Simplification
Concepts & Definitions
- Simplification: Reducing a complex mathematical expression to its simplest form by following a fixed order of operations.
- VBODMAS: The order in which operations must be solved: Vinculum, Brackets, Orders (powers, roots) / Of, Division, Multiplication, Addition, Subtraction.
- Regional variations: BODMAS (India/UK), BIDMAS / BEDMAS (UK/Canada) and PEMDAS (USA) follow exactly the same logic.
The VBODMAS Order
| Letter | Operation | What to do |
|---|---|---|
| V | Vinculum (Bar) | Solve the terms under a horizontal bar first, treating them as one bracket. |
| B | Brackets | Work from the innermost to the outermost: ( ) → { } → [ ]. |
| O | Orders / Of | Evaluate powers, roots and the word “of” (which means multiplication). |
| D, M | Division, Multiplication | Equal priority. Solve left to right, in the order they appear. |
| A, S | Addition, Subtraction | Equal priority. Solve left to right, in the order they appear. |
Essential Formulas
Algebraic Identities
| Identity | Use / Logic |
|---|---|
| (a + b)² = a² + b² + 2ab | Expand a sum squared; find a² + b² from a + b and ab. |
| (a − b)² = a² + b² − 2ab | Expand a difference squared; find a² + b² from a − b and ab. |
| (a + b)² = (a − b)² + 4ab | Convert between sum and difference forms. |
| a² − b² = (a − b)(a + b) | Quick squares, e.g. 87² − 13² = 74 × 100. |
| a³ + b³ = (a + b)(a² − ab + b²) | Sum of cubes. |
| a³ − b³ = (a − b)(a² + ab + b²) | Difference of cubes. |
| a² + b² = ½ [(a + b)² + (a − b)²] | Find a² + b² when both a + b and a − b are known. |
Sums of Counting Numbers
| Series | Formula |
|---|---|
| Sum of first n natural numbers | n(n + 1) / 2 |
| Sum of first n odd natural numbers | n² |
| Sum of first n even natural numbers | n(n + 1) |
Laws of Indices (Exponents)
| Law | Example |
|---|---|
| aᵐ × aⁿ = aᵐ⁺ⁿ | 2³ × 2⁴ = 2⁷ |
| aᵐ ÷ aⁿ = aᵐ⁻ⁿ | 3⁵ ÷ 3² = 3³ |
| (aᵐ)ⁿ = aᵐⁿ | (2³)² = 2⁶ |
| (ab)ⁿ = aⁿ bⁿ | (2 × 3)² = 2² × 3² |
| (a/b)ⁿ = aⁿ / bⁿ | (2/3)² = 4/9 |
| a⁰ = 1 | 7⁰ = 1 |
Laws of Surds
| Law | Example |
|---|---|
| ⁿ√a = a(1/n) | √9 = 9(1/2) = 3 |
| ⁿ√(ab) = ⁿ√a × ⁿ√b | √12 = √4 × √3 = 2√3 |
| ⁿ√(a/b) = ⁿ√a / ⁿ√b | √(50/2) = √50 / √2 = 5 |
| (ⁿ√a)ⁿ = a | (√5)² = 5 |
| (ⁿ√a)ᵐ = ⁿ√(aᵐ) | (√3)³ = √27 |
Bracket Sign Rules
- x + (y + z) = x + y + z
- x − (y + z) = x − y − z (the minus sign changes every sign inside)
- x(y + z) = xy + xz
Essential Shortcuts & Solution Methods
A. Direct VBODMAS Evaluation
- Procedure: Clear the innermost bracket or bar first, then orders/of, then division and multiplication (left to right), then addition and subtraction (left to right).
- Example: 16 + (10 − 3 × 2). Bracket first: 10 − 6 = 4. Then 16 + 4 = 20.
- Pitfall 1: Division does not beat multiplication. 8 ÷ 2 × 4 = 4 × 4 = 16 (not 8 ÷ 8 = 1).
- Pitfall 2: Addition does not beat subtraction. 1 − 3 + 4 = −2 + 4 = 2 (not 1 − 7 = −6).
B. Symbol Substitution
- Procedure: Replace each coded symbol with the operation it stands for, rewrite the whole expression, then apply VBODMAS. Substitute all signs at once, never one by one.
- Example: If ‘+’ means ‘÷’, ‘−’ means ‘×’, ‘×’ means ‘+’, ‘÷’ means ‘−’, find 45 + 9 − 3 × 15 ÷ 2.
- Substitute: 45 ÷ 9 × 3 + 15 − 2 → 5 × 3 + 15 − 2 → 15 + 15 − 2 = 28.
C. Interchanging Signs and Numbers
- Procedure: Swap each pair of signs (or numbers) given in the options, one option at a time, and evaluate the left side. The correct option is the one that makes both sides equal.
- Time saver: First compare the current left-hand value with the right-hand value to see whether you need a bigger or smaller result, then eliminate options that go the wrong way.
D. Clearing Decimals and Fractions in Linear Equations
- Decimals: Multiply both sides by 10, 100, … according to the maximum number of decimal places.
2.3x + 2.8 = −1.2x + 9.8 → (×10) → 23x + 28 = −12x + 98 → 35x = 70 → x = 2. - Fractions: Multiply both sides by the LCM of all denominators.
x/3 + 1/5 = x/5 − 1 → (×15) → 5x + 3 = 3x − 15 → 2x = −18 → x = −9.
E. Repeating Decimals to Fractions
- Pure repeating (0.818181…): Let x = 0.818181…. Two digits repeat, so multiply by 100: 100x = 81.8181… = 81 + x.
99x = 81, so x = 81/99 = 9/11. - Shortcut: For a pure repeating decimal, the fraction is (repeating digits) / (as many 9s as digits). So 0.(27) = 27/99 = 3/11.
Some Solved Problems on Simplification
1. Simplify: 36 ÷ 4 × 3 + 12 − 8 ÷ 2
Answer: (B) 35
Solution: Division and multiplication have equal priority, so go left to right: 36 ÷ 4 = 9, then 9 × 3 = 27.
Next, 8 ÷ 2 = 4.
Expression becomes 27 + 12 − 4 = 35.
2. Simplify: 30 − [8 + {15 − (9 − 2)}]
Answer: (A) 14
Solution: Innermost bracket: (9 − 2) = 7.
Curly bracket: {15 − 7} = 8.
Square bracket: [8 + 8] = 16.
30 − 16 = 14.
3. If ‘+’ means ‘×’, ‘−’ means ‘÷’, ‘×’ means ‘−’ and ‘÷’ means ‘+’, find the value of 12 + 6 − 3 × 4 ÷ 2.
Answer: (C) 22
Solution: Substitute the signs: 12 × 6 ÷ 3 − 4 + 2.
Left to right: 12 × 6 = 72, then 72 ÷ 3 = 24.
24 − 4 + 2 = 22.
4. Which interchange of signs makes the equation correct? 4 + 5 × 6 − 12 ÷ 3 = 22
Answer: (A) + and ×
Solution: Interchange + and ×: 4 × 5 + 6 − 12 ÷ 3.
= 20 + 6 − 4 = 22, which equals the right-hand side.
The other three interchanges give 3.5, −22 and −15, so the answer is + and ×.
5. Solve for x: 0.6x + 1.5 = 0.2x + 3.1
Answer: (C) 4
Solution: Maximum decimal places = 1, so multiply both sides by 10: 6x + 15 = 2x + 31.
4x = 16, so x = 4.
6. Solve for x: x/3 + 1/4 = x/6 + 5/4
Answer: (D) 6
Solution: LCM of 3, 4, 6 = 12. Multiply every term by 12: 4x + 3 = 2x + 15.
2x = 12, so x = 6.
7. Find the value of 0.(36) + 0.(63), where the digits in brackets repeat.
Answer: (B) 1
Solution: 0.(36) = 36/99 and 0.(63) = 63/99 (two repeating digits, so divide by 99).
Sum = (36 + 63)/99 = 99/99 = 1.
8. If a + b = 10 and ab = 21, find a² + b².
Answer: (A) 58
Solution: Use (a + b)² = a² + b² + 2ab.
100 = a² + b² + 42, so a² + b² = 58.
9. If a + b = 5 and ab = 6, find a³ + b³.
Answer: (D) 35
Solution: First, a² + b² = (a + b)² − 2ab = 25 − 12 = 13.
a³ + b³ = (a + b)(a² − ab + b²) = 5 × (13 − 6) = 5 × 7 = 35.
10. Evaluate: 87² − 13²
Answer: (B) 7400
Solution: Use a² − b² = (a − b)(a + b).
(87 − 13)(87 + 13) = 74 × 100 = 7400.
11. Find the sum of the first 20 odd natural numbers and the first 15 even natural numbers.
Answer: (C) 640
Solution: Sum of first n odd numbers = n², so 20² = 400.
Sum of first n even numbers = n(n + 1), so 15 × 16 = 240.
Total = 400 + 240 = 640.
12. Simplify: (2⁷ × 4³) ÷ 8²
Answer: (A) 128
Solution: Write everything with base 2: 4³ = 2⁶ and 8² = 2⁶.
(2⁷ × 2⁶) ÷ 2⁶ = 2^(7 + 6 − 6) = 2⁷ = 128.
13. Simplify: √12 × √3 + √50 ÷ √2
Answer: (D) 11
Solution: √12 × √3 = √36 = 6.
√50 ÷ √2 = √(50/2) = √25 = 5.
6 + 5 = 11.
14. Using the digital root check, which of the following products is wrong?
Answer: (C) 52 × 19 = 998
Solution: DR(52) = 7 and DR(19) = 1, so DR of the product should be 7 × 1 = 7.
DR(998) = 9 + 9 + 8 = 26, then 2 + 6 = 8. This does not match 7, so the product is wrong (the correct value is 988).
The other three products all match.