid.

Solution:

1. Remaining Principal (P) = ₹60,000 - ₹10,000 = ₹50,000.

2. Total SI Percentage = R * T = 15% * 2 = 30%.

3. SI = 30% of ₹50,000 = (30/100) * 50,000 = ₹15,000.