Text Book Solutions All Class

 Multiplication of Fractions

Multiplication involving fractions can arise in various scenarios, such as finding a part of a whole or scaling quantities.

 We begin by examining the multiplication of a fraction by a whole number.

Consider a situation where we need to find the total distance covered if someone walks a certain fractional distance multiple times. 

For instance, if Aaron walks 3 kilometres in 1 hour, how far can he walk in 5 hours? 

This is a simple multiplication of whole numbers: 5 hours × 3 km/hour = 15 km. The calculation involves repeated addition: 3 + 3 + 3 + 3 + 3 = 15 km.

Now, let’s adapt this to fractions.

 Suppose Aaron’s pet tortoise walks only 1/4 kilometre in 1 hour. How far can it walk in 3 hours? 

The distance covered is a fraction, but the principle remains the same: multiplication.

 We need to calculate 3 × (1/4) km.

Visually, we can represent this as adding 1/4 three times: 1/4 + 1/4 + 1/4. 

Since the denominators are the same, we add the numerators: (1+1+1)/4 = 3/4. 

Therefore, the tortoise walks 3/4 kilometre in 3 hours.

Now, we need to figure out how far Aaron can walk in a fraction of an hour, for example, in 1/5 of an hour and 2/5 of an hour.

How Do We Calculate Distance in Fractional Time?

When the time is a fraction of an hour, like 1/5 or 2/5 of an hour, we calculate the distance by multiplying the fraction by the distance Aaron covers in 1 full hour.

1. How Far Can Aaron Walk in 1/5 of an Hour?

We know that in 1 hour, Aaron walks 3 kilometers.

To find out how far Aaron can walk in 1/5 hour, we multiply:

=15×3km

This means we need to divide 3 kilometers into 5 equal parts because the time is 1/5 of an hour.

Step-by-Step:

1. Divide 3 kilometers by 5 to get the distance Aaron walks in 1/5 of an hour.

35=0.653 = 0.6 km

So, Aaron can walk 0.6 kilometers in 1/5 of an hour.

2. How Far Can Aaron Walk in 2/5 of an Hour?

Now, let’s figure out how far Aaron can walk in 2/5 of an hour.

We know that in 1 hour, Aaron walks 3 kilometers.

To find out how far Aaron can walk in 2/5 of an hour, we multiply:

=25×3km

This means we need to find the distance covered in 1/5 of an hour first, and then multiply that by 2 (since 2/5 is double 1/5).

Step-by-Step:

1. First, calculate the distance Aaron covers in 1/5 of an hour. We already know from the previous part that in 1/5 of an hour, Aaron covers 0.6 kilometers.
2. Now, multiply the result by 2 (since we need to find the distance in 2/5 of an hour):

2×0.6=1.2 km

So, Aaron can walk 1.2 kilometers in 2/5 of an hour.

Discussion

We did this multiplication as follows: 

  First, we divided the multiplicand, 3 , by the denominator of the multiplier, 5, to get 3/5.

We then multiplied the result by the numerator of the multiplier, that is 2, to get 6 /5 . Thus, whenever we need to multiply a fraction and a whole number, we follow the steps above.

Multiplying Two Fractions

We know, that Aaron’s pet tortoise can walk only 1/4 km in 1 hour. How far can it walk in half an hour? 

Following our approach of using multiplication to solve such problems, we have, Distance covered in 1/2 hour = 1/2 × 1/4 km.

Finding the product:

Imagine a whole unit, say a square. First, divide it into four equal vertical strips; one strip represents 1/4.

Now, we need to find half of this shaded 1/4. We can do this by dividing the entire square horizontally into two equal parts. 

The portion that is both shaded (part of the original 1/4) and falls within the bottom half represents (1/2) × (1/4).

Since the whole is divided into 8 equal parts  and one of the parts is shaded, we can say that 1/ 8 of the whole is shaded. So, the distance covered by the tortoise in half an hour is 1/ 8 km. 

This tells us that 1/2 × 1/4 = 1/8 .

Example: If the tortoise walks faster and it can cover 2/ 5 km in 1 hour, how far will it walk in 3/ 4 of an hour?

Sol: We want to calculate the distance covered in 3/4 of an hour, which can be written as:

=34×25

Now, let’s break this problem into two smaller parts to make it easier to understand:

Since the tortoise covers 2/5 km in 1 hour, we first need to find out how much it walks in 1/4 of an hour (since 3/4 of an hour is 3 times 1/4 of an hour).

To find the distance in 1/4 of an hour, we need to divide 2/5 km into 4 equal parts (because we are looking for the distance in just 1/4 of an hour).

How to divide 2/5 into 4 equal parts:

• Start with the distance the tortoise covers in 1 hour: 2/5 km.
• How much of the whole is it? 
  The whole is divided into 5 rows and 4 columns, creating 5 × 4 = 20 equal parts. Number of these parts shaded = 2. 
  So, the distance covered in 1/ 4 of an hour = 2/ 20 .

Now, we need to multiply 2/ 20 by 3. Distance covered in 3/ 4 of an hour = 

Try yourself:If Aaron walks at a speed of 3 kilometers per hour, how far can he walk in 1/5 of an hour?

• A.0.6 kilometers
• B.1.2 kilometers
• C.1.0 kilometers
• D.0.8 kilometers

View Solution

Connection between the Area of a Rectangle and Fraction Multiplication

We are given a unit square with side length 1 unit. A rectangle is drawn inside the unit square. The length and breadth of the rectangle are both fractions:

• Length = 1/2 unit
• Breadth = 1/4 unit

We need to calculate the area of this rectangle, and also understand the relationship between the product of the length and breadth and the area of the rectangle.

The area of any rectangle can be found by multiplying its length by its breadth.

So, the area of the rectangle is:

=12×14

To multiply fractions, we multiply the numerators together and the denominators together:

So, the area of the rectangle is 1/8 square units.

This means that the area of the rectangle is the product of the length and breadth of the rectangle, even when they are fractions.

Let us find the product by observing what we did in the above cases. 

In each case, the whole is divided into rows and columns. The number of rows is the denominator of the multiplicand, which is 18 in this case. The number of columns is the denominator of the multiplier, which is 12 in this case.

 Thus, the whole is divided into 18 × 12 equal parts.

So, 

Thus, when two fractional units are multiplied, their product is

We express this as:

Multiplying Numerators and Denominators

When you multiply two fractions, you multiply the numerators (the top numbers) together, and the denominators (the bottom numbers) together.

In the example below:

• We are multiplying 5/12 by 7/18.
• First, you multiply the numerators:
  5×7=35.
• Then, you multiply the denominators:
  12×18=216.

So, the result of multiplying these fractions is:

The rule you are learning here is that when you multiply fractions, the product is:

This formula was first stated in this general form by Brahmagupta in his Brāhmasphuṭasiddhānta in 628 CE. 

Whole Numbers as Fractions

If you are multiplying a fraction by a whole number, you can turn the whole number into a fraction by giving it a denominator of 1.

For example:

Multiplication of Fractions — Simplifying to Lowest Form 

• Multiply the following fractions and express the product in its lowest form:  
• Instead of multiplying the numerators (12 and 5) and denominators (7 and 24) first and then simplifying, we could do the following:

We see that both the circled numbers have a common factor of 12. We know that a fraction remains the same when the numerator and denominator are divided by the common factor. In this case, we can divide them by 12.

Let us use the same technique to do one more multiplication.

When multiplying fractions, we can first divide the numerator and denominator by their common factors before multiplying the numerators and denominators. This is called cancelling the common factors.

Is the Product Always Greater than the Numbers Multiplied?

Since, we know that when a number is multiplied by 1, the product remains unchanged, we will look at multiplying pairs of numbers where neither of them is 1. When we multiply two counting numbers greater than 1, say 3 and 5, the product is greater than both the numbers being multiplied.

3 × 5 = 15 

The product, 15, is more than both 3 and 5. But what happens when we multiply 1 /4 and 8?

In the above multiplication the product, 2, is greater than 1/4 , but less than 8. 

What happens when we multiply 3/4 and 2/5 ?

  Let us compare this product 6/20 with the numbers 3/4 and 2/5 . For this,

let us express 3 /4 as 15/ 20 and 2 /5 as 8/ 20 . From this we can see that the product is less than both the numbers.

Order of Multiplication

Just like with whole numbers, the order in which we multiply fractions does not affect the result (commutative property). 

(a/b) × (c/d) = (c/d) × (a/b).

 For example, 

Division of Fractions

When you’re dividing fractions, it might look tricky at first, but we can turn it into a multiplication problem, which is easier to handle. Here’s how you do it.

The Idea Behind Division:

First, think about regular division. For example:

12÷4=3

You can also think about it this way:

What number times 4 equals 12?

4×3=12

This is how we can think of division as multiplication: you find out what number you need to multiply by the divisor to get the dividend.

Now, Let’s Do the Same with Fractions:

When we divide fractions, we’ll use the same idea. Instead of dividing directly, we turn it into a multiplication problem by using the reciprocal of the second fraction.

Example 1:

Let’s try this: 

Let us rewrite this as a multiplication problem

What should be multiplied by 2/3 to get the product 1?

 If we somehow cancel out the 2 and the 3, we are left with 1.

So,

Let us try another problem:

This is the same as 

Can you find the answer?

 We know what to multiply 2/3 by to get 1. We just need to multiply that by 3 to get 3. So,

So,

What is  1/5 ÷ 1/2 ?

Rewriting it as a multiplication problem, we have

How do we solve this?

Rearranging this as multiplication, we have

How will we solve this?

Try yourself:

What is the area of a rectangle with length 1/2 unit and breadth 1/4 unit?

• A.1 square unit
• B.1/4 square units
• C.1/2 square units
• D.1/8 square units

View Solution

Discussion

The process of dividing by a fraction is equivalent to multiplying by its reciprocal. This rule applies whether you are dividing a whole number by a fraction, a fraction by a whole number, or a fraction by another fraction.

Rule: a ÷ (c/d) = a × (d/c) 

Rule: (a/b) ÷ c = (a/b) ÷ (c/1) = (a/b) × (1/c) = a / (b × c) 

Rule: (a/b) ÷ (c/d) = (a/b) × (d/c) = (a × d) / (b × c)

 Dividend, Divisor and the Quotient

When dividing positive numbers:

• If the divisor is a proper fraction (between 0 and 1), the quotient will be greater than the dividend. Example: 5 ÷ (1/2) = 5 × 2 = 10. Here, 10 > 5.
• If the divisor is an improper fraction (greater than 1), the quotient will be less than the dividend. Example: 5 ÷ (3/2) = 5 × (2/3) = 10/3 (approx 3.33). Here, 10/3 < 5.

This makes sense because dividing by a number less than 1 means finding out how many times that small piece fits into the dividend, resulting in a larger number. 

Dividing by a number greater than 1 means splitting the dividend into larger chunks, resulting in a smaller number of chunks.

Some Problems Involving Fractions

This section applies the concepts of fraction multiplication and division to solve practical word problems.

Example: Leena made 5 cups of tea and used a total of 1/4 litre of milk. How much milk is there in each cup of tea, assuming it was distributed equally?

Sol: To find the amount of milk per cup, we need to divide the total amount of milk (1/4 litre) by the number of cups (5). Calculation: (1/4) ÷ 5

Using the rule for division (multiply by the reciprocal): (1/4) ÷ 5 = (1/4) ÷ (5/1) = (1/4) × (1/5) Multiply numerators and denominators: (1 × 1) / (4 × 5) = 1/20.

So, each cup of tea has 1/20 litre of milk.

Example: Some of the oldest examples of working with non-unit fractions occur in humanity’s oldest geometry texts, the Śhulbasūtra. Here is an example from Baudhāyana’s Śhulbasūtra (c. 800 BCE). 

 Cover an area of 7 ½ square units with square bricks, each having sides of 1/5 units. How many such square bricks are needed?

Sol: First, find the area of one square brick: Area = side × side = (1/5) × (1/5) = 1/25 square units.

Next, convert the total area to an improper fraction: 7 ½ = (7 × 2 + 1) / 2 = 15/2 square units.

To find the number of bricks, divide the total area by the area of one brick: Number of bricks = (Total Area) ÷ (Area of one brick) Number of bricks = (15/2) ÷ (1/25)

Using the rule for division (multiply by the reciprocal): 

Number of bricks = (15/2) × (25/1)
 Multiply numerators and denominators: 
(15 × 25) / (2 × 1) = 375 / 2.

Example: This problem is from Chaturveda Prithudakasvami’s commentary (c. 860 CE) on Brahmagupta’s work. 

Four fountains fill a cistern. The first fills it in a day (rate = 1 cistern/day). The second fills it in half a day (rate = 1 ÷ (1/2) = 2 cisterns/day). The third fills it in a quarter of a day (rate = 1 ÷ (1/4) = 4 cisterns/day). The fourth fills it in one-fifth of a day (rate = 1 ÷ (1/5) = 5 cisterns/day). 

If they all flow together, how much time will they take to fill the cistern?

Sol: First, find the combined rate by adding the individual rates: Combined Rate = Rate1 + Rate2 + Rate3 + Rate4 Combined Rate = 1 + 2 + 4 + 5 = 12 cisterns/day.

This means together, the fountains fill 12 cisterns in one day. To find the time to fill one cistern, we take the reciprocal of the combined rate: Time = 1 / (Combined Rate) Time = 1 / 12 days.

So, together, the four fountains will fill the cistern in 1/12 of a day.

Fractional Relations

In the image, we have a square with lines drawn inside it, and part of the square is shaded. The problem asks, “What fraction of the whole square does the shaded region occupy?”

To solve this, the idea is to divide the whole square into smaller, manageable parts and use these smaller parts to calculate the fraction of the area taken by the shaded region.

Now, the top right square is colored red. It occupies 1/4 of the area of the whole square. This is because the entire square is divided into four equal smaller squares, and the red square is one of them. So, if the whole square is considered to have an area of 1 square unit, the area of the red square is 1/4 of that.

Now, A yellow triangle inside the red square. We are told that the yellow triangle occupies half of the red square’s area. So, the area of the yellow triangle is:

12×14=18 of the whole square.

Now, the yellow triangle is shaded. The shaded region inside the yellow triangle occupies 3/4 of the area of the triangle. So, the shaded area in the triangle is:

34×18=332 of the whole square.

A Dramma-tic Donation

A Dramma-tic Donation (from Līlāvatī): “O wise one! A miser gave to a beggar 1/5 of 1/16 of 1/4 of 1/2 of 2/3 of 3/4 of a dramma. If you know the mathematics of fractions well, tell me O child, how many cowrie shells were given by the miser to the beggar?”

A dramma was a silver coin. 1 dramma = 1280 cowrie shells (the lowest value coin). 

Fraction given =  part of a dramma.

Evaluating it gives 6 / 7680.

Upon simplifying to its lowest form, we get

So, one cowrie shell was given to the beggar.
You can see in the answer Bhāskarāchārya’s humour! The miser had given the beggar only one coin of the least value (cowrie).

Around the 12th century, several types of coins were in use in different kingdoms of the Indian subcontinent. Most commonly used were gold coins (called dinars/gadyanas and hunas), silver coins (called drammas/tankas), copper coins (called kasus/panas and mashakas), and cowrie shells. The exact conversion rates between these coins varied depending on the region, time period, economic conditions, weights of coins and their purity.

Gold coins had high value and were used in large transactions and to store wealth. Silver coins were more commonly used in everyday transactions. Copper coins had low value and were used in smaller transactions. Cowrie shells were the lowest denomination and were used in very small transactions and as change.

If we assume:
1 gold dinar = 12 silver drammas
1 silver dramma = 4 copper panas
1 copper pana = 30 cowrie shells

Then:
1 cowrie shell = 1/30 copper pana
1 copper pana = 1/(12 × 4) = 1/48 gold dinar
Therefore,
1 cowrie shell = 1/30 × 1/48 = 1/1440 gold dinar

A Pinch of History

Fractions have played a crucial role throughout history, particularly in problems involving sharing and dividing quantities. The general concept of non-unit fractions and their arithmetic operations (addition, subtraction, multiplication, division) developed significantly in India.

• Ancient Texts (Shulbasutra, c. 800 BCE): These texts, concerned with constructing fire altars for rituals, used general non-unit fractions extensively, including performing division as seen in Example 4.
• Early Commonplace Use (c. 150 BCE): Evidence suggests fractions became common in popular culture, indicated by references to reducing fractions to lowest terms (apavartana) in the philosophical work of the Jain scholar Umasvati.
• Codification by Brahmagupta (c. 628 CE): General rules for arithmetic operations on fractions, similar to modern methods, were first codified by Brahmagupta in his work Brahmasphutasiddhanta.
  • Multiplication: “Multiplication of two or more fractions is obtained by taking the product of the numerators divided by the product of the denominators.” (Verse 12.1.3).
• 
  • Division: “The division of fractions is performed by interchanging the numerator and denominator of the divisor; the numerator of the dividend is then multiplied by the (new) numerator, and the denominator by the (new) denominator.” 
• Clarification by Bhaskara II (Līlāvatī, c. 1150 CE): Bhaskara II further clarified Brahmagupta’s statement on division using the explicit notion of the reciprocal: “Division of one fraction by another is equivalent to multiplication of the first fraction by the reciprocal of the second.” (Verse 2.3.40).
• Geometric Interpretations (Bhaskara I, c. 629 CE): In his commentary on Aryabhata’s work, Bhaskara I described geometric interpretations of fraction multiplication and division using the division of a square into rectangles (similar to the area model discussed earlier).
• Other Indian Mathematicians: Many others like Shridharacharya (c. 750 CE), Mahaviracharya (c. 850 CE), Caturveda Prithudakasvami (c. 860 CE), and Bhaskara II further developed the use of arithmetic fractions.
• Transmission: The Indian theory of fractions and arithmetic operations was transmitted to and developed further by Arab and African mathematicians (like al-Hassar, c. 1192 CE, Morocco). It was then transmitted to Europe via the Arabs over subsequent centuries, coming into general use around the 17th century and becoming indispensable in modern mathematics.

Table of contents

Introduction

Have you ever looked closely at the world around you? From the slice of cheese you eat to the towering bridges you cross, shapes are everywhere! 

One of the most fundamental and fascinating shapes is the triangle. It might seem simple, just three lines meeting at three points, but this basic shape holds a surprising amount of depth and is the building block for many complex structures.

A triangle, in its essence, is the most basic closed shape you can make with straight lines. It’s defined by three key components:

• Vertices: These are the three corner points where the lines meet.
• Sides: These are the three straight line segments that connect pairs of vertices, forming the boundary of the triangle.
• Angles: Formed at each vertex where two sides meet, a triangle has three internal angles.

Let’s dive deeper into the world of triangles!

Understanding Triangles

As we discovered, a triangle is the most fundamental closed shape formed by three straight lines. Let’s break down its core components:

A triangle is defined by:

• Three Vertices: These are the corner points where the sides intersect. Think of them as the ‘tips’ of the triangle.
• Three Sides: These are the straight line segments connecting each pair of vertices. They form the boundary of the triangle.

Triangles can look very different! Some are tall and skinny, some are wide, and some have all sides the same length. Here are a few examples:

Naming Triangles

We usually name triangles using their vertices. For example, if a triangle has vertices labeled A, B, and C, we call it ∆ABC. The symbol ∆ is shorthand for “triangle”.

When naming a triangle, the order of the vertices doesn’t matter. So, ∆ABC is the same triangle as ∆BCA, ∆CAB, ∆CBA, ∆BAC, or ∆ACB. We just need to use the three vertex letters.

Angles of a Triangle

The three sides meeting at the vertices also create three angles inside the triangle. For ∆ABC, the angles are ∠CAB (the angle at vertex A), ∠ABC (the angle at vertex B), and ∠BCA (the angle at vertex C). Often, we simplify this and just call them ∠A, ∠B, and ∠C.

A triangle is a shape with three straight sides and three corners (vertices), where the sides connect to form a closed shape.

Now imagine this:

• You have three points (vertices).
• But all three points lie on a single straight line (this is called being collinear).

If you try to connect these points, you’ll just get a line segment — not a closed shape.

No, you cannot form a triangle.

Even if you connect all three points, the “shape” you make is just a straight line.

Try yourself:

What is a triangle defined by?

• A.Three vertices
• B.One vertex
• C.Four sides
• D.Two angles

View Solution

Equilateral Triangles

An equilateral triangle is a triangle in which all three sides are of equal length.

Among all triangles, equilateral triangles stand out for their perfect symmetry.

Because all sides are equal, all angles in an equilateral triangle are also equal (each measuring 60 degrees).

Constructing an Equilateral Triangle

Let’s try constructing one! How would you construct a triangle where all sides are exactly 4 cm long?

You could try using just a ruler, drawing a 4 cm base (say, AB) and then trying to position the third vertex (C) so that AC and BC are both 4 cm. However, this often takes a lot of trial and error!

A much more efficient and accurate method uses a compass, similar to constructions you might have done before.

Steps using Ruler and Compass:

• Draw the Base: Use your ruler to draw a line segment AB that is 4 cm long.
• Draw the First Arc: Place the compass point on vertex A. Adjust the compass width to 4 cm (the desired side length). Draw a long arc above the base line. The third vertex, C, must lie somewhere on this arc because every point on this arc is exactly 4 cm away from A
  
  
• Draw the Second Arc: Now, without changing the compass width (keep it at 4 cm), place the compass point on vertex B. Draw another arc that intersects the first arc. Let the point where the arcs intersect be C.
  
  
  
• Join the Vertices: Use your ruler to draw straight lines connecting A to C and B to C.
  
   You have constructed an equilateral triangle ∆ABC with all sides equal to 4 cm. The intersection point C is guaranteed to be 4 cm from A (because it’s on the first arc) and 4 cm from B (because it’s on the second arc).

Constructing a Triangle When its Sides are Given

Make a triangle with sides 4 cm, 5 cm, and 6 cm.

Steps to Construct the Triangle:

Step 1: Draw the base

• Pick one of the side lengths (say 4 cm) and draw a straight line.
• Label the ends as A and B.

Step 2: Use your compass

• From point A, draw an arc with radius 5 cm. This arc shows all the places where point C (the third corner of the triangle) could be, if it’s 5 cm from A.
• Step 3: Draw another arc
  • From point B, draw another arc with radius 6 cm. This arc shows where C could be if it’s 6 cm from B.
• Step 4: Find the point where the arcs meet
  • The place where the two arcs cross is the point C. This point is exactly 5 cm from A and 6 cm from B — just like we need.
• Step 5: Join the points
  • Now connect A to C and B to C to complete the triangle.

You’ve just drawn a triangle with the exact sides: 4 cm, 5 cm, and 6 cm!

If you try to make a triangle using three points (called vertices), but all of them lie on the same straight line, then you cannot form a triangle. Why?

Because a triangle is a closed shape with three sides. If the points are all on one straight line, they’re said to be collinear, and you can only draw a straight line through them — not a shape with corners and angles. That means you don’t get a “triangle,” just a line.

You need the three points to be in different directions — not just lying flat on the same path — so that the lines connecting them can turn and close up the shape.

Are Triangles Possible for Any Lengths?

This section explores when triangle construction is possible or impossible, introducing the Triangle Inequality Theorem.

Examples given:

• Try to make a triangle with side lengths:
  (1) 3 cm, 4 cm, and 8 cm
  (2) 2 cm, 3 cm, and 6 cm
  (3) 10 cm, 15 cm, and 30 cm

When you try these, you’ll notice that:

• In all these cases, the sum of the two shorter sides is equal to or less than the third side.
• For example: 3 + 4 = 7 < 8 → Not possible
  2 + 3 = 5 < 6 → Not possible
  10 + 15 = 25 < 30 → Not possible

Triangle Inequality 

This is a beautiful real-world illustration:

• Imagine walking from a tent to a tree directly (shortest path).
• Or walking from the tent to a pole, then from the pole to the tree (longer path).Triangle Inequality Example
• This demonstrates: The direct path is always shorter than any indirect one.
• So the sum of two sides of a triangle must always be greater than the third side, or the triangle collapses into a line.

Let’s have a look on the below given example –

A triangle is possible only if the sum of the lengths of any two sides is greater than the third side.

For example:

• Side lengths: 10 cm, 15 cm, 30 cm
  Check:
  • 10 + 15 = 25 < 30 → ❌ Not valid
    So, a triangle is not possible with these lengths.

Try yourself:

What is the measure of each angle in an equilateral triangle?

• A.90 degrees
• B.60 degrees
• C.30 degrees
• D.45 degrees

View Solution

Does everything look right with this triangle?
 If this triangle were possible, then the direct path between any two vertices should be shorter than the roundabout path via the third vertex. Is this true for our rough diagram? 

1. From B to C

• Direct path: BC = 10 cm
• Roundabout path via A: BA + AC = 15 cm + 30 cm = 45 cm
  ✔️ The direct path (10 cm) is shorter than the roundabout path (45 cm) → This is OK.

2. From A to B

• Direct path: AB = 15 cm
• Roundabout path via C: AC + CB = 30 cm + 10 cm = 40 cm
  ✔️ The direct path (15 cm) is shorter than the roundabout path (40 cm) → This is OK too.

3. From C to A

• Direct path: CA = 30 cm
• Roundabout path via B: CB + BA = 10 cm + 15 cm = 25 cm
  ⛔️ The direct path (30 cm) is longer than the roundabout path (25 cm) → This is not possible in a real triangle!

This situation is not possible in real life. You cannot make a triangle where one side is longer than the sum of the other two sides.

In our example:

30 cm > 10 cm + 15 cm
30 cm > 25 cm → ❌ This breaks the triangle rule!

Visualising the construction of circles

We want to construct a triangle with side lengths:

• AB = 8 cm (the longest side),
• AX = 4 cm,
• BX = 5 cm.

Step 1: Draw the base AB = 8 cm

• Use a ruler to draw a straight line 8 cm long.
• Label the ends as A and B.

Step 2: Use compass to draw circle with centre A and radius 4 cm

• Open your compass to 4 cm.
• Place the needle on A, and draw a full circle.
• Every point on this circle is 4 cm away from A. This circle shows all the places where point X could be, such that AX = 4 cm.

Step 3: Use compass to draw circle with centre B and radius 5 cm

• Open your compass to 5 cm.
• Place the needle on B, and draw another full circle.

Every point on this circle is 5 cm away from B. This circle shows all the possible positions of X such that BX = 5 cm.Circles intersecting each other at two points

What happens when you draw both circles?

The two circles intersect at two points. This means there are two possible positions for point C, which will satisfy both:

• AC = 4 cm (from A’s circle), and
• BC = 5 cm (from B’s circle).

So, we can now join A to C and B to C to form the triangle ABC.

Why does this work?

The point of intersection of the two circles is exactly the point that is 4 cm from A and 5 cm from B. So, the triangle ABC will have:

• AB = 8 cm (the base),
• AC = 4 cm (from the first circle),
• BC = 5 cm (from the second circle).

Q: Will triangles always exist when a set of lengths satisfies the triangle inequality? How can we be sure?

Ans: Case 1:  Imagine you have two points, A and B, on a piece of paper, and the distance between them is one of the sides of your triangle (let’s call it AB). This side AB is going to be the longest side of the triangle you’re trying to make.Case 1: Circles touch each other

Case 2: The other two sides of the triangle (let’s call them “radius 1” and “radius 2”) are represented by two circles:

• One circle is centered at point A, and its radius (the distance from A to the edge of the circle) is “radius 1.”
• The other circle is centered at point B, and its radius is “radius 2.”
• These two radii (radius 1 and radius 2) are the lengths of the other two sides of the triangle you’re trying to make.Case 2: Circles do not intersect

Can these two circles intersect (cross each other) at a third point, C?
 If they do, then point C can be the third corner of your triangle, and you can draw the triangle ABC. If they don’t intersect, you can’t make a triangle.

Case 3: Circles Intersect Internally (Cross Each Other at Two Points)

• This is the only case where a triangle can be formed!
• The two intersection points of the circles give possible positions for point C, creating the triangle ABC.

Let us study each of these cases by finding the relation between the radii (the smaller two lengths) and AB (longest length).

Case 1: Circles Touch Each Other at a Point

In this case, the two circles just barely touch each other at one spot. 

• When the circles touch like this, the distance between A and B (which is the length of side AB) is exactly equal to the sum of the two radii.
• In math terms: radius 1 + radius 2 = AB.
• If we think about the triangle inequality, this means the sum of the two smaller sides (radius 1 and radius 2) is equal to the longest side (AB).

But here’s the problem: when the sum of the two smaller sides is exactly equal to the longest side, the “triangle” you get isn’t really a triangle—it’s a straight line! Imagine points A, B, and C (where the circles touch) all lining up perfectly. That’s not a triangle because a triangle needs to have an inside area, not just be a flat line. So, in this case, a triangle cannot form.

Case 2: Circles do not Intersect internally

In this case, the two circles are too far apart to touch each other. 

• When the circles don’t intersect, the distance between A and B (the length of AB) is bigger than the sum of the two radii.
• In math terms: radius 1 + radius 2 < AB.
• If we think about the triangle inequality, this means the sum of the two smaller sides (radius 1 and radius 2) is less than the longest side (AB).

When this happens, the two smaller sides are too short to connect and form a triangle. Imagine trying to build a triangle with two short sticks that can’t even reach the ends of the longest stick—it just won’t work. So, in this case, a triangle cannot form either.

Case 3: Circles intersect each other

In this case, the two circles overlap and cross each other at two points. It’s like the balloons at A and B are big enough to overlap a bit.

• When the circles intersect, the distance between A and B (the length of AB) is less than the sum of the two radii.
• In math terms: radius 1 + radius 2 > AB.
• If we think about the triangle inequality, this means the sum of the two smaller sides (radius 1 and radius 2) is greater than the longest side (AB).

When this happens, the circles intersect at two points, and you can pick one of those points (let’s call it C) to be the third corner of your triangle. Now you can draw a triangle ABC, with sides AB, AC (which is radius 1), and BC (which is radius 2). Since the sum of the two smaller sides is bigger than the longest side, the triangle inequality is satisfied, and a triangle can form!

Conclusion 

The triangle inequality says that for any three lengths to form a triangle, the sum of any two sides must be greater than the third side.

From the cases we looked at:

• In Case 1, the sum of the two smaller lengths equals the longest length, so a triangle doesn’t form (it’s a straight line).
• In Case 2, the sum of the two smaller lengths is less than the longest length, so a triangle doesn’t form.
• In Case 3, the sum of the two smaller lengths is greater than the longest length, and a triangle does form.

So, triangles only exist when the sum of the two smaller lengths is greater than the longest length (like in Case 3). If the sum is equal to or less than the longest length (like in Cases 1 and 2), a triangle can’t form.

Construction of Triangles When Some Sides and Angles are Given

Constructing triangles — that means drawing a triangle correctly using some given information.

In this case, we are given:

• Two sides of the triangle (AB and AC),
• And the angle between them (∠A).

This is called “Two Sides and the Included Angle” because the angle is between the two given sides.

Steps to draw:

We are given:

• AB = 5 cm
• AC = 4 cm
• ∠A = 45°

We will draw triangle ABC using these steps:

1. Draw AB = 5 cm – This is one side of the triangle.
2. At point A, draw a 45° angle – Use a protractor to do this.
3. From point A, measure 4 cm along the arm of the angle – Mark this point as C. This is the second side AC = 4 cm.
4. Join point B to point C – Now you have triangle ABC.

Minie is confused

Not always! Here’s why:

• A triangle must close — the third side (BC) must be able to connect the two arms.
• If the angle is too small, and the sides are not long enough, the two lines might not meet.
• If the angle is too wide, and one side is short, again the lines may not reach each other.

So, even if you have two sides and an angle, sometimes it’s not possible to make a triangle.

Two Angles and the Included Side

What does this mean?

You are given:

• Two angles (like ∠A and ∠B),
• And the side between them (AB).

This side is called the included side.Triangles with one side and two angles

How to draw the triangle?

Suppose you’re given:

• AB = 5 cm
• ∠A = 45°
• ∠B = 80°

Follow these simple steps:

1. Draw AB = 5 cm (the given side).
2. At point A, use a protractor to draw ∠A = 45°.
3. At point B, draw ∠B = 80°.
4. The two lines you draw will meet — mark the meeting point as C.
5. Now, join AC and BC — you have your triangle ABC!

Do triangles always exist with any two angles and a side?

No! Not always. Here’s why:

• If the two given angles are too big, they might never meet to form a triangle.
• For example, if both angles are 90° or more, then the two sides just go straight and never intersect — so, no triangle is formed!

Example:

If one angle is 40°, the second angle (at the other end) must be less than 140°, or else the two lines won’t meet.

Why 140°?

Because in a triangle, the sum of all three angles must be 180°.

So if angle A is 40°, and you try to make angle B = 140°, the third angle becomes 0°, which is impossible!

And if angle B is more than 140°, the lines go too wide and never meet.

Conclusion:

A triangle does not exist if the sum of the two given angles is greater than or equal to 180°.

Even if you have a side between the angles, it doesn’t help if the lines never meet.

Using Parallel Lines

In the diagram given below, a parallel line (XY) is drawn through vertex A to help show that the angles formed with the parallel lines are related to the original angles. This helps in understanding the alternate angles that are equal.

When you draw a line parallel to one side of a triangle, it creates certain relationships between the angles. 
For example, alternate angles (the angles on opposite sides of a transversal) are equal.

Step-by-step Example:
In the triangle △ABC, we are given two angles:

• ∠ABC=50°
• ∠ACB=70°
• We want to find ∠BAC.

How to Find ∠BAC:

First, add the known angles: 
50°+70°=120°.

Since the sum of the angles in any triangle is 180°, subtract 120° from 180° to get the third angle:
180°−120°=60°.

Therefore, ∠BAC=60°.

Angle Sum Property

The sum of the angles of any triangle is always 180°. 

This means that if you know two of the angles in a triangle, you can find the third angle by subtracting the sum of the known angles from 180°.

We need to find ∠A + ∠B + ∠C. 

We know that ∠B = ∠XAB, ∠C = ∠YAC. 

So, ∠A + ∠B + ∠C = ∠A + ∠XAB + ∠YAC 

= 180° as together they form a straight angle. 

Thus we have proved that the sum of the three angles in any triangle is 180°! This rather surprising result is called the angle sum property of triangles.

• Parallel Lines to Find Angles: One method to show that the sum of the angles of a triangle is 180° is by using parallel lines. When you draw a line parallel to one side of the triangle, it helps form relationships between the angles. 
  In the example, the line XY is parallel to the base of the triangle BC, and by using alternate angles, we can prove that the sum of the angles in a triangle is always 180°.

This rule is known as the Angle Sum Property of Triangles. It is a very important property in geometry that helps in solving for unknown angles in a triangle.

Here is an image demonstrates a creative way to verify the Angle Sum Property of a triangle using a paper cut-out. Here’s how it works:

• Cut out a Triangle: Start by cutting out a triangle from a piece of paper. This triangle represents any triangle, and its angles are the focus.
• Folding the Angles: Next, you fold the triangle along its edges. The key idea is to fold each corner (or vertex) of the triangle so that each angle aligns with one another. When you do this, all three angles of the triangle will meet at a single point.
• Forming a Straight Line: Once the folds are made, you’ll see that the three angles of the triangle come together to form a straight line. This is the key observation: the total amount of space along the straight line is 180°.

 This simple folding method shows that the sum of the three angles in a triangle is always 180°. No matter what triangle you use, when you fold the corners together, the angles will always add up to a straight line, which equals 180°.

Exterior Angles

This is the angle formed when you extend one side of a triangle. The exterior angle is always connected to the two interior angles of the triangle.

In the example, ∠ACD is the exterior angle. The key idea here is that the exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles. So, if you know the two interior angles that are not next to the exterior angle, you can find the exterior angle by adding them together.

Example:

If ∠A = 50° and ∠B = 60°, you can find the exterior angle ∠ACD by adding these two angles:

∠ACD = 50° + 60° = 110°

This shows that the exterior angle is equal to the sum of the two opposite interior angles.

Example: Can we make a triangle with sides of length 3 cm, 4 cm, and 8 cm?

Sol: We will use the Triangle Inequality Theorem, which states that the sum of the lengths of any two sides of a triangle must always be greater than the length of the third side.

Let’s check the triangle inequality for the given side lengths:

• 3 cm + 4 cm = 7 cm (which is less than 8 cm)

Since 7 cm is less than 8 cm, it’s impossible to form a triangle with sides of 3 cm, 4 cm, and 8 cm.

Constructions Related to Altitudes of Triangles

An altitude is a line drawn from any vertex of a triangle that is perpendicular (at a right angle) to the opposite side (or the line that extends the opposite side).

Height of a Triangle: In everyday life, we talk about the height of things, like the height of a building or a tree. In geometry, the term “height” refers to the perpendicular distance from a vertex of a triangle to the opposite side. This perpendicular line is called the altitude.

How to Draw an Altitude in a Triangle?

Altitude from Vertex A: In the first image, we have triangle ABC, and we are interested in the altitude from vertex A to side BC.

• The line AD is drawn from vertex A perpendicular to BC.
• The length of this line, AD, is the altitude of vertex A from the side BC.

Altitude from Other Vertices: Similarly, we can draw altitudes from the other vertices of the triangle. In the image given below, we can see how altitudes are also drawn from vertices B and C to their respective opposite sides:

BE is the altitude from vertex B to side AC. CF is the altitude from vertex C to side AB.

How Altitudes Change with Different Triangles

• In right-angled triangles, the altitude from the right angle vertex will be the line along the height of the triangle.
• In acute triangles (all angles less than 90°), altitudes are drawn inside the triangle.
• In obtuse triangles (one angle greater than 90°), altitudes might need to be extended outside the triangle, as seen in the figure given below. This is shown by extending the side BC and drawing a perpendicular from A to this extended line.

Construction of the Altitudes of a Triangle

Let’s imagine we have a triangle, and we want to find the altitude from vertex A to the base BC. Here’s how we can do it:

Start with the triangle: Label the vertices as A, B, and C, with BC being the base.

Using a ruler and set square: A set square is a tool that helps you make sure you have a perfect right angle (90°). If you just use a ruler, the line may not be exactly 90°, and that’s not good for our altitude. The set square helps us be accurate!

Aligning the ruler and set square:

• Place the ruler along the base BC.
• Put the set square on the ruler so that one of its right-angle edges is touching the ruler. This way, we’re guaranteed that one side of the set square forms a perfect 90° angle.

Sliding the set square: Now, move the set square along the ruler until the vertical edge of the set square touches vertex A. This will be the point from which the altitude is drawn.

Draw the altitude: Using the vertical edge of the set square, draw a straight line from A straight down to the base BC. This line is your altitude!

What Happens if One Side is the Altitude?

Now, you might wonder, “Can a side of a triangle also be an altitude?” The answer is yes, but this happens only in a right-angled triangle. In a right triangle, one of the sides already makes a 90° angle, so that side can also be the altitude. This is special because in other triangles, the altitude is always a separate line.

Types of Triangles

Types of Triangles Based on Sides

1. Equilateral Triangle:
   • All three sides are equal in length.
   • All three angles are also equal, and each angle measures 60°.
2. Isosceles Triangle:
   • Two sides are equal in length.
   • The two angles opposite these equal sides are also equal.
3. Scalene Triangle:
   • All three sides have different lengths.
   • All three angles are different from each other.
4. Right-Angled Triangle:
   • One of the angles is exactly 90° (a right angle).
   • The other two angles are acute (less than 90°).Right-angled Triangle

Try yourself:

What is the Angle Sum Property of a triangle?

• A.The sum of the angles is always 360°.
• B.The sum of the angles is always 270°.
• C.The sum of the angles is always 180°.
• D.The sum of the angles is always 90°.

View Solution

Types of Triangles Based on Angles

Triangles can also be classified based on their angle measures. These are:

1. Acute-Angled Triangle:
   • All three angles are acute (less than 90°).
   • In other words, every angle in this triangle is greater than 0° but less than 90°.
2. Right-Angled Triangle:
   • One of the angles is exactly 90° (a right angle).
3. Obtuse-Angled Triangle:
   • One of the angles is obtuse (greater than 90° but less than 180°).
   • The other two angles will be acute because the total sum of angles in any triangle must always be 180°.

Is There a Connection Between the Two Classifications?

Yes, there is a connection between the two classifications!

Equilateral triangles are always acute-angled triangles because all their angles are 60°, which is an acute angle.

Isosceles triangles can be of any type based on angles:

• If the two equal angles are both less than 90°, it’s an acute-angled isosceles triangle.
• If one angle is 90°, it’s a right-angled isosceles triangle.
• If one angle is greater than 90°, it’s an obtuse-angled isosceles triangle.

Scalene triangles can also be classified by angles:

• If all three angles are less than 90°, it’s an acute-angled scalene triangle.
• If one angle is 90°, it’s a right-angled scalene triangle.
• If one angle is greater than 90°, it’s an obtuse-angled scalene triangle.

What about Acute-Angled Triangles?

In an acute-angled triangle, all three angles are acute (less than 90°). So, it’s not just a triangle with one acute angle – it has all three angles being acute. This is important because it means every angle is small, but the sum of them still equals 180°.

Try yourself:

What type of triangle has one angle that is exactly 90°?

• A.Equilateral Triangle
• B.Obtuse-Angled Triangle
• C.Right-Angled Triangle
• D.Acute-Angled Triangle

View Solution

Numbers Tell us Things

Numbers aren’t just for counting; they can represent relationships and follow specific rules. 

Imagine a group of children standing in a line. Each child is given a number, but the number they say does not match the order in which they are standing.

Let’s understand this using both pictures shown.Children are standing in a straight line and saying numbers: (0,1,0,2,2,2,1)

So the pattern seems unclear for now. But now, let’s look at what happens in the second image.

Now the same children are in a new order, and they say:  (0,0, 1, 2, 2, 5, 6)

If you observe carefully, this isn’t just random! Something is happening based on their positions.

What Do These Numbers Actually Mean?

Guess 1: The numbers show how many people are in front who are taller.

Imagine the height of each child is fixed.

Let’s say the red child is tallest. So in the first arrangement, the children call out the number of children taller than them in front of them.

Guess 2: The numbers are showing how many children are taller or shorter around them or how many children came before them in line.

So the number each child says seems to reflect how many children are taller and ahead of them in line.

It’s like each child is counting how many kids ahead of them are taller.

Picking Parity

Parity refers to the property of a number being either even or odd. 

This section explores the rules and patterns related to parity.

Kishor’s Puzzle:

Kishor has number cards (1, 3, 5, 7, 9, 11, 13) and 5 empty boxes. He needs to place exactly one card in each box such that the numbers sum to 30. Can he do it?

Let’s investigate the properties of even and odd numbers to solve this.

Properties of Even and Odd Numbers:

• Even numbers: Can be grouped in pairs with no leftovers. Example: 2, 4, 6, 8, 10…No pair is left
• Odd numbers: Always leave one unpaired object. Example: 1, 3, 5, 7, 9…One unpaired leftWhat happens when you add two odd numbers?
•      The sum of two odd numbers can always be arranged in pairs, meaning the sum is even.
• What about adding three odd numbers? Can the sum be arranged in pairs?

 No. The sum of three odd numbers is odd.

Try yourself:

What does the term ‘parity’ refer to?

• A.The sum of two numbers
• B.The height comparison of children
• C.The property of a number being even or odd
• D.The arrangement of numbers in a line

View Solution

The Puzzle:

Kishor has some number cards. He needs to place 5 cards in boxes so that their total equals 30.

The number cards are:

13, 9, 7, 11, 5, 3

 All are odd numbers.

But here’s the question: Can the sum of 5 odd numbers be 30 (an even number)?

Let’s try:

• Add 3 + 5 + 7 + 9 + 11 = 35 ❌
• Try different combinations – Still doesn’t give you 30!

Let’s use parity logic.

• When you add even numbers, the result is always even.
• When you add two odd numbers, the result is even.
• BUT, when you add an odd number of odd numbers (like 3, 5, or 7 odd numbers), the result is always odd.

So: ✅ 2 odd numbers → even
✅ 4 odd numbers → even
❌ 5 odd numbers → odd

So it’s impossible to make 30 (which is even) by adding 5 odd numbers. That’s the trick of parity!

So the answer of Kishore’s Puzzle is no.

Real-Life Example:

Martin and Maria are siblings born one year apart. If their total age is 112, is that possible?

Try:

• Martin = 55, Maria = 56 → 55 + 56 = 111 (odd)
• Martin = 56, Maria = 57 → 113 (odd)

You can try any two consecutive numbers, and they always add up to an odd number. So, 112 is not possible.

Why?

Because one of them will be odd and the other even → odd + even = odd.

Small Squares in Grids:

• A 3 × 3 grid has 9 squares (odd).
• A 3 × 4 grid has 12 squares (even).

Can you tell the parity of the total squares just from the dimensions?

Rule: The product (total squares) is odd only if both dimensions are odd.

• Odd × Odd = Odd
• Odd × Even = Even
• Even × Odd = Even
• Even × Even = Even

Parity of Expressions:

Consider 3n + 4. Its parity depends on n:

• If n is odd, 3n is odd, 3n + 4 (odd + even) is odd.
• If n is even, 3n is even, 3n + 4 (even + even) is even.
• Expressions that are always even: Any expression where the result must have a factor of 2. Examples: 2n, 100p, 48w – 2, 6k + 2, n + n.
• Expressions that are always odd: An even expression plus or minus 1. Examples: 2n + 1, 2n – 1.

Finding the nth Even and Odd Number:

• The nth even number is given by the formula 2n.
  • Example: The 100th even number is 2 × 100 = 200.
• The nth odd number is one less than the nth even number. The formula is 2n – 1.
  • Example: The 100th odd number is (2 × 100) – 1 = 200 – 1 = 199.

Some Explorations in Grids

Let’s explore number patterns within grids.

The 3×3 Sum Grid Puzzle:

Observe the 3×3 grid below.It’s filled with numbers 1 to 9, with no repeats. The circled numbers outside represent the sum of the numbers in the corresponding row or column.

Can you verify the row and column sums?

Challenge Grids:

Fill the grids below using numbers 1-9 without repetition, ensuring the row and column sums match the circled numbers.

Grid 1:Ans:

Grid 2:

Ans: 

Impossible Grid?

Consider this grid setup:

Is it possible to find a solution? Why or why not?

• Reasoning: The smallest possible sum for a row/column using three distinct numbers from 1-9 is 1 + 2 + 3 = 6. The largest possible sum is 7 + 8 + 9 = 24. Any circled sum must be between 6 and 24, inclusive. In the grid above, the first column requires a sum of 5, which is impossible. Therefore, this grid has no solution.

Try yourself:

Can the sum of 5 odd numbers equal an even number?

• A.Only with specific numbers
• B.Only sometimes
• C.No, it cannot
• D.Yes, it can

View Solution

Total Sum Check:

What is the sum of all numbers from 1 to 9? (1 + 2 + … + 9 = 45).

What is the sum of all the row sums (or column sums) in a completed 3×3 grid puzzle? 

It must also be 45, as each number in the grid is counted exactly once in the row sums and exactly once in the column sums.

What is a Magic Square?

A magic square is a special grid of numbers where:

• The sum of numbers in every row, every column, and both diagonals is the same.
• This common total is called the magic sum.
• In a 3 × 3 magic square, we use numbers 1 to 9, without repetition.

Look at this 3 × 3 grid from the image:

But here’s the catch—this grid is not a magic square, because:

• Row sums: 16, 9, and 20 (not equal)
• Column sums: 4+6+3=13, 7+1+9=17, 5+2+8=15 (also not equal)
• Diagonals: Not the same sum either

So, this is just a filled 3×3 grid, not a magic square.

So, what makes it a real magic square?

Let’s take the correct magic square example:

Now, check the magic:

Each row adds to 15

• 2 + 7 + 6 = 15
• 9 + 5 + 1 = 15
• 4 + 3 + 8 = 15

Each column adds to 15

• 2 + 9 + 4 = 15
• 7 + 5 + 3 = 15
• 6 + 1 + 8 = 15

Both diagonals add to 15

• 2 + 5 + 8 = 15
• 6 + 5 + 4 = 15

This is a true magic square with magic sum = 15.

Why is the magic sum always 15 in a 3×3 magic square?

The total sum of all numbers from 1 to 9 is:

1 + 2 + 3 + … + 9 = 45

Now, this total of 45 is split evenly across 3 rows (or 3 columns), so:

45 ÷ 3 = 15

That’s why the magic sum for every row/column/diagonal must be 15.

Can the magic sum be any number?

No, not when using 1 to 9 without repetition. The magic sum is fixed because:

• You’re using the same numbers (1 to 9).
• The total is always 45.
• So, magic sum = total ÷ number of rows = 45 ÷ 3 = 15

If you use a different set of numbers (like 2 to 10), then the total and magic sum will change.

Let us focus, for the moment, only on the row sums. We have seen that in a 3 × 3 grid with numbers 1 – 9, the total of row sums will always be 45. Since in a magic square the row sums are all equal, and they add up to 45, they have to be 15 each. Thus, we have the following observation.

Observation 1:

• All rows, columns, and diagonals must add up to 15.

Can 9 be at the centre?

Let’s try putting 9 in the centre:

Now imagine one of the rows has 8 and 9.
So:
8 + 9 + ? = 15
17 + ? = 15

That’s impossible! 17 is already more than 15. So no matter what the third number is, the total will not be 15.

So, 9 cannot be at the centre.

Can 1 be at the centre?

Now try 1 in the centre. Imagine another number in the same row: 2.

1 + 2 + ? = 15
3 + ? = 15
⇒ ? = 12

But we are only allowed to use numbers from 1 to 9, and 12 is not allowed.

So, 1 cannot be at the centre either.

Try All Numbers from 1 to 9

If you keep trying this way for every number from 1 to 9, you’ll find:

✅ Only 5 can be at the centre, because only with 5 in the centre, we can find four different combinations of two numbers that add with 5 to make 15:

• 5 + 1 + 9 = 15
• 5 + 2 + 8 = 15
• 5 + 3 + 7 = 15
• 5 + 4 + 6 = 15

Observation 2:

• The number 5 must be at the center of the magic square.
• Why? Because it’s the middle value of numbers 1–9 and balances the other values.

What About Numbers 1 and 9?

Now let’s think: where should the smallest number (1) and the largest number (9) go?

They cannot go in the centre (we just saw why).
So, they must go in one of the outer positions — either in a corner or on the sides.

Let’s check if 1 can be placed in a corner.

We need to find combinations like:

1+?+?=15

Some possible ones are:

• 1 + 5 + 9 = 15
• 1 + 6 + 8 = 15

That’s two valid combinations.

Now try 9 in a corner.

Possible combinations:

• 9 + 1 + 5 = 15
• 9 + 2 + 4 = 15

Again, two valid combinations.

Observation 3:

• Numbers 1 and 9 cannot be placed in the corners. If you place them there, you cannot find enough combinations to make 15.
• So, they must go in the middle positions on the boundary (not in the corners or center).

How to Start Filling the Magic Square

Let’s say we place 1, 5, and 9 in one row or column like this:

That already adds up to 15. Now we can fill the other positions in a way that each row, column, and diagonal still adds up to 15.

Try completing the square from here by using:

• Remaining numbers: 2, 3, 4, 6, 7, 8
• Make sure each number is used only once
• Check each row, column, and diagonal

A Classic 3 × 3 Magic Square:

• Rows: 2+7+6 = 15, 9+5+1 = 15, 4+3+8 = 15
• Columns: 2+9+4 = 15, 7+5+3 = 15, 6+1+8 = 15
• Diagonals: 2+5+8 = 15, 6+5+4 = 15

✅ Every direction adds up to the magic sum of 15!

Generalising a 3 × 3 Magic Square

When we generalise something in Maths, we try to find a pattern or rule that works for all similar cases, not just one example.

In this case, we want to look at how the numbers in a 3 × 3 magic square are related to the centre number, and then write that using algebra.

Let’s say we are working with a 3 × 3 magic square made using consecutive numbers, like 1 to 9. Here’s one example:

The middle number is 5, and the magic sum is 15.

Now let’s call the middle number m. So in this example, m = 5.

We now want to describe every other number in terms of how much more or less it is than m.

Let’s place m in the centre of the grid:

Now let’s figure out how the other numbers relate to m. We’ll use + or − to show how far they are from m.

In a typical 3 × 3 magic square, if m is the centre, then the numbers around it are arranged as follows:

But we need to use a pattern that always gives us the magic square.

Here is the correct general form of a 3 × 3 magic square using m as the centre:

When we generalise the magic square, we find a formula or pattern for where each number goes in relation to the centre number m.

The First-ever 4 × 4 Magic Square

This is a wonderful example of a 4 × 4 magic square from ancient India, called the Chautīśā Yantra, found in the Parshvanath Jain temple in Khajuraho. Let’s explore and understand this amazing pattern!The first ever recorded 4 × 4 magic square, the Chautīsā Yantra, at Khajuraho, India

A magic square is a grid of numbers where:

• Each row, column, and both diagonals add up to the same total.
• In this case, it’s a 4 × 4 grid, meaning it has 4 rows and 4 columns.
• The sum of each row/column/diagonal is 34.

That’s why it’s called the Chautīśā Yantra — “Chautīśā” means 34 in Hindi.

Let’s check some sums of the given grid:

• Row 1: 7 + 12 + 1 + 14 = 34
• Column 1: 7 + 2 + 16 + 9 = 34
• Diagonal (top-left to bottom-right): 7 + 13 + 10 + 4 = 34
• Diagonal (top-right to bottom-left): 14 + 8 + 3 + 9 = 34

✅ Every row, column, and diagonal gives the same magic sum = 34.

Can We Find Other Patterns That Also Add Up to 34?

Yes! The Chautīśā Yantra is special because even more combinations of 4 numbers add up to 34. Try looking at:

Corners:

• 7 (top-left) + 14 (top-right) + 4 (bottom-right) + 9 (bottom-left)
• 7 + 14 + 4 + 9 = 34

2×2 squares:

• 13 + 8 + 3 + 10 (center 2×2) = 34

Same positions in different rows:

• 12 (1st row), 13 (2nd), 3 (3rd), 6 (4th)
• 12 + 13 + 3 + 6 = 34

This magic square has a lot of hidden combinations!

Try yourself:

What is a magic square?

• A.A grid where all rows, columns, and diagonals add to the same total.
• B.A grid filled with random numbers.
• C.A grid that can have any sum.
• D.A grid that uses only negative numbers.

View Solution

Magic Squares in History and Culture

Lo Shu Square – The Oldest Known Magic Square

The first recorded magic square is called the Lo Shu Square. It came from ancient China, more than 2000 years ago. Here’s the story:

The Lo River flooded and the gods sent a turtle to help. The turtle had a special 3 × 3 grid on its back, made up of numbers 1 to 9. The pattern looked like this:

If you add:

• Any row: 2 + 7 + 6 = 15
• Any column: 2 + 9 + 4 = 15
• Any diagonal: 2 + 5 + 8 = 15

So the magic sum = 15 in this case.

Where Else Are Magic Squares Found?

• Magic squares have been studied in many countries like India, Japan, China, Central Asia, and Europe.
• In India, ancient mathematicians created not only 3 × 3 but also 4 × 4, 5 × 5, and even bigger magic squares.
• They found smart methods to build these squares without just guessing.

Ancient Magic Squares in Indian Temples

• One such 3 × 3 magic square is found on a pillar in a temple in Palani, Tamil Nadu, dating back to the 8th century CE.
• Magic squares are also found in homes and shops, especially in the Navagraha Yantra, which connects magic squares with planets like:

Each of these has a magic square inside it.

What is a Kubera Yantra?

• The Kubera Yantra is a sacred diagram (or symbol) used in Indian tradition.
• It is associated with Lord Kubera, the deity of wealth.
• This Yantra often contains a magic square inside it.

The Magic Square in Kubera Yantra

Here’s the 3 × 3 grid from the image:

Let’s check if this is a magic square by calculating the sums of each row, column, and diagonal:

Rows:

• Row 1: 27 + 20 + 25 = 72
• Row 2: 22 + 24 + 26 = 72
• Row 3: 23 + 28 + 21 = 72

Columns:

• Column 1: 27 + 22 + 23 = 72
• Column 2: 20 + 24 + 28 = 72
• Column 3: 25 + 26 + 21 = 72

Diagonals:

• 27 + 24 + 21 = 72
• 25 + 24 + 23 = 72

✅ Yes! Every row, column, and diagonal adds up to 72. So, 72 is the magic sum here.

Why Is This Special?

Just like the Lo Shu Square with magic sum 15, the Kubera Yantra is a magic square—but with different numbers and a different magic sum.

• It still follows the magic square rules: all rows, columns, and diagonals total the same.
• But the numbers used are bigger: from 20 to 28.
• This shows that magic squares can be built using different sets of numbers, not just 1 to 9.

Nature’s Favourite Sequence: The Virahāṅka- Fibonacci Numbers!

It’s a special number pattern that looks like this:

1, 2, 3, 5, 8, 13, 21, 34, …

Each number (after the first two) is the sum of the previous two numbers:

• 1 + 2 = 3
• 2 + 3 = 5
• 3 + 5 = 8
• 5 + 8 = 13
• and so on.

This sequence is also called the Fibonacci sequence, and it appears all around us—in plants, music, art, and nature!

Discovery of the Virahāṅka Numbers

Surprisingly, it first appeared in poetry, especially in Sanskrit, Tamil, Telugu, and other Indian languages.

In poetry:

• A short syllable = 1 beat
• A long syllable = 2 beats

Ancient poets used different combinations of these syllables to make up rhythms. So, they started asking:

“In how many ways can we fill a line of poetry with 8 beats using short (1) and long (2) syllables?”

Let’s Take an Example (n = 4 beats):

You want to write 4 beats using only 1s and 2s. Let’s list the combinations:

There are 5 ways to write 4 beats using 1s and 2s. This is the Fibonacci number at position 4 (starting from n=1).

How to Get the Next Fibonacci Number?

You use a systematic method:

• Start with all the combinations for the previous number.
• Add a 1 in front of all of them (for one short syllable)
• Add a 2 in front of the combinations two steps before (for one long syllable)

This builds up the next number in the sequence!

Virahāṅka–Fibonacci Numbers 

What is this sequence?

It’s a famous list of numbers:1, 2, 3, 5, 8, 13, 21, 34, 55…

Each number is found by adding the two numbers before it.

• 1 + 1 = 2
• 1 + 2 = 3
• 2 + 3 = 5
• 3 + 5 = 8
• 5 + 8 = 13
• 8 + 13 = 21
• and so on…

This sequence is called the Fibonacci Sequence in the West, but Virahāṅka Numbers in India, after the Indian scholar Virahāṅka who discovered them around 700 CE, much before Fibonacci!

How did these numbers come up?

They came from poetry, not math!

In ancient Indian poetry:

• A short syllable = 1 beat
• A long syllable = 2 beats

Poets wanted to know:
How many ways can we make a poem line with 8 beats?
Each line can be filled with a mix of short (1 beat) and long (2 beat) syllables.

 Let’s take an example:

How many rhythms are possible for 5 beats?

Use only 1’s and 2’s. Let’s find all the ways to add up to 5:

• 1 + 1 + 1 + 1 + 1
• 1 + 1 + 1 + 2
• 1 + 1 + 2 + 1
• 1 + 2 + 1 + 1
• 2 + 1 + 1 + 1
• 1 + 2 + 2
• 2 + 1 + 2
• 2 + 2 + 1

That’s 8 rhythms.

So, the number of ways to form rhythms for 5 beats is the 5th number in the Virahāṅka–Fibonacci sequence, which is 8!

Why does this work?

Because each rhythm either:

• Starts with 1, then the rest must make (n – 1) beats
• Starts with 2, then the rest must make (n – 2) beats

So: Ways to write n = Ways to write (n – 1) + Ways to write (n – 2)

That’s exactly how Fibonacci numbers grow!

History Fun Fact:

• Virahāṅka introduced the numbers in India around 700 CE
• Fibonacci used them in Europe only in 1202 CE
• These numbers are in nature (pinecones, flowers), art, music, and poetry!

What’s the Rule for the Fibonacci Sequence?

To get the next number in the sequence, add the two numbers before it:

1, 2, 3, 5, 8, 13, 21, 34, 55

Next is: 34 + 55 = 89

Then:   55 + 89 = 144

Then:   89 + 144 = 233

Look at flowers like daisies.Many have 13, 21, or 34 petals.
These are Fibonacci numbers!

Nature uses this pattern in:

• Sunflower spirals
• Pinecones
• Pineapples
• Flowers

This shows how deeply math and nature are connected.

Digits in Disguise

There is a math puzzle where numbers are replaced by letters. It’s called a cryptarithm or alphametic puzzle. The idea is to figure out which digit (from 0 to 9) each letter stands for so that the math equation makes sense.

That means:

• We are adding T + T + T, which is the same as 3 × T.
• The answer is a 2-digit number: UT (U is the tens digit, T is the ones digit again!).

So we look for a number T such that 3 × T = UT, and the last digit of the answer is the same as T.

Try T = 5:

• 3 × 5 = 15 → Yes! The answer is 15 (U = 1, T = 5)

Next, we have:

• “K2” means a 2-digit number where the first digit is K and the second is 2.
• Add it to itself: K2 + K2 = HMM
• The answer is a 3-digit number HMM, where both M digits are the same.

Let’s try K = 6:

• K2 = 62 → 62 + 62 = 124 → Not the right format.
  Try K = 7:
• 72 + 72 = 144 → H = 1, M = 4

⇒ HMM = 144

Try yourself:

What is the magic sum of the Lo Shu Square?

• A.72
• B.30
• C.15
• D.9

View Solution

In this chapter, we will explore the fascinating relationship between lines when they are drawn on a flat surface, like a piece of paper, a table top, or a blackboard. Think about the lines you see around you. Some lines cross each other, while others seem to run alongside each other forever without touching. 

We will learn the special names for these lines and discover some interesting properties about the angles they form.

 Across the Line

• Imagine taking a square piece of paper and folding it in different ways. 
• If you draw lines along the creases, you’ll notice various lines on the paper. What happens if you look at any two of these lines? Do they cross each other? If they don’t cross on the paper, would they meet if you could extend them infinitely?

When two lines meet or cross each other at a single point on a flat surface, we say they intersect. The point where they cross is called the point of intersection.

Let’s look closer at what happens when two lines intersect. Consider line l intersecting line m as shown below:

How many angles are formed at the point of intersection? 

We can see four angles, labeled a, b, c, and d.

Thinking about the Angles:

Look back at Figure above . If we know the measure of one angle, say ∠a = 120°, can we figure out the measures of the other angles without measuring?

• Angles ∠a and ∠b form a straight line together. A straight angle measures 180°. So, ∠a + ∠b = 180°. If ∠a = 120°, then ∠b must be 180° – 120° = 60°.
• Similarly, ∠b and ∠c also form a straight line. So, ∠b + ∠c = 180°. If ∠b = 60°, then ∠c must be 180° – 60° = 120°.
• And, ∠c and ∠d form a straight line. So, ∠c + ∠d = 180°. If ∠c = 120°, then ∠d must be 180° – 120° = 60°.

Notice something interesting? ∠a = ∠c = 120° and ∠b = ∠d = 60°. 

The angles opposite each other at the intersection point are equal!

Key Concepts:

Linear Pair: Adjacent angles formed by two intersecting lines that add up to 180° (like ∠a and ∠b, or ∠b and ∠c). They form a straight line together.

Vertically Opposite Angles: Angles that are opposite each other when two lines intersect (like ∠a and ∠c, or ∠b and ∠d). Vertically opposite angles are always equal to each other. This is a fundamental concept, a proof in mathematics!

 Perpendicular Lines

 If all four angles are equal and they form a complete circle (360°), what would each angle measure? 

That’s right, each angle would be 360° / 4 = 90°.

Perpendicular Lines: A pair of lines that intersect each other at a perfect right angle (90°). We use a small square symbol at the intersection to show that lines are perpendicular.

In the figure above, lines l and m are perpendicular to each other.

Try yourself:

What do we call the angles that are opposite each other when two lines intersect?

• A.Adjacent Angles
• B.Vertically Opposite Angles
• C.Complementary Angles
• D.Supplementary Angles

View Solution

Between Lines

Let’s look at different ways lines can relate to each other. Observe the line segments in the figure below.

Some lines meet or intersect (like FG and FH meeting at F). But what about lines ST and UV? Or OP and QR? If you imagine extending them forever, would they ever meet?

​Here are some examples of lines we notice around us.

These lines seem to run alongside each other, always keeping the same distance apart, and never intersecting. These are called parallel lines.

Parallel Lines: A pair of lines on the same flat surface (plane) that never meet or intersect, no matter how far they are extended in either direction.

Can you spot some parallel lines in your classroom or home?

Parallel lines are often used in art and design, like for shading.

 Parallel and Perpendicular Lines in Paper Folding

Let’s explore parallel and perpendicular lines using a fun paper folding activity!

Activity: Folding Fun

Grab a plain square sheet of paper (newspaper works well!).

• Edges: Look at the opposite edges of the square. 
  How would you describe them? 
  They are parallel to each other. Now look at the adjacent edges (edges next to each other). 
  How do they relate? 
  They are perpendicular to each other, meeting at right angles (corners).
• First Fold: Fold the sheet horizontally exactly in half. You’ve created a new line (crease) in the middle.
  How many parallel lines do you see now?
   (You should see the original top/bottom edges and the new middle crease – 3 parallel lines). 
  How does this new crease relate to the vertical sides of the original square? 
  It is perpendicular to them.
• More Folds: Make another horizontal fold in the already folded sheet. 
  How many parallel lines now? 
  You should have 5.What if you fold it horizontally again? 
  You’d get 9
  Do you see a pattern? 
  ​Each fold adds more parallel lines
• Vertical Fold: Now, unfold the paper back to the original square. Make a vertical fold down the middle. How does this new vertical line relate to the horizontal lines you made earlier? 
  It is perpendicular to them.
• Diagonal Fold: Fold the sheet along a diagonal (corner to opposite corner). 
  Can you make another fold that creates a line parallel to this diagonal?

Answer: No, you cannot make a fold that is truly parallel to a diagonal using standard folding techniques — that is, by folding corner to corner or edge to edge.

 Notations

How do we show lines are parallel or perpendicular in diagrams without writing it out every time?

• Parallel Lines: We use arrow marks (>) on the lines. If there’s more than one set of parallel lines, we use double arrows (>>), then triple arrows (>>>), and so on, for each different set.
• Perpendicular Lines: We draw a small square symbol at the point of intersection where the right angle (90°) is formed.Representation of Parallel and Perpendicular Lines

Transversals

Imagine a road crossing two or more parallel railway tracks. That road acts like a transversal.

Transversal: A line that intersects (crosses) two or more other lines (which may or may not be parallel) at distinct points.

Let’s look at line t crossing lines l and m:

When a transversal cuts two lines, it creates several angles. These angles have special names based on their positions:

Types of Angles Formed by a Transversal:

1. Interior Angles: Angles that lie between the two lines l and m. (In Fig: ∠3, ∠4, ∠5, ∠6)

2. Exterior Angles: Angles that lie outside the two lines l and m. (In Fig: ∠1, ∠2, ∠7, ∠8)

3. Pairs of Corresponding Angles: Angles that are in the same relative position at each intersection point. Think of them as being in the “same corner”. (Pairs: (∠1, ∠5), (∠2, ∠6), (∠3, ∠7), (∠4, ∠8))

4. Pairs of Alternate Interior Angles: Angles that are between the two lines but on opposite sides of the transversal. (Pairs: (∠3, ∠6), (∠4, ∠5))

5. Pairs of Alternate Exterior Angles: Angles that are outside the two lines and on opposite sides of the transversal. (Pairs: (∠1, ∠8), (∠2, ∠7))

6. Pairs of Interior Angles on the Same Side of the Transversal: Angles that are between the two lines and on the same side of the transversal. (Pairs: (∠3, ∠5), (∠4, ∠6)). These are sometimes called consecutive interior angles or co-interior angles.

Try yourself:What are parallel lines?A.Lines that meet at a pointB.Lines that cross each otherC.Lines that never meet or intersectD.Lines that form right anglesView Solution

Corresponding Angles

1. Corresponding Angles are Equal: If a transversal intersects two parallel lines, then each pair of corresponding angles is equal. (e.g., ∠1 = ∠5, ∠2 = ∠6, ∠3 = ∠7, ∠4 = ∠8)

• Converse is also true: If a transversal intersects two lines such that a pair of corresponding angles is equal, then the two lines are parallel.

2. Alternate Interior Angles are Equal: If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal. (e.g., ∠3 = ∠6, ∠4 = ∠5)

• Converse is also true: If a transversal intersects two lines such that a pair of alternate interior angles is equal, then the two lines are parallel.

3. Interior Angles on the Same Side are Supplementary: If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary (they add up to 180°). (e.g., ∠3 + ∠5 = 180°, ∠4 + ∠6 = 180°)

• Converse is also true: If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal is supplementary, then the two lines are parallel.

Drawing Parallel Lines

Parallel lines are two lines that never meet, no matter how long you extend them. They are always the same distance apart.

Let’s draw a pair of parallel lines using a ruler and a set square.

Tools You Need:

• Ruler
• Set square (a triangle-shaped tool with right angles)

Steps:

Step 1: Draw a line l using a ruler.

This line will be our reference or base line.

Step 2: Use your set square to draw a perpendicular line to l.

Place the set square so that one of its sides is touching line l and forms a 90° angle. Now draw a line along the other side of the set square.

Step 3: Slide the set square.

Keep the ruler still and carefully slide the set square up or down while keeping the same angle. Draw a new line the same way.

Now you’ve drawn two lines that are both perpendicular to the same line (line l).

Are These Lines Parallel?

Yes! Here’s why:

• You used the set square to ensure both lines made 90° angles with the same line (line l).
• When a line cuts two other lines and makes equal corresponding angles (in this case 90°), the two lines must be parallel.

This is a rule in geometry:
If corresponding angles are equal, the lines are parallel.

Making Parallel Lines through Paper Folding

Imagine you have a piece of paper with a crease (fold) line labeled as line l. You are also given a point A not on this line. Your task is to draw a line through A that is parallel to l using only folds.

• First Fold (Perpendicular to line l through A):
  • Fold the paper so that the crease you make is perpendicular to line l, and it passes through point A.
  • Call this new line t.
• Second Fold (Perpendicular to t through A):
  • Now fold the paper again, but this time make the crease perpendicular to line t, and ensure it passes through A again.
  • Call this new crease m.

Since m is perpendicular to t, and t is perpendicular to l, the new line m is parallel to l.

Why This Works:

This is because two lines that are both perpendicular to the same line are always parallel to each other.

Alternate Angles

We have two lines l and m cut by a third line t (this third line is called a transversal).

Look at the diagram with angles a, b, c, d, e, f, g, h.

• Alternate angles are on opposite sides of the transversal but between the two lines.
• Example: ∠f and ∠d are alternate angles. Similarly, ∠e and ∠c are alternate angles.

Rule:

 If the two lines are parallel, then alternate angles are equal.

If ∠f = 120°, then:

• ∠b = 120° (corresponding angle)
• ∠d = 120° (vertically opposite angle of ∠b)

So, ∠f = ∠d → which proves they are alternate angles and equal.

This proves lines l and m must be parallel if alternate angles are equal.

Summary: 

Parallel Illusions

A parallel illusion happens when your eyes and brain are tricked into thinking lines are not parallel, even though they actually are!

In the picture, look carefully at all three designs. At first, it looks like nothing is parallel, right?

But let’s find out what’s really going on.

1. It looks like a burst of lines coming out from a center point.

• At a glance, your brain thinks the lines are going in many directions, not parallel.
• But guess what? → Some pairs of lines here are actually straight and parallel!
• The radial design (lines from the center) makes your eyes focus on the middle, making it harder to see which lines are straight or equidistant.

2. This one is super tricky.

• It looks like the black and white patterns are tilted or slanted.
• But if you focus only on the horizontal black lines, you’ll notice: → They are actually all perfectly parallel!
• The staircase or zig-zag shapes in between confuse your brain and create the illusion of a slope.

3. It seems like all the lines are curving or pointing in different directions.

• But these lines are often used to trick your sense of depth (as if you’re looking into a tunnel).
• Some of these lines may be parallel, but because they are made to converge toward a center, it fools your eyes.

Why Do These Illusions Work?

Illusions like these work because of how our brain processes patterns and space:

• Distractions: Other shapes (like zigzags or curves) distract your eyes.
• Angles: Sharp angles make you think lines are changing direction.
• Context: Your brain compares lines to nearby objects.
• Depth Perception: Some images make flat things look 3D (like the tunnel effect).

The straight lines start looking curved or tilted, even though you used a ruler!

Try yourself:What are corresponding angles when a transversal intersects two parallel lines?

• A.They are alternate.
• B.They are supplementary.
• C.They are adjacent.
• D.They are equal.

View Solution

Some Solved Examples

Q1: You look at an image that has horizontal black lines between zigzag patterns. The lines appear slanted. Are they actually slanted?

Sol: No. The lines are perfectly parallel, but the design around them creates an optical illusion that tricks your eyes.

Q2: Two angles form a straight line. One angle measures 122°. What is the other?

Sol: Angles on a straight line add up to 180° (linear pair).
So,
180° − 122° = 58°

Q3: A transversal crosses two lines. Angles on opposite sides of the transversal between the lines are alternate interior angles. One angle is 65°. What is the other?

Sol: If the lines are parallel, then alternate interior angles are equal.
So the other angle is also 65°.

The Notion of Letter-Numbers

“Why do we need to use letters in math? Numbers make sense, but letters?” Ravi asked his friend Meena as they walked home from school. Meena smiled, remembering how she had the same question last year.Ravi is asking to Meena

“Let me show you something interesting,” Meena replied. “Think about Shabnam and Aftab from our class. Shabnam is 3 years older than Aftab, right?”

Shabnam’s age = Aftab’s age + 3.

Ravi nodded, unsure where this was going.

“So if Aftab is 12 years old now, Shabnam is…”

“15,” Ravi answered quickly.

“And if Aftab were 20, Shabnam would be…”

“23,” said Ravi.

“What if we don’t know Aftab’s exact age, but still want to talk about Shabnam’s age?” 

Meena asked. “That’s where letters come in! 

If we call Aftab’s age ‘a’, then Shabnam’s age is always ‘a + 3’. It’s a way to write a relationship that’s always true, no matter what the actual numbers are.”

s = a + 3

Ravi’s eyes widened with understanding. “So letters are like… placeholders for numbers that can change or that we don’t know yet?”Ravi is happy now.

“Exactly!” Meena exclaimed. “And when we use letters with numbers and operations, we create what mathematicians call ‘algebraic expressions’.”

In this chapter, we’ll explore how letters can represent numbers and how we can use them to write expressions that describe patterns, relationships, and real-world situations. We’ll learn how to work with these expressions, evaluate them when we know the values of the letters, and simplify them to make them easier to understand and use.

Example: Shabnam is 3 years older than Aftab. If we represent Aftab’s age with the letter ‘a’, then Shabnam’s age can be written as ‘a + 3’. This simple expression tells us the relationship between their ages, no matter how old they actually are.

Sol: If Aftab is 12 years old, then a = 12, and Shabnam’s age is:

a + 3 = 12 + 3 = 15 years

If Aftab is 20 years old, then a = 20, and Shabnam’s age is:

a + 3 = 20 + 3 = 23 years

The expression ‘a + 3’ captures the relationship perfectly, regardless of the specific values.

Example: Parthiv was playing with matchsticks and noticed something interesting when he made L-shaped patterns:

• 1 L needs 2 matchsticks.
• 2 Ls need 2 × 2 = 4 matchsticks.
• 5 Ls need 2 × 5 = 10 matchsticks.
• 45 Ls need 2 × 45 = 90 matchsticks.

We notice that the number of matchsticks is always 2 times the number of Ls.

To write this in math language, we use a variable. A variable is just a letter that stands for a number. Let’s use:

• n = number of Ls
  Then,
• Number of matchsticks = 2 × n

Example: Rani went to the market to buy coconuts and jaggery. Each coconut costs ₹35, and jaggery costs ₹60 per kilogram. If she buys ‘c’ coconuts and ‘j’ kilograms of jaggery, her total cost can be expressed as:

If you want to buy:

• 10 coconuts → cost = 10 × ₹35 = ₹350
• 5 kg jaggery → cost = 5 × ₹60 = ₹300
  So, the total cost = ₹350 + ₹300 = ₹650

Instead of calculating each time, you can create a formula (algebraic expression):

Let:

• c = number of coconuts
• j = number of kilograms of jaggery

Expression:

Total cost = c × 35 + j × 60

This formula lets you calculate the total cost quickly by just plugging in different values for c and j.

Example:

• For 7 coconuts and 4 kg jaggery:
  • c = 7, j = 4
  • Total cost = 7 × 35 + 4 × 60 = ₹245 + ₹240 = ₹485

Example: The perimeter of a shape is the total length around it. For a square, all four sides are equal in length.

So, if one side is q cm, then:

• The perimeter = q + q + q + q = 4 × q

This is written as an algebraic expression:
Perimeter = 4 × q, where q is the length of one side.

If the side of the square is 7 cm, then:

• Perimeter = 4 × 7 = 28 cm

Revisiting Arithmetic Expressions

Before we dive deeper into algebraic expressions, let’s quickly review what we know about arithmetic expressions.

An arithmetic expression is a combination of numbers and operations (addition, subtraction, multiplication, division). For example:

• 5 + 3
• 10 – 4
• 6 × 2
• 15 ÷ 3
• (8 + 2) × 3

When evaluating arithmetic expressions, we follow the order of operations (sometimes remembered as BODMAS or PEMDAS):

1. Brackets (or Parentheses)
2. Orders (or Exponents)
3. Division and Multiplication (from left to right)
4. Addition and Subtraction (from left to right)

For example, to evaluate 2 + 3 × 4:

1. First, we do the multiplication: 3 × 4 = 12
2. Then, we do the addition: 2 + 12 = 14

But if we have (2 + 3) × 4:

1. First, we evaluate what’s in the brackets: (2 + 3) = 5
2. Then, we do the multiplication: 5 × 4 = 20

Algebraic expressions follow these same rules, but they include letter-numbers (variables) as well as regular numbers.

Evaluating Algebraic Expressions

When we know the value of the letter-numbers in an algebraic expression, we can substitute those values and evaluate the expression just like an arithmetic expression.

Let’s evaluate the expression 3n + 5 when n = 4:

1. Substitute n = 4 into the expression: 3(4) + 5
2. Multiply: 12 + 5
3. Add: 17

Let’s try another one. Evaluate 2p – 7 when p = 10:

1. Substitute p = 10 into the expression: 2(10) – 7
2. Multiply: 20 – 7
3. Subtract: 13

Try yourself:

What does the expression ‘a + 3’ represent?

• A.The total cost of coconuts
• B.Aftab’s age only
• C.Shabnam’s age compared to Aftab’s age
• D.The perimeter of a square

View Solution

Omission of the Multiplication Symbol in Algebraic Expressions

Look at the pattern:

4, 8, 12, 16, 20, 24, 28, …

This is a list of multiples of 4, which means each number is 4 multiplied by another number.

• 3rd term = 4 × 3 = 12
• 29th term = 4 × 29 = 116

To get the nth term of this pattern, the expression is: 4 × n

But in algebra, we don’t usually write the × sign. Instead, we just write them together like this:

➡️ 4n

This is a shorter and standard way.

More Examples:

1. 7k means 7 × k
   • If k = 4, then 7k = 7 × 4 = 28
2. 5m + 3 means:
   • First, multiply 5 × m
   • Then add 3
   • If m = 2, then:
   • 5 × 2 + 3 = 10 + 3 = 13

Mind the Mistake, Mend the Mistake

This activity is called “Mind the Mistake, Mend the Mistake”. It helps you practice algebra by identifying and fixing mistakes in evaluating expressions when values are substituted.

Let’s go through each and spot the mistakes:

1. If a = 4, then 10 – a = 6.
Correct! → 10 – 4 = 6

2. If d = 6, then 3d = 36.
Mistake!
3d = 3 × 6 = 18, not 36.

3. If s = 7, then 3s – 2 = 15.
Incorrect!
3 × 7 – 2 = 21 – 2 = 19, not 15.

4. If r = 8, then 2r + 1 = 29.
Wrong!
2 × 8 + 1 = 16 + 1 = 17

5. If j = 5, then 2j = 10.
Correct!

6. If m = -6, then 3(m + 1) = 19.
Mistake!
m + 1 = -6 + 1 = -5
3 × (-5) = -15, not 19.

7. If f = 3, g = 1 then 2f – 2g = 2.
But there is no value for f, and z is unused.
Confusing/wrong variables used. Needs correction.

8. If t = 4, b = 3 then 2t + b = 24.
Mistake!
2 × 4 + 3 = 8 + 3 = 11, not 24.

9. If h = 5, n = 6, then h – (3 – n) = 4
Let’s solve it:
3 – n = 3 – 6 = -3
h – (-3) = 5 + 3 = 8, not 4
❌ Mistake in subtracting a negative number.

Simplification of Algebraic Expressions

1. Perimeter of a Rectangle Using Expressions

• Perimeter formula: Add all sides of the rectangle
  p = l + b + l + b
• We can rearrange it using properties of addition:
  p = 2l + 2b (This is the simplified form)

Example:  If length (l) = 3 cm and breadth (b) = 4 cm

Sol: p = l + b + l + b = 3 + 4 + 3 + 4 = 14

p = 2l + 2b = 2 × 3 + 2 × 4 = 6 + 8 = 14

Both give the same result, so they are equivalent expressions.

Expressions from a Word Problem

Example: Here is a table showing the number of pencils and erasers sold in a shop. The price per pencil is c, and the price per eraser is d. Find the total money earned by the shopkeeper during these three days.

• Suppose the price of a pencil is ₹c and that of an eraser is ₹d.
• A table gives the number of pencils and erasers sold over 3 days:

Expression for pencils:

Day 1: 5c, Day 2: 3c, Day 3: 10c
Total = 5c + 3c + 10c

Using distributive property:
= (5 + 3 + 10) × c = 18c

If c = ₹50, then total = 18 × 50 = ₹900

Expression for erasers:

Day 1: 4d, Day 2: 6d, Day 3: 1d
Total = 4d + 6d + 1d = 11d

Combined expression for total sales:
18c + 11d (This is already simplified)

Area of a Big Rectangle

Example: A large rectangle is split into two smaller ones with widths 4 and 3, and the same height (v).

• Area of big rectangle:
  • Way 1: Area = v × (4 + 3) = 7v
  • Way 2: Area = 4v + 3v = 7v
• This shows the distributive property:
  v(4+3) = 4v + 3v

AEFD Rectangle Area Expression

• A large rectangle ABCD is split into AEFD and EBCF.
• Total length AB = 12 units, EB = 4 units, so AE = 12 – 4 = 8 units.
• Height = n
• Area of AEFD:
  • Way 1: Area = n × 8 = 8n
  • Way 2: Area = n × 12 – n × 4 = 12n – 4n = 8n

Conclusion: Both methods give the same answer using different algebraic approaches.

Try yourself:

What does the expression 7k mean in algebra?

• A.7 divided by k
• B.7 minus k
• C.7 plus k
• D.7 multiplied by k

View Solution

Like Terms vs. Unlike Terms

• Like Terms: Same variable (e.g., 5c, 10c, 18c)
• Unlike Terms: Different variables or powers (e.g., 18c, 11d)
• Like terms can be added together, unlike terms cannot.

Combining Like Terms

To simplify an expression, we combine like terms by adding or subtracting their coefficients (the numbers in front of the letter-numbers).

Example: Simplify 5c + 3c + 10c

All terms have the letter c, so they are like terms.

5c + 3c + 10c = (5 + 3 + 10)c = 18c

Example: Simplify 4v + 3v

Both terms have the letter v, so they are like terms.

4v + 3v = (4 + 3)v = 7v

Example: Simplify l + b + l + b (perimeter of a rectangle)

The like terms are l and l, and b and b.

l + b + l + b = (l + l) + (b + b) = 2l + 2b

Simplification Involving Subtraction

Example: Simplify 8x – 3x

Both terms have the letter x, so they are like terms.

8x – 3x = (8 – 3)x = 5x

Example: Simplify 12y – 5y + 2y

All terms have the letter y, so they are like terms.

12y – 5y + 2y = (12 – 5 + 2)y = 9y

Simplification Involving Brackets

When a minus sign is placed before brackets, it changes the sign of each term inside the brackets when we remove the brackets.

Example: Simplify 7a – (3a + 2)

7a – (3a + 2) = 7a – 3a – 2 = (7 – 3)a – 2 = 4a – 2

Example: Simplify 5b – (b – 4)

5b – (b – 4) = 5b – b + 4 = (5 – 1)b + 4 = 4b + 4

Mind the Mistake, Mend the Mistake

Pick Patterns and Reveal Relationships

What is a Number Machine?

A number machine takes two input numbers and performs a specific rule or operation on them to produce an output. You need to observe the pattern or rule used to calculate the result and write that as a mathematical expression (also called a formula).

Formula Detective

In the top row of machines, you are given two inputs and one output. Find out the formula of this number machine.

Sol: 

Formula:

Two times the first number minus the second number
In algebra:
Expression = 2a − b
Where:

• a = first input
• b = second input

Let’s apply this formula to each set of inputs and write the expressions:

Formula:
The number machine uses this expression:
2 × (first number) − (second number)
or
2a − b

So the missing output for the last set (6, 4) is:
2 × 6 − 4 = 12 − 4 = 8

Algebraic Expressions to Describe Patterns

Somjit saw a repeating pattern along the border of a saree. The designs appear in the following order:

These three designs (A, B, C) are repeated again and again in the same sequence. This is called a repeating pattern.

We can see that:

• Every 3rd position is Design C
• Every position 1 less than a multiple of 3 is Design B
• Every position 2 less than a multiple of 3 is Design A

Design C:
It appears at positions: 3, 6, 9, 12, …

These are multiples of 3, which means:

C appears at position:
3n (where n is 1, 2, 3, …)

Design B:
It appears at positions: 2, 5, 8, 11, …

That is always 1 less than C, so:

B appears at position:
3n − 1

Design A:
It appears at positions: 1, 4, 7, 10, …

That is 2 less than C, so:

A appears at position:
3n − 2

Try yourself:

What are like terms?

• A.Same constants
• B.Different variables
• C.Different powers
• D.Same variable

View Solution

Using Division to Quickly Find the Design at Any Position

If someone gives you a random number like 99, how will you know which design it is?

You use division and remainders:

• Divide the number by 3
• Look at the remainder

Example: Position 99

99 ÷ 3 = 33 remainder 0
→ Remainder is 0 → So, it’s Design C

Example: Position 122

122 ÷ 3 = 40 remainder 2
→ Remainder is 2 → So, it’s Design B

Example: Position 148

148 ÷ 3 = 49 remainder 1
→ Remainder is 1 → So, it’s Design A

Patterns in a Calendar

Here is the calendar of November 2024. Consider 2 × 2 squares, as marked in the calendar. The numbers in this square show an interesting property.

Let us take the marked 2 × 2 square.    2 × 2

These are positioned in the calendar with 7-day weeks. So:

• The number to the right = +1
• The number directly below = +7
• The diagonal number = +8

Diagonal Sum Property

Take the diagonals of this 2×2 square:

• First diagonal: 12 + 20 = 32
• Second diagonal: 13 + 19 = 32

Observation: Both diagonals add up to the same total!

Now check the diagonals:

• First diagonal = a + (a + 8) = 2a + 8
• Second diagonal = (a + 1) + (a + 7) = 2a + 8

Both diagonals are equal. This pattern will always hold!

Let’s prove it works for any number. Let’s call the top-left number a. Then the square is:

Now check the diagonals:

• First diagonal = a + (a + 8) = 2a + 8
• Second diagonal = (a + 1) + (a + 7) = 2a + 8

Both diagonals are equal. This pattern will always hold!

A Plus-Shaped Pattern in the Calendar

Let’s take another shape like a plus sign, using this pattern of numbers:

Add up all the numbers:
8 + 14 + 15 + 16 + 22 = 75

Now look at the center number: 15
5 × 15 = 75

Observation: The sum of all 5 numbers is always 5 times the center number.

Why Does This Happen?

Let’s say the center number is a. Then the other numbers are:

• One above: a − 7
• One below: a + 7
• One left: a − 1
• One right: a + 1

Now sum:

(a – 7) + (a + 7) + (a – 1) + (a + 1) + a

= 5a

The pattern is algebraically proven — no matter which number is in the center, the total will be 5 × center number!

Matchstick Patterns

You are given a series of patterns made by arranging triangles using matchsticks. Each triangle shares a side with the next one.

Look at the image given above:

• Step 1 has 1 triangle.
• Step 2 has 2 triangles.
• Step 3 has 3 triangles.
• And so on…

So what’s changing? The number of matchsticks.

How many matchsticks are used?

• Step 1 → 3 matchsticks
• Step 2 → 5 matchsticks
• Step 3 → 7 matchsticks
• Step 4 → 9 matchsticks
• Step 5 → 11 matchsticks

What do we observe?

Each new triangle adds 2 more matchsticks.
So the number of matchsticks increases by 2 every time.

 Let’s use algebra now

Let’s say the step number is y. We want a rule (or formula) to find the number of matchsticks at any step.

From the pattern:

• Step 1 → 3 matchsticks = 1 + 2
• Step 2 → 3 + 2 = 5
• Step 3 → 3 + 2 + 2 = 7
• Step 4 → 3 + 2 + 2 + 2 = 9

So the general formula becomes:

Matchsticks = 3 + 2 × (y − 1)

Now let’s simplify this expression:

3 + 2 × (y − 1)

= 3 + 2y − 2

= 2y + 1

Let’s use this formula for:

• Step 33
  2 × 33 + 1 = 66 + 1 = 67 matchsticks
• Step 84
  2 × 84 + 1 = 168 + 1 = 169 matchsticks
• Step 108
  2 × 108 + 1 = 216 + 1 = 217 matchsticks

Two orientations of matchsticks

In each triangle:

• There are horizontal matchsticks (top and bottom)
• And diagonal matchsticks (middle)

For example:

• In Step 2:
  • Horizontal: 2
  • Diagonal: 3

Check Step 3 and 4 to see how this changes.

You can write two expressions—one for horizontal matchsticks and one for diagonal, and their sum should be 2y + 1.

Some Solved Examples

Example 1: Evaluate the expression 4x + 3y when x = 5 and y = 2.

Sol:

Step 1: Substitute the values of x and y into the expression.

4x + 3y = 4(5) + 3(2)

Step 2: Multiply.

4(5) + 3(2) = 20 + 6

Step 3: Add.

20 + 6 = 26

Therefore, when x = 5 and y = 2, the value of 4x + 3y is 26.

Example 2: Simplify the expression 9p – (3p – 4q + 2).

Sol:

Step 1: Remove the brackets, changing the sign of each term inside.

9p – (3p – 4q + 2) = 9p – 3p + 4q – 2

Step 2: Combine like terms.

9p – 3p + 4q – 2 = (9 – 3)p + 4q – 2

= 6p + 4q – 2

Therefore, the simplified expression is 6p + 4q – 2.

Example 3: A rectangular field has length ‘l’ meters and width ‘w’ meters. A path of width 2 meters runs around the inside of the field. Write an expression for the area of the path.

Sol:

Step 1: Find the area of the entire field.

Area of field = l × w

Step 2: Find the area of the inner rectangle (field minus path).

Length of inner rectangle = l – 2 – 2 = l – 4 (2 meters less on each side)

Width of inner rectangle = w – 2 – 2 = w – 4 (2 meters less on each side)

Area of inner rectangle = (l – 4) × (w – 4)

= lw – 4l – 4w + 16

Step 3: Find the area of the path by subtracting the inner area from the total area.

Area of path = Area of field – Area of inner rectangle

= lw – (lw – 4l – 4w + 16)

= lw – lw + 4l + 4w – 16

= 4l + 4w – 16

Therefore, the area of the path is 4l + 4w – 16 square meters.

 The Need for Smaller Units

Have you ever tried to fix something, maybe a toy or a small gadget, and found that a tiny screw just wouldn’t fit? That’s exactly what happened to Sonu one day!

•  He was watching his mother trying to fix his favourite toy car. She had a small screw, but it just wasn’t the right size to join two pieces together. 
• Sonu watched with wide eyes, full of curiosity. Why wouldn’t it work?

Sonu’s mother trying to fix his toy with small screw

• His mother explained, “This screw isn’t quite the right size, Sonu.” She rummaged through her toolbox and found another screw.
•  This time, it worked perfectly, and the toy was fixed! To Sonu, both screws looked almost exactly the same. He picked them up and looked very, very closely. 
• Ah! He could see it now – one screw was just a tiny bit longer than the other.Yes, one screw was bit longer than the other
• Sonu was amazed. How could such a small, almost invisible difference in length make such a big difference in fixing his toy?
•  He became really curious. How much longer was one screw than the other? 
• And how could he even measure such a tiny difference when they looked so similar?

This little adventure with the screws got Sonu thinking about measurement. Sometimes, just saying something is ‘long’ or ‘short’ isn’t enough. We need to know exactly how long or short it is. 

Measuring with Greater PrecisionLook at the screws below placed on different rulers. Can you see how the bottom ruler, with more marks, helps us measure more accurately?Screws placed on different Ruler to measure the length

• Look at the top ruler first. It only has markings for whole centimeters (cm). Using this ruler, we can only say that both screws are between 2 cm and 3 cm long. That’s not very helpful when we need to know which one is longer!
• Now look at the middle ruler. This one has more markings – it divides each centimeter into 10 equal parts. With this ruler, we can be more precise. The first screw measures 2 7/10 cm (or 2.7 cm). That means it’s 2 whole centimeters plus 7 tenths of a centimeter.
• The bottom ruler is even more detailed. It divides each centimeter into many tiny parts, allowing for even more precise measurements.

A Tenth Part

Have you ever wondered how we can measure things more precisely? Let’s explore this with a simple example.

Look at the pencil shown in the figure below:

Notice how the pencil’s length isn’t exactly 3 units or 4 units. It’s somewhere in between. When we look at the ruler with more detailed markings (the bottom one), we can see that the pencil measures 3 and 4 tenths of a unit, which we write as 

But what does  actually mean? Let’s break it down:

• 3 represents the whole units (3 × 1)
• 4/10 represents four one-tenths (4 × 1/10)

So the total length is (3 × 1) + (4 × 1/10) units.

This length is the same as 34 one-tenths units because 10 one-tenths units make one unit.

There are different ways to express measurements with tenths:

•  is read as “four and one-tenth”
• 4/10 is read as “four-tenths”
• 41/10 is read as “forty-one tenths”
•  is read as “forty-one and one-tenth”

 Example 1: Sonu was measuring parts of his arm. The length of his lower arm was units, and his upper arm was  units. What is the total length of his arm?

Sol: To find the total length, we need to add the measurements:

Method 1:

• Lower arm: 2 units and 7 tenths
• Upper arm: 3 units and 6 tenths
• Total units: 2 + 3 = 5 units
• Total tenths: 7 + 6 = 13 tenths

But 13 tenths equals 1 unit and 3 tenths (since 10 tenths = 1 unit).

So the total length is 5 + 1 + 3/10 =  units.

Method 2:

We can convert both measurements to tenths and then add:

So Sonu’s complete arm measures units long.

A Hundredth Part

We learned about tenths, which help us measure more precisely than whole units. But what if we need to be even more precise? Let’s explore!

• Imagine you have a sheet of paper that measures 8 whole 9/10 units long. That’s 8 whole units and 9 tenths of a unit. 
• Now, what happens if you fold this paper exactly in half along its length? What would the new length be?

If we try to measure the folded paper using a ruler marked only in tenths, we might find that the length falls between two tenth markings. For example, it might be between  and  units. We can estimate, but we can’t get an exact measurement with just tenths.

Try yourself:

What did Sonu’s mother find in her toolbox to fix the toy car?

• A.A bigger screw
• B.A longer screw
• C.A smaller screw
• D.A different tool

View Solution

Introducing Hundredths

To measure even smaller lengths, we can do something similar to what we did before: we can divide each tenth into 10 smaller, equal parts.

Think about it:

• 1 unit is divided into 10 tenths.
• Each tenth is divided into 10 smaller parts.

So, how many of these tiny parts make up one whole unit? That’s right!

 10 tenths × 10 parts per tenth = 100 parts.

Each of these tiny parts is called one-hundredth (1/100) of the unit.

Measuring with Hundredths

Now, let’s go back to our folded paper. Using a ruler with hundredth markings, we can measure its length much more accurately. Let’s say the folded edge lands exactly on the 5th small mark after .

We can express this length in a few ways:

• 4 units, 4 tenths, and 5 hundredths: 4 + 4/10 + 5/100
• 4 units and 45 hundredths: 4 + 45/100 (since 4 tenths = 40 hundredths, so 40/100 + 5/100 = 45/100)

Reading and Writing Hundredths

Just like with tenths, we have different ways to read and write measurements involving hundredths:

These all represent the same length!

Example 2: What is the sum of ?

Sol: Method 1 (Adding units, tenths, hundredths separately):

Method 2 (Converting everything to hundredths):

Method 3:

Method 4:

Example 3: What is the difference:?

Sol: 

Example 4: What is the difference ?

Sol:

Decimal Place Value

We’ve seen how dividing units into tenths and then hundredths helps us measure more precisely. But why do we always divide by 10?  Why not 4, or 5, or 8 equal parts?

Yes, we can.

Look at the pencils in the image:

• One ruler is marked in tenths (0.1, 0.2, 0.3…)
• The other is marked in quarters (¼, ½, ¾…)

If we split each quarter further into 4 smaller parts, we get sixteenths (1⁄16), which are even more accurate.

So, why do we usually split into 10 parts?

That’s because of our number system – the Indian place value system (and the international system too) is based on 10s.

Each place value is 10 times:

• Bigger than the one to its right
• Smaller than the one to its left

For example:

• 10 ones = 1 ten
• 10 tens = 1 hundred
• 10 hundreds = 1 thousand

That’s why decimal place values go like:

• 0.1 = 1⁄10 (one-tenth)
• 0.01 = 1⁄100 (one-hundredth)
• 0.001 = 1⁄1000 (one-thousandth)

Notice a pattern?

 Each place value is 10 times bigger than the one immediately to its right. Or, looking the other way, each place value is 10 times smaller (or one-tenth) than the one immediately to its left.

To extend this system to numbers smaller than one, it makes perfect sense to continue this pattern: divide the Ones place by 10!

• Dividing 1 by 10 gives one-tenth (1/10).
• Dividing one-tenth (1/10) by 10 gives one-hundredth (1/100).
• Dividing one-hundredth (1/100) by 10 gives one-thousandth (1/1000).
• …and so on!
• What will the fraction be when 1/ 100 is split into 10 equal parts? 

It will be 1/ 1000 , i.e., a thousand such parts make up a unit.

This system, based on the number 10, is called the decimal system.

Try yourself:

What is each tiny part called when a unit is divided into 100 parts?

• A.One-quarter
• B.One-tenth
• C.One-hundredth
• D.One-thousandth

View Solution

Notation, Writing and Reading of Numbers

We need a way to clearly separate the whole number part from the fractional part (tenths, hundredths, etc.). If we just wrote 42, how would we know if it means “4 tens and 2 ones” (42) or “4 ones and 2 tenths” ()?

To solve this, we use a special symbol: the decimal point (.) .

 It acts as a separator between the Ones place and the Tenths place.

• Whole Numbers are to the left of the decimal point.
• Fractional Parts (tenths, hundredths, etc.) are to the right of the decimal point.

Let’s look at some examples:

These numbers, when shown through place value, are as follows:

Reading Decimal Numbers

How do we read these numbers aloud?

• 705: “Seven hundred five” (No decimal part)
• 70.5: “Seventy point five” (or “Seventy and five tenths”)
• 7.05: “Seven point zero five” (or “Seven and five hundredths”)

Important: When reading the digits after the decimal point, we usually say each digit individually (like “point two seven four”, not “point two hundred seventy-four”). This helps avoid confusion about place value.

Example 5: How can we write 234 tenths in decimal form?

Sol: 234 tenths = 234/10

We can break this down:

234/10 = (200 + 30 + 4) / 10

= 200/10 + 30/10 + 4/10

= 20 + 3 + 4/10

= 23.4

In decimal notation, this is 23.4.

Units of Measurement

Now that we understand decimal numbers, let’s see how they help us work with different units of measurement, especially for length.

Length Conversion: Millimeters (mm) and Centimeters (cm)

You’ve probably used a ruler marked in centimeters (cm) and millimeters (mm). You might already know that:

⇒ 1 cm = 10 mm

This means that each centimeter is divided into 10 equal parts, and each part is a millimeter. 

So, how much of a centimeter is one millimeter?

Since 10 mm make 1 cm, then:

⇒ 1 mm = 1/10 cm

Using our new decimal notation, we can write this as:

⇒ 1 mm = 0.1 cm (one-tenth of a centimeter)

Let’s practice converting between mm and cm using decimals:

Q: How many cm is 5 mm?

5 mm = 5/10 cm = 0.5 cm

Q: How many cm is 12 mm?

12 mm = 10 mm + 2 mm

= 1 cm + 2/10 cm

=  1.2 cm

Q: How many mm is 5.6 cm?

5.6 cm = 5 cm and 6/10 cm

= (5 × 10 mm) + (6/10 × 10 mm)

= 50 mm + 6 mm = 56 mm

Length Conversion: Centimeters (cm) and Meters (m)

We also know the relationship between centimeters and meters:

⇒ 1 m = 100 cm

This means one meter is divided into 100 equal parts, and each part is a centimeter. So, how much of a meter is one centimeter?

Since 100 cm make 1 m, then:

⇒ 1 cm = 1/100 m

Using decimal notation:

⇒ 1 cm = 0.01 m (one-hundredth of a meter)

These tiny measurements show how important tenths and hundredths (and even smaller units!) are in science and everyday life.

Weight ConversionIt shows how weights can be expressed in decimal form and why that’s useful in real life.

• 1 kilogram (kg) = 1000 grams (g)
• So, 1 gram = 1/1000 kg = 0.001 kg

Q: How many kilograms is 5 g?

5 g = 5 ÷ 1000 = 0.005 kg

Q: How many kilograms is 10 g?

10 g = 10 ÷ 1000 = 0.010 kg

As each gram is one-thousandth of a kg, 254 g can be written as

This can be broken as:
200 g → 0.2 kg
50 g → 0.05 kg
4 g → 0.004 kg
Total = 0.254 kg

Rupee ─ Paise conversion

In India, the currency system is based on rupees and paise.

Basic Conversion Rule:

• 1 rupee = 100 paise
• So, 1 paise = 1/100 rupee = ₹ 0.01

That means if you divide 1 rupee into 100 equal parts, each part is called 1 paise.

How many rupees is 75 paise?

Locating and Comparing Decimals

What are Decimals?

Decimals are numbers that show values smaller than 1. They help us measure things more accurately, like length, weight, and money.

How to Show Decimals on a Number Line?

To place a decimal like 1.4 on a number line:

• First, see which two whole numbers it lies between.
  1.4 lies between 1 and 2.
• Then divide the gap between 1 and 2 into 10 equal parts.
• The 4th division after 1 is 1.4.

So, 1.4 = 1 unit and 4 tenths.

Does Adding Zeros Change the Value?

Let’s compare:

• 0.2 = 0.20 = 0.200 — All same (2 tenths)

But:

• 0.02 = 2 hundredths
  0.002 = 2 thousandths
  These are different.

Adding zeros to the right after decimal digits doesn’t change the value. But changing the place value (tenths, hundredths, thousandths) does.

Try yourself:

What does a decimal point separate in a number?

• A.Large numbers from small numbers
• B.Whole numbers from fractional parts
• C.Even numbers from odd numbers
• D.Positive numbers from negative numbers

View Solution

Ordering Decimal Numbers

Look at these numbers:

• 4.5, 4.05, 4.005, 4.050, 4.50

Let’s convert them all to the same format (3 decimal places):

• 4.500
• 4.050
• 4.005
• 4.050
• 4.500

Now, it’s easier to compare:

• Smallest = 4.005
• Largest = 4.500
• 4.5 = 4.50
• 4.05 = 4.050 

Closest Decimals

This concept teaches you how to compare decimal numbers and figure out which decimal is closest to a given number.

Let’s compare the numbers:
0.9, 1.1, 1.01, and 1.11

Which one is closest to 1?

Step 1: Arrange the numbers in order:
0.9 < 1 < 1.01 < 1.1 < 1.11

Step 2: See how far each number is from 1:

• 0.9 is 0.1 away (10/100)
• 1.1 is also 0.1 away
• 1.01 is just 0.01 away (1/100)

So, 1.01 is closest to 1.

This improves your understanding of how place value affects the size of decimal numbers.

Addition and Subtraction of Decimals

This concept teaches how to add and subtract numbers with decimals (like 2.7, 3.5, etc.), just like you do with whole numbers. The only difference is that you have to align the decimal points and work carefully with tenths, hundredths, or thousandths.

Example 6: Priya needs 2.7 m of cloth and Shylaja needs 3.5 m of cloth. How much cloth is needed in total?

Sol: We add:
2.7 + 3.5 = 6.2 m

How?

• 2.7 = 2 + 7/10
• 3.5 = 3 + 5/10
• Add: (2 + 3) + (7/10 + 5/10) = 5 + 12/10
• 12/10 = 1 whole and 2/10
• Total = 6 + 2/10 = 6.2 m

Example 7: How much longer is Shylaja’s cloth compared to Priya’s?

Sol: We subtract:
3.5 – 2.7 = 0.8 m

In fractions:

• 3.5 = 3 + 5/10
• 2.7 = 2 + 7/10So, difference = 8/10 = 0.8

Detailed Decimal Addition (Place Value Style)

Let’s add:
75.345 + 86.691

Each digit is added based on place value:

Decimal Sequences

A sequence is a list of numbers that follow a pattern.

In a decimal sequence, each number includes a decimal point and the numbers change in a regular way.

What is the pattern?

Let’s look at this example:4.4, 4.8, 5.2, 5.6, 6.0…

• What’s happening here?
• Each number is increasing by 0.4.

So, to continue the sequence, we keep adding 0.4:

• 6.0 + 0.4 = 6.4
• 6.4 + 0.4 = 6.8
• 6.8 + 0.4 = 7.2

The next 3 terms are: 6.4, 6.8, 7.2.

Try yourself:Which number is closest to 1?A.0.9B.1.1C.1.01D.1.11View Solution

Estimating Sums and Differences

Sometimes, when we add or subtract decimal numbers, we want to quickly estimate the answer before calculating exactly. Estimating helps us check if our final answer is reasonable.

Sonu noticed an interesting pattern and made a smart observation:

“If we add two decimal numbers, the answer will always be more than the sum of their whole number parts, but less than 2 more than that.”

Example:

1. Let’s take the numbers 25.936 and 8.202

• Whole number parts: 25 and 8
• Their sum is: 25 + 8 = 33
• So, Sonu says the final sum should be:
  • More than 33
  • Less than 33 + 2 = 35

Let’s check:

• Actual sum = 25.936 + 8.202 = 34.138

Yes! It lies between 33 and 35 – so Sonu’s idea works here.

2. Let’s say you want to subtract two decimals, like:

9.6 – 2.3

• Whole number parts are 9 and 2
• The answer will lie between 9 – 2 = 7 and 9 – 2 + 1 = 8

Actual answer: 7.3, which lies between 7 and 8 

Conclusion:

Think of the whole number part first.

Then remember:

• For addition: the answer is a little more than just adding the whole numbers.
• For subtraction: the answer is a little less than just subtracting the whole numbers.

More on the Decimal System

Decimal numbers use a dot (called the decimal point) to separate the whole part of a number from the fractional part. For example:

• 1.5 means one and a half
• 0.05 means five hundredths

Sometimes, small mistakes in placing or reading this decimal point can lead to huge real-world problems which are as follows:

1. Amsterdam’s Money Mistake (2013):

• They meant to send €1.8 million.
• But due to a decimal error, they sent €188 million.
• Why? Because the amount was entered in euro-cents instead of euros (1 euro = 100 cents).
• That’s 100 times more money!

2. The Plane Fuel Disaster (1983):

• Ground staff gave fuel in pounds instead of kilograms.
• The plane had half the fuel it actually needed.
• The decimal error could’ve cost lives, but thankfully, everyone survived.

3. Medical Mistakes:

• If a doctor reads 0.05 mg as 0.5 mg, that’s 10 times the correct amount.
• This could be dangerous when giving medicine.

Deceptive Decimal Notation

1. Decimal Confusion in Time (Deceptive Decimal Notation):

• When you see 4.5 hours, don’t think it’s 4 hours and 5 minutes.
• 0.5 hours actually means 30 minutes, not 5 minutes!
• So, 4.5 hours after noon is 4:30 PM, not 4:05 PM.

2. Real-Life Measurement Mistake:

• If someone says 2.5 feet, it means 2 feet 6 inches (because 0.5 ft = 6 inches).
• But if they actually meant 2 feet 5 inches, the door or object won’t fit correctly.
• This shows why it’s important to know what decimals mean in real measurements.

3. Decimal Notation in Sports (Like Cricket):

• In cricket, 5.5 overs means 5 overs + 5 balls (because 1 over = 6 balls).
• So, 5.5 ≠ 5 overs 50%; it means 5 overs and 5 balls = 5.833 overs in true decimal form!

Try yourself:

What does Sonu observe about adding two decimal numbers?

• A.The answer is exactly the sum of whole numbers.
• B.The answer is equal to two times the whole numbers.
• C.The answer is more than the sum of whole numbers.
• D.The answer is less than the sum of whole numbers.

View Solution

A Pinch of History – Decimal Notation Over Time

• Ancient mathematicians like Shridhara and Abu’l Hassan used fractions like 1/10, 1/100.
• In 15th century, people marked decimal parts in different colors or added little superscript numbers (like 3⁶).
• In 16th century, John Napier and Christopher Clavius introduced the use of a dot (.) to separate whole numbers and decimals (like 2.5).
• Some countries today use comma  instead of a dot (e.g., 1,000.5 becomes 1.000,5).

Introduction

Have you ever noticed how we combine numbers and operations like addition (+), subtraction (-), multiplication (×), and division (÷) to represent situations or solve problems? 
Phrases like “13 + 2” (13 plus 2), “20 – 4” (20 minus 4), or “12 × 5” (12 times 5) are common in mathematics. These combinations are called arithmetic expressions.

In this chapter, we will delve deeper into the world of arithmetic expressions. 

 Simple Expressions

• Every arithmetic expression has a specific value, which is the single number it represents. For instance, the expression “13 + 2” has a value of 15. 
• We use the equals sign (=) to show this relationship: 13 + 2 = 15. 
• Think about Mallika spending ₹25 each weekday (Monday to Friday) for lunch. To find her total weekly spending, we can write the expression 5 × 25. This expression represents “5 times 25,” and its value tells us the total amount spent.

An important point is that different expressions can result in the same value. Consider the number 12. It can be represented by various expressions:

• 10 + 2 = 12
• 15 – 3 = 12
• 3 × 4 = 12
• 24 ÷ 2 = 12

This flexibility allows us to express the same mathematical idea in multiple ways, which can be useful in different contexts or for simplifying problems. 

Comparing Expressions

Just as we compare individual numbers using symbols like equals (=), greater than (>), and less than (<), we can also compare arithmetic expressions. This comparison is based on the values that the expressions evaluate to.

For example, let’s compare the expressions 10 + 2 and 7 + 1:

• The value of 10 + 2 is 12.
• The value of 7 + 1 is 8.
• Since 12 is greater than 8, we can write: 10 + 2 > 7 + 1.

Similarly, let’s compare 13 – 2 and 4 × 3:

• The value of 13 – 2 is 11.
• The value of 4 × 3 is 12.
• Since 11 is less than 12, we write: 13 – 2 < 4 × 3.

 Reading and Evaluating Complex Expressions

Simple expressions usually involve just one operation. But what happens when an expression combines multiple operations, like 30 + 5 × 4? 
Without a clear context or rule, how do we know whether to add first or multiply first? This ambiguity can lead to different answers.

Consider the language example from the textbook:

• Sentence (a): “Shalini sat next to a friend with toys”. (Meaning: The friend has toys, Shalini sat next to her).
• Sentence (b): “Shalini sat next to a friend, with toys”. (Meaning: Shalini has the toys, and she sat next to her friend).

The comma in sentence (b) acts like punctuation, clarifying the meaning. Without it, the sentence could be interpreted in two ways. 
Similarly, in mathematics, we need rules and tools to ensure everyone evaluates a complex expression the same way.

Let’s look at this example: Mallesh brought 30 marbles, and Arun brought 5 bags with 4 marbles each. The total number of marbles can be written as 30 + 5 × 4.

• Purna’s calculation: Added 30 and 5 first (30 + 5 = 35), then multiplied by 4 (35 × 4 = 140).
• Mallesh’s calculation: Multiplied 5 and 4 first (5 × 4 = 20), then added 30 (30 + 20 = 50).

⇒ In the context of the story, Mallesh’s calculation (50 marbles) makes sense. Purna’s calculation (140 marbles) doesn’t fit the situation.
⇒ This highlights that just looking at the expression 30 + 5 × 4 isn’t enough; we need a standard order of operations.

⇒ To resolve such confusion and ensure consistent evaluation, mathematics uses specific tools and conventions, primarily brackets and the concept of terms, which we will explore next.

Try yourself:

What is the value of the expression 13 + 2?

• A.10
• B.20
• C.15
• D.12

View Solution

Brackets in Expressions

• One of the primary tools used in mathematics to clarify the order of operations in complex expressions is brackets ( ).
• When an expression contains brackets, the part of the expression inside the brackets must be evaluated first, before performing operations outside the brackets.

Let’s revisit the expression for the total number of marbles Mallesh and Arun brought: 30 + 5 × 4. 
We determined that multiplication should happen before addition based on the context. We can make this explicit using brackets:

30 + (5 × 4)

To evaluate this:

1. First, calculate the value inside the brackets: 5 × 4 = 20.
2. Then, perform the remaining operation: 30 + 20 = 50.

This use of brackets removes ambiguity and ensures the expression correctly represents the intended calculation.

Example 1: Irfan buys biscuits for ₹15 and toor dal for ₹56. He pays with ₹100. How much change does he get?

Ans: Irfan spent ₹15 on a biscuit packet and ₹56 on toor dal. 

So, the total cost in rupees is 15 + 56.

 He gave ₹100 to the shopkeeper. So, he should get back 100 minus the total cost.

 Can we write that expression as— 100 – 15 + 56 ? 

Can we first subtract 15 from 100 and then add 56 to the result?

 We will get 141. 

It is absurd that he gets more money than he paid the shopkeeper!

 We can use brackets in this case:

 100 – (15 + 56). 

Evaluating the expression within the brackets first, we get 100 minus 71, which is 29. 

So, Irfan will get back ₹29.

Example 2: Rani went to a stationery shop. She bought a notebook for ₹40 and a pen for ₹25. She gave the shopkeeper ₹100. How much money will she get back?

Step 1: Total cost of the items
₹40 + ₹25 = ₹65

Step 2: Amount given to the shopkeeper
₹100

Step 3: Expression to find the balance
100 – (40 + 25)

Step 4: Solve using brackets
= 100 – 65
= ₹35

Ans: Rani will get back ₹35.

Terms in Expressions

What if an expression has multiple operations but no brackets to specify the order, like 30 + 5 × 4? While brackets are one way to clarify order, another fundamental concept used is that of terms.

Terms are the parts of an expression separated by addition (+) signs.

To identify terms correctly, we first need to handle subtraction. Remember that subtracting a number is the same as adding its inverse (the number with the opposite sign). So, before identifying terms, we convert all subtractions into additions of negative numbers.

Let’s see some examples:

1. Expression:12 + 7
   • Already in addition form.
   • Terms: 12 and 7.
   • Marked: 12 + 7
2. Expression:83 – 14
   • Convert subtraction: 83 + (–14)
   • Terms: 83 and –14.
   • Marked: 83 + (–14)
3. Expression:–18 – 3
   • Convert subtraction: –18 + (–3)
   • Terms: –18 and –3.
   • Marked: –18 + (–3)
4. Expression:6 × 5 + 3
   • Already in addition form.
   • Notice that 6 × 5 does not contain an addition sign separating the 6 and 5. It represents a single value obtained through multiplication.
   • Terms: 6 × 5 and 3.
   • Marked: (6 × 5) + 3
5. Expression:2 – 10 + 4 × 6
   • Convert subtraction: 2 + (–10) + 4 × 6
   • Terms: 2, –10, and 4 × 6.
   • Marked: 2 + (–10) + (4 × 6)

Identifying terms is crucial for the standard order of operations:

1. Evaluate each term first: Perform all multiplications and divisions within each term.

2. Add the resulting values of the terms: Once each term has been simplified to a single number, perform the additions (including the additions that came from converted subtractions).

Let’s re-evaluate 30 + 5 × 4 using terms:

1. Identify terms: 30 and 5 × 4.
2. Evaluate each term: The term 30 is already evaluated. The term 5 × 4 evaluates to 20.
3. Add the term values: 30 + 20 = 50.

This process of identifying and evaluating terms provides a systematic way to handle expressions with multiple operations, even without brackets, ensuring a consistent result.

Swapping and Grouping 

Once we have identified the terms in an expression (after converting all subtractions to additions), does the order in which we add these terms matter?
 Let’s investigate.

Consider the expression 6 – 4. Converting to addition gives 6 + (–4).

• The terms are 6 and –4.
• The value is 6 + (–4) = 2.

What if we swap the terms? –4 + 6.

• The value is –4 + 6 = 2.

The value remains the same! This isn’t just true for positive numbers; it holds even when negative numbers are involved. 
Swapping any two terms in an addition expression does not change the final value.

This property is formally known as the Commutative Property of Addition.

Now, consider an expression with three terms, like (–7) + 10 + (–11).

Let’s try adding them in different groups:

1. Group the first two terms:( (–7) + 10 ) + (–11)= ( 3 ) + (–11)= –8
2. Group the last two terms:(–7) + ( 10 + (–11) )= (–7) + ( –1 )= –8

Again, the value is the same regardless of how we group the terms for addition. This also holds true for expressions with more than three terms and when negative numbers are involved.

This property is formally known as the Associative Property of Addition.

Try yourself:

What is the name of the property that states the grouping of terms does not affect the sum?

• A.Identity Property of Addition
• B.Distributive Property
• C.Commutative Property of Addition
• D.Associative Property of Addition

View Solution

Conclusion: Order Doesn’t Matter for Addition

Because of the commutative and associative properties, when an expression only involves addition (after converting subtractions), we can add the terms in any order or grouping we find convenient, and the result will always be the same.

For example, in (–7) + 10 + (–11), we could add the negative terms first: (–7) + (–11) = –18, and then add the positive term: –18 + 10 = –8.

In mathematics we use the phrase commutative property of addition instead of saying “swapping terms does not change the sum”. Similarly, “grouping does not change the sum” is called the associative property of addition.

Swapping the Order of Things in Everyday Life 

Manasa’s mom says:
“Wear your hat and shoes.”

Manasa can do either:

• Wear her hat first, then her shoes, or
• Wear her shoes first, then her hat.

Result:
Manasa will still be ready to go out and look the same either way.
➡ Here, the order does not matter.

In math, this is like:
2 + 3 = 5 and 3 + 2 = 5
Addition can be done in any order — it won’t change the answer

Order Does Matter

Now her mom says:
“Wear your socks and shoes.”

This time:

• If Manasa wears socks first, then shoes — ✅ that’s the correct way.
• But if she wears shoes first, then tries to wear socks — ❌ it’s wrong and uncomfortable!

Result:
She’ll look odd and it won’t work properly.
➡ Here, the order matters.

In math, this is like:
8 – 5 = 3, but 5 – 8 = -3
So in subtraction, changing the order changes the answer.

More Expressions and Their Terms

Let’s look at a few more examples from the textbook to solidify our understanding of writing expressions and identifying their terms in different scenarios.

Example: Amu, Charan, Madhu, and John went to a hotel and ordered four dosas. Each dosa cost ₹23, and they wish to thank the waiter by tipping ₹5. Write an expression describing the total cost.

• Situation: 4 friends order 4 dosas at ₹23 each and want to leave a ₹5 tip.
• Expression for total cost: The cost of the dosas is 4 × 23. The tip is 5. The total cost is the sum: 4 × 23 + 5.
• Identifying Terms: The expression is already a sum. The terms are 4 × 23 and 5.
• Evaluation:
  1. Evaluate terms: 4 × 23 = 92. The term 5 is already evaluated.
  2. Add term values: 92 + 5 = 97.
• Total Cost: ₹97.

Example : Children in a class are playing “Fire in the mountain, run, run, run!”. Whenever the teacher calls out a number, students are supposed to arrange themselves in groups of that number. Whoever is not part of the announced group size, is out. Ruby wanted to rest and sat on one side. The other 33 students were playing the game in the class. The teacher called out ‘5’. 

• Situation: 33 students are playing. The teacher calls out ‘5’. Students form groups of 5. Ruby observes.
• Observation: Ruby sees 6 full groups of 5, with 3 students left over.
• Expression: Ruby writes 6 × 5 + 3 (representing 6 groups of 5, plus the 3 remaining).
• Identifying Terms: The terms are 6 × 5 and 3.
• Evaluation:(6 × 5) + 3 = 30 + 3 = 33 (the total number of students playing).
• Variations:
  • If the teacher called ‘4’: There would be 8 groups of 4 with 1 left over. Expression: 8 × 4 + 1. Terms: 8 × 4 and 1.
  • If the teacher called ‘7’: There would be 4 groups of 7 with 5 left over. Expression: 4 × 7 + 5. Terms: 4 × 7 and 5.

Example: Raghu bought 100 kg of rice from the wholesale market and packed them into 2 kg packets. He already had four 2 kg packets. Write an expression for the number of 2 kg packets of rice he has now and identify the terms.  

• Situation: Raghu buys 100 kg rice, packs it into 2 kg bags. He already had 4 such bags.
• Expression for total bags: Number of new bags = 100 ÷ 2 (or 100/2). Total bags = 4 + 100 ÷ 2.
• Identifying Terms: The terms are 4 and 100 ÷ 2.
• Evaluation:
  1. Evaluate terms: 4 is evaluated. 100 ÷ 2 = 50.
  2. Add term values: 4 + 50 = 54.
• Total Bags: 54 bags.

Example: Kannan has to pay ₹432 to a shopkeeper using coins of ₹1 and ₹5, and notes of ₹10, ₹20, ₹50 and ₹100. How can he do it?

• Situation: Paying ₹432 using various denominations.
• Possibility 1 Expression:4 × 100 + 1 × 20 + 1 × 10 + 2 × 1
  • Terms:4 × 100, 1 × 20, 1 × 10, 2 × 1.
  • Evaluation:400 + 20 + 10 + 2 = 432.
• Possibility 2 Expression:8 × 50 + 1 × 10 + 4 × 5 + 2 × 1
  • Terms:8 × 50, 1 × 10, 4 × 5, 2 × 1.
  • Evaluation:400 + 10 + 20 + 2 = 432.
• This shows how expressions can represent real-world combinations, and identifying terms helps understand the structure.

Example: Here are two pictures. Which of these two arrangements matches with the expression 5 × 2 + 3?

Which image is correct?

• Expression:5 × 2 + 3
• Identifying Terms:5 × 2 and 3.
• Evaluation: 
• Interpretation: The expression means “3 more than 5 groups of 2”. This matches the picture showing 5 pairs of items plus 3 individual items.This Image is correct

Removing Brackets — I

Let us find the value of this expression, 

200 – (40 + 3). 

We first evaluate the expression inside the bracket to 43 and then subtract it from 200. But it is simpler to first subtract 40 from 200:

 200 – 40 = 160.

 And then subtract 3 from 160: 

160 – 3 = 157. 

What we did here was 200 – 40 – 3. Notice, that we did not do

 200 – 40 + 3. 

So, 

200 – (40 + 3) = 200 – 40 – 3.

Removing Brackets – II

Example: Lhamo and Norbu went to a hotel. Each of them ordered a vegetable cutlet and a rasgulla. A vegetable cutlet costs ₹43 and a rasgulla costs ₹24. Write an expression for the amount they will have to pay.

Situation:
Lhamo and Norbu each buy:

• 1 vegetable cutlet = ₹43
• 1 rasgulla = ₹24

So, the amount one person pays is:
43 + 24

There are two people, so together they pay:
(43 + 24) + (43 + 24)
Instead of writing it this way, we can simplify using brackets and multiplication:
2 × (43 + 24)

Why use brackets?
Brackets tell us to first add the items, and then multiply the total by 2 (for two people).
So,
2 × (43 + 24) = 2 × 67 = ₹134

This is much simpler than adding 43 + 24 twice!

What if another friend, Sangmu  joins?

If a third person, Sangmu, joins and buys the same items, then the expression becomes:
3 × (43 + 24)
That means: 3 × 67 = ₹201

Example: In the Republic Day parade, there are boy scouts and girl guides marching together. The scouts march in 4 rows with 5 scouts in each row. The guides march in 3 rows with 5 guides in each row (see the figure below). How many scouts and guides are marching in this parade?

Boy scouts: 4 rows × 5 boys = 20
Girl guides: 3 rows × 5 girls = 15

So total people = 20 + 15 = 35

But we can do this smarter:
Instead of calculating separately, first add the rows:
(4 + 3) × 5 

Computing these expressions, we get

(4 + 3) × 5 = 7 × 5 = 35

What’s the lesson here?

Using brackets helps us to group numbers and make multiplication or subtraction easier.
For example:

Distributive Property:

• (a + b) × c = a × c + b × c
• (a – b) × c = a × c – b × c

Example:
(10 + 3) × 98 = 10 × 98 + 3 × 98 = 13 × 98

This makes solving faster and more organized.

The multiple of a sum (or difference) = sum (or difference) of the multiples.

Solved Examples

Example 1:  Evaluate 30 + 5 × 4

Expression:30 + 5 × 4Identify Terms: The terms are 30 and 5 × 4.
Evaluate Terms:

• 30 is already evaluated.
• 5 × 4 = 20.

Add Term Values:30 + 20 = 50.
Answer: 50

Example 2: Evaluate 100 – (15 + 56)

Expression:100 – (15 + 56)

Evaluate Inside Brackets First:15 + 56 = 71.

Perform Remaining Operation:100 – 71 = 29.

Answer: 29

Example 3: Evaluate 4 × 23 + 5

Expression:4 × 23 + 5

Identify Terms:4 × 23 and 5.

Evaluate Terms:

• 4 × 23 = 92.
• 5 is already evaluated.

Add Term Values:92 + 5 = 97.

Answer: 97