Profit and Loss

Introduction to Profit and Loss

Profit and Loss is a fundamental concept in arithmetic that deals with the financial gain or loss incurred in a transaction. Every business, big or small, aims to make a profit. Understanding how to calculate profit and loss is crucial for making sound financial decisions, whether you are a business owner or a consumer.

Key Terms in Profit and Loss

Before diving into calculations, let's define the essential terms:

• Cost Price (CP): The price at which an article is purchased. This is the amount of money spent to acquire an item.
• Selling Price (SP): The price at which an article is sold. This is the amount of money received when an item is sold.
• Profit: When the Selling Price (SP) is greater than the Cost Price (CP), the difference is the profit. Profit = SP - CP.
• Loss: When the Cost Price (CP) is greater than the Selling Price (SP), the difference is the loss. Loss = CP - SP.
• Overheads: Additional expenses incurred on an article, such as transportation, repair, labor, etc. These are added to the original Cost Price to get the Total Cost Price.

Calculating Profit and Loss

The basic formulas for profit and loss are straightforward:

• If SP > CP, there is a Profit. Profit = SP - CP.
• If CP > SP, there is a Loss. Loss = CP - SP.

Calculating Profit and Loss Percentages

Often, profit or loss is expressed as a percentage of the Cost Price. This gives a standardized way to compare profitability across different transactions.

• Profit Percentage: (Profit / CP) × 100%
• Loss Percentage: (Loss / CP) × 100%

It is important to note that profit and loss percentages are always calculated on the Cost Price, unless stated otherwise.

Formulas Relating CP, SP, Profit %, and Loss %

We can rearrange the basic formulas to find any unknown value if others are known.

• To find SP when Profit % is known: SP = CP × (100 + Profit %) / 100
• To find SP when Loss % is known: SP = CP × (100 - Loss %) / 100
• To find CP when Profit % is known: CP = SP × 100 / (100 + Profit %)
• To find CP when Loss % is known: CP = SP × 100 / (100 - Loss %)

Example Calculation

Suppose a shopkeeper buys a toy for ₹200 (CP) and sells it for ₹250 (SP).

1. Calculate Profit: SP - CP = ₹250 - ₹200 = ₹50
2. Calculate Profit Percentage: (Profit / CP) × 100 = (₹50 / ₹200) × 100 = (1/4) × 100 = 25%.

Now, suppose the shopkeeper buys a book for ₹300 (CP) and sells it for ₹270 (SP).

1. Calculate Loss: CP - SP = ₹300 - ₹270 = ₹30
2. Calculate Loss Percentage: (Loss / CP) × 100 = (₹30 / ₹300) × 100 = (1/10) × 100 = 10%.

Impact of Overheads

Overheads increase the actual cost incurred for an item.

Example: A vendor buys an old radio for ₹500. He spends ₹100 on its repairs. He then sells it for ₹750.

1. Total Cost Price (TCP): Original CP + Repair Cost = ₹500 + ₹100 = ₹600
2. Selling Price (SP): ₹750
3. Profit: SP - TCP = ₹750 - ₹600 = ₹150
4. Profit Percentage: (Profit / TCP) × 100 = (₹150 / ₹600) × 100 = (1/4) × 100 = 25%.

Without considering the repair cost, the profit would appear to be ₹250 (₹750 - ₹500), and the profit percentage would be (₹250 / ₹500) × 100 = 50%. This highlights the importance of including all associated costs.

Marked Price (MP) and Discount

In retail, shopkeepers often display goods with a Marked Price (MP), which is higher than the cost price. They then offer a Discount on this Marked Price to attract customers.

• Marked Price (MP): The price tagged on the product by the manufacturer or seller.
• Discount: A reduction offered on the Marked Price. It can be a percentage or a fixed amount.
• Discount Amount: MP - SP
• Discount Percentage: (Discount Amount / MP) × 100%

The Selling Price (SP) is calculated after deducting the discount from the Marked Price:

• SP = MP - Discount Amount
• SP = MP × (100 - Discount %) / 100

Example with Marked Price and Discount

A shirt has a Marked Price of ₹800. The shopkeeper offers a 10% discount.

1. Discount Amount: 10% of ₹800 = (10/100) × ₹800 = ₹80
2. Selling Price (SP): MP - Discount Amount = ₹800 - ₹80 = ₹720

If the Cost Price (CP) of the shirt was ₹600:

1. Profit: SP - CP = ₹720 - ₹600 = ₹120
2. Profit Percentage: (Profit / CP) × 100 = (₹120 / ₹600) × 100 = 20%.

Relationship between Profit, Loss, MP, and Discount

We can also relate CP, SP, MP, Profit %, and Discount % in a single formula.

Let CP be the Cost Price and Profit % be the profit percentage. SP = CP × (100 + Profit %) / 100

Let MP be the Marked Price and Discount % be the discount percentage. SP = MP × (100 - Discount %) / 100

Equating the two expressions for SP: CP × (100 + Profit %) / 100 = MP × (100 - Discount %) / 100

This simplifies to: CP × (100 + Profit %) = MP × (100 - Discount %)

This formula is very useful for solving problems where CP, MP, Profit %, and Discount % are involved.

Example using the combined formula

A shopkeeper marks his goods at 40% above the Cost Price and then offers a 20% discount. What is his profit percentage?

1. Let CP = ₹100 (for simplicity).
2. MP is 40% above CP, so MP = ₹100 + 40% of ₹100 = ₹100 + ₹40 = ₹140.
3. Discount is 20% on MP. Discount Amount = 20% of ₹140 = (20/100) × ₹140 = ₹28.
4. SP = MP - Discount Amount = ₹140 - ₹28 = ₹112.
5. Profit = SP - CP = ₹112 - ₹100 = ₹12.
6. Profit Percentage = (Profit / CP) × 100 = (₹12 / ₹100) × 100 = 12%.

Alternatively, using the formula: CP × (100 + Profit %) = MP × (100 - Discount %) Let Profit % be P. Let CP = 100. Then MP = 100 × (100 + 40)/100 = 140. 100 × (100 + P) = 140 × (100 - 20) 100 × (100 + P) = 140 × 80 10000 + 100P = 11200 100P = 11200 - 10000 100P = 1200 P = 12%

Successive Discounts

Sometimes, two or more discounts are offered one after another. These are called successive discounts. The second discount is applied to the price after the first discount has been deducted.

Example: A shopkeeper offers two successive discounts of 10% and 20% on an article marked at ₹500.

1. First Discount = 10% of ₹500 = (10/100) × ₹500 = ₹50.
2. Price after first discount = ₹500 - ₹50 = ₹450.
3. Second Discount = 20% of ₹450 = (20/100) × ₹450 = ₹90.
4. Final Selling Price = ₹450 - ₹90 = ₹360.

A common mistake is to add the percentages (10% + 20% = 30%) and calculate a single discount. A 30% discount on ₹500 would be ₹150, leading to an SP of ₹350, which is incorrect.

Formula for Equivalent Single Discount

If two successive discounts are $d_1\%$ and $d_2\%$, the equivalent single discount $D\%$ is: $D\% = (d_1 + d_2 - (d_1 \times d_2) / 100)\%$

Using the previous example: $D\% = (10 + 20 - (10 \times 20) / 100)\% = (30 - 200 / 100)\% = (30 - 2)\% = 28\%$.

Let's verify: 28% discount on ₹500 = (28/100) × ₹500 = ₹140. SP = ₹500 - ₹140 = ₹360. This matches our earlier calculation.

Ratio and Proportion

Introduction to Ratio

A ratio is a way of comparing two or more quantities by division. It shows the relative size of these quantities. Ratios are typically expressed in their simplest form.

For example, if there are 5 apples and 10 oranges, the ratio of apples to oranges is 5:10, which simplifies to 1:2. This means for every 1 apple, there are 2 oranges.

Expressing a Ratio

A ratio of two quantities $a$ and $b$ can be written as $a:b$ or as a fraction $a/b$.

Terms of a Ratio: In a ratio $a:b$, $a$ is called the first term or antecedent, and $b$ is called the second term or consequent.

Types of Ratios

• Ratio of Equality: When both terms are equal (e.g., 5:5, which simplifies to 1:1).
• Ratio of Greater Inequality: When the first term is greater than the second term (e.g., 5:3).
• Ratio of Less Inequality: When the first term is less than the second term (e.g., 3:5).
• Duplicate Ratio: The ratio of the squares of the terms (e.g., the duplicate ratio of $a:b$ is $a^2:b^2$).
• Sub-duplicate Ratio: The ratio of the square roots of the terms (e.g., the sub-duplicate ratio of $a:b$ is $\sqrt{a}:\sqrt{b}$).
• Triplicate Ratio: The ratio of the cubes of the terms (e.g., the triplicate ratio of $a:b$ is $a^3:b^3$).
• Sub-triplicate Ratio: The ratio of the cube roots of the terms (e.g., the sub-triplicate ratio of $a:b$ is $\sqrt[3]{a}:\sqrt[3]{b}$).
• Inverse Ratio: The ratio obtained by reversing the order of the terms (e.g., the inverse ratio of $a:b$ is $b:a$).

Compound Ratio

The compound ratio of two or more ratios is the product of the antecedents divided by the product of the consequents.

Example: The compound ratio of $a:b$, $c:d$, and $e:f$ is $(a \times c \times e) : (b \times d \times f)$.

Proportion

A proportion is an equation stating that two ratios are equal. If $a:b$ and $c:d$ are two ratios, and they are equal, then $a:b = c:d$. This is read as "$a$ is to $b$ as $c$ is to $d$".

Terms of a Proportion: In a proportion $a:b = c:d$:

• $a$ and $d$ are called the extremes.
• $b$ and $c$ are called the means.

Rule of Proportion: The product of the extremes is equal to the product of the means. $a \times d = b \times c$

Types of Proportion

• Direct Proportion: Two quantities are said to be in direct proportion if they increase or decrease together in the same ratio. For example, if the price of 5 kg of sugar is ₹200, then the price of 10 kg of sugar will be ₹400. (Quantity increases, Price increases).
• Inverse Proportion: Two quantities are said to be in inverse proportion if, as one quantity increases, the other quantity decreases in the same ratio, and vice versa. For example, if 4 men can complete a work in 6 days, then 8 men will complete the same work in 3 days. (Number of men increases, Days decrease).

Continued Proportion

Three quantities $a, b, c$ are said to be in continued proportion if $a:b = b:c$. In this case, $b$ is called the mean proportional between $a$ and $c$.

From $a:b = b:c$, we get $b^2 = ac$, so $b = \sqrt{ac}$.

Applications of Ratio and Proportion

Ratio and proportion are widely used in various contexts:

• Sharing: Dividing an amount among people in a given ratio.
• Mixtures: Determining the proportions of different ingredients in a mixture.
• Scale Drawings and Maps: Representing real-world objects or areas at a reduced scale.
• Time and Work Problems: Relating the number of workers to the time taken.
• Speed, Distance, and Time Problems: Relating these quantities.

Example Problems

1. Sharing Money: A sum of money is divided between A, B, and C in the ratio 2:3:7. If C's share is ₹4200, what is A's share?
   • Let the shares be 2x, 3x, and 7x.
   • C's share is 7x = ₹4200.
   • So, x = ₹4200 / 7 = ₹600.
   • A's share is 2x = 2 × ₹600 = ₹1200.
2. Mixture Problem: In what ratio must rice at ₹6.20 per kg be mixed with rice at ₹7.30 per kg so that by selling the mixture at ₹8.00 per kg, there may be a gain of 20%?
   • First, find the Cost Price (CP) of the mixture.
   • SP = ₹8.00, Profit = 20%.
   • CP = SP × 100 / (100 + Profit %) = 8.00 × 100 / (100 + 20) = 8.00 × 100 / 120 = 800 / 120 = ₹6.67 (approx).
   • Now, use the rule of alligation (a method for solving mixture problems using ratios).
   • Let the quantities of the two types of rice be $q_1$ and $q_2$.
   • CP of first rice = ₹6.20
   • CP of second rice = ₹7.30
   • CP of mixture = ₹6.67
   • Using alligation:
   • Quantity 1 / Quantity 2 = (CP of mixture - CP of second rice) / (CP of first rice - CP of mixture)
   • $q_1 / q_2 = (7.30 - 6.67) / (6.67 - 6.20)$
   • $q_1 / q_2 = 0.63 / 0.47$
   • The ratio is approximately 63:47.
3. Inverse Proportion: If 12 men can build a wall in 30 days, how many men are needed to build the same wall in 20 days?
   • Let the number of men be $M$ and the number of days be $D$.
   • Since it's inverse proportion, $M \times D$ is constant.
   • $M_1 \times D_1 = M_2 \times D_2$
   • 12 men × 30 days = $M_2$ × 20 days
   • $M_2 = (12 \times 30) / 20 = 360 / 20 = 18$ men.

Average

Introduction to Average

The average, or arithmetic mean, is a measure of central tendency. It represents a typical value for a set of numbers. It is calculated by summing up all the values in a dataset and then dividing by the count of those values.

Formula for Average

Average = (Sum of all observations) / (Number of observations)

Let's say we have a set of observations $x_1, x_2, x_3, \dots, x_n$. The average (denoted by $\bar{x}$) is: $\bar{x} = (x_1 + x_2 + x_3 + \dots + x_n) / n$

Calculating the Sum from the Average

If the average and the number of observations are known, we can find the sum of observations:

Sum of observations = Average × Number of observations

Example Calculation

Find the average of the numbers: 10, 15, 20, 25, 30.

1. Sum of observations = 10 + 15 + 20 + 25 + 30 = 100
2. Number of observations = 5
3. Average = 100 / 5 = 20

If the average of 5 numbers is 20, their sum is 5 × 20 = 100.

Types of Problems involving Average

1. When a new item is added to the group

Example: The average weight of 5 students is 40 kg. If a new student joins, the average weight becomes 42 kg. What is the weight of the new student?

1. Sum of weights of 5 students = 5 × 40 kg = 200 kg.
2. After a new student joins, there are 6 students.
3. New average weight = 42 kg.
4. Sum of weights of 6 students = 6 × 42 kg = 252 kg.
5. Weight of the new student = (Sum of weights of 6 students) - (Sum of weights of 5 students)
6. Weight of new student = 252 kg - 200 kg = 52 kg.

2. When an item is removed from the group

Example: The average score of 10 players is 65 runs. If one player's score is removed, the average score of the remaining 9 players becomes 60 runs. What was the score of the player who was removed?

1. Sum of scores of 10 players = 10 × 65 = 650 runs.
2. Sum of scores of 9 players = 9 × 60 = 540 runs.
3. Score of the removed player = (Sum of scores of 10 players) - (Sum of scores of 9 players)
4. Score of removed player = 650 - 540 = 110 runs.

3. When one item is replaced by another

Example: The average age of 7 persons in a group is 30 years. If a person aged 55 years is replaced by another person, the average age of the group increases by 2 years. What is the age of the new person?

1. Initial sum of ages of 7 persons = 7 × 30 = 210 years.
2. New average age = 30 + 2 = 32 years.
3. New sum of ages of 7 persons = 7 × 32 = 224 years.
4. Age of the new person = (New sum of ages) - (Initial sum of ages)
5. Age of the new person = 224 - 210 = 14 years.
6. Alternatively: The increase in total age is 7 persons × 2 years/person = 14 years. The new person's age = Old person's age + Increase in total age = 55 + 14 = 69 years.

Wait, there's a mistake in my calculation above. Let's re-evaluate the alternative method. The increase in average age is 2 years. This means the total age increased by $7 \times 2 = 14$ years. The old person's age was 55. The new person's age must be $55 + 14 = 69$ years. Let's recheck the first method: Initial sum = 210. New average = 32. New sum = $7 \times 32 = 224$. The difference in sum is $224 - 210 = 14$. This difference is the net change. Let the age of the old person be $O = 55$ and the age of the new person be $N$. New Sum = Old Sum - $O + N$. $224 = 210 - 55 + N$ $224 = 155 + N$ $N = 224 - 155 = 69$ years. Both methods yield 69 years. My previous calculation was incorrect.

4. Average of Consecutive Numbers

The average of consecutive numbers (e.g., 1, 2, 3, 4, 5) is the middle number. Example: Average of 1, 2, 3, 4, 5 is 3. Average of 10, 11, 12, 13, 14, 15 is (12 + 13) / 2 = 12.5.

For an even number of consecutive terms, the average is the average of the two middle terms.

The average of an Arithmetic Progression (AP) is (First Term + Last Term) / 2.

Example: Find the average of the first 20 natural numbers.

1. First term = 1, Last term = 20.
2. Average = (1 + 20) / 2 = 21 / 2 = 10.5.

Example: Find the average of all odd numbers between 1 and 100.

1. The odd numbers are 1, 3, 5, ..., 99. This is an AP.
2. First term = 1, Last term = 99.
3. Average = (1 + 99) / 2 = 100 / 2 = 50.

5. Average in Weighted Cases

Sometimes, different observations have different weights or frequencies.

Formula: Weighted Average = (Sum of (Weight × Observation)) / (Sum of Weights)

Example: In an examination, a student secured marks in 5 subjects as follows: English (50 marks), Hindi (60 marks), Maths (75 marks), Science (80 marks), Social Science (55 marks). Find the average marks secured by the student.

1. Assuming equal weightage (frequency = 1 for each subject):
2. Sum of marks = 50 + 60 + 75 + 80 + 55 = 320
3. Number of subjects = 5
4. Average marks = 320 / 5 = 64.

Example with different weights: The average marks of 30 students is 70. The average marks of another 20 students is 80. Find the overall average marks.

1. Sum of marks for the first group = 30 × 70 = 2100.
2. Sum of marks for the second group = 20 × 80 = 1600.
3. Total sum of marks = 2100 + 1600 = 3700.
4. Total number of students = 30 + 20 = 50.
5. Overall average = 3700 / 50 = 370 / 5 = 74.