Maths for Competitive Exams: Concept, Syllabus, & Questions

While preparing for a competitive exam like GRE, GMAT or SAT, it is common to feel panicked about the mathematical section which is a complex one to crack. The main purpose of adding the category of Quantitative Aptitude to a competitive exam is to test your ability to solve basic mathematical problems logically. For this, a good grasp of quantitative concepts with sufficient practice is needed.

If you are planning to appear for a competitive test, you have come to the right place. In this blog, we will elaborate on the key concepts of Maths for competitive exams along with helpful tips, tricks and math books for competitive exams you can employ to score better.

Syllabus Coverage for Competitive Exams 2026-27

Mathematics is one of the core sections in almost every competitive exam, including SSC, Banking, Railways, Defence, Insurance, and Management Entrance Tests. The syllabus is designed to test a candidate’s conceptual clarity, logical reasoning, and problem-solving speed. Each chapter plays a vital role in building a strong foundation and developing strategies to handle moderate to advanced-level questions.

The syllabus broadly covers the Number System, Divisibility Rules, HCF & LCM, Algebra, Quadratic Equations, Average, Profit and Loss, Ratio and Proportion, Time-Speed-Distance, Time and Work, Simple and Compound Interest, Geometry and Mensuration, Trigonometry, Statistics, and Percentage. These topics not only appear individually but are also integrated into data interpretation, arithmetic reasoning, and higher-order application-based questions. To understand these topics better, let us list them and then explore each in detail.

• Average
• Chain Rule
• Logarithm
• Boats and Streams
• Races and Games
• Bankers Discount
• Percentage
• Partnership
• Probability
• Time and Work
• Simple Interest
• Profit and Loss
• H.C.F and L.C.M
• Problems on Ages
• Stocks and Shares
• Area and Perimeter
• Alligation or Mixture
• Time and Distance
• Problems on Trains
• Height and Distance
• Volume and Surface Area
• Square Root and Cube Root
• Average Value and Calculation
• Verification of Truth Reasoning
• Odd Man Out Questions and Answers

Explore: General Aptitude Syllabus for all Competitive Exams

Concepts and Examples for Important Chapters of Maths in Competitive Exams

Concept and examples for the important chapter of the Maths competitive exams section covers the fundamental and advanced concepts of mathematics that are frequently tested in competitive exams. Each chapter is explained with clear definitions, formulas, and step-by-step examples to build strong problem-solving skills. The topics included range from the basics of the number system and divisibility rules to advanced chapters like algebra, quadratic equations, trigonometry, and statistics.

1. Number System

The number system forms the foundation of mathematics for competitive exams. It includes natural numbers, integers, rational and irrational numbers, and real numbers. Questions often test concepts such as unit digits, remainders, cyclicity, and divisibility. Strong knowledge of number properties helps solve complex problems quickly in exams like SSC, Banking, UPSC, and CAT. Let us now explore the core concepts of the number system in detail.

Classification of Numbers	Definition	Examples
Natural Numbers (N)	These are the basic counting numbers that start from 1.	1, 2, 3, 4, …
Whole Numbers (W)	Natural numbers along with zero.	0, 1, 2, 3, …
Integers (Z)	Whole numbers and their negative counterparts. They include positive, negative, and zero.	…, -3, -2, -1, 0, 1, 2, …
Rational Numbers (Q)	Numbers that can be expressed as a fraction p/q, where q ≠ 0. They include terminating and repeating decimals.	2/3, -5, 0.75, 0.333…
Irrational Numbers	Numbers that cannot be written as a fraction. Their decimal expansion is non-terminating and non-repeating.	√2, π, e
Real Numbers (R)	All rational and irrational numbers together. They represent every possible number on the number line.	All numbers on the number line

How number system is useful for competitive examination:

• Most exam questions in arithmetic, algebra, data interpretation, and simplification are built on number system concepts. Without this base, solving advanced problems becomes difficult.
• Competitive exams are time-bound. Knowing properties of numbers, divisibility rules, and shortcuts in fractions or decimals reduces calculation time.
• Questions related to LCM, HCF, percentages, ratios, probability, and equations use number system principles. Mastering this section makes other areas easier.
• Many exams include direct number system questions such as identifying even/odd numbers, prime numbers, or remainder-based problems. These can be quick marks if concepts are clear.
• A strong number system helps avoid common mistakes in decimals, fractions, and negative values, which improves overall accuracy in competitive tests.

2. Divisibility Rules for Competitive Exams

Divisibility rules provide shortcuts to identify whether a number can be divided by another without performing actual division. In competitive exams, these rules are applied in problems involving missing digits, remainders, and simplification. Mastering divisibility improves both speed and accuracy in quantitative aptitude sections. Since these rules form the base for many arithmetic problems, understanding them thoroughly is essential. Let us now go through the key rules of divisibility in detail.

Number	Divisibility Rules	Examples
2	A number is divisible if its last digit is even (0, 2, 4, 6, 8).	746 → last digit 6 → divisible by 2
3	A number is divisible if the sum of its digits is divisible by 3.	345 → 3+4+5=12 → divisible by 3
4	A number is divisible if the last two digits form a number divisible by 4.	3216 → last two digits 16 → divisible by 4
5	A number is divisible if its last digit is 0 or 5.	8905 → last digit 5 → divisible by 5
6	A number is divisible if it is divisible by both 2 and 3.	354 → even number and 3+5+4=12 divisible by 3
9	A number is divisible if the sum of its digits is divisible by 9.	837 → 8+3+7=18 → divisible by 9
11	A number is divisible if the difference of the sum of digits at odd and even places is divisible by 11.	2728 → (2+2) – (7+8) = -11 → divisible by 11

3. HCF and LCM 

Highest Common Factor (HCF) and Lowest Common Multiple (LCM) are widely used in questions related to ratios, remainders, and problem-solving involving cycles and repetitions. Competitive exams often include real-life applications such as arranging events, dividing items, or synchronisation problems.

Concept of HCF and LCM	Explanation	Examples
HCF (Highest Common Factor)	The greatest number that divides two or more numbers exactly. Also called Greatest Common Divisor (GCD).	HCF of 12 and 18 = 6
LCM (Least Common Multiple)	The smallest number that is a multiple of two or more given numbers.	LCM of 12 and 18 = 36
Important Formula	For two numbers:HCF × LCM = Product of the numbers	For 12 and 18:6 × 36 = 216 = 12 × 18

4. Algebra

Algebra is one of the most scoring topics in competitive mathematics. It covers expressions, equations, factorisation, and identities. Questions may range from simple substitutions to advanced simplifications and logical applications. A strong grasp of algebraic manipulation is crucial for exams like SSC CGL, RRB, CAT, and GRE. Algebra becomes easier to master when the basic identities are well understood. The following are the fundamental algebraic identities along with their expansions and examples.

Basic Algebraic Identities	Expansion	Examples
(a + b)²	a² + 2ab + b²	(x + 5)² = x² + 10x + 25
(a – b)²	a² – 2ab + b²	(2y – 3)² = 4y² – 12y + 9
a² – b²	(a + b)(a – b)	49 – 16 = (7 + 4)(7 – 4) = 33
(a + b)³	a³ + b³ + 3ab(a + b)	(x + 2)³ = x³ + 6x² + 12x + 8
(a – b)³	a³ – b³ – 3ab(a – b)	(p – 4)³ = p³ – 12p² + 48p – 64
a³ + b³	(a + b)(a² – ab + b²)	x³ + 8 = (x + 2)(x² – 2x + 4)
a³ – b³	(a – b)(a² + ab + b²)	27 – y³ = (3 – y)(9 + 3y + y²)

5. Quadratic Equations

Quadratic equations test both conceptual clarity and problem-solving ability. Questions may ask about the nature of roots, relationships between roots, or applications in word problems. In exams, quadratic equations are often framed to test both algebraic and logical thinking.

Concept of Quadratic Equations	List of Formulas	Questions	Solutions
Standard Form	ax² + bx + c = 0	2x² + 5x – 3 = 0	Already in standard form
Roots (solutions)	x = [−b ± √(b² – 4ac)] / 2a	Solve x² – 5x + 6 = 0	a = 1, b = –5, c = 6; x = [5 ± √(25 – 24)] / 2 = [5 ± 1]/2 x = 3, 2
Sum of roots	–b / a	x² – 5x + 6 = 0	Sum = –(–5)/1 = 5
Product of roots	c / a	x² – 5x + 6 = 0	Product = 6 / 1 = 6
Discriminant (D)	D = b² – 4ac	x² – 5x + 6 = 0	D = (–5)² – 4(1)(6) = 25 – 24 = 1
Nature of Roots	D > 0 → Real & distinctD = 0 → Real & equalD < 0 → Imaginary	x² – 4x + 4 = 0	D = 16 – 16 = 0 Roots real & equal

6. Average 

The concept of average (arithmetic mean) is commonly tested in competitive exams through questions on results, weights, and salaries. Many problems involve using averages to find missing values or to compare data sets. Candidates must learn shortcuts and properties to solve these efficiently.

1. Simple Average

Average is a commonly tested concept in competitive exams. It helps simplify a set of numbers to a single representative value. Questions on average appear in exams like SSC, Banking, Railways, and other aptitude sections.

Formula: Average = (Sum of Observations) ÷ (Number of Observations)

Example: Find the average of 10, 20, 30, 40.

Sum = 10 + 20 + 30 + 40 = 100, n = 4

Average = 100 ÷ 4 = 25

Importance for Exams:

• Directly tested in quantitative aptitude sections.
• Used in problems on speed, work, profit, and data interpretation.
• Quick calculation techniques save time in time-bound exams.

2. Weighted Average

Weighted average is used when observations have unequal importance. It is frequently tested in competitive exams in topics like marks distribution, mixtures, and population problems.

Formula: Weighted Average = (w₁x₁ + w₂x₂ + … + wₙxₙ) ÷ (w₁ + w₂ + … + wₙ)

Example: A student scores 80 in Maths (weight 3) and 70 in English (weight 2). Find the weighted average.

Weighted Average = (80×3 + 70×2) ÷ (3+2)

= (240 + 140) ÷ 5

= 380 ÷ 5 = 76

Importance for Exams:

• Directly applied in marks, population, and mixture-based questions.
• Helps solve real-life quantitative problems quickly.
• Saves time by avoiding repeated manual calculations of averages.

3. New Average (When a Number is Added or Removed)

New average is calculated when a number is added to or removed from a set of observations. This concept is frequently tested in competitive exams in questions related to marks, scores, or data adjustment.

Formulas:

• If the average of n numbers is A, and a new number x is added:

New Average = A + (x – A) ÷ (n + 1)

• If the average of n numbers is A, and a number x is removed:

            New Average = A – (x – A) ÷ (n – 1)

Example: The average of 5 numbers is 20. A new number 30 is added. Find the new average.

New Average = 20 + (30 – 20) ÷ 6

= 20 + 10 ÷ 6

= 21.67

7. Percentage

In competitive exams, questions on percentages test the ability to calculate proportions and changes quickly. These are widely asked in exams like SSC, Banking, Railway, CAT, and State PSC.

Concept of Percentage	List of Formulas	Questions	Solution
Basic Percentage	Percentage = (Value / Total) × 100	A student scored 45 out of 60 marks.	(45 / 60) × 100 = 75%
Percentage Change	Percentage Change = (New Value – Old Value) / Old Value × 100	Price of a book increased from ₹200 to ₹250.	(250 – 200) / 200 × 100 = 25%
Successive Percentage Change	Effective Change % = a + b + (a × b) / 100	Price increased by 10% in January and 20% in February.	10 + 20 + (10 × 20)/100 = 32%
Equal Increase & Decrease	Net Effect % = x² / 100	Price increased by 10% and then decreased by 10%.	(10²)/100 = 1% decrease

8. Profit, Loss, and Discount 

Profit, Loss and Discount topic has direct applications in commercial mathematics. Competitive exams test questions on cost price, selling price, discounts, successive profits, and marked price adjustments. Questions often combine percentage concepts, requiring quick calculations for accurate results.

Concept of Profit, Loss and Discount	List of Formulas	Questions	Solutions
Profit (%)	Profit (%) = [(S.P. – C.P.) / C.P.] × 100	C.P. = ₹500, S.P. = ₹600	[(600 – 500) / 500] × 100 = 20%
Loss (%)	Loss (%) = [(C.P. – S.P.) / C.P.] × 100	C.P. = ₹800, S.P. = ₹700	[(800 – 700) / 800] × 100 = 12.5%
Selling Price from Profit/Loss	S.P. = C.P. × (100 ± Profit%/Loss%) / 100	C.P. = ₹400, Profit = 25%	S.P. = 400 × (100 + 25)/100 = 400 × 1.25 = ₹500
Discount	Discount = M.P. – S.P.	M.P. = ₹1200, S.P. = ₹1000	Discount = 1200 – 1000 = ₹200
Discount %	Discount % = (Discount / M.P.) × 100	Discount = ₹200, M.P. = ₹1200	(200 / 1200) × 100 = 16.67%
S.P. after Discount	S.P. = M.P. × (100 – Discount%) / 100	M.P. = ₹1500, Discount = 20%	S.P. = 1500 × (100 – 20)/100 = 1500 × 0.8 = ₹1200

9. Ratio, Proportion, and Variation

Ratio and proportion are the backbone of many aptitude problems. Applications include mixtures, partnerships, direct and inverse variations, and real-life problem solving. Exams frequently include questions that integrate ratios with speed, time, and work problems, making it an essential topic for practice.

Concept of Ratio, Proportion and Variation	List of Formulas	Questions	Solutions
Ratio	a : b = a / b	Ratio of boys to girls in a class is 3:2, total students = 25. Find boys and girls.	Let total parts = 3 + 2 = 5; Boys = 25 × 3/5 = 15,  Girls = 25 × 2/5 = 10
Proportion	a : b :: c : d ⇒ a/b = c/d ⇒ a × d = b × c	If 2 : 3 :: x : 12, find x	2 × 12 = 3 × x ⇒ 24 = 3x ⇒ x = 8
Direct Variation	x ∝ y ⇒ x = k × y	y varies directly with x. If y = 12 when x = 4, find y when x = 10	k = y / x = 12 / 4 = 3;  y = 3 × 10 = 30
Inverse Variation	x ∝ 1/y ⇒ x = k / y	y varies inversely with x. If x = 8 when y = 6, find x when y = 12	k = x × y = 8 × 6 = 48;  x = 48 / 12 = 4

10. Time, Speed, and Distance

Time, Speed and Distance evaluates logical thinking and mathematical accuracy. It includes questions on trains, boats and streams, relative speed, and circular tracks. In competitive exams, problems are often framed in tricky ways to test concepts of average speed and time efficiency.

Concept of Time, Speed and Distance	List of Formulas	Questions	Solutions
Speed	Speed = Distance / Time	A car travels 150 km in 3 hours. Find speed.	Speed = 150 / 3 = 50 km/h
Time	Time = Distance / Speed	How long will it take to cover 200 km at 40 km/h?	Time = 200 / 40 = 5 hours
Distance	Distance = Speed × Time	A train travels at 60 km/h for 4 hours. Find distance.	Distance = 60 × 4 = 240 km
Relative Speed (same direction)	Relative Speed = S1 – S2	Two trains moving in same direction at 70 km/h and 50 km/h. Find relative speed.	Relative Speed = 70 – 50 = 20 km/h
Relative Speed (opposite direction)	Relative Speed = S1 + S2	Two trains moving opposite at 60 km/h and 40 km/h. Find relative speed.	Relative Speed = 60 + 40 = 100 km/h
Average Speed (same distance)	Average Speed = (2 × S1 × S2) / (S1 + S2)	A car covers two equal distances at 50 km/h and 70 km/h. Find average speed.	Average Speed = (2 × 50 × 70) / (50 + 70) = 7000 / 120 ≈ 58.33 km/h
Conversion (km/h → m/s)	Multiply by 5/18	Convert 90 km/h to m/s	90 × 5/18 = 25 m/s

Explore: Questions of Verification of Truth Reasoning

11. Time and Work

Time and Work is closely connected with efficiency and ratio concepts. Competitive exam questions involve men, women, and machines working together, pipes and cisterns, or alternate working patterns. Candidates need to master shortcut approaches for faster solving.

Concept of Time and Work	List of Formulas	Questions	Solutions
Work	Work = Rate × Time	A worker completes 120 units of work at 30 units/day. Find time taken.	Time = Work / Rate = 120 / 30 = 4 days
1 Day’s Work	If A can do work in n days → 1 day’s work = 1/n	A can finish a task in 6 days. 1 day’s work = ?	1 day’s work = 1/6 of total work
Time to Complete Work	If 1 day’s work = 1/n → Time = n days	A does 1/8 of work per day. Total time?	Time = 1 ÷ (1/8) = 8 days
Efficiency Ratio	If A is x times as efficient as B → Work Ratio = x:1	A is 3 times as efficient as B. Work done in a day ratio = ?	Work ratio A:B = 3:1
Time Ratio	If A is x times as efficient as B → Time Ratio = 1:x	A is 3 times as efficient as B. Time ratio = ?	Time ratio A:B = 1:3

12. Simple & Compound Interest

Interest-based questions are highly practical and frequently asked in banking, insurance, and finance-related exams. Simple Interest (SI) problems are straightforward, while Compound Interest (CI) requires understanding exponential growth, installments, and differences between SI and CI.

Concept of Simple & Compoun Interest	List of Formulas	Questions	Solutions
Simple Interest (SI)	SI = (P × R × T) / 100	Find SI on ₹10,000 at 5% per annum for 3 years.	SI = (10000 × 5 × 3)/100 = ₹1,500
Amount (SI)	A = P + SI = P[1 + (R × T)/100]	Amount for above example	A = 10000 + 1500 = ₹11,500
Compound Interest (CI)	A = P[1 + (R/100)]ᵀ	Find amount for ₹5,000 at 10% per annum for 2 years.	A = 5000[1 + 10/100]² = 5000 × 1.21 = ₹6,050
CI	CI = A – P = P{[1 + (R/100)]ᵀ – 1}	CI for above example	CI = 6,050 – 5,000 = ₹1,050
Half-Yearly Compounding	Rate = R/2%, Time = 2T	₹8,000 at 12% p.a. compounded half-yearly for 2 years	Rate = 12/2 = 6%, Time = 4 periods; A = 8000 × (1 + 0.06)⁴ = ₹10,118.30
Quarterly Compounding	Rate = R/4%, Time = 4T	₹10,000 at 8% p.a. compounded quarterly for 1 year	Rate = 8/4 = 2%, Time = 4; A = 10000 × (1 + 0.02)⁴ ≈ ₹10,824.32

13. Geometry & Mensuration

Two-dimensional (2D) shapes are flat figures with only length and breadth. In exams, questions focus on their area, perimeter, and properties, making them essential for geometry and mensuration. Following is the list of 2-dimensional shapes, their area, perimeter, and related question-answers.

Shape	Area	Perimeter	Questions	Solutions
Triangle	(1/2) × base × height	a + b + c	Triangle with base = 10 cm, height = 5 cm, sides = 10, 6, 8	Area = 0.5 × 10 × 5 = 25 cm², Perimeter = 10 + 6 + 8 = 24 cm
Rectangle	length × width	2(l + b)	l = 8 m, b = 5 m	Area = 8 × 5 = 40 m², Perimeter = 2(8+5) = 26 m
Square	side²	4 × side	side = 6 cm	Area = 6² = 36 cm², Perimeter = 4 × 6 = 24 cm
Parallelogram	base × height	2(a + b)	base = 12 m, height = 5 m, sides a=12, b=8	Area = 12 × 5 = 60 m², Perimeter = 2(12+8) = 40 m
Circle	πr²	2πr	r = 7 cm	Area = 3.1416 × 7² ≈ 153.94 cm², Circumference = 2 × 3.1416 × 7 ≈ 43.98 cm
Rhombus	(1/2) × d₁ × d₂	4 × side	d₁=10 cm, d₂=8 cm, side=6	Area = 0.5 × 10 × 8 = 40 cm², Perimeter = 4 × 6 = 24 cm

3D Shapes

Three-dimensional (3D) shapes are solid figures that have length, breadth, and height. In exams, questions focus on their surface area, volume, and properties, making them important for geometry and mensuration. Following is the list of 3-dimensional shapes, their surface area, volume, and related question-answers.

Shape	Volume	Curved or Lateral Surface Area (LSA)	Total Surface Area (TSA)	Examples	Solutions
Cube	a³	4a²	6a²	a = 5 cm	Volume = 5³ = 125 cm³, LSA = 4 × 25 = 100 cm², TSA = 6 × 25 = 150 cm²
Cuboid	l × b × h	2h(l + b)	2(lb + bh + hl)	l=4, b=3, h=5	Volume = 60, LSA =2×5(4+3)=70, TSA=2(12+15+20)=94
Sphere	(4/3)πr³	–	4πr²	r = 7 cm	Volume ≈ 1436.76 cm³, TSA ≈ 615.75 cm²
Cylinder	πr²h	2πrh	2πr(r + h)	r=3, h=10	Volume ≈ 282.74 cm³, LSA ≈ 188.4 cm², TSA ≈ 282.74 cm²
Cone	(1/3)πr²h	πrl	πr(r + l)	r=3, h=4, l=5	Volume ≈ 37.7 cm³, LSA ≈ 47.1 cm², TSA ≈ 75.4 cm²

14. Trigonometry 

Trigonometry applies to both theoretical and practical questions in competitive exams. It involves trigonometric identities, heights and distances, and simplifications. Exams often frame questions requiring a mix of identities, ratios, and applications in real-world contexts like angles and elevations.

Basic Ratios (SOH CAH TOA)

To solve trigonometric problems effectively, it is essential to understand the relationship between the sides of a right-angled triangle and its angles. This is explained through the basic ratios, commonly remembered by the term SOH CAH TOA.

Ratio	Formulas	Questions	Solutions
Sin θ	Opposite / Hypotenuse	Right triangle: Opposite = 3, Hypotenuse = 5	Sin θ = 3/5 = 0.6
Cos θ	Adjacent / Hypotenuse	Adjacent = 4, Hypotenuse = 5	Cos θ = 4/5 = 0.8
Tan θ	Opposite / Adjacent = Sin θ / Cos θ	Opposite = 3, Adjacent = 4	Tan θ = 3/4 = 0.75

Reciprocal Identities

Once the basic ratios are clear, the next step is to understand their reciprocal forms. Reciprocal identities establish direct relationships between trigonometric functions and make solving equations and simplifying expressions much easier.

Identity	Formulas	Questions	Solution
Cosec θ	1 / Sin θ	Sin θ = 0.6	Cosec θ = 1 / 0.6 ≈ 1.667
Sec θ	1 / Cos θ	Cos θ = 0.8	Sec θ = 1 / 0.8 = 1.25
Cot θ	1 / Tan θ	Tan θ = 0.75	Cot θ = 1 / 0.75 ≈ 1.333

Important Identities

Let us dive into the list of important identities with related questions and solutions.

Identity	Questions and Solutions
Sin²θ + Cos²θ = 1	If Sin θ = 3/5 → Cos²θ = 1 – (3/5)² = 1 – 9/25 = 16/25 → Cos θ = 4/5
1 + Tan²θ = Sec²θ	Tan θ = 3/4 → Sec²θ = 1 + (3/4)² = 1 + 9/16 = 25/16 → Sec θ = 5/4
1 + Cot²θ = Cosec²θ	Cot θ = 4/3 → Cosec²θ = 1 + (4/3)² = 1 + 16/9 = 25/9 → Cosec θ = 5/3

15. Statistics 

Statistics involves the collection, analysis, and interpretation of numerical data. In competitive exams, questions focus on mean, median, mode, variance, and standard deviation. Candidates must know shortcuts to calculate quickly, as data-based questions often come with time constraints.

Measures of Central Tendency & Dispersion

After building a foundation for an important chapter in basic mathematics, the next focus is on measures of central tendency and dispersion. These concepts help summarise data using averages, medians, and modes while also analysing how values spread around the central point.

Concept	Formula / Explan	Questions	Solution
Mean (Average)	Sum of all items ÷ Number of items	Data: 5, 8, 10, 7, 10	Mean = (5 + 8 + 10 + 7 + 10)/5 = 40/5 = 8
Median	Middle value after arranging in ascending/descending order	Data: 3, 7, 5, 9, 11 → Ascending: 3, 5, 7, 9, 11	n = 5 (odd), Median = (5+1)/2 = 3ᵗʰ term = 7
	For even n: Median = 1/2 [ (n/2)ᵗʰ term + ((n/2)+1)ᵗʰ term ]	Data: 4, 8, 6, 10 → Ascending: 4, 6, 8, 10	Median = 1/2 (6 + 8) = 7
Mode	Value that occurs most frequently	Data: 2, 5, 7, 5, 9	Mode = 5
Range	Highest value − Lowest value	Data: 4, 9, 15, 7, 12	Range = 15 − 4 = 11

Explore: Mastering Fractions and Decimals Questions

Maths Practice Questions for Competitive Exams

Practice is the most effective way to master mathematics for competitive exams. The ‘Maths Practice Questions’ section provides a wide range of questions covering all important topics such as Number System, Algebra, Geometry, Trigonometry, Statistics, Time and Work, and more. The questions are designed for all levels, entry-level, moderate, and advanced, to match the pattern of exams like SSC, Banking, Railway, and similar exams.

1. Number System 

1. Find the sum of all prime numbers between 1 and 50.
2. Find the remainder when 2^10 is divided by 7.
3. Determine whether 1729 is divisible by 7.
4. Find the sum of the first 20 natural numbers.
5. Express 0.625 as a fraction in simplest form.
6. Find the greatest number that divides 84 and 126 exactly.
7. If a number leaves a remainder 5 when divided by 9, find the remainder when twice the number is divided by 9.
8. If a number leaves remainder 4 when divided by 7, what is the remainder when the square of that number is divided by 7?
9. Determine the greatest number which divides 144, 216, and 288 leaving remainder 0.

2. Divisibility Rules

1. Is 354 divisible by 2, 3, 4, 5, 6, 9, or 11?
2. Find whether 2728 is divisible by 11.
3. Check if 837 is divisible by 9.
4. Find the smallest number divisible by both 4 and 5.
5. Determine if 8905 is divisible by 5.
6. Verify if 3216 is divisible by 4.
7. Find a number divisible by 2, 3, and 6.
8. The sum of digits of a number is 27. Is it divisible by 9?
9. Check whether 2468 is divisible by 4 and 8.
10. Find a number which is divisible by both 3 and 11, and the sum of its digits is 15.

3. HCF and LCM 

1. Find HCF and LCM of 12 and 18.
2. If HCF of two numbers is 6 and LCM is 72, find the numbers.
3. Find HCF of 24, 36, and 60.
4. Find LCM of 8, 12, and 20.
5. Two numbers are in the ratio 3:4, and LCM is 36. Find the numbers.
6. Verify that HCF × LCM = Product of numbers for 15 and 20.
7. Find HCF of 56, 98, and 210.
8. If the LCM of two numbers is 360 and one number is 45, find the other number given their HCF is 15.
9. Three numbers are in the ratio 3:4:5. Their HCF is 7. Find their LCM.
10. Two numbers are 56 and 98. Find their HCF and LCM using prime factorisation.

4. Algebra 

1. Solve for x: 2x + 5 = 17.
2. If a + b = 10 and a – b = 2, find a and b.
3. Solve: 3x – 7 = 2x + 5.
4. FactoriSe x^2 + 5x + 6.
5. Factorise x^2 – 9.
6. Solve x^2 + 3x – 10 = 0.
7. If x = 3 + 2*sqrt(2), find x – 1/x.
8. Simplify (a + b)^2 – (a – b)^2.
9. Solve 2x^2 – 7x + 3 = 0.
10. If a + 1/a = 5, find a^2 + 1/a^2.

5. Quadratic Equations 

1. Solve x^2 – 5x + 6 = 0.
2. Solve 3x^2 + 7x – 6 = 0.
3. Find roots of 2x^2 – x – 3 = 0.
4. Find sum and product of roots for x^2 + 4x + 3 = 0.
5. If roots of x^2 + px + q = 0 are equal, find the relation between p and q.
6. Solve x^2 + 6x + 9 = 0.
7. Find the roots of the quadratic equation 3x^2 – 8x + 4 = 0.
8. The sum of the squares of two numbers is 100, and their product is 21. Form a quadratic equation whose roots are these numbers.
9. If one root of the equation x^2 – 5x + k = 0 is twice the other, find the value of k.
10. Find the values of m for which the quadratic equation x^2 + (m + 2)x + m = 0 has equal roots.

6. Average

1. Find the average of 12, 15, 18, 20, and 25.
2. Average of five numbers is 20. Find the sum.
3. Find the missing number if average of 10 numbers is 15 and sum of 9 numbers is 135.
4. Average marks of 40 students is 50. If one student scored 70, find the new average.
5. Find the average of first 50 natural numbers.
6. The average of 12 numbers is 25. If one number is excluded, the average becomes 24. Find the excluded number.
7. The average age of 30 students in a class is 15 years. If the teacher’s age is included, the average becomes 16 years. Find the teacher’s age.
8. The average marks of a class of 50 students is 72. Later it was found that a mistake of 5 marks was given to each student. Find the correct average.
9. The average weight of 10 men is 60 kg. If one man weighing 75 kg is replaced by another man, the new average becomes 59 kg. Find the weight of the new man.
10. The average score of 8 students in an exam is 75. If the highest score is 95 and the lowest is 55, find the average of the remaining 6 students.

7. Profit, Loss, and Discount 

1. A shopkeeper buys an article for 500 and sells for 600. Find profit %.
2. Cost price of an article is 800. Sold at 20% discount on marked price of 1000. Find profit/loss.
3. Selling price is 450 and loss is 10%. Find cost price.
4. An article is marked at 1200. If discount of 25% is given, find selling price.
5. A trader makes 20% profit on cost price of 500. Find selling price.
6. A shopkeeper sells an article at a profit of 20%. If he had bought it at 10% less and sold at ₹50 more, his profit would have been 40%. Find the cost price of the article.
7. A trader marks his goods at 25% above the cost price. He allows a discount of 12%. Find his profit or loss percentage.
8. A man buys two articles for ₹6,000 each. He sells one at a loss of 10% and the other at a profit of 20%. Find his overall gain or loss percentage.
9. An article is sold at a profit of 20%. If the cost price were 20% more, the profit would have been 25%. Find the cost price.
10. A shopkeeper sells an article at a profit of 12%. If the selling price had been ₹48 less, he would have incurred a loss of 8%. Find the cost price of the article.

8. Ratio, Proportion, and Variation

1. Simplify the ratio 24:36.
2. If a:b = 3:4, find b:a.
3. If x:y = 5:7 and y:z = 2:3, find x:z.
4. If y varies directly as x and y = 10 when x = 2, find y when x = 5.
5. If y varies inversely as x and y = 6 when x = 2, find y when x = 3.
6. The ratio of the ages of A and B is 5:7. Five years hence, the sum of their ages will be 78. Find their present ages.
7. A can do a piece of work in 12 days, and B in 18 days. They work together for 4 days. How much work is left, and in how many additional days can B complete it alone?
8. Three numbers are in the ratio 2:3:5. If the sum of the numbers is 200, find the numbers.
9. The salaries of A and B are in the ratio 4:5. If A gets a 20% increase and B gets a 10% increase, find the new ratio of their salaries.
10. A sum of money is divided among A, B, and C in the ratio 2:3:5. If C receives ₹200 more than B, find the total sum.

9. Time, Speed, and Distance 

1. A car travels 120 km in 3 hours. Find speed.
2. Time taken to cover 240 km at 60 km/h.
3. A train moves 360 km in 4 hours. Find average speed.
4. Two cars move in opposite directions at 50 km/h and 60 km/h. Find relative speed.
5. Two cars move in the same direction at 60 km/h and 40 km/h. Find relative speed.
6. Convert 72 km/h into m/s.
7. A train 240 m long crosses a platform 360 m long in 30 seconds. Find the speed of the train in km/h.

8. Two trains, one 120 m long and the other 180 m long, run in opposite directions on parallel tracks. If they cross each other in 12 seconds, and the faster train runs at 10 km/h more than the slower train, find the speed of each train.

9. A boat goes 30 km downstream in 2 hours and returns the same distance upstream in 3 hours. Find the speed of the boat in still water and the speed of the current.

10. A man walks at 5 km/h in still water. How far will he go downstream in 2 hours if the river has a current of 2 km/h?

10. Time and Work

1. A can do a work in 12 days. Find 1 day’s work.
2. A and B can do a work in 10 days and 15 days respectively. Find days to complete together.
3. If A is twice as efficient as B, find ratio of their time taken.
4. A can do a job in 8 days, B in 12 days. How long together?
5. If 3 men or 5 women can do a work in 12 days, find how long 2 men and 3 women will take.
6. A can complete a work in 12 days and B in 16 days. They work together for 4 days. Then C joins them and the work is completed in 2 more days. Find the time taken by C alone to complete the work.
   
7. A, B, and C can complete a work in 15, 20, and 30 days respectively. They work on alternate days in the order A → B → C → A → … How long will it take to complete the work?
   
8. A and B can do a piece of work in 18 days and 24 days respectively. They start working together but after 6 days, B leaves. How many more days will A take to finish the work?
   
9. A alone can do a work in 10 days, but working with B, the same work can be done in 6 days. In how many days can B alone do the work?
   
10. Pipes A and B can fill a tank in 12 hours and 15 hours respectively. Pipe C can empty the full tank in 20 hours. If all three pipes are opened together, how long will it take to fill the tank?

11. Simple & Compound Interest

1. Find SI on 5000 at 10% for 2 years.
2. Find amount on 6000 at 5% SI for 3 years.
3. Find CI on 4000 at 5% for 2 years.
4. CI on 5000 for 2 years at 10% compounded half-yearly.
5. If SI for 3 years is 900 at 5%, find principal.
6. Find amount after 2 years on 10000 at 10% CI annually.
7. The difference between the compound interest and simple interest on a sum for 2 years at 10% per annum is ₹50. If the rate of interest were 12%, the difference would increase by ₹6. Find the principal.

8. A sum of money amounts to ₹13,310 in 2 years and ₹14,641 in 3 years at compound interest compounded annually. If the same sum were invested at simple interest, the amount after 3 years would have been ₹14,610. Find the principal and rate of interest.

9. A sum of money becomes ₹7,290 in 2 years at compound interest compounded annually. If it had been invested at 12% per annum compounded half-yearly, the amount would have been ₹7,324. Find the principal and the original rate.

10. A sum of money doubles itself in 15 years at compound interest. Determine in how many years it will become eight times the principal if the interest continues at the same rate.

12. Geometry & Mensuration 

1. Area of triangle with base 10 and height 5.
2. TSA of cuboid with l=5, b=4, h=3.
3. LSA of cylinder with r=3, h=7.
4. Find hypotenuse if base=6, height=8.
5. Area of parallelogram with base 10 and height 6.
6. The diagonals of a rhombus are 16 cm and 12 cm. A circle is inscribed inside the rhombus. Find the radius of the circle and the perimeter of the rhombus.

7. The area of a circle is 154 cm². Another semicircle has twice the radius of the first circle. Find the perimeter of the semicircle and the ratio of the areas of the two figures.

8. A metallic sphere of radius 6 cm is melted and recast into small cones of radius 2 cm and height 3 cm. Find the number of cones, the total curved surface area of all cones, and the leftover metal, if any.

9. The length of a rectangle is increased by 40%. By what percent should its breadth be reduced to keep the area the same? Also, calculate the change in the perimeter.

10. The sides of a triangle are 12 cm, 16 cm, and 20 cm. Find its area, the length of the altitude on the longest side, and the radius of the inscribed circle.

13. Trigonometry 

1. Find sin(θ) if opposite=3, hypotenuse=5.
2. Find cos(θ) if adjacent=4, hypotenuse=5.
3. Find tan(θ) if opposite=3, adjacent=4.
4. Find cosec(θ) if sin(θ)=1/2.
5. If sin^2(θ) + cos^2(θ) = ?, verify identity.
6. Find sec^2(θ) if tan^2(θ) = 3.
7. A ladder 10 m long leans against a wall making an angle of 60° with the ground. Find the height reached on the wall.
8. Solve for θ in 2 cos²θ – sin²θ = 1, where 0° ≤ θ ≤ 180°.
9. If sin A + cos A = 1, find sin 2A.
10. Find the height of a tower if its angle of elevation from a point 50 m away is 30°.

14. Statistics 

1. Find mean of 5, 10, 15, 20, 25.
2. Find median of 3, 7, 5, 9, 11.
3. Find mode of 2, 3, 2, 5, 3, 2.
4. Find range of 10, 20, 15, 25, 30.
5. Average of 6 numbers is 12. Find sum.
6. The scores of 15 students are given. Find the range: 23, 27, 30, 32, 35, 36, 40, 41, 45, 50, 52, 55, 57, 60, 65.

7. For the data 2, 5, 7, 10, 12, calculate the mean deviation from the mean.
8. Find the mean of the data set: 12, 15, 18, 20, 25.
9. The marks of 10 students are 5, 7, 8, 10, 12, 12, 15, 18, 20, 25. Find the median.
10. Find the mode of the data set: 3, 7, 7, 7, 10, 12, 12, 15.

Maths Books for Competitive Exams

Maths is often considered the trickiest section in competitive exams, and solving maths questions can feel overwhelming and intimidating for many candidates. To make the subject easier and more approachable, we have compiled a list of the best maths books for competitive exams. These books are widely recommended for banking exams, management entrance tests, and UPSC. Candidates should refer to this list as the books include practice questions, shortcuts, and strategies to strengthen the quantitative section and boost exam performance.

• Fast Track Objective Arithmetic by Rajesh Verma
• Handbook for Mathematics by Arihant Experts
• For Competitive Exams Vedic Mathematics by Ramnandan Shastri
• Quantitative Aptitude for Competitive Examinations by R.S. Aggarwal 
• Mathematics for All Competitive Exams SSC (Pre./Mains) by Ramniwas Mathuriya 
• Arithmetic Subjective and Objective for Competitive Examination by R.S. Aggarwal
• Objective Arithmetic (SSC & Railway Exam Special) by R.S Aggarwal
• Shortcuts in Quantitative Aptitude for Competitive Exams by Disha Publication
• Teach Yourself Quantitative Aptitude by Arun Sharma
• The Pearson Guide To Quantitative Aptitude For Competitive Examination by Dinesh Khattar
• Quantitative Aptitude for all Competitive Exam by Abhijit Gupta
• NCERT Math Books for 10th,11th and 12th

Tips to Improve Maths for Competitive Exams

Mathematics in competitive exams demands both accuracy and speed. Success comes from consistent practice, concept clarity, and smart strategies. To help aspirants strengthen their preparation, here are some effective tips:

1. Always Keep a List of Important Formulas on Your Desk

Maths without formulas is like Pizza without cheese. Instead of cramming up the formulas at the last moment, it is better to go through them as many times as you can. Also, try to memorise only few important formulas and not the entire list. 

2. Master Important Topics First

Study smart, not hard. Almost all the topics in Maths for competitive exams are important but there are a few concepts that are commonly asked. Identify those areas and practise them thoroughly. 

3. Break the Shapes

In the section of Maths for competitive exams, you are bound to get confused in questions with complex and intertwined shapes. Simply redraw the shapes separately on a rough sheet and then answer the question.

4. Memorise the Right Stuff

Remember multiplication tables at least up to 30 and learn the square as well as cube roots for numbers till 40. These hacks can help you save time from difficult calculations.

5. Follow the Mantra of Practice

When you take competitive tests that are set in a specific time frame, it improves your speed and helps you identify the areas where you still lag. So, keep practising the key concepts until you feel confident enough to crack them.

Mathematics in competitive exams is less about memorisation and more about the application of concepts under time pressure. With consistent practice, a clear understanding of fundamentals, and the use of smart solving strategies, candidates can turn this section into a scoring 

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