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Classwise Concept with Examples
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Class 7th Chapters
1. Integers 2. Fractions and Decimals 3. Data Handling
4. Simple Equations 5. Lines and Angles 6. The Triangle and its Properties
7. Congruence of Triangles 8. Comparing Quantities 9. Rational Numbers
10. Practical Geometry 11. Perimeter and Area 12. Algebraic Expressions
13. Exponents and Powers 14. Symmetry 15. Visualising Solid Shapes

Content On This Page
Ratio & its Related Terms Proportion & its Related Terms Unitary Method
Percentage and its Conversion Finding the Percentage of a Given Quantity Profit and Loss
Simple Interest


Chapter 8 Comparing Quantities (Concepts)

This chapter embarks on a practical exploration of various essential methods used for comparing quantities, a fundamental skill applicable across numerous real-life scenarios, from everyday shopping to financial planning. Mastering these techniques allows for informed decision-making and a clearer understanding of relationships between different values. We begin with the foundational concept of Ratio. A ratio provides a means to compare two quantities through division, expressing how many times one quantity contains, or is contained within, another. Ratios are often written using a colon, such as $a:b$, or as a fraction, $\frac{a}{b}$, and are typically simplified to their lowest terms. For example, comparing 6 apples to 8 oranges yields a ratio of $6:8$, which simplifies to $3:4$. Understanding equivalent ratios is also key; these are different ratios that represent the exact same proportional relationship, like $1:2$ being equivalent to $2:4$ or $5:10$.

Building directly on ratios is the concept of Proportion. A proportion is essentially a statement declaring the equality of two ratios. It signifies that two pairs of quantities maintain the same comparative relationship. Proportions are often expressed using a double colon ($a:b :: c:d$, read as "a is to b as c is to d") or more commonly as an equation involving fractions: $\frac{a}{b} = \frac{c}{d}$. Many practical problems involve solving proportions, such as finding an unknown term when the other three are given, or verifying if four given quantities truly form a proportion.

A major focus shifts towards Percentages, denoted by the symbol $\%$. Percentages offer a highly standardized and universally understood method for comparing quantities by expressing them as a fraction out of one hundred. This common baseline makes comparisons highly intuitive. We will practice converting between different forms: transforming fractions and decimals into percentages (e.g., $\frac{3}{4} = 0.75 = 75\%$) and converting percentages back into fractions or decimals (e.g., $40\% = \frac{40}{100} = \frac{2}{5} = 0.4$). Key applications rigorously covered include:

These foundational comparison tools are then applied to essential practical business and financial scenarios, particularly involving Profit and Loss. We define the core terminology:

While knowing the absolute profit or loss amount is useful, calculating the profit percentage or loss percentage provides a better measure of profitability relative to the initial investment. Crucially, these percentages are always calculated based on the Cost Price (CP):

Finally, the chapter introduces the concept of Simple Interest (SI), a fundamental method for calculating interest, often applied to short-term loans, deposits, or basic investments. The calculation involves three key components:

The formula to calculate Simple Interest is given by: $SI = \frac{P \times R \times T}{100}$. The total Amount (A) accumulated or due after the time period is the sum of the initial Principal and the calculated Simple Interest: $A = P + SI$. This chapter equips learners with indispensable mathematical tools for everyday financial literacy, comparison, and practical problem-solving.



Ratio & its Related Terms

In many real-life situations, we need to compare different quantities. For example, we might compare the heights of two people, the cost of two items, or the number of boys and girls in a class. Simple comparison might tell us which quantity is larger or smaller. However, sometimes we need to understand the comparison in a relative sense, for instance, how many times one quantity is of another, or what part of one quantity is of another. This is where the concept of Ratio is useful.

In this chapter, we will learn about ratios and related terms, extend this idea to proportion, and use a technique called the unitary method to solve problems involving comparison of quantities.


What is a Ratio?

A Ratio is a comparison of two quantities of the same kind and in the same units. This comparison is done by division. It tells us how many times one quantity is contained within or contains another quantity.

If $a$ and $b$ are two quantities of the same kind and in the same units (and $b$ is not zero), then the ratio of $a$ to $b$ is written as:

Example: Suppose there are 10 red apples and 15 green apples in a basket. We can find the ratio of the number of red apples to the number of green apples.

Number of red apples = 10.

Number of green apples = 15.

The quantities are of the same kind (number of apples) and in the same unit (count). The ratio of red apples to green apples is:

Ratio $= \frac{\text{Number of red apples}}{\text{Number of green apples}} = \frac{10}{15} $

This fraction can be simplified by dividing the numerator and denominator by their HCF, which is 5.

Ratio $= \frac{10 \div 5}{15 \div 5} = \frac{2}{3} $

So, the ratio of red apples to green apples is $2 : 3$. This means for every 2 red apples, there are 3 green apples in the basket.

Terms of a Ratio:

In the ratio $a : b$, $a$ and $b$ are called the Terms of the ratio.

Important Points about Ratios:

  1. Quantities of the Same Kind and Unit: This is a fundamental requirement. You must compare quantities of the same type (lengths with lengths, volumes with volumes, counts with counts) and ensure they are in the same unit before forming the ratio.
  2. Ratio Has No Units: Since the units of the quantities being compared are the same, they cancel out during the division. Hence, a ratio is a unitless number. Example: $\frac{50 \text{ cm}}{200 \text{ cm}} = \frac{50}{200} = \frac{1}{4}$ or $1:4$.
  3. Order Matters: The ratio $a : b$ is generally different from $b : a$ (unless $a=b$). The order in which the quantities are stated determines the order of the terms in the ratio.
  4. Simplest Form: Ratios should typically be expressed in their simplest form (lowest terms) by dividing both the antecedent and the consequent by their HCF. This is the same as reducing the fraction form of the ratio to its lowest terms.
  5. Terms are Usually Positive: In the context of comparing quantities like lengths, weights, counts, etc., the terms of a ratio are generally positive numbers.

Example of Simplifying a Ratio:

Find the ratio of 50 cm to 2 m in simplest form.

Step 1: Ensure same units. Convert 2 m to cm. $1 \text{ m} = 100 \text{ cm}$, so $2 \text{ m} = 2 \times 100 \text{ cm} = 200 \text{ cm}$.

Step 2: Write the ratio. Ratio $= 50 \text{ cm} : 200 \text{ cm} = 50:200$.

Step 3: Simplify the ratio. Find HCF(50, 200). HCF is 50.

Divide both terms by 50: $50 \div 50 = 1$, $200 \div 50 = 4$.

The ratio in simplest form is $1 : 4$.


Equivalent Ratios

Equivalent Ratios are ratios that represent the same comparison or proportion, even though the numbers (terms) might be different. They are similar to equivalent fractions.

You can obtain an equivalent ratio by multiplying or dividing both terms of a given ratio by the same non-zero number.

If $a : b$ is a ratio, and $k$ is any non-zero number, then $a \times k : b \times k$ is equivalent to $a : b$.

If $a : b$ is a ratio, and $m$ is a common factor of $a$ and $b$ ($m \neq 0$), then $a \div m : b \div m$ is equivalent to $a : b$.

Example: The ratio $2 : 3$ is equivalent to:

All these ratios ($2:3, 4:6, 6:9, 20:30$) represent the same proportional relationship. If you write them as fractions ($\frac{2}{3}, \frac{4}{6}, \frac{6}{9}, \frac{20}{30}$), they all simplify to $\frac{2}{3}$.

Checking for Equivalence:

To check if two ratios $a : b$ and $c : d$ are equivalent, you can compare their fractional forms. They are equivalent if $\frac{a}{b} = \frac{c}{d}$.

Using cross-multiplication (as learned with fractions), $\frac{a}{b} = \frac{c}{d}$ if and only if $a \times d = b \times c$. So, you can check for equivalence by comparing the product of the extremes ($a \times d$) and the product of the means ($b \times c$).

Example: Are the ratios $2 : 5$ and $6 : 15$ equivalent?

Method 1 (Simplify to lowest terms):

$2 : 5$ is already in simplest form.

$6 : 15$. Find HCF(6, 15) = 3. Divide both terms by 3: $6 \div 3 = 2$, $15 \div 3 = 5$. The simplified ratio is $2:5$.

Since both ratios simplify to the same ratio $2:5$, they are equivalent.

Method 2 (Cross-product):

For $a:b :: c:d$, check if $a \times d = b \times c$. Here $a=2, b=5, c=6, d=15$.

Product of extremes $= a \times d = 2 \times 15 = 30$.

Product of means $= b \times c = 5 \times 6 = 30$.

Since the product of extremes equals the product of means ($30 = 30$), the ratios are equivalent.


Comparing Ratios

To compare two or more ratios (to see which one is larger or smaller), we convert them into a format that allows for easy comparison.

  1. Convert to Fractions with Common Denominator: Convert each ratio into its fractional form ($\frac{a}{b}$). Then, find the LCM of the denominators and convert all the fractions into equivalent like fractions (fractions with the same denominator). Compare the numerators of these like fractions; the fraction with the larger numerator corresponds to the larger ratio.
  2. Convert to Decimals: Convert each ratio to its fractional form and then perform the division to get a decimal value (terminating or repeating). Compare the decimal values.

Example: Compare the ratios $2 : 3$ and $4 : 5$.

Method 1 (Common Denominator):

Ratio $2 : 3 = \frac{2}{3}$. Ratio $4 : 5 = \frac{4}{5}$.

LCM of the denominators 3 and 5 is $3 \times 5 = 15$. Convert both fractions to have denominator 15.

$\frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15} $.

$\frac{4}{5} = \frac{4 \times 3}{5 \times 3} = \frac{12}{15} $.

Now compare $\frac{10}{15}$ and $\frac{12}{15}$. Since $12 > 10$, we have $\frac{12}{15} > \frac{10}{15}$.

Therefore, $4 : 5 > 2 : 3$.

Method 2 (Decimals):

Convert ratios to decimal values:

$2 : 3 = \frac{2}{3}$. Perform division: $2 \div 3 \approx 0.666...$

$4 : 5 = \frac{4}{5}$. Perform division: $4 \div 5 = 0.8$.

Compare decimal values: $0.8$ and $0.666...$. Since $0.8 > 0.666...$, we have $4 : 5 > 2 : 3$.


Dividing a Quantity in a Given Ratio

A common application of ratios is to divide a total quantity into parts according to a given ratio. For example, sharing an amount of money or a quantity of something in a specific ratio among a group of people.

Steps to Divide a Given Quantity in a Given Ratio ($m : n$):

  1. Find the sum of the terms of the ratio ($m + n$). This sum represents the total number of equal shares the quantity is divided into.
  2. Divide the given total quantity by the sum of the ratio terms (from Step 1). This gives you the value of one "share" or "part" in the ratio.
  3. Multiply the value of one share (from Step 2) by each term of the ratio ($m$ and $n$) to find the value of each individual part.

Example: Divide $ \textsf{₹} 720 $ between Ram and Shyam in the ratio $4 : 5$.

The given ratio is Ram : Shyam = $4 : 5$. The total quantity to be divided is $ \textsf{₹} 720 $.

Step 1: Find the sum of the terms of the ratio. Sum $= 4 + 5 = 9$. This means the total amount is divided into 9 equal shares.

Step 2: Find the value of one share. Value of one share $= \frac{\text{Total Amount}}{\text{Sum of ratio terms}} = \frac{\textsf{₹} 720}{9}$.

Perform the division: $720 \div 9 = 80$. So, the value of one share is $ \textsf{₹} 80 $.

Step 3: Calculate the individual parts (Ram's share and Shyam's share) by multiplying the value of one share by each term of the ratio.

Ram's share $= 4 \times \textsf{₹} 80 = \textsf{₹} 320$.

Shyam's share $= 5 \times \textsf{₹} 80 = \textsf{₹} 400$.

Check: Add the shares to ensure they sum up to the total quantity: $ \textsf{₹} 320 + \textsf{₹} 400 = \textsf{₹} 720 $. The division is correct.

Example 1. Find the ratio of 3 km to 300 m.

Answer:

To find the ratio of 3 km to 300 m, the two quantities must be in the same unit. We have kilometers (km) and meters (m). Let's convert kilometers to meters.

We know that $1 \text{ km} = 1000 \text{ m}$.

So, $3 \text{ km} = 3 \times 1000 \text{ m} = 3000 \text{ m}$.

Now, we find the ratio of 3000 m to 300 m.

Ratio $= 3000 \text{ m} : 300 \text{ m}$.

Write the ratio as a fraction and simplify:

Ratio $= \frac{3000}{300} $.

Cancel zeros or divide directly:

$\frac{3000}{300} = \frac{30 \times \cancel{100}}{3 \times \cancel{100}} = \frac{30}{3} = 10 $.

The ratio is $10$, which is written as $10:1$.

So, the ratio of 3 km to 300 m is $10 : 1$.

Example 2. Are the ratios $1:2$ and $2:3$ equivalent?

Answer:

To check if the ratios $1:2$ and $2:3$ are equivalent, we can use the cross-product method (which checks if their fractional forms are equal).

The first ratio is $1:2$, which is the fraction $\frac{1}{2}$.

The second ratio is $2:3$, which is the fraction $\frac{2}{3}$.

We check if $\frac{1}{2} = \frac{2}{3}$.

Using the cross-product rule ($a \times d = b \times c$ for $\frac{a}{b} = \frac{c}{d}$):

Here, $a=1, b=2, c=2, d=3$.

Calculate the product of the extremes ($a \times d$): $1 \times 3 = 3$.

Calculate the product of the means ($b \times c$): $2 \times 2 = 4$.

Compare the two products: $3$ and $4$.

$3 \neq 4 $.

Since the cross-products are not equal, the fractional forms are not equal, and thus the ratios are not equivalent.

Alternatively, we could compare them using decimals:

$1:2 = \frac{1}{2} = 0.5$.

$2:3 = \frac{2}{3} \approx 0.666...$

Since $0.5 \neq 0.666...$, the ratios are not equivalent.

Therefore, the ratios $1:2$ and $2:3$ are not equivalent.



Proportion & its Related Terms

In the previous section, we learned about Ratio as a way to compare two quantities using division. Now, we will extend this idea to compare two ratios. When two ratios are equal, we say they are in Proportion. Proportion is a powerful tool for solving problems involving quantities that are related to each other in a consistent way.


What is Proportion?

Proportion is a statement that asserts the equality of two ratios. If two ratios are equal, the four quantities involved are said to be in proportion. This means that the relationship between the first two quantities is the same as the relationship between the last two quantities.

If four quantities $a, b, c,$ and $d$ are such that the ratio $a : b$ is equal to the ratio $c : d$, then $a, b, c,$ and $d$ are in proportion.

This is formally written using the symbol '::' (read as "as") between the two ratios:

$a : b :: c : d $

(read as "a is to b as c is to d")

Alternatively, since it's a statement of equality between ratios, we can use the equals sign:

$\frac{a}{b} = \frac{c}{d} $

In the proportion $a : b :: c : d$:

In $a : b :: c : d$, we have:

$\underbrace{a}_{\text{Extreme}} : \underbrace{b}_{\text{Mean}} :: \underbrace{c}_{\text{Mean}} : \underbrace{d}_{\text{Extreme}}$

For $a:b::c:d$ to be a valid proportion, $a, b$ must be of the same kind and unit (and $b \neq 0$), and $c, d$ must be of the same kind and unit (and $d \neq 0$). The quantities ($a, b$) can be of a different kind than ($c, d$), as long as the two ratios are numerically equal (e.g., ratio of lengths = ratio of costs).

Fundamental Property of Proportion:

A key property derived from the definition of proportion is that the product of the extreme terms is equal to the product of the middle terms.

If $a : b :: c : d$, which is equivalent to $\frac{a}{b} = \frac{c}{d}$, then by cross-multiplication:

$a \times d = b \times c $

This is known as the Property of Proportion:

$\text{Product of Extremes} = \text{Product of Means} $

This property is very useful for checking if four quantities are in proportion or for finding an unknown term when the other three are known.

Example: Are the numbers 2, 4, 6, 12 in proportion?

We need to check if $2:4 :: 6:12$ is a true statement. We can use the Product of Extremes and Product of Means test.

The extremes are the 1st and 4th terms: $a=2, d=12$. Product of extremes $= 2 \times 12 = 24$.

The means are the 2nd and 3rd terms: $b=4, c=6$. Product of means $= 4 \times 6 = 24$.

Since Product of extremes = Product of means ($24 = 24$), the numbers 2, 4, 6, and 12 are in proportion in that order.

We can write this as $2 : 4 :: 6 : 12$. This is true because $\frac{2}{4} = \frac{1}{2}$ and $\frac{6}{12} = \frac{1}{2}$, and $\frac{1}{2} = \frac{1}{2}$.


Continued Proportion

Sometimes, we have a relationship between three quantities of the same kind, where the ratio of the first to the second is equal to the ratio of the second to the third. This is called Continued Proportion.

Three quantities $a, b,$ and $c$ (of the same kind) are in continued proportion if $a : b :: b : c$.

This can be written fractionally as $\frac{a}{b} = \frac{b}{c}$.

In this continued proportion $a : b :: b : c$:

Applying the property of proportion ($a \times d = b \times c$) to $a : b :: b : c$, we get $a \times c = b \times b$, which is $ac = b^2$.

So, the mean proportional $b$ is the square root of the product of $a$ and $c$: $b = \sqrt{ac}$.

Example: Are the numbers 4, 6, 9 in continued proportion?

We need to check if $4 : 6 :: 6 : 9$ is true. Check if the ratio $4:6$ is equal to the ratio $6:9$.

Ratio $4 : 6 = \frac{4}{6}$. Simplify: $\frac{4 \div 2}{6 \div 2} = \frac{2}{3}$.

Ratio $6 : 9 = \frac{6}{9}$. Simplify: $\frac{6 \div 3}{9 \div 3} = \frac{2}{3}$.

Since both ratios are equal to $\frac{2}{3}$, the numbers 4, 6, and 9 are in continued proportion.

Here, 6 is the mean proportional between 4 and 9. Check $b^2 = ac$: $6^2 = 36$, and $4 \times 9 = 36$. The property holds.


Finding an Unknown Term in a Proportion

If three of the four terms in a proportion are known, we can use the property "Product of extremes = Product of means" to find the value of the unknown term. We represent the unknown term with a variable and solve the resulting equation.

Example: Find the value of $x$ if $3 : 4 :: x : 8$.

The four terms in proportion are $a=3, b=4, c=x, d=8$.

Extremes are 3 and 8. Means are 4 and $x$.

Using the property: Product of extremes = Product of means.

$3 \times 8 = 4 \times x $

Calculate the products:

$24 = 4x $

We have a simple linear equation $4x = 24$. To find $x$, divide both sides by 4 (or transpose 4 which is multiplying $x$).

$x = \frac{24}{4} $.

Perform the division:

$x = 6 $.

So, the missing term is 6. The proportion is $3 : 4 :: 6 : 8$.

Check: $3:4 = \frac{3}{4}$. $6:8 = \frac{6 \div 2}{8 \div 2} = \frac{3}{4}$. Since $\frac{3}{4} = \frac{3}{4}$, the proportion is correct.


Example 1. Do the ratios 15 cm to 2 m and 10 sec to 3 minutes form a proportion?

Answer:

To check if two ratios form a proportion, we need to compare them. First, ensure that quantities within each ratio are in the same units, and then simplify each ratio.

First ratio: 15 cm to 2 m.

The quantities are lengths, but units are different (cm and m). Convert 2 m to cm:

$2 \text{ m} = 2 \times 100 \text{ cm} = 200 \text{ cm} $.

The first ratio is 15 cm : 200 cm. Write as a fraction and simplify:

Ratio 1 $= \frac{15}{200} $.

Find HCF(15, 200). Factors of 15: 1, 3, 5, 15. Factors of 200: 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 200. HCF is 5.

Ratio 1 $= \frac{15 \div 5}{200 \div 5} = \frac{3}{40} $.

The first ratio is $3 : 40$.

Second ratio: 10 sec to 3 minutes.

The quantities are time, but units are different (sec and minutes). Convert 3 minutes to seconds:

$3 \text{ minutes} = 3 \times 60 \text{ seconds} = 180 \text{ seconds} $.

The second ratio is 10 sec : 180 sec. Write as a fraction and simplify:

Ratio 2 $= \frac{10}{180} $.

Find HCF(10, 180). HCF is 10.

Ratio 2 $= \frac{10 \div 10}{180 \div 10} = \frac{1}{18} $.

The second ratio is $1 : 18$.

Now, we need to check if the ratios $3 : 40$ and $1 : 18$ are equal. That is, is $\frac{3}{40} = \frac{1}{18}$?

We can use the cross-product test. Check if $3 \times 18 = 40 \times 1$.

Product of extremes ($3 \times 18$) $= 54 $.

Product of means ($40 \times 1$) $= 40 $.

Since $54 \neq 40$, the product of extremes is not equal to the product of means. Therefore, the ratios do not form a proportion.

Therefore, No, the given ratios do not form a proportion.

Example 2. If the cost of 6 cans of juice is $\textsf{₹} 210$, what will be the cost of 4 cans of juice? (Assuming the rate is proportional).

Answer:

This problem assumes a direct proportionality between the number of cans and their cost. This means the ratio of the number of cans is equal to the ratio of their costs.

Let $N_1$ be the initial number of cans (6) and $C_1$ be their cost ($\textsf{₹} 210$).

Let $N_2$ be the new number of cans (4) and $C_2$ be their unknown cost (let $C_2 = x$).

Since the number of cans and the cost are proportional, we can set up a proportion:

Ratio of initial cans to new cans = Ratio of initial cost to new cost.

$N_1 : N_2 :: C_1 : C_2 $

Substitute the values:

$6 : 4 :: 210 : x $

Using the property "Product of extremes = Product of means":

$6 \times x = 4 \times 210 $

Calculate the product on the right side:

$6x = 840 $

We have a linear equation $6x = 840$. To find $x$, divide both sides by 6 (or transpose 6):

$x = \frac{840}{6} $.

Perform the division:

$ \begin{array}{r} 140 \\ 6{\overline{\smash{\big)}\,840}} \\ \underline{-~\phantom{(}6}\downarrow\phantom{00} \\ 24\phantom{0} \\ \underline{-~24}\downarrow \\ 00\phantom{)} \\ \underline{-~0}\phantom{)} \\ 0\phantom{)} \end{array} $

$x = 140 $.

Since $x$ represents the cost in Rupees, the cost of 4 cans of juice is $ \textsf{₹} 140 $.

Therefore, the cost of 4 cans of juice will be $\textsf{₹} 140$.

(This problem can also be solved using the Unitary Method, which is directly based on the idea of finding the value per unit, which is inherent in proportion).



Unitary Method

In the previous sections, we learned about ratios and proportions. Ratio is a comparison of two quantities, and proportion is an equality between two ratios. The concept of proportion is very useful for solving problems where quantities are related in a consistent way (e.g., if the cost per item is constant). The Unitary Method is a widely used technique that leverages the idea of finding the value of 'one unit' to solve such proportional problems efficiently.


What is the Unitary Method?

The word 'unitary' comes from 'unit', which means 'one'. The Unitary Method is a method used to solve problems by first finding the value of a single unit and then using that value to find the value of the required number of units.

This method is particularly applicable to problems involving direct proportion, where an increase in one quantity leads to a proportional increase in another quantity, and a decrease in one leads to a proportional decrease in the other (e.g., more items cost more, less time covers less distance at constant speed).

The process essentially breaks down the problem into two manageable steps.

Steps in the Unitary Method:

Suppose you are given the total value (cost, distance, quantity, etc.) corresponding to a certain number of units, and you need to find the total value for a different number of units.

  1. Step 1: Find the value of one unit. To do this, divide the given total value by the given number of units.

    Value of 1 unit $= \frac{\text{Total value of given number of units}}{\text{Given number of units}} $

    This step determines the rate or value associated with a single item or unit quantity (e.g., cost per pen, distance per hour).

  2. Step 2: Find the value of the required number of units. Once you know the value of one unit, multiply this unit value by the required number of units to get the final answer.

    Value of required units $=$ (Value of 1 unit) $\times$ (Required number of units)

    This step scales up the unit value to the desired quantity.

The unitary method provides a clear and logical way to solve many real-life problems involving proportional relationships.


Examples using Unitary Method (Direct Proportion)

Example 1. If the cost of 6 pens is $\textsf{₹} 90$, what is the cost of 10 such pens?

Answer:

Given: Cost of 6 pens = $\textsf{₹} 90$.

To Find: Cost of 10 pens.

This is a problem involving direct proportion (more pens mean more cost). We use the unitary method.

Step 1: Find the cost of 1 pen (the unit).

The cost of 6 pens is $\textsf{₹} 90$. To find the cost of one pen, divide the total cost by the number of pens.

Cost of 1 pen $= \frac{\text{Total cost of 6 pens}}{\text{Number of pens}} = \frac{\textsf{₹} 90}{6} $

Perform the division:

$ \begin{array}{r} 15 \\ 6{\overline{\smash{\big)}\,90}} \\ \underline{-~6}\downarrow \\ 30 \\ \underline{-~30} \\ 0 \end{array} $

Cost of 1 pen $= \textsf{₹} 15 $.

Step 2: Find the cost of 10 pens (the required number of units).

Now that we know the cost of a single pen is $\textsf{₹} 15$, we multiply this unit cost by the required number of pens (10).

Cost of 10 pens $= (\text{Cost of 1 pen}) \times 10$

Cost of 10 pens $= \textsf{₹} 15 \times 10 = \textsf{₹} 150 $.

Therefore, the cost of 10 such pens is $\textsf{₹} 150$.

Example 2. A car travels 165 km in 3 hours. Assume the speed is uniform. (a) How long will it take to travel 440 km? (b) How far will it travel in 7 hours?

Answer:

Given: Distance travelled = 165 km, Time taken = 3 hours. The speed is uniform, implying direct proportion between distance and time.

We can find either the speed (distance per hour) or the time taken per km. Finding speed is often more intuitive.

Finding the unit value: Speed (distance per unit time).

In 3 hours, the car travels 165 km.

To find the distance travelled in 1 hour, divide the total distance by the time taken.

Distance travelled in 1 hour $= \frac{\text{Total distance}}{\text{Time taken}} = \frac{165 \text{ km}}{3 \text{ hours}} $

Perform the division:

$ \begin{array}{r} 55 \\ 3{\overline{\smash{\big)}\,165}} \\ \underline{-~15}\downarrow \\ 15 \\ \underline{-~15} \\ 0 \end{array} $

Distance travelled in 1 hour $= 55 \text{ km/hour} $.

The car's speed is 55 km/hour. This is our unit rate.

(a) How long will it take to travel 440 km?

We know the car travels 55 km in 1 hour.

This means that to travel 55 km, the time taken is 1 hour.

To find the time taken for 1 km, divide the time taken for 55 km by 55.

Time taken to travel 1 km $= \frac{1 \text{ hour}}{55 \text{ km}} = \frac{1}{55} \text{ hours/km} $.

Now, find the time taken to travel 440 km. Multiply the time taken per km by the total distance.

Time taken to travel 440 km $= (\text{Time per km}) \times 440 \text{ km} = \frac{1}{55} \times 440 $ hours.

Perform the multiplication:

Time $= \frac{440}{55} $ hours.

Simplify the fraction $\frac{440}{55}$. Both are divisible by 5 ($440 \div 5 = 88$, $55 \div 5 = 11$).

Time $= \frac{88}{11} $ hours.

Divide 88 by 11:

Time $= 8 $ hours.

Therefore, it will take 8 hours to travel 440 km.

(b) How far will it travel in 7 hours?

We know the distance travelled in 1 hour (our unit rate): 55 km.

To find the distance travelled in 7 hours, multiply the distance per hour by the required number of hours (7).

Distance travelled in 7 hours $= (\text{Distance per hour}) \times 7 \text{ hours}$

Distance travelled in 7 hours $= 55 \text{ km/hour} \times 7 \text{ hours} $.

Perform the multiplication:

$ \begin{array}{cc} & 55 \\ \times & 7 \\ \hline & 385 \\ \hline \end{array} $

Distance travelled in 7 hours $= 385 \text{ km} $.

Therefore, it will travel 385 km in 7 hours.


Inverse Proportion and Unitary Method (Brief Mention)

The unitary method as described above is designed for problems of direct proportion. However, there are also situations involving inverse proportion, where an increase in one quantity causes a proportional decrease in another quantity (e.g., if more people work on a task, the time taken to complete the task decreases). For inverse proportion problems, the product of the two quantities is constant ($x_1 y_1 = x_2 y_2$).

While a full treatment of inverse proportion with the unitary method is often covered in higher classes, the basic idea of finding the 'unit' value can be adapted:

For inverse proportion, if 'n' units of the first quantity correspond to a total value 'V' of the second quantity, then 1 unit of the first quantity corresponds to $V \times n$ units of the second quantity (e.g., 1 worker takes $10 \times 12$ days if 12 workers take 10 days).

This 'unit value' is then used to find the value for the required number of units, typically by division.

Example (Inverse Proportion): If 12 workers can build a wall in 10 days, how many days will 15 workers take to build the same wall? (Assuming all workers work at the same rate, this is inverse proportion: more workers, less time).

Total work done is constant. Work = Number of workers $\times$ Number of days.

Total work $= 12 \text{ workers} \times 10 \text{ days} = 120$ 'worker-days'.

Now, 15 workers need to complete 120 worker-days of work. Let D be the number of days 15 workers take.

$15 \text{ workers} \times D \text{ days} = 120 \text{ worker-days} $

$15D = 120 $

Divide by 15:

$D = \frac{120}{15} = 8 $ days.

So, 15 workers will take 8 days.

Using Unitary Method logic for Inverse Proportion:

12 workers take 10 days.

1 worker would take $10 \times 12 = 120$ days (If one worker does the whole job alone, it takes much longer). This is the value for 'one unit' (one worker).

15 workers would take $\frac{120 \text{ days}}{15} = 8$ days (If 15 workers share the work, it takes less time).

While the term 'unitary method' in Class 7 usually refers to direct proportion, it's good to be aware that similar logic applies with an adjustment for inverse relationships.



Percentage and its Conversion

In the previous sections, we looked at comparing quantities using ratios and proportions. Percentage is another very common and useful way to compare quantities, especially when the comparison needs to be made on a standard base of 100. The concept of percentage is widely used in various real-life scenarios, such as calculating discounts, expressing interest rates, reporting survey results, or comparing performance.


What is Percentage?

The word "percent" comes from the Latin phrase "per centum", which means "by the hundred" or "for every hundred". A Percentage is a way of expressing a number or a ratio as a fraction of 100. The symbol used for percent is %.

When we say "$x$ percent" (written as $x\%$), it means $x$ parts out of 100 equal parts of a whole.

So, $x\% = \frac{x}{100}$.

Percentage is essentially a ratio with the consequent (second term) being 100.

Example: $50\%$ means 50 per hundred, which can be written as the fraction $\frac{50}{100}$.

Example: $75\%$ means 75 per hundred, or $\frac{75}{100}$.

Example: $100\%$ means 100 per hundred, or $\frac{100}{100} = 1$ (one whole).

Example: $10\%$ means 10 per hundred, or $\frac{10}{100} = \frac{1}{10}$.


Converting Percentages

We can convert percentages into fractions or decimals, and vice-versa.

1. Converting Percentage to Fraction

To convert a percentage into a fraction, remember that $x\%$ means $\frac{x}{100}$. So, divide the number preceding the percent sign by 100, and then simplify the resulting fraction.

Steps:

  1. Remove the % sign.
  2. Write the number as the numerator and 100 as the denominator.
  3. Simplify the fraction to its lowest terms, if possible.

Examples:

(a) Convert 50% to a fraction.

$50\% = \frac{50}{100} $

Simplify the fraction $\frac{50}{100}$ (divide numerator and denominator by HCF = 50).

$ \frac{50 \div 50}{100 \div 50} = \frac{1}{2} $.

So, $50\% = \frac{1}{2}$.

(b) Convert 75% to a fraction.

$75\% = \frac{75}{100} $.

Simplify $\frac{75}{100}$ (divide by HCF = 25).

$ \frac{75 \div 25}{100 \div 25} = \frac{3}{4} $.

So, $75\% = \frac{3}{4}$.

(c) Convert $12.5\%$ to a fraction.

$12.5\% = \frac{12.5}{100} $.

To remove the decimal from the numerator, multiply both the numerator and denominator by 10:

$ \frac{12.5 \times 10}{100 \times 10} = \frac{125}{1000} $.

Simplify $\frac{125}{1000}$ (HCF is 125).

$ \frac{125 \div 125}{1000 \div 125} = \frac{1}{8} $.

So, $12.5\% = \frac{1}{8}$.

(d) Convert $33\frac{1}{3}\%$ to a fraction.

First, convert the mixed number percentage to an improper fraction:

$33\frac{1}{3}\% = \frac{(33 \times 3) + 1}{3}\% = \frac{99+1}{3}\% = \frac{100}{3}\% $.

Now, convert this percentage fraction to a standard fraction by dividing by 100:

$\frac{100}{3}\% = \frac{(\frac{100}{3})}{100} = \frac{100}{3} \div 100 = \frac{100}{3} \times \frac{1}{100} $.

Multiply and simplify (cancel 100):

$ \frac{\cancel{100}^1}{3} \times \frac{1}{\cancel{100}_1} = \frac{1 \times 1}{3 \times 1} = \frac{1}{3} $.

So, $33\frac{1}{3}\% = \frac{1}{3}$.

2. Converting Percentage to Decimal

To convert a percentage to a decimal, remember that $x\% = \frac{x}{100}$. Dividing by 100 involves shifting the decimal point two places to the left.

Steps:

  1. Remove the % sign.
  2. Move the decimal point two places to the left. If the number is a whole number, imagine the decimal point is at the end (e.g., $45 = 45.$). Add leading zeros if needed.

Examples:

(a) Convert 45% to a decimal.

$45\% = \frac{45}{100} $. Shift decimal in $45.$ two places left $\to .45$. Add leading zero $\to 0.45$.

So, $45\% = 0.45$.

(b) Convert 8% to a decimal.

$8\% = \frac{8}{100} $. Shift decimal in $8.$ two places left. $8. \to .8 \to .08$. Add leading zero $\to 0.08$.

So, $8\% = 0.08$.

(c) Convert 125% to a decimal.

$125\% = \frac{125}{100} $. Shift decimal in $125.$ two places left $\to 1.25$.

So, $125\% = 1.25$.

(d) Convert $0.5\%$ to a decimal.

$0.5\% = \frac{0.5}{100} $. Shift decimal in $0.5$ two places left. $0.5 \to .05 \to .005$. Add leading zero $\to 0.005$.

So, $0.5\% = 0.005$.


Converting to Percentages

1. Converting Fraction to Percentage

To convert a fraction into a percentage, we need to express it as an equivalent fraction with a denominator of 100. A simple way to do this is to multiply the fraction by 100 and add the % sign.

Formula:

Fraction to Percentage $= (\text{Given Fraction} \times 100)\% $

Examples:

(a) Convert $\frac{1}{4}$ to a percentage.

$(\frac{1}{4} \times 100)\% = (\frac{100}{4})\% = 25\% $.

(b) Convert $\frac{3}{5}$ to a percentage.

$(\frac{3}{5} \times 100)\% = (3 \times \frac{100}{5})\% = (3 \times 20)\% = 60\% $.

(c) Convert $\frac{7}{20}$ to a percentage.

$(\frac{7}{20} \times 100)\% = (7 \times \frac{100}{20})\% = (7 \times 5)\% = 35\% $.

(d) Convert $1\frac{1}{2}$ to a percentage.

First, convert the mixed fraction to an improper fraction: $1\frac{1}{2} = \frac{(1 \times 2) + 1}{2} = \frac{3}{2}$.

Now, convert the improper fraction to a percentage:

$(\frac{3}{2} \times 100)\% = (3 \times \frac{100}{2})\% = (3 \times 50)\% = 150\% $.

Percentages can indeed be greater than 100%.

2. Converting Decimal to Percentage

To convert a decimal number to a percentage, we multiply the decimal by 100 and add the % sign. Multiplying by 100 means shifting the decimal point two places to the right.

Formula:

Decimal to Percentage $= (\text{Given Decimal} \times 100)\% $

Steps:

  1. Move the decimal point two places to the right. Add trailing zeros if needed.
  2. Add the % sign at the end.

Examples:

(a) Convert $0.65$ to a percentage.

Move decimal in $0.65$ two places right: $65. \to 65$. Add % sign.

$(0.65 \times 100)\% = 65\% $.

(b) Convert $0.9$ to a percentage.

Move decimal in $0.9$ two places right. Add a zero: $0.90 \to 90$. Add % sign.

$(0.9 \times 100)\% = 90\% $.

(c) Convert $0.02$ to a percentage.

Move decimal in $0.02$ two places right: $0.02 \to 00.2 \to 2$. Add % sign.

$(0.02 \times 100)\% = 2\% $.

(d) Convert $1.75$ to a percentage.

Move decimal in $1.75$ two places right: $1.75 \to 17.5 \to 175$. Add % sign.

$(1.75 \times 100)\% = 175\% $.


Interpreting Percentages

Percentages help us understand proportions and compare quantities easily by providing a common base of 100.

Example 1. Convert the following:

(a) $25\%$ to a fraction and a decimal.

(b) $\frac{2}{5}$ to a percentage.

(c) $0.07$ to a percentage.

Answer:

(a) Convert $25\%$ to a fraction and a decimal.

Percentage to Fraction:

Remove % sign and write over 100: $25\% = \frac{25}{100}$.

Simplify the fraction (HCF of 25 and 100 is 25): $\frac{25 \div 25}{100 \div 25} = \frac{1}{4}$.

So, $25\% = \frac{1}{4}$.

Percentage to Decimal:

Remove % sign and divide by 100 (shift decimal 2 places left): $25\% = \frac{25}{100}$. $25. \to .25 \to 0.25$.

So, $25\% = 0.25$.

(b) Convert $\frac{2}{5}$ to a percentage.

Multiply the fraction by 100 and add % sign:

$(\frac{2}{5} \times 100)\% = (\frac{2 \times 100}{5})\% $.

Calculate the value inside the parenthesis: $\frac{200}{5} = 40$.

$ = 40\% $.

So, $\frac{2}{5} = 40\%$.

(c) Convert $0.07$ to a percentage.

Multiply the decimal by 100 and add % sign:

$(0.07 \times 100)\% $.

Calculate the value inside the parenthesis (shift decimal 2 places right): $0.07 \times 100 = 7$.

$ = 7\% $.

So, $0.07 = 7\%$.

Example 2. Out of 40 children in a class, 10 are boys. What is the percentage of boys?

Answer:

Total number of children in the class = 40.

Number of boys = 10.

To find the percentage of boys, first find the fraction of children who are boys. The fraction is the number of boys divided by the total number of children.

Fraction of boys $= \frac{\text{Number of boys}}{\text{Total number of children}} = \frac{10}{40} $.

Simplify the fraction: $\frac{10}{40} = \frac{1}{4}$.

Now, convert this fraction to a percentage by multiplying by 100 and adding the % sign.

Percentage of boys $= (\frac{1}{4} \times 100)\% $.

Calculate the value: $\frac{1}{4} \times 100 = \frac{100}{4} = 25$.

Percentage of boys $= 25\% $.

Therefore, the percentage of boys in the class is 25%.



Finding the Percentage of a Given Quantity

In the previous section, we learned what percentage is and how to convert between fractions, decimals, and percentages. A very practical application of percentages is to calculate a specific percentage of a given number or quantity. This is commonly used in calculating discounts, taxes, parts of a whole amount, or solving problems related to marks, money, and measurements.


Method to Find Percentage of a Quantity

To find $x\%$ of a quantity $Q$, you can follow these steps:

  1. Convert the percentage $x\%$ into its fractional form, which is $\frac{x}{100}$.
  2. Multiply this fraction $\frac{x}{100}$ by the given quantity $Q$. Remember that the word "of" in such contexts means multiplication.

Formula:

$x\% \text{ of } Q = \frac{x}{100} \times Q $

Alternatively, you can convert the percentage to its decimal form and then multiply by the quantity:

$x\% \text{ of } Q = (\text{Decimal form of } x\%) \times Q $

The choice between using the fractional or decimal form depends on the numbers involved and which calculation seems easier.

Understanding "of" in Mathematics:

As mentioned, in mathematical phrases like "$x\%$ of $Q$" or "a fraction of a quantity", the word "of" is a keyword that indicates the operation of multiplication.

Example: $\frac{1}{2}$ of 100 = $\frac{1}{2} \times 100 = 50$.

Example: $25\%$ of 160 means $25\% \times 160$.


Examples

Example 1. Find 25% of 160.

Answer:

We need to calculate 25% of the quantity 160.

Method 1: Using fraction conversion

Convert the percentage to a fraction: $25\% = \frac{25}{100}$. Simplify this fraction: $\frac{25}{100} = \frac{1}{4}$.

Now, multiply the fraction by the quantity 160:

$25\% \text{ of } 160 = \frac{1}{4} \times 160 $.

Write 160 as $\frac{160}{1}$ and multiply (using cancellation):

$ = \frac{1}{\cancel{4}_1} \times \frac{\cancel{160}^{40}}{1} = \frac{1 \times 40}{1 \times 1} = 40 $.

So, 25% of 160 is 40.

Method 2: Using decimal conversion

Convert the percentage to a decimal: $25\% = 0.25$.

Now, multiply the decimal by the quantity 160:

$25\% \text{ of } 160 = 0.25 \times 160 $.

Perform the multiplication:

$ \begin{array}{@{}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c} & & & 1&6&0 \\ \times & & 0&.&2&5 \\ \hline & & & 8&0&0 & \leftarrow & 160 \times 5 \\ & & 3&2&0 & \times & \leftarrow & 160 \times 2 \\ \hline & 4&0&.&0&0 \\ \hline \end{array} $

(When multiplying decimals, multiply as whole numbers, then place decimal in product - 160 has 0 decimal places, 0.25 has 2 decimal places. Total 2 decimal places in the product. $160 \times 25 = 4000$. Placing decimal: $40.00$).

$0.25 \times 160 = 40.00 = 40 $.

Both methods yield the same result.

Therefore, 25% of 160 is 40.

Example 2. A school team won 6 games this year against 4 games won last year. What is the percentage increase in the number of games won?

Answer:

Given:

Number of games won this year = 6.

Number of games won last year (This is the original amount for comparison) = 4.

To Find: Percentage increase.

First, calculate the actual increase in the number of games won.

Amount of Increase = (Number of games won this year) $-$ (Number of games won last year)

Amount of Increase $= 6 - 4 = 2$ games.

Now, to find the percentage increase, we compare this increase to the original amount (games won last year) and express it as a percentage.

Formula for Percentage Increase:

Percentage Increase $= (\frac{\text{Amount of Increase}}{\text{Original Amount}} \times 100)\% $

Substitute the values: Amount of Increase = 2, Original Amount = 4.

Percentage Increase $= (\frac{2}{4} \times 100)\% $.

Simplify the fraction $\frac{2}{4}$: $\frac{2}{4} = \frac{1}{2}$.

Percentage Increase $= (\frac{1}{2} \times 100)\% $.

Calculate the value inside the parenthesis: $\frac{1}{2} \times 100 = 50$.

Percentage Increase $= 50\% $.

Therefore, the percentage increase in the number of games won is 50%.

Example 3. The price of a scooter was $\textsf{₹} 34,000$ last year. It has increased by 20% this year. What is the price now?

Answer:

Given:

Original price of the scooter = $\textsf{₹} 34,000$.

Percentage increase in price = 20%.

To Find: The new price of the scooter this year.

Method 1: Calculate the amount of increase and add it to the original price.

The amount of increase is 20% of the original price.

Amount of Increase = 20% of $\textsf{₹} 34,000$.

Convert 20% to a fraction: $20\% = \frac{20}{100} = \frac{1}{5}$.

Amount of Increase $= \frac{1}{5} \times \textsf{₹} 34,000 = \frac{34000}{5} \textsf{₹}$.

Perform the division: $34000 \div 5$.

$ \begin{array}{r} 6800 \\ 5{\overline{\smash{\big)}\,34000}} \\ \underline{-~30}\downarrow\phantom{0} \\ 40\phantom{00} \\ \underline{-~40}\downarrow\phantom{} \\ 00\phantom{)} \\ \underline{-~0}\phantom{)} \\ 0\phantom{)} \end{array} $

Amount of Increase $= \textsf{₹} 6,800$.

The new price is the original price plus the amount of increase.

New Price = Original Price + Amount of Increase

New Price $= \textsf{₹} 34,000 + \textsf{₹} 6,800 $.

Perform the addition:

$ \begin{array}{cc} & 34000 \\ + & \phantom{0}6800 \\ \hline & 40800 \\ \hline \end{array} $

New Price $= \textsf{₹} 40,800$.

Alternate Method: Calculate the new price directly as a percentage of the original price.

If the price increases by 20%, the new price will be $100\%$ (original price) $+ 20\%$ (increase) $= 120\%$ of the original price.

New Price = 120% of Original Price

Convert 120% to a fraction: $120\% = \frac{120}{100} = \frac{12}{10} = \frac{6}{5}$.

New Price $= \frac{6}{5} \times \textsf{₹} 34,000$.

New Price $= \frac{6}{5} \times 34000 $.

Multiply (using cancellation):

New Price $= 6 \times \frac{34000}{5} = 6 \times 6800 $.

Calculate the product: $6 \times 6800 = 40800$.

New Price $= \textsf{₹} 40,800 $.

Therefore, the price of the scooter now is $\textsf{₹} 40,800$.


Finding What Percentage One Quantity is of Another

Sometimes, you are given two quantities and need to express the first quantity as a percentage of the second quantity (which serves as the base or the whole). To find what percentage a quantity A is of another quantity B:

  1. Ensure that both quantities A and B are in the same units.
  2. Form a fraction with quantity A as the numerator and quantity B as the denominator: $\frac{A}{B}$.
  3. Multiply this fraction by 100.
  4. Add the % sign at the end of the result.

Formula:

Percentage $= (\frac{\text{Quantity A}}{\text{Quantity B (Base)}} \times 100)\% $

Example 4. What percentage of 200 is 50?

Answer:

We want to find what percentage the quantity 50 is of the quantity 200. Here, the quantity A is 50, and the base quantity B is 200. We assume they are quantities of the same kind and unit.

Using the formula: Percentage $= (\frac{A}{B} \times 100)\% $.

Percentage $= (\frac{50}{200} \times 100)\% $.

Simplify the fraction $\frac{50}{200}$: $\frac{50}{200} = \frac{1}{4}$.

Percentage $= (\frac{1}{4} \times 100)\% $.

Calculate the value inside the parenthesis: $\frac{1}{4} \times 100 = 25$.

Percentage $= 25\% $.

Therefore, 50 is 25% of 200.



Profit and Loss

In the world of business and trade, buying and selling goods is a fundamental activity. Whenever an article is purchased and then sold, the prices involved determine whether the transaction results in a financial gain or a financial loss. Understanding these concepts of Profit and Loss is very important in mathematics and for managing personal finances.


Basic Terms in Profit and Loss

1. Cost Price (C.P.)

The Cost Price (C.P.) is the total amount of money paid to acquire an article or commodity. This includes not only the price at which the article was initially bought, but also any additional expenses incurred to bring the article into a condition and location suitable for selling. These additional expenses are called Overhead Expenses.

Effective C.P. = Purchase Price + Overhead Expenses.

Examples of overhead expenses include transportation charges, packaging costs, repair costs to make an item usable, or installation charges.

2. Selling Price (S.P.)

The Selling Price (S.P.) is the price at which an article or commodity is sold to a customer.

3. Profit (or Gain)

When an article is sold for a price higher than its cost price, the difference is called Profit or Gain. A profit occurs when the seller earns more money than they spent to acquire the item.

Profit = S.P. $-$ C.P.

(This is when S.P. $>$ C.P.)

4. Loss

When an article is sold for a price lower than its cost price, the difference is called a Loss. A loss occurs when the seller receives less money than they spent to acquire the item.

Loss = C.P. $-$ S.P.

(This is when C.P. $>$ S.P.)

5. No Profit, No Loss

If the Selling Price (S.P.) of an article is exactly equal to its Cost Price (C.P.), then there is neither any profit nor any loss in the transaction. S.P. = C.P.


Calculating Profit Percentage and Loss Percentage

Profit and loss are often expressed as a percentage to provide a standardised measure of performance relative to the initial investment (the cost price). The Cost Price (C.P.) is always used as the base for calculating profit or loss percentage, unless a different base (like selling price) is specifically mentioned (which is rare in basic problems).

1. Profit Percentage (Profit %)

The Profit Percentage (Profit %) tells us how much profit was made for every $\textsf{₹} 100$ of the cost price. It is calculated as the ratio of the profit to the cost price, multiplied by 100.

Profit % = $(\frac{\text{Profit}}{\text{C.P.}} \times 100)\% $

2. Loss Percentage (Loss %)

The Loss Percentage (Loss %) tells us how much loss was incurred for every $\textsf{₹} 100$ of the cost price. It is calculated as the ratio of the loss to the cost price, multiplied by 100.

Loss % = $(\frac{\text{Loss}}{\text{C.P.}} \times 100)\% $


Formulas to Find S.P. or C.P.

Using the definitions of profit and loss percentages, we can derive formulas to find the Selling Price (S.P.) if the Cost Price (C.P.) and percentage profit or loss are known, or to find the C.P. if the S.P. and percentage profit or loss are known.

1. Finding S.P. when C.P. and Profit % are given:

If there is a profit of Profit %, the selling price is the cost price plus the profit amount. Profit amount is Profit % of C.P.

Profit Amount = Profit % of C.P. $= \frac{\text{Profit %}}{100} \times \text{C.P.}$.

S.P. = C.P. + Profit Amount = C.P. + $(\frac{\text{Profit %}}{100} \times \text{C.P.})$.

We can factor out C.P.:

S.P. = C.P. $\times (1 + \frac{\text{Profit %}}{100}) $

Combine the terms inside the parenthesis by finding a common denominator (100):

S.P. = C.P. $\times (\frac{100}{100} + \frac{\text{Profit %}}{100}) = \text{C.P.} \times (\frac{100 + \text{Profit %}}{100}) $

2. Finding S.P. when C.P. and Loss % are given:

If there is a loss of Loss %, the selling price is the cost price minus the loss amount. Loss amount is Loss % of C.P.

Loss Amount = Loss % of C.P. $= \frac{\text{Loss %}}{100} \times \text{C.P.}$.

S.P. = C.P. $-$ Loss Amount = C.P. $-$ $(\frac{\text{Loss %}}{100} \times \text{C.P.})$.

We can factor out C.P.:

S.P. = C.P. $\times (1 - \frac{\text{Loss %}}{100}) $

Combine the terms inside the parenthesis:

S.P. = C.P. $\times (\frac{100}{100} - \frac{\text{Loss %}}{100}) = \text{C.P.} \times (\frac{100 - \text{Loss %}}{100}) $

3. Finding C.P. when S.P. and Profit % are given:

We can rearrange the formula for S.P. when profit is made to find C.P.

S.P. = C.P. $\times (\frac{100 + \text{Profit %}}{100})$.

To isolate C.P., divide both sides by $(\frac{100 + \text{Profit %}}{100})$. Dividing by a fraction is the same as multiplying by its reciprocal.

C.P. = S.P. $\div (\frac{100 + \text{Profit %}}{100}) = \text{S.P.} \times (\frac{100}{100 + \text{Profit %}}) $

4. Finding C.P. when S.P. and Loss % are given:

We rearrange the formula for S.P. when a loss is incurred to find C.P.

S.P. = C.P. $\times (\frac{100 - \text{Loss %}}{100})$.

To isolate C.P., divide both sides by $(\frac{100 - \text{Loss %}}{100})$. Multiply by its reciprocal.

C.P. = S.P. $\div (\frac{100 - \text{Loss %}}{100}) = \text{S.P.} \times (\frac{100}{100 - \text{Loss %}}) $


Examples

Example 1. A shopkeeper bought a chair for $\textsf{₹} 375$ and sold it for $\textsf{₹} 400$. Find his gain percentage.

Answer:

Given:

Cost Price (C.P.) of the chair = $\textsf{₹} 375$.

Selling Price (S.P.) of the chair = $\textsf{₹} 400$.

To Find: Gain percentage.

Since S.P. ($\textsf{₹} 400$) is greater than C.P. ($\textsf{₹} 375$), the shopkeeper made a profit (gain).

Calculate the amount of profit:

Profit = S.P. $-$ C.P. $= \textsf{₹} 400 - \textsf{₹} 375 = \textsf{₹} 25$.

Now, calculate the gain percentage using the formula: Profit % = $(\frac{\text{Profit}}{\text{C.P.}} \times 100)\%$.

Gain % $= (\frac{\textsf{₹} 25}{\textsf{₹} 375} \times 100)\% $.

The units cancel out, so we simplify the fraction $\frac{25}{375}$ and multiply by 100.

Simplify $\frac{25}{375}$ by dividing numerator and denominator by their HCF, which is 25.

$25 \div 25 = 1 $.

$375 \div 25$. We can do this by long division or recognising that $375 = 3 \times 125 = 3 \times 5 \times 25$. So $375 \div 25 = 15$.

$ \begin{array}{r} 15 \\ 25{\overline{\smash{\big)}\,375}} \\ \underline{-~25}\downarrow \\ 125 \\ \underline{-~125} \\ 0 \end{array} $

So, $\frac{25}{375} = \frac{1}{15}$.

Gain % $= (\frac{1}{15} \times 100)\% = \frac{100}{15}\%$.

Simplify the fraction $\frac{100}{15}$ by dividing numerator and denominator by 5: $\frac{100 \div 5}{15 \div 5} = \frac{20}{3}$.

Gain % $= \frac{20}{3}\% $.

As a mixed fraction, $\frac{20}{3} = 6 \frac{2}{3}$. So, Gain % $= 6\frac{2}{3}\%$.

As a decimal, $20 \div 3 \approx 6.666...$. So, Gain % $\approx 6.67\%$ (rounded to two decimal places).

Therefore, the gain percentage is $6\frac{2}{3}\%$.

Example 2. An article was sold for $\textsf{₹} 250$ with a profit of 5%. What was its cost price?

Answer:

Given:

Selling Price (S.P.) = $\textsf{₹} 250$.

Profit % = 5%.

To Find: Cost Price (C.P.).

We can use the formula for finding C.P. when S.P. and Profit % are given:

C.P. = S.P. $\times (\frac{100}{100 + \text{Profit %}}) $.

Substitute the given values:

C.P. = $250 \times (\frac{100}{100 + 5}) $.

C.P. = $250 \times \frac{100}{105} $.

Simplify the fraction $\frac{100}{105}$ by dividing numerator and denominator by their HCF, which is 5.

$\frac{100 \div 5}{105 \div 5} = \frac{20}{21} $.

Substitute the simplified fraction back into the calculation for C.P.:

C.P. = $250 \times \frac{20}{21} = \frac{250 \times 20}{21} = \frac{5000}{21} $.

Now, perform the division $\frac{5000}{21}$ to get the numerical value.

$ \begin{array}{r} 238.095...\phantom{0} \\ 21{\overline{\smash{\big)}\,5000.000\phantom{)}}} \\ \underline{-~42}\downarrow\phantom{00000} \\ 80\phantom{0000} \\ \underline{-~63}\downarrow\phantom{000} \\ 170\phantom{00} \\ \underline{-~168}\downarrow\phantom{0} \\ 20\phantom{)} \\ \underline{-~\phantom{()}0}\downarrow \\ 200\phantom{)} \\ \underline{-~189}\downarrow \\ 110 \\ \underline{-~105} \\ 5 \end{array} $

The result is a non-terminating repeating decimal. For practical purposes with money, we round to two decimal places.

C.P. $\approx \textsf{₹} 238.095...$. Rounding to the nearest paise (two decimal places), the cost price is approximately $\textsf{₹} 238.10$.

Therefore, the cost price was approximately $\textsf{₹} 238.10$.

Check: If C.P. is $\textsf{₹} 238.10$, Profit = 5% of $\textsf{₹} 238.10 = \frac{5}{100} \times 238.10 = 0.05 \times 238.10 = 11.905$. S.P. = $238.10 + 11.905 = \textsf{₹} 250.005 \approx \textsf{₹} 250$. (The check works with rounding).

Example 3. Juhi sells a washing machine for $\textsf{₹} 13,500$. She loses 20% in the bargain. What was the price at which she bought it?

Answer:

Given:

Selling Price (S.P.) of the washing machine = $\textsf{₹} 13,500$.

Loss % = 20%.

To Find: Cost Price (C.P.).

We use the formula for finding C.P. when S.P. and Loss % are given:

C.P. = S.P. $\times (\frac{100}{100 - \text{Loss %}}) $.

Substitute the given values:

C.P. = $13500 \times (\frac{100}{100 - 20}) $.

C.P. = $13500 \times \frac{100}{80} $.

Simplify the fraction $\frac{100}{80}$. Divide numerator and denominator by their HCF, which is 20.

$\frac{100 \div 20}{80 \div 20} = \frac{5}{4} $.

Substitute the simplified fraction back into the calculation for C.P.:

C.P. = $13500 \times \frac{5}{4} = \frac{13500 \times 5}{4} $.

Calculate the numerator: $13500 \times 5 = 67500$.

C.P. = $\frac{67500}{4} $.

Now, perform the division $\frac{67500}{4}$:

$ \begin{array}{r} 16875 \\ 4{\overline{\smash{\big)}\,67500}} \\ \underline{-~4}\downarrow\phantom{0000} \\ 27\phantom{000} \\ \underline{-~24}\downarrow\phantom{00} \\ 35\phantom{0} \\ \underline{-~32}\downarrow \\ 30 \\ \underline{-~28} \\ 20 \\ \underline{-~20} \\ 0 \end{array} $

C.P. $= 16875 $.

Since C.P. is in Rupees, the cost price was $\textsf{₹} 16,875$.

Therefore, the price at which Juhi bought the washing machine was $\textsf{₹} 16,875$.

Check: If C.P. is $\textsf{₹} 16,875$, Loss = 20% of $\textsf{₹} 16,875 = \frac{20}{100} \times 16875 = \frac{1}{5} \times 16875 = 3375$. Loss amount is $\textsf{₹} 3375$. S.P. = C.P. $-$ Loss = $16875 - 3375 = \textsf{₹} 13500$. This matches the given S.P.



Simple Interest

In financial transactions involving borrowing or lending money, an additional amount is usually involved as a charge for using the money. This additional amount is called Interest. When you borrow money, you pay interest; when you lend money (or deposit it in a bank), you receive interest. There are different ways to calculate interest, and the most basic one is called Simple Interest (S.I.).


Basic Terms in Simple Interest

1. Principal (P)

The Principal is the original sum of money that is borrowed, lent, or deposited. It is the initial amount on which interest is calculated. It is commonly denoted by the letter P.

2. Rate of Interest (R)

The Rate of Interest is the percentage at which the interest is calculated on the principal amount. It is usually specified as a percentage per annum (p.a.), which means "per year". It is denoted by R (or sometimes r).

Example: If the rate of interest is 10% p.a., it means that for every $\textsf{₹} 100$ of the principal amount, an interest of $\textsf{₹} 10$ will be charged or earned over a period of one year.

3. Time (T)

The Time is the duration for which the money is borrowed, lent, or deposited. For the Simple Interest formula, the time period is usually required in years. It is denoted by T (or sometimes n).

If the time is given in units other than years (like months or days), you must convert it into years before using the formula:

4. Simple Interest (S.I.)

Simple Interest (S.I.) is the interest calculated solely on the original principal amount for the entire duration of the loan or deposit. In simple interest calculations, the principal amount remains constant over the entire time period.

5. Amount (A)

The Amount is the total money that is repaid by the borrower to the lender (or received by the lender/depositor from the bank) at the end of the agreed time period. It is the sum of the original Principal and the calculated Simple Interest.

Amount (A) = Principal (P) + Simple Interest (S.I.)


Formula for Simple Interest

The Simple Interest (S.I.) is directly proportional to the Principal (P), the Rate of Interest (R), and the Time (T). The formula for calculating Simple Interest is:

S.I. $= \frac{P \times R \times T}{100} $

... (i)

Where:

Derivation of the S.I. Formula:

Let P be the principal amount, R be the rate of interest per annum (as R%), and T be the time in years.

By the definition of rate R%, the interest on $\textsf{₹} 100$ for 1 year is $\textsf{₹} R$.

Interest on $\textsf{₹} 1$ for 1 year $= \textsf{₹} \frac{R}{100}$.

Interest on $\textsf{₹} P$ for 1 year $$= (\text{Interest on } \textsf{₹} 1 \text{ for 1 year}) \times P = \textsf{₹} \frac{R}{100} \times P = \textsf{₹} \frac{PR}{100}$$

Simple Interest is calculated on the original principal for the entire time T. So, the interest on $\textsf{₹} P$ for T years $$= (\text{Interest on } \textsf{₹} P \text{ for 1 year}) \times T = \textsf{₹} (\frac{PR}{100}) \times T = \textsf{₹} \frac{P \times R \times T}{100}$$

Thus, S.I. $= \frac{P \times R \times T}{100}$.


Calculating P, R, or T when other values are known

The Simple Interest formula $S.I. = \frac{P \times R \times T}{100}$ is an equation with four variables (S.I., P, R, T). If we know any three of these variables, we can use the formula to find the fourth one by rearranging the equation.

From $S.I. \times 100 = P \times R \times T$, we can derive:

  1. To find Principal (P): Divide $(S.I. \times 100)$ by $(R \times T)$.

    P $= \frac{S.I. \times 100}{R \times T} $

  2. To find Rate of Interest (R): Divide $(S.I. \times 100)$ by $(P \times T)$.

    R $= \frac{S.I. \times 100}{P \times T} $ (Result is in % p.a.)

  3. To find Time (T): Divide $(S.I. \times 100)$ by $(P \times R)$.

    T $= \frac{S.I. \times 100}{P \times R} $ (Result is in years)

These derived formulas can be used directly, or you can always substitute the known values into the main formula $S.I. = \frac{P \times R \times T}{100}$ and solve the resulting equation for the unknown variable using algebraic methods (like transposition).


Examples

Example 1. Find the simple interest on $\textsf{₹} 5,000$ for 2 years at the rate of 8% per annum. Also, find the amount.

Answer:

Given:

Principal (P) = $\textsf{₹} 5,000$.

Time (T) = 2 years.

Rate of Interest (R) = 8% per annum.

To Find: Simple Interest (S.I.) and Amount (A).

Calculate Simple Interest (S.I.):

Use the formula S.I. $= \frac{P \times R \times T}{100}$.

Substitute the given values (P=5000, R=8, T=2):

S.I. $= \frac{5000 \times 8 \times 2}{100} $.

Perform the multiplication in the numerator and division by 100. It's easy to cancel the two zeros in the denominator with two zeros in the numerator (from 5000).

S.I. $= \frac{50 \times \cancel{100} \times 8 \times 2}{\cancel{100}} $.

S.I. $= 50 \times 8 \times 2 $.

Perform the multiplication: $50 \times 8 = 400$. Then $400 \times 2 = 800$.

S.I. $= \textsf{₹} 800 $.

The simple interest earned on $\textsf{₹} 5,000$ for 2 years at 8% p.a. is $\textsf{₹} 800$.

Calculate Amount (A):

The Amount is the sum of the Principal and the Simple Interest.

Amount (A) = Principal (P) + Simple Interest (S.I.)

Substitute the values of P and S.I.:

A $= \textsf{₹} 5,000 + \textsf{₹} 800 $.

A $= \textsf{₹} 5,800 $.

Therefore, the amount after 2 years will be $\textsf{₹} 5,800$.

Example 2. Anita takes a loan of $\textsf{₹} 5,000$ at 15% per year as rate of interest. Find the interest she has to pay at the end of one year.

Answer:

Given:

Principal (P) = $\textsf{₹} 5,000$.

Rate of Interest (R) = 15% per year.

Time (T) = 1 year.

To Find: Simple Interest (S.I.) at the end of one year.

Use the formula S.I. $= \frac{P \times R \times T}{100}$.

Substitute the given values (P=5000, R=15, T=1):

S.I. $= \frac{5000 \times 15 \times 1}{100} $.

Cancel the two zeros in the denominator with two zeros in the numerator:

S.I. $= \frac{50 \times \cancel{100} \times 15 \times 1}{\cancel{100}} $.

S.I. $= 50 \times 15 \times 1 $.

Perform the multiplication: $50 \times 15 = 750$.

S.I. $= \textsf{₹} 750 $.

Therefore, the interest Anita has to pay at the end of one year is $\textsf{₹} 750$.

Example 3. If Meena gives an interest of $\textsf{₹} 45$ for one year at 9% rate p.a. What is the sum she has borrowed?

Answer:

Given:

Simple Interest (S.I.) = $\textsf{₹} 45$.

Time (T) = 1 year.

Rate of Interest (R) = 9% p.a.

To Find: Principal (P), which is the sum borrowed.

We can use the rearranged formula for Principal: P $= \frac{S.I. \times 100}{R \times T}$.

Substitute the given values (S.I.=45, R=9, T=1):

P $= \frac{45 \times 100}{9 \times 1} $.

Simplify the expression. We can cancel out common factors before multiplying and dividing.

P $= \frac{\cancel{45}^5 \times 100}{\cancel{9}_1 \times 1} $.

P $= \frac{5 \times 100}{1 \times 1} = 5 \times 100 = 500 $.

Since P represents the principal amount borrowed, the unit is Rupees.

P $= \textsf{₹} 500 $.

Therefore, the sum Meena has borrowed is $\textsf{₹} 500$.

Check: Calculate the interest on $\textsf{₹} 500$ for 1 year at 9% p.a. S.I. $= \frac{500 \times 9 \times 1}{100} = \frac{5 \times 9 \times 1}{1} = 45$. This matches the given S.I.