Algebra Basics
Key Concepts & Formulas
Provide 5-7 essential concepts for Algebra Basics:
| # | Concept | Quick Explanation |
|---|---|---|
| 1 | Linear Equations | Equations with degree 1 (highest power of variable is 1) e.g., 2x + 3 = 7 |
| 2 | Quadratic Equations | Equations with degree 2 (highest power of variable is 2) e.g., x² - 5x + 6 = 0 |
| 3 | Algebraic Identities | Equations true for all values of variables, used for quick calculations |
| 4 | Factorization | Breaking expressions into simpler multiplicative components |
| 5 | Simultaneous Equations | Two or more equations with multiple variables solved together |
| 6 | Degree of Equation | Highest power of the variable in the equation determines its type |
Essential Formulas
| Formula | Usage |
|---|---|
| (a+b)² = a² + 2ab + b² | When squaring binomials or expanding expressions like (x+3)² |
| (a-b)² = a² - 2ab + b² | When squaring binomials with subtraction or expanding (x-5)² |
| a² - b² = (a+b)(a-b) | When factorizing difference of squares or simplifying 16-9 |
| (a+b)³ = a³ + 3a²b + 3ab² + b³ | For cubic expansions and volume calculations |
| (a-b)³ = a³ - 3a²b + 3ab² - b³ | For cubic expansions with subtraction |
10 Practice MCQs
Q1. If (x+4)² = 49, find the value of x. A) 3 B) -11 C) 3 or -11 D) 7 or -7
Answer: C) 3 or -11
Solution: (x+4)² = 49 Taking square root: x+4 = ±7 Case 1: x+4 = 7 → x = 3 Case 2: x+4 = -7 → x = -11
Shortcut: Remember that √49 = ±7 (both positive and negative)
Concept: Algebra Basics - Solving quadratic equations using square root method
Q2. Factorize: x² - 9 A) (x-3)² B) (x+3)² C) (x+3)(x-3) D) (x-9)(x+1)
Answer: C) (x+3)(x-3)
Solution: Using identity: a² - b² = (a+b)(a-b) Here, x² - 9 = x² - 3² = (x+3)(x-3)
Shortcut: Difference of squares always factors as (sum)(difference)
Concept: Algebra Basics - Factorization using identities
Q3. A train travels at (x+20) km/hr. If it covers 300 km in 5 hours, find x. A) 40 B) 50 C) 60 D) 70
Answer: A) 40
Solution: Speed = Distance/Time x+20 = 300/5 = 60 x = 60-20 = 40
Concept: Algebra Basics - Linear equations in speed-distance problems
Q4. If x + 1/x = 5, find x² + 1/x² A) 23 B) 25 C) 27 D) 29
Answer: A) 23
Solution: Square both sides: (x + 1/x)² = 5² x² + 2(x)(1/x) + 1/x² = 25 x² + 2 + 1/x² = 25 x² + 1/x² = 25-2 = 23
Shortcut: Remember (a+b)² = a² + 2ab + b²
Concept: Algebra Basics - Algebraic manipulation and identities
Q5. The sum of two consecutive odd numbers is 84. Find the larger number. A) 41 B) 43 C) 45 D) 47
Answer: B) 43
Solution: Let numbers be x and x+2 x + (x+2) = 84 2x + 2 = 84 2x = 82 x = 41 Larger number = 41+2 = 43
Concept: Algebra Basics - Linear equations in word problems
Q6. If (3x-2)(2x+5) = ax² + bx + c, find a+b+c A) 25 B) 30 C) 35 D) 40
Answer: C) 35
Solution: (3x-2)(2x+5) = 6x² + 15x - 4x - 10 = 6x² + 11x - 10 Therefore: a=6, b=11, c=-10 a+b+c = 6+11-10 = 7
Correction: Let me recalculate a+b+c = 6+11+(-10) = 7 Answer should be: None of these (7)
Concept: Algebra Basics - Expansion and coefficient identification
Q7. A platform of length 200m has two trains. Train A (x meters) and Train B (x+50 meters). If their total length is 550m, find x. A) 200 B) 250 C) 300 D) 350
Answer: B) 250
Solution: x + (x+50) = 550 2x + 50 = 550 2x = 500 x = 250
Concept: Algebra Basics - Linear equations in train length problems
Q8. If x² - 5x + 6 = 0 and y² - 5y + 6 = 0, where x ≠ y, find x+y A) 5 B) 6 C) 7 D) 8
Answer: A) 5
Solution: Both equations are identical: x² - 5x + 6 = 0 Factorizing: (x-2)(x-3) = 0 Therefore: x = 2 or 3, y = 2 or 3 Since x ≠ y, we have x=2, y=3 or x=3, y=2 In both cases: x+y = 5
Concept: Algebra Basics - Quadratic equations and roots
Q9. If (x+y)² = 36 and xy = 8, find x² + y² A) 20 B) 28 C) 36 D) 44
Answer: B) 20
Solution: (x+y)² = x² + 2xy + y² = 36 Given xy = 8, so 2xy = 16 x² + y² + 16 = 36 x² + y² = 36-16 = 20
Shortcut: x² + y² = (x+y)² - 2xy
Concept: Algebra Basics - Algebraic identities application
Q10. A train's speed is reduced by 10 km/hr due to fog. It takes 3 hours more to cover 270 km. Find original speed. A) 45 B) 50 C) 55 D) 60
Answer: A) 45
Solution: Let original speed = x km/hr Original time = 270/x hours New speed = (x-10) km/hr New time = 270/(x-10) hours Given: 270/(x-10) - 270/x = 3 Solving: 270x - 270(x-10) = 3x(x-10) 2700 = 3x² - 30x x² - 10x - 900 = 0 (x-45)(x+40) = 0 x = 45 (speed can’t be negative)
Concept: Algebra Basics - Quadratic equations in time-speed problems
5 Previous Year Questions
PYQ 1. If a+b = 10 and a-b = 4, find a² + b² RRB NTPC 2021 CBT-1
Answer: 58
Solution: From a+b = 10 and a-b = 4: Adding: 2a = 14 → a = 7 Subtracting: 2b = 6 → b = 3 a² + b² = 7² + 3² = 49 + 9 = 58
Alternative: a² + b² = ½[(a+b)² + (a-b)²] = ½[100 + 16] = 58
Exam Tip: Remember the identity for a² + b² in terms of sum and difference
PYQ 2. Factorize completely: 4x² - 25 RRB Group D 2022
Answer: (2x+5)(2x-5)
Solution: 4x² - 25 = (2x)² - 5² = (2x+5)(2x-5)
Exam Tip: Always look for perfect squares in factorization problems
PYQ 3. If x + 1/x = 3, find x³ + 1/x³ RRB ALP 2018
Answer: 18
Solution: Using identity: a³ + b³ = (a+b)³ - 3ab(a+b) x³ + 1/x³ = (x + 1/x)³ - 3(x)(1/x)(x + 1/x) = 3³ - 3(1)(3) = 27 - 9 = 18
Exam Tip: Memorize the identity for a³ + b³ in terms of (a+b)
PYQ 4. Solve: 3(x-2) + 5 = 2(x+1) - 1 RRB JE 2019
Answer: x = 2
Solution: 3x - 6 + 5 = 2x + 2 - 1 3x - 1 = 2x + 1 3x - 2x = 1 + 1 x = 2
Exam Tip: Always expand brackets first, then collect like terms
PYQ 5. The product of two consecutive even numbers is 168. Find their sum. RPF SI 2019
Answer: 26
Solution: Let numbers be x and x+2 x(x+2) = 168 x² + 2x - 168 = 0 (x+14)(x-12) = 0 x = 12 (taking positive value) Numbers: 12 and 14 Sum = 12 + 14 = 26
Exam Tip: For consecutive number problems, always consider both positive and negative roots
Speed Tricks & Shortcuts
| Situation | Shortcut | Example |
|---|---|---|
| Finding a² + b² given a+b and ab | Use: a² + b² = (a+b)² - 2ab | If a+b=7, ab=10, then a² + b² = 49-20 = 29 |
| Difference of squares | a² - b² = (a+b)(a-b) | 49-36 = (7+6)(7-6) = 13×1 = 13 |
| Squaring numbers ending in 5 | (x5)² = x(x+1) hundred + 25 | 35² = 3×4 hundred + 25 = 1225 |
| Finding value of symmetric expressions | Use identities directly | If x + 1/x = 4, then x² + 1/x² = 16-2 = 14 |
| Solving (x-a)(x-b) = 0 | Roots are directly a and b | (x-3)(x-7) = 0 → x = 3 or 7 |
Common Mistakes to Avoid
| Mistake | Why Students Make It | Correct Approach |
|---|---|---|
| Forgetting ± sign when taking square root | Assuming only positive root | Always consider both +ve and -ve roots: √9 = ±3 |
| Incorrect expansion of (a+b)² | Writing a² + b² (missing 2ab) | Remember: (a+b)² = a² + 2ab + b² |
| Sign errors in factorization | Making (a-b)² = a² - b² | Remember: (a-b)² = a² - 2ab + b² |
| Not checking solutions | Substituting without verification | Always verify solutions by plugging back |
| Cancelling variables incorrectly | Dividing by variable that could be zero | Factor out common terms instead of dividing |
Quick Revision Flashcards
| Front (Question/Term) | Back (Answer) |
|---|---|
| (a+b)² | a² + 2ab + b² |
| (a-b)² | a² - 2ab + b² |
| a² - b² | (a+b)(a-b) |
| If x + 1/x = k, then x² + 1/x² | k² - 2 |
| Sum of roots of ax² + bx + c = 0 | -b/a |
| Product of roots of ax² + bx + c = 0 | c/a |
| If a+b and a-b are given, find ab | Use: ab = ¼[(a+b)² - (a-b)²] |
| Degree of linear equation | 1 |
| Degree of quadratic equation | 2 |
| Number of solutions for quadratic | 2 (can be real or complex) |
Topic Connections
How Algebra Basics connects to other RRB exam topics:
- Direct Link: Simplification - Algebraic techniques are used to simplify complex numerical expressions
- Direct Link: Number System - Properties of numbers help in solving algebraic equations
- Combined Questions: Algebra + Profit & Loss - Finding cost prices using linear equations
- Combined Questions: Algebra + Time & Work - Solving work rate problems using equations
- Foundation For: Advanced Algebra - Quadratic equations, polynomials, and progressions build on these basics
- Foundation For: Data Interpretation - Setting up equations from given data sets
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