Number Series

Number series questions are a common feature in competitive exams. The goal is to identify the pattern in a given sequence of numbers and then find the missing number or the next number in the series. These patterns can be arithmetic, geometric, based on squares or cubes, or a combination of various operations.

Types of Number Series Patterns

1. Arithmetic Progression (AP)

In an AP, the difference between consecutive terms is constant. This constant difference is called the common difference (d).

Example: 2, 5, 8, 11, 14, ?

Here, the common difference is 3 (5-2=3, 8-5=3, etc.). The next term will be 14 + 3 = 17.

2. Geometric Progression (GP)

In a GP, the ratio between consecutive terms is constant. This constant ratio is called the common ratio (r).

Example: 3, 6, 12, 24, 48, ?

Here, the common ratio is 2 (6/3=2, 12/6=2, etc.). The next term will be 48 * 2 = 96.

3. Squares and Cubes

The series might be based on the squares or cubes of consecutive numbers.

Example (Squares): 1, 4, 9, 16, 25, ?

These are the squares of 1, 2, 3, 4, 5. The next term is 62 = 36.

Example (Cubes): 1, 8, 27, 64, 125, ?

These are the cubes of 1, 2, 3, 4, 5. The next term is 63 = 216.

4. Alternating Series

In this type, two or more different patterns are interleaved.

Example: 1, 5, 3, 7, 5, 9, ?

The pattern here alternates between adding 2 and subtracting 2, or it can be viewed as two separate series: 1, 3, 5, ? (adding 2) and 5, 7, 9, ? (adding 2). The next term in the first series is 5 + 2 = 7.

5. Difference of Differences

If the first difference between terms is not constant, check the difference between these differences. This is also known as a second-order difference.

Example: 3, 7, 13, 21, 31, ?

First differences: 4, 6, 8, 10. (7-3=4, 13-7=6, 21-13=8, 31-21=10)

Second differences: 2, 2, 2. (6-4=2, 8-6=2, 10-8=2)

Since the second difference is constant (2), the next first difference will be 10 + 2 = 12. The next term in the series is 31 + 12 = 43.

6. Combination Patterns

These involve a mix of operations like addition, subtraction, multiplication, division, squares, cubes, prime numbers, etc.

Example: 5, 10, 20, 35, 55, ?

First differences: 5, 10, 15, 20. (10-5=5, 20-10=10, 35-20=15, 55-35=20)

The differences are increasing by 5 each time. The next difference will be 20 + 5 = 25. The next term is 55 + 25 = 80.

How to Solve Number Series Questions

  1. Analyze the Series: Look at the numbers. Are they increasing or decreasing? Is the change gradual or rapid?
  2. Calculate Differences: Find the difference between consecutive terms. If the difference is constant, it's an AP.
  3. Calculate Ratios: If differences don't reveal a pattern, try finding the ratio between consecutive terms. If constant, it's a GP.
  4. Check for Squares/Cubes: See if the numbers are perfect squares or cubes, or related to them.
  5. Look for Alternating Patterns: Try splitting the series into two or more sub-series.
  6. Calculate Second Differences: If the first difference isn't constant, check the differences of the differences.
  7. Consider Mixed Operations: Think about combinations of addition, subtraction, multiplication, division, and powers.
  8. Prime Numbers: Sometimes series are based on prime numbers (2, 3, 5, 7, 11, ...).
  9. Fibonacci Sequence: In a Fibonacci sequence, each number is the sum of the two preceding ones (e.g., 0, 1, 1, 2, 3, 5, 8, ...).
Memory Trick: When faced with number series, always start with the simplest patterns (AP, GP) and gradually move to more complex ones (squares, cubes, differences, combinations). Don't get stuck on one pattern if it's not working.

Practice Examples

Example 1: 7, 15, 31, 63, 127, ?

Pattern: Multiply by 2 and add 1.

7 * 2 + 1 = 15

15 * 2 + 1 = 31

31 * 2 + 1 = 63

63 * 2 + 1 = 127

Next term: 127 * 2 + 1 = 254 + 1 = 255

Example 2: 4, 9, 20, 43, ?

Pattern: Multiply by 2, add 1; Multiply by 2, add 2; Multiply by 2, add 3...

4 * 2 + 1 = 9

9 * 2 + 2 = 18 + 2 = 20

20 * 2 + 3 = 40 + 3 = 43

Next term: 43 * 2 + 4 = 86 + 4 = 90

Example 3: 100, 95, 80, 55, 20, ?

Pattern: Subtract 5, subtract 15, subtract 25, subtract 35...

100 - 5 = 95

95 - 15 = 80

80 - 25 = 55

55 - 35 = 20

The differences are decreasing by 10 each time (5, 15, 25, 35). The next difference will be 35 + 10 = 45. The next term is 20 - 45 = -25.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree. This means they contain at least one term that is squared. The standard form of a quadratic equation is:

ax2 + bx + c = 0

where 'a', 'b', and 'c' are coefficients (constants), and 'x' is the variable. The coefficient 'a' cannot be zero (a ≠ 0), otherwise, it would be a linear equation.

Solving Quadratic Equations

There are several methods to solve quadratic equations, each suitable for different types of problems. The most common methods are:

1. Factoring Method

This method involves rewriting the quadratic expression as a product of two linear factors. It's usually the quickest method if the expression is easily factorable.

Steps:

  1. Ensure the equation is in the standard form ax2 + bx + c = 0.
  2. Find two numbers that multiply to 'ac' (the product of the coefficient of x2 and the constant term) and add up to 'b' (the coefficient of x).
  3. Rewrite the middle term (bx) using these two numbers.
  4. Factor by grouping.
  5. Set each factor equal to zero and solve for x.

Example: Solve x2 + 5x + 6 = 0

Here, a = 1, b = 5, c = 6. So, ac = 1 * 6 = 6.

We need two numbers that multiply to 6 and add up to 5. These numbers are 2 and 3.

Rewrite the equation: x2 + 2x + 3x + 6 = 0

Factor by grouping: x(x + 2) + 3(x + 2) = 0

Factor out the common term (x + 2): (x + 2)(x + 3) = 0

Set each factor to zero:

x + 2 = 0 => x = -2

x + 3 = 0 => x = -3

The solutions are x = -2 and x = -3.

2. Quadratic Formula Method

This is a universal method that works for all quadratic equations. The formula is derived using the process of completing the square.

For an equation ax2 + bx + c = 0, the solutions for x are given by:

x = [-b ± sqrt(b2 - 4ac)] / 2a

The term inside the square root, 'b2 - 4ac', is called the discriminant (Δ or D).

Discriminant (Δ):

  • If Δ > 0, there are two distinct real roots.
  • If Δ = 0, there is exactly one real root (or two equal real roots).
  • If Δ < 0, there are no real roots (two complex conjugate roots).

Example: Solve 2x2 - 7x + 3 = 0

Here, a = 2, b = -7, c = 3.

Calculate the discriminant: Δ = b2 - 4ac = (-7)2 - 4 * 2 * 3 = 49 - 24 = 25.

Since Δ > 0, there are two distinct real roots.

Apply the quadratic formula:

x = [-(-7) ± sqrt(25)] / (2 * 2)

x = [7 ± 5] / 4

Two solutions:

x1 = (7 + 5) / 4 = 12 / 4 = 3

x2 = (7 - 5) / 4 = 2 / 4 = 1/2

The solutions are x = 3 and x = 1/2.

3. Completing the Square Method

This method involves manipulating the equation to create a perfect square trinomial on one side. It's the basis for deriving the quadratic formula.

Steps:

  1. Ensure the equation is in the form ax2 + bx + c = 0.
  2. If a ≠ 1, divide the entire equation by 'a'.
  3. Move the constant term 'c/a' to the right side of the equation.
  4. Take half of the coefficient of the x term (b/a), square it [(b/2a)2], and add it to both sides of the equation.
  5. The left side will now be a perfect square trinomial. Factor it as (x + b/2a)2.
  6. Take the square root of both sides (remembering the ± sign).
  7. Solve for x.

Example: Solve x2 + 6x - 7 = 0

a = 1, b = 6, c = -7.

  1. Equation is already in standard form with a=1.
  2. Move constant: x2 + 6x = 7
  3. Half of the coefficient of x is 6/2 = 3. Square it: 32 = 9. Add 9 to both sides:

    4. x2 + 6x + 9 = 7 + 9

  4. Factor the left side: (x + 3)2 = 16
  5. Take square root: x + 3 = ±sqrt(16)

    6. x + 3 = ±4

  6. Solve for x:

    7. x + 3 = 4 => x = 1

    x + 3 = -4 => x = -7

    The solutions are x = 1 and x = -7.

Comparison of x and y (for Data Interpretation)

In data interpretation questions, you might be asked to compare the values of x and y derived from two different quadratic equations. The approach is to solve both equations and then compare their roots.

Example:

Equation I: x2 - 7x + 10 = 0

Equation II: y2 - 9y + 20 = 0

Solving Equation I:

ac = 10, b = -7. We need two numbers that multiply to 10 and add to -7. These are -2 and -5.

x2 - 2x - 5x + 10 = 0

x(x - 2) - 5(x - 2) = 0

(x - 2)(x - 5) = 0

So, x = 2 or x = 5.

Solving Equation II:

ac = 20, b = -9. We need two numbers that multiply to 20 and add to -9. These are -4 and -5.

y2 - 4y - 5y + 20 = 0

y(y - 4) - 5(y - 4) = 0

(y - 4)(y - 5) = 0

So, y = 4 or y = 5.

Comparison:

We have x values {2, 5} and y values {4, 5}.

Compare each x value with each y value:

  • If x = 2, then x < y (since y can be 4 or 5).
  • If x = 5, then x = y (since y can be 5).

Since x can be less than y and also equal to y, the relationship cannot be definitively stated. The answer would be "Cannot be determined" or "x ≤ y".

Exam Tip: Always check the signs of 'b' and 'c' coefficients.
  • If 'c' is positive, both roots have the same sign as 'b'.
  • If 'c' is negative, the roots have opposite signs.
  • If 'b' is positive and 'c' is positive, both roots are negative.
  • If 'b' is negative and 'c' is positive, both roots are positive.

Simplification and Approximation

These questions test your ability to perform calculations quickly and accurately, often involving large numbers, fractions, decimals, percentages, and square roots. The key is to use shortcuts and estimation techniques rather than brute-force calculation.

Simplification

Simplification problems require you to find the exact value of a given mathematical expression. The most important rule to follow is the order of operations.

BODMAS/PEMDAS Rule

This rule dictates the sequence in which operations should be performed:

  • Brackets (or Parentheses)
  • Orders (or Exponents/Indices/Roots)
  • Division
  • Multiplication
  • Addition
  • Subtraction

Division and Multiplication have the same priority and should be performed from left to right. Similarly, Addition and Subtraction have the same priority and should also be performed from left to right.

Example: Simplify 18 + [16 - (5 * 2) + 8] / 4

  1. Brackets: First, solve the innermost bracket (5 * 2) = 10.

    2. Expression becomes: 18 + [16 - 10 + 8] / 4

  2. Brackets: Now solve the remaining part inside the square brackets from left to right.

    3. 16 - 10 = 6

    6 + 8 = 14

    Expression becomes: 18 + 14 / 4

  3. Division: Perform the division.

    4. 14 / 4 = 3.5

    Expression becomes: 18 + 3.5

  4. Addition: Perform the addition.

    5. 18 + 3.5 = 21.5

    The simplified value is 21.5.

Key Techniques for Simplification

  • Fractions: Convert mixed fractions to improper fractions, find a common denominator for addition/subtraction, and simplify before multiplying.
  • Decimals: Align decimal points for addition/subtraction. For multiplication, multiply as whole numbers and then place the decimal point. For division, make the divisor a whole number by multiplying both numerator and denominator by a power of 10.
  • Percentages: Convert percentages to fractions (e.g., 20% = 1/5, 25% = 1/4) or decimals (e.g., 35% = 0.35) for easier calculation. Remember 'of' means multiplication.
  • Square Roots: Simplify square roots by finding perfect square factors (e.g., sqrt(72) = sqrt(36 * 2) = 6*sqrt(2)).

Approximation

Approximation questions ask for a value that is close to the exact answer. The goal is to make calculations easier by rounding numbers to the nearest convenient values.

Rounding Rules

  • If the digit to the right of the rounding place is 5 or greater, round up.
  • If the digit to the right is less than 5, round down (keep the digit as it is).

Example: Approximate the value of sqrt(4096) * (34.9% of 801) / (19.8 + 25.2)

Step 1: Approximate each term

  • sqrt(4096): We know 602 = 3600 and 702 = 4900. Since 4096 ends in 6, the square root might end in 4 or 6. Let's try 642 = (60+4)2 = 3600 + 2*60*4 + 16 = 3600 + 480 + 16 = 4096. So, sqrt(4096) = 64. (Exact value here, but usually it's an approximation).
  • 34.9% of 801: Round 34.9% to 35% and 801 to 800. So, 35% of 800.

    • 35% of 800 = (35/100) * 800 = 35 * 8 = 280.

  • 19.8 + 25.2: Round 19.8 to 20 and 25.2 to 25.

    • 20 + 25 = 45.

Step 2: Substitute the approximated values

The expression becomes: 64 * 280 / 45

Step 3: Perform the calculation

64 * 280 = 17920

17920 / 45

Approximate 17920 / 45. Let's estimate 18000 / 45.

18000 / 45 = 18000 / (9 * 5) = 2000 / 5 = 400.

Let's do the actual division for a closer approximation: 17920 / 45 ≈ 398.22

The approximate value is 398 or 400, depending on the options provided.

Approximation Strategy:
  • Round numbers judiciously to simplify calculations.
  • Convert percentages and decimals to fractions where convenient.
  • Perform calculations step-by-step, approximating at each stage if necessary, but be careful not to over-approximate too early.
  • Look at the options provided; they often guide the level of approximation needed.

Common Simplification/Approximation Techniques

  • Percentage Shortcuts:
    • a% of b = b% of a
    • 12.5% = 1/8, 25% = 1/4, 37.5% = 3/8, 50% = 1/2, 62.5% = 5/8, 75% = 3/4, 87.5% = 7/8
    • To find x% of Y, calculate (x/100) * Y.
  • Squaring Numbers:
    • Numbers ending in 5: (N5)2 = N*(N+1) followed by 25. (e.g., 352 = 3*4 followed by 25 = 1225).
    • Using (a+b)2 = a2 + 2ab + b2 or (a-b)2 = a2 - 2ab + b2. (e.g., 582 = (60-2)2 = 602 - 2*60*2 + 22 = 3600 - 240 + 4 = 3364).
  • Multiplication Shortcuts:
    • Multiplying by 5: Multiply by 10 and divide by 2.
    • Multiplying by 11: Write down the last digit, then add consecutive pairs of digits from right to left, carrying over if necessary, and finally write down the first digit plus any carry-over. (e.g., 345 * 11 = 3 (3+4) (4+5) 5 = 3795).
    • Multiplying numbers close to a base (e.g., 98 * 97): (100-2)*(100-3) = 10000 - 500 + 6 = 9506. Or, (98-3) | (2*3) = 95 | 06 = 9506.
  • Division Shortcuts:
    • Dividing by 5: Multiply by 2 and divide by 10.
    • Dividing by 25: Multiply by 4 and divide by 100.
    • Dividing by 125: Multiply by 8 and divide by 1000.
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