Partnership

Partnership is a business relationship where two or more individuals agree to share in the profits or losses of a business. The key elements of a partnership are agreement, business, mutual agency, and profit sharing.

Types of Partnerships

• Partnership-in-Partnership (or Simple Partnership): Here, capital is invested for the same duration by all partners. The profit is shared in the ratio of their capitals.
• Partnership-in-Continual (or Compound Partnership): Here, capital is invested for different durations by different partners. The profit is shared in the ratio of (Capital × Time).

Key Concepts and Formulas

Let's consider two partners, A and B, investing capital C_A and C_B for time T_A and T_B respectively. The profits P_A and P_B are shared accordingly.

Case 1: Investment for the same duration

If T_A = T_B, then the ratio of profits is equal to the ratio of their capitals.

P_A : P_B = C_A : C_B

Case 2: Investment for different durations

If T_A ≠ T_B, then the ratio of profits is equal to the ratio of their equivalent capitals.

Equivalent Capital of A = C_A × T_A

Equivalent Capital of B = C_B × T_B

P_A : P_B = (C_A × T_A) : (C_B × T_B)

Example

A starts a business by investing ₹40,000. After 3 months, B joins with an investment of ₹60,000. After 6 months from B's joining, C joins with an investment of ₹1,20,000. If the total profit at the end of the year is ₹45,000, what is the share of A?

Solution:

A's investment = ₹40,000 for 12 months. Equivalent capital = 40,000 × 12 = 4,80,000

B's investment = ₹60,000 for (12 - 3) = 9 months. Equivalent capital = 60,000 × 9 = 5,40,000

C's investment = ₹1,20,000 for (12 - 3 - 6) = 3 months. Equivalent capital = 1,20,000 × 3 = 3,60,000

Ratio of profits = 4,80,000 : 5,40,000 : 3,60,000

Simplify by dividing by 10,000: 48 : 54 : 36

Further simplify by dividing by 6: 8 : 9 : 6

Total parts of the ratio = 8 + 9 + 6 = 23

A's share = (8 / 23) × 45,000

A's share = ₹87,000 / 23 ≈ ₹3,782.61

Mixture and Alligation

Mixture and Alligation deals with problems involving the mixing of two or more ingredients or commodities to produce a mixture of a desired quality or price. Alligation is a rule that enables us to find the ratio in which two or more ingredients of known price/quality are to be mixed to produce a mixture of a desired price/quality.

Key Concepts

Mixture: A combination of two or more substances. In these problems, we usually deal with mixing two types of commodities (e.g., tea, rice, milk) with different prices to obtain a mixture with an intermediate price.

Alligation: A method to find the ratio of quantities of two or more ingredients that must be mixed to obtain a mixture of a desired mean value.

The Rule of Alligation

This rule is based on the principle that the ratio of the quantities of the ingredients is inversely proportional to the differences of their prices from the mean price.

Let C1 be the cost price of the cheaper ingredient and C2 be the cost price of the dearer ingredient. Let M be the mean price (the desired price of the mixture). Then, the ratio of the quantity of the cheaper ingredient (Q1) to the quantity of the dearer ingredient (Q2) is given by:

Q1 / Q2 = (C2 - M) / (M - C1)

This can be visualized using a diagram:

Cheaper Quantity (Q1) : Dearer Quantity (Q2)

C1 (Cost of Cheaper) C2 (Cost of Dearer)
\ /
\ /
\ /
M (Mean Price)
/
/
/
(C2 - M) (M - C1)

Example 1

In what ratio should rice costing ₹15 per kg be mixed with rice costing ₹20 per kg so that the mean price of the mixture is ₹18 per kg?

Solution:

Cheaper cost (C1) = ₹15/kg

Dearer cost (C2) = ₹20/kg

Mean price (M) = ₹18/kg

Ratio of cheaper quantity to dearer quantity = (C2 - M) : (M - C1)

= (20 - 18) : (18 - 15)

= 2 : 3

So, the rice should be mixed in the ratio 2:3.

Example 2: Mixing with Water

A shopkeeper sells milk at cost price but mixes water with it. What is the percentage of water in the mixture if he gains 25%?

Solution:

The shopkeeper gains 25% by selling milk at cost price. This gain is due to the addition of water.

Gain % = (Quantity of water / Quantity of milk) × 100

25% = (Quantity of water / Quantity of milk) × 100

Quantity of water / Quantity of milk = 25 / 100 = 1/4

This means for every 4 parts of milk, there is 1 part of water.

Total quantity of mixture = Quantity of milk + Quantity of water = 4 + 1 = 5 parts.

Percentage of water = (Quantity of water / Total quantity of mixture) × 100

= (1 / 5) × 100 = 20%

Mensuration

Mensuration is the branch of mathematics concerned with the measurement of geometric figures. It involves calculating the area, volume, perimeter, and surface area of various shapes.

I. Mensuration of Plane Figures (2D Shapes)

1. Square

• Side = a
• Area = a²
• Perimeter = 4a
• Diagonal = a√2

2. Rectangle

• Length = l, Breadth = b
• Area = l × b
• Perimeter = 2(l + b)
• Diagonal = √(l² + b²)

3. Triangle

• Equilateral Triangle: Side = a
• Area = (√3 / 4) × a²
• Height = (√3 / 2) × a
• Right-angled Triangle: Base = b, Height = h, Hypotenuse = p
• Area = (1/2) × b × h
• By Pythagoras Theorem: p² = b² + h²
• General Triangle: Sides = a, b, c
• Semi-perimeter (s) = (a + b + c) / 2
• Area (Heron's Formula) = √[s(s-a)(s-b)(s-c)]

4. Parallelogram

• Base = b, Height = h
• Area = b × h
• If sides are a and b, and angle between them is θ: Area = ab sin(θ)

5. Rhombus

• Diagonals = d₁, d₂
• Area = (1/2) × d₁ × d₂
• Side = a = √[(d₁/2)² + (d₂/2)²]
• Perimeter = 4a

6. Trapezium

• Parallel sides = a, b; Height = h
• Area = (1/2) × (a + b) × h

7. Circle

• Radius = r
• Diameter = d = 2r
• Area = πr² = π(d/2)²
• Circumference = 2πr = πd
• Area of a Semicircle = (1/2)πr²
• Circumference of a Semicircle = πr + 2r
• Area of a Sector (angle θ in degrees) = (θ/360) × πr²
• Length of an Arc (angle θ in degrees) = (θ/360) × 2πr

II. Mensuration of Solid Figures (3D Shapes)

1. Cuboid

• Length = l, Breadth = b, Height = h
• Volume = l × b × h
• Surface Area = 2(lb + bh + hl)
• Length of Diagonal = √(l² + b² + h²)

2. Cube

• Side = a
• Volume = a³
• Surface Area = 6a²
• Length of Diagonal = a√3

3. Cylinder

• Radius = r, Height = h
• Volume = πr²h
• Curved Surface Area (CSA) = 2πrh
• Total Surface Area (TSA) = 2πrh + 2πr² = 2πr(h + r)

4. Cone

• Radius = r, Height = h, Slant Height = l
• l = √(r² + h²)
• Volume = (1/3)πr²h
• Curved Surface Area (CSA) = πrl
• Total Surface Area (TSA) = πrl + πr² = πr(l + r)

5. Sphere

• Radius = r
• Volume = (4/3)πr³
• Surface Area = 4πr²

6. Hemisphere

• Radius = r
• Volume = (2/3)πr³
• Curved Surface Area (CSA) = 2πr²
• Total Surface Area (TSA) = 2πr² + πr² = 3πr²

Example: Conversion of Solids

A solid metallic sphere of radius 6 cm is melted and recast into a cylinder of height 12 cm. Find the radius of the cylinder.

Solution:

Volume of sphere = (4/3)π(6)³ = (4/3)π(216) = 288π cubic cm.

Let the radius of the cylinder be R.

Volume of cylinder = πR²h = πR²(12) cubic cm.

Since the volume remains the same after melting and recasting:

Volume of sphere = Volume of cylinder

288π = 12πR²

R² = 288 / 12 = 24

R = √24 = 2√6 cm.

Probability

Probability is the measure of the likelihood that an event will occur. It is quantified as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

Basic Concepts

• Experiment: An action or process that produces an outcome.
• Outcome: A possible result of an experiment.
• Sample Space (S): The set of all possible outcomes of an experiment.
• Event (E): A subset of the sample space, representing a specific outcome or a set of outcomes we are interested in.

Calculating Probability

The probability of an event E occurring is given by the formula:

P(E) = (Number of favorable outcomes for E) / (Total number of possible outcomes in the sample space S)

P(E) = n(E) / n(S)

Properties of Probability:

• 0 ≤ P(E) ≤ 1 (Probability is always between 0 and 1, inclusive).
• P(S) = 1 (The probability of the entire sample space occurring is 1).
• P(impossible event) = 0.

Types of Events

• Equally Likely Events: Events that have the same chance of occurring.
• Mutually Exclusive Events: Two events are mutually exclusive if they cannot occur at the same time. If A and B are mutually exclusive, then P(A and B) = 0.
• Independent Events: The occurrence of one event does not affect the probability of the other event. If A and B are independent, then P(A and B) = P(A) × P(B).
• Dependent Events: The occurrence of one event affects the probability of the other event.

Key Formulas for Probability

• Probability of Union of Two Events: P(A or B) = P(A) + P(B) - P(A and B)
• If A and B are mutually exclusive: P(A or B) = P(A) + P(B)
• Complementary Events: The probability of an event not happening is 1 minus the probability of it happening. P(not E) = 1 - P(E).

Examples

1. Coin Toss

A fair coin is tossed once. What is the probability of getting a head?

Solution:

Sample Space (S) = {H, T}. Total outcomes = 2.

Event (E) = Getting a head = {H}. Favorable outcomes = 1.

P(Head) = n(E) / n(S) = 1 / 2.

2. Dice Roll

A fair die is rolled. What is the probability of getting a number greater than 4?

Solution:

Sample Space (S) = {1, 2, 3, 4, 5, 6}. Total outcomes = 6.

Event (E) = Getting a number greater than 4 = {5, 6}. Favorable outcomes = 2.

P(Number > 4) = n(E) / n(S) = 2 / 6 = 1 / 3.

3. Drawing Cards

A card is drawn from a well-shuffled deck of 52 cards. What is the probability that the card is a King or a Queen?

Solution:

Total outcomes = 52.

Event A = Drawing a King. Number of Kings = 4.

Event B = Drawing a Queen. Number of Queens = 4.

Are these events mutually exclusive? Yes, a card cannot be both a King and a Queen.

P(King) = 4/52

P(Queen) = 4/52

P(King or Queen) = P(King) + P(Queen) = 4/52 + 4/52 = 8/52 = 2/13.

4. Independent Events

The probability of A solving a problem is 1/2, and the probability of B solving the same problem is 1/3. What is the probability that both solve the problem?

Solution:

Assume their solving abilities are independent.

P(A solves) = 1/2

P(B solves) = 1/3

P(Both solve) = P(A solves) × P(B solves) = (1/2) × (1/3) = 1/6.

Permutation and Combination

Permutation and Combination are two important concepts in counting techniques used to determine the number of ways certain arrangements or selections can be made.

I. Permutation

Permutation deals with arrangements where the order of items matters. The number of permutations of *n* distinct objects taken *r* at a time is denoted by ⁿP_r or P(n, r).

Formula for Permutation

ⁿP_r = n! / (n - r)!

Where '!' denotes the factorial (e.g., 5! = 5 × 4 × 3 × 2 × 1).

Note: n! / n! = 1, and 0! = 1.

Example 1

In how many ways can the letters of the word 'CAT' be arranged?

Solution:

Here, n = 3 (letters C, A, T) and we are arranging all 3 letters, so r = 3.

³P₃ = 3! / (3 - 3)! = 3! / 0! = 3! / 1 = 3 × 2 × 1 = 6.

The arrangements are: CAT, CTA, ACT, ATC, TCA, TAC.

Example 2

How many different 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5 without repetition?

Solution:

Here, n = 5 (digits 1, 2, 3, 4, 5) and we need to form 3-digit numbers, so r = 3.

⁵P₃ = 5! / (5 - 3)! = 5! / 2! = (5 × 4 × 3 × 2 × 1) / (2 × 1) = 5 × 4 × 3 = 60.

There are 60 such 3-digit numbers.

II. Combination

Combination deals with selections where the order of items does not matter. The number of combinations of *n* distinct objects taken *r* at a time is denoted by ⁿC_r or C(n, r) or (ⁿ_r).

Formula for Combination

ⁿC_r = n! / [r! × (n - r)!]

Note that ⁿC_r = ⁿC_(n-r).

Example 3

How many ways can a committee of 3 people be selected from a group of 5 people?

Solution:

Here, n = 5 (people) and we need to select 3, so r = 3. The order of selection does not matter for a committee.

⁵C₃ = 5! / [3! × (5 - 3)!] = 5! / (3! × 2!) = (5 × 4 × 3 × 2 × 1) / [(3 × 2 × 1) × (2 × 1)]

= (5 × 4) / (2 × 1) = 20 / 2 = 10.

There are 10 ways to form the committee.

Example 4

From a group of 6 men and 4 women, how many ways can a committee of 3 men and 2 women be formed?

Solution:

Number of ways to select 3 men from 6 = ⁶C₃ = 6! / (3! × 3!) = (6 × 5 × 4) / (3 × 2 × 1) = 20.

Number of ways to select 2 women from 4 = ⁴C₂ = 4! / (2! × 2!) = (4 × 3) / (2 × 1) = 6.

Total number of ways to form the committee = (Ways to select men) × (Ways to select women)

= 20 × 6 = 120.