You must have observed your teacher recording the attendance of students in your class every, or recording marks obtained by you after every test or examination. Similarly, you must have also seen a cricket scoreboard. Two score boars have been illustrated here.
You know that in a game of cricket, the information recorded is not simply about who won and who lost. In the scoreboard, you will also find some equally important information about the game. For instance, you may find out the time taken and some balls faced by the highest run-scorer.
Similarly, in your day-to-day life, you must have seen several kinds of tables consisting of numbers, figures, names, etc.
These tables provide ‘Data’. Data is a collection of numbers gathered to give some information.
Let us take an example of a class which is preparing to go for a picnic. The teacher asked the students to give their choice of fruits out of banana, apple, orange, or guava. Uma is asked to prepare the list.
She prepared a list of all the children and wrote the choice of fruit against each name. This list would help the teacher to distribute fruits according to the choice.
If the teacher wants to know the number of bananas required for the class, she has to read the names in the list one by one and count the total number of bananas required.
To know the number of apples, guavas, and oranges separately she has to repeat the same process for each of these fruits. How tedious and time-consuming it is! It might become more tedious if the list has, say, 50 students.
So, Uma writes only the names of these fruits one by one like banana, apple, guava, orange, apple, banana, orange, guava, banana, banana, apple, banana, apple, banana, orange, guava, apple, banana, guava, banana.
Do you think this makes the teacher’s work easier? She still has to count the fruits in the list one by one as she did earlier.
Salma has another idea. She makes four squares on the floor. Every square is kept for fruit of one kind only. She asks the students to put one pebble in the square that matches their choices, i.e. a student opting for banana will put a square marked for banana and so on.
By counting the pebbles in each square, Salma can quickly tell the number of each kind of fruit required. She can get the required information quickly by systematically placing the pieces in different squares.
Try to perform this activity for 40 students with the names of any four fruits. Instead of pebbles, you can also use bottle caps or some other tokens.
To get the same information that Salma got, Ronald needs only a pen and a piece of paper. He does not need pebbles. He also does not ask to come and place the pebbles. He prepares the following table.
Do you understand Ronald’s table?
What does one (✓) mark indicate?
Four students preferred guava. How many (✓) marks are there against guava? How many students were there in the class? Find all this information. Discuss these methods. Which is the best? Why? Which method is more useful when information from much larger data is required?
Example 1. A teacher wants to know the choice of food of each student as part of the mid-day meal program. The teacher assigns the task of collecting this information to Maria. Maria does so using paper and a pencil. After arranging the choices in a column, she puts against a choice of food one (I) mark for every student making that choice.
Solution: Umesh, after seeing the table suggested a better method to count the students. He asked Maria to organize the marks (l) in a group of ten as shown below:
Rajan made it simpler by asking her to make groups of five instead of ten, as shown below:
The teacher suggested that the fifth mark in a group of five marks should be used as a cross, as shown by 
These are tally marks. Thus, II shows the count to be five plus two (i.e. seven) and shows five plus five (i.e. ten).
With this, the table looks like:
Example 2. Ekta is asked to collect data for the size of the shoes of students in her Class 6. Her findings are recorded in the manner shown below:
Solution:
Javed wanted to know
Ekta prepared a table using tally marks.
Now the questions asked earlier could be answered easily.
You may also do some such activity in your class using tally marks.
A cupboard has five compartments. In each compartment, a row of books is arranged.
The details are indicated in the adjoining table
Which row has the greatest number of books? Which row has the least number of books? Is there any row that does not have books?
You can answer these questions by just studying the diagram. The picture visually helps you to understand the data. It is a pictograph.
A pictograph represents data through pictures of objects. It helps answer the questions on the data at a glance.
Pictographs are often used by dailies and magazines to attract readers’ attention.
Example 1. The following pictograph shows the number of absentees in a class of 30 students during the previous week:
Solution:
Example 2. The colors of fridges preferred by people living in a locality are shown by the following pictograph.
Solution:
(1) Blue color is preferred by 50 people.
[
= 10, so 5 pictures indicate 5 x 10 people].
(2) Deciding the number of people like red color needs more care.
For 5 complete pictures, we get 5 x 10 = 50 people.
For the last incomplete picture, we may roughly take it as 5.
So, number of people preferring red color is nearly 55.
In the above example, the number of people who like the red color was taken as 50 + 5. If your friend wishes to take it as 50 + 8, is it acceptable?
Example 3. A survey was carried out on 30 students of class 6 in a school. Data about the different modes of transport used by them to travel to school was displayed as pictographs. What can you conclude from the pictograph?
Solution: From the pictograph, we find that:
Geometry has a long and rich history. The term ‘Geometry’ is the English equivalent of the Greek word ‘Geometron’. ‘Geo’ means Earth and ‘metron’ means Measurement.
According to historians, geometrical ideas were shaped in ancient times, probably due to the need for art, architecture, and measurement.
These include occasions when the boundaries of cultivated lands had to be marked without giving room for complaints. Construction of magnificent palaces, temples, lakes, dams, and cities, art, and architecture propped up these ideas.
Even today geometrical ideas are reflected in all forms of art, measurements, architecture, engineering, cloth designing, etc. You observe and use different objects like boxes, tables, books, the tiffin box you carry to your school for lunch, the ball with which you play, and so on.
All such objects have different shapes. The ruler which you use, and the pencil with which you write are straight. The pictures of a bangle, the one rupee coin, or a ball appear round.
Here, you will learn some interesting facts that will help you know more about the shapes around you.
With the sharp tip of the pencil, mark a dot on the paper. The sharper the tip, the thinner will be the dot. This almost invisible tiny dot will give you an idea of a point.
A point determines a location.
These are some models for a point:
If you mark three points on a paper, you will be required to distinguish them. For this, they are denoted by a single capital letter like A, B, C.
These points will be read as point A, point B, and point C.
Of course, the dots have to be invisibly thin.
Fold a piece of paper and unfold it. Do you see a fold? This gives the idea of a line segment. It has two end points A and B.
Take a thin thread. Hold its two ends and stretch it without a slack. It represents a line segment. The ends held by hands are the end B points of the line segment.
The following are some models for a line segment:
Try to find more examples of line segments from your surroundings.
Mark any two points A and B on a sheet of paper. Try to connect A to B by all possible routes.
What is the shortest route from A to B?
This shortest join of points A to B (including A and B) shown here is a line segment. It is denoted by \(\overrightarrow{A B}\) or \(\overrightarrow{B A}\) . Points A and B are called the endpoints of the segment.
Imagine that the line segment from A to B (i.e. \(\overrightarrow{A B}\) ) is extended beyond A in one direction and beyond B in the other direction without any end (see figure). You now get a model for a line.
Do you think you can draw a complete picture of a line? No. (Why?)
A line through two points A and B is written as \(\overrightarrow{A B}\)• It extends indefinitely in both directions. So it contains a countless number of points. (Think about this).
Two points are enough to fix a line. We say ‘Two points determine a line’.
The adjacent diagram is that of a line PQ written as \(\overrightarrow{P Q}\). Sometimes a line is denoted by a letter like l, m.
Look at the diagram. Two lines l1 and l2 are shown. Both the lines pass through point P.
We say l1 and l2 intersect at P. If two lines have one common point, they are called intersecting lines.
The following are some models of a pair of intersecting lines:
Try to find some more models for a pair of intersecting lines.
Let us look at this table. The top ABCD is flat. Are you able to see some points and line segments?
Are there intersecting line segments?
Yes, AS and BQ intersect at the point B.
Which fine segments intersect at A? at C? or at D?
Do the lines AD and CD intersect?
Do the lines \(\overrightarrow{A D}\) and \(\overrightarrow{B C}\) intersect?
You find that on the table’s surface, there are line segments that will not meet, however far they are extended. \(\overrightarrow{A D}\) and \(\overrightarrow{B C}\) form one such pair. Can you identify one more such pair of lines (which do not meet) on the top of the table?
Lines like these that do not meet are said to be parallel and are called parallel lines.
The following are some models of a ray:
A ray is a portion of a line. It starts at one point (called the starting point or initial point) and goes endlessly in a direction.
Look at the diagram of the ray shown here. Two points are shown on the ray.
They are
We denote it by \(\overrightarrow{\mathrm{AP}}\)
You can draw some of these drawings without lifting the pencil from the paper and without die use of a ruler. These are all curves.
‘Curve’ in everyday usage means “not straight’’. In Mathematics, a curve can be straight like the one shown.
Observe that curves (3) and (4) in Figure cross themselves, whereas curves (1), (2), (5), and (6) do not. If a curve does not cross itself, then it is called a simple curve.
Draw five more simple curves and five curves that are not simple.
Consider these now.
What is the difference between these two? The first i.e. (1) is an open curve and the second i.e. (2) is a closed curve. Can you identify some closed and open curves from the above? Draw five curves each that are open and closed.
Position in a figure
A court line in a tennis court is divided into three parts: inside the line, on the line, and outside the line. You cannot enter inside without crossing the line.
A compound wall separates your house from the road. You talk about ‘inside’ the compound, ‘on’ the boundary of the compound, and ‘outside’ the compound.
In a closed curve, thus, there are three parts.
The below is in the interior, C is in the exterior and B is on the curve.
The interior of a curve together with its boundary is called its “region”.
Look at these below (1), (2), (3), (4), (5) and (6).
What can you say? Are they closed? How does each one of them differ from die other? (1), (2), (3), (4), and (5) are special because they are made up entirely of line segments.
Out of these (1), (2), (3), and (4) are also simple closed curves. They are called polygons.
So, a figure is a polygon if it is a simple closed figure made up entirely of line segments. Draw ten differently shaped polygons.
Sides, vertices, and diagonals
Examine the figure given here. Give justification to call it a polygon.
The line segments forming a polygon are called its sides.
What are the sides of polygon ABCDE? (Note how the comers are named in order.)
Sides are \(\overline{\mathrm{AB}}\), \(\overline{\mathrm{BC}}\), \(\overline{\mathrm{CD}}\), \(\overline{\mathrm{DE}} \text { and } \overline{\mathrm{EA}}\).
The meeting point of a pair of sides is called its vertex.
Sides \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{ED}}\) meet at E, so E is a vertex of the polygon ABCDE. Points B and C are its other vertices. Can you name the sides that meet at these points? Can you name the other vertices of the above polygon ABCDE?
Any two sides with a common endpoint are called the adjacent sides of the polygon.
Are the sides \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{BC}}\) adjacent? How about \(\overline{\mathrm{AE}}\) and \(\overline{\mathrm{DC}}\)? The endpoints of the same side of a polygon are called the adjacent vertices.
Vertices E and D are adjacent, whereas vertices A and D are not adjacent vertices. Do you see why?
Consider the pairs of vertices which are not adjacent. The joins of these vertices are called the diagonals of the polygon.
In the given \(\overline{\mathrm{AC}}, \overline{\mathrm{AD}}, \overline{\mathrm{BD}}, \overline{\mathrm{BE}} \text { and } \overline{\mathrm{CE}}\) are diagonals. Is \(\overline{\mathrm{BC}}\) a diagonal, Why or why not?
If you try to join adjacent vertices, will the result be a diagonal?
Name all the aides, adjacent aides, and adjacent vertices of the figure ABCDE.
Draw q polygon ABCDEFGH and name all the aides, adjacent sides, and vertices as well as the diagonals of the polygon.
Angles are made when comers are formed.
Here is a picture where the top of a box is like a hinged lid. The edges AD of the box and AP of the door can be imagined as two rays \(\overrightarrow{\mathrm{AD}}\) and \(\overrightarrow{\mathrm{AD}}\).
These two rays have a common initial point A. The two rays here together are said to form an angle.
An angle is made up of two rays starting from a common initial point. The two rays forming the angle are called the arms or sides of the angle. The common initial point is the vertex of the angle.
This is an angle formed by rays \(\overrightarrow{\mathrm{OP}} \text { and } \overline{\mathrm{OQ}}\). To show this we use a small curve at the vertex. 0 is the vertex. What are the sides? Are they not \(\overrightarrow{\mathrm{OP}} \text { and } \overline{\mathrm{OQ}}\)?
How can we name this angle? We can simply say that is an angle at O. To be more specific we identify two points, one on each side and the vertex to; tame the angle. Angle POQ is thus a better way of naming the angle. We denote this by ∠POQ.
Take any angle, say ∠ABC.
Shade that portion of the paper bordering \(\overrightarrow{\mathrm{BA}}\) and where \(\overrightarrow{\mathrm{BC}}\) lies.
Now shade in a different color the portion of the paper bordering \(\overrightarrow{\mathrm{BC}}\) and where \(\overrightarrow{\mathrm{AB}}\) lies.
The portion common to both shadings is called the interior of ABC. (Note that the interior is not a restricted area; it extends indefinitely since the two sides extend indefinitely).
In this diagram, X is in the interior of the angle, Z is not in the interior but in the exterior of the angle, and S is on the ∠PQR. Thus, the angle also has three parts associated with it.
Understanding Elementary Shapes Introduction
All the shapes we see around us are formed using curves or lines. We can see comers, edges, planes, open curves, and closed curves in our surroundings.
We organize them into line segments, angles, triangles, polygons, and circles. We find that they have different sizes and measures. Let us now try to develop tools to compare their sizes.
We have drawn and seen so many line segments. A triangle is made of three, and a quadrilateral of four line segments. A line segment is a fixed portion of a line.
This makes it possible to measure a line segment. This measure of each line segment is a unique number called its “length”. We use this idea to compare line segments.
To compare any two line segments, we find a relation between their lengths. This can be done in several ways.
1. Comparison by observation:
By just looking at them can you tell which one is longer?
You can see that \(\overline{\mathrm{AB}}\) is longer.
But you cannot always be sure about your usual judgment.
For example, look at the adjoining segments:
The difference in lengths between these two may not be obvious. This makes other ways of comparing necessary.
In this adjacent figure, \(\overline{\mathrm{AB}} \text { and } \overline{\mathrm{PQ}}\) have the same lengths. This is not quite obvious.
So, we need better methods of comparing line A segments.
2. comparison by tracing
To compare \(\overline{\mathrm{AB}} \text { and } \overline{\mathrm{CD}}\), we use a tracing paper, trace \(\overline{\mathrm{CD}}\), and place the traced segment on AB.
Can you decide now which one among \(\overline{\mathrm{AB}} \text { and } \overline{\mathrm{CD}}\), is longer?
The method depends upon the accuracy of tracing the line segment. Moreover, if you want to compare with another length, you have to trace another line segment.
This is difficult and you cannot trace the lengths every time you want to compare them.
3. Comparison Using Ruler and a Divider
Have you seen or can you recognize all the instruments in your instrument box? Among other things, you have a miler and a divider.
Note how the meter is marked along one of its edges. It is divided into 15 parts. Each of these 15 parts is 1cm in length. Each centimeter is divided into subparts. Each subpart of the division of a cm is 1mm.
How many millimeters make one centimeter? Since 1cm = 10 mm, how will we write 2 cm? 3mm? What do we mean by 7.7 cm?
Place the zero mark of the meter at A.
Read the mark against. This gives the length of \(\overline{\mathrm{AB}}\). Suppose the length is 5.8 cm, we may write, Length AB = 5.8 cm or more simply as AB = 5.8 cm.
There is room for errors even in this procedure. The thickness of the meter may cause difficulties in reading off the marks on it.
Let us use the divider to measure length.
Open the divider. Place the end point of one of its arms at A and the end point of the second arm at B.
Taking care that the opening ofthe divider is not disturbed, lift the divider and place it on the ruler. Ensure that one endpoint is at the zero mark of the ruler. Now read the mark against the other endpoint.
You have heard of directions in Geography. We know that China is to the north of India, and Sri Lanka is to the south. We also know that the Sun rises in the east and sets in the west. There are four main directions.
They are North (N), South (S), East (E) and West (W). Do you know which direction is opposite to the north? Which direction is opposite to the west? Just recollect what you know already.
We now use this knowledge to learn a few properties about angles. Stand facing north.
Turn clockwise to the east.
We say you have turned through a right angle. Follow this with a ‘right-angle-tum’, clockwise.
You now face south. If you turn by a right angle in the anti-clockwise direction, which direction will you face? It is east again! (Why?) Study the following positions
From facing north to facing south, you have turned by two right angles. Is not this the same as a single turn by two right angles? The turn from north to east is at a right angle.
The turn from north to south is by two right angles; it is called a straight angle. (NS is a straight line) Stand facing south.
Turn by a straight angle. Which direction do you face now? You face north! To turn from north to south, you took a straight-angle turn, again to turn from south to north.
You took another straight-angle turn in the same direction. Thus, turning by two straight angles you reach your original position.
By how many right, angles should you turn in the same direction to reach your original position? Turning by two straight angles (or four right angles) in the same direction makes a full turn.
This one complete turn is called one revolution. The angle for one revolution is a complete angle. We can see such revolutions on clock faces.
When the hand of a clock moves from one position to another, it turns through an angle.
Suppose the hand of a clock starts at 12 and goes round until it reaches 12 again. Has it not made one revolution? So, how many right angles has it moved? Consider these examples
We saw what we mean by a right angle and a straight angle. However, not all the angles we come across are one of these two kinds.
The angle made by a ladder with the wall (or with the floor) is neither a right angle nor a straight angle.
Are there angles smaller than a right angle? Are there angles greater than a right angle? Have you seen a carpenter’s square? It looks like the letter “L” of the English alphabet.
He uses it to check right angles. Let us also make a similar ‘tester’ for a right angle.
Other Names
1. An angle smaller than a right angle is called an acute angle. These are acute angles
Do you see that each one of them is less than one-fourth of a revolution? Examine them with your RA tester.
2. If an angle is larger than a right angle but less than a straight angle, it is called an obtuse angle. These are obtuse angles.
Do you see that each one of them is greater than one-fourth of a revolution but less than half a revolution? Your RA tester may help to examine.
Identify the obtuse angles in the previous examples too.
3. A reflex angle is larger than a straight angle. It looks like this. (See the angle mark) Were there any reflex angles in the shapes you made earlier? How would you check for them?
The improvised ‘Right-angle tester’ we made is helpful to compare angles with a right angle. We were able to classify the angles as acute, obtuse, or reflex.
But this does not give a precise comparison. It cannot find which one of the two obtuse angles is greater. So in order to be more precise in comparison, we need to ‘measure’ the angles. We can do it with a ‘protractor’.
The measure of angle
We call our measure, ‘degree measure’. One complete revolution is divided into 360 equal parts. Each part is a degree. We write 360° to say ‘three hundred sixty degrees’.
The Protractor
You can find a readymade protractor in your ‘instrument box’. The curved edge is divided into 180 equal parts.
Each part is equal to a ‘degree’. The markings start from 0° on the right side and end at 180° on the left side, and vice versa.
Suppose you want to measure an angle ABC.
We write m ∠ABC= 40°, or simply ∠ABC= 40°.
When two lines intersect and the angle between them is a right angle, then the lines are said to be perpendicular. If a line AB is perpendicular to CD, we write \(A B \perp C D \text {. }\) .
If \(A B \perp C D\), then should we say that \(\mathrm{CD} \perp \mathrm{AB}\) also?
Perpendiculars around us!
You can give plenty of examples from things around you for perpendicular lines (or line segments). The English alphabet T is one. Is there any other alphabet that illustrates perpendicularity?
Consider the edges of a postcard. Are the edges perpendicular?
Let \(\overline{\mathrm{AB}}\) be a line segment. Mark its midpoint as M. Let MN be a line perpendicular to \(\overline{\mathrm{AB}}\) through M.
Does MN divide \(\overline{\mathrm{AB}}\) into two equal parts? MN bisects \(\overline{\mathrm{AB}}\) (that is, divides \(\overline{\mathrm{AB}}\) into two equal parts) and is also perpendicular to \(\overline{\mathrm{AB}}\).
So we say MN is the perpendicular bisector of \(\overline{\mathrm{AB}}\). You will learn to construct it later.
Quadrilaterals
A quadrilateral, if you remember, is a polygon that has four sides.
So far you studied polygons of 3 o r4 sides (known as triangles and quadrilaterals respectively). We now try to extend the idea of polygon to figures with more sides. We may classify polygons according to the number of their sides.
image-
You can find many of these shapes in everyday life. Windows, doors, walls, almirahs, blackboards, notebooks are all usually rectanglular in shape. Floor tiles are rectangles. The sturdy nature of a triangle makes it the most useful shape in engineering constructions.
Our study so far has been with numbers and shapes. We have learned numbers, operations on numbers, and properties of numbers. We applied our knowledge of numbers to various problems in our life.
The branch of mathematics in which we studied numbers is arithmetic. We have also learned about figures in two and three dimensions and their properties.
The branch of mathematics in which we studied shapes is geometry. Now we begin the study of another branch of mathematics. It is called algebra.
The main feature of the new branch which we are going to study is the use of letters. The use of letters will allow us to write rules and formulas in a general way. By using letters, we can talk about any number and not just a particular number.
Secondly, letters may stand for unknown quantities. By learning methods of determining unknowns, we develop powerful tools for solving puzzles and many problems from daily life.
Thirdly, since letters stand for numbers, operations can be performed on them as on numbers. This leads to the study of algebraic expressions and their properties.
You will find algebra interesting and useful. It is very useful in solving problems. Let us begin our study with simple examples.
Ameena and Sarita are making patterns with matchsticks. They decide to make simple patterns of the letters of the English alphabet. Ameena takes two matchsticks and forms the letter L as shown.
Then Sarita also picks two sticks, forms another letter L, and puts it next to the one made by Ameena.
Then Ameena adds one more L and this goes on as shown by the dots.
Their friend Appu comes in. He looks at the pattern. Appu always asks questions. He asks the girls, “How many matchsticks will be required to make seven Ls”?
Ameena and Sarita are systematic. They go on forming the patterns with 1L, 2Ls, 3Ls, and so on, and prepare a table.
Appu gets the answer to his question from the Table, 7Ls require 14 matchsticks.
While writing the table, Ameena realizes that the number of matchsticks required is twice the number of Ls formed.
Number of matchsticks required = 2 x number of Ls.
For convenience, let us write the letter n for the number of Ls.
If one L is made, n = 1; if two Ls are made, n = 2, and so on; thus, n can be any natural number 1,2,3,4,5,….
We then write, Number of matchsticks required = 2 x n.
Instead of writing 2 x n, we write 2n.
Note that 2 n is the same as 2 x n.
Ameena tells her friends that her rule gives the number of matchsticks required for forming any number of Ls.
Thus, For n = 1, the number of matchsticks required = 2×1=2
For n = 2, the number of matchsticks required = 2×2 = 4
For = 3, the number of matchsticks required = 2×3 = 6, etc.
These numbers agree with those from Table.
Santa says, “Tito rule is very powerful! Using the rule, I can say how litany matchsticks are required to form even 100 Ls. I do not need to draw the pattern or make a table, once the rule is known”.
In the above example, we found a rule to give the number of matchsticks required to make a pattern of Ls. The rule was:
Number of matchsticks required = 2n
Here, n is the number of Ls in the pattern, and n takes values 1, 2,3,4,…
Let us look at Table once again.
In the table, the value of n goes on changing (increasing). As a result, the number of matchsticks required also changed (increasing).
n is an example of a variable. Its value is not fixed; it can take any value 1, 2,3,4,… We wrote the rule for the number of matchsticks required using the variable n.
The word ‘variable’ means something that can vary, i.e. change. The value of a variable is not fixed. It can take different values.
We shall look at another example of matchstick patterns to learn more about variables.
Ameena and Sarita have become quite interested in matchstick patterns. They now want to try a pattern of the letter C. To make one C, they use three matchsticks as shown.
The table gives the number of matchsticks required to make a pattern of Cs.
Santa conies up with the rule:
Number of matchsticks required = 3n
She has used the letter n for the number of Cs; is a variable taking on values 1, 2, 3,4,…
Do you agree with Sarita?
Remember 3n is the same as 3 x n.
Next, Ameena and Sarita wish to make a pattern of Fs.
They make one F using 4 matchsticks as shown.
Can you now write the rule for making patterns of F?
Think of other letters of the alphabet and other shapes that can be made from matchsticks. For example, U (U), V (V), triangle (A), square (□), etc.
We have used the letter n to show a variable. Raju asks, “Why not m”?
There is nothing special about n, any letter can be used.
One may use any letter, such as m, l, p, x, y, z, etc. to show a variable. Remember, a variable is a number which does not have a fixed value.
For example, the number 5 the number 100, or any other given number is not a variable.
They have fixed values. Similarly, the number of angles of a triangle has a fixed value i.e. 3. It is not a variable. The number of corners of a quadrilateral (4) is fixed; it is also not a variable.
But n in the examples we have looked at is a variable. It takes on various values 1,2,3,4,…
Let us now consider variables in a more familiar situation.
Students went to buy notebooks from the school bookstore. The price of one notebook is ₹5. Munnu wants to buy 5 notebooks, Appu wants to buy 7 notebooks, Sara wants to buy 4 notebooks, and so on.
How much money should a student carry when she or he goes to the bookstore to buy notebooks?
This will depend on how many notebooks the student wants to buy. The students work together to prepare a table.
The letter m stands for the number of notebooks a student wants to buy; m is a variable, which can take any value 1, 2, 3, 4, …..
The total cost of m notebooks is given by the rule:
The total cost in rupees = 5 x number of notebooks required = 5m
If Munnu wants to buy 5 notebooks, then taking m = 5, we say that Munnu should carry ₹5 x 5 or ₹ 25 with him to the school bookstore.
Let us take one more example. For the Republic Day celebration in the school, children are going to perform mass drills in the presence of the chief guest. They stand 10 in a row. How many children can there be in the drill?
The number of children will depend on the number of rows. If there is 1 row, there will be 10 children. If there are 2 rows, there will be 2 x 10 or 20 children, and so on.
If there are r rows, there will be 1 Or children in the drill; here, r is a variable that stands for the number of rows and so takes on values 1,2,3,4…..
In all the examples seen so far, the variable was multiplied by a number. There can be different situations as well in which numbers are added to or subtracted from the variable as seen below.
Sarita says that she has 10 more marbles in her collection than Ameena. If Ameena has 20 marbles, then Sarita has 30. If Ameena has 30 marbles, then Sarita has 40, and so on.
We do not know exactly how many marbles Ameena has. She may have any number of marbles.
But we know that, Sarita’s marbles = Ameena’s marbles + 10.
We shall denote Ameena’s marbles by the letter x. Here, x is a variable, which can take any value 1, 2, 3, 4,…,10,…,20,…,30,…….
Using x, we write Sarita’s marbles = x + 10. The expression (x + 10) is read as ‘x plus ten’.
It means 10 added to x. If x is 20, (x + 10) is 30. If x is 30, (x + 10) is 40 and so on.
The expression (x + 10) cannot be simplified further.
Do not confuse x + 10 with 10x, they are different.
In 10x, x is multiplied by 10. In (x + 10), 10 is added to x. We may check this for some values of x.
For example,
If x = 2, 10x = 10 x 2 = 20 and x + 10 = 2 + 10 = 12.
If x= 10, 10x= 10 x 10= 100 and x + 10= 10+ 10 = 20.
Raju and Balu are brothers. Balu is younger than Raju by 3 years. When Raju is 12 years old, Balu is 9 years old. When Raju is 15 years old, Balu is 12 years old.
We do not know Raju’s age exactly. It may have any value. Let x denote Raju’s age in years, x is a variable.
If Raju’s age in years is x, then Balu’s age in years is (x – 3). The expression (x – 3) is read as x minus three. As you would expect, when x is 12, (x – 3) is 9, and when x is 15, (x – 3) is 12.
In our daily life, many times we compare two quantities of the same type. For example, Avnee and Shari collected flowers for scrap notebooks. Avnee collected 30 flowers and Shari collected 45 flowers.
So, we may say that Shari collected 45 – 30 = 15 flowers more than Avnee.
Also; if the height of Rahim is 150 cm and that of Avnee is l40 cm then, we may say that the height of Rahim is 150 cm – 140 cm = 10 cm more than Avnee. This is one way of comparison by taking differences.
If we wish to compare the lengths of an ant and a grasshopper, taking the difference does not express the comparison. The grasshopper’s length, typically 4 cm to 5 cm is too long as compared to the ant’s length which is a few mm.
Comparison will be better if we try to find how many ants can be placed one behind the other to match the length of the grasshopper. So, we can say that 20 to 30 ants have the same length as a grasshopper.
Consider another example.
The cost of a car is ₹ 2,50,000 and that of a motorbike is ₹ 50,000. If we calculate the difference between the costs, it is ₹ 2,00,000 and if we compare by division; ie.., \(\frac{2,50,000}{50,000}=\frac{5}{1}\)
We can say that the cost of the car is five times the cost of the motorbike.
Thus, in certain situations, comparison by division makes better sense than comparison by taking the difference. The comparison by division is the Ratio. In the next section, we shall learn more about ‘Ratios’.
Consider The Following:
Isha’s weight is 25 kg and her father’s weight is 75 kg. How many times Father’s weight is of Isha’s weight? It is three times.
The cost of a pen is ₹ 10 and the cost of a pencil is ₹ 2. How many times the cost of a pen is that of a pencil? Obviously, it is five times.
In the above examples, we compared the two quantities in terms of ‘how many times’. This comparison is known as the Ratio. We denote the ratio using the symbol ‘:’
Consider the earlier examples again. We can say,
The ratio of father’s weight to Isha’s weight = \(\frac{75}{25}=\frac{3}{1}=3: 1\)
The ratio of the cost of a pen to the cost of a pencil = \(\frac{10}{2}=\frac{5}{1}=5: 1\)
Let us look at this problem.
In a class, there are 20 boys and 40 girls. What is the ratio of
First, we need to find the total number of students, which is,
Number of girls + Number of boys = 20 + 40 = 60.
Then, the ratio of the number of girls to the total number of students is \(\frac{40}{60}=\frac{2}{3}=2: 3\)
Find the answer to part (2) in a similar manner.
Now consider the following example.
The length of a house lizard is 20 cm and the length of a crocodile is 4 m.
“I am 5 times bigger thank you”. says the lizard. As we can see this is really absurd. Alizard’s length cannot be 5 times the length of a crocodile.
So, what is wrong? Observe that the length of the lizard is in centimetres and the length of the crocodile is in metres. So, we have to cover the same unit.
Length of the crocodile = 4 m = 4 x 100 = 400 cm.
Therefore, the ratio of the length of the crocodile to the length of the lizard = \(\frac{400}{20}=\frac{20}{1}=20: 1.\)
Two quantities can be compared only if they are in the same unit.
Now what is the ratio of the length of the lizard to the length of the crocodile?
It is \(\frac{20}{400}=\frac{1}{20}=1: 20.\)
Observe that the two ratios 1:20 and 20:1 are different from each other.
The ratio 1: 20 is the ratio of the length of the lizard to the length of the crocodile whereas, 20: 1 is the ratio of the length of the crocodile to the length of the lizard.
Now consider another example.
The length of a pencil is 18 cm and its diameter is 8 mm.
What is the ratio of the diameter of the pencil to that of its length?
Since the length and the diameter of the pencil are given in different units, we first need to convert them into the same unit.
Thus, the length of the pencil = 18 cm = 18 x 10 mm = 180 mm.
The ratio of the diameter of the pencil to that of the length of the pencil = \(\frac{8}{180}=\frac{2}{45}=20: 45.\)
Think of some more situations where you compare two quantities of the same type in different units.
We use the concept of ratio in many situations of our daily life without realising that we do so.
Compare the drawings A and B. B looks more natural than A. Why?
The legs in picture A are too long in comparison to the other body parts. This is because we normally expect a certain ratio of the length of the legs to the length of the whole body.
Compare the two pictures of a pencil. Is the first one looking like a full pencil? No.
Why not? The reason is that the thickness and the length of the pencil are not in the correct ratio.
Same Ratio In Different Situations:
Consider the following:
Then, the ratio of marbles that Sheena and Shabnam have is 2:3.
Can you write some more situations for this ratio? Give any ratio to your friends and ask them to frame situations.
Ravi and Rani started a business and invested money in the ratio 2:3. After one year the total profit was ₹ 4,00,000.
Ravi said “We would divide it equally”, Rani said, “I should get more as I have invested more”.
It was then decided that profit would be divided by the ratio of their investment.
Here, the two terms of the ratio 2 : 3 are 2 and 3.
Some of these terms = 2 + 3 = 5 What does this mean?
This means if the profit is ₹ 5 then Ravi should get ₹ 2 and Rani should get ₹ 3.
Or, we can say that Ravi gets 2 parts and Rani gets 3 parts out of the 5 parts.
i. e., Ravi should get \(\frac{2}{5}[latex] of the total profit and Rani should get [latex]\frac{3}{5}[latex] of the total profit.
If the total profit were ₹ 500
Ravi would get ₹ [latex]\frac{2}{5}[latex] x 500 = ₹ 200
and Rani would get [latex]\frac{3}{5}[latex] x 500 = ₹ 300
Now, if the profit were ₹ 4,00,000 could you find the share of each?
Ravi’s share = ₹ [latex]\frac{2}{5}[latex] x 4,00,000 = ₹ 1,60,000
And Rani’s share = ₹ [latex]\frac{3}{5}[latex] x 4,00,000 = ₹ 2,40,000
Can you think of some more examples where you have to divide a number of things in some ratio? Frame three such examples and ask your friends to solve them.
Let us look at the kind of problems we have solved so far.
Example 1. The length and breadth of a rectangular field are 50 m and 15 m respectively. Find the ratio of the length to the breadth of the field.
Solution: Length of the rectangular field = 50 m
The breadth of the rectangular field = 15 m
The ratio of the length to the breadth is 50: 15
The ratio can be written as [latex]\frac{50}{15}=\frac{50 \div 5}{15 \div 5}=\frac{10}{3}=10: 3\)
Thus, the required ratio is 10 : 3.
Example 2. Find the ratio of 90 cm to 1.5 m.
Solution: The two quantities are not in the same units. Therefore, we have to convert them into the same units.
1.5 m = 1.5 x 100cm = 150cm.
Therefore, the required ratio is 90: 150.
= \(\frac{90}{150}=\frac{90 \times 30}{150 \times 30}=\frac{3}{5}\)
The required ratio is 3: 5.
Example 3. There are 45 people working in the office. If the number of females is 25 and the remaining are males, find the ratio of:
Solution: Number of females = 25
Total number of workers = 45
Number of males = 45 – 25 = 20
Therefore, the ratio of the number of females to the number of males = 25: 20 = 5: 4
The ratio of a number of males to the number of females = 20:25 = 4:5.
(Notice that there is a difference between the two ratios 5: 4 and 4: 5).
Example 4. Give two equivalent ratios of 6: 4.
Solution: Ratio \(6: 4=\frac{6}{4}=\frac{6 \times 2}{4 \times 2}=\frac{12}{8} \text {. }\)
Therefore, 12: 8 is an equivalent ratio of 6: 4
Similarly, the ratio \(6: 4=\frac{6}{4}=\frac{6 \times 2}{4 \times 2}=\frac{3}{2}\)
So, 3:2 is another equivalent ratio of 6: 4.
Therefore, we can get equivalent ratios by multiplying or dividing the numerator and denominator by the same number.
Write two more equivalent ratios of 6: 4.
Example 5. Fill in the missing numbers: \(\frac{14}{21}=\frac{\square}{3}=\frac{6}{\square}\)
Solution: In order to get the first missing number, we consider the fact that 21=3×7.
i.e. when we divide 21 by 7 we get 3.
This indicates that to get the missing number of the second ratio, 14 must also be divided by 7
When we divide, we have, 14 ÷ 7 = 2
Hence, the second ratio is \(\frac{2}{3}\).
Similarly, to get the third ratio we multiply both terms of the second ratio by 3. (Why?)
Hence, the third ratio is \(\frac{6}{9}\).
Therefore, \(\frac{14}{21}=\frac{2}{3}=\frac{6}{9}\) [These are all equivalent ratios.]
Example 6. The ratio of the distance of the school from Mary’s home to the distance of the school from John’s home is 2: 1.
(1)Who lives nearer to the school?
(2)Complete the following table which shows some possible distances that Mary and John could live from the school
(3)If the ratio of the distance of Mary’s home to the distance of Kalam’s home from school is 1: 2, then who lives nearer to the school?
Solution:
(1) John lives nearer to the school (As the ratio is 2: 1).
(2)
(3) Since the ratio is 1: 2, so Mary lives nearer to the school.
Example 7. Divide ₹ 60 in the ratio 1: 2 between Kriti and Kiran.
Solution: The two parts are 1 and 2.
Therefore, the sum of the parts =1+2 = 3.
This means if there are ₹ 3, Kriti will get ₹ 1 and Kiran will get ₹ 2.
Or, we can say that Kriti gets 1 part and Kiran gets 2 parts out of every 3 parts.
Therefore, Kriti’s share = \(\frac{1}{3} \times 60=₹ 20\)
And Kiran’s share = \(\frac{2}{3} \times 60=₹ 40\)
Consider This Situation:
Raju went to the market to purchase tomatoes. One shopkeeper tells him that the cost of tomatoes is ₹ 40 for 5 kg. Another shopkeeper gives the cost as 6 kg for ₹ 42. Now, what should Raju do?
Should he purchase tomatoes from the first shopkeeper or from the second? Will the comparison by taking the difference help him decide? No. Why not?
Think of some way to help him. Discuss with your friends.
Consider Another Example.
Bhavika has 28 marbles and Vini has 180 flowers. They want to share these among themselves. Bhavika gave 14 marbles to Vini and Vini gave 90 flowers to Bhavika. But Vini is not satisfied.
She felt she had given more flowers to Bhavika than the marbles given by Bhavika to her.
What do you think? Is Vini correct?
To solve this problem both went to Vini’s mother Pooja.
Pooja explained that out of 28 marbles, Bahvika gave 15 marbles to Vini.
Therefore, the ratio is 14: 28 = 1: 2
And out of 180 flowers, Vini had given 90 flowers to Bhavika.
Therefore, the ratio is 90: 180 = 1: 2
Since both the ratios are the same, the distribution is fair.
Two friends Ashma and Pankhuri went to market to purchase hair clips. They purchased 20 chair clips for ₹ 30.
Ashma gave ₹ 12 and Pankhuri gave ₹ 18. After they came back home, Ashma asked Pankhuri to give 10 hair clips to her. But Pankhuri said, “Since I have given money I should get more clips. You should get 8 hair clips and I should get 12”.
Can you tell who is correct? Ashma or Phankhuri? Why?
The ratio of money given by Ashma to the money given by Pankhuri = ₹ 12 : ₹ 18 = 2 : 3
According to Ashma’s suggestion, the ratio of the number of hair clips for Ashma to the number of clips for Pankhuri = 10: 10 = 1: 1
According to Pankhuru’s suggestion, the ratio of the number of hair clips for Ashma to the number of clips for Pankhuri = 8: 12 = 2 : 3
Now, notice that according to Ashma’s distribution, the ratio of hair clips and the ratio of money given by them is not the same. However, according to Pankhuri’s distribution, the two ratios are the same.
Hence, we can say that Pankhuri’s distribution is correct.
Does sharing a ratio mean something?
Consider the following examples:
Raj purchased 3 pens for ₹ 15 and Anu purchased 10 pens for ₹ 50. Whose pens are more expensive?
The ratio of the number of pens purchased by Raj to the number of pnes purchased by Anu = 3:10
Ratio of their costs = 15 : 50 = 3 : 10
Both the ratios 3: 10 and 15: 50 are equal. Therefore, the pens were purchased for the same price by both.
Rahim sells 2 kg of apples for ₹ 180 and Roshan sells 4 kg of apples for ₹ 360.
Whose apples are more expensive?
Ratio of the weight of apples = 2 kg : 4 kg = 1 : 2
Ratio of their cost = ₹ 180 : ₹ 360 = 6 : 12 = 1 : 2
So, the ratio of the weight of apples = ratio of their cost.
Since both the ratios are equal, hence, we say that they are in proportion. They are selling apples at the same rate.
If two ratios are equal, we say that they are in proportion and use the symbol ‘::’ or ‘=’ to equate the two ratios.
For the first example, we can say 3, 10, 15 and 50 are in proportion which is written as 3: 10:: 15: 50 and is read as 3 is to 10 as 15 is to 50 or it is written as 3: 10 = 15: 50.
For the second example, we can say 2,4, 180 and 360 are in proportion which is written as 2: 4:: 180: 360 and is read as 2 is to 4 as 180 is to 360.
Let us consider another example.
A man travels 35 km in 2 hours. With the same speed would he be able to travel 70 km in 4 hours?
Now, the ratio of the two distances travelled by the man is 35 to 70 =1:2 and the ratio of the time taken to cover these distances is 2 to 4 = 1: 2.
Hence, the two ratios are equal i.e. 35: 70 = 2: 4.
Therefore, we can say that the four numbers 35,70,2 and 4 are in proportion.
Hence, we can write it as 35: 70:: 2: 4 and read it as 35 is to 70 as 2 is to 4. Hence, he can travel 70 km in 4 hours with that speed.
Now, consider this example.
What cost of 2 kg of apples is ₹ 180 and a 5 kg watermelon costs ₹ 45.
Now, the ratio of the weight of apples to the weight of watermelon is 2: 5.
And ratio of the cost of apples to the cost of the watermelon is 180: 45 = 4: 1.
Here, the two ratios 2: 5 and 180:45 are not equal,
i.e. 2: 5 ≠ 180: 45
Therefore, the four quantities 2,5,180 and 45 are not in proportion.
If two ratios are not equal, then we say that they are not in proportion.
In a statement of proportion, the four quantities involved when taken in order are known as respective terms. The first and fourth terms are known as extreme terms. The second and third terms are known as middle terms.
For example, in 35: 70:: 2: 4;
35,70,2,4 are the four terms. 35 and 4 are the extreme terms. 70 and 2 are the middle terms.
Example 8. Are the ratios 25g: 30g and 40 kg: 48 kg in proportion?
Solution: 25g: 30g = \(\frac{25}{30}\) = 5:6
40 kg : 48 kg = \(\frac{40}{48}\) = 5:6
So, 25: 30 = 40: 48.
Therefore, the ratios 25 g: 30 g and 40 kg: 48 kg are in proportion, i.e. 25: 30:: 40: 48
The middle terms in this are 30,40 and the extreme terms are 25,48.
Example 9. Are 30, 40,45 and 60 in proportion?
Solution: Ratio of 30 to 40 = \(\frac{30}{40}\) = 3 : 4.
Ratio of 45 to 60 = \(\frac{45}{60}\) = 3 : 4.
Since, 30: 40 = 45: 60.
Therefore, 30, 40, 45, and 60 are in proportion.
Example 10. Do the ratios 15 cm to 2 m and 10 sec to 3 minutes form a proportion?
Solution: Ratio of 15 cm to 2 m = 15 : 2 x 100 (1 m = 100 cm) = 3:40
Ratio of 10 sec to 3 min = 10 : 3 x 60 (1 min = 60 sec) = 1:18
Since, 3.40 ≠ 1.18, therefore, the given ratios do not form a proportion.
Consider the following situations:
These are examples of the kinds of situations that we face in our daily lives. How would you solve these?
Reconsider the first example: What cost of 2 notebooks is ₹ 24.
Therefore, the cost of 1 notebook = ₹ 24 ÷ 2 = ₹ 12.
Now, if you were asked to find the cost of 5 such notebooks. It would be = ₹ 12 x 5 = ₹ 60
Reconsider the second example: We want to know how many litres are needed to travel 1 km.
For 80 km, petrol needed = 2 litres.
Therefore, to travel 1 km, petrol needed = \(\frac{2}{80}\) = \(\frac{1}{40}\) litres.
Now, if you are asked to find out how many litres of petrol are required to cover 120 km?
Then petrol needed = \(\frac{1}{40}\) x 120 litres = 3 litres
The method in which first we find the value of one unit and then the value of the required number of units is known as the Unitary Method.
We see that,
Distance travelled by Karan in 2 hours = 8 km
Distance travelled by Karan in 1 hour = \(\frac{8}{2}\) km = 4 km
Therefore, the distance travelled by Karan in 4 hours = 4 x 4 = 16 km
Similarly, to find the distance travelled by Kriti in 4 hours, first find the distance travelled by her in 1 hour.
Example 11. If the cost of 6 cans of juice is ₹ 210, then what will be the cost of 4 cans of juice?
Solution: Cost of 6 cans of juice = ₹ 210
Therefore, cost of one can of juice = \(\frac{210}{6}\) = ₹ 35
Therefore, the cost of 4 cans of juice = ₹ 35 x 4 = ₹ 140.
Thus, the cost of 4 cans of juice is ₹ 140.
Example 12. A motorbike travels 220 km in 5 litres of petrol. How much distance will it cover in 1.5 litres of petrol?
Solution: With 5 litres of petrol, a motorbike can travel 220 km.
Therefore, in 1 litre of petrol, motorbike travels = \(\frac{220}{5}\) km
Therefore, in 1.5 litres, motorbike travels = \(\frac{220}{5}\) x 1.5 km
= \(\frac{220}{5}\) x \(\frac{15}{10}\) km = 66 km.
Thus, the motorbike can travel 66 km in 1.5 litres of petrol.
Example 13. If the cost of a dozen soaps is ₹ 153.60, what will be the cost of 15 such soaps?
Solution:
We know that 1 dozen = 12
Since, the cost of 12 soaps = ₹ 153.60
Therefore, cost of 1 soap = \(\frac{153.60}{12}\) = ₹ 12.80
Therefore, the cost of 15 soaps = ₹ 12.80 x 15 = ₹ 192
Thus, the cost of 15 soaps is ₹ 192.
Example 14. The cost of 105 envelopes is ₹ 350. How many envelopes can be purchased for ₹ 100?
Solution: In ₹ 350, the number of envelopes that can be purchased = 105
Therefore, in ₹ 1, the number of envelopes that can be purchased = \(\frac{105}{350}\)
Therefore, in ₹ 100, the number of envelopes that can be purchased = \(\frac{105}{350}\) x 100 = 30
Thus, 30 envelopes can be purchased for ₹ 100.
Example 15. A car travels 90 km in 2 \(\frac{1}{2}\) hours.
Solution:
(1) In this case, time is unknown and distance is known. Therefore, we proceed as follows:
2 \(\frac{1}{2}\) hours = \(\frac{5}{2}\) hours = \(\frac{5}{2}\) x 60 minutes = 150 minutes.
90 km is covered in 150 minutes
Therefore, 1 km can be covered in \(\frac{150}{90}\) minutes
Therefore, 30 km can be covered in \(\frac{150}{90}\) x 30 minutes i.e. 50 minutes
Thus, 30 km can be covered in 50 minutes.
(2) In this case, distance is unknown and time is known. Therefore we proceed as follows:
Distance covered in 2 \(\frac{1}{2}\) hours (i.e. \(\frac{5}{2}\) hours) = 90 km
Therefore, distance covered in 1 hour = 90 ÷ \(\frac{5}{2}\) km = 90 x \(\frac{2}{5}\) = 36 km
Therefore, distance covered in 2 hours = 36 x 2 = 72 km
Thus, in 2 hours, the distance covered is 72 km
When we talk about some plane figures as shown below we think of their regions and their boundaries. We need some measures to compare them. We are looking into these now.
Look at the following figures. You can make them with a wire or a string.
If you start from point S in each case and move along the line segments then you again reach point S. You have made a complete round of the shape in each ease (1), (2) and (3). The distance covered is equal to the length of the wire used to draw the figure.
This distance is known as the perimeter of the closed figure. It is the length of the wire needed to form the figures.
The idea of perimeter is widely used in our daily life.
All these people use the idea of ‘perimeter’.
Give five examples of situations where you need to know the perimeter.
Perimeter is the distance covered along the boundary forming a closed figure when you go around the figure once.
So, how will you find the perimeter of any closed figure made up entirely of line segments? Simply find the sum of the lengths of all the sides (which are line segments).
Let us consider a rectangle ABCD whose length and breadth are 15 cm and 9 cm g respectively.
What will be its perimeter?
The perimeter of the rectangle is = Sum of the lengths of its four sides.
= AB + BC + CD + DA
= AB + BC + AB + BC
= 2 x AB + 2 x BC
= 2 x (AB + BC)
= 2 x (15cm + 9cm)
= 2 x (24cm)
= 48 cm
Hence, from the said example, we notice that
Perimeter of a rectangle = length + breadth + length + breadth
i.e. Perimeter of a rectangle = 2 x (length + breadth)
Let us now see the practical applications of this idea :
Example 1. Shabana wants to put a lace border all around a rectangular table cover, 3 m long and 2 m wide. Find the length of the lace required by Shabana.
Solution:
Length of the rectangular table cover = 3 m
The breadth of the rectangular table cover = 2 m
Shabana wants to put a lace border all around the table cover.
Therefore, the length of the lace required will be equal to the perimeter of the rectangular table cover.
Now, perimeter of the rectangular table cover = 2 x (length + breadth) = 2x(3m + 2m) = 2x5m=10m
So, the length of the lace required is 10 m.
Example 2. An athlete takes 10 rounds of a rectangular park, 50 m long and 25 m wide. Find the total distance covered by him.
Solution: Length of the rectangular park = 50 m
The breadth of the rectangular park = 25 m
The total distance covered by the athlete in one round will be the perimeter of the park.
Now, perimeter of the rectangular park = 2 x (length + breadth) = 2 x (50 m + 25 m)
= 2 x 75 m =150 m
So, the distance covered by the athlete in one round is 150 m.
Therefore, the distance covered in 10 rounds =10xl50m = 1500m The total distance covered by the athlete is 1500 m.
Example 3. Find the perimeter of a rectangle whose length and breadth are 150 cm and 1 m respectively.
Solution: Length = 150 cm
Breadth = lm = 100 cm
Perimeter of the rectangle = 2 x (length + breadth)
= 2 x (150 cm + 100 cm)
= 2 x (250 cm) = 500 cm = 5 m
Example 4. A farmer has a rectangular field of length and breadth of 240 m and 180 m respectively. He wants to fence it with 3 rounds of rope as shown. What is the total length of rope he must use?
Solution: The farmer has to cover three times the perimeter of that field.
Therefore, the total length of rope required is thrice its perimeter.
Perimeter of the field = 2 x (length + breadth)
= 2 x (240 m+ 180 m)
= 2 x 420 m = 840 m
Total length of rope required = 3 x 840 m = 2520 m
Example 5. Find the cost of fencing a rectangular park of length 250 m and breadth 175 ra at the rate of ₹ 12 per m
Solution: Length of the rectangular park = 250 m
The breadth of the rectangular park = 175 m
To calculate the cost of fencing we require a perimeter
Perimeter of the rectangle = 2 x (length + breadth)
= 2 x (250 m + 175 m)
= 2 x (425 m) = 850 m
Cost of fencing lm of park = ₹ 12
Therefore, the total cost of fencing the park = ₹12 x 850 = ₹ 10200
Consider this example.
Biswamitra wants to put coloured tape all around a square picture of the side lm as shown. What will be the length of the coloured tape he requires?
Since Biswamitra wants to put the coloured tape all around the square picture, he needs to find the perimeter of the picture frame.
Thus, the length of the tape required
= Perimeter of square = 1m + 1m + 1m + 1m = 4m
Now, we know that all the four sides of a square are equal, therefore, in place of adding it four times, we can multiply the length of one side by 4.
Thus, the length of the tape required = 4 x 1m = 4m
From this example, we see that
The perimeter of a square = 4 x length of a side
Draw more such squares and find the perimeters.
Now, look at an equilateral triangle with each side equal to 4 cm.
Can we find its perimeter?
Perimeter of this equilateral triangle = 4 + 4 + 4 cm = 3 x 4 cm = 12 cm
So, we find that
The perimeter of an equilateral triangle = 3 x the length of a side
What is similar between a square and an equilateral triangle?
They are figures having all the sides of equal length and all angles of equal measure Such figures are known as regular closed figures. Thus, a square and an equilateral triangle are regular closed figures,
You found that,
Perimeter of a square = 4 x length of one side
The perimeter of an equilateral triangle = 3 x the length of one side
Example 6. Find the distance travelled by Shaina if she takes three rounds of a square park of side 70 m.
Solution: Perimeter of the square park = 4 x length of a side = 4 x 70 m = 280 m
Distance covered in one round = 280 m
Therefore, distance travelled in three rounds=3 x 280m = 840m
Example 7. Pinky runs around a square field of side 75 m, Bob runs around a rectangular field with a length of 160 m and a breadth of 105 m. Who covers more distance and by how much?
Solution: Distance covered by Pinky in one round
= Perimeter of the square
= 4 x length of a side
= 4×75 m = 300m
Distance covered by Bob in one round
= Perimeter of the rectangle
= 2 x (length + breadth) = 2 x (160 m + 105 m)
= 2 x 265 m = 530 m
Difference in the distance covered = 530 m – 300 m = 230 m.
Therefore, Bob covers more distance by 230 m.
Example 8. Find the perimeter of a regular pentagon with each side measuring 3 cm.
Solution: This regular closed figure has 5 sides, each with a length of 3 cm.
Thus, we get Perimeter of the regular pentagon = 5 x 3 cm = 15 cm
Example 9. The perimeter of a regular hexagon is 18 cm. How long is it on one side?
Solution: Perimeters 18 cm
A regular hexagon has 6 sides, so we can divide the perimeter by 6 to get the length of one side.
One side of the hexagon = 18 cm -f 6 = 3 cm
Therefore, the length of each side of the regular hexagon is 3 cm.
Look at the closed figures given below. All of them occupy some region of a flat surface.
Can you tell which one occupies more region?
The amount of surface enclosed by a closed figure is called its area.
So, can you tell, which of the above figures has more area?
Now, look at the adjoining figures.
Which one of these has a larger area? It is difficult to tell just by looking at these figures. So, what do you do?
Place them on a squared paper or graph paper where every square measures 1 cm x 1 cm.
Make an outline of the figure.
Look at the squares enclosed by the figure. Some of them are completely enclosed, some half, some less than half and some more than half.
The area is the number of centimetre squares that are needed to cover it.
But there is a small problem the squares do not always fit exactly into the area you measure. We get over this difficulty by adopting a convention.
Such a convention gives a fair estimate of the desired area.
Example 1. Find the area of the shape shown in the figure.
Solution: This figure is made up of line segments.
Moreover, it is covered by full squares and half squares only. This makes our job simple.
Fully-filled squares = 3
Half-filled squares = 3
Area covered by full squares = 3 x 1 sq units = 3 sq units
Total area = 4 1/2 sq units.
Example 2. By counting squares, estimate the area of the figure.
Solution: Make an outline of the figure on a graph sheet.
Total area = 11 + 3 x \(\frac{1}{2}\) + 7 = 19 \(\frac{1}{2}\) = sq units.
Example 3. By counting squares, estimate the area of the figure.
Solution: Make an outline of the figure on a graph sheet. This is how the squares cover the figure.
Total area = 1 + 7 = 8 sq units.
With the help of the squared paper, can we tell, what will be the area of a rectangle whose length is 5 cm and breadth is 3 cm?
Draw the rectangle on a graph paper having 1 cm x 1 cm, squares. The rectangle covers 15 squares completely.
The area of the rectangle = 15 sq which can be written as 5 x 3 sq cm (length x breadth).
The measures of the sides of some of the rectangles are given. Find their areas by placing them on graph paper and counting the number of squares.
What do we infer from this?
We find,
Area of a rectangle = (length x breadth)
Without using graph paper, can we find the area of a rectangle whose length is 6 cm and breadth is 4cm?
Yes, it is possible.
What do we infer from this?
We find that,
Area of the rectangle = length x breadth = 6 cm x 4 cm = 24 sq cm.
Let us now consider a square of side 4 cm.
What will be its area?
If we place it on a centimetre of graph paper, then what do we observe?
It covers 16 squares i.e. the area of the square = 16 sq cm = 4 x 4 sq cm
Calculate areas of a few squares by assuring the length of one side of the squares by yourself.
Find their areas using graph paper.
What do we infer from this?
We find that in each case,
Area of the square = side x side
You may use this as a formula for doing problems.
Example 1. Find the area of a rectangle whose length and breadth are 12 cm and 4 cm respectively.
Solution: Length of the rectangle =12 cm
Breadth of the rectangle = 4 cm
Area of the rectangle = length x breadth
= 12 cm x 4 cm = 48 sq cm.
Example 2. Find the area of a square plot of side 8 m.
Solution: Side of the square = 8 m
Area of the square = side x side
= 8 m x 8 m = 64 sq m.
Example 3. The area of a rectangular piece of cardboard is 36 sq cm and its length is 9 cm. What is the width of the cardboard?
Solution: Area of the rectangle = 36 sq cm
Length = 9 cm
Width =?
Area of a rectangle = length x width
So, width = \(\frac{\text { Area }}{\text { Length }}=\frac{36}{9}=4 \mathrm{~cm}\)
Thus, the width of the rectangular cardboard is 4 cm.
Example 4. Bob wants to cover the floor of a room 3 m wide and 4 m long with squared tiles. If each square tile is of side 0.5 m, then find the number of tiles required to cover the floor of the room.
Solution: The total area of tiles must be equal to the area of the floor of the room.
Length of the room = 4 m Breadth of the room = 3 m
Area of the floor = length x breadth = 4mx3m=12sqm
Area of one square tile = side x side = 0.5 m x 0.5 m = 0.25 sq m
Number of tiles required = \(\frac{\text { Area of the floor }}{\text { Area of one tile }}=\frac{12}{0.25}=\frac{1200}{25}=48 \text { tiles. }\)
Example 5. Find the area in square metres of a piece of cloth 1 m 25 cm wide and 2 m long.
Solution: Length of the cloth = 2 m
Breadth of the cloth = 1 m 25 cm = 1 m + 0. 25 m = 1.25 m (since 25 cm = 0.25m)
Area of the cloth = length of the cloth x breadth of the cloth = 2 m x 1.25 m = 2.50 sq m
Ramesh has 6 marbles with him. He wants to arrange them in rows in such a way that each row has the same number of marbles. He arranges them in the following ways and matches the total number of marbles.
1. 1 Marble in each row
Number of rows = 6
Total number of marbles = 1 x 6 = 6.
2. 2 marbles in each row
Number of rows = 3
Total number of marbles = 2 x 3 = 6.
3. 3 marbles in each row
Number of rows = 2
Total number of marbles = 3 x 2 = 6.
4. He could not think of any arrangement in which each row had 4 marbles or 5 marbles. So, the only possible arrangement left was with all the 6 marbles in a row.
Number of rows = 1
Total number of marbles = 6 x 1 = 6
From these calculations, Ramesh observes that 6 can be written as a product of two numbers in different ways as
6 = 1 x 6; 6 = 2 x 3; 6 = 3 x 2; 6 = 6 x 1;
From 6 = 2 x 3, it can be said that 2 and 3 exactly divide 6. So, 2 and 3 are exact divisors of 6. From the other product 6 = 1 x 6, the exact divisors of 6 are found to be 1 and 6.
Thus, 1, 2, 3, and 6 are exact divisors of 6. They are called the factors of 6. found to be 1 and 6. Try arranging 18 marbles in rows and find the factors of 18.
Mary wants to find those numbers that are exactly divided by 4. She divides 4 by numbers less than 4 this way.
She finds that the number 4 can be written as 4 = 1 x 4; 4 = 2 x 2; 4 = 4×1 and knows that the numbers 1,2 and 4 are exact divisors of 4.
These numbers are called factors of 4.
A factor of a number is an exact divisor of that number.
Observe each of the factors of 4 is less than or equal to 4.
Game-1: This is a game to be played by two persons say A and B. It is
about spotting factors.
It requires 50 pieces of cards numbered 1 to 50.
Arrange the cards on the table like this.
Steps
When we write the number 20 as 20 = 4 x 5, we say 4 and 5 are factors of 20. We also say that 20 is a multiple of 4 and 5.
The representation 24 = 2 x 12 shows that 2 and 12 are factors of 24, whereas 24 is a multiple of 2 and 12.
We can say that a number is a multiple of each of its factors
Let us now see some interesting facts about factors and multiples.
The length ofthe strip at the top is 3 =1 x 3 units.
The length ofthe strip below it is 3 +3 =6 units.
Also, 6=2×3. The length ofthe next strip is 3 + 3 + 3 =9 units, and 9 =3×3. Continuing this way we can express the other lengths as,
12 = 3 x 4; 15 = 5 x 3
We say that the numbers 3, 6, 9, 12, 15 are multiples of 3.
The list of multiples of 3 can be continued as 18, 21, 24, …
Each of these multiples is greater than or equal to 3.
The multiples of the number 4 are 4, 8, 12, 16, 20, 24, …
The list is endless. Each of these numbers is greater than or equal to 4
Let us see what we conclude about factors and multiples:
1. Is there any number that occurs as a factor of every number? Yes. It is 1. For example 6 = 1 x 6, 18 = 1 x 18, and so on. Check it for a few more numbers.
We say 1 is a factor of every number.
2. Can 7 be a factor of itself? Yes. You can write 7 as 7 = 7 x 1. What about 10? and 15?
You will find that every number can be expressed in this way. We say that every number is a factor of itself.
3. What are the factors of 16? They are 1, 2,4, 8, 16. Out of these factors do you find any factor which does not divide 16? Try it for 20; 36.
You will find that every factor of a number is an exact divisor of that number.
4. What are the factors of 34? They are 1,2, 17 and 34. Out of these which is the greatest factor? It is 34 itself.
The other factors 1,2 and 17 are less than 34. Try to check this for 64, 81, and 56.
We say that every factor is less than or equal to the given number.
5. The number 76 has 5 factors. How many factors does 136 or 96 have? You will find that you are able to count the number of factors of each of these.
Even if the numbers are as large as 10576, 25642, etc., or larger, you can still count the number of factors of such numbers, (though you may find it difficult to factorize such numbers).
We say that the number of factors of a given number is finite.
6. What are the multiples of 7? Obviously, 7,14,21,28,… You will find that each of these multiples is greater than or equal to 7. Will it happen with each number? Check this for the multiples of 6, 9, and 10.
We find that every multiple of a number is greater than or equal to that number.
7. Write the multiples of 5. They are 5, 10, 15, 20, … Do you think this list will end anywhere? No! The list is endless. Try it with multiples of
6,7 etc.
We find that the number of multiples of a given number is infinite.
8. Can 7 be a multiple of itself? Yes, because 7 = 7×1. Will it be true for other numbers also? Try it with 3,12 and 16.
You will find that every number is a multiple of itself.
The factors of 6 are 1,2,3 and 6. Also, 1+2+3+6 = 12 = 2 x 6. We find that the sum of the factors of 6 is twice the number 6.
All the factors of 28 are 1,2, 4,7,14, and 28. Adding these we have, 1 + 2 + 4 + 7 + 14 + 28 = 56 = 2 x 28. The sum of the factors of 28 is equal to twice the number 28.
A number for which the sum of all its factors is equal to twice the number is called a perfect number. The numbers 6 and 28 are perfect numbers.
Is 10 a perfect number?
Example 1: Write all the factors of 68.
Solution :
We note that
68 = 1 x 68 68 = 2 x 34
68 = 4 x 17 68 = 17 x 4
Stop here, because 4 and 17 have occurred earlier.
Thus, all the factors of 68 are 1,2,4,17, 34 and 68.
Example 2: Find the factors of 36.
Solution:
36 = 1 x 36 36 = 2 x 18 36 = 3 x 12
36 = 4 x 9 36 = 6 x 6
Stop here, because both factors (6) are the same. Thus, the factors are 1,2, 3,4, 6, 9, 12,18 and 36.
Example 3: Write the first five multiples of 6.
Solution:
The required multiples are 6 x 1= 6, 6 x 2 = 12, 6 x 3 = 18, 6 x 4 = 24, 6 x 5 = 30 i.e. 6, 12, 18, 24, and 30.
We are now familiar with the factors of a number. Observe the number of factors of a few numbers arranged in this table.
We find that
Try to find some more prime numbers other than these.
There are numbers having more than two factors like 4,6,8,9,10 and so on. These numbers are composite numbers.
Numbers having more than two factors are called Composite numbers. Is 15 a composite number? Why? What about 18? 25?
Without actually checking the factors of a number, we can find prime numbers from 1 to 100 with an easier method.
This method was given by a Greek Mathematician Eratosthenes, in the third century B.C. Let us sec the method. List all numbers from 1 to 100, as shown below.
Step 1: Cross out 1 because it is not a prime number.
Step 2: Encircle 2, cross out all the multiples of 2, other than 2 itself, i.e. 4,6, 8, and so on.
Step 3: You will find that the next uncrossed number is 3. Encircle 3 and cross out all the multiples of 3, other than 3 itself.
Step 4: The next uncrossed number is 5. Encircle 5 and cross out all the multiples of 5 other than 5 itself.
Step 5: Continue this process till all the numbers in the list are either encircled or crossed out.
All the encircled numbers are prime numbers. All the crossed-out numbers, other than 1 are composite numbers.
This method is called the Sieve of Eratosthenes.
Example 4: Write all the prime numbers less than 15.
Solution :
By observing the Sieve Method, we can easily write the required prime numbers as 2,3, 5,7,11, and 13.
even and odd numbers
Do you observe any pattern in the numbers 2,4, 6, 8, 10, 12, 14,…? You will find that each of them is a multiple of 2.
These are called even numbers. The rest of the numbers 1, 3, 5, 7, 9,11,… are called odd numbers.
You can verify whether a two-digit number or a three-digit number is even or not. How will you know whether a number like 756482 is even? By dividing it by 2. Will it not be tedious?
We say that a number with 0,2,4,6, 8 at the one’s place is an even number. So, 350,4862,59246 are even numbers. The numbers 457,2359,8231 are all odd.
Let us try to find some interesting facts:
Thus, we can say that every prime number except 2 is odd.
Is the number 38 divisible by 2? by 4? by 5?
By actually dividing 38 by these numbers we find that it is divisible by 2 but not by 4 and by 5.
Let us see whether we can find a pattern that can tell us whether a number is divisible by 2,3,4,5,6, 8,9,10, or 11. Do you think such patterns can be easily seen?
Divisibility by 10: Charu was looking at the multiples of 10. The multiples are 10, 20, 30,40, 50, 60,… She found something common in these numbers. Can you tell me what? J Each of these numbers has 0 in one place.
She thought of some more numbers with 0 at one place ^ like 100,1000,3200,7010. She also found that all such numbers are divisible by 10.
She finds that if a number has 0 in the ones place then it is divisible by 10. Can you find out the divisibility rule for 100?
Divisibility by 5: Mani found some interesting patterns in the numbers 5,10, 15,20,25,30,35,… Can you tell the pattern? Look at the unit’s place.
All these numbers have either 0 or 5 in their ones place. We know that these numbers are divisible by 5.
Mani took up some more numbers that are divisible by 5, like 105, 215, 6205, 3500. Again these numbers have either 0 or 5 in their places.
He tried to divide the numbers 23, 56, 97 by 5. Will he be able to do that? Check it. He observes that a number that has either 0 or 5 in its place is divisible by 5, other numbers leave a remainder. Is 1750125 divisible 5?
Divisibility by 2: Charu observes a few multiples of 2 to be 10, 12,14,16… and also numbers like 2410,4356, 1358, 2972, 5974.
She finds some patterns in the place of these numbers. Can you tell that? These numbers have only the digits 0, 2,4, 6, and 8 in one place.
She divides these numbers by 2 and gets the remainder 0. She also finds that the numbers 2467, and 4829 are not divisible by 2. These numbers do not have 0, 2, 4, 6, or 8 in their one place.
Looking at these observations she concludes that a number is divisible by 2 if it has any of the digits 0,2,4, 6, or 8 in its ones place.
Divisibility by 3: Are the numbers 21, 27, 36, 54, 219 divisible by 3? Yes, they are.
Are the numbers 25, 37, 260 divisible by 3? No.
Can you see any pattern in the one place? We cannot, because numbers with the same digit in the ones places can be divisible by 3, like 27, or may not be divisible by 3 like 17, 37.
Let us now try to add the digits of 21, 36, 54, and 219. Do you observe anything special? 2+1=3,3+6=9,5+4=9,2+1+9=12.
All these additions are divisible by 3.
Add the digits 25, 37, 260. We get 2+5=7, 3+7=10, 2+6+0 = 8.
These are not divisible by 3.
We say that if the sum of the digits is a multiple of 3, then the number is divisible by 3. Is 7221 divisible by 3?
Divisibility by 6: Can you identify a number that is divisible by both 2 and 3? One such number is 18. Will 18 be divisible by 2×3=6? Yes, it is.
Find some more numbers like 18 and check if they are divisible by 6 also.
Can you quickly think of a number that is divisible by 2 but not by 3?
Now for a number divisible by 3 but not by 2, one example is 27. Is 27 divisible by 6? No. Try to find numbers like 27.
From these observations, we conclude that if a number is divisible by 2 and 3 then it is divisible by 6 also.
Divisibility by 4: Can you quickly give five 3-digit numbers divisible by 4? One such number is 212. Think of such 4-digit numbers. One example is 1936.
Observe the number formed by the ones and tens places of 212. It is 12; which is divisible by 4. For 1936 it is 36, again divisible by 4.
Try the exercise with other such numbers, for example, 4612; 3516; 9532. Is the number 286 divisible by 4? No. Is 86 divisible by 4? No.
So, we see that a number with 3 or more digits is divisible by 4 if the number formed by its last two digits (i.e. ones and tens) is divisible by 4. Check this rule by taking ten more examples.
Divisibility for 1 or 2-digit numbers by 4 has to be checked by actual division.
Divisibility by 8: Are the numbers 1000, 2104, and 1416 divisible by 8? You can check that they are divisible by 8. Let us try to see the pattern.
Look at the digits at ones, tens, and hundreds in place of these numbers. These are 000,104 and 416 respectively. These too are divisible by 8.
Find some more numbers in which the number formed by the digits at units, tens, and hundreds place (i.e. last 3 digits) is divisible by 8.
For example, 9216,8216,7216,10216, 9995216, etc. You will find that the numbers themselves are divisible by 8.
We find that a number with 4 or more digits is divisible by 8 if the number formed by the last three digits is divisible by 8. Is 73512 divisible by 8?
The divisibility for numbers with 1,2 or 3 digits by 8 has to be checked by actual division.
Divisibility by 9: The multiples of 9 are 9, 18, 27, 36, 45, 54,… There are other numbers like 4608, and 5283 that are also divisible by 9.
Do you find any pattern when the digits of these numbers are added?
1 + 8 = 9, 2 + 7 = 9, 3 + 6 = 9, 4 + 5 = 9 4 + 6 + 0 + 8= 18,5 + 2 + 8 + 3 = 18 All these sums are also divisible by 9. Is the number 758 divisible by 9?
No. The sum of its digits 7 + 5 + 8 = 20 is also not divisible by 9.
These observations lead us to say that if the sum of the digits of a number is divisible by 9, then the number itself is divisible by 9.
Divisibility by 11: The numbers 308,1331 and 61809 are all divisible by 11. We form a table and see if the digits in these numbers lead us to some pattern.
We observe that in each case the difference is either 0 or divisible by 11. All these numbers are also divisible by 11.
For the number 5081, the difference between the digits is (5+8) – (1+0) = 12 which is not divisible by 11. The number 5081 is also not divisible by 11.
Thus, to check the divisibility of a number by 11, the rule is, to find the difference between the sum of the digits at odd places (from the right) and the sum of the digits at even places (from the right) of the number.
If the difference is either 0 or divisible by 11, then the number is divisible by 11.
Observe the factors of some numbers taken in pairs.
1. What are the factors of 4 and 18?
The factors of 4 are 1, 2, and 4.
The factors of 18 are 1,2,3,6,9 and 18.
The numbers 1 and 2 are the factors of both 4 and 18.
They are the common factors of 4 and 18.
2. What are the common factors of 4 and 15?
These two numbers have only 1 as the common factor.
What about 7 and 16?
Two numbers having only 1 as a common factor are called co-prime numbers. Thus, 4 and 15 are co-prime numbers.
Are 7 15,12 49,18 and 23 co-prime numbers?
3. Can we find the common factors of 4,12 and 16?
Factors of 4 are 1, 2, and 4.
Factors of 12 are 1, 2, 3, 4, 6 and 12.
Factors of 16 are 1, 2, 4, 8 and 16.
Clearly, 1, 2, and 4 are the common factors of 4,12, and 16.
Find the common factors of (1) 8, 12, 20 (2) 9,15, 21.
Let us now look at the multiples of more than one number taken at a time.
1. What are the multiples of 4 and 6?
The multiples of 4 are 4, 8,12,16, 20,24,… (write a few more)
The multiples of 6 are 6,12,18, 24, 30, 36,… (write a few more)
Out of these, are there any numbers that occur in both lists?
We observe that 12,24, 36,… are multiples of both 4 and 6.
Can you write a few more?
They are called the common multiples of 4 and 6.
2. Find the common multiples of 3,5 and 6.
Multiples of 3 are 3, 6, 9,12,15, 18,21, 24, 27, 30, 33, 36,… Multiples of 5 are 5,10,15, 20, 25, 30, 35,…
Multiples of 6 are 6, 12, 18, 24, 30,…
Common multiples of 3,5 and 6 are 30, 60,…
Write a few more common multiples of 3,5 and 6.
Example 5: Find the common factors of 75,60 and 210.
Solution:
Factors of 75 are 1, 3, 5, 15, 25 and 75.
Factors of 60 are 1, 2, 3,4, 5, 6,10,12, 15, 30 and 60.
Factors of 210 are 1,2,3,5,6,7,10,14,15,21,30,35,42,70,105 and 210.
Thus, common factors of 75,60 and 210 are 1,3,5 and 15.
Example 6: Find the common multiples of 3,4 and 9.
Solution:
Multiples of 3 are 3,6,9,12,15,18, 21, 24,27,30, 33,36, 39,42, 45,48,….
Multiples of 4 are 4, 8,12,16, 20, 24, 28, 32, 36, 40,44,48,…
Multiples of 9 are 9,18, 27, 36,45, 54,63, 72, 81,…
Clearly, common multiples of 3,4, and 9 are 36,72,108,…
When a number is expressed as a product of its factors we say that the number has been factorised. Thus, when we write 24 = 3×8, we say that 24 has been factorized. This is one of the factorizations of 24.
The others are:
24 = 2 x 12
= 2 x 2 x 6
= 2 x 2 x 2 x 3
24 = 4 x 6
= 2 x 2 x 6
= 2 x 2 x 2 x 3
24 = 3 x 8
= 3 x 2 x 2 x 2
= 2 x 2 x 2 x 3
In all the above factorisations of 24, we ultimately arrive at only one factorisation 2 x 2 x 2 x 3. In this factorization, the only factors 2 and 3 are prime numbers.
Such a factorization of a number is called a prime factorization. Let us check this for the number 36.
The prime factorization of 36 is 2 x 2 x 3 x 3. i.e. the only prime factorization of 36.
Example 7: Find the prime factorization of 980.
Solution:
We proceed as follows:
We divide the number 980 by 2,3,5,7 etc. in this order repeatedly so long as the quotient is divisible by that number. Thus, the prime factorization of 980 is 2 x 2 x 5 x 7 x 7.
⇒ \(\begin{array}{r|r}
2 & 980 \\
\hline 2 & 490 \\
\hline 5 & 245 \\
\hline 7 & 49 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
We can find the common factors of any two numbers. We now try to find the highest of these common factors.
What are the common factors of 12 and 16? They are 1,2 and 4.
What is the highest of these common factors? It is 4.
What are the common factors of 20, 28 and 36? They are 1, 2 and 4 and again 4 is the highest of these common factors.
The Highest Common Factor (HCF) of two or more given numbers is the highest (or greatest) of their common factors.
It is also known as the Greatest Common Divisor (GCD).
The HCF of 20, 28, and 36 can also be found by prime factorization of these numbers as follows:
⇒ \(\begin{array}{l|l}
2 & 20 \\
\hline 2 & 10 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
⇒ \(\begin{array}{l|l}
2 & 28 \\
\hline 2 & 14 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
⇒ \(\begin{array}{l|l}
2 & 36 \\
\hline 2 & 18 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
Thus,
20 = 2 x 2 x 5
28 = 2 x 2 x 7
36 = 2 x 3 x 3
The common factor of 20,28 and 36 is 2(occurring twice). Thus, HCF of 20, 28 and 36 is 2 x 2 = 4.
What are the common multiples of 4 and 6? They are 12,24,36,… Which is the lowest of these? It is 12.
We say that the lowest common multiple of 4 and 6 is 12. It is the smallest number and both the numbers are factors of this number.
The Lowest Common Multiple (LCM) of two or more given numbers is the lowest (or smallest or least) of their common multiples. What will be the LCM of 8 and 12? 4 and 9? 6 and 9?
Example 8: Find the LCM of 12 and 18.
Solution:
We know that common multiples of 12 and 18 are 36,72,108 etc. The lowest of these is 36. Let us see another method to find the LCM of two numbers.
The prime factorizations of 12 and 18 are:
12 = 2 x 2 x 3; 18 = 2 x 3 x 3
In these prime factorizations, the maximum number of times the prime factor 2 occurs is two; this happens for 12. Similarly, the maximum number of times the factor 3 occurs is two; this happens for 18.
The LCM of the two numbers is the product of the prime factors counted the maximum number of times they occur in any of the numbers. Thus, in this case, LCM = 2 x 2 x 3 x 3 = 36.
Example 9: Find the LCM of 24 and 90.
Solution:
The prime factorizations of 24 and 90 are:
24 = 2 x 2 x 2 x 3; 90 = 2 x 3 x 3 x 5
In these prime factorizations, the maximum number of times the prime factor 2 occurs is three; this happens for 24.
Similarly, the maximum number of times the prime factor 3 occurs is two; this happens for 90. The prime factor 5 occurs only once in 90.
Thus, LCM = (2 x 2 x 2) x (3 x 3) x 5 = 360
Example 10: Find the LCM of 40, 48 and 45.
Solution:
The prime factorizations of 40,48 and 45 are;
40 = 2 x 2 x 2 x 5
48 = 2 x 2 x 2 x 2 x 3
45 = 3 x 3 x 5
The prime factor 2 appears a maximum number of four times in the prime factorization of 48, the prime factor 3 occurs a maximum number of two times in the prime factorization of 45, The prime factor 5 appears one time in the prime factorizations of 40 and 45, we take it only ponce.
Therefore, required LCM = (2 x 2 x 2 2) x (3 x 3) x 5 = 720
LCM can also be found in the following way:
Example 11: Find the LCM of 20, 25 and 30.
Solution:
We write the numbers as follows in a row:
So, LCM = 2 x 2 x 3 x 5 x 5.
We come across a number of situations in which we make use of the concepts of HCF and LCM. We explain these situations through a few examples.
Example 12: Two tankers contain 850 liters and 680 liters of kerosene oil respectively. Find the maximum capacity of a container that can measure the kerosene oil of both tankers when used an exact number of times.
Solution:
The required container has to measure both tankers in a way that the count is an exact number of times.
So its capacity must be an exact divisor of the capacities of both the tankers. Moreover, this capacity should be maximum. Thus the maximum capacity of such a container will be the HCF of 850 and 680.
⇒ \(\begin{array}{l|l}
2 & 850 \\
\hline 5 & 425 \\
\hline 5 & 85 \\
\hline 17 & 17 \\
\hline & 1
\end{array}\)
⇒ \(\begin{array}{l|l}
2 & 680 \\
\hline 2 & 340 \\
\hline 2 & 170 \\
\hline 5 & 85 \\
\hline 17 & 17 \\
\hline & 1
\end{array}\)
Hence,
850 = 2 x 5 x 5 x 17 = 2 x 5 x 17 x 5 and
680 = 2 x 2 x 2 x 5 x 17 = 2 x 5 x 17 x 2 x 2
The common factors of 850 and 680 are 2, 5 and 17.
Thus, the HCF of 850 and 680 is 2 x 5 x 17 = 170.
Therefore, the maximum capacity of the required container is 170 liters.
It will fill the first container in 5 and the second in 4 refills.
Example 13: In a morning walk, three persons step off together. Their steps measure 80 cm, 85 cm, and 90 cm respectively. What is the minimum distance each should walk so that all can cover the same distance in complete steps?
Solution:
The distance covered by each one of them is required to be the same as well as minimum. The required minimum distance each should walk would be the lowest common multiple of the measures of their steps.
Can you describe why? Thus, we find the LCM of 80, 85, and 90. The LCM of 80, 85, and 90 is 1 2240.
The required minimum distance is 12240 cm.
Example 14: Find the least number which when divided by 12, 16, 24, and 36 leaves a remainder of 7 in each case.
Solution:
We first find the LCM of 12, 16, 24, and 36 as follows:
∴ \(\begin{array}{c|cccc}
2 & 12 & 16 & 24 & 36 \\
\hline 2 & 6 & 8 & 12 & 18 \\
\hline 2 & 3 & 4 & 6 & 9 \\
\hline 2 & 3 & 2 & 3 & 9 \\
\hline 3 & 3 & 1 & 3 & 9 \\
\hline 3 & 1 & 1 & 1 & 3 \\
\hline & 1 & 1 & 1 & 1
\end{array}\)
Thus, LCM = 2 x 2 x 2 x 2 x 3 x 3 = 144
144 is the least number which when divided by the given numbers will leave the remainder 0 in each case. But we need the least number that leaves the remainder 7 in each case.
Therefore, the required number is 7 more than 144. The required least number = 144 + 7 = 151.
Savita and Shama were going to the market to buy some stationary items. Savita said, “I have 5 rupees and 75 paise”. Shama said, “I have 7 rupees and 50 paise”.
They knew how to write rupees and paise using decimals.
So Savita said, I have ₹ 5.75 and Shama said, “I have ₹ 7.50”.
Have they written correctly?
We know that the dot represents a decimal point.
In this chapter, we will learn more about working with decimals.
Can you tell which is greater, 0.07 or 0.1?
Take two pieces of square paper of the same size. Divide them into 100 equal parts.
For 0.07 we have to shade 7 parts out of 100.
Now, 0.1 = \(\frac{1}{10}\) = \(\frac{10}{100}\), so, for 0.1, shade 10 parts out 100.
This means 0.1 >0.07
Let us now compare the numbers 32.55 and 32.5. In this case, we first compare the whole part. We see that the whole part for both the numbers is 32 and, hence, equal.
We, however, know that the two numbers are not equal. So, we now compare the tenth part. We find that for 32.55 and 32.5, the tenth part is also equal, and then we compare the hundredth part.
We find, 32.55 = 32 + \(\frac{5}{10}\) + \(\frac{5}{100}\) and 32.5 = 32 + \(\frac{5}{100}\) + \(\frac{0}{100}\), therefore, 32.55>32.5 as the hundredth part of 32.55 is more.
Example 1. Which is greater?
Solution:
(1) 1 = 1 + \(\frac{0}{10}\) + \(\frac{0}{100}\); 0.99 = 0 + \(\frac{9}{10}\) + \(\frac{9}{100}\)
The whole part of 1 is greater than that of 0.99.
Therefore, 1 > 0.99
(2) 1.09 = 1 + \(\frac{0}{10}\) + \(\frac{9}{100}\) + \(\frac{0}{1000}\); 1.093 = 1 + \(\frac{0}{10}\) + \(\frac{9}{100}\) + \(\frac{3}{1000}\)
In this case, the two numbers have the same parts up to a hundredth.
But the thousandth part of 1.093 is greater than that of 1.09.
Therefore, 1.093 > 1.09.
Money
We know that 100 paise = ₹ 1
Therefore, 1 paise = ₹ \(\frac{1}{100}\) = ₹ 0.01
So, 65 paise = ₹ \(\frac{65}{100}\) = ₹ 0.65
and 5 paise = ₹ \(\frac{5}{100}\) = ₹ 0.05
What is 105 paise? It is ₹ 1 and 5 paise = ₹ 1.05
Length
Mahesh wanted to measure the length of his tabletop in metres. He had a 50 cm scale. He found that the length of the tabletop was 156 cm. What will be its length in metres?
Mahesh knew that 1 cm = \(\frac{1}{100}\) m or 0.01 m
Therefore, 56 cm = \(\frac{56}{100}\) m = 0.56 m
Thus, the length of the tabletop is 156 cm = 100 cm + 56 cm
= 1 m + \(\frac{56}{100}\) m = 1.56 m
Mahesh also wants to represent this length pictorially. He took squared papers of equal size and divided them into 100 equal parts. He considered each small square as one cm.
Weight
Nandu bought 500g potatoes, 250g capsicum, 700g onions, 500g tomatoes, 100g ginger and 300g radish. What is the total weight of the vegetables in the bag? Let us add the weight of all the vegetables in the bag.
500 g + 250 g + 700 g + 500 g + 100 g + 300 g = 2350 g
We know that 1000 g = 1 kg
Therefore, 1 g = \(\frac{1}{1000}\) = 0.001 kg
Thus, 2350 g = 2000 g +350 g
= \(\frac{2000}{1000}\) kg + \(\frac{350}{1000}\) kg
= 2 kg+ 0.350 kg = 2.350 kg
i. e. 2350 g = 2 kg 350 g = 2.350 kg
Thus, the weight of vegetables in Nandu’s bag is 2.350 kg.
Add 0.35 and 0.42
Take a square and divide it into 100 equal parts
Mark 0.35 in this square by shading
3 tenths and colouring 5 hundredths.
Mark 0.42 in this square by shading
4 tenths and colouring 2 hundredths.
Now count the total number of tenths in the square and the total number of hundredths in the square.
Therefore, 0.35 + 0.42 = 0.77
Thus, we can add decimals in the same way as whole numbers.
Can you now add 0.68 and 0.54?
Thus, 0.68 + 0.54= 1.22
Example 1. Lata spent 19.50 for buying a pen and ₹ 2.50 for one pencil. How much money did she spend?
Solution: Money spent for pen = ₹ 9.50
Money spent for pencil = ₹ 2.50
Total money spent = ₹ 9.50 +₹2.50
Total money spent = ₹ 12.00
Example 2. Samson travelled 5 km 52 m by bus, 2 km 265 m by car and the rest 1km 30 m he walked. How much distance did he travel in all?
Solution: Distance travelled by bus = 5 km 52 m = 5.052 km
Distance travelled by car = 2 km 265 m = 2.265 km
Distance travelled on foot = 1 km 30 m = 1.030 km
Therefore, the total distance travelled is \(\begin{array}{r}
5.052 \mathrm{~km} \\
2.265 \mathrm{~km} \\
+\quad 1.030 \mathrm{~km} \\
\hline 8.347 \mathrm{~km} \\
\hline
\end{array}\)
Therefore, total distance travelled = 8.347 km
Example 3. Rahul bought 4 kg 90 g of apples, 2 kg 60 g of grapes and 5 kg 300 g of mangoes. Find the total weight of all the fruits he bought.
Solution: Weight of apples = 4 kg 90 g = 4.090 kg
Weight of grapes = 2 kg 60 g = 2.060 kg
Weight of mangoes = 5 kg 300 g = 5.300 kg
Therefore, the total weight of the fruits bought is \(\begin{array}{r}
4.090 \mathrm{~kg} \\
2.060 \mathrm{~kg} \\
+\quad 5.300 \mathrm{~kg} \\
\hline 11.450 \mathrm{~kg} \\
\hline
\end{array}\)
The total weight of the fruits bought = 11.450 kg.
Subtract 1.32 from 2.58
This can be shown in the table.
Thus, 2.58-1.32 = 1.26
Therefore, we can say that, subtraction of decimals can be done by subtracting hundredths from hundredths, tenths from tenths, ones from ones and so on, just as we did in addition.
Sometimes while subtracting decimals, we may need to regroup like we did in addition.
Let us subtract 1.74 from 3.5.
Subtract in the hundredth place.
Can’t subtract! so regroup \(\begin{array}{r}
{ }^2 3 .{ }^{14} 5{ }^{10} 0 \\
-1.74 \\
\hline 1.76 \\
\hline
\end{array}\)
Example 1. Abhishek had ₹ 7.45. Did he buy toffees for ₹ 5.30? hind the balance amount left with Abhishek.
Solution: Total amount of money = ₹ 7.45
Amount spent on toffees = ₹ 5.30
Balance amount of money = ₹ 7.45 – ₹ 5.30 = ₹ 2.15
Example 2. Urmila’s school is at a distance of 5 km 350 m from her house. She travels 1 km 70 m on foot and the rest by bus. How much distance does she travel by bus?
Solution: Total distance of the school from the house = 5.350 km
Distance travelled on foot = 1.070 km
Therefore, distance travelled by bus = 5.350 – 1.070 km
Thus, distance travelled by bus = 4.280 km or 4 km 280 m
Example 3. Kanchan bought a watermelon weighing 5 kg 200 g. Out of this, she gave 2 kg 750 g to her neighbour. What is the weight of the watermelon left with Kanchan?
Solution: Total weight of the watermelon = 5.200 kg
Watermelon given to the neighbour = 2.750 kg
Therefore, weight of the remaining watermelon = 5.200 kg-2.750kg =2.450kg
Subhash had learnt about fractions in Classes 4 and 5, so whenever possible he would try to use fractions.
One occasion was when he forgot his lunch at home. His friend Farida invited him to share her lunch. She had five pooris in her lunch box.
So, Subhash and Farida took two pooris each. Then Farida made two equal halves of the fifth poori and gave one half to Subhash and took the other half herself. Thus, both Subhash and Farida had 2 full pooris and one-half poori.
Where do you come across situations with fractions in your life?
Subhash knew that one-half is written as \(\frac{1}{2}\) eating he further divided his half poori into two equal parts and asked Farida what fraction of the whole poori was that piece.
Without answering, Farida also divided her portion of the half puri into two equal parts and kept them beside Subhash’s shares.
She said that these four equal parts together make one whole. So, each equal part is one-fourth of one whole poori and 4 parts together will be \(\frac{4}{4}\) or 1 whole poori.
When they ate, they discussed what they had learnt earlier. Three parts out of 4 equal parts is \(\frac{3}{4}\).
Similarly, \(\frac{3}{7}\) is obtained when we divide a whole into seven equal parts and take three parts. For \(\frac{1}{8}\), we divide a whole into eight equal parts and take one part out of it.
Farida said that we have learnt that a fraction is a number representing part of a whole. The whole may be a single object or a group of objects. Subhash observed that the parts have to be equal.
Let us recapitulate the discussion.
A fraction means a part of a group or of a region.
∴ \(\frac{5}{2}\) is a fraction. We read it as “five-twelfths”.
What does “12” stand for? It is the number of equal parts into which the whole has been divided.
What does “5” stand for? It is the number of equal parts which have been taken out.
Here 5 is called the numerator and 12 is called the denominator.
Name the numerator of \(\frac{3}{7}\) and the denominator of \(\frac{4}{15}\).
You have learnt to show whole numbers like 0,1,2… on a number line.
We can also show fractions on a number line. Let us draw a number line and 1 try to mark \(\frac{1}{2}\) on it.
We know that \(\frac{1}{2}\) is greater than 0 and less than 1, so it should lie between 0 and 1.
Since we have to show b, we divide the gap between 0 and 1 into two equal parts and show 1 part as \(\frac{1}{2}\).
Suppose we want to show \(\frac{1}{3}\) on a number line. Into how many equal parts should the length between 0 and 1 be divided? We divide the length between 0 and 1 into 3 equal parts and show one part as \(\frac{1}{3}\)
Can we show \(\frac{2}{3}\) on this number line? \(\frac{2}{3}\) means 2 parts out of 3 parts as shown.
Similarly, how would you show \(\frac{0}{3}\) and \(\frac{3}{3}\) on this number line?
⇒ \(\frac{0}{3}\) is the point zero whereas since \(\frac{3}{3}\) is 1 whole, it can be shown by the point 1.
So if we have to show \(\frac{3}{7}\) on a number line, then, into how many equal parts should the length between 0 and 1 be divided? If P shows \(\frac{3}{7}\) then how many equal divisions lie between 0 and P? Where do \(\frac{0}{7}\) and \(\frac{7}{7}\) lie?
You have now learnt how to locate fractions on a number line. Locate the fractions \(\frac{3}{4}, \frac{1}{2}, \frac{9}{10}, \frac{0}{3}, \frac{5}{8}\) on separate number lines.
Does any one of the fractions lie beyond 1?
All these fractions lie to the left of las they are less than 1.
In fact, all the fractions we have learnt so far are less than 1. These are proper fractions. A proper fraction as Farida said, is a number representing part of a whole.
In a proper fraction, the denominator shows the number of parts into which the whole is divided and the numerator shows the number of parts which have been considered. Therefore, in a proper fraction, the numerator is always less than the denominator.
Anagha, Ravi, Reshma and John shared their tiffin. Along with their food, they had also, brought 5 apples. After eating the other food, the four friends wanted to eat apples.
How can they share five apples among four of them?
Anagha said, ‘Let each of us have one full apple and a quarter of the fifth apple.’
Reshma said, ‘That is fine, but we can also divide each of the five apples into 4 equal parts and take one-quarter from each apple.’
Ravi said, ‘In both the ways of sharing each of us would get the same share, i.e., 5 quarters. Since 4 quarters make one whole, we can also say that each of us would get 1 whole and one quarter.
The value of each share would be five divided by four. Is it written as 5 -r 4?’ John said, ‘Yes the same as \(\frac{5}{4}\). Reshma added that in \(\frac{5}{4}\), the numerator is bigger than the denominator.
The fractions, where the numerator is bigger than the denominator are called improper fractions.
Thus, fractions like \(\frac{3}{2}, \frac{12}{7}, \frac{18}{5}\) are all improper fractions.
Ravi reminded John, ‘What is the other way of writing the share? Does it follow from Anagha’s way of dividing 5 apples?’
John nodded, ‘Yes, It indeed follows from Anagha’s way. In her way, each share is one whole and one quarter. It is 1 + \(\frac{1}{4}\) and written in short as 1+\(\frac{1}{4}\) and written in short as 1\(\frac{1}{4}\). Remember, 1\(\frac{1}{4}\) is the same as \(\frac{5}{4}\).
Recall the pooris eaten by Farida, She got 2\(\frac{1}{2}\) poories.
How many shaded halves are there in 2\(\frac{1}{2}\)? There are 5 shaded halves.
So, the fraction can also be written as \(\frac{5}{2}\), 2\(\frac{1}{2}\) is the Same as \(\frac{5}{2}\).
Fractions such as 1\(\frac{1}{4}\) and 2\(\frac{1}{2}\) are called Mixed Fractions. A mixed fraction is a combination of a whole and a part.
Where do you come across mixed fractions? Give some examples.
Example 1. Express the following mixed fractions as improper fractions:
Solution:
Thus, we can express a mixed fraction as an improper fraction as \(\frac{\text { (Whole } \times \text { Denominator })+ \text { Numerator }}{\text { Denominator }} \text {. }\)
Look at all these representations of fractions.
These fractions are \(\frac{1}{2}, \frac{2}{4}, \frac{3}{6}\), representing the parts taken from the total number of parts.
If we place the pictorial representation of one over the other they are found to be equal. Do you agree?
These fractions are called equivalent fractions. Think of three more fractions that are equivalent to the above fractions.
Understanding Equivalent Fractions \(\frac{1}{2}, \frac{2}{4}, \frac{3}{6}, \ldots, \frac{36}{72} \ldots\) are all equivalent fractions. They represent the same part of a whole.
Think, Discuss And Write Why do the equivalent fractions represent the same part of a whole? How can we obtain one from the other?
We note \(\frac{1}{2}=\frac{2}{4}=\frac{1 \times 2}{2 \times 2}\) Similarly, \(\frac{1}{2}=\frac{3}{6}=\frac{1 \times 3}{2 \times 3}=\frac{1}{2}\) and \(\frac{1}{2}=\frac{4}{8}=\frac{1 \times 4}{2 \times 4}\)
To find an equivalent fraction of a given fraction, you may multiply both the numerator and the denominator of the given fraction by the same number.
Rajni says that equivalent fractions of \(\frac{1}{3}\) are: \(\frac{1 \times 2}{3 \times 2}=\frac{2}{6}, \quad \frac{1 \times 3}{3 \times 3}=\frac{3}{9}, \quad \frac{1 \times 4}{3 \times 4}=\frac{4}{12}\) and many more.
Another Way Is there any other way to obtain equivalent fractions?
These include an equal number of shaded things i.e. \(\frac{4}{6}=\frac{2}{3}=\frac{4 \div 2}{6 \div 2}\)
To find an equivalent fraction, we may divide both the numerator and the denominator by the same number.
One equivalent fraction of \(\frac{12}{15} \text { is } \frac{12 \div 3}{15 \div 3}=\frac{4}{5}\)
Can you find an equivalent fraction of \(\frac{9}{15}\) having denominator 5?
Example 1. Find the equivalent fraction of \(\frac{2}{5}\) with numerator 6.
Solution: We know 2×3 = 6. This means we need to multiply both the numerator and the denominator by 3 to get the equivalent fraction.
Hence, \(\frac{2}{5}=\frac{2 \times 3}{5 \times 3}=\frac{6}{15} ; \frac{6}{15}\) is the required equivalent fraction.
Example 2. Find the equivalent fraction of \(\frac{15}{35}\) with denominator 7.
Solution: We have \(\frac{15}{35}=\frac{□}{7}\)
We observe the denominator and find 35 – 5 = 7. We, therefore, divide both the numerator and the denominator \(\frac{15}{35}\) by 5.
Thus, \(\frac{15}{35}=\frac{15 \div 5}{35 \div 5}=\frac{3}{7}\)
An interesting fact
Let us now note an interesting fact about equivalent fractions. For this, complete the given table. The first two rows have already been completed for you.
What do we infer? The product of the numerator of the first and the denominator of the second is equal to the product of the denominator of the first and the numerator of the second in all these cases. These two products are called cross-products.
Work out the cross-products for other pairs of equivalent fractions. Do you find any pair of fractions for which cross products are not equal? This rule is helpful in finding equivalent fractions.
Example 3. Find the equivalent fraction of \(\frac{2}{9}\) with denominator 63.
Solution: We have \(\frac{2}{9}=\frac{□}{63}\)
For this, we should have, 9 x □ = 2 x 63.
But 63 = 7 x 9, so 9 x □ = 2 x 7 x 9 = 14 x 9 = 9 x 14 or 9 x □ = 9 x 14
By comparison, □ = 14.
Therefore, \(\frac{2}{9}=\frac{14}{63}\)
Given the fraction \(\frac{36}{54}\), let us try to get an equivalent fraction in which the numerator and the denominator have no common factor except 1.
How do we do it? We see that both 36 and 54 are divisible by 2.
∴ \(\frac{36}{54}=\frac{36 \div 2}{54 \div 2}=\frac{18}{27}\)
But 18 and 27 also have common factors other than one. The common factors are 1, 3, 9; the highest is 9.
Therefore, \(\frac{18}{27}=\frac{18 \div 9}{27 \div 9}=\frac{2}{3}\)
Now 2 and 3 have no common factor except 1; we say that the fraction – is in the simplest form.
A fraction is said to be in the simplest (or lowest) form if its numerator and denominator have no common factor except 1.
The Shortest Way The shortest way to find the equivalent fraction in the simplest form is to find the HCF of the numerator and denominator, and then divide both of them by the HCF.
Consider \(\frac{36}{24}\).
The HCF of 36 and 24 is 12.
Therefore, \(\frac{36}{24}=\frac{36 \div 12}{24 \div 12}=\frac{3}{2}\).
The fraction \(\frac{3}{2}\) is in the lowest form.
Thus, HCF helps us to reduce a fraction to its lowest form.
Fractions with the same denominators are called fractions.
Thus, \(\frac{1}{15}, \frac{2}{15}, \frac{3}{15}, \frac{8}{15}\) are all like fractions. Are \(\frac{7}{27}\) and \(\frac{7}{28}\) like fractions?
Their denominators are different. Therefore, they are not like fractions. They are called unlike fractions.
Sohni has 3\(\frac{1}{2}\) rotis in her plate and Rita has 2\(\frac{3}{4}\) rods in her plate. Who has more rotis on her plate? Clearly, Sohni has 3 full rotis and more and Rita has less than 3 rotis. So, Sohni has more rotis.
Consider \(\frac{1}{2}\) and \(\frac{1}{3}\) as shown. The portion of the whole corresponding to \(\frac{1}{2}\) is clearly larger than the portion of the same whole corresponding to \(\frac{1}{3}\).
So \(\frac{1}{2}\) is greater than \(\frac{1}{3}\).
But often it is not easy to say which one out of a pair of fractions is larger. For example, which is greater, \(\frac{1}{4}\) or \(\frac{3}{10}\)?
For this, we may wish to show the fractions using figures, but drawing figures may not be easy, especially with denominators like 13.
We should therefore like to have a systematic procedure to compare fractions. It is particularly easy to compare fractions. We do this first.
Comparing Like Fractions Fractions are fractions with the same denominator. Which of these are like fractions?
∴ \(\frac{2}{5}, \frac{3}{4}, \frac{1}{5}, \frac{7}{2}, \frac{3}{5}, \frac{4}{5}, \frac{4}{7}\)
Let us compare two like fraction, \(\frac{3}{8}\) and \(\frac{5}{8}\)
In both the fractions the whole is divided into 8 equal parts. For and \(\frac{3}{8}\) and \(\frac{5}{8}\), we take 3 and 5 parts respectively out of the 8 equal parts.
Clearly, out of 8 equal parts, the portion corresponding to 5 parts is larger than the portion corresponding to 3 parts. Hence, \(\frac{5}{8}\) > \(\frac{3}{8}\).
Note the number of the parts taken is given by the numerator. It is, therefore, clear that for two fractions with the same denominator, the fraction with the greater numerator is greater.
Between \(\frac{4}{5}\) and \(\frac{3}{5}\), \(\frac{4}{5}\) is greater. Between \(\frac{11}{20}\) and \(\frac{13}{20}\), \(\frac{13}{20}\) is greater and so on.
Comparing Unlike Fractions Two fractions are unlike if they have different denominators. For example, \(\frac{1}{3}\) and \(\frac{1}{5}\) are unlike fractions. So are \(\frac{2}{3}\) and \(\frac{3}{5}\).
Unlike Fractions With The Same Numerator: Consider a pair of unlike fractions \(\frac{1}{3}\) and \(\frac{1}{5}\), in which the numerator is the same.
Which is greater \(\frac{1}{3}\) or \(\frac{1}{5}\)?
In \(\frac{1}{3}\), we divide the whole into 3 equal parts and take one. In \(\frac{1}{5}\), we divide the whole into 5 equal parts and take one. Note that in \(\frac{1}{3}\), the whole is divided into a smaller number of parts than in \(\frac{1}{5}\).
The equal part that we get in \(\frac{1}{3}\) is, therefore, larger than the equal part we get in \(\frac{1}{5}\).
Since in both cases, we take the same number of parts (i.e. one), the portion of the whole showing \(\frac{1}{3}\) is larger than the portion showing \(\frac{1}{5}\), and therefore \(\frac{1}{3}\) > \(\frac{1}{5}\).
In the same way we can say \(\frac{2}{3}\) > \(\frac{2}{5}\). In this case, the situation is the same as in the case above, except that the common numerator is 2, not 1.
The whole is divided into a large number of equal parts for \(\frac{2}{5}\) than for \(\frac{2}{3}\).
Therefore, each equal part of the whole in case of \(\frac{2}{3}\) is larger than that in case of \(\frac{2}{5}\). Therefore, the portion of the whole showing \(\frac{2}{3}\) is larger than the portion showing \(\frac{2}{5}\) and hence, \(\frac{2}{3}\) > \(\frac{2}{5}\).
We can see from the above example that if the numerator is the same in two fractions, the fraction with the smaller denominator is greater of the two.
Thus \(\frac{1}{8}>\frac{1}{10}, \frac{3}{5}>\frac{3}{7}, \frac{4}{9}>\frac{4}{11}\) and so on.
Let us arrange \(\frac{2}{1}, \frac{2}{13}, \frac{2}{9}, \frac{2}{5}, \frac{2}{7}\) in increasing order. All these fractions are unlike, but their numerator is the same.
Hence, in such a case, the larger the denominator, the smaller is the fraction. The smallest is \(\frac{2}{13}\), as it has the largest denominator.
The next three fractions in order are \(\frac{2}{9}, \frac{2}{7}, \frac{2}{5}\). The greatest fraction is \(\frac{2}{1}\)(It is with the smallest denominator).
The arrangement in increasing order, therefore, is \(\frac{2}{13}, \frac{2}{9}, \frac{2}{7}, \frac{2}{5}, \frac{2}{1}\).
Suppose we want to compare \(\frac{2}{3}\) and \(\frac{3}{4}\). Their numerators are different and so are their denominators. We know how to compare fractions, i.e. fractions with the same denominator.
We should, therefore, try to change the denominators of the given fractions, so that they become equal. For this purpose, we can use the method of equivalent fractions which we already know. Using this method we can change the denominator of a fraction without changing its value.
Let us find equivalent fractions of both \(\frac{2}{3}\) and \(\frac{3}{4}\).
∴ \(\frac{2}{3}=\frac{4}{6}=\frac{6}{9}=\frac{8}{12}=\frac{10}{15}=\ldots\).
Similarly, \(\frac{3}{4}=\frac{6}{8}=\frac{9}{12}=\frac{12}{16}=\ldots\).
The equivalent fractions of \(\frac{2}{3}\) and \(\frac{3}{4}\) with the same denominator 12 are \(\frac{8}{12}\) and \(\frac{9}{12}\) respectively.
i.e. \(\frac{2}{3}=\frac{8}{12}\) and \(\frac{3}{4}=\frac{9}{12}\)
Since, \(\frac{9}{12}>\frac{8}{12}\) we have, \(\frac{3}{4}>\frac{2}{3} \text {. }\)
Example 1. Compare \(\frac{4}{5}\) and \(\frac{5}{6}\)
Solution:
The fractions are unlike fractions. Their numerator is different too. Let us write their equivalent fractions.
⇒ \(\frac{4}{5}=\frac{8}{10}=\frac{12}{15}=\frac{16}{20}=\frac{20}{25}=\frac{24}{30}=\frac{28}{35}=\ldots \ldots \ldots . .\)
and \(\frac{5}{6}=\frac{10}{12}=\frac{15}{18}=\frac{20}{24}=\frac{25}{30}=\frac{30}{36}=\ldots \ldots \ldots . .\)
The equivalent fractions with the same denominator are: \(\frac{4}{5}=\frac{24}{30}\) and \(\frac{5}{6}=\frac{25}{30}\)
Since, \(\frac{25}{30}>\frac{24}{30}\) so, \(\frac{5}{6}>\frac{4}{5}\)
Note that the common denominator of the equivalent fractions is 30 which is 5 x 6. It is a common multiple of both 5 and 6.
So, when we compare two unlike fractions, we first get their equivalent fractions with a denominator which is a common multiple of the denominators of both fractions.
Example 2. Compare \(\frac{5}{6}\) and \(\frac{13}{15}\).
Solution: The fractions are unlike. We should first get their equivalent fractions with a denominator which is a common multiple of 6 and 15.
Now, \(\frac{5 \times 5}{6 \times 5}=\frac{25}{30}, \frac{13 \times 2}{15 \times 2}=\frac{26}{30}\)
Since \(\frac{26}{30}>\frac{25}{30}\) we have \(\frac{13}{15}>\frac{5}{6}\).
Why LCM?
The product of 6 and 15 is 90; obviously 90 is also a common multiple of 6 and 15. We may use 90 instead of 30; it will not be wrong. But we know that it is easier and more convenient to work with smaller numbers.
So the common multiple that we take is as small as possible. This is why the LCM of the denominators of the fractions is preferred as the common denominator.
So far in our study, we have learnt about natural numbers, whole numbers and then integers. In the present chapter, we are learning about fractions and different types of numbers.
Whenever we come across new types of numbers, we want to know how to operate with them.
Can we combine and add them? If so, how? Can we take away some number from another? i.e., can we subtract one from the other? and so on.
Which of the properties learnt earlier about the numbers hold now? Which are the new properties? We also see how these help us deal with our daily life situations.
Adding Or Subtracting Like Fractions
All fractions cannot be added orally. We need to know how they can be added in different situations and learn the procedure for it. We begin by looking at the addition of like fractions.
Take a 7 x 4 grid sheet. The sheet has seven boxes in each row and four boxes in each column.
How many boxes are there in total?
Colour five of its boxes in green.
What fraction of the whole is the green region?
Now colour another four of its boxes in yellow.
What fraction of the whole is this yellow region?
What, fraction of the whole is coloured altogether?
Does this explain that \(\frac{5}{28}+\frac{4}{28}=\frac{9}{28} ?\)
Look At More Examples
(1) we have 2 quarter parts of the figure shaded. This means we have 2 parts out of 4 shaded or \(\frac{1}{2}\) of the figure shaded.
That is, \(\frac{1}{4}+\frac{1}{4}=\frac{1+1}{4}=\frac{2}{4}=\frac{1}{2}\).
Look at (2) Demonstrates \(\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{1+1+1}{9}=\frac{3}{9}=\frac{1}{3}\).
What do we learn from the above examples? The sum of two or more like fractions can be obtained as follows:
Step 1 Add the numerators.
Step 2 Retain the (common) denominator.
Step 3 Write the fraction as: \(\frac{\text { Result of Step } 1}{\text { Result of Step } 2}\)
Let us, thus, add \(\frac{3}{5}\) and \(\frac{1}{5}\).
We have \(\frac{3}{5}+\frac{1}{5}=\frac{3+1}{5}=\frac{4}{5}\)
So, what will be the sum of \(\frac{7}{12}\) and \(\frac{3}{12}\)?
Finding The Balance
Sharmila had \(\frac{5}{6}\) of a cake. She gave \(\frac{2}{6}\) out of that to her younger brother. How much cake is left with her?
A diagram can explain the situation. (Note that, here the given fractions are like fractions).
We find that \(\frac{5}{6}-\frac{2}{6}=\frac{5-2}{6}=\frac{3}{6} \text { or } \frac{1}{2}\)
(Is this not similar to the method of adding like fractions?)
Thus, we can say that the difference of two like fractions can be obtained as follows:
Step 1 Subtract the smaller numerator from the bigger numerator.
Step 2 Retain the (common) denominator.
Step 3 Write the fraction as: \(\frac{\text { Result of Step } 1}{\text { Result of Step } 2}\)
Can we now subtarct \(\frac{3}{10}\) from \(\frac{8}{10}\)?
Adding And Subtracting Fractions
We have learnt to add and subtract like fractions. It is also not very difficult to add fractions that do not have the same denominator. When we have to add or subtract fractions we first find equivalent fractions with the same denominator and then proceed.
What added to \(\frac{1}{5}\) gives \(\frac{1}{2}\)? This means subtract \(\frac{1}{5}\) from \(\frac{1}{2}\) to get the required number.
Since \(\frac{1}{5}\) and \(\frac{1}{2}\) are unlike fractions, in order to subtract them, we first find their equivalent fractions with the same denominator. These are \(\frac{2}{10}\) and \(\frac{5}{10}\) respectively.
This is because \(\frac{1}{2}=\frac{1 \times 5}{2 \times 5}=\frac{5}{10}\)
and \(\frac{1}{5}=\frac{1 \times 2}{5 \times 2}=\frac{2}{10}\)
Therefore, \(\frac{1}{2}=\frac{1}{5}=\frac{5}{10}-\frac{2}{10}=\frac{5-2}{10}=\frac{3}{10}\)
Note that 10 is the least common multiple (LCM) of 2 and 5
Example 1. Subtract \(\frac{3}{4}\) from \(\frac{5}{6}\)
Solution: We need to find equivalent fractions of \(\frac{3}{4}\) and \(\frac{5}{6}\), which have the same denominator. This denominator is given by the LCM of 4 and 6. The required LCM is 12.
Therefore, \(\frac{5}{6}-\frac{3}{4}=\frac{5 \times 2}{6 \times 2}-\frac{3 \times 3}{4 \times 3}=\frac{10}{12}-\frac{9}{12}=\frac{1}{12}\)
Example 2. Add \(\frac{2}{5}\) to \(\frac{1}{3}\)
Solution: The LCM of 5 and 3 is 15.
Therefore, \(\frac{2}{5}+\frac{1}{3}=\frac{2 \times 3}{5 \times 3}+\frac{1 \times 5}{3 \times 5}=\frac{6}{15}+\frac{5}{15}=\frac{11}{15}\)
Example 3. Simplify: \(\frac{3}{5}\) – \(\frac{7}{20}\)
Solution: The LCM of 5 and 20 is 20.
Therefore, \(\frac{3}{5}-\frac{7}{20}=\frac{3 \times 4}{5 \times 4}-\frac{7}{20}=\frac{12}{20}-\frac{7}{20}\)
= \(\frac{12-7}{20}=\frac{5}{20}=\frac{1}{4}\)
How do we add or subtract mixed fractions? Mixed fractions can be written either as a whole part plus a proper fraction or entirely as an improper fraction.
One way to add (or subtract) mixed fractions is to do the operation separately for the whole parts and the other way is to write the mixed fractions as improper fractions and then directly add (or subtract) them.
Example 4. Add \(2 \frac{4}{5}\) and \(3 \frac{5}{6}\)
Solution: \(2 \frac{4}{5}+3 \frac{5}{6}=2+\frac{4}{5}+3+\frac{5}{6}=5+\frac{4}{5}+\frac{5}{6}\)
Now \(\frac{4}{5}+\frac{5}{6}=\frac{4 \times 6}{5 \times 6}+\frac{5 \times 5}{6 \times 5}\)(Since LCM of 5 and 6=30)
= \(\frac{24}{30}+\frac{25}{30}=\frac{49}{30}=\frac{30+19}{30}=1+\frac{19}{30}\)
Thus, \(5+\frac{4}{5}+\frac{5}{6}=5+1+\frac{19}{30}=6+\frac{19}{30}=6 \frac{19}{30}\)
And, therefore, \(2 \frac{4}{5}+3 \frac{5}{6}=6 \frac{19}{30}\)
Example 5. Find \(4 \frac{2}{5}-2 \frac{1}{5}\)
Solution: The whole numbers 4 and 2 and the fractional numbers \(\frac{2}{5}\) and \(\frac{1}{5}\) can be subtracted separately. (Note that 4>2 and \(\frac{2}{5}>\frac{1}{5}\))
So, \(4 \frac{2}{5}-2 \frac{1}{5}=(4-2)+\left(\frac{2}{5}-\frac{1}{5}\right)=2+\frac{1}{5}=2 \frac{1}{5}\)
Example 6. Simplify: \(8 \frac{1}{4}-2 \frac{5}{6}\)
Solution: Here 8>2 but \(\frac{1}{4}<\frac{5}{6}\).
We proceed as follows: \(8 \frac{1}{4}=\frac{(8 \times 4)+1}{4}=\frac{33}{4}\) and \(2 \frac{5}{6}=\frac{2 \times 6+5}{6}=\frac{17}{6}\)
Now, \(\frac{33}{4}-\frac{17}{6}=\frac{33 \times 3}{12}-\frac{17 \times 2}{12}\) (Since LCM of 4 and 6=12)
= \(\frac{99-34}{12}=\frac{65}{12}=5 \frac{5}{12}\)