Mathematics: Same Topics as CBT-1 at a Higher Difficulty Level

Welcome to the advanced mathematics section for your RRB ALP CBT-2 Part A preparation. This section covers the same core topics as CBT-1, but with increased complexity and depth. Mastering these concepts is crucial for success. We will break down each topic, providing detailed explanations, examples, and strategies to tackle challenging problems.

1. Number System

The number system is the foundation of all mathematical operations. In CBT-2, you'll encounter more complex problems involving different types of numbers and their properties.

1.1 Types of Numbers

Understanding the classification of numbers is essential.

• Natural Numbers (N): {1, 2, 3, ...}
• Whole Numbers (W): {0, 1, 2, 3, ...}
• Integers (Z): {..., -3, -2, -1, 0, 1, 2, 3, ...}
• Rational Numbers (Q): Numbers that can be expressed as p/q, where p and q are integers and q ≠ 0. (e.g., 1/2, -3/4, 5)
• Irrational Numbers (I): Numbers that cannot be expressed as p/q. Their decimal representation is non-terminating and non-repeating. (e.g., √2, π, e)
• Real Numbers (R): The set of all rational and irrational numbers.
• Prime Numbers: Natural numbers greater than 1 that have only two divisors: 1 and themselves. (e.g., 2, 3, 5, 7, 11)
• Composite Numbers: Natural numbers greater than 1 that are not prime. (e.g., 4, 6, 8, 9, 10)
• Even Numbers: Integers divisible by 2.
• Odd Numbers: Integers not divisible by 2.

1.2 Divisibility Rules

These rules help quickly determine if a number is divisible by another without performing division.

• Divisibility by 2: The last digit is even.
• Divisibility by 3: The sum of the digits is divisible by 3.
• Divisibility by 4: The number formed by the last two digits is divisible by 4.
• Divisibility by 5: The last digit is 0 or 5.
• Divisibility by 6: The number is divisible by both 2 and 3.
• Divisibility by 8: The number formed by the last three digits is divisible by 8.
• Divisibility by 9: The sum of the digits is divisible by 9.
• Divisibility by 10: The last digit is 0.
• Divisibility by 11: The alternating sum of the digits is divisible by 11.

1.3 HCF and LCM

Highest Common Factor (HCF) and Lowest Common Multiple (LCM) are crucial for problems involving ratios and proportions, and time and work.

• HCF (GCD): The largest positive integer that divides two or more integers without leaving a remainder.
• LCM: The smallest positive integer that is a multiple of two or more integers.

Relationship: For two numbers 'a' and 'b', HCF(a, b) * LCM(a, b) = a * b.

Methods: Prime factorization and division method.

Example: Find HCF and LCM of 12 and 18.

• Prime Factorization: 12 = 2² * 3, 18 = 2 * 3²
• HCF = 2¹ * 3¹ = 6
• LCM = 2² * 3² = 4 * 9 = 36
• Check: 6 * 36 = 216. 12 * 18 = 216. Correct.

1.4 Problems on Numbers

These problems often involve algebraic representation of numbers and solving equations based on digit manipulation.

Example: A two-digit number is such that the sum of its digits is 9. If 27 is added to the number, the digits are reversed. Find the number.

Let the two-digit number be 10x + y, where x is the tens digit and y is the units digit.

Given: x + y = 9 (Equation 1)

Given: (10x + y) + 27 = 10y + x

10x + y + 27 = 10y + x

9x - 9y = -27

x - y = -3 (Equation 2)

Adding Equation 1 and Equation 2:

(x + y) + (x - y) = 9 + (-3)

2x = 6 => x = 3

Substitute x = 3 in Equation 1: 3 + y = 9 => y = 6

The number is 10(3) + 6 = 36.

2. Decimals and Fractions

This topic tests your ability to convert between decimals and fractions, perform operations, and compare values.

2.1 Types of Fractions

• Proper Fraction: Numerator < Denominator (e.g., 3/5)
• Improper Fraction: Numerator ≥ Denominator (e.g., 7/4)
• Mixed Fraction: An integer part and a proper fraction part (e.g., 1 3/4)
• Equivalent Fractions: Fractions that represent the same value (e.g., 1/2 = 2/4)
• Recurring Decimals: Decimals that have a repeating pattern of digits.

2.2 Conversion between Fractions and Decimals

To convert a fraction to a decimal, divide the numerator by the denominator.

To convert a terminating decimal to a fraction, write the decimal as the numerator and a power of 10 as the denominator, then simplify.

Example: Convert 0.625 to a fraction.

0.625 = 625/1000. Simplifying by dividing numerator and denominator by 125 gives 5/8.

2.3 Operations on Fractions and Decimals

Addition, subtraction, multiplication, and division require careful attention to rules, especially with mixed numbers and recurring decimals.

Recurring Decimals:

• 0.abc (repeating): abc / 999
• 0.ab (repeating): ab / 99
• 0.a(bc) (repeating): (abc - a) / 990

2.4 Comparison of Fractions and Decimals

To compare fractions, either convert them to decimals or find a common denominator.

Example: Which is greater, 5/8 or 7/12?

Method 1: Convert to decimals. 5/8 = 0.625, 7/12 ≈ 0.583. So, 5/8 is greater.

Method 2: Common denominator (LCM of 8 and 12 is 24). 5/8 = 15/24, 7/12 = 14/24. Since 15 > 14, 15/24 (or 5/8) is greater.

3. Simplification

This section involves simplifying complex mathematical expressions using various rules and BODMAS/PEMDAS.

3.1 BODMAS/PEMDAS Rule

This rule dictates the order of operations:

• Brackets (Parentheses)
• Orders (Exponents, Roots)
• Division
• Multiplication
• Addition
• Subtraction

Multiplication and Division have equal precedence, solved from left to right. Addition and Subtraction also have equal precedence, solved from left to right.

3.2 Algebraic Simplification

This involves simplifying expressions with variables, often using factorization and identities.

Key Identities:

• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²
• a² - b² = (a + b)(a - b)
• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
• (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)

3.3 Surds and Indices

Simplifying expressions involving roots and powers.

Laws of Indices:

• a^m * aⁿ = a^m+n
• a^m / aⁿ = a^m-n
• (a^m)ⁿ = a^mn
• (ab)ⁿ = aⁿbⁿ
• (a/b)ⁿ = aⁿ/bⁿ
• a⁰ = 1
• a^-n = 1/aⁿ

Laws of Surds:

• √ab = √a * √b
• √(a/b) = √a / √b
• Rationalizing the denominator is a common technique.

Example: Simplify (√5 + √3)².

(√5 + √3)² = (√5)² + 2(√5)(√3) + (√3)²

= 5 + 2√15 + 3

= 8 + 2√15

4. Percentage

Percentages are fundamental for understanding profit/loss, interest, and discounts.

4.1 Conversion

• To convert a percentage to a fraction: Divide by 100. (e.g., 25% = 25/100 = 1/4)
• To convert a fraction to a percentage: Multiply by 100. (e.g., 1/4 = (1/4) * 100 = 25%)
• To convert a decimal to a percentage: Multiply by 100. (e.g., 0.75 = 0.75 * 100 = 75%)

4.2 Percentage Increase and Decrease

Formula for Percentage Increase: [(New Value - Original Value) / Original Value] * 100

Formula for Percentage Decrease: [(Original Value - New Value) / Original Value] * 100

Example: A price increased from 500 to 600. Find the percentage increase.

Increase = 600 - 500 = 100

Percentage Increase = (100 / 500) * 100 = (1/5) * 100 = 20%

4.3 Successive Percentages

When a quantity is increased or decreased by successive percentages, the final result is not simply the sum of the percentages.

Formula: If a value is increased by x% and then by y%, the net percentage increase is [x + y + (xy/100)]%.

If there is a decrease, use a negative sign for that percentage.

Example: A price is increased by 10% and then by 20%. Find the net percentage increase.

Net Increase = [10 + 20 + (10 * 20 / 100)]% = [30 + (200 / 100)]% = (30 + 2)% = 32%.

4.4 Percentage and its Applications (Profit/Loss, Discount, Simple/Compound Interest)

These are direct applications of percentage concepts, covered in detail in their respective sections.

5. Ratio and Proportion

This topic deals with the comparison of quantities and the equality of ratios.

5.1 Ratio

A ratio compares two quantities of the same kind. It is usually expressed in the simplest form.

Example: The ratio of boys to girls in a class is 3:5. If there are 15 boys, how many girls are there?

Let the number of boys be 3x and girls be 5x.

Given 3x = 15 => x = 5.

Number of girls = 5x = 5 * 5 = 25.

5.2 Proportion

A proportion states that two ratios are equal. If a:b = c:d, then ad = bc.

Example: Find the value of x if 4:x :: 12:24.

4/x = 12/24

4 * 24 = 12 * x

96 = 12x => x = 96/12 = 8.

5.3 Partnership

Problems involving partners investing capital for different periods. The profit is shared in the ratio of their effective investments (Capital * Time).

Example: A invests ₹10,000 for 4 months and B invests ₹8,000 for 6 months in a business. If the total profit is ₹44,000, find A's share.

A's effective investment = 10,000 * 4 = 40,000

B's effective investment = 8,000 * 6 = 48,000

Ratio of profits = 40,000 : 48,000 = 40 : 48 = 5 : 6.

Total parts = 5 + 6 = 11.

A's share = (5/11) * 44,000 = 5 * 4,000 = ₹20,000.

5.4 Variation (Direct and Inverse)

Direct Variation: If y varies directly as x (y ∝ x), then y = kx, or y1/x1 = y2/x2.

Inverse Variation: If y varies inversely as x (y ∝ 1/x), then y = k/x, or y1*x1 = y2*x2.

6. Time and Work

This topic involves calculating the time taken to complete a task individually or collectively.

6.1 Basic Concepts

• If a person can do a piece of work in 'x' days, they complete 1/x of the work per day.
• If a person completes 1/y of the work per day, they can complete the whole work in 'y' days.

6.2 Combined Work

If A can do a work in x days and B can do it in y days, then in one day, A does 1/x work and B does 1/y work. Together, they do (1/x + 1/y) work per day. The time taken by both to complete the work is 1 / (1/x + 1/y) = xy / (x+y) days.

Example: A can do a piece of work in 10 days, and B can do it in 15 days. How long will they take to complete the work together?

A's daily work = 1/10

B's daily work = 1/15

Combined daily work = 1/10 + 1/15 = (3 + 2) / 30 = 5/30 = 1/6.

Time taken together = 6 days.

6.3 Work, Wages, and Efficiency

Wages are generally distributed in proportion to the work done. Efficiency is inversely proportional to the time taken.

Example: A is twice as good a workman as B. Together they can finish a piece of work in 18 days. In how many days can A alone finish the work?

Efficiency ratio A:B = 2:1.

Time ratio is inversely proportional to efficiency, so A:B = 1:2.

Let A take x days and B take 2x days.

Combined work = 1 / (1/x + 1/2x) = 1 / (3/2x) = 2x/3.

Given combined work takes 18 days: 2x/3 = 18 => 2x = 54 => x = 27.

A alone can finish the work in 27 days.

6.4 Pipes and Cisterns

This is a variation of Time and Work, where pipes fill (positive work) and leaks empty (negative work) a cistern.

• If an inlet pipe can fill a cistern in x hours, it fills 1/x of the cistern per hour.
• If an outlet pipe can empty a cistern in y hours, it empties 1/y of the cistern per hour.

Example: Pipe A can fill a tank in 10 hours, and Pipe B can empty it in 15 hours. If both pipes are opened, how long will it take to fill the tank?

Work done by A in 1 hour = +1/10

Work done by B in 1 hour = -1/15

Combined work in 1 hour = 1/10 - 1/15 = (3 - 2) / 30 = 1/30.

Time to fill the tank = 30 hours.

7. Time, Speed, and Distance

This topic involves calculations related to motion.

7.1 Basic Formulas

• Distance = Speed × Time
• Speed = Distance / Time
• Time = Distance / Speed

7.2 Unit Conversions

Crucial for consistency in calculations.

• km/h to m/s: Multiply by 5/18. (e.g., 36 km/h = 36 * 5/18 = 10 m/s)
• m/s to km/h: Multiply by 18/5. (e.g., 20 m/s = 20 * 18/5 = 72 km/h)

7.3 Average Speed

Average Speed = Total Distance / Total Time.

Important Note: Average speed is NOT the average of the speeds, unless the time intervals are equal.

Example: A car travels the first 100 km at 50 km/h and the next 100 km at 40 km/h. Find the average speed.

Time taken for the first part = 100 km / 50 km/h = 2 hours.

Time taken for the second part = 100 km / 40 km/h = 2.5 hours.

Total Distance = 100 + 100 = 200 km.

Total Time = 2 + 2.5 = 4.5 hours.

Average Speed = 200 km / 4.5 hours = 2000 / 45 = 400 / 9 ≈ 44.44 km/h.

(Note: Average of speeds = (50+40)/2 = 45 km/h, which is incorrect).

7.4 Relative Speed

Used when two objects are moving.

• Same Direction: Relative Speed = |Speed1 - Speed2|
• Opposite Direction: Relative Speed = Speed1 + Speed2

Example: Two trains moving in the same direction at 60 km/h and 40 km/h. How long will the faster train take to pass the slower train if they are 1 km apart?

Relative Speed = 60 - 40 = 20 km/h.

Distance to cover = 1 km.

Time = Distance / Relative Speed = 1 km / 20 km/h = 1/20 hours.

1/20 hours = (1/20) * 60 minutes = 3 minutes.

7.5 Problems on Trains

Key considerations:

• When a train passes a pole or a standing person, the distance covered is the length of the train itself.
• When a train passes a platform or a bridge, the distance covered is the length of the train + length of the platform/bridge.

Example: A train 150m long passes a platform 250m long in 10 seconds. Find the speed of the train.

Total distance covered = Length of train + Length of platform = 150m + 250m = 400m.

Time taken = 10 seconds.

Speed = Distance / Time = 400m / 10s = 40 m/s.

Convert to km/h: 40 m/s * (18/5) = 8 * 18 = 144 km/h.

7.6 Problems on Boats and Streams

• Speed in Still Water: Let this be 'u'.
• Speed of the Stream: Let this be 'v'.
• Speed Downstream (with the stream): u + v
• Speed Upstream (against the stream): u - v

Example: A boat takes 3 hours to travel downstream and 5 hours to travel upstream a distance of 15 km. Find the speed of the boat in still water.

Speed Downstream = 15 km / 3 hours = 5 km/h. So, u + v = 5.

Speed Upstream = 15 km / 5 hours = 3 km/h. So, u - v = 3.

Adding the two equations: (u + v) + (u - v) = 5 + 3 => 2u = 8 => u = 4 km/h.

The speed of the boat in still water is 4 km/h.

8. Simple and Compound Interest

These concepts are vital for financial aptitude.

8.1 Simple Interest (SI)

Interest calculated only on the principal amount.

Formula: SI = (P × R × T) / 100

• P = Principal (initial amount)
• R = Rate of interest per annum
• T = Time period in years

Amount (A) = P + SI

Example: Calculate the simple interest on ₹5,000 at 8% per annum for 3 years.

SI = (5000 × 8 × 3) / 100 = 50 × 8 × 3 = 400 × 3 = ₹1,200.

Amount = 5000 + 1200 = ₹6,200.

8.2 Compound Interest (CI)

Interest calculated on the principal amount plus the accumulated interest from previous periods.

Formula for Amount (A): A = P (1 + R/100)^T

Compound Interest (CI) = A - P

When interest is compounded half-yearly: A = P (1 + (R/2)/100)^2T

When interest is compounded quarterly: A = P (1 + (R/4)/100)^4T

Example: Calculate the compound interest on ₹5,000 at 8% per annum for 3 years, compounded annually.

A = 5000 (1 + 8/100)³

A = 5000 (1 + 0.08)³

A = 5000 (1.08)³

A = 5000 × 1.259712 ≈ 6298.56

CI = A - P = 6298.56 - 5000 = ₹1298.56.

8.3 Difference between CI and SI

The difference arises because CI earns interest on interest, while SI does not.

• For 1 year: Difference = 0
• For 2 years: Difference = P(R/100)²
• For 3 years: Difference = P(R/100)²(3 + R/100)

9. Mensuration

This topic involves calculating the area, perimeter, volume, and surface area of various geometric shapes.

9.1 2D Shapes (Area and Perimeter)

• Square: Side 'a'. Area = a², Perimeter = 4a.
• Rectangle: Length 'l', Breadth 'b'. Area = lb, Perimeter = 2(l + b).
• Triangle: Base 'b', Height 'h'. Area = (1/2)bh. For equilateral triangle (side 'a'), Area = (√3/4)a². For right-angled triangle, Area = (1/2) * product of perpendicular sides.
• Circle: Radius 'r'. Area = πr², Circumference = 2πr. Diameter 'd'. Area = π(d/2)², Circumference = πd.
• Parallelogram: Base 'b', Height 'h'. Area = bh.
• Trapezium: Parallel sides 'a', 'b', Height 'h'. Area = (1/2)(a + b)h.

Example: The radius of a circle is 7 cm. Find its circumference and area.

Circumference = 2πr = 2 * (22/7) * 7 = 44 cm.

Area = πr² = (22/7) * 7² = 22 * 7 = 154 cm².

9.2 3D Shapes (Volume and Surface Area)

• Cuboid: Length 'l', Breadth 'b', Height 'h'. Volume = lbh. Surface Area = 2(lb + bh + hl).
• Cube: Side 'a'. Volume = a³. Surface Area = 6a².
• Cylinder: Radius 'r', Height 'h'. Volume = πr²h. Curved Surface Area = 2πrh. Total Surface Area = 2πrh + 2πr² = 2πr(h + r).
• Cone: Radius 'r', Height 'h', Slant height 'l'. Volume = (1/3)πr²h. Curved Surface Area = πrl. Total Surface Area = πrl + πr² = πr(l + r). (Note: l = √(r² + h²))
• Sphere: Radius 'r'. Volume = (4/3)πr³. Surface Area = 4πr².
• Hemisphere: Radius 'r'. Volume = (2/3)πr³. Curved Surface Area = 2πr². Total Surface Area = 3πr².

Example: Find the volume and total surface area of a cylinder with radius 3 cm and height 7 cm.

Volume = πr²h = (22/7) * 3² * 7 = 22 * 9 = 198 cm³.

Total Surface Area = 2πr(h + r) = 2 * (22/7) * 3 * (7 + 3) = 2 * (22/7) * 3 * 10 = (1320 / 7) cm² ≈ 188.57 cm².

10. Algebra

This section tests your understanding of algebraic expressions, equations, and inequalities.

10.1 Basic Algebraic Expressions

Simplifying, expanding, and factorizing expressions. Operations like addition, subtraction, multiplication, and division of polynomials.

10.2 Linear Equations in One Variable

Equations of the form ax + b = c, where x is the variable.

Example: Solve for x: 3(x - 2) + 5 = 2x + 7

3x - 6 + 5 = 2x + 7

3x - 1 = 2x + 7

3x - 2x = 7 + 1

x = 8

10.3 Quadratic Equations

Equations of the form ax² + bx + c = 0.

Methods for solving: Factorization, completing the square, quadratic formula.

Quadratic Formula: x = [-b ± √(b² - 4ac)] / 2a

Discriminant (Δ): Δ = b² - 4ac. Determines the nature of roots:

• If Δ > 0, two distinct real roots.
• If Δ = 0, two equal real roots.
• If Δ < 0, no real roots (complex roots).

Example: Solve x² - 5x + 6 = 0.

Using factorization: (x - 2)(x - 3) = 0. So, x = 2 or x = 3.

Using quadratic formula: a=1, b=-5, c=6.

x = [ -(-5) ± √((-5)² - 4*1*6) ] / (2*1)

x = [ 5 ± √(25 - 24) ] / 2

x = [ 5 ± √1 ] / 2

x = (5 + 1) / 2 = 6/2 = 3

x = (5 - 1) / 2 = 4/2 = 2

10.4 Arithmetic Progression (AP) and Geometric Progression (GP)

AP: Sequence where the difference between consecutive terms is constant (common difference, 'd').

• nth term (a_n) = a + (n-1)d
• Sum of n terms (S_n) = (n/2) * [2a + (n-1)d] or (n/2) * (a + a_n)

GP: Sequence where the ratio between consecutive terms is constant (common ratio, 'r').

• nth term (a_n) = ar^n-1
• Sum of n terms (S_n) = a(rⁿ - 1) / (r - 1) (if r > 1)
• Sum of n terms (S_n) = a(1 - rⁿ) / (1 - r) (if r < 1)

Example: Find the sum of the first 10 terms of the AP: 2, 5, 8, ...

a = 2, d = 3, n = 10.

S₁₀ = (10/2) * [2*2 + (10-1)*3]

S₁₀ = 5 * [4 + 9*3]

S₁₀ = 5 * [4 + 27]

S₁₀ = 5 * 31 = 155.

11. Probability

The measure of the likelihood of an event occurring.

11.1 Basic Concepts

• Sample Space (S): Set of all possible outcomes.
• Event (E): A subset of the sample space.
• Probability of an Event P(E): P(E) = Number of favorable outcomes / Total number of outcomes = n(E) / n(S).

Range of Probability: 0 ≤ P(E) ≤ 1.

• P(E) = 0 means the event is impossible.
• P(E) = 1 means the event is certain.

11.2 Coins, Dice, and Cards

Problems typically involve calculating probabilities related to tossing coins, rolling dice, or drawing cards from a standard deck.

Example: What is the probability of getting a sum of 7 when two dice are rolled?

Total possible outcomes n(S) = 6 * 6 = 36.

Favorable outcomes for a sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). So, n(E) = 6.

P(Sum of 7) = n(E) / n(S) = 6 / 36 = 1/6.

11.3 Conditional Probability

The probability of an event occurring given that another event has already occurred. P(A|B) = P(A ∩ B) / P(B).

12. Statistics

This involves the collection, analysis, interpretation, and presentation of data.

12.1 Mean, Median, and Mode

• Mean: Average of the data set. Sum of all observations / Number of observations.
• Median: The middle value in a sorted data set. If the number of observations (n) is odd, median is the ((n+1)/2)th term. If n is even, median is the average of the (n/2)th and ((n/2)+1)th terms.
• Mode: The value that appears most frequently in the data set.

Example: Find the mean, median, and mode of the data: 3, 5, 2, 5, 8, 5, 9.

Sorted data: 2, 3, 5, 5, 5, 8, 9.

Mean = (2+3+5+5+5+8+9) / 7 = 37 / 7 ≈ 5.29

Median: There are 7 observations (odd). The middle term is the (7+1)/2 = 4th term, which is 5.

Mode: The number 5 appears most frequently (3 times).

12.2 Data Interpretation (Charts and Graphs)

Problems often involve interpreting information presented in tables, bar graphs, pie charts, line graphs, etc. Focus on identifying trends, comparing values, and calculating percentages or ratios from the data.