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Quantitative Aptitude

Simplification & Surds

Divisibility & Remainders4 / 51Time & Work
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Concepts (2)

Master BODMAS for quick arithmetic simplification. Prioritize Brackets, Orders, Division/Multiplication, Addition/Subtraction to boost speed and accuracy in SSC CGL.

Core Formula

Arithmetic simplification questions are foundational for SSC CGL. The key to solving them quickly and accurately is strictly adhering to the BODMAS (or PEMDAS) rule. This rule dictates the order of operations:

B - Brackets (Parentheses) O - Orders (Exponents, Roots, Powers) D - Division M - Multiplication A - Addition S - Subtraction

Important Note: Division and Multiplication have equal precedence. Perform them from left to right. Similarly, Addition and Subtraction have equal precedence; perform them from left to right.

Order of Operations:

  1. Solve expressions inside Brackets first (innermost to outermost).
  2. Evaluate Powers, Roots, or Exponents.
  3. Perform Division and Multiplication from left to right.
  4. Perform Addition and Subtraction from left to right.

Worked Example 1

Question: Simplify: 15 - [ 6 + { 12 ÷ (4 - 2) } ]

Solution:

  1. Innermost Bracket (4 - 2): 15 - [ 6 + { 12 ÷ 2 } ]
  2. Curly Bracket (12 ÷ 2): 15 - [ 6 + 6 ]
  3. Square Bracket (6 + 6): 15 - 12
  4. Subtraction: 3

Answer: 3

Worked Example 2

Question: Simplify: 2/3 of 45 + 18 ÷ 3 - 4 × 2.5

Solution:

  1. 'of' (Multiplication within context): 2/3 × 45 = 30 Expression becomes: 30 + 18 ÷ 3 - 4 × 2.5
  2. Division (18 ÷ 3): 18 ÷ 3 = 6 Expression becomes: 30 + 6 - 4 × 2.5
  3. Multiplication (4 × 2.5): 4 × 2.5 = 10 Expression becomes: 30 + 6 - 10
  4. Addition (30 + 6): 36 - 10
  5. Subtraction (36 - 10): 26

Answer: 26

Shortcuts & Tricks

  • Digit Sum Method: For questions with integer answers and multiple-choice options, calculate the digit sum of the question and each option. The correct option's digit sum will match the question's. This can quickly eliminate incorrect choices, especially when calculations are complex. (e.g., 123 = 1+2+3=6).
  • Approximation: If the options are widely spaced, approximate decimal values or fractions to the nearest whole number to get a rough estimate. This can often narrow down the answer or even give the exact one.
  • Work Backwards: For questions involving a missing number (e.g., ? + 5 × 3 = 20), start from the known result and reverse the operations.
  • Look for Cancellations/Patterns: In complex fraction or series problems, often terms cancel out or follow a discernible pattern. Don't rush into full multiplication/division.

Common Mistakes

  1. Incorrect Order of Operations: The most frequent error is performing addition before multiplication or subtraction before division. Always stick to BODMAS.
  2. Sign Errors: Mismanaging positive and negative signs, especially when dealing with brackets or multiple subtractions.
  3. Calculation Mistakes: Simple arithmetic errors due to rushing. Double-check mental calculations, especially with decimals or large numbers.
  4. Ignoring 'of': The word 'of' implies multiplication and has higher precedence than standard multiplication/division, often treated after brackets and powers but before D/M/A/S.

Derivation (brief)

The need for an agreed-upon order of operations arose to ensure that mathematical expressions yield a unique and consistent result. Without rules like BODMAS, an expression like 2 + 3 × 4 could be (2 + 3) × 4 = 20 or 2 + (3 × 4) = 14. To avoid ambiguity, mathematicians established a standard hierarchy, which is universally applied.

Advanced Examples

Question: Simplify: [ (1/2 + 1/3) ÷ (1/4 - 1/5) ] × 60

Solution:

  1. First Bracket (1/2 + 1/3): Find common denominator (6). 3/6 + 2/6 = 5/6
  2. Second Bracket (1/4 - 1/5): Find common denominator (20). 5/20 - 4/20 = 1/20
  3. Expression becomes: [ (5/6) ÷ (1/20) ] × 60
  4. Division of Fractions: (5/6) × (20/1) = 100/6 = 50/3
  5. Expression becomes: (50/3) × 60
  6. Multiplication: 50 × (60/3) = 50 × 20 = 1000

Answer: 1000

Variation Types

  • Missing Number Problems: Questions like 12 + ? × 3 = 30. Here, you'd work backwards: ? × 3 = 30 - 12 = 18, so ? = 18 ÷ 3 = 6.
  • Comparison Problems: Two expressions, A and B, are given, and you need to compare their simplified values (e.g., A > B, A < B, A = B, Cannot be determined).
  • Equation-based Simplification: Problems where an equation needs to be solved for an unknown variable, often involving complex arithmetic on both sides.
  • Series Simplification: Questions involving a pattern in a series of additions, subtractions, or multiplications, often requiring recognition of telescoping series or specific formulas.

Time-Saving Methods

  • Prioritize Mental Math: For small calculations (e.g., 12 ÷ 4, 7 × 3), do them mentally to save time writing steps.
  • Convert Decimals to Fractions (and vice versa) Wisely: Sometimes, converting 0.25 to 1/4 simplifies multiplication. Other times, keeping 0.5 as a decimal is easier for addition/subtraction.
  • Look for Common Factors: Before multiplying large numbers, see if you can cancel out common factors in fractions or division steps.
  • Break Down Complex Expressions: Don't try to solve everything at once. Tackle one BODMAS step at a time, rewriting the simplified expression after each step to avoid errors.
  • Practice with Timers: Regular practice under timed conditions is crucial. Identify your weak spots (e.g., fraction arithmetic, decimal multiplication) and focus on improving speed there.
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Master Surds & Indices by understanding exponent rules and rationalization. Focus on quick application of formulas and shortcuts for rapid problem-solving in SSC CGL.

Core Formulas

Surds and Indices are fundamental for simplification. Memorize these laws for quick recall.

Laws of Indices (Exponents):

  1. a^m × a^n = a^(m+n)
  2. a^m / a^n = a^(m-n)
  3. (a^m)^n = a^(mn)
  4. (ab)^m = a^m × b^m
  5. (a/b)^m = a^m / b^m
  6. a^0 = 1 (where a ≠ 0)
  7. a^(-m) = 1/a^m
  8. a^(m/n) = (n√a)^m or n√(a^m)

Laws of Surds (Radicals):

  1. √a × √b = √(ab)
  2. √a / √b = √(a/b)
  3. (√a)^2 = a
  4. n√a × n√b = n√(ab)
  5. n√a / n√b = n√(a/b)

Worked Example 1

Question: The simplified form of (27)^(2/3) is:

Solution: Step 1: Recognize 27 as a power of 3. 27 = 3^3. Step 2: Substitute this into the expression: (3^3)^(2/3). Step 3: Apply the index law (a^m)^n = a^(mn). 3^(3 * 2/3) = 3^2. Step 4: Calculate the final value: 3^2 = 9.

Worked Example 2

Question: If √(x + 5) = 3, then x = ?

Solution: Step 1: To remove the square root, square both sides of the equation. (√(x + 5))^2 = 3^2 Step 2: Simplify both sides. x + 5 = 9 Step 3: Isolate x by subtracting 5 from both sides. x = 9 - 5 x = 4

Shortcuts & Tricks

  1. Rationalization Pattern: For 1/(√a ± √b), multiply numerator and denominator by the conjugate (√a ∓ √b). E.g., 1/(√5 + √3) = (√5 - √3) / (5-3) = (√5 - √3)/2. This is (conjugate) / (a-b).
  2. Nested Surds:
    • If √(x + √(x + √(x + ...))) = k, then k(k-1) = x. If x can be written as n(n+1), then k = n+1.
    • If √(x - √(x - √(x - ...))) = k, then k(k+1) = x. If x can be written as n(n+1), then k = n.
    • If √(a ± √b) can be written as √(X) ± √(Y), then X+Y = a and 4XY = b. Find X and Y. A common form is √(a ± 2√b) = √x ± √y where x+y=a and xy=b.

Common Mistakes

  1. Incorrectly adding/subtracting exponents: a^m + a^n ≠ a^(m+n). This rule only applies to multiplication a^m × a^n.
  2. Distributing square roots: √(a+b) ≠ √a + √b. For example, √(9+16) = √25 = 5, but √9 + √16 = 3+4 = 7.
  3. Sign errors during rationalization: When multiplying by the conjugate (√a - √b), remember the denominator becomes a - b, not a + b or b - a (unless b > a). Always (first term)^2 - (second term)^2.

Derivation (brief)

Let's briefly look at why a^m × a^n = a^(m+n). Consider a^m as 'a' multiplied by itself 'm' times, and a^n as 'a' multiplied by itself 'n' times. So, a^m × a^n = (a × a × ... (m times)) × (a × a × ... (n times)). Combining these, we have 'a' multiplied by itself (m+n) times. Hence, a^(m+n). This fundamental understanding helps in recalling the rules under pressure.

Advanced Examples

Question: If x = (√3 + √2) / (√3 - √2) and y = (√3 - √2) / (√3 + √2), find the value of x + y.

Solution: Step 1: Rationalize x. x = (√3 + √2) / (√3 - √2) × (√3 + √2) / (√3 + √2) x = (√3 + √2)^2 / ((√3)^2 - (√2)^2) x = (3 + 2 + 2√6) / (3 - 2) x = (5 + 2√6) / 1 = 5 + 2√6

Step 2: Rationalize y. Notice y is the reciprocal of x. y = (√3 - √2) / (√3 + √2) × (√3 - √2) / (√3 - √2) y = (√3 - √2)^2 / ((√3)^2 - (√2)^2) y = (3 + 2 - 2√6) / (3 - 2) y = (5 - 2√6) / 1 = 5 - 2√6

Step 3: Find x + y. x + y = (5 + 2√6) + (5 - 2√6) x + y = 5 + 5 + 2√6 - 2√6 x + y = 10

Variation Types

  1. Direct Simplification: Problems like (64)^(1/3) + (27)^(1/3).
  2. Equation Solving: Finding 'x' in (2^x)^3 = 8^4.
  3. Comparison of Surds: Determining which surd is larger, e.g., 3√2 vs 2√3. Convert them to the same root order or square them.
  4. Rationalization of Complex Expressions: Involving multiple terms in the denominator or nested surds.
  5. Algebraic Identities with Surds: Problems like (a+b)^2 or (a-b)^2 where 'a' and 'b' are surds.

Time-Saving Methods

  1. Reciprocal Recognition: If x = (a+b)/(a-b), then 1/x = (a-b)/(a+b). If asked for x + 1/x or x - 1/x, rationalize only one and use the reciprocal property. For x = (√A + √B) / (√A - √B), then 1/x = (√A - √B) / (√A + √B). Then x + 1/x = 2(A+B)/(A-B) and x - 1/x = 4√AB/(A-B).
  2. Instant Nested Surds: For √(a ± 2√b), look for two numbers whose sum is 'a' and product is 'b'. If found (say x, y), the answer is √x ± √y.
  3. Comparing Surds: To compare n√a and m√b, raise both to the power of LCM(n, m). For example, to compare 3√2 and 2√3, raise both to power 6: (3√2)^6 = (2^(1/3))^6 = 2^2 = 4 and (2√3)^6 = (3^(1/2))^6 = 3^3 = 27. Since 27 > 4, 2√3 > 3√2.
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Start Lesson: BODMAS & Arithmetic Simplification