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Quantitative Aptitude

Algebraic Identities & Expressions

Coordinate Geometry20 / 51Probability
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Concepts (3)

Master value substitution by replacing variables with given numbers to quickly evaluate algebraic expressions. Use smart assumptions for variables when equations are fewer than unknowns to save time.

The fundamental principle of value substitution involves directly replacing variables (like x, y, a, b) in an algebraic expression with their given numerical values. This technique is crucial for evaluating expressions, solving equations, and simplifying complex algebraic problems quickly in competitive exams. It's not a single formula, but a direct application of arithmetic operations after variable replacement.

Method Steps:

  1. Identify the expression to be evaluated.
  2. Note the given values for the variables.
  3. Carefully substitute each variable with its corresponding numerical value.
  4. Perform the arithmetic operations (addition, subtraction, multiplication, division, powers) in the correct order (BODMAS/PEMDAS) to arrive at the final answer.

Worked Example 1

Q: If x = 3 and y = -2, find the value of 2x^2 + 3xy - y^2. A:

  1. Given expression: 2x^2 + 3xy - y^2
  2. Given values: x = 3, y = -2
  3. Substitute the values: 2(3)^2 + 3(3)(-2) - (-2)^2
  4. Perform calculations: 2(9) + 3(-6) - (4) 18 - 18 - 4 0 - 4 = -4 Therefore, the value of the expression is -4.

Worked Example 2

Q: If a + b = 7 and ab = 10, find the value of a^2 + b^2. A:

  1. We know the identity: (a + b)^2 = a^2 + b^2 + 2ab
  2. Rearrange to find a^2 + b^2: a^2 + b^2 = (a + b)^2 - 2ab
  3. Given values: a + b = 7, ab = 10
  4. Substitute the values into the rearranged identity: a^2 + b^2 = (7)^2 - 2(10) a^2 + b^2 = 49 - 20 a^2 + b^2 = 29 Therefore, the value of a^2 + b^2 is 29.

Shortcuts & Tricks

  1. Smart Value Assumption (when variables > equations): If you have an algebraic equation or expression with more variables than independent equations, you can often assume convenient values (like 0, 1, or -1) for the "extra" variables. This is valid as long as your assumptions don't violate any given conditions (e.g., making a denominator zero) or lead to contradictions. This drastically simplifies calculations.
    • Example: If x + y + z = 0, find x^3 + y^3 + z^3. By identity, x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx). Since x+y+z=0, then x^3+y^3+z^3 = 3xyz. You could assume x=1, y=-1, z=0 to quickly verify this and apply it.
  2. Using Options: For objective questions, sometimes substituting the answer options back into the given equation or expression can be quicker than solving it algebraically, especially if the options are simple numbers.
  3. Symmetry/Cyclicity: Recognize symmetric expressions (e.g., x^2+y^2+z^2) or cyclic expressions (e.g., x(y-z)+y(z-x)+z(x-y)). For cyclic expressions, if the expression evaluates to 0 for specific values (e.g., x=y=z), it might be 0 generally. For symmetric expressions, assuming x=y=z can often lead to the correct answer quickly, or at least eliminate options.

Common Mistakes

  1. Sign Errors: Careless handling of negative numbers, especially when squaring or multiplying, is a frequent mistake. Always use parentheses for negative values during substitution, e.g., (-2)^2 is 4, not -4.
  2. Order of Operations (BODMAS/PEMDAS): Incorrectly applying the order of operations (Brackets/Parentheses, Orders/Exponents, Division/Multiplication, Addition/Subtraction) leads to wrong answers. Exponents must be calculated before multiplication, etc.
  3. Invalid Value Assumption: Assuming values that contradict the problem's conditions (e.g., assuming x=0 when x is in the denominator) or lead to undefined expressions. Always check constraints.

Underlying Principle & Advanced Applications

Value substitution is essentially evaluating a polynomial or rational function at specific points. Its power in competitive exams comes from its flexibility. When direct algebraic manipulation is complex or time-consuming, smart substitution can bypass intricate steps.

Advanced Examples

Q: If a + b + c = 6, a^2 + b^2 + c^2 = 14, and a^3 + b^3 + c^3 = 36, find the value of abc. A:

  1. We know the identity: (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
  2. Substitute given values: (6)^2 = 14 + 2(ab + bc + ca) 36 = 14 + 2(ab + bc + ca) 22 = 2(ab + bc + ca) ab + bc + ca = 11
  3. We also know the identity: a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - (ab + bc + ca))
  4. Substitute all known values: 36 - 3abc = (6)(14 - 11) 36 - 3abc = 6(3) 36 - 3abc = 18 3abc = 36 - 18 3abc = 18 abc = 6 Therefore, abc = 6.

Variation Types

  1. Conditional Substitution: Problems where variables are related by a condition (e.g., x + 1/x = 5) and you need to find an expression's value (e.g., x^2 + 1/x^2). Here, you often manipulate the given condition to directly find the expression's value rather than solving for x first.
  2. Implicit Values: Sometimes, values aren't explicitly given but can be deduced. For instance, if (x-2)^2 + (y-3)^2 + (z-4)^2 = 0, then x=2, y=3, z=4 must hold (since squares of real numbers cannot be negative, and their sum is zero only if each term is zero).
  3. Variable-to-Constant Conversion: In problems like (a/b) + (b/a) = 1, find a^3 + b^3. You can multiply by ab to get a^2 + b^2 = ab, or a^2 - ab + b^2 = 0. Then use the identity a^3 + b^3 = (a+b)(a^2 - ab + b^2). Substituting a^2 - ab + b^2 = 0 directly gives a^3 + b^3 = (a+b)(0) = 0. This avoids finding a and b explicitly.

Time-Saving Methods

  1. Pattern Recognition: Many SSC CGL questions are based on standard algebraic identities. Recognizing these patterns allows direct application of results or quick manipulation.
  2. Elimination by Substitution: If an expression has variables and the options are numerical, try substituting simple values (e.g., x=0, x=1) into the original expression and then into the options. The option that matches is likely correct. This is especially useful for complex expressions that simplify to a constant.
  3. Focus on the Target Expression: Before substituting, look at the expression you need to find. Can it be simplified or rewritten using the given conditions before plugging in numbers? Often, this reduces the complexity of calculations. For example, if you need x^2 + 1/x^2 and have x + 1/x = k, directly square the given equation: (x + 1/x)^2 = k^2 => x^2 + 1/x^2 + 2 = k^2 => x^2 + 1/x^2 = k^2 - 2.
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Master Remainder and Factor Theorems for quick solutions. Substitute x-value from divisor into polynomial to find remainder or unknown coefficients efficiently.

Core Formula

Remainder Theorem: If a polynomial P(x) is divided by (x - a), the remainder is P(a).

Factor Theorem: If P(a) = 0, then (x - a) is a factor of P(x). Conversely, if (x - a) is a factor of P(x), then P(a) = 0.

Division Algorithm: P(x) = D(x) * Q(x) + R(x), where P(x) is the dividend, D(x) is the divisor, Q(x) is the quotient, and R(x) is the remainder. For a linear divisor (x-a), R(x) is a constant.

Worked Example 1

Q: Find the remainder when P(x) = x^3 - 4x^2 + 5x - 2 is divided by (x - 1).

A:

  1. Set the divisor (x - 1) to zero to find the value of x: x - 1 = 0 => x = 1.
  2. Substitute x = 1 into the polynomial P(x): P(1) = (1)^3 - 4(1)^2 + 5(1) - 2 P(1) = 1 - 4 + 5 - 2 P(1) = 0 The remainder is 0.

Worked Example 2

Q: If (x - 2) is a factor of P(x) = x^3 - kx^2 + 3x - 2, find the value of k.

A:

  1. Since (x - 2) is a factor, by the Factor Theorem, P(2) must be 0.
  2. Set the divisor (x - 2) to zero: x - 2 = 0 => x = 2.
  3. Substitute x = 2 into P(x) and set the result to 0: P(2) = (2)^3 - k(2)^2 + 3(2) - 2 = 0 8 - 4k + 6 - 2 = 0 12 - 4k = 0 4k = 12 k = 3 The value of k is 3.

Shortcuts & Tricks

  • Direct Substitution: For linear divisors (ax + b), set ax + b = 0 to find x = -b/a. Directly substitute this value into the polynomial P(x) to get the remainder. This method is significantly faster than polynomial long division for competitive exams.
  • Divisibility Rules for x^n ± a^n:
    • x^n - a^n is always divisible by (x - a) for any positive integer n.
    • x^n - a^n is divisible by (x + a) if n is an even positive integer.
    • x^n + a^n is divisible by (x + a) if n is an odd positive integer.
    • x^n + a^n is never divisible by (x - a).
  • Mental Math: Practice quick mental calculations for substitution, especially with small integer values of x (like 1, -1, 2, -2).

Common Mistakes

  1. Sign Errors: Incorrectly substituting the value of x. For a divisor (x - a), substitute x = a. For (x + a), substitute x = -a. Always set the divisor to zero to get the correct x value.
  2. Calculation Errors: Simple arithmetic mistakes during the substitution process, particularly with negative numbers or higher powers. Double-check each step.
  3. Confusing Remainder with Factor: Remember that if (x - a) is a factor, the remainder P(a) must be 0. If it's just a divisor, P(a) will be the specific remainder value given.

Derivation (brief)

The Remainder Theorem is a direct application of the Division Algorithm. When a polynomial P(x) is divided by a linear polynomial (x - a), the Division Algorithm states: P(x) = (x - a) * Q(x) + R, where Q(x) is the quotient and R is a constant (because the divisor (x - a) has degree 1, so the remainder must have a degree less than 1, i.e., degree 0, which is a constant). If we substitute x = a into this equation: P(a) = (a - a) * Q(a) + R P(a) = 0 * Q(a) + R P(a) = R This proves that the remainder R is simply P(a). The Factor Theorem follows directly: if R = 0, then P(a) = 0, meaning (x - a) is a factor of P(x).

Advanced Examples

Q: When x^3 + ax^2 + bx + 6 is divided by (x - 2), the remainder is 0. When divided by (x - 1), the remainder is 4. Find the values of a and b.

A:

  1. From the first condition: Divided by (x - 2), remainder is 0. By the Remainder Theorem, P(2) = 0. (2)^3 + a(2)^2 + b(2) + 6 = 0 8 + 4a + 2b + 6 = 0 4a + 2b = -14 2a + b = -7 (Equation 1)

  2. From the second condition: Divided by (x - 1), remainder is 4. By the Remainder Theorem, P(1) = 4. (1)^3 + a(1)^2 + b(1) + 6 = 4 1 + a + b + 6 = 4 a + b + 7 = 4 a + b = -3 (Equation 2)

  3. Solve the system of linear equations: Subtract Equation 2 from Equation 1: (2a + b) - (a + b) = -7 - (-3) a = -4

  4. Substitute a = -4 into Equation 2: -4 + b = -3 b = 1

    Thus, a = -4 and b = 1.

Variation Types

  • Finding Unknown Coefficients: Problems like the advanced example, requiring solving simultaneous equations based on multiple remainder conditions.
  • Divisor of the form (ax + b): Remember to set ax + b = 0 to find x = -b/a for substitution. E.g., for (2x + 1), substitute x = -1/2.
  • Higher Degree Polynomials: The principles remain the same, but calculations might involve larger numbers or more steps.
  • Exact Divisibility: This is a direct application of the Factor Theorem, where the remainder is explicitly stated as zero.

Time-Saving Methods

  • Prioritize Substitution: Always use the Remainder Theorem (direct substitution) over polynomial long division for finding remainders. Long division is a time sink in competitive exams.
  • Systematic Equation Setup: For problems with multiple unknowns, clearly set up your simultaneous equations. This prevents errors and speeds up solving.
  • Practice with Fractions/Negatives: Get comfortable substituting fractional or negative values of x quickly and accurately, as these are common traps.
  • Recognize Common Divisibility Patterns: Instantly identify when x^n ± a^n forms are divisible by (x ± a) using the rules mentioned in shortcuts.
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Mastering standard algebraic identities like (a+b)², a²-b², a³+b³ is crucial for SSC CGL. Quick recall and application save significant time, boosting speed and accuracy.

Core Formulas

Standard algebraic identities are fundamental for solving algebra problems quickly in SSC CGL. Memorize these key formulas:

  1. (a + b)² = a² + b² + 2ab
  2. (a - b)² = a² + b² - 2ab
  3. a² - b² = (a - b)(a + b)
  4. (a + b)³ = a³ + b³ + 3ab(a + b)
  5. (a - b)³ = a³ - b³ - 3ab(a - b)
  6. a³ + b³ = (a + b)(a² - ab + b²)
  7. a³ - b³ = (a - b)(a² + ab + b²)
  8. (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

Worked Example 1

Question: If x + 1/x = 4, find the value of x² + 1/x². Solution: We know (a + b)² = a² + b² + 2ab. Let a = x and b = 1/x.

  1. Square both sides of x + 1/x = 4: (x + 1/x)² = 4²
  2. Expand the left side using the identity: x² + (1/x)² + 2 * x * (1/x) = 16
  3. Simplify: x² + 1/x² + 2 = 16
  4. Isolate x² + 1/x²: x² + 1/x² = 16 - 2 x² + 1/x² = 14

Worked Example 2

Question: Simplify (103)² - (97)². Solution: This problem uses the identity a² - b² = (a - b)(a + b). Let a = 103 and b = 97.

  1. Substitute the values into the identity: (103)² - (97)² = (103 - 97)(103 + 97)
  2. Perform the subtractions and additions: = (6)(200)
  3. Multiply the results: = 1200

Shortcuts & Tricks

  • For x + 1/x = k:
    • x² + 1/x² = k² - 2
    • x³ + 1/x³ = k³ - 3k
  • For x - 1/x = k:
    • x² + 1/x² = k² + 2
    • x³ + 1/x³ = k³ + 3k
  • Relating squares:
    • (a + b)² - (a - b)² = 4ab
    • (a + b)² + (a - b)² = 2(a² + b²)

Common Mistakes

  1. Sign Errors: Frequently misplacing signs, especially in (a - b)² (often written as a² - b² - 2ab instead of a² + b² - 2ab) or (a - b)³. Pay close attention to the -3ab(a-b) term.
  2. Confusing a³ + b³ with (a + b)³: These are distinct. (a + b)³ is the cube of the sum, while a³ + b³ is the sum of the cubes. Remember (a + b)³ = a³ + b³ + 3ab(a + b).
  3. Incorrect application of a² - b²: While simple, sometimes students try to expand (a - b)(a + b) instead of directly using a² - b² for simplification, losing valuable time.

Derivation (brief)

Let's quickly derive (a + b)² = a² + b² + 2ab. By definition, (a + b)² means (a + b) * (a + b). Using the distributive property (FOIL method): a * (a + b) + b * (a + b) = a*a + a*b + b*a + b*b = a² + ab + ab + b² = a² + b² + 2ab Understanding the derivation helps in recalling the formula correctly and even deriving it if you forget under pressure.

Advanced Identities & Conditional Cases

  1. a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca) This is a very important identity for SSC CGL. Many questions revolve around it.
  2. Conditional Identity: If a + b + c = 0, then a³ + b³ + c³ = 3abc. This is a direct consequence of the previous identity. If a + b + c = 0, then the right side (a + b + c)(...) becomes 0 * (...) = 0, leading to a³ + b³ + c³ - 3abc = 0, hence a³ + b³ + c³ = 3abc.

Variation Types

Many problems involve expressions like x + 1/x, x - 1/x, x² + 1/x², x³ + 1/x³, etc. The shortcuts mentioned earlier are invaluable here. Also, be prepared for problems where a and b are complex expressions, not just single variables.

Time-Saving Methods

  • Value Putting/Substitution: For certain problems, especially those with multiple variables and options, try substituting simple values for variables (e.g., a=1, b=2, c=0) that satisfy any given conditions. Then check which option matches your calculated result. Be careful not to pick values that make denominators zero or lead to ambiguous results.
  • Observation & Pattern Recognition: Many SSC CGL questions are designed to be solved quickly if you spot the underlying identity. For instance, an expression like (x+y)² - (x-y)² should immediately trigger 4xy in your mind, not a full expansion.
  • Practice with x + 1/x type problems: These are extremely common. Practice converting between x + 1/x, x - 1/x, x² + 1/x², x³ + 1/x³, x⁵ + 1/x⁵, etc., using the derived shortcuts.
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