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SSC MTS Algebraic Identities

Study Material — 16 PYQs (2018–2018) · Concept Notes · Shortcuts

SSC MTS Algebraic Identities is a frequently tested subtopic — 16 previous year questions from 2018–2018 papers are included below with concept notes, key rules and shortcut tricks.

16 PYQs
2018–2018
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Previous Year Questions

SSC MTS Algebraic Identities — Past Exam Questions

16 questions from actual SSC MTS papers · all shown free · click option to reveal solution

Exam Q 12018Previous Year Pattern

If x + 1/x = 5, find the value of x² + 1/x².

Exam Q 22018Previous Year Pattern

Expand (2x + 3y)² using the algebraic identity.

Exam Q 32018Previous Year Pattern

Simplify: (p + q)² + (p - q)² using algebraic identities.

Exam Q 42018Previous Year Pattern

If (a + b)² = 49 and ab = 12, find the value of a² + b².

Exam Q 52018Previous Year Pattern

If a - b = 7 and ab = 18, find the value of a² + b².

Exam Q 62018Previous Year Pattern

Simplify: (x + 5)² - (x - 5)² using algebraic identities.

Exam Q 72018Previous Year Pattern

If a² + b² = 34 and ab = 15, find (a + b)².

Exam Q 82018Previous Year Pattern

If (a + b + c) = 12 and (ab + bc + ca) = 47, find a² + b² + c².

Exam Q 92018Previous Year Pattern

Simplify: (2x + 3y)² - (2x - 3y)²

Exam Q 102018Previous Year Pattern

If (a + b)² = 81 and (a - b)² = 25, find the value of ab.

Exam Q 112018Previous Year Pattern

If x + 1/x = 5, find the value of x³ + 1/x³.

Exam Q 122018Previous Year Pattern

If p - q = 7 and p² - q² = 147, find the value of p + q.

Exam Q 132018Previous Year Pattern

If m + n = 12 and m² + n² = 90, find the value of (m - n)².

Exam Q 142018Previous Year Pattern

If (a + b)² = 81 and (a - b)² = 49, find the value of ab.

Exam Q 152018Previous Year Pattern

If a² + b² + c² = 50 and ab + bc + ca = 25, find the value of (a + b + c)².

Exam Q 162018Previous Year Pattern

If (2x + 3y)² - (2x - 3y)² = k·xy, find the value of k.

Concept Notes

Algebraic Identities— Rules & Concept

Core ConceptRead this first — the foundation of the topic

Algebraic identities are pre-proven mathematical formulas that help solve complex algebraic expressions quickly. These are your shortcuts to crack algebra problems in SSC CGL without lengthy calculations. Think of them as ready-made formulas that work every time.

Key RulesCore rules you must know cold

Algebraic identities are universally true for all values of variables. They help expand, factorize, and simplify expressions. The most important ones follow specific patterns that SSC loves to test.

Formula BlockMemorise — at least one formula appears in every paper
• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²
• (a + b)(a - b) = a² - b²
• (a + b)³ = a³ + 3a²b + 3ab² + b³
• (a - b)³ = a³ - 3a²b + 3ab² - b³
• a³ + b³ = (a + b)(a² - ab + b²)
• a³ - b³ = (a - b)(a² + ab + b²)
• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Exam PatternsWhat examiners ask — read before attempting PYQs

SSC CGL typically asks direct substitution problems, value finding questions, and simplification using identities. Common question types include finding the value of expressions like x² + 1/x² when x + 1/x is given, or simplifying complex algebraic fractions.

ShortcutsUse these to save 30–60 seconds per question

- The 'Recognition Method': Instead of expanding everything, learn to recognize patterns. If you see (something)² - (something else)², immediately think a² - b² = (a+b)(a-b). This saves 2-3 minutes per question.

Worked ExampleSolve this step-by-step before moving on
1
Step 1

Start with the given condition x + 1/x = 5

2
Step 2

Square both sides: (x + 1/x)² = 5² = 25

3
Step 3

Apply the identity (a + b)² = a² + 2ab + b² Here, a = x and b = 1/x

4
Step 4

(x + 1/x)² = x² + 2(x)(1/x) + (1/x)²

5
Step 5

25 = x² + 2(1) + 1/x²

6
Step 6

25 = x² + 2 + 1/x²

7
Step 7

x² + 1/x² = 25 - 2 = 23 Therefore, x² + 1/x² = 23.

Exam TrapsCommon mistakes students make — avoid these

Students often forget the middle term in expansions. In (a + b)², many write a² + b² and miss the 2ab term. Always count terms: square of binomial has THREE terms, cube has FOUR terms.

Also, students confuse signs in (a - b) expansions. Remember: alternate signs in (a - b)³.

Key Points to Remember

  • (a + b)² = a² + 2ab + b² - never forget the middle term 2ab
  • (a - b)² = a² - 2ab + b² - note the minus sign before 2ab
  • (a + b)(a - b) = a² - b² - difference of squares identity
  • a³ + b³ = (a + b)(a² - ab + b²) - sum of cubes factorization
  • a³ - b³ = (a - b)(a² + ab + b²) - difference of cubes factorization
  • (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca - three variable expansion
  • When x + 1/x is given, find x² + 1/x² by squaring both sides
  • Recognize patterns first, then apply appropriate identity to save time

Exam-Specific Tips

  • (a + b)² has exactly 3 terms: a², 2ab, and b²
  • (a - b)³ = a³ - 3a²b + 3ab² - b³ with alternating signs
  • x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)
  • If a + b + c = 0, then a³ + b³ + c³ = 3abc
  • (a + b)³ - (a - b)³ = 2b(3a² + b²)
  • The coefficient of middle term in (a + b)² is always 2
  • a⁴ - b⁴ = (a² + b²)(a + b)(a - b)
  • (x + 1/x)² = x² + 1/x² + 2

60-Second Revision — Algebraic Identities

  • Remember: Square of binomial has 3 terms, cube has 4 terms
  • Formula: (a + b)(a - b) = a² - b² for quick factorization
  • Trap: Don't forget middle terms in expansions - very common error
  • Trick: If x + 1/x given, square it to find x² + 1/x²
  • Pattern: Recognize a² - b² form immediately for difference of squares
  • Sign Rule: (a - b)³ has alternating positive-negative signs
  • Quick Check: Count terms in your final answer to avoid missing any
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