# NCERT Solutions Class 6th Mathematics

At the beginning of the new academic session, students are very excited about new books and are also curious to know what they're going to learn in Class 6th. They must have a clear idea of their curriculum. Mathematics is one of the most important and challenging subjects in class 6th and students must refer to the NCERT Solutions for Class 6 Mathematics for guidance in any subject because NCERT solutions follow the basic and simpler approach to concepts and avoid any crafty methods for students. The concepts included in the NCERT Solutions are simplified so that students can optimise their potential and be happy and satisfied with their results.

## NCERT Class 6 Maths Solutions PDF Download

Students who want to have a comprehensive understanding of each chapter can refer to the concepts of Mathematics provided on the  Extramarks website. With the class 6 Maths Solutions on Extramarks, students can easily access everything they are looking for.  They can get all the chapters provided in  Mathematics for free.

### NCERT Solutions for Class 6 Maths

Mathematics:

• Knowing our numbers
• Whole numbers
• Playing with numbers
• Basic geometrical ideas
• Understanding elementary shapes
• Integers
• Fractions
• Decimals
• Data handling
• Mensuration
• Algebra
• Ratio and Proportion
• Symmetry
• Practical Geometry

NCERT Class 6 Maths Chapter wise Solutions in Hindi Medium

Students looking for NCERT Class 6 Maths chapter-wise solutions in Hindi Medium can refer to Extramarks. We do not want language to become a barrier in your success, therefore, experts at Extramarks have prepared solutions in Hindi for NCERT Class 6 Mathematics. Download them from the website.

### Chapter 1: Knowing Our Numbers

The chapter covers interesting topics like the formation of large numbers, arranging the numbers in ascending and descending order, shifting digits to create new numbers, and place value tables. The chapter has solved and unsolved exercises along with real-life examples.

### Chapter 2: Whole Numbers

The chapter covers whole numbers, their properties, and the number line. The last topic in the chapter discusses patterns in whole numbers that are arranged in the form of dots or numbers. There are a total of three exercises in this chapter.

### Chapter 3: Playing with Numbers

An interesting chapter that covers tests for divisibility of numbers, multiples and factors, prime factorization, prime and composite numbers, highest common factor and problems on LCM and HCF.

### Chapter 4: Basic Geometrical Ideas

The chapter covers points, line segment, interesting lines, curves, polygons, and parallel lines. Students will also be introduced to angles and shapes. NCERT books have diagrams and examples to explain the chapter in a better way.

### Chapter 5: Understanding Elementary Shapes

Students will learn about concepts related to shapes that we see around in daily life. The chapter starts with an introduction to Measuring Line Segments, and covers different methods of comparison. Measuring angles, quadrilaterals, perpendicular lines, three-dimensional shapes, and polygons are some other topics covered in Chapter 5.

### Chapter 6: Integers

The chapter familiarises students with Integers and how they are represented on a number line. A detailed explanation of addition of integers, additive inverse, and subtraction of integers, is also given in the chapter.

### Chapter 7: Fractions

Fractions are studied in higher classes as well, which makes this introductory chapter very important for students. Fractions on a number line, different types of fractions, and the simplest form of a fraction are some of the topics covered in Chapter 7.

### Chapter 8: Decimals

Chapter 8 teaches students how to work with decimals. In the first half of the chapter, students are introduced to concepts like hundredths, tenths, and comparing decimals, while the second half deals with the addition and subtraction of numbers with decimals. There are six unsolved exercises at the end of the chapter.

### Chapter 9: Data Handling

The chapter teaches how to deal with different types of data. Concepts related to recording and organising data are covered in the chapter. Other topics include bar graphs, pictographs, and tally marks.

### Chapter 10: Mensuration

The Chapter 10 covers concepts related to perimeter and area. The two methods of calculating the area - using graph paper and applying the direct formula - are explained in the chapter. There are three unsolved exercises at the end of the chapter.

### Chapter 11: Algebra

The chapter Algebra starts with the concept of Matchsticks Pattern and the idea of a variable. It also includes a detailed explanation of an equation, and the solution of the equation done through the trial and error method.

### Chapter 12: Ratio and Proportion

The chapter explains ratios in detail by providing many solved examples. It also covers proportion followed by the unitary method. Students must understand this chapter in a detailed way because Ratio and Proportion are topics that are helpful in our daily life.

### Chapter 13: Symmetry

The chapter Symmetry will introduce you to the concept of Figures with Two Lines of Symmetry, Reflection and Symmetry, and Figures with Multiple Lines of Symmetry. The concepts are explained with the help of images so that students have a clearer understanding of the chapter.

### Chapter 14: Practical Geometry

The Chapter 14 Practical Geometry covers all the fundamentals of drawing different shapes. It introduces students to geometrical tools and their uses. They will also learn about construction of angles, circles, and line segments. Bisectors of an angle and perpendicular bisector are also explained in the chapter.

NCERT Maths Book Class 6 Chapter 1 Solutions Knowing Our Numbers

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers

NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes

NCERT Solutions for Class 6 Maths Chapter 6 Integers

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

NCERT Solutions for Class 6 Maths Chapter 8 Decimals

NCERT Solutions for Class 6 Maths Chapter 9 Data Handling

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion

NCERT Solutions for Class 6 Maths Chapter 13 Symmetry

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry

### Last-Minute Exam Tips for Maths

Here are a few tips  for students to speed up their preparation at the eleventh hour:

• Set a particular time when you're going to revise topic-wise problems.
•  Instead of solving the questions right away, try to read the question properly and then try to solve the question not vice-versa. It saves time.
• If there is a possibility of drawing a diagram in the question, draw the diagram as it will help you answer the question properly. Try to practice the diagram questions properly.
• Understand the logic behind the answer rather than memorising it at the last moment. Avoid cramming in Mathematics except for the formulas.
• Practise as many previous year question papers as possible.
• Do not take the basic calculations lightly and don't answer the questions in a hurry, especially the easy ones.

### Benefits of Studying from Class 6 NCERT Books and Solutions

The benefits of studying from NCERT solutions are-

• NCERT solutions provide detailed explanations of every concept which makes it easier for students to understand the concepts and work on them. It also helps them to follow the same pattern in the exam, to avoid those silly errors and improve on their score.
• NCERT books are enough for students to practise for the class 6 Mathematics. It has plenty of examples along with in-text and end-text exercises that help students in practising the concepts and preparing ahead of the examination.
• Students of class 6 do not have to refer to any other guide. They can solely refer to the NCERT solutions book and it will be enough to prepare them for their exam thoroughly.
• Students who refer to the class 6 NCERT solutions will be able to practise the diagrams questions properly.

Why Choose Extramarks For Class 6 Maths Solutions?

Extramarks provide detailed explanations of every concept. Students of Class 6 can access Extramarks website to solve any queries in any chapter in Mathematics. We aim at improving the fundamentals concepts of students which  has a positive impact on studies in higher classes in all subjects. Students can take the help from subject matter experts as well.

Why Should One Download NCERT Solutions of Class 6 Mathematics ??

Mathematics is all about practice. The concepts of students will become clear in their head depending upon how much practice they have done on a regular basis. To understand a mathematical equation thoroughly, students must know the steps involved in the question which can be easily found in the class 6 Maths Solutions of NCERT. Yet another thing that is important Mathematics is accuracy. Students will have to be precise with their answers to score well in the exam, which is taught in the NCERT books of class 6 Mathematics. Students must learn how to solve the question step by step to get the right answer.

Q.1 Find the measure of the angle shown in each figure. (First estimate with your eyes and then find the actual measure with a protractor).

Ans

Measure of given angle is 40°

Measure of given angle is 120°

Measure of given angle is 60°

Measure of given angle is 130°

Q.2 Find the angle measure between the hands of the clock in each figure:

Ans

(a) Right angle; 90°

(b) Acute angle; 30°

(c) Straight angle; 180°

Q.3 Investigate

In the given figure, the angle measures 30°.
Look at the same figure through a magnifying glass.
Does the angle becomes larger? Does the size of the angle change?

Ans

No, there is no change in the measurement of the Angle, if looked through a magnifying glass.

Q.4 Measure and classify each angle:

 Angle Measure Type ∠AOB ∠AOC ∠BOC ∠DOC ∠DOA ∠DOB

Ans

 Angle Measure Type ∠AOB 40° Acute angle ∠AOC 130° Obtuse angle ∠BOC 90° Right angle ∠DOC 90° Right angle ∠DOA 140° Obtuse angle ∠DOB 180° Straight angle

Q.5 Which of the following are models for perpendicular lines :
(a) The adjacent edges of a table top.
(b) The lines of a railway track.
(c) The line segments forming the letter ‘L’.
(d) The letter V.

Ans

(a) The adjacent edges of a table top are perpendicular lines as they form a right angle.

(b) The lines of a railway track do not form a right angle. As they are parallel lines and forms a zero angle.

(c) The line segments of the letter ‘L’ are perpendicular lines as they form a right angle.

(d) The letter V do not form a right angle but forms an acute angle.

Q.6

$\begin{array}{l}\mathrm{Let}\overline{\mathrm{PQ}}\mathrm{be}\mathrm{the}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{line}\mathrm{segment}\overline{\mathrm{XY}}.\\ \mathrm{Let}\overline{\mathrm{PQ}}\mathrm{and}\overline{\mathrm{XY}}\mathrm{intersect}\mathrm{in}\mathrm{the}\mathrm{point}\mathrm{A}.\mathrm{What}\mathrm{is}\mathrm{the}\\ \mathrm{measure}\mathrm{of}\angle \mathrm{PAY}?\end{array}$

Ans

The measure of angle PAY is 90°.

Q.7 There are two set-squares in your box. What are the measures of the angles that are formed at their corners? Do they have any angle measure that is common?

Ans

One set square is of 30°- 60°- 90° and another set square is of 45°- 45°- 90°. The right angle i.e. 90° is common between them.

Q.8 Study the diagram. The line l is perpendicular to line m

(a) Is CE = EG?
(b) Does PE bisect CG?
(c) Identify any two line segments for which PE is the perpendicular bisector.
(d) Are these true?
(i) AC > FG
(ii) CD = GH
(iii) BC < EH.

Ans

(a) Yes CE = EG; because C and G both are at equal distance from E.

(b) Yes; because E lies exactly half way between C and G.

(c)

$\overline{\mathrm{BH}}\text{and}\overline{\text{DF}}$

(d) (i) True; AC > FG

(ii) True; CD = GH

(iii) True; BC < EH

Q.9 Name the types of following triangles:
(a) Triangle with lengths of sides 7 cm, 8 cm and 9 cm.
(b) ΔABC with AB = 8.7 cm, AC = 7 cm and BC = 6 cm.
(c) ΔPQR such that PQ = QR = PR = 5 cm.
(d) ΔDEF with m∠D= 90°
(e) ΔXYZ with m∠Y= 90° and XY = YZ.
(f) ΔLMN with m∠L = 30°, m∠M = 70° and m∠N= 80°.

Ans

(a) All the sides of the triangle are unequal, so it is a scalene triangle.

(b) All the sides of triangle ABC are unequal, so triangle ABC is a scalene triangle.

(c) All the sides of triangle PQR are equal, so it is an equilateral triangle.

(d) In triangle DEF, angle D is of 90°, so it is a right-angled triangle.

(e) In triangle XYZ, two sides XY and YZ are equal and angle Y is of 90°, so it is an isosceles right triangle.

(f) In triangle LMN, all the angles are acute, so it is an acute-angled triangle.

Q.10 Match the following :
Measures of Triangle Type of Triangle
(i) 3 sides of equal length (a) Scalene
(ii) 2 sides of equal length (b) Isosceles right angled
(iii) All sides are of different length (c) Obtuse angled
(iv) 3 acute angles (d) Right angled
(v) 1 right angle (e) Equilateral
(vi) 1 obtuse angle (f) Acute angled
(vii) 1 right angle with two sides of equal length (g) Isosceles

Ans

Measures of Triangle Type of Triangle
(i) 3 sides of equal length (e) Equilateral triangle
(ii) 2 sides of equal length (g)Isosceles
(iii) All sides are of different length (a)Scalene
(iv) 3 acute angles (f)Acute angled
(v) 1 right angle (d)Right angled
(vi) 1 obtuse angle (c)Obtuse angled
(vii) 1 right angle with two sides of equal length (b)Isosceles right angled

Q.11 Name each of the following triangles in two different ways: (you may judge the nature of the angle by observation)

Ans

(a) This triangle has two sides of equal length and all angles are acute. So, it is an isosceles acute triangle.

(b) This triangle has all sides unequal and one right angle. So, it is a scalene-right triangle

(c) This triangle has two equal sides and one obtuse angle. So, it is an obtuse-isosceles triangle.

(d) This triangle has two equal sides and one right angle. So, it is an isosceles-right triangle.

(e) This triangle has all sides equal. So, it is an equilateral acute triangle.

(f) This triangle has all sides unequal and one obtuse angle. So, it is a scalene-obtuse triangle.

Q.12 Try to construct triangles using match sticks. Some are shown here. Can you make a triangle with
(a) 3 matchsticks?
(b) 4 matchsticks?
(c) 5 matchsticks?
(d) 6 matchsticks?
(Remember you have to use all the available matchsticks in each case) Name the type of triangle in each case. If you cannot make a triangle, think of reasons for it.

Ans

(a)

(b) It is not possible to make triangle using four match sticks because sum of any two sides of a triangle is always greater than the third side.

(c)

(d)

Q.13 Say True or False:
(a) Each angle of a rectangle is a right angle.
(b) The opposite sides of a rectangle are equal in length.
(c) The diagonals of a square are perpendicular to one another.
(d) All the sides of a rhombus are of equal length.
(e) All the sides of a parallelogram are of equal length.
(f) The opposite sides of a trapezium are parallel.

Ans

(a) True

(b) True

(c) True

(d) True

(e) False

(f) False

Q.14 Give reasons for the following:
(a) A square can be thought of as a special rectangle.
(b) A rectangle can be thought of as a special parallelogram.
(c) A square can be thought of as a special rhombus.
(d) Squares, rectangles, parallelograms are all quadrilaterals.
(e) Square is also a parallelogram.

Ans

(a) A square is a special rectangle because a rectangle with all sides equal is a square.
(b) A rectangle is a special parallelogram because a parallelogram with all angle of measure 90° becomes a rectangle.
(c) A square is a special rhombus because a rhombus with all angles of measure 90° becomes a square.
(d) Squares, rectangles and parallelograms are quadrilaterals because they all have four sides.
(e) Square is also a parallelogram because its opposite sides are equal and parallel.

Q.15 A figure is said to be regular if its sides are equal in length and angles are equal in measure. Can you identify the regular quadrilateral?

Ans

A square has all sides equal and all angles of equal measure. So, square is a regular quadrilateral.

Q.16 Examine whether the following are polygons. If any one among them is not, say why?

Ans

(a) The given figure is not closed, so it is not a polygon.
(b) The given figure is closed and has six line segments, so it is a polygon.
(c) The given figure is closed but has no line segment, so it is not a polygon.
(d) The given figure is closed but is joined by a curve, so it is not a polygon.

Q.17 Name each polygon.

Make two more examples of each of these.

Ans

1. The given figure has four sides, so it is a quadrilateral. Example : A table top, book
2. The given figure has three sides, so it is a triangle. Example : Pizza slice, Sandwich
3. The given figure has five sides, so it is a pentagon. Example : Road signs, Rangoli patterns
4. The given figure has eight sides, so it is an octagon. Example : Bolt, Some clock designs.

Q.18 Draw a rough sketch of a regular hexagon. Connecting any three of its vertices, draw a triangle. Identify the type of the triangle you have drawn.

Ans

This is the rough sketch of a regular hexagon.

The triangle drawn on connecting any three vertices of the hexagon is an isosceles obtuse triangle.

Q.19 Draw a rough sketch of a regular octagon. (Use squared paper if you wish). Draw a rectangle by joining exactly four of the vertices of the octagon.

Ans

Q.20 A diagonal is a line segment that joins any two vertices of the polygon and is not a side of the polygon. Draw a rough sketch of a pentagon and draw its diagonals.

Ans

Q.21 Match the following:

(a) Cone (i)

(b) Sphere (ii)

(c) Cylinder (iii)

(d) Cuboid (iv)

(e) Pyramid (v)

Give two new examples of each shape.

Ans

 (a) Cone (b) Sphere (c) Cylinder (d) Cuboid (e) Pyramid

Q.22 What shape is
(a) Your instrument box?
(b) A brick?
(c) A match box?
(e) A sweet laddu?

Ans

(a) Cuboid
(b) Cuboid
(c) Cuboid
(d) Cylinder
(e) Sphere

Q.23 Find the perimeter of each of the following figures:

Ans

(a) Perimeter of figure = 4 cm + 2 cm + 1 cm
+ 5 cm
= 12 cm
(b) Perimeter of figure = 23 cm + 35 cm + 35 cm
+ 40 cm
= 133 cm
(c) Perimeter of figure = 15 cm + 15 cm + 15 cm
+ 15 cm
= 60 cm
(d) Perimeter of figure = 4 cm + 4 cm + 4 cm
+ 4 cm + 4 cm
= 20 cm

(e) Perimeter of figure = 2.5 cm + 0.5 cm + 4 cm
+ 1 cm + 4 cm + 0.5 cm
+ 2.5 cm
= 15 cm
(f) Perimeter of figure = 4 cm + 1 cm + 3 cm
+ 2 cm+ 3 cm+ 4 cm+ 1 cm + 3 cm
+ 2 cm + 3 cm + 4 cm + 1 cm + 3 cm
+ 2 cm + 3 cm + 4 cm + 1 cm + 3 cm
+ 2 cm + 3 cm
= 52 cm

Q.24 The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?

Ans

Since, lid of rectangular box is like a rectangle. So,

Length of tape = 2(length + Breadth)
= 2(40 + 10)
= 100 cm
= 1 m

Q.25 Find the perimeter of each of the following figures:

Ans

(a) Perimeter of figure = 4 cm + 2 cm + 1 cm
+ 5 cm
= 12 cm
(b) Perimeter of figure = 23 cm + 35 cm + 35 cm
+ 40 cm
= 133 cm
(c) Perimeter of figure = 15 cm + 15 cm + 15 cm
+ 15 cm
= 60 cm
(d) Perimeter of figure = 4 cm + 4 cm + 4 cm
+ 4 cm + 4 cm
= 20 cm

(e) Perimeter of figure = 2.5 cm + 0.5 cm + 4 cm
+ 1 cm + 4 cm + 0.5 cm
+ 2.5 cm
= 15 cm
(f) Perimeter of figure = 4 cm + 1 cm + 3 cm
+ 2 cm+ 3 cm+ 4 cm+ 1 cm + 3 cm
+ 2 cm + 3 cm + 4 cm + 1 cm + 3 cm
+ 2 cm + 3 cm + 4 cm + 1 cm + 3 cm
+ 2 cm + 3 cm
= 52 cm

Q.26 The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?

Ans

Since, lid of rectangular box is like a rectangle. So,

Length of tape = 2(length + Breadth)
= 2(40 + 10)
= 100 cm
= 1 m

Q.27 A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?

Ans

Length of table-top = 2 m 25 cm

= 2.25 m
Breadth of table-top = 1 m 50 cm

= 1.5 m
Perimeter of table-top = 2(Length + Breadth)

= 2(2.25 + 1.50)
= 2(3.75)
= 7.50 m

Q.28 What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?

Ans

Length of photograph = 32 cm
Breadth of photograph = 21 cm
Perimeter of photograph = 2(Length + Breadth)
= 2(32 + 21)
= 106 cm

Q.29 A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?

Ans

Length of rectangular piece of land = 0.7 km
Breadth of rectangular piece of land = 0.5 km
Perimeter of rectangular piece of land = 2(0.7 + 0.5)
= 2(1.2) km
= 2.4 km

Length of required wire = 4(2.4) km
= 9.6 km
[Since, Length of required wire = 4(Perimeter of land)]

Q.30 Find the perimeter of each of the following shapes:
(a) A triangle of sides 3 cm, 4 cm and 5 cm.
(b) An equilateral triangle of side 9 cm.
(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.

Ans

(a) Perimeter of triangle = 3 cm + 4 cm + 5 cm
= 12 cm
(b) Perimeter of equilateral triangle
= 9 cm + 9 cm + 9 cm
= 27 cm
(c) Perimeter of isosceles triangle
= 8 cm + 8 cm + 6 cm
= 22 cm

Q.31 Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.

Ans

Perimeter of triangle = 10 cm + 14 cm + 15 cm
= 39 cm

Q.32 Find the perimeter of a regular hexagon with each side measuring 8 m.

Ans

Perimeter of regular hexagon = 6(Length of side)
= 6(8 cm)
= 48 cm

Q.33 Find the side of the square whose perimeter is 20 m.

Ans

$\begin{array}{l}\mathrm{Perimeter}\text{of square}=\text{4}×\text{side}\\ \text{\hspace{0.17em}\hspace{0.17em}Side of square}=\frac{\mathrm{Perimeter}}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}Side of square}=\frac{20\text{\hspace{0.17em}}\mathrm{m}}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5\text{\hspace{0.17em}}\mathrm{m}\end{array}$

Q.34 The perimeter of a regular pentagon is 100 cm. How long is its each side?

Ans

$\begin{array}{l}\mathrm{Perimeter}\text{of regular pentagon}=\text{5}×\text{side}\\ \text{\hspace{0.17em}Side of regular pentagon}=\frac{\mathrm{Perimeter}}{5}\\ \text{\hspace{0.17em}\hspace{0.17em}Side of regular pentagon}=\frac{100\text{\hspace{0.17em}}}{4}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=25\text{\hspace{0.17em}}\mathrm{cm}\end{array}$

Q.35 A piece of string is 30 cm long. What will be the length of each side if the string is used to form :
(a) a square?
(b) an equilateral triangle?
(c) a regular hexagon?

Ans

$\begin{array}{l}\text{Length of a piece of string}=\text{3}0\text{m}\mathrm{}\\ \left(\text{a}\right)\mathrm{Perimeter}\text{of square}=\text{Length of a piece of string}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 4}×\mathrm{side}\text{of square}=\text{30 m}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}side of square}=\frac{30}{4}\text{\hspace{0.17em}}\mathrm{m}\\ =7.5\text{\hspace{0.17em}}\mathrm{m}\\ \left(\text{b}\right)\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Perimeter}\text{of an equilateral triangle}\\ =\text{Length of a piece of string}\\ \text{3}×\mathrm{side}\text{of equilateral triangle}\\ =\text{30 m}\\ \text{Side of equilateral triangle}\\ =\frac{30}{3}\text{\hspace{0.17em}}\mathrm{m}\\ =10\text{\hspace{0.17em}}\mathrm{m}\\ \left(\text{c}\right)\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Perimeter}\text{of a regular hexagon}\\ =\text{Length of a piece of string}\\ \text{6}×\mathrm{side}\text{of regular hexagon}\\ =\text{30 m}\\ \text{Side of regular hexagon}\\ =\frac{30}{6}\text{\hspace{0.17em}}\mathrm{m}\\ =5\text{\hspace{0.17em}}\mathrm{m}\end{array}$

Q.36 Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?

Ans

Let third side of triangle be x cm.
Perimeter of triangle = 12 cm + 14 cm + x cm
36 cm = 26 cm + x
x = 36 cm – 26 cm
= 10 cm
Thus, third side of triangle is 10 cm.

Q.37 Sweety runs around a square park of side 75m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance?

Ans

Side of square park = 75 m
Perimeter of square park = 4 (75 m)
= 300 m
Distance covered by Sweety =300 m
Length of park = 60 m
Breadth of park = 45 m
Perimeter of rectangular park = 2 (60 m + 45 m)
= 210 m
Distance covered by Bulbul = 210 m
Thus, Bulbul covers less distance.

Q.38 What is the perimeter of each of the following figures? What do you infer from the answers?

Ans

(a) Perimeter of square = 4(25 cm)
= 100 cm
(b) Perimeter of rectangle = 2(30 cm + 20 cm)
= 2(50 cm)
= 100 cm
(c) Perimeter of rectangle = 2(40 cm + 10 cm)
= 2(50 cm)
= 100 cm
(d) Perimeter of triangle = 30 cm + 30 cm+ 40 cm
= 100 cm

Q.39

Ans

$\begin{array}{l}\mathrm{Length}\text{of each side of square paving}=\frac{1}{2}\text{\hspace{0.17em}}\mathrm{m}\\ \mathrm{Number}\text{of square paving}=\text{9}\\ \text{When square paving laid in the form of square,}\\ \text{Number of square paving in each side of square form}\\ \text{3}\\ \text{So, the length of each side of new square}\\ \text{\hspace{0.17em}\hspace{0.17em}}=3×\left(\mathrm{side}\mathrm{of}\mathrm{square}\mathrm{paving}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=3×\frac{1}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{3}{2}\\ \left(\mathrm{a}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Perimeter}\text{of new square}=4×\mathrm{side}\\ \text{\hspace{0.17em}\hspace{0.17em}}=4×\frac{3}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}}=6\text{\hspace{0.17em}}\mathrm{m}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Perimeter}\text{of fig}\left(\mathrm{ii}\right)=4×\left(\mathrm{side}\mathrm{having}\mathrm{single}\mathrm{paving}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}8×\left(\text{side having double}\mathrm{paving}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=4×\frac{1}{2}+8×\frac{1}{2}×2\\ \text{\hspace{0.17em}\hspace{0.17em}}=2+8\\ \text{\hspace{0.17em}\hspace{0.17em}}=10\text{\hspace{0.17em}}\mathrm{m}\\ \left(\mathrm{c}\right)\text{Cross figure i.e., fig}\left(\mathrm{ii}\right)\text{has greater perimeter.}\\ \left(\mathrm{d}\right)\text{Yes, if we put all paving in a straight line. It forms}\\ \text{a rectangle whose dimensions are:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Length}=9×\frac{1}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}}=4.5\text{\hspace{0.17em}}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Breadth}=\frac{1}{2}\text{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\text{0.5\hspace{0.17em}cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Perimeter}\text{of rectangle}=2\left(\mathrm{L}+\mathrm{B}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=2\left(4.5+0.5\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=10\text{â€‹\hspace{0.17em}}\mathrm{cm}\end{array}$

Q.40 Find the areas of the following figures by counting square:

Ans

(a) Number of squares in given figure = 9
So, area of given figure = 9 Sq. unit

(b) Number of squares in given figure = 5
So, area of given figure = 5 Sq. unit

(c) Number of squares in given figure = 2 + 4(1/2)

= 4
So, area of given figure = 4 Sq. unit

(d) Number of squares in given figure = 8
So, area of given figure = 8 Sq. unit

(e) Number of squares in given figure = 10
So, area of given figure = 10 Sq. unit

(f) Number of squares in given figure = 2 + 4(1/2)

= 4
So, area of given figure = 4 Sq. unit

(g) Number of squares in given figure = 4 + 4(1/2)

= 6
So, area of given figure = 6 Sq. unit

(h) Number of squares in given figure = 5
So, area of given figure = 5 Sq. unit

(i) Number of squares in given figure = 9
So, area of given figure = 9 Sq. unit

(j) Number of squares in given figure = 2 + 4(1/2)

= 4
so, area of given figure = 4 Sq. unit

(k) Number of squares in given figure = 4 + 2(1/2)

= 5
So, area of given figure = 5 Sq. unit

(l)

 Covered area Number Area estimate (sq units) Fully filled squares 2 2 Half filled squares 2 2(1/2) = 1 More than filled squares 5 5 Less than filled squares 2 0

So, the area of given figure = 2+1+5
= 8 square cm.

(m)

 Covered area Number Area estimate (sq units) Fully filled squares 7 2 Half filled squares 2 2(1/2) = 1 More than filled squares 6 6 Less than filled squares 8 0

So, the area of given figure = 7+1+6
=14 square cm.

(n)

 Covered area Number Area estimate (sq units) Fully filled squares 9 9 Half filled squares 4 4(1/2) = 2 More than filled squares 7 7 Less than filled squares 8 0

So, the area of given figure = 9+2+7
=18 square cm.

Q.41 Find the areas of the rectangles whose sides are:
(a) 3 cm and 4 cm (b) 12 m and 21 m
(c) 2 km and 3 km (d) 2 m and 70 cm

Ans

$\begin{array}{l}âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Area}\text{of rectangle}=\mathrm{Length}×\mathrm{Breadth}\\ \left(\mathrm{a}\right)\mathrm{Area}\text{of rectangle}=3\text{\hspace{0.17em}}\mathrm{cm}×4\text{\hspace{0.17em}}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=12{\mathrm{cm}}^{2}\\ \left(\mathrm{b}\right)\mathrm{Area}\text{of rectangle}=12\text{\hspace{0.17em}}\mathrm{cm}×21\text{\hspace{0.17em}}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=252\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \left(\mathrm{c}\right)\mathrm{Area}\text{of rectangle}=2\text{\hspace{0.17em}}\mathrm{km}×3\text{\hspace{0.17em}}\mathrm{km}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6\text{\hspace{0.17em}}{\mathrm{km}}^{2}\\ \left(\mathrm{d}\right)\mathrm{Length}\text{of rectangle}=2\text{\hspace{0.17em}}\mathrm{m}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=200\text{\hspace{0.17em}}\mathrm{cm}\\ \mathrm{Area}\text{of rectangle}=200\text{\hspace{0.17em}}\mathrm{cm}×70\text{\hspace{0.17em}}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=14000\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\end{array}$

Q.42 Find the areas of the squares whose sides are:

(a) 10 cm (b) 14 cm (c) 5 m

Ans

Area of square = (side of square)2
(a) Area of square = (10)2

= 100 cm2
(b) Area of square = (14)2
= 196 cm2
(c) Area of square = (5)2

= 25 cm2

Q.43 The length and breadth of three rectangles are as given below:
(a) 9 m and 6 m
(b) 17 m and 3 m
(c) 4 m and 14 m
Which one has the largest area and which one has the smallest?

Ans

$\begin{array}{l}\text{\hspace{0.17em}}\mathrm{Area}\text{of rectangle}=\mathrm{Length}×\mathrm{Breadth}\\ \left(\mathrm{a}\right)\mathrm{Area}\text{of rectangle}=9\text{\hspace{0.17em}}\mathrm{m}×6\text{\hspace{0.17em}}\mathrm{m}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=54\text{\hspace{0.17em}}{\mathrm{m}}^{2}\\ \left(\mathrm{b}\right)\mathrm{Area}\text{of rectangle}=17\text{\hspace{0.17em}}\mathrm{m}×3\text{\hspace{0.17em}}\mathrm{m}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=51\text{\hspace{0.17em}}{\mathrm{m}}^{2}\\ \left(\mathrm{c}\right)\mathrm{Area}\text{of rectangle}=4\text{\hspace{0.17em}}\mathrm{m}×14\text{\hspace{0.17em}}\mathrm{m}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=56\text{\hspace{0.17em}}{\mathrm{m}}^{2}\\ \text{Rectangle}\left(\text{c}\right)\text{has the largest area and rectangle}\left(\mathrm{b}\right)\text{has the}\\ \text{smallest area.}\end{array}$

Q.44 The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.

Ans

$\begin{array}{c}\text{Area of rectangular garden}=\text{3}00\text{sq m}\\ \text{Length of rectangular garden}=50\text{\hspace{0.17em}}\mathrm{m}\\ \mathrm{Breadth}\text{of rectangular garden}=\frac{\mathrm{Area}\text{of garden}}{\mathrm{Length}\text{of garden}}\\ =\frac{300}{50}\\ =6\text{\hspace{0.17em}}\mathrm{m}\end{array}$

Q.45 A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?

Ans

Length of table top = 2 m

Breadth of table top = 1 m 50 cm

= (1 + 0.50) m

= 1.5 m

Area of table-top = Length x Breadth

= 2 m x 1.5 m

= 3 sq m

Q.46 A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is needed to cover the floor of the room?

Ans

Length of room = 4 m

Breadth of room = 3 m 50 cm

= (3 + 0.50) m

= 3.5 m

Area of floor = Length x Breadth

= 4 m x 3.5 m

= 14 sq m
Therefore, 14 sq m carpet is required to cover the floor of the room.

Q.47 A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.

Ans

$\begin{array}{c}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Area}\text{of rectangular floor}=5\mathrm{m}×4\mathrm{m}\\ =20\text{\hspace{0.17em}}{\mathrm{m}}^{2}\\ \text{\hspace{0.17em}}\mathrm{Length}\text{of side of square carpet}=3\text{\hspace{0.17em}}\mathrm{m}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Area}\text{of square carpet}={\left(3\text{\hspace{0.17em}}\mathrm{m}\right)}^{2}\\ =9\text{\hspace{0.17em}}{\mathrm{m}}^{2}\\ \mathrm{Area}\text{of floor that is not carpeted}=20-9\\ =11{\mathrm{m}}^{2}\end{array}$

Q.48 Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land?

Ans

Area of rectangular land = 5m x 4m

= 20 m2
Area of 1 flower bed = (1m)2
= 1m2
Area of 5 flower beds = 5m2
Area of remaining part of land= 20 – 5
= 15 m2

Q.49 By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).

Ans

(a)Area of given figure = 3 × 3 + 1 × 2 +3 × 3
+ 2 × 4
= 9 + 2 + 9 + 8

= 28 cm2
(b) Area of given figure = 3 × 1 + 3 × 1 +3 × 1
= 3 + 3 + 3

= 9 cm2

Q.50 Split the following shapes into rectangles and find their areas. (The measures are given in centimeters)

Ans

(a) Area of given figure = 12 x 2 + 8 x 2

= 24 + 16

= 40 cm2
(b) Area of given figure = 7 x 7 + 7 x 21 +7 x 7
= 49 + 147 + 49

= 245 cm2

(c) Area of given figure = 1 x 5 + 4 x 1
= 5 + 4

= 9 cm2

Q.51 How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively:

(a) 100 cm and 144 cm

(b) 70 cm and 36 cm.

Ans

$\begin{array}{l}\left(\mathrm{a}\right)\text{Number of bricks in wall}=\frac{\mathrm{Area}\text{of rectangular region}}{\mathrm{Area}\text{of surface of brick}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{100×144}{12×5}\\ \text{\hspace{0.17em}\hspace{0.17em}}=20×12\\ \text{\hspace{0.17em}\hspace{0.17em}}=240\\ \left(\mathrm{b}\right)\text{Number of bricks in wall}=\frac{\mathrm{Area}\text{of rectangular region}}{\mathrm{Area}\text{of surface of brick}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{70×36}{12×5}\\ \text{\hspace{0.17em}\hspace{0.17em}}=14×3\\ \text{\hspace{0.17em}\hspace{0.17em}}=42\end{array}$

Q.52 Find the rule which gives the number of matchsticks required to make the following matchstick patterns. Use a variable to write the rule.

$\begin{array}{l}\left(a\right)\mathrm{A}\mathrm{pattern}\mathrm{of}\mathrm{letter}\mathrm{T}\mathrm{as}\overline{\text{\hspace{0.17em}}|\text{\hspace{0.17em}}}\\ \left(b\right)\text{\hspace{0.17em}}\mathrm{A}\mathrm{pattern}\mathrm{of}\mathrm{letter}\mathrm{Z}\mathrm{as}\overline{\underset{¯}{\text{\hspace{0.17em}}\overline{)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}}}\text{\hspace{0.17em}}}\\ \left(c\right)\text{\hspace{0.17em}}\mathrm{A}\mathrm{pattern}\mathrm{of}\mathrm{letter}\mathrm{U}\mathrm{as}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{¯}{|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|}\text{\hspace{0.17em}}\\ \left(d\right)\text{\hspace{0.17em}}\mathrm{A}\mathrm{pattern}\mathrm{of}\mathrm{letter}\mathrm{V}\mathrm{as}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}/\\ \left(e\right)\text{\hspace{0.17em}}\mathrm{A}\mathrm{pattern}\mathrm{of}\mathrm{letter}\mathrm{E}\mathrm{as}\text{\hspace{0.17em}}\frac{|\text{\hspace{0.17em}}\overline{\text{â€‹}\text{â€‹}\text{â€‹}\text{â€‹}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}}\text{\hspace{0.17em}}}{|\text{\hspace{0.17em}}\underset{¯}{\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}}\\ \left(f\right)\text{\hspace{0.17em}}\mathrm{A}\mathrm{pattern}\mathrm{of}\mathrm{letter}\mathrm{S}\mathrm{as}\frac{|\text{\hspace{0.17em}}\overline{\text{â€‹}\text{â€‹}\text{â€‹}\text{â€‹}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}}\text{\hspace{0.17em}}}{\underset{¯}{\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|}\\ \left(g\right)\text{\hspace{0.17em}}\mathrm{A}\mathrm{pattern}\mathrm{of}\mathrm{letter}\mathrm{A}\mathrm{as}\frac{|\text{\hspace{0.17em}}\overline{\text{â€‹}\text{â€‹}\text{â€‹}\text{â€‹}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}}{|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}}\end{array}$

Ans

(a) In the formation of pattern of letter T, we can see that it will require two matchsticks.

 Number of T 1 2 3 4 5 6 – Required matchsticks 2 4 6 8 10 12 –

Therefore, the pattern is 2n.

(b) In the formation of pattern of letter Z, we can see that it will require three matchsticks.

 Number of Z 1 2 3 4 5 6 – Required matchsticks 3 6 9 12 15 18 –

Therefore, the pattern is 3n.

(c) In the formation of pattern of letter U, we can see that it will require three matchsticks.

 Number of U 1 2 3 4 5 6 – Required matchsticks 3 6 9 12 15 18 –

Therefore, the pattern is 3n.

(d) In the formation of pattern of letter V, we cansee that it will require two matchsticks.

 Number of V 1 2 3 4 5 6 – Required matchsticks 2 4 6 8 10 12 –

Therefore, the pattern is 2n.

(e) In the formation of pattern of letter E, we can see that it will require five matchsticks.

 Number of E 1 2 3 4 5 6 – Required matchsticks 5 10 15 20 25 30 –

Therefore, the pattern is 5n.
(f) In the formation of pattern of letter S, we can see that it will require five matchsticks.

 Number of S 1 2 3 4 5 6 – Required matchsticks 5 10 15 20 25 30 –

Therefore, the pattern is 5n.

(g) In the formation of pattern of letter A, we can see that it will require six matchsticks.

 Number of A 1 2 3 4 5 6 – Required matchsticks 6 12 18 24 30 36 –

Therefore, the pattern is 6n.

Q.53 We already know the rule for the pattern of letters L, C and F. Some of the letters from Q.1 (given above) give us the same rule as that given by L. Which are these? Why does this happen?

Ans

In the formation of pattern of letter L, we can see that it will require two matchsticks.

 Number of T 1 2 3 4 5 6 – Required matchsticks 2 4 6 8 10 12 –

Therefore, the pattern is 2n. V and T are the letters in Q.1, which give us the same rule as that given by L. This is happened because in the formation of letters L, T and V, we require only two matchsticks in each.

Q.54 Cadets are marching in a parade. There are 5 cadets in a row. What is the rule which gives the number of cadets, given the number of rows? (Use n for the number of rows.)

Ans

Let number of rows be n.
Number of cadets in each row = 5
Number of cadets in n rows = 5n
The general rule to get the number of cadets in n rows is 5n.

Q.55 If there are 50 mangoes in a box, how will you write the total number of mangoes in terms of the number of boxes? (Use b for the number of boxes.)

Ans

Let number of boxes b.
Number of mangoes in each box = 50
Number of mangoes in b boxes = 50b
The general rule to get the number of mangoes in n boxes is 5n.

Q.56 The teacher distributes 5 pencils per student. Can you tell how many pencils are needed, given the number of students? (Use s for the number of students.)

Ans

Let number of students be s.
Each student got pencils = 5
s students got pencils = 5s
Therefore, 5s pencils are needed to distribute in s students.

Q.57 A bird flies 1 kilometer in one minute. Can you express the distance covered by the bird in terms of its flying time in minutes? (Use t for flying time in minutes.)

Ans

Let the flying time of a bird be t minutes.
Distance covered by the bird in one minute = 1 km
Distance covered by the bird in t minutes = t km
Therefore, the distance covered by a bird in t minutes is t km.

Q.58 Radha is drawing a dot Rangoli (a beautiful pattern of lines joining dots) with chalk powder. She has 9 dots in a row. How many dots will her Rangoli have for r rows?
How many dots are there if there are 8 rows?
If there are 10 rows?

Ans

Number of dots in a row = 9
Number of rows = r
Number of dots in r rows = 9r
Number of dots in 8 rows = 9 x 8
= 72
Number of dots in 10 rows = 9 x 10 = 90

Q.59 Leela is Radha’s younger sister. Leela is 4 years younger than Radha. Can you write Leela’s age in terms of Radha’s age? Take Radha’s age to be x years.

Ans

Let age of Radha = x years
Difference between the age of Leela and Radha = 4 years
Therefore, age of Leela = Age of Radha – 4
= (x – 4) years

Q.60 Mother has made laddus. She gives some laddus to guests and family members; still 5 laddus remain. If the number of laddus mother gave away is l, how many laddus did she make?

Ans

Number of laddus gave away by mother = l
Remaining laddus with mother = 5
Total laddus made by mother = l + 5

Q.61 Oranges are to be transferred from larger boxes into smaller boxes. When a large box is emptied, the oranges from it fill two smaller boxes and still 10 oranges remain outside. If the number of oranges in a small box are taken to be x, what is the number of oranges in the larger box?

Ans

Number of oranges in smaller box = x
Number of boxes filled by oranges = 2
Remaining oranges after filling 2 smaller boxes = 10
Total oranges in a big box = 2x + 10

Q.62 (a) Look at the following matchstick pattern of squares (Fig 11.6). The squares are not separate. Two neighbouring squares have a common matchstick. Observe the patterns and find the rule that gives the number of matchsticks in terms of the number of squares.

(b) Fig 11.7 gives a matchstick pattern of triangles. As in Exercise 11 (a) above, find the general rule that gives the number of matchsticks in terms of the number of triangles.

Ans

(a) We see that number of matchsticks in given pattern are 4, 7, 10 and 13, which are 1 more than thrice of the number of square. Therefore, the pattern is (3n + 1), where n is the number of squares in pattern.

(b) We see that number of matchsticks in given pattern are 3, 5, 7 and 9, which are 1 more than twice of the number of triangles. Therefore, the pattern is (2n + 1), where n is the number of triangles in pattern.

Q.63 The side of an equilateral triangle is shown by l. Express the perimeter of the equilateral triangle using l.

Ans

Length of side of equilateral triangle = l
Perimeter of equilateral triangle = l + l + l = 3 l

Q.64 The side of a regular hexagon is denoted by l. Express the perimeter of the hexagon using l.

Ans

Length of each side of regular hexagon = l
Perimeter of regular hexagon = l + l + l + l + l+ l

= 6 l

Q.65 A cube is a three-dimensional figure as shown in Fig 11.11. It has six faces and all of them are identical squares. The length of an edge of the cube is given by l. Find the formula for the total length of the edges of a cube.

Ans

Length of each side of cube = l
Number of edges in cube = 12
Total length of the edges of cube = Number of edges × Length of one edge

= 12 l

Q.66 The diameter of a circle is a line which joins two points on the circle and also passes through the centre of the circle. (In the adjoining figure (Fig 11.12) AB is a diameter of the circle; C is its centre.) Express the diameter of the circle (d) in terms of its radius(r).

Ans

Radius of circle = r
Diameter of a circle is double of its radius.
So, diameter (d) of circle = 2r

Q.67 To find sum of three numbers 14, 27 and 13, we can have two ways:
(a) We may first add 14 and 27 to get 41 and then add 13 to it to get the total sum 54 or
(b) We may add 27 and 13 to get 40 and then add 14 to get the sum 54.
Thus, (14 + 27) + 13 = 14 + (27 + 13)This can be done for any three numbers.
This property is known as the associativity of addition of numbers.
Express this property which we have already studied in the chapter on Whole Numbers, in a general way, by using variables a, b and c.

Ans

For three whole numbers a, b and c. We, have (a + b) + c = a + (b + c)

Q.68 Make up as many expressions with numbers (no variables) as you can from three numbers 5, 7 and 8. Every number should be used not more than once. Use only addition, subtraction and multiplication.

Ans

By using addition, subtraction and multiplication, we can make many expressions by using 5, 7 and 8. Some of these are as given below:
(5 + 7) – 8
(5 – 7) + 8
(8 – 7) + 5
(8 – 5) + 7
(7 – 5) + 7
(5 – 7) × 8
(5 – 8) × 7
(8 – 7) × 5
(5 + 7) × 8
(8 + 7) × 5
(5 + 8) × 7; etc.

Q.69 Which out of the following are expressions with numbers only?
(a) y + 3 (b) (7 × 20) – 8z
(c) 5(21 – 7) + 7 × 2 (d) 5
(e) 3x (f) 5 – 5n
(g) (7 × 20) –(5 × 10) – 45 + p

Ans

Expressions (c) and (d) have only numbers.

Q.70 Identify the operations (addition, subtraction, division, multiplication) in forming the following expressions and tell how the expressions have been formed.

$\begin{array}{l}\left(a\right)z+1,z-1,y+17,y–17\text{}\left(b\right)17y,\frac{y}{17},5z\\ \left(c\right)2y+17,2y-17\left(d\right)7m,-7\text{\hspace{0.17em}}m+3,-7\text{\hspace{0.17em}}m-3\end{array}$

Ans

(a) z + 1, here is addition because 1 is added to z.
z – 1, here is subtraction because 1 is subtracted from z.
y + 17, here is addition because 17 is added to y.
y – 17, here is subtraction because 17 is subtracted from y.

(b) 17y, here is multiplication because y is multiplied by 17.
(y/17), here is division because y is divided by 17.
5z, here is multiplication because z is multiplied by 5.

(c) 2y + 17, here is multiplication and addition because 17 is added to twice of y.
2y – 17, here is multiplication and subtraction because 17 is subtracted from twice of y.

(d) 7m, here is multiplication because m is multiplied by 7.
–7m + 3, here is multiplication and addition because 3 is added to –7 times of m.
– 7m – 3, here is multiplication and subtraction because 3 is subtracted from – 7 times of m.

Q.71 Give expressions for the following cases.
(a) 7 added to p (b) 7 subtracted from p
(c) p multiplied by 7 (d) p divided by 7
(e) 7 subtracted from – m (f) – p multiplied by 5
(g) – p divided by 5 (h) p multiplied by – 5

Ans

(a) 7 added to p = p + 7
(b) 7 subtracted from p = p – 7
(c) p multiplied by 7 = 7p
(d) p divided by 7 = p/7
(e) 7 subtracted from – m =– m – 7
(f) – p multiplied by 5 = 5(– p) = – 5p
(g) – p divided by 5 = (– p/ 5)
(h) p multiplied by – 5 = – 5p

Q.72 Give expressions in the following cases.
(a) 11 added to 2m
(b) 11 subtracted from 2m
(c) 5 times y to which 3 is added
(d) 5 times y from which 3 is subtracted
(e) y is multiplied by – 8
(f) y is multiplied by – 8 and then 5 is added to the result
(g) y is multiplied by 5 and the result is subtracted from 16
(h) y is multiplied by – 5 and the result is added to16.

Ans

(a)11 added to 2m = 2m + 11
(b) 11 subtracted from 2m = 2m – 11
(c)5 times y to which 3 is added = 5y + 3
(d) 5 times y from which 3 is subtracted = 5y – 3
(e) y is multiplied by – 8 = – 8y
(f) y is multiplied by – 8 and then 5 is added to the result = – 8y + 5
(g) y is multiplied by 5 and the result is subtracted from 16 = 16 – 5y
(h) y is multiplied by – 5 and the result is added to 16 = 16 + (– 5y) = 16 – 5y

Q.73 With the same centre O, draw two circles of radii 4 cm and 2.5 cm.

Ans

Steps of construction:

1. Take a measurement of 4 cm by the compass on a scale.

2. Draw a circle with centre O and radius 4 cm.

3. Now take a measurement of 2.5 cm by the compass on a scale.

4. Draw a circle with centre O and radius 2.5 cm.

Q.74 (a) Form expressions using t and 4. Use not more than one number operation. Every expression must have t in it.
(b) Form expressions using y, 2 and 7. Every expression must have y in it. Use only two number operations. These should be different.

Ans

(a) The expressions formed by using t and 4 are: 4 + t, 4 – t, t – 4, 4t, (4/t), (t/4)
(b) The different expressions formed by using y, 2 and 7 are: 2y + 7, 2y – 7, 7y + 2, 7y – 2, …

Q.75 Answer the following:
(a) Take Sarita’s present age to be y years
(i) What will be her age 5 years from now?
(ii) What was her age 3 years back?
(iii) Sarita’s grandfather is 6 times her age. What is the age of her grandfather?
(iv) Grandmother is 2 years younger than grandfather. What is grandmother’s age?
(v) Sarita’s father’s age is 5 years more than 3 times Sarita’s age. What is her father’s age?

(b) The length of a rectangular hall is 4 meters less than 3 times the breadth of the hall. What is the length, if the breadth is b meters?

(c) A rectangular box has height h cm. Its length is 5 times the height and breadth is 10 cm less than the length. Express the length and the breadth of the box in terms of the height.

(d) Meena, Beena and Leena are climbing the steps to the hill top. Meena is at step s, Beena is 8 steps ahead and Leena 7 steps behind. Where are Beena and Meena? The total number of steps to the hill top is 10 less than 4 times what Meena has reached. Express the total number of steps using s.

(e) A bus travels at v km per hour. It is going from Daspur to Beespur. After the bus has travelled 5 hours, Beespur is still 20 km away. What is the distance from Daspur to Beespur? Express it using v.

Ans

(a) Present age of Sarita = y years
(i) Age of Sarita after 5 years = (y + 5) years
(ii) Age of Sarita 3 years back = (y – 5) years
(iii) Age of Sarita’s grandfather = 6 times of y = 6y years
(iv)The Grandmother is 2 years younger than grandfather. So, Age of the grandmother = (6x – 2) years
(v) Sarita’s father’s age is 5 years more than 3 times Sarita’s age. So, father’s age = (3y + 5) years

(b) The breadth of rectangular hall = b m The length of rectangular hall = (3b – 4) m

(c) Height of the box = h cm
Length of the box = 5h cm
Breadth of the box = (h – 10) cm

(d) Meena is at steps = s
Since, Beena is 8 steps ahead. Then Beena is at steps = (s + 8)
Since, Leena is 7 steps behind to Meena. So, Leena is at steps = (s – 7)
Total number of steps = (4s – 10)

e) Speed of bus = v km/hr
Distance covered by bus in 5 hours = time x speed
= 5v
Remaining distance = 20 km
Total distance between Daspur to Beespur = (5v + 20) km

Q.76 Change the following statements using expressions into statements in ordinary language.
(For example, Given Salim scores r runs in a cricket match, Nalin scores (r + 15) runs. In ordinary language – Nalin scores 15 runs more than Salim.)
(a) A notebook costs â‚¹ p. A book costs â‚¹ 3p.
(b) Tony puts q marbles on the table. He has 8q marbles in his box.
(c) Our class has n students. The school has 20n students.
(d) Jaggu is z years old. His uncle is 4z years old and his aunt is (4z – 3) years old.
(e) In an arrangement of dots there are r rows. Each row contains 5 dots.

Ans

(a) The cost of a book is three times the cost of a notebook.
(b) The box of tony contains 8 times the number of marbles on the table.
(c) The total students in the school is 20 times of the students of our class.
(d) Jaggu’s uncle is 4 times older than Jaggu. Jaggu’s unty is 3 years younger than his uncle.
(e) Number of rows = r
Number of dots in a row = 5
Total number of dots = 5r
Therefore, total number of dots is 5 times of the total number of rows.

Q.77 Draw a circle of radius 3.2 cm.

Ans

Steps of construction:

1. Take a measurement of 3.2 cm by the compass on a scale.

2. Draw a circle with centre O and radius 3.2 cm.

Q.78 (a) Given Munnu’s age to be x years, can you guess what (x – 2) may show? Can you guess what (x + 4) may show? What (3 x + 7) may show?

$\begin{array}{l}\text{(b) Given Sara’s age today to be y years. Think of her}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}age in the future or in the past.}\\ \text{What will the following expression indicate?}\\ y+7,y–3,y+4\frac{1}{2},\text{â€„}y–2\frac{1}{2}.\\ \text{(c) Given n students in the class like football, what}\\ \text{may 2n show? What mayâ€„}\frac{\text{n}}{\text{2}}\text{â€„show?}\end{array}$

Ans

(a) The present age of Mannu is x years. Age of any person is 2 years less than the age of Mannu i.e., (x – 2) years.
Age of any person is 4 years more than the age of Mannu i.e., (x + 4) years.
Age of any person is 7 years more than thrice of the age of Mannu i.e., (3x + 7) years.
(b) The present age of Sara = y years
(y + 7) is the age of Sara after 7 years.
(y – 3) is the age of Sara before 3 years.

$\begin{array}{l}\left(y+4\frac{1}{2}\right)\text{is the age of Sara after 4}\frac{1}{2}\text{years}\text{.}\\ \left(y-2\frac{1}{2}\right)\text{is the age of Sara before 2}\frac{1}{2}\text{years}\text{.}\end{array}$

(c) n students like football, but 2n students may like other game or football. Similarly, (n/2) students may like any other game out of the number of students who like football.

Q.79 State which of the following are equations (with a variable). Give reason for your answer. Identify the variable from the equations with a variable.

$\begin{array}{l}\left(a\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}17=x+7\left(\mathrm{b}\right)\left(t-7\right)>5\left(c\right)\frac{4}{2}=2\\ \left(d\right)\left(7×3\right)-19=8\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(e\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}5×4-8=2x\left(f\right)x-2=0\\ \left(g\right)\text{\hspace{0.17em}}2m<30\left(h\right)\text{\hspace{0.17em}}2n+1=11\left(i\right)\text{\hspace{0.17em}}7=\left(11×5\right)-\left(12×4\right)\\ \left(j\right)\text{\hspace{0.17em}}7=\left(11×2\right)+p\left(k\right)\text{\hspace{0.17em}}20=5y\left(l\right)\text{\hspace{0.17em}}\frac{3q}{2}<5\\ \left(m\right)\text{z}+\text{12>24}\left(n\right)20-\left(10-5\right)=3×5\\ \left(0\right)\text{\hspace{0.17em}}7-x=5\end{array}$

Ans

$\begin{array}{l}\left(a\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}17=x+7\text{, it is an equation in variable x because equal to sign is there}\text{.}\\ \left(b\right)\left(t-7\right)>5,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{it is an not equation in variable t because equal}\text{\hspace{0.17em}}\text{to sign is not there}\text{.}\\ \left(c\right)\frac{4}{2}=2,\text{it is numerical equation because there is no variable}\text{.}\\ \left(d\right)\left(7×3\right)-19=8,\text{it is numerical equation because there is}\text{\hspace{0.17em}}\text{no variable}\text{.}\\ \left(e\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}5×4-8=2x,\text{It is an equation in variable x because equal}\text{\hspace{0.17em}}\text{to sign is there}\text{.}\\ \left(f\right)x-2=0,\text{\hspace{0.17em}}\text{It is an equation in variable x because equal}\text{\hspace{0.17em}}\text{to sign is there}\text{.}\\ \left(g\right)\text{\hspace{0.17em}}2m<30,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{it is not an equation because equal to sign is not there}\text{.}\\ \left(h\right)\text{\hspace{0.17em}}2n+1=11,\text{\hspace{0.17em}}\text{It is an equation in variable n because equal}\text{\hspace{0.17em}}\text{to sign is there}\text{.}\\ \left(i\right)\text{\hspace{0.17em}}7=\left(11×5\right)-\left(12×4\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{it is numerical equation because there is no variable}\text{.}\\ \left(j\right)\text{\hspace{0.17em}}7=\left(11×2\right)+p,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{It is an equation in variable p because equal}\text{\hspace{0.17em}}\text{to sign is there}\text{.}\\ \left(k\right)\text{\hspace{0.17em}}20=5y,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{It is an equation in variable y because equal}\text{\hspace{0.17em}}\text{to sign is there}\text{.}\\ \left(l\right)\text{\hspace{0.17em}}\frac{3q}{2}<5,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{it is not an equation because equal to sign is not there}\text{.}\\ \left(m\right)\text{z}+\text{12>24,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{it is not an equation because equal to sign is not there}\text{.}\\ \left(n\right)20-\left(10-5\right)=3×5,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{it is numerical equation because there is no variable}\text{.}\\ \left(o\right)\text{\hspace{0.17em}}7-x=5,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{It is an equation in variable x because equal to sign is there}\text{.}\end{array}$

Q.80 Complete the entries in the third column of the table.

 S.No. Equation Value of variable Equation satisfied Yes/No (a) 10y = 80 y = 10 (b) 10y = 80 y = 8 (c) 10y = 80 y = 5 (d) 4l = 20 l = 20 (e) 4l = 20 l = 80 (f) 4l = 20 l = 5 (g) b + 5 = 9 b = 5 (h) b + 5 = 9 b = 9 (i) b + 5 = 9 b = 4 (j) h – 8 = 5 h = 13 (k) h – 8 = 5 h = 8 (l) h – 8 = 5 h = 0 (m) p + 3 = 1 p = 3 (n) p + 3 = 1 p = 1 (o) p + 3 = 1 p = 0 (p) p + 3 = 1 p = – 1 (q) p + 3 = 1 p = – 2

Ans

 S.No. Equation Value of variable Equation satisfied Yes/No (a) 10y = 80 y = 10 No (b) 10y = 80 y = 8 yes (c) 10y = 80 y = 5 No (d) 4l = 20 l = 20 No (e) 4l = 20 l = 80 No (f) 4l = 20 l = 5 Yes (g) b + 5 = 9 b = 5 No (h) b + 5 = 9 b = 9 No (i) b + 5 = 9 b = 4 Yes (j) h – 8 = 5 h = 13 Yes (k) h – 8 = 5 h = 8 No (l) h – 8 = 5 h = 0 No (m) p + 3 = 1 p = 3 No (n) p + 3 = 1 p = 1 No (o) p + 3 = 1 p = 0 No (p) p + 3 = 1 p = – 1 No (q) p + 3 = 1 p = – 2 Yes

Q.81 Pick out the solution from the values given in the bracket next to each equation. Show that the other values do not satisfy the equation.

$\begin{array}{l}\left(\mathrm{a}\right)\text{\hspace{0.17em}}5\mathrm{m}=60\left(10,5,12,15\right)\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}}\mathrm{n}+12=20\left(12,8,20,0\right)\\ \left(\mathrm{c}\right)\mathrm{p}-5=5\left(0,10,5,-5\right)\\ \left(\mathrm{d}\right)\frac{\mathrm{q}}{2}=7\left(7,2,10,14\right)\\ \left(\mathrm{e}\right)\mathrm{r}-4=0\left(4,-4,8,0\right)\\ \left(\mathrm{f}\right)\mathrm{x}+4=2\left(-2,0,2,4\right)\end{array}$

Ans

$\begin{array}{l}\left(\text{a}\right)\text{We have},\text{5m}=\text{6}0\dots \left(\text{i}\right)\\ \text{Putting m}=\text{1}0\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{5m}=\text{5}\left(\text{1}0\right)\\ \mathrm{}=\text{5}0\ne 60\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{m}=\text{10 is not the solution of equation 5m}=\text{6}0.\\ \mathrm{Again},\\ \text{Putting m}=5\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{5m}=\text{5}\left(5\right)\\ =\text{}25\ne 60\\ ⇒\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{m}=\text{5 is not the solution of equation 5m}=\text{6}0.\\ \mathrm{Again},\text{\hspace{0.17em}Putting m}=12\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{5m}=\text{5}\left(12\right)\\ =60\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{m}=\text{12 is the solution of equation 5m}=\text{6}0.\\ \mathrm{Again},\\ \text{Putting m}=12\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \mathrm{}\text{5m}=\text{5}\left(12\right)\\ =\text{}60\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{m}=\text{12 is the solution of equation 5m}=\text{6}0.\\ \mathrm{Again},\\ \text{Putting m}=15\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \mathrm{}\text{5m}=\text{5}\left(15\right)\\ =\text{}75\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{m}=\text{15 is not the solution of equation 5m}=\text{6}0.\\ \left(\text{b}\right)\text{We have},\text{n}+\text{12}=\text{2}0\dots \left(\text{i}\right)\\ \text{Putting n}=\text{1}2\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}n}+\text{12}=\text{12}+\text{12}\\ =24\ne 20\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{n}=\text{12 is not the solution of equation n}+\text{12}=\text{2}0.\\ \mathrm{Again},\\ \text{Putting n}=8\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}n}+\text{12}=\text{8}+\text{12}\\ =20\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{n}=\text{8 is the solution of equation n}+\text{12}=\text{2}0.\\ \mathrm{Again},\\ \text{Putting n}=20\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}n}+\text{12}=\text{20}+\text{12}\\ =32\ne 20\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{n}=\text{20 is not the solution of equation n}+\text{12}=\text{2}0.\\ \mathrm{Again},\\ \text{Putting n}=0\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}n}+\text{12}=\text{0}+\text{12}\\ =12\ne 20\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{n}=\text{0 is not the solution of equation n}+\text{12}=\text{2}0.\\ \left(\text{c}\right)\text{We have},\text{p}-\text{5}=5\dots \left(\text{i}\right)\\ \text{Putting p}=0\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}p}-\text{5}=\text{0}-\text{5}\\ =-5\ne 5\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{p}=\text{0 is not the solution of equation p}-\text{5}=5.\\ \mathrm{Again},\\ \text{Putting p}=10\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}p}-\text{5}=1\text{0}-\text{5}\\ =5\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{p}=1\text{0 is the solution of equation p}-\text{5}=5.\\ \mathrm{Again},\\ \text{Putting p}=5\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}p}-\text{5}=\text{5}-\text{5}\\ =0\ne 5\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{p}=\text{5 is not the solution of equation p}-\text{5}=5.\\ \mathrm{Again},\\ \text{Putting p}=-\text{\hspace{0.17em}}5\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}p}-\text{5}=-\text{5}-\text{5}\\ =-10\ne 5\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{p}=-\text{\hspace{0.17em}5 is not the solution of equation p}-\text{5}=5.\\ \left(\text{d}\right)\text{We have},\text{}\frac{\mathrm{q}}{2}=7\dots \left(\text{i}\right)\\ \text{Putting q}=7\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{q}}{2}=\text{\hspace{0.17em}}\frac{7}{2}\\ =3.5\ne 7\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{q}=\text{7 is not the solution of equation}\frac{\mathrm{q}}{2}=7.\\ \mathrm{Again},\\ \text{Putting q}=2\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{q}}{2}=\text{\hspace{0.17em}}\frac{2}{2}\\ =1\ne 7\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{q}=\text{2 is not the solution of equation}\frac{\mathrm{q}}{2}=7.\\ \mathrm{Again},\\ \text{Putting q}=10\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{q}}{2}=\text{\hspace{0.17em}}\frac{10}{2}\\ =5\ne 7\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{q}=\text{10 is not the solution of equation}\frac{\mathrm{q}}{2}=7.\\ \mathrm{Again},\\ \text{Putting q}=14\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{q}}{2}=\text{\hspace{0.17em}}\frac{14}{2}\\ =7\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{q}=\text{14 is the solution of equation}\frac{\mathrm{q}}{2}=7.\\ \left(\text{e}\right)\text{We have},\text{}\mathrm{r}-4=0\dots \left(\text{i}\right)\\ \text{Putting r}=4\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}r}-4=4-4\\ =0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{r}=\text{4 is the solution of equation}\mathrm{r}-4=0.\\ \mathrm{Again},\\ \text{Putting r}=-\text{\hspace{0.17em}}4\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}r}-4=-\text{\hspace{0.17em}}4-4\\ =-\text{\hspace{0.17em}}8\ne 0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{r}=-\text{\hspace{0.17em}4 is not the solution of equation}\mathrm{r}-4=0.\\ \mathrm{Again},\\ \text{Putting r}=8\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}r}-4=8-4\\ =4\ne 0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{r}=\text{8 is not the solution of equation}\mathrm{r}-4=0.\\ \mathrm{Again},\\ \text{Putting r}=0\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}r}-4=0-4\\ =-\text{\hspace{0.17em}}4\ne 0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{r}=\text{0 is not the solution of equation}\mathrm{r}-4=0.\\ \left(\text{f}\right)\text{We have},\text{}\mathrm{x}+4=2\dots \left(\text{i}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Putting x}=-\text{\hspace{0.17em}}2\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+4=-\text{\hspace{0.17em}}2+4\\ =2\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{x}=-\text{\hspace{0.17em}}2\text{is the solution of equation x}+\text{4}=\text{\hspace{0.17em}\hspace{0.17em}}2.\\ \mathrm{Again},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Putting x}=0\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+4=0+4\\ =4\ne 2\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{x}=0\text{is not the solution of equation x}+\text{4}=\text{\hspace{0.17em}\hspace{0.17em}}2.\\ \mathrm{Again},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Putting x}=2\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+4=2+4\\ =6\ne 2\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{x}=2\text{is not the solution of equation x}+\text{4}=\text{\hspace{0.17em}\hspace{0.17em}}2.\\ \mathrm{Again},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Putting x}=4\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+4=4+4\\ =8\ne 2\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{x}=4\text{is not the solution of equation x}+\text{4}=\text{\hspace{0.17em}\hspace{0.17em}}2.\end{array}$

Q.82 (a) Complete the table and by inspection of the table find the solution to the equation m + 10 = 16.

 m 1 2 3 4 5 6 7 8 9 10 – – – m+10

(b) Complete the table and by inspection of the table, find the solution to the equation 5t = 35.

 t 3 4 5 6 7 8 9 10 11 – – – – 5t

(c) Complete the table and find the solution of the equation z/3 =4 using the table.

 z 8 9 10 11 12 13 14 15 16 – – – z/3 $2\frac{2}{3}$ 3 $3\frac{1}{3}$

(d) Complete the table and find the solution to the equation m – 7 = 3.

 m 5 6 7 8 9 10 11 12 13 – – – m-7

Ans

(a) Table having solution of m+10 = 16, is given below

 m m+10 1 1+10 = 11 2 2 + 10 = 12 3 3 + 10 = 13 4 4 + 10 = 14 5 5 + 10 = 15 6 6 + 10 = 16 7 7 + 10 = 17 8 8 + 10 = 18 9 9 + 10 = 19 10 10 + 10 = 20

By inspection of the table, the solution to the equation m + 10 = 16 is m = 6.

(b) Table having solution of 5t = 35, is given below:

 t 5t 3 5(3) = 15 4 5(4) = 20 5 5(5) = 25 6 5(6) = 30 7 5(7) = 35 8 5(8) = 40 9 5(9) = 45 10 5(10) = 50 11 5(11) = 55 12 5(12) = 60

By inspection of the table, the solution to the equation 5t = 35 is t = 7.

(c) Table having solution of z/3 = 4, is given below:

 z z/3 8 $\frac{\text{8}}{\text{3}}=2\frac{2}{3}$ 9 9/3 = 3 10 $\frac{\text{10}}{\text{3}}=3\frac{1}{3}$ 11 $\frac{\text{11}}{\text{3}}=3\frac{2}{3}$ 12 12/3 = 4 13 $\frac{\text{13}}{\text{3}}=4\frac{1}{3}$ 14 $\frac{\text{14}}{\text{3}}=4\frac{2}{3}$ 15 15/3 = 5 16 $\frac{\text{16}}{\text{3}}=5\frac{1}{3}$ 17 $\frac{\text{17}}{\text{3}}=5\frac{2}{3}$

By inspection of the table, the solution to the equation z/3 = 4 is z = 12.

(d) Table having solution of m – 7 = 3, is given below:

 m m-7 5 5 – 7 = – 2 6 6 – 7 = – 1 7 7 – 7 = 0 8 8 – 7 = 1 9 9 – 7 = 2 10 10 – 7 = 3 11 11 – 7 = 4 12 12 – 7 = 5 13 13 – 7 = 6

By inspection of the table, the solution to the equation m – 7 = 3 is m = 10.

Q.83 Solve the following riddles, you may yourself construct such riddles. Who am I?
(i) Go round a square
Counting every corner
Thrice and no more!
Add the count to me
To get exactly thirty four!

(ii) For each day of the week
Make an upcount from me
If you make no mistake
You will get twenty three!

(iii) I am a special number
Take away from me a six!
A whole cricket team
You will still be able to fix!

(iv) Tell me who I am
I shall give a pretty clue!
You will get me back
If you take me out of twenty two!

Ans

(i) Let value of me be x.
Number of corners in a square = 4
Thrice of 4 = 12
Then, x + 12 = 24
x = 24 – 12
= 22
Therefore, the value of me is 22.

(ii) Let value of me be x.
Upcount on Sunday from me = x + 1
Upcount on Monday from me = x + 2
Upcount on Tuesday from me = x + 3
Upcount on Wednesday from me = x + 4
Upcount on Thursday from me = x + 5
Upcount on Friday from me = x + 6
Upcount on Saturday from me = x + 7
In this way, x + 7 = 23
x = 16
Therefore, the value of me is 16.

(iii) Let the special number be x
Number of player in cricket team
= 11
Subtracted number = 6
Then, according to condition,
x – 6 = 11

(iv) Let the value of me be x.
Then, according to condition,
22 – x = x
22 = 2x
x = 11
Therefore, the special number is 11.

Q.84 Draw a circle and any two of its diameters. If you join the ends of these diameters, what is the figure obtained? What figure is obtained if the diameters are perpendicular to each other? How do you check your answer?

Ans

When we join the end points of diameters that are not perpendicular to each other, we get a rectangle.

When diameters are at right angle and we join end points of these, we get square.

Since diagonals of a rectangle are equal and bisect each other at a specific angle. and diagonals of square are equal and bisect each other at right angle.

Q.85 Draw any circle and mark points A, B and C such that

(a) A is on the circle.

(b) B is in the interior of the circle.

(c) C is in the exterior of the circle.

Ans

(a) A is on the circle with centre O.

(b) B is in the interior of the circle with centre O.

(c) C is in the exterior of the circle with centre O.

Q.86

$\begin{array}{l}\text{Letâ€„A,â€„Bâ€„beâ€„theâ€„centresâ€„ofâ€„twoâ€„circlesâ€„ofâ€„equalâ€„radii;â€„draw}\\ \text{themâ€„soâ€„thatâ€„eachâ€„oneâ€„ofâ€„themâ€„passesâ€„throughâ€„theâ€„centre}\\ \text{ofâ€„theâ€„other.â€„Letâ€„themâ€„intersectâ€„atâ€„Câ€„andâ€„D.â€„Examine}\\ \text{whetherâ€„}\overline{\text{AB}}\text{â€„andâ€„}\overline{\text{CD}}\text{â€„areâ€„atâ€„rightâ€„angles.}\end{array}$

Ans

$\begin{array}{l}\text{Draw two circles of same radius which are passing through the centres of the other circle.}\\ \text{Here,â€„pointâ€„Aâ€„andâ€„Bâ€„areâ€„theâ€„centresâ€„ofâ€„theseâ€„circlesâ€„andâ€„theseâ€„circlesâ€„areâ€„intersectingâ€„eachâ€„other}\\ \text{atâ€„pointâ€„Câ€„andâ€„D}.\\ \text{In quadrilateral ADBC,}\\ \text{AD}=\text{AC (Radius of circle centered at A)}\\ \text{BC}=\text{BD (Radius of circle centered at B)}\\ \text{As radius of both circles are equal, therefore, ADâ€„=â€„ACâ€„=â€„BCâ€„=â€„BD}\\ \text{Hence, ADBC is a rhombus and in rhombus, the diagonals bisect each other at}90°\text{.}\\ \text{Hence,}\overline{\text{AB}}\text{and}\overline{\text{CD}}\text{are at right angles.}\end{array}$

Q.87 Draw a line segment of length 7.3 cm using a ruler.

Ans

Steps of construction:

1. Draw a line l. Mark a point A on line l.

2. Take a measurement of 7.3 cm from the scale with the help of compass.

3. Mark an arc of radius 7.3 with centre A on line l which intersect at point B on line l.

4. AB is the required line segment.

Q.88 Construct a line segment of length 5.6 cm using ruler and compasses.

Ans

Steps of construction:

1. Draw a line l. Mark a point A on line l.

2. Take a measurement of 5.6 cm from the scale with the help of compass.

3. Mark an arc of radius 5.6 with centre A on line l which intersect at point B on line l.

4. AB is the required line segment.

Q.89

$\begin{array}{l}\text{Constructâ€„}\overline{\text{AB}}\text{â€„ofâ€„lengthâ€„7.8â€„cm.â€„Fromâ€„this,â€„cutâ€„offâ€„}\overline{\text{AC}}\text{â€„of}\\ \text{lengthâ€„4.7â€„cm.â€„Measureâ€„}\overline{\text{BC}}\text{.}\end{array}$

Ans

Steps of construction:
1. Draw a line l. Mark a point A on line l.
2. Take a measurement of 7.8 cm from the scale with the help of compass.
3. Mark an arc of radius 7.8 with centre A on line l which intersect at point B on line l.
4. AB is the required line segment.
5. Similarly, we cut line segment AC = 4.7 from line segment AB.

The length of the remaining line segment BC is 3.1 cm.

Q.90

$\begin{array}{l}\text{Givenâ€„}\overline{\text{AB}}\text{â€„ofâ€„lengthâ€„3.9â€„cm,â€„constructâ€„}\overline{\text{PQ}}\text{â€„suchâ€„thatâ€„the}\\ \text{lengthâ€„ofâ€„}\overline{\text{PQ}}\text{â€„isâ€„twiceâ€„thatâ€„ofâ€„}\overline{\text{AB}}\text{.â€„Verifyâ€„byâ€„measurement.}\end{array}$

Ans

Steps of construction:
1. Draw a line l and cut a line segment AB of 3.9 cm from it.
2. Mark an arc of length 3.9 cm with centre B, which intersect given line l at the point C such that AC = 2AB.

3. Now, again draw a line l’.
4. Take distance AC in compass and mark an arc with centre P and radius AC on line l’.

5. PQ is the required line segment.

Q.91

$\begin{array}{l}\text{Givenâ€„ABâ€„ofâ€„lengthâ€„7.3â€„cmâ€„andâ€„CDâ€„ofâ€„lengthâ€„3.4â€„cm,â€„construct}\\ \text{aâ€„lineâ€„segmentâ€„XYâ€„suchâ€„thatâ€„theâ€„lengthâ€„ofâ€„XYâ€„isâ€„equalâ€„toâ€„the}\\ \text{differenceâ€„betweenâ€„theâ€„lengthsâ€„ofâ€„ABâ€„andâ€„CD.}\\ \text{Verifyâ€„byâ€„measurement.}\end{array}$

Ans

Steps of construction:

1. Draw a line l and take a point X on it.

2. Construct a line segment XM such that XM = AB = 7.3 cm

3. Then cut off a line segment YM such that YM = CD = 3.4 cm

4. Thus, length of XY = Length of AB – Length of CD

By measurement, we find that XY = 3.9 cm = 7.3 cm –3.4 cm = AB –CD

Q.92

$\begin{array}{l}\text{Drawâ€„anyâ€„lineâ€„segmentâ€„}\overline{\text{PQ}}\text{.â€„Withoutâ€„measuringâ€„}\overline{\text{PQ}}\text{,}\\ \text{constructâ€„aâ€„copyâ€„ofâ€„}\overline{\text{PQ}}\text{.}\end{array}$

Ans

Steps of construction:
(i) Draw a line segment PQ of any length.
(ii) Fix the compass pointer at P and pencil pointer at Q. The opening of the instrument gives the length of PQ.
(iii) Draw a line l’.
(iv) Choose a point R on l’. Mark an arc of radius equal to PQ and with centre R which intersects the line at point S.
So, RS is a copy of line segment PQ.

Q.93

$\begin{array}{l}\text{Givenâ€„someâ€„lineâ€„segmentâ€„}\overline{\text{AB}}\text{,â€„whoseâ€„lengthâ€„youâ€„doâ€„not}\\ \text{know,â€„constructâ€„}\overline{\text{PQ}}\text{â€„suchâ€„thatâ€„theâ€„lengthâ€„ofâ€„}\overline{\text{PQ}}\text{â€„isâ€„twice}\\ \text{thatâ€„ofâ€„}\overline{\text{AB}}\text{.}\end{array}$

Ans

Step of construction:
(i) AB is a line segment of any length.

(ii) Take the measurement of AB with the help of compass.
(iii) Draw a line l’.
(iv) Mark an arc of radius equal to AB and with centre P which intersects line at point R.
(v) Again mark another arc with radius AB and centre R which cuts line at point Q.
(vi) Therefore, PQ is required line segment.

Q.94

$\begin{array}{l}\text{Drawâ€„anyâ€„lineâ€„segmentâ€„}\overline{\text{AB}}\text{.â€„Markâ€„anyâ€„pointâ€„Mâ€„onâ€„it.â€„Through}\\ \text{M,â€„drawâ€„aâ€„perpendicularâ€„toâ€„}\overline{\text{AB}}\text{.â€„(useâ€„rulerâ€„andâ€„compasses)}\end{array}$

Ans

Steps of construction:
1. Draw a line l and mark points A, B and M on it.
2. With the help of compasses mark an arc of any radius with centre M, which intersects AB at X and Y.

3. Mark arcs of same radius with centre X and Y. These arcs intersect each other at N.
4. Join MN. Therefore, TM is the required perpendicular to the line segment AB.

Q.95

$\begin{array}{l}\text{Drawâ€„anyâ€„lineâ€„segmentâ€„}\overline{\text{PQ}}\text{.â€„Takeâ€„anyâ€„pointâ€„Râ€„notâ€„onâ€„it.}\\ \text{Throughâ€„R,â€„drawâ€„aâ€„perpendicularâ€„toâ€„}\overline{\text{PQ}}\text{.}\\ \text{â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„(useâ€„rulerâ€„andâ€„set-square)}\end{array}$

Ans

Steps of construction:
Step 1: Draw a line segment PQ and take a point R outside it.

Step 2: Put a set square on line segment PQ such that one side of its right angle and line PQ should be in same direction.

Step 3: Place a ruler along the edge opposite to the right angle of the set-square.

Step 4: Hold the ruler fixed. Slide the set square along the ruler till the point R touches the other arm of the set square.

Step 5: Join RS along the edge through R meeting PQ at S. Then RS is perpendicular to PQ.

Q.96

$\begin{array}{l}\text{Drawâ€„aâ€„lineâ€„lâ€„andâ€„aâ€„pointâ€„Xâ€„onâ€„it.â€„Throughâ€„X,â€„drawâ€„aâ€„line}\\ \text{segmentâ€„XYâ€„perpendicularâ€„toâ€„l.â€„Nowâ€„drawâ€„aâ€„perpendicular}\\ \text{toâ€„XYâ€„atâ€„Y.â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„â€„(useâ€„rulerâ€„andâ€„compasses)}\end{array}$

Ans

Steps of construction:
1. Draw a line l and mark any point X on it.
2. With the help of compass mark an arc of any radius on the line with centre X. It intersects the line at the points A and B.
3. Mark another arcs of same radius with centres A and B respectively which intersect each other at Y.
4. Join line segment XY.
5. Therefore, XY is required perpendicular to line l.

6. In the same way, line ZY is perpendicular on XY.

Q.97 Draw

$\overline{\mathrm{AB}}$

of length 7.3 cm and find its axis of symmetry.

Ans

Steps of construction:
1. Draw a line segment

$\overline{\mathrm{AB}}$

of length 7.3 cm.
2. Mark two arcs of same radii and centres A and B, which intersect each other at C and D respectively.
3. CD is the axis of symmetry i.e., perpendicular bisector of line segment AB.

Q.98 Draw a line segment of length 9.5 cm and construct its perpendicular bisector.

Ans

Steps of construction:
1. Draw a line segment PQ of length 9.5 cm.
2. Mark two arcs of same radii and centres P and Q, which intersect each other at R and S respectively.
3. RS is the axis of symmetry i.e., perpendicular bisector of line segment PQ.

Q.99

$\begin{array}{l}\text{Drawâ€„theâ€„perpendicularâ€„bisectorâ€„ofâ€„}\overline{\text{XY}}\text{â€„whoseâ€„lengthâ€„is}\\ \text{10.3â€„cm.â€„}\left(\text{a}\right)\text{â€„Takeâ€„anyâ€„pointâ€„Pâ€„on theâ€„bisectorâ€„drawn.â€„Examine}\\ \text{whetherâ€„}\overline{\text{PX}}\text{â€„=â€„}\overline{\text{PY}}\text{.â€„}\left(\text{b}\right)\text{â€„Ifâ€„Mâ€„isâ€„theâ€„midpointâ€„ofâ€„XY,â€„whatâ€„can}\\ \text{youâ€„sayâ€„aboutâ€„theâ€„lengthsâ€„}\overline{\text{MX}}\text{â€„andâ€„}\overline{\text{XY}}\text{?}\end{array}$

Ans

Steps of construction:
1. Draw a line segment XY = 10.3 cm.

2. Draw perpendicular bisector of line segment XY, which intersect each other at R and S.

3. Join RS. RS is the axis of symmetry.

(a) Here, by measuring we get: PX = PY

(b) If M is the midpoint of line segment XY, then MX = MY. This means MX = (1/2) XY.

Q.100 Draw a line segment of length 12.8 cm. Using compasses, divide it into four equal parts.Verify by actual measurement.

Ans

Step of construction:
1. Draw a line segment XY = 12.8 cm.
2. Mark two arcs with radius more than 6.4 cm and centres X and Y respectively which intersect each other at R and S.
3. Join RS which intersect XY at P.
4. Similarly draw perpendicular bisectors of XP and PY.
5. By measuring with scale, we find that XA= 3.2 cm, AP = 3.2 cm, PB = 3.2 cm and BY = 3.2 cm.

Q.101

$\mathrm{With}\stackrel{\to }{\mathrm{PQ}}\mathrm{of}\mathrm{length}6.1\mathrm{cm}\mathrm{as}\mathrm{diameter},\mathrm{draw}\mathrm{a}\mathrm{circle}.$

Ans

Step of construction:
1. Draw a line segment PQ of length 6.1 cm.
2. Mark arcs of radius more than half of 6.1 cm with centre P and Q respectively. These arcs intersect each other at points C and D.
3. Join C and D. CD intersects PQ at the point O.
4. Draw a circle of radius OP or OQ with centre O.Thus, required circle passes through P and Q.

Q.102

$\begin{array}{l}\mathrm{Draw}\mathrm{a}\mathrm{circle}\mathrm{with}\mathrm{centre}\mathrm{C}\mathrm{and}\mathrm{radius}3.4\mathrm{cm}.\mathrm{Draw}\mathrm{any}\\ \mathrm{chord}\overline{\mathrm{AB}}.\mathrm{Construct}\mathrm{the}\mathrm{perpendicular}\mathrm{bisector}\mathrm{of}\overline{\mathrm{AB}}\mathrm{and}\\ \mathrm{examine}\mathrm{if}\mathrm{it}\mathrm{passes}\mathrm{through}\mathrm{C}.\end{array}$

Ans

Steps of construction:

1. Draw a circle with centre C and radius 3.4 cm.
2. Draw any chord AB.
3. Taking A and B as centres and radius more than half of AB, draw two arcs which cut each other at S and T.
4. Join ST. Then ST is the perpendicular bisector of AB.

This perpendicular bisector of AB passes through the centre C of the circle.

Q.103

$\begin{array}{l}\mathrm{Draw}\mathrm{a}\mathrm{circle}\mathrm{with}\mathrm{centre}\mathrm{C}\mathrm{and}\mathrm{diameter}\overline{\mathrm{AB}}3.4\mathrm{cm}.\\ \mathrm{Construct}\mathrm{the}\mathrm{perpendicular}\mathrm{bisector}\mathrm{of}\overline{\mathrm{AB}}\mathrm{and}\\ \mathrm{examine}\mathrm{if}\mathrm{it}\mathrm{passes}\mathrm{through}\mathrm{C}.\end{array}$

Ans

Steps of construction:

1. Draw a circle with radius 3.4/2 = 1.7 cm with
centre C.
2. Draw diameter AB.

3. Draw perpendicular bisector of AB with centre A
and B and any radius more than 1.7 cm.
4. The bisector PQ of AB intersects AB at centre C. Therefore, it can be observed that perpendicular bisector of AB passes through centre C.

Q.104 Draw a circle of radius 4 cm. Draw any two of its chords. Construct the perpendicular bisectors of these chords. Where do they meet?

Ans

Step of construction:
1. Draw a circle of radius 4 cm with centre C.
2. Draw two chords AB and PQ.
3. Draw perpendicular bisectors of AB and PQ, which intersect at C. Therefore, it can be observed that bisectors of both chords intersect at the centre C of the circle.

Q.105 Draw any angle with vertex O. Take a point A on one of its arms and B on another such that OA = OB. Draw the perpendicular bisectors of OA and OB. Let them meet at P. Is PA = PB?

Ans

Steps of construction:
1. Make any angle with vertex O.
2. Mark two arcs on both rays with any radius and centre O.
3. Draw perpendicular bisectors of OA and OB. These perpendicular bisectors intersect at the point P.
4. Join PA and PB. By measuring PA and PB, we see that PA = PB.

Q.106

$\mathrm{Draw}\angle \mathrm{POQ}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{measure}\text{\hspace{0.17em}}75\mathrm{°}\mathrm{and}\text{\hspace{0.17em}}\mathrm{find}\text{\hspace{0.17em}}\mathrm{its}\text{\hspace{0.17em}}\mathrm{line}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{symmetry}.$

Ans

Steps of construction:
1. Draw a ray OP.

2. Mark an arc of any radius with centre O. This arc intersects OP at the point A.
3. Mark another arc of the same radius with centre A, it intersects earlier arc at the point B.
4. Arc of same radius with centre B intersects the earliest arc at R.
5. The arcs drawn of any radius with centre B and R respectively intersect at the point S.
6. Draw ray OC through the point S.
6. Mark another arc of any radius with centre U (intersection point of arc and OC) and B which intersect at T.
7. Draw ray OQ through the point T
8. Draw two arcs of any radius with centre W (intersection point of arc and OQ) and A. These arcs intersect at the point V. Therefore, line OV is the line of symmetry for angle POQ.

Q.107 Draw an angle of measure 147° and construct its bisector.

Ans

Steps of construction:
1. Draw a ray OP.

2. Draw an angle of 147° with the help of protractor at point O.

3. Draw an arc of any radius with centre O which intersects OP at A and OQ at B.
4. Mark two arcs of any radius with centre A and B, which intersect each other at C.
5. Draw ray OR through C. Therefore, OR is bisector of angle POQ.

Q.108 Draw a right angle and construct its bisector.

Ans

Steps of construction:

1. Draw a ray OP.
2. Mark an arc of any radius with centre O, which intersects ray OP at the point A.
3. Mark another arc of same radius with centre A which intersects earlier arc at B.
4. Mark an arc of same radius with centre B which intersects the earliest arc at the point C.
5. Mark two arcs of any radius with centre B and C respectively which intersect at the point R.
6. Draw ray OQ through the point R. In this way we get right angle POQ.
7. Again mark two arcs of any radius with centre A and D (intersection point of arc and OQ), which intersect each other at E.
8. Draw ray OS through E. Therefore, OS is bisector of angle POQ.

Q.109 Draw an angle of measure 153° and divide it into four equal parts.

Ans

Steps of construction:
1. Draw a ray OP.

2. Draw an angle of 153° with the help of protractor at point O.

3. Taking O as centre draw an arc of any radius which cuts OP at A and OQ at B.
4. Mark two arcs of any radius with centre A and B which intersect each other at C.
5. Draw ray OR through C. Ray OR is bisector of angle POQ.

6. Mark two arcs of any radius with centre A and D which intersect at E.
7. Draw ray OS through the point E. OS is bisector of angle POR.
6. Draw two arcs of any radius with centre D and B, which intersect at F.
8. Draw ray OT through the point F. Therefore, by observation we see rays OS, OR and OT divide the angle POQ into four equal parts.

Q.110 Construct with ruler and compasses, angles of following measures:
(a) 60° (b) 30° (c) 90°
(d) 120° (e) 45° (f) 135°

Ans

(a) Steps of construction:
1. Draw a ray OP.
2. Mark an arc of any radius with centre O. This arc intersects OP at the point A.
3. Mark another arc of the same radius with centre A, it intersects earlier arc at the point B.
4. Draw ray OQ through the point B.Therefore, angle POQ is required angle of 60°.

(b) Steps of construction:
1. Draw a ray OP.
2. Mark an arc of any radius with centre O. This arc intersects OP at the point A.
3. Mark another arc of the same radius with centre A, it intersects earlier arc at the point B.
4. Draw a ray OQ through the point B.
5. Mark two arcs of any radius with centre A and B, which intersect at point C.
6. Draw ray OR through point C. Therefore, angle POR is required angle of 30°.

(c) Steps of construction:
1. Draw a ray OP.
2. Mark an arc of any radius with centre O, which intersects ray OP at the point A.
3. Mark another arc of same radius with centre A which intersects earlier arc at B.
4. Mark an arc of same radius with centre B which intersects the earliest arc at the point C.
5. Mark two arcs of any radius with centre B and C respectively which intersect at the point R.
6. Draw ray OQ through the point R. In this way we get right angle POQ.

(d) Steps of construction:
1. Draw a ray OP.
2. Mark an arc of any radius with centre O. This arc intersects OP at the point A.
3. Mark another arc of the same radius with centre A, it intersects earlier arc at the point B.
4. Draw an arc with centre B of same radius. This arc intersects the earliest arc at the point C.
5. Draw ray OQ through the point C. Therefore, angle POQ is required angle of 120°.

(e) Steps of construction:
1. Draw a ray OP.
2. Mark an arc of any radius with centre O, which intersects ray OP at the point A.
3. Mark another arc of same radius with centre A which intersects earlier arc at B.
4. Mark an arc of same radius with centre B which intersects the earliest arc at the point C.
5. Mark two arcs of any radius with centre B and C respectively which intersect at the point R.
6. Draw ray OQ through the point R. In this way we get right angle POQ.
7. Mark two arcs of any radius with centre A and D, which intersect at E.
8. Draw a ray OS through the point E. Therefore, POS is required angle of 45°.

(f) Steps of construction:
1. Draw a line PR.
2. Take a point O on the line.
3. Draw a semi-circle of any radius with centre O.
4. Mark two arcs of any radius with centre A and B respectively which intersect at R.
5. Draw a ray OQ through the point D.
6. Mark two arcs with centre D and B which intersect at E.
7. Draw ray OS through the point E. Therefore, angle POS is required angle of 135°.

Q.111 Draw an angle of measure 45° and bisect it.

Ans

Steps of construction:
1. Draw a ray OP.
2. Mark an arc of any radius with centre O, which intersects ray OP at the point A.
3. Mark another arc of same radius with centre A which intersects earlier arc at B.
4. Mark an arc of same radius with centre B which intersects the earliest arc at the point C.
5. Mark two arcs of any radius with centre B and C respectively which intersect at the point R.
6. Draw ray OQ through the point R. In this way we get right angle POQ.
7. Mark two arcs of any radius with centre A and D (intersection point of arc and OQ), which intersect at E.
8. Draw a ray OS through the point E.
9. Mark two arcs of any radius with centre A and G (intersection point of arc and OS), which intersect at F.
10. Draw ray OT through the point F. Therefore, angle POS is required angle of measure 45° and OT is bisector of angle POS.

Q.112 Draw an angle of measure 135° and bisect it.

Ans

Steps of construction:
1. Draw a line MN.
2. Take a point O on the line.
3. Draw an arc of any radius with centre O which cuts the line MN at A and B.
4. Mark two arcs of radius more than half of AB with centre A and B respectively which intersect at R.
5. Draw a ray OQ through the point R. Let the ray OQ intersects the arc in (3) at D.
6. Mark two arcs with centre D and B of radius more than half of DB which intersect at E.
7. Draw ray OS through the point E. Therefore, angle NOS is required angle of 135°.
8. Mark two arcs of radius more than half of AC with centre A and C respectively which intersect at F.
9. Draw ray OT through the intersection point F. Therefore, angle NOT is the bisector of angle NOS.

Q.113 Draw an angle of 70°. Make a copy of it using only a straight edge and compasses.

Ans

Steps of construction:
1. Draw a ray OA.
2. Draw an angle of 70° at point O with the help of protractor.

3. Mark an arc with centre O which cuts OA and OB at C and D respectively.
4. Draw a ray QP and mark an arc with the same radius and centre Q on ray QP.
5. Set your compasses to the length CD with the same radius and mark an arc with centre T which cuts earlier arc at the point R.
6. Draw ray QS through the point R. Therefore, angle PQS is required angle of 70°.

Q.114 Draw an angle of 40°. Copy its supplementary angle.

Ans

Steps of construction:
Draw a line PA and mark a point O on it.
Draw an angle of 40° at point O with the help of protractor.

3. Angle POB is supplementary of 40°.
4. Taking O as centre, draw an arc between the rays of angle POB which cuts PO and OB at D and C respectively.

5. Now draw a line m and take a point S on it. Taking same radius and S as centre draw an arc. This arc cuts line m at point T.
6. Set your compasses to the length CD with the same radius and mark an arc with centre T which cuts earlier arc at the point R.
7. Join S and R. SR is the required ray which makes a supplementary angle of 40° with line m or makes an angle of 140°.

Q.115 Write the fraction representing the shaded portion.

Ans

$\begin{array}{l}\left(\text{i}\right)\frac{2}{4}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{ii}\right)\frac{8}{9}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{iii}\right)\frac{4}{8}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{iv}\right)\frac{1}{4}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \left(\text{v}\right)\frac{3}{7}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{vi}\right)\frac{3}{12}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{vii}\right)\frac{10}{10}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \left(\text{viii}\right)\frac{4}{9}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{ix}\right)\frac{4}{8}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{x}\right)\frac{1}{2}\end{array}$

Q.116 Colour the part according to the given fraction.

(a) (b)

(c)

(e)

Ans

(a) (b)

(c) (d)

(e)

Q.117 Identify the error, if any
(a)

$\text{This is}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{2}$

(b)

$\text{This is}\frac{1}{4}$

(c)

$\text{This is}\frac{3}{4}$

Ans

(a)

This is

$\frac{1}{2}$

This is incorrect as the shaded portion does not represent the required fraction.
(b)

This is

$\frac{1}{4}$

This is incorrect as the shaded portion does not represent the required fraction.
(c)

This is

$\frac{3}{4}$

This is incorrect as the shaded portion does not represent the required fraction.

Q.118 What fraction of a day is 8 hours?

Ans

Number of hours in a day = 24 hours

Fraction =

$\frac{8}{24}$

Q.119 What fraction of an hour is 40 minutes?

Ans

Number of minutes in an hour = 60 minutes

Fraction =

$\frac{40}{60}$

Q.120 Arya, Abhimanyu, and Vivek shared lunch. Arya has brought two sandwiches, one made of vegetable and one of jam. The other two boys forgot to bring their lunch. Arya agreed to share his sandwiches so that each person will have an equal share of each sandwich.
(a) How can Arya divide his sandwiches so that each person has an equal share?
(b) What part of a sandwich will each boy receive?

Ans

(a) each sandwich will be divided into 3 equal parts, so that each one of them get equal share.

(b) Number of parts of sandwich = 6
Every one of them will get 2 pieces each, out of 6 pieces. Each boy will get part =

$\frac{2}{6}=\frac{1}{3}$

Q.121 Kanchan dyes dresses. She had to dye 30 dresses. She has so far finished 20 dresses. What fraction of dresses has she finished?

Ans

Total dresses to dye = 30

Dresses done dying = 20

Fraction

$=\frac{20}{30}$

Q.122 Write the natural numbers from 2 to 12. What fraction of them are prime numbers?

Ans

Natural numbers from 2 to 12 = 11

Prime numbers = 2, 3, 5, 7, 11

Fraction

$=\frac{5}{11}$

Q.123 Write the natural numbers from 102 to 113. What fraction of them are prime numbers?

Ans

Natural numbers from 102 to 113 = 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113.

Prime numbers among them = 103, 107, 109, 113

Fraction =

$\frac{4}{12}$

Q.124 What fraction of these circles have X’s in them?

Ans

Number of circles having X’s in them = 4

Fraction =

$\frac{4}{8}$

Q.125 Kristin received a CD player for her birthday. She bought 3 CDs and received 5 others as gifts. What fraction of her total CDs did she buy and what fraction did she receive as gifts?

Ans

CDs bought by Kristin = 3

Total = 8

Fraction of CDs she buy =

$\frac{3}{8}$

Fraction of CDs gifted =

$\frac{5}{8}$

Q.126 Draw number lines and locate the points on them :

$\begin{array}{l}\left(\mathrm{a}\right)\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{2},\frac{1}{4},\frac{3}{4},\frac{4}{4}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}}\frac{1}{8},\frac{2}{8},\frac{3}{8},\frac{7}{8}\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}}\frac{2}{5},\frac{3}{5},\frac{8}{5},\frac{4}{5}\end{array}$

Ans

(a)

$\frac{1}{2},\frac{1}{4},\frac{3}{4},\frac{4}{4}$

(b)

$\frac{1}{8},\frac{2}{8},\frac{3}{8},\frac{7}{8}$

(c)

$\frac{2}{5},\frac{3}{5},\frac{8}{5},\frac{4}{5}$

Q.127 Express the following as mixed fractions :

$\begin{array}{l}\left(\mathrm{a}\right)\text{\hspace{0.17em}\hspace{0.17em}}\frac{20}{3}\text{\hspace{0.17em} \hspace{0.17em}}\left(\mathrm{b}\right)\text{\hspace{0.17em}}\frac{11}{5}\text{\hspace{0.17em}\hspace{0.17em}}\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}\hspace{0.17em}}\frac{17}{7}\text{\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{d}\right)\frac{28}{5}\text{\hspace{0.17em}}\\ \left(\mathrm{e}\right)\text{\hspace{0.17em}}\frac{19}{6}\text{\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{f}\right)\frac{35}{9}\end{array}$

Ans

$\begin{array}{l}\left(a\right)\frac{20}{3}=3\begin{array}{c}\hfill 6\\ \hfill \overline{)\begin{array}{l}\text{}20\\ \underset{¯}{-18}\\ \underset{¯}{\text{}2}\end{array}}\end{array}\\ \text{â€‹}\text{â€‹}\text{â€‹}\text{â€‹}\text{â€‹}\text{â€‹}\text{â€‹}\text{â€‹}\text{â€‹}\text{â€‹}\text{â€‹}\text{â€‹}\text{â€‹}\text{â€‹}\text{â€‹}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=6\frac{2}{3}\\ \left(b\right)\text{\hspace{0.17em}}\frac{11}{5}=5\begin{array}{c}\hfill 2\\ \hfill \overline{)\begin{array}{l}\text{11}\\ \underset{¯}{-10}\\ \underset{¯}{\text{1}}\end{array}}\end{array}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\frac{1}{5}\\ \left(c\right)\text{\hspace{0.17em}}\frac{17}{7}=7\begin{array}{c}\hfill 2\\ \hfill \overline{)\begin{array}{l}\text{17}\\ \underset{¯}{-14}\\ \underset{¯}{\text{3}}\end{array}}\end{array}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\frac{3}{7}\\ \left(d\right)\text{\hspace{0.17em}}\frac{28}{5}=5\begin{array}{c}\hfill 5\\ \hfill \overline{)\begin{array}{l}\text{28}\\ \underset{¯}{-25}\\ \underset{¯}{\text{3}}\end{array}}\end{array}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=5\frac{3}{5}\\ \left(e\right)\text{\hspace{0.17em}}\frac{19}{6}=6\begin{array}{c}\hfill 3\\ \hfill \overline{)\begin{array}{l}\text{19}\\ \underset{¯}{-18}\\ \underset{¯}{\text{1}}\end{array}}\end{array}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=3\frac{1}{6}\\ \left(f\right)\text{\hspace{0.17em}}\frac{35}{9}=9\begin{array}{c}\hfill 3\\ \hfill \overline{)\begin{array}{l}\text{35}\\ \underset{¯}{-27}\\ \underset{¯}{\text{8}}\end{array}}\end{array}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=3\frac{8}{9}\end{array}$

Q.128 Express the following as improper fractions :

$\begin{array}{l}\left(\mathrm{a}\right)\text{\hspace{0.17em}}7\frac{3}{4}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{b}\right)5\frac{6}{7}\\ \left(\mathrm{c}\right)2\frac{5}{6}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{d}\right)10\frac{3}{5}\\ \left(\mathrm{e}\right)9\frac{3}{7}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{f}\right)8\frac{4}{9}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{a}\right)\text{\hspace{0.17em}}7\frac{3}{4}=7+\frac{3}{4}=\frac{7×4+3}{4}=\frac{31}{4}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}}5\frac{6}{7}=5+\frac{6}{7}=\frac{5×7+6}{7}=\frac{41}{7}\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}}2\frac{5}{6}=2+\frac{5}{6}=\frac{2×6+5}{6}=\frac{17}{6}\\ \left(\mathrm{d}\right)\text{\hspace{0.17em}}10\frac{3}{5}=10+\frac{3}{5}=\frac{10×5+3}{5}=\frac{53}{5}\\ \left(\mathrm{e}\right)\text{\hspace{0.17em}}9\frac{3}{7}=9+\frac{3}{7}=\frac{9×7+3}{7}=\frac{66}{7}\\ \left(\mathrm{f}\right)\text{\hspace{0.17em}}8\frac{4}{9}=8+\frac{4}{9}=\frac{8×9+4}{9}=\frac{76}{9}\end{array}$

Q.129 Write the fractions. Are all these fractions equivalent ?
(a)

(b)

Ans

(a)

$\text{Here,}\frac{1}{2},\frac{\overline{)2}}{\overline{)4}}=\frac{1}{2},\frac{\overline{)3}}{\overline{)6}}=\frac{1}{2},\frac{\overline{)4}}{\overline{)8}}=\frac{1}{2}$

So, all these fractions are equivalent.

(b)

$\text{Here,}\frac{4}{12}=\frac{1}{3},\frac{3}{9}=\frac{1}{3},\frac{2}{6}=\frac{1}{3},\frac{6}{15}=\frac{2}{5}.$

Q.130 </strWrite the fractions and pair up the equivalent fractions from each row.

Ans

Equivalent fractions are shown below :

Q.131 Write the fractions and pair up the equivalent fractions from each row.

Ans

Equivalent fractions are shown below :

Q.132 Write the fractions and pair up the equivalent fractions from each row.

Ans

Equivalent fractions are shown below :

Q.133 Replace in each of the following by the correct number :

$\left(\text{a}\right)\frac{2}{7}=\frac{8}{}\left(\text{b}\right)\frac{5}{8}=\frac{10}{}\left(\text{c}\right)\frac{3}{5}=\frac{}{20}\left(\text{d}\right)\frac{45}{60}=\frac{15}{}\left(\text{e}\right)\frac{18}{24}=\frac{}{4}$

Ans

$\begin{array}{l}\text{(a)}\frac{2}{7}=\frac{8}{}\\ ⇒\frac{2}{7}=\frac{2×4}{7×4}=\frac{8}{28}\\ \text{(b)}\frac{5}{8}=\frac{10}{}\\ ⇒\frac{5}{8}=\frac{2×5}{8×2}=\frac{10}{16}\\ \left(c\right)\frac{3}{5}=\frac{}{20}\\ ⇒\frac{3}{5}=\frac{3×4}{5×4}=\frac{12}{20}\\ \left(d\right)\frac{45}{60}=\frac{15}{}\\ ⇒\frac{45÷3}{60÷3}=\frac{15}{20}\\ \left(e\right)\frac{18}{24}=\frac{}{4}\\ ⇒\frac{18÷6}{24÷6}=\frac{3}{4}\end{array}$

Q.134 Find the equivalent fraction of having
(a) denominator 20 (b) numerator 9
(c) denominator 30 (d) numerator 27

Ans

$\begin{array}{l}\left(\text{a}\right)\text{Denominator 2}0\\ \frac{3}{5}=\frac{}{20}\\ ⇒\frac{3}{5}=\frac{3×4}{5×4}=\frac{12}{20}\\ \left(\text{b}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Numerator 9}\\ \text{}\frac{3}{5}=\frac{9}{}\\ ⇒\frac{3}{5}=\frac{3×3}{5×3}=\frac{9}{15}\\ \left(\text{c}\right)\text{Denominator 3}0\\ \text{}\frac{3}{5}=\frac{}{30}\\ ⇒\frac{3}{5}=\frac{3×6}{5×6}=\frac{18}{30}\\ \left(\text{d}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Numerator 27}\\ \text{}\frac{3}{5}=\frac{27}{}\\ ⇒\frac{3}{5}=\frac{3×9}{5×9}=\frac{27}{45}\end{array}$

Q.135 Find the equivalent fraction of

$\frac{36}{48}$

having
(a) numerator 9 (b) denominator 4

Ans

$\begin{array}{l}\left(\text{a}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Numerator 9}\\ \frac{36}{48}=\frac{9}{}\\ ⇒\frac{36÷4}{48÷4}=\frac{9}{12}\\ \left(\text{b}\right)\text{\hspace{0.17em}}\text{Denominator 4}\\ \text{}\frac{36}{48}=\frac{}{4}\\ ⇒\frac{36÷12}{48÷12}=\frac{3}{4}\end{array}$

Q.136 Check whether the given fractions are equivalent :

$\begin{array}{l}\left(\text{a}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{5}{9},\frac{30}{54}\text{\hspace{0.17em}}\\ \left(\text{b}\right)\frac{3}{10},\frac{12}{50}\\ \left(\text{c}\right)\frac{7}{13},\frac{5}{11}\end{array}$

Ans

(a)

$\frac{5}{9},\frac{30}{54}$

The given fractions will be equivalent if the product of numerator of one fraction with denominator of another fraction is equal.

Here, 5 × 54 = 9 × 30. So, the fractions are equivalent.

(b)

$\frac{3}{10},\frac{12}{50}$

The given fractions will be equivalent if the product of numerator of one fraction with denominator of another fraction is equal.

$\text{Here,}50×3\ne 12×10\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{So},\text{these fractions are not equivalent}.$ $\left(c\right)$ $\frac{7}{13},\frac{5}{11}$ $\text{The given fractions will be equivalent if the product of numerator of one fraction with denominator of another fraction is equal}.\phantom{\rule{0ex}{0ex}}\text{\hspace{0.17em}}\text{Here, 11}×7\ne 13×5\text{\hspace{0.17em}}\text{So},\text{these fractions are not equivalent}.$

Q.137 Reduce the following fractions to simplest form :

$\begin{array}{l}\left(\mathrm{a}\right)\text{\hspace{0.17em}}\frac{48}{60}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{b}\right)\frac{150}{60}\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}}\frac{84}{98}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{d}\right)\frac{12}{52}\\ \left(\mathrm{e}\right)\text{\hspace{0.17em}}\frac{7}{28}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{a}\right)\text{\hspace{0.17em}}\frac{48}{60}=\frac{12×4}{12×5}=\frac{4}{5}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}}\frac{150}{60}=\frac{15×10}{6×10}=\frac{15}{6}=\frac{5}{2}\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}}\frac{84}{98}=\frac{14×6}{14×7}=\frac{6}{7}\\ \left(\mathrm{d}\right)\text{\hspace{0.17em}}\frac{12}{52}=\frac{4×3}{4×13}=\frac{3}{13}\\ \left(\mathrm{e}\right)\text{\hspace{0.17em}}\frac{7}{28}=\frac{1×7}{4×7}=\frac{1}{4}\end{array}$

Q.138 Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her/his pencils ?

 Total pencils Used pencils Fraction Equivalent fraction Ramesh 20 10 $\frac{10}{20}$ $\frac{1}{2}$ Sheelu 50 25 $\frac{25}{50}$ $\frac{1}{2}$ Jamaal 80 40 $\frac{40}{80}$ $\frac{1}{2}$

Here, we can see that Ramesh, Sheelu and Jamaal has used up an equal fraction of his/her pencils.

Ans

 Total pencils Used pencils Fraction Equivalent fraction Ramesh 20 10 $\frac{10}{20}$ $\frac{1}{2}$ Sheelu 50 25 $\frac{25}{50}$ $\frac{1}{2}$ Jamaal 80 40 $\frac{40}{80}$ $\frac{1}{2}$

Here, we can see that Ramesh, Sheelu and Jamaal has used up an equal fraction of his/her pencils.

Q.139 Match the equivalent fractions and write two more for each :

 (i) $\frac{250}{400}$ (a) $\frac{2}{3}$ (ii) $\frac{180}{200}$ (b) $\frac{2}{5}$ (iii) $\frac{660}{990}$ (c) $\frac{1}{2}$ (iv) $\frac{180}{360}$ (d) $\frac{5}{8}$ (v) $\frac{220}{550}$ (e) $\frac{9}{10}$

Ans

 (i) $\frac{250}{400}$ (d) $\frac{5}{8},\text{}\frac{10}{16},\text{}\frac{15}{24}$ (ii) $\frac{180}{200}$ (e) $\frac{9}{10},\text{}\frac{18}{20},\text{}\frac{27}{30}$ (iii) $\frac{660}{990}$ (a) $\frac{2}{3},\text{}\frac{4}{6},\text{}\frac{6}{9}$ (iv) $\frac{180}{360}$ (c) $\frac{1}{2},\text{}\frac{2}{4},\text{}\frac{3}{6}$ (v) $\frac{220}{550}$ (b) $\frac{2}{5},\text{}\frac{4}{10},\text{}\frac{6}{15}$

Q.140 Write shaded portions as fraction. Arrange them in ascending or descending order using the correct sign ‘<’ ‘=’ ‘>’ between the fractions.

$\begin{array}{l}\text{(c)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \text{Show}\frac{2}{6},\frac{4}{6},\frac{8}{6}\text{and}\frac{6}{6}\text{on the number line}\text{. Put appropriate}\\ \text{signs between the fractions given :}\\ \frac{5}{6}__\frac{2}{6},\text{}\frac{3}{6}__0,\text{}\frac{1}{6}__\frac{6}{6},\text{}\frac{8}{6}__\frac{5}{6}\end{array}$

Ans

Since, they all are like fractions. They can be arranged in increasing order by observing numerator only.

So,

$\frac{1}{8}<\frac{3}{8}<\frac{4}{8}<\frac{6}{8}$

(b)

Since, they all are like fractions. They can be arranged in increasing order by observing numerator only.

So,

$\frac{3}{9}<\frac{4}{9}<\frac{6}{9}<\frac{8}{9}$

(c)

$\frac{5}{6}\begin{array}{|c|}\hline >\\ \hline\end{array}\frac{2}{6},\frac{3}{6}\begin{array}{|c|}\hline >\\ \hline\end{array}0,\frac{1}{6}\begin{array}{|c|}\hline <\\ \hline\end{array}\frac{6}{6},\frac{8}{6}\begin{array}{|c|}\hline >\\ \hline\end{array}\frac{5}{6}$

Q.141 Write shaded portions as fraction. Arrange them in ascending or descending order using the correct sign ‘<’ ‘=’ ‘>’ between the fractions.

$\begin{array}{l}\text{(c)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \text{Show}\frac{2}{6},\frac{4}{6},\frac{8}{6}\text{and}\frac{6}{6}\text{on the number line}\text{. Put appropriate}\\ \text{signs between the fractions given :}\\ \frac{5}{6}__\frac{2}{6},\text{}\frac{3}{6}__0,\text{}\frac{1}{6}__\frac{6}{6},\text{}\frac{8}{6}__\frac{5}{6}\end{array}$

Ans

Since, they all are like fractions. They can be arranged in increasing order by observing numerator only.

So,

$\frac{1}{8}<\frac{3}{8}<\frac{4}{8}<\frac{6}{8}$

(b)

Since, they all are like fractions. They can be arranged in increasing order by observing numerator only.

So,

$\frac{3}{9}<\frac{4}{9}<\frac{6}{9}<\frac{8}{9}$

(c)

$\frac{5}{6}\overline{)>}\frac{2}{6},\text{}\frac{3}{6}\overline{)>}0,\text{}\frac{1}{6}\overline{)<}\frac{6}{6},\text{}\frac{8}{6}\overline{)>}\frac{5}{6}$

Q.142 Compare the fractions and put an appropriate sign.

$\begin{array}{l}\left(\mathrm{a}\right)\text{\hspace{0.17em}\hspace{0.17em}}\frac{3}{6}\left[\text{\hspace{0.17em}}\right]\frac{5}{6}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(b)\hspace{0.17em}}\frac{1}{7}\left[\text{\hspace{0.17em}}\right]\frac{1}{4}\text{}\\ \text{(c)\hspace{0.17em}}\frac{4}{5}\left[\text{\hspace{0.17em}}\right]\frac{5}{5}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{d}\right)\frac{3}{5}\left[\text{\hspace{0.17em}}\right]\frac{3}{7}\text{}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{a}\right)\text{\hspace{0.17em}}\frac{3}{6}\left[\right]\frac{5}{6}\text{}\\ \text{Since},\text{these are like fractions and 3}<\text{5}.\text{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{3}{6}\left[<\right]\frac{5}{6}\text{}\\ \text{(b)\hspace{0.17em}}\frac{1}{7}\left[\right]\frac{1}{4}\text{}\\ \text{Here},\text{numerator of both the fractions are same}.\phantom{\rule{0ex}{0ex}}\text{So},\text{larger the denominator is},\text{smaller is the value of the fraction}.\\ \mathrm{So},\text{\hspace{0.17em}}\frac{1}{7}\left[<\right]\frac{1}{4}\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}}\frac{4}{5}\left[\right]\frac{5}{5}\text{}\\ \text{Since},\text{these are like fractions and 4}<\text{5}.\text{So},\text{\hspace{0.17em}}\frac{4}{5}\left[<\right]\frac{5}{5}\text{}\\ \text{(d)\hspace{0.17em}}\frac{3}{5}\left[\right]\frac{3}{7}\text{}\\ \text{Here},\text{numerator of both the fractions are same}.\phantom{\rule{0ex}{0ex}}\text{So},\text{larger the denominator is},\text{smaller is the value of the fraction}.\\ \mathrm{So},\text{\hspace{0.17em}}\frac{3}{5}\left[>\right]\frac{3}{7}\text{}\end{array}$

Q.143 Compare the fractions and put an appropriate sign.

$\begin{array}{l}\left(\mathrm{a}\right)\text{\hspace{0.17em}\hspace{0.17em}}\frac{3}{6}\left[\text{\hspace{0.17em}}\right]\frac{5}{6}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\phantom{\rule{0ex}{0ex}}\left(\text{b}\right)\text{\hspace{0.17em}}\frac{1}{7}\left[\text{\hspace{0.17em}}\right]\frac{1}{4}\text{}\\ \left(\text{c}\right)\text{\hspace{0.17em}}\frac{4}{5}\left[\text{\hspace{0.17em}}\right]\frac{5}{5}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{d}\right)\frac{3}{5}\left[\text{\hspace{0.17em}}\right]\frac{3}{7}\text{}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{a}\right)\text{\hspace{0.17em}}\frac{3}{6}\left[\right]\frac{5}{6}\text{}\\ \text{Since},\text{these are like fractions and 3}<\text{5}.\text{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{3}{6}\left[<\right]\frac{5}{6}\text{}\\ \left(\text{b}\right)\text{\hspace{0.17em}}\frac{1}{7}\left[\right]\frac{1}{4}\text{}\\ \text{Here},\text{numerator of both the fractions are same}.\text{So},\text{larger the denominator is},\text{}\phantom{\rule{0ex}{0ex}}\text{smaller is the value of the fraction}.\\ \\ \mathrm{So},\text{\hspace{0.17em}}\frac{1}{7}\left[<\right]\frac{1}{4}\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}}\frac{4}{5}\left[\right]\frac{5}{5}\text{}\\ \text{Since},\text{these are like fractions and 4}<\text{5}.\text{So},\text{\hspace{0.17em}}\frac{4}{5}\left[<\right]\frac{5}{5}\text{}\\ \text{(d)\hspace{0.17em}}\frac{3}{5}\left[\right]\frac{3}{7}\text{}\\ \text{Here},\text{numerator of both the fractions are same}.\text{So},\text{larger the denominator is},\phantom{\rule{0ex}{0ex}}\text{smaller is the value of the fraction}.\\ \mathrm{So},\text{\hspace{0.17em}}\frac{3}{5}\left[>\right]\frac{3}{7}\text{}\end{array}$

Q.144 et the pairs of fractions to compare is

$\begin{array}{l}\left(a\right)\text{\hspace{0.17em}}\frac{3}{7}\left[\right]\frac{5}{7}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{(b)}\text{\hspace{0.17em}}\frac{2}{7}\left[\right]\frac{2}{4}\text{}\\ \left(c\right)\text{\hspace{0.17em}}\frac{4}{9}\left[\right]\frac{5}{9}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(d\right)\frac{1}{5}\left[\right]\frac{1}{7}\text{}\\ \left(e\right)\text{\hspace{0.17em}}\frac{13}{15}\left[\right]\frac{23}{15}\text{}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{a}\right)\frac{3}{7}\left[\right]\frac{5}{7}\text{}\\ \text{Since},\text{these are like fractions and 3}<\text{5}.\text{So},\\ \frac{3}{7}\left[<\right]\frac{5}{7}\text{}\\ \text{(b)}\frac{2}{7}\left[\right]\frac{2}{4}\text{}\\ \text{Here},\text{numerator of both the fractions are same}.\phantom{\rule{0ex}{0ex}}\text{So},\text{larger the denominator is},\text{smaller is the value of the fraction}.\\ \mathrm{So},\text{\hspace{0.17em}}\frac{2}{7}\left[<\right]\frac{2}{4}\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}}\frac{4}{9}\left[\right]\frac{5}{9}\text{}\\ \text{Since},\text{these are like fractions and 4}<\text{5}.\text{So},\\ \frac{4}{9}\left[<\right]\frac{5}{9}\text{}\\ \text{(d)\hspace{0.17em}}\frac{1}{5}\left[\right]\frac{1}{7}\text{}\\ \text{Here},\text{numerators of both the fractions are same}.\phantom{\rule{0ex}{0ex}}\text{So},\text{larger the denominator is},\text{smaller is the value of the fraction}.\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{5}\left[>\right]\frac{1}{7}\text{}\\ \text{(e)\hspace{0.17em}}\frac{13}{15}\left[\right]\frac{23}{15}\text{}\\ \text{Since},\text{these are like fractions and 13}<\text{23}.\text{So},\\ \frac{13}{23}\left[<\right]\frac{15}{23}\text{}\end{array}$

Q.145 Look at the figures and write ‘<’ or ‘>’, ‘=’ between the given pairs of fractions.

$\left(\text{a}\right)\frac{1}{6}\left[\right]\frac{1}{3}\phantom{\rule{0ex}{0ex}}\left(\text{b}\right)\frac{3}{4}\left[\right]\frac{2}{6}\phantom{\rule{0ex}{0ex}}\left(\text{c}\right)\frac{2}{3}\left[\right]\frac{2}{4}\phantom{\rule{0ex}{0ex}}\left(\text{d}\right)\frac{5}{6}\left[\right]\frac{5}{5}$

Ans

$\left(\text{a}\right)\frac{1}{6}\left[<\right]\frac{1}{3}\phantom{\rule{0ex}{0ex}}\left(\text{b}\right)\frac{3}{4}\left[>\right]\frac{2}{6}\phantom{\rule{0ex}{0ex}}\left(\text{c}\right)\frac{2}{3}\left[>\right]\frac{2}{4}\phantom{\rule{0ex}{0ex}}\left(\text{d}\right)\frac{5}{6}\left[<\right]\frac{5}{5}$

Q.146 How quickly can you do this ? Fill appropriate sign. ( ‘<’, ‘=’, ‘>’)

$\begin{array}{l}\left(\text{a}\right)\frac{1}{2}\begin{array}{|c|}\hline \\ \hline\end{array}\frac{1}{5}\left(\text{b}\right)\frac{2}{4}\begin{array}{|c|}\hline \\ \hline\end{array}\frac{3}{6}\left(\text{c}\right)\frac{3}{5}\begin{array}{|c|}\hline \\ \hline\end{array}\frac{2}{3}\\ \left(\text{d}\right)\frac{3}{4}\begin{array}{|c|}\hline \\ \hline\end{array}\frac{2}{8}\left(\text{e}\right)\frac{3}{5}\begin{array}{|c|}\hline \\ \hline\end{array}\frac{6}{5}\left(\text{f}\right)\frac{7}{9}\begin{array}{|c|}\hline \\ \hline\end{array}\frac{3}{9}\\ \left(\text{g}\right)\frac{1}{4}\begin{array}{|c|}\hline \\ \hline\end{array}\frac{2}{8}\left(\text{h}\right)\frac{6}{10}\begin{array}{|c|}\hline \\ \hline\end{array}\frac{4}{5}\left(\text{i}\right)\frac{3}{4}\begin{array}{|c|}\hline \\ \hline\end{array}\frac{7}{8}\\ \left(\text{j}\right)\frac{6}{10}\begin{array}{|c|}\hline \\ \hline\end{array}\frac{4}{5}\left(\text{k}\right)\frac{5}{7}\begin{array}{|c|}\hline \\ \hline\end{array}\frac{15}{21}\end{array}$

Ans

$\begin{array}{l}\left(\text{a}\right)\frac{1}{2}\begin{array}{|c|}\hline >\\ \hline\end{array}\frac{1}{5}\left(\text{b}\right)\frac{2}{4}\begin{array}{|c|}\hline =\\ \hline\end{array}\frac{3}{6}\left(\text{c}\right)\frac{3}{5}\begin{array}{|c|}\hline <\\ \hline\end{array}\frac{2}{3}\\ \left(\text{d}\right)\frac{3}{4}\begin{array}{|c|}\hline >\\ \hline\end{array}\frac{2}{8}\left(\text{e}\right)\frac{3}{5}\begin{array}{|c|}\hline <\\ \hline\end{array}\frac{6}{5}\left(\text{f}\right)\frac{7}{9}\begin{array}{|c|}\hline >\\ \hline\end{array}\frac{3}{9}\\ \left(\text{g}\right)\frac{1}{4}\begin{array}{|c|}\hline =\\ \hline\end{array}\frac{2}{8}\left(\text{h}\right)\frac{6}{10}\begin{array}{|c|}\hline <\\ \hline\end{array}\frac{4}{5}\left(\text{i}\right)\frac{3}{4}\begin{array}{|c|}\hline <\\ \hline\end{array}\frac{7}{8}\\ \left(\text{j}\right)\frac{6}{10}\begin{array}{|c|}\hline <\\ \hline\end{array}\frac{4}{5}\left(\text{k}\right)\frac{5}{7}\begin{array}{|c|}\hline =\\ \hline\end{array}\frac{15}{21}\end{array}$

Q.147 How quickly can you do this ? Fill appropriate sign. ( ‘<’, ‘=’, ‘>’)

$\begin{array}{l}\left(\mathrm{a}\right)\frac{1}{2}\overline{)}\frac{1}{5}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{b}\right)\text{\hspace{0.17em}}\frac{2}{4}\overline{)}\frac{3}{6}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{c}\right)\frac{3}{5}\overline{)}\frac{2}{3}\\ \left(\mathrm{d}\right)\frac{3}{4}\overline{)}\frac{2}{8}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{e}\right)\frac{3}{5}\overline{)}\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{f}\right)\frac{7}{9}\overline{)}\frac{3}{9}\\ \left(\mathrm{g}\right)\frac{1}{4}\overline{)}\frac{2}{8}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{h}\right)\frac{6}{10}\overline{)}\frac{4}{5}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{i}\right)\frac{3}{4}\overline{)}\frac{7}{8}\\ \left(\mathrm{j}\right)\frac{6}{10}\overline{)}\frac{4}{5}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{k}\right)\frac{5}{7}\overline{)}\frac{15}{21}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{a}\right)\frac{1}{2}\overline{)>}\frac{1}{5}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{b}\right)\text{\hspace{0.17em}}\frac{2}{4}\overline{)=}\frac{3}{6}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{c}\right)\frac{3}{5}\overline{)<}\frac{2}{3}\\ \left(\mathrm{d}\right)\frac{3}{4}\overline{)>}\frac{2}{8}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{e}\right)\frac{3}{5}\overline{)<}\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{f}\right)\frac{7}{9}\overline{)>}\frac{3}{9}\\ \left(\mathrm{g}\right)\frac{1}{4}\overline{)=}\frac{2}{8}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{h}\right)\frac{6}{10}\overline{)<}\frac{4}{5}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{i}\right)\frac{3}{4}\overline{)<}\frac{7}{8}\\ \left(\mathrm{j}\right)\frac{6}{10}\overline{)<}\frac{4}{5}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{k}\right)\frac{5}{7}\overline{)=}\frac{15}{21}\end{array}$

Q.148 Kirti bookstore sold books worth â‚¹ 2,85,891 in the first week of June and books worth â‚¹ 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?

Ans

Worth of Books sold in 1st week = â‚¹2,85,891

Worth of books sold in 2nd week = â‚¹4,00,768

Total sale = Sale in 1st week + Sale in 2nd week

= â‚¹2,85,891 + â‚¹4,00,768 = â‚¹6,86,659

Thus, the sale for the two weeks together was â‚¹6,86,659.

Also, 4,00,768 – 2,85,891 = 1,14,877

Thus, the sale in 2nd week was greater than the sale in 1st week by â‚¹1,14,877.

Q.149 A merchant had â‚¹78,592 with her. She placed an order for purchasing 40 radio sets at â‚¹1200 each. How much money will remain with her after the purchase?

Ans

Cost of one radio set = â‚¹1200

So, cost of 40 radio sets = â‚¹1200 × 40 = â‚¹48000

Merchant had â‚¹78,592 with her before purchase.

So, after purchase, she will have â‚¹(78,592 – 48,000),i.e. â‚¹30,592.

Therefore, â‚¹30,592 will remain with her after the purchase.

Q.150 A taxi driver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litres of petrol. If the petrol costs â‚¹44 per litre, how much did he spend in all on petrol?

Ans

The cost of one litre petrol = â‚¹44
Petrol filled in tank on Monday = 40 litres

Petrol filled in tank on Tuesday = 50 litres

Driver spend money on petrol = 44 × (40+50)

= 44 × 90

= â‚¹3960

Thus, a taxi driver spent â‚¹3960 on petrol.

Q.151 Find the cost of fencing a square park of side 250 m at the rate of â‚¹20 per metre.

Ans

Side of square park = 250 m
Rate of fencing the park = â‚¹20 per metre
Perimeter of square park = 4(250 m)
= 1000 m
Cost of fencing the square park

= â‚¹20 x 1000 m
= â‚¹20,000
Thus, the cost of fencing the square park is â‚¹20,000

Q.152 Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of â‚¹12 per metre.

Ans

Length of a rectangular park = 175 m
Breadth of a rectangular park = 125 m
Rate of fencing per metre = â‚¹12
Length of fencing of park = 2(L + B)
= 2(175+ 125) m
= 2(300) m
= 600 m
Cost of fencing the park = â‚¹12(600 m)
= â‚¹7200

Q.153 What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of â‚¹8 per hundred sq m.?

Ans

Area of rectangular plot = 500 m × 200 m

= 1, 00, 000 sq m
Rate of tilling per sq m = â‚¹8 per hundred
= â‚¹(8/100)
Cost of tilling the plot = â‚¹(8/100) × 1,00,000

= â‚¹8000

Q.154 Write the following as numbers in the given table:

Ans

 Hundreds(100) Tens(10) Ones(1) Tenths(1/10) (a) 0 3 1 2 (b) 1 1 0 4

Q.155 Write the following decimals in the place value table.

(a) 19.4 (b) 0.3 (c) 10.6 (d) 205.9

Ans

 Hundreds(100) Tens(10) Ones(1) Tenths(1/10) 19.4 0 1 9 4 0.3 0 0 0 3 10.6 0 1 0 6 205.9 2 0 5 9

Q.156 Complete the table with the help of these boxes and use decimals to write the number.

Ans

 Ones Tenths Hundredths Number (a) 0 2 6 0.26 (b) 1 3 8 1.38 (c) 1 2 8 1.28

Q.157 Write the following decimals in the place value table.
(a) 0.29 (b) 2.08 (c) 19.60 (d) 148.32 (e) 200.812

Ans

 Hundreds Tens Ones Tenths Hundredths Thousandths a 0 0 0 2 9 9 b 0 0 2 0 8 0 c 0 1 9 6 0 0 d 1 4 8 3 2 0 e 2 0 0 8 1 2

Q.158 Express as rupees using decimals.
(a) 5 paise (b) 75 paise (c) 20 paise
(d) 50 rupees 90 paise (e) 725 paise

Ans

$\begin{array}{l}\left(\mathbf{a}\right)\text{}\mathbf{5}\text{}\mathbf{paise}\\ \text{We know that 1}00\text{paise}=\text{1}\\ ⇒\text{1 paise}=â‚¹\frac{1}{100}\text{}\\ ⇒5\text{}\mathrm{paise}=â‚¹\frac{5}{100}\text{\hspace{0.17em}}\\ =\text{â€‹}â‚¹\text{}0.05\text{}\\ \end{array}$ $\begin{array}{l}\left(\mathbf{b}\right)\text{}\mathbf{75}\text{}\mathbf{paise}\\ \text{We know that 1}00\text{paise}=â‚¹\text{1}\\ ⇒\text{1 paise}=â‚¹\frac{1}{100}\text{}\\ ⇒75\text{}\mathrm{paise}=â‚¹\frac{75}{100}\text{\hspace{0.17em}}\\ =\text{â€‹}â‚¹\text{}0.75\text{}\\ \\ \left(\mathbf{c}\right)\text{}\mathbf{20}\text{}\mathbf{paise}\\ \text{We know that 1}00\text{paise}=â‚¹\text{1}\\ ⇒\text{1 paise}=â‚¹\frac{1}{100}\text{}\\ ⇒20\text{}\mathrm{paise}=â‚¹\frac{20}{100}\text{\hspace{0.17em}}\\ =\text{â€‹}â‚¹\text{}0.20\\ \text{}\\ \left(\mathbf{d}\right)\text{}\mathbf{50}\text{}â‚¹\text{}\mathbf{90}\text{}\mathbf{paise}\\ \text{We know that 1}00\text{paise}=â‚¹\text{1}\\ ⇒\text{1 paise}=â‚¹\frac{1}{100}\text{}\\ ⇒90\text{}\mathrm{paise}=â‚¹\frac{90}{100}\text{\hspace{0.17em}}\\ =\text{â€‹}â‚¹\text{}0.90\\ \\ \text{}\mathrm{}\therefore 50â‚¹90\mathrm{paise}=50â‚¹+\text{}â‚¹\text{}0.90\\ =\text{\hspace{0.17em}}â‚¹\text{}50.90\end{array}$ $\begin{array}{l}\left(\mathbf{e}\right)\text{}\mathbf{725}\text{}\mathbf{paise}\\ \text{We know that 1}00\text{paise}=â‚¹\text{1}\\ ⇒\text{1 paise}=â‚¹\frac{1}{100}\text{}\\ ⇒725\text{}\mathrm{paise}=â‚¹\frac{725}{100}\text{\hspace{0.17em}}\\ =\text{â€‹}â‚¹\text{}7.25\\ \end{array}$

Q.159 Rashid spent â‚¹ 35.75 for Maths book and â‚¹ 32.60 for Science book. Find the total amount spent by Rashid.

Ans

$\begin{array}{l}\text{Amount spent for Maths book}=\text{}â‚¹\text{35}.\text{75}\\ \text{Amount spent for Science book}=\text{}â‚¹\text{32}.\text{6}0\\ \\ \text{Total amount spent by Rashid is:}\\ \text{}\\ \text{35}\text{.75}\\ \text{}\underset{¯}{\text{+ 32}\text{.60}}\\ \text{\hspace{0.17em}}\underset{¯}{\text{68}\text{.35}}\\ \\ \text{Therefore},\text{the amount spent by Rashid is â‚¹ 68}.\text{35}.\end{array}$

Q.160 Radhika’s mother gave her â‚¹ 10.50 and her father gave her â‚¹ 15.80, find the total amount given to Radhika by the parents.

Ans

$\begin{array}{l}\text{Amount given by mother}=\text{â‚¹ 1}0.\text{5}0\\ \text{Amount given by father}=\text{â‚¹ 15}.\text{8}0\end{array}$

$\begin{array}{l}\text{Total amount given by parents is:}\\ \text{}\\ \text{10}\text{.50}\\ \text{}\underset{¯}{\text{+ 15}\text{.80}}\\ \text{\hspace{0.17em}}\underset{¯}{\text{26}\text{.30}}\\ \\ \text{Therefore},\text{the amount given by her parents is â‚¹ 26}.\text{3}0.\end{array}$

Q.161 Subtract :
(a) â‚¹ 18.25 from â‚¹20.75
(b) 202.54 m from 250 m
(c) â‚¹ 5.36 from â‚¹ 8.40
(d) 2.051 km from 5.206 km
(e) 0.314 kg from 2.107 kg

Ans

(a) â‚¹ 18.25 from â‚¹20.75

$\begin{array}{l}\text{20}\text{.75}\\ \text{}-\text{}\underset{¯}{\text{18}\text{.25}}\\ \text{}\underset{¯}{\text{}2.50\text{}}\text{\hspace{0.17em}}\text{}\end{array}$

Therefore, â‚¹ 20.75 − â‚¹ 18.25 = â‚¹ 2.50
(b) 202.54 m from 250 m

$\begin{array}{l}\text{250}\text{.00}\\ \text{}-\text{}\underset{¯}{\text{202}\text{.54}}\\ \text{}\underset{¯}{\text{47}\text{.4}6\text{}}\end{array}$

Therefore, 250 m – 202.54 m = 47.46 m
(c) â‚¹ 5.36 from â‚¹ 8.40

$\begin{array}{l}\text{}\text{}\\ \text{8}\text{.40}\\ \text{}-\text{}\underset{¯}{\text{5}\text{.36}}\\ \text{}\underset{¯}{\text{}3.04\text{\hspace{0.17em}}}\end{array}$

Therefore, â‚¹ 8.40 – â‚¹ 18.25 = â‚¹ 3.04
(d) 2.051 km from 5.206 km

$\begin{array}{l}\text{5}\text{.206}\\ \text{}-\text{}\underset{¯}{\text{2}\text{.051}}\\ \text{}\underset{¯}{\text{}3.155\text{}}\end{array}$

Therefore, 5.206 km – 2.051 km = 3.155 km
(e) 0.314 kg from 2.107 kg

$\begin{array}{l}\text{}\\ \text{2}\text{.107}\\ \text{}-\text{}\underset{¯}{\text{0}\text{.314}}\\ \text{}\underset{¯}{\text{}1.793\text{\hspace{0.17em}}\text{}}\\ \text{â€‹}\text{}\\ \text{}\end{array}$

Therefore, 2.107 kg – 0.314 kg = 1.793 kg

Q.162 Raju bought a book for â‚¹ 35.65. He gave â‚¹ 50 to the shopkeeper. How much money did he get back from the shopkeeper?

Ans

Cost price of the book = â‚¹ 35.65
Money given to the shopkeeper = â‚¹ 50
So, the money that Raju will get back is

$\begin{array}{l}\text{}\\ \text{50}\text{.00}\\ \text{}-\underset{¯}{\text{35}\text{.65}}\\ \text{}\underset{¯}{\text{}14.35\text{}}\text{â€‹}\text{}\\ \text{}\end{array}$

Therefore, the money that Raju will get back is â‚¹ 14.35.

Q.163 Rani had â‚¹ 18.50. She bought one ice-cream for â‚¹ 11.75. How much money does she have now?

Ans

Total money with Rani = â‚¹ 18.50

The money spent for an ice-cream= â‚¹ 11.75

So, the money left with Rani =

$\begin{array}{l}\text{}\\ \text{18}\text{.50}\\ \text{}-\underset{¯}{\text{11}\text{.75}}\\ \text{}\underset{¯}{\text{}6.75\text{}}\text{}\\ \text{}\end{array}$

Therefore, the money left with Rani is â‚¹ 6.75.

Q.164 Name the angles in the given figure.

Ans

Four angles in the given figure are:
∠DAB, ∠ABC, ∠BCD and ∠CDA.

Q.165 In the given diagram, name the point(s)
(a) In the interior of ∠DOE
(b) In the exterior of ∠EOF
(c) On ∠EOF

Ans

(a) Point A is in the interior of ∠DOE.
(b) Points A, C and D are in the exterior of ∠EOF.
(c) Points E, B, O and F lie on ∠EOF.

Q.166 Draw rough diagrams of two angles such that they have
(a) One point in common.
(b) Two points in common.
(c) Three points in common.
(d) Four points in common.
(e) One ray in common.

Ans

(a) ∠ABC and ∠CBD have B as a common point.

(b) ∠ABC and ∠ABD have A and B as two common points.

(c) ∠ABC and ∠ACB have three common points.

(d) ∠DAC and ∠EPD have four common points i.e. B, C, D and E.

Q.167 (a) Identify three triangles in the figure.
(b) Write the names of seven angles.
(c) Write the names of six line segments.
(d) Which two triangles have ∠B as common?

Ans

(a) Three triangles are ΔABD, ΔADC and ΔABC.
(c) Six line segments are:

$\overline{\text{AB}},\text{}\overline{\text{AC}},\text{}\overline{\text{BD}},\text{}\overline{\text{DC}},\text{}\overline{\text{BC}}\text{and}\overline{\text{AD}}.$

(d) ΔABC and ΔABD have ∠B as common.

Q.168 Draw a rough sketch of a quadrilateral KLMN. State,
(a) two pairs of opposite sides,
(b) two pairs of opposite angles,
(c) two pairs of adjacent sides,
(d) two pairs of adjacent angles.

Ans

(a) Two pairs of opposite sides are:
KL and MN, LM and KN.
(b) Two pairs of opposite angles are:
∠LKN and ∠LMN,
∠KLM and ∠KNM.
(c) Two pairs of adjacent sides are:
KL and LM, LM and MN.
(d) Two pairs of adjacent angles are:
∠LMN and ∠MNK,
∠KLM and ∠LMN.

Q.169 The following fractions represent just three different numbers. Separate them into three groups of equivalent fractions, by changing each one to its simplest form.

$\begin{array}{l}\left(\mathrm{a}\right)\frac{2}{12}\left(\mathrm{b}\right)\frac{3}{15}\left(\mathrm{c}\right)\frac{8}{50}\\ \left(\mathrm{d}\right)\frac{16}{100}\left(\mathrm{e}\right)\frac{10}{60}\left(\mathrm{f}\right)\frac{15}{75}\\ \left(\mathrm{g}\right)\frac{12}{60}\left(\mathrm{h}\right)\frac{16}{96}\left(\mathrm{i}\right)\frac{12}{75}\\ \left(\mathrm{j}\right)\frac{12}{72}\left(\mathrm{k}\right)\frac{3}{18}\left(\mathrm{l}\right)\frac{4}{25}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{a}\right)\frac{2}{12}=\frac{1×2}{6×2}=\frac{1}{6}\left(\mathrm{b}\right)\frac{3}{15}=\frac{1×3}{5×3}=\frac{1}{5}\left(\mathrm{c}\right)\frac{8}{50}=\frac{4×2}{25×2}=\frac{4}{25}\\ \left(\mathrm{d}\right)\frac{16}{100}=\frac{4×4}{25×4}=\frac{4}{25}\left(\mathrm{e}\right)\frac{10}{60}=\frac{1×10}{6×10}=\frac{1}{6}\left(\mathrm{f}\right)\frac{15}{75}=\frac{1×15}{5×15}=\frac{1}{5}\\ \left(\mathrm{g}\right)\frac{12}{60}=\frac{1×12}{5×12}=\frac{1}{5}\left(\mathrm{h}\right)\frac{16}{96}=\frac{1×16}{6×16}=\frac{1}{6}\left(\mathrm{i}\right)\frac{12}{75}=\frac{3×4}{3×25}=\frac{4}{25}\\ \left(\mathrm{j}\right)\frac{12}{72}=\frac{1×12}{6×12}=\frac{1}{6}\left(\mathrm{k}\right)\frac{3}{18}=\frac{1×3}{6×3}=\frac{1}{6}\left(\mathrm{l}\right)\frac{4}{25}=\frac{1×4}{1×25}=\frac{4}{25}\end{array}$

The three different groups representing same fraction are
Fractions representing

$\frac{1}{6}$

:

$\frac{2}{12},\frac{10}{60},\frac{16}{96},\frac{12}{72},\frac{3}{18}$

Fractions representing

$\frac{1}{5}$

:

$\frac{3}{15},\frac{15}{75},\frac{12}{60}$

Fractions representing

$\frac{4}{25}$

:

$\frac{8}{50},\frac{16}{100},\frac{12}{75},\frac{4}{25}$

Q.170 Find answers to the following. Write and indicate how you solved them.

$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{Is}\frac{5}{9}\mathrm{equal}\text{}\mathrm{to}\frac{4}{5}?\phantom{\rule{0ex}{0ex}}\left(\mathrm{b}\right)\mathrm{Is}\frac{9}{16}\mathrm{equal}\text{}\mathrm{to}\frac{5}{9}?\\ \left(\mathrm{c}\right)\mathrm{Is}\frac{4}{5}\mathrm{equal}\text{}\mathrm{to}\frac{16}{20}?\phantom{\rule{0ex}{0ex}}\left(\mathrm{d}\right)\mathrm{Is}\frac{1}{15}\mathrm{equal}\text{}\mathrm{to}\frac{4}{30}?\end{array}$

Ans

(a) Let us convert both the fractions into like fractions.

$\begin{array}{l}\frac{5}{9}\text{and}\frac{4}{5}\\ \frac{5×5}{9×5}\text{and}\frac{4×9}{5×9}\\ \frac{25}{45}\text{and}\frac{36}{45}\\ \text{Since, 25 < 36}\text{. So,}\frac{5}{9}<\frac{4}{5}.\end{array}$

The fractions are not equal.
(b)
Let us convert both the fractions into like fractions.

$\begin{array}{l}\frac{9}{16}\text{and}\frac{5}{9}\\ \frac{9×9}{16×9}\text{and}\frac{5×16}{16×9}\\ \frac{81}{144}\text{and}\frac{80}{144}\\ \text{Since, 81>80}\text{. So,}\frac{9}{16}>\frac{5}{9}.\end{array}$

The fractions are not equal.
(c)
Let us convert both the fractions into like fractions.

$\begin{array}{l}\frac{16}{20}\text{and}\frac{4}{5}\\ \frac{16}{20}\text{and}\frac{4×4}{5×4}\\ \frac{16}{20}\text{and}\frac{16}{20}\\ \text{Since, 16 = 16}\text{. So,}\frac{16}{20}=\frac{4}{5}.\end{array}$

The fractions are equal.
(d)
Let us convert both the fractions into like fractions.

$\begin{array}{l}\frac{1}{15}\text{and}\frac{4}{30}\\ \frac{1×2}{15×2}\text{and}\frac{4}{30}\\ \frac{2}{30}\text{and}\frac{4}{30}\\ \text{Since, 2 < 4}\text{. So,}\frac{1}{15}<\frac{4}{30}.\end{array}$

Q.171 Ila read 25 pages of a book containing 100 pages. Lalita read

$\frac{2}{5}$

of the same book. Who read less?

Ans

Pages read by Ila = 25

Fraction of pages read by Ila =

$\frac{25}{100}=\frac{1}{4}$

Fraction of pages read by Lalita =

$\frac{2}{5}$

Comparing both the fractions :

$\begin{array}{l}\frac{1}{4}\text{and}\frac{2}{5}\\ \frac{1×5}{4×5}\text{and}\frac{2×4}{4×5}\\ \frac{5}{20}\text{and}\frac{8}{20}\\ \text{Since, 5 < 8}\text{.}\\ \end{array}$

So, Ila read less.

Q.172 Rafiq exercised for

$\frac{3}{6}$

of an hour, while Rohit exercised for

$\frac{3}{4}$

of an hour. Who exercised for a longer time?

Ans

Rafiq exercised for

$\frac{3}{6}$

of an hour
Rohit exercised for

$\frac{3}{4}$

of an hour
Comparing both the fractions :
Since, numerator of both the fractions are same and denominators 4 < 6.
So, Rohit exercised longer.

Q.173 In a class A of 25 students, 20 passed in first class. In another class B of 30 students, 24 passes in first class. In which class was a greater fraction of students getting first class?

Ans

Fraction of students of class A who passed in first class =

$\frac{20}{25}=\frac{4}{5}$

Fraction of students of class B who passed in first class =

$\frac{24}{30}=\frac{4}{5}$

So, equal fraction of students got first class.

Q.174 Write these fractions appropriately as additions or subtractions :

Ans

Q.175 Solve :

$\begin{array}{l}\left(\text{a}\right)\text{â€„}\frac{\text{1}}{\text{18}}\text{+}\frac{\text{1}}{\text{18}}\\ \left(\text{b}\right)\text{â€„}\frac{\text{8}}{\text{15}}\text{+}\frac{\text{3}}{\text{15}}\\ \left(\mathrm{c}\right)\text{â€„}\frac{7}{5}-\frac{7}{5}\\ \left(\mathrm{d}\right)\text{â€„}\frac{1}{22}+\frac{21}{22}\\ \left(\mathrm{e}\right)\text{â€„}\frac{12}{15}-\frac{7}{15}\\ \left(\mathrm{f}\right)\text{â€„}\frac{5}{8}+\frac{3}{8}\\ \left(\mathrm{g}\right)\text{â€„}1-\frac{2}{3}\left(1=\frac{3}{3}\right)\\ \left(\mathrm{h}\right)\text{â€„}\frac{1}{4}+\frac{0}{4}\\ \left(\mathrm{i}\right)\text{â€„}3-\frac{12}{5}\end{array}$

Ans

$\begin{array}{l}\text{(a)}\frac{1}{18}+\frac{1}{18}\\ \text{}=\frac{1+1}{18}\\ \text{}=\frac{2}{18}\end{array}$ $\begin{array}{l}\text{(b)}\frac{8}{15}+\frac{3}{15}\\ \text{}=\frac{8+3}{15}\\ \text{}=\frac{11}{15}\end{array}$ $\begin{array}{l}\text{}\end{array}$

$\begin{array}{l}\text{(c)}\frac{7}{7}-\frac{5}{7}\\ \text{}=\frac{7-5}{7}\\ \text{}=\frac{2}{7}\end{array}$ $\begin{array}{l}\text{}\end{array}$ $\begin{array}{l}\text{(d)}\frac{1}{22}+\frac{21}{22}\\ \text{}=\frac{1+21}{22}\\ \text{}=\frac{22}{22}=1\end{array}$ $\begin{array}{l}\text{}\end{array}$ $\begin{array}{l}\text{(e)}\frac{12}{15}-\frac{7}{15}\\ \text{}=\frac{12-7}{15}\\ \text{}=\frac{5}{15}=\frac{1}{3}\end{array}$

$\begin{array}{l}\begin{array}{l}\text{(f)}\frac{5}{8}+\frac{3}{8}\\ \text{â€„}=\frac{8}{8}=1\end{array}\\ \text{â€‹â€‹}\left(\text{g}\right)\text{}1–\frac{2}{3}\text{â€‹}\\ \text{â€„â€„â€„â€„â€„}=\frac{3}{3}–\frac{1}{3}\text{â€‹}\\ \text{â€„â€„â€„â€„â€„}=\frac{2}{3}\end{array}$ $\begin{array}{l}\text{}\end{array}$ $\begin{array}{l}\text{(h)}\frac{1}{4}+\frac{0}{4}\\ \text{}=\frac{1}{4}+0\\ \text{}=\frac{1}{4}\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}\left(\text{i) 3}-\frac{12}{5}\\ \text{}=\frac{3×5}{5}-\frac{12}{5}\\ \text{}=\frac{15}{5}-\frac{12}{5}\\ \text{}=\frac{3}{5}\end{array}$

Q.176 Shubham painted

$\frac{2}{3}$

of the wall space in his room. His sister Madhavi helped and painted

$\frac{1}{3}$

of the wall space. How much did they paint together?

Ans

$\begin{array}{l}\text{Space painted by shubham =}\frac{2}{3}\text{of the room}\\ \text{Space painted by Madhvi =}\frac{1}{3}\text{of the room}\\ \text{Hence, together they painted =}\frac{2}{3}+\frac{1}{3}=1\\ \text{}\mathrm{i}.\mathrm{e}.\text{the complete wall.}\end{array}$

Q.177 Fill in the missing fractions :

$\begin{array}{l}\left(\text{a}\right)\text{â€„}\frac{\text{7}}{\text{10}}-\overline{)}\text{â€„=â€„}\frac{\text{3}}{\text{10}}\text{â€„}\\ \left(\text{b}\right)\text{â€„}\overline{)}\text{â€„}-\frac{\text{3}}{\text{21}}\text{=}\frac{\text{5}}{\text{21}}\\ \left(\text{c}\right)\text{â€„}\overline{)}\text{â€„}-\frac{\text{3}}{\text{6}}\text{â€„=â€„}\frac{\text{3}}{\text{6}}\\ \left(\text{d}\right)\text{â€„}\overline{)}\text{â€„+â€„}\frac{\text{5}}{\text{27}}\text{â€„=â€„}\frac{\text{12}}{\text{27}}\end{array}$

Ans

$\begin{array}{l}\left(\text{a)}\frac{7}{10}-\overline{)}=\frac{3}{10}\\ ⇒\overline{)}=\frac{7}{10}-\frac{3}{10}\\ ⇒\overline{)}=\frac{4}{10}\\ \left(\text{b)}\overline{)}-\frac{3}{21}=\frac{5}{21}\\ ⇒\overline{)}=\frac{3}{21}+\frac{5}{21}\\ ⇒\overline{)}=\frac{8}{21}\\ \text{(c)}\overline{)}-\frac{3}{6}=\frac{3}{6}\\ ⇒\overline{)}=\frac{3}{6}+\frac{3}{6}\\ ⇒\overline{)}=\frac{6}{6}=1\\ \text{(d)}\overline{)}+\frac{5}{27}=\frac{12}{27}\\ ⇒\overline{)}=\frac{12}{27}-\frac{5}{27}\\ ⇒\overline{)}=\frac{12-5}{27}=\frac{7}{27}\end{array}$

Q.178 Javed was given

$\frac{5}{7}$

of a basket of oranges. What fraction of oranges was left in the basket?

Ans

Fractions given to Javed =

$\frac{5}{7}$

Fractions left in basket

$=1-\frac{5}{7}\phantom{\rule{0ex}{0ex}}=\frac{7}{7}-\frac{5}{7}\phantom{\rule{0ex}{0ex}}=\frac{2}{7}$

Q.179

$\begin{array}{l}\left(\mathrm{a}\right)\frac{2}{3}+\frac{1}{7}\\ \left(\mathrm{b}\right)\frac{3}{10}+\frac{7}{15}\\ \left(\mathrm{c}\right)\frac{4}{9}+\frac{2}{7}\\ \left(\mathrm{d}\right)\frac{5}{7}+\frac{1}{3}\\ \left(\mathrm{e}\right)\frac{2}{5}+\frac{1}{6}\\ \left(\mathrm{f}\right)\frac{4}{5}+\frac{2}{3}\\ \left(\mathrm{g}\right)\frac{3}{4}-\frac{1}{3}\\ \left(\mathrm{h}\right)\frac{5}{6}-\frac{1}{3}\\ \left(\mathrm{i}\right)\frac{2}{3}+\frac{3}{4}+\frac{1}{2}\\ \left(\mathrm{j}\right)\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\\ \left(\mathrm{k}\right)1\frac{1}{3}+3\frac{2}{3}\\ \left(\mathrm{l}\right)4\frac{2}{3}+3\frac{1}{4}\\ \left(\mathrm{m}\right)\frac{16}{5}-\frac{7}{5}\\ \left(\mathrm{n}\right)\frac{4}{3}-\frac{1}{2}\end{array}$

Ans

$\begin{array}{l}\text{(a)}\frac{2}{3}+\frac{1}{7}\\ \text{The L}\text{.C}\text{.M}\text{. of 3 and 7 is 21}\text{.}\\ \text{=}\frac{7×2+1×3}{21}\\ =\frac{14+3}{21}\\ =\frac{17}{21}\end{array}$

$\begin{array}{l}\text{(b)}\frac{3}{10}+\frac{7}{15}\\ \text{The L}\text{.C}\text{.M}\text{. of 10 and 15 is 30}\text{.}\\ \text{=}\frac{3×3+2×7}{30}\\ =\frac{9+14}{30}\\ =\frac{23}{30}\end{array}$

$\begin{array}{l}\text{(c)}\frac{4}{9}+\frac{2}{7}\\ \text{The L}\text{.C}\text{.M}\text{. of 9 and 7 is 63}\text{.}\\ \text{=}\frac{7×4+9×2}{63}\\ =\frac{28+18}{63}\\ =\frac{46}{63}\end{array}$

$\begin{array}{l}\text{(d)}\frac{5}{7}+\frac{1}{3}\\ \text{The L}\text{.C}\text{.M}\text{. of 3 and 7 is 21}\text{.}\\ \text{=}\frac{5×3+1×7}{21}\\ =\frac{15+7}{21}\\ =\frac{22}{21}\end{array}$

$\begin{array}{l}\text{(e)}\frac{2}{5}+\frac{1}{6}\\ \text{The L}\text{.C}\text{.M}\text{. of 5 and 6 is 30}\text{.}\\ \text{=}\frac{6×2+5×1}{30}\\ =\frac{12+5}{30}\\ =\frac{17}{30}\end{array}$

$\begin{array}{l}\text{(f)}\frac{4}{5}+\frac{2}{3}\\ \text{The L}\text{.C}\text{.M}\text{. of 5 and 3 is 15}\text{.}\\ \text{=}\frac{4×3+5×2}{15}\\ =\frac{12+10}{15}\\ =\frac{22}{15}\end{array}$

$\begin{array}{l}\text{(g)}\frac{3}{4}-\frac{1}{3}\\ \text{The L}\text{.C}\text{.M}\text{. of 4 and 3 is 12}\text{.}\\ \text{=}\frac{3×3-4×1}{12}\\ =\frac{9-4}{12}\\ =\frac{5}{12}\end{array}$

$\begin{array}{l}\text{(h)}\frac{5}{6}-\frac{1}{3}\\ \text{The L}\text{.C}\text{.M}\text{. of 6 and 3 is 6}\text{.}\\ \text{=}\frac{1×5-2×1}{6}\\ =\frac{5-2}{6}\\ =\frac{3}{6}=\frac{1}{2}\end{array}$

$\begin{array}{l}\text{(i)}\frac{2}{3}+\frac{3}{4}+\frac{1}{2}\\ \text{The L}\text{.C}\text{.M}\text{. of 2, 3 and 4 is 12}\text{.}\\ \text{=}\frac{2×4+3×3+6×1}{12}\\ =\frac{8+9+6}{12}\\ =\frac{23}{12}\end{array}$

$\begin{array}{l}\text{(j)}\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\\ \text{The L}\text{.C}\text{.M}\text{. of 2, 3 and 6 is 6}\text{.}\\ \text{=}\frac{1×3+1×2+1×1}{6}\\ =\frac{3+2+1}{6}\\ =\frac{6}{6}=1\end{array}$

$\begin{array}{l}\text{(k) 1}\frac{1}{3}+3\frac{2}{3}\\ \text{Converting the given fraction into improper fraction}\\ \text{=}\frac{4}{3}+\frac{11}{3}\\ =\frac{11+4}{3}\\ =\frac{15}{3}=5\end{array}$

$\begin{array}{l}\text{(l) 4}\frac{2}{3}+3\frac{1}{4}\\ \text{Converting the given fraction into improper fraction}\\ \text{=}\frac{14}{3}+\frac{13}{4}\\ \text{The L}\text{.C}\text{.M}\text{. of 3 and 4 is 12}\\ =\frac{14×4+3×13}{12}\\ =\frac{56+39}{12}=\frac{95}{12}\end{array}$

$\begin{array}{l}\text{(m)}\frac{16}{5}-\frac{7}{5}\\ \text{=}\frac{16-7}{5}\\ =\frac{9}{5}\end{array}$

$\begin{array}{l}\text{(n)}\frac{4}{3}-\frac{1}{2}\\ \text{The L}\text{.C}\text{.M}\text{. of 3 and 2 is 6}\\ \text{=}\frac{4×2-1×3}{6}\\ =\frac{8-3}{6}\\ =\frac{5}{6}\end{array}$

Q.180 Sarita bought

$\frac{2}{5}$

metre of ribbon and Lalita bought

$\frac{3}{4}$

metre of ribbon. What is the total length of the ribbon they bought?

Ans

Length of ribbon bought by Sarita =

$\frac{2}{5}$

Length of ribbon bought by Lalita =

$\frac{3}{4}$

Total length of ribbon

$\begin{array}{l}\text{=}\frac{2}{5}+\frac{3}{4}\\ =\frac{2×4+5×3}{20}\\ =\frac{8+15}{20}\\ =\frac{23}{20}\mathrm{m}\end{array}$

Q.181 Naina was given

$1\frac{1}{2}$

piece of cake and Najma was given

$1\frac{1}{3}$

piece of cake. Find the total amount of cake was given to both of them.

Ans

Piece of cake given to Naina =

$1\frac{1}{2}=\frac{3}{2}$

Piece of cake given to Najma =

$1\frac{1}{3}=\frac{4}{3}$

Total amount of cake given

$\begin{array}{l}\text{=}\frac{3}{2}+\frac{4}{3}\\ =\frac{3×3+4×2}{6}\\ =\frac{9+8}{6}\\ =\frac{17}{6}=2\frac{5}{6}\end{array}$

Q.182

$\left(\mathrm{a}\right)\text{\hspace{0.17em}}\overline{)}-\frac{5}{8}=\frac{1}{4}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{b}\right)\text{\hspace{0.17em}\hspace{0.17em}}\overline{)}-\frac{1}{5}=\frac{1}{2}\text{\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{c}\right)\frac{1}{2}-\overline{)}=\frac{1}{6}$

Ans

$\begin{array}{l}\text{(a)}\overline{)}-\frac{5}{8}=\frac{1}{4}\\ ⇒\overline{)}=\frac{1}{4}+\frac{5}{8}\\ ⇒\overline{)}=\frac{2×1+5×1}{8}\\ ⇒\overline{)}=\frac{7}{8}\\ \text{(b)}\overline{)}-\frac{1}{5}=\frac{1}{2}\\ ⇒\overline{)}=\frac{1}{5}+\frac{1}{2}\\ ⇒\overline{)}=\frac{1×2+1×5}{10}\\ ⇒\overline{)}=\frac{7}{10}\\ \text{(c)}\frac{1}{2}-\overline{)}=\frac{1}{6}\\ ⇒\overline{)}=\frac{1}{2}-\frac{1}{6}\\ ⇒\overline{)}=\frac{3-1}{6}=\frac{2}{6}\end{array}$

Q.183 Complete the addition-subtraction box.

Ans

$\begin{array}{l}\text{(a)}\frac{2}{3}+\frac{4}{3}=\frac{6}{3}=2\\ \text{}\frac{1}{3}+\frac{2}{3}=\frac{3}{3}=1\\ \text{}\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\\ \text{}\frac{4}{3}-\frac{2}{3}=\frac{2}{3}\\ \text{}2-1=1\\ \text{}\end{array}$

$\begin{array}{l}\text{(b)}\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\\ \text{}\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\\ \text{}\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\\ \text{}\frac{1}{3}-\frac{1}{4}=\frac{1}{12}\\ \text{}\frac{5}{6}-\frac{7}{12}=\frac{3}{12}=\frac{1}{4}\\ \text{}\end{array}$

Q.184 A piece of wire

$\frac{7}{8}$

metre long broke into two pieces. One piece was

$\frac{1}{4}$

metre long. How long is the other piece?

Ans

Length of one piece =

$\frac{1}{4}$

Length of other piece is equal to the difference of the length of the wire and the length of the first piece of wire.

So, length of other piece

$\begin{array}{l}=\frac{7}{8}-\frac{1}{4}\\ =\frac{7-2}{8}\\ =\frac{5}{8}\mathrm{m}\end{array}$

Q.185 Nandini’s house is

$\frac{9}{10}$

km from her school. She walked some distance and then took a bus for

$\frac{1}{2}$

km to reach the school. How far did she walk?

Ans

Distance walked by Nandini
= Total distance – Distance for which she took the bus

$\begin{array}{l}\text{=}\frac{9}{10}-\frac{1}{2}\\ =\frac{9-5}{10}\\ =\frac{4}{10}\\ =\frac{2}{5}\text{}km\end{array}$

Q.186 Asha and Samuel have bookshelves of the same size partly filled with books. Asha’s shelf is

$\frac{5}{6}$

th full and Samuel’s shelf is

$\frac{2}{5}$

th full. Whose bookshelf is more full? By what fraction

Ans

Fraction of Asha’s shelf that is full =

$\frac{5}{6}$

Fraction of Samuel’s shelf that is full =

$\frac{2}{5}$

Converting both fractions into like fractions :

$\begin{array}{l}\frac{5}{6}=\frac{5×5}{6×5}=\frac{25}{30}\\ \frac{2}{5}=\frac{2×6}{5×6}=\frac{12}{30}\\ \text{Here,}\frac{5}{6}>\frac{2}{5}\end{array}$

Hence, Asha’s bookshelf is more full.

$\begin{array}{l}\text{Difference}=\text{}\frac{25}{30}-\frac{12}{30}\\ \text{}=\frac{25-12}{30}\\ \text{}=\frac{13}{30}\end{array}$

Q.187 Jaidev takes

$2\frac{1}{5}$

minutes to walk across the school ground. Rahul takes

$\frac{7}{4}$

minutes to do the same. Who takes less time and by what fraction?

Ans

Time taken by Jaidev =

$2\frac{1}{5}\mathrm{minutes}=\frac{11}{5}\mathrm{minutes}$

Time taken by Rahul =

$\frac{7}{4}\mathrm{minutes}$

Converting both fractions into like fractions :

$\begin{array}{l}\frac{11}{5}=\frac{11×4}{5×4}=\frac{44}{20}\\ \frac{7}{4}=\frac{7×5}{4×5}=\frac{35}{20}\\ \text{Here,}\frac{11}{5}>\frac{7}{4}\end{array}$

Hence, Rahul takes lesser time.

$\begin{array}{l}\text{Difference}=\text{}\frac{44}{20}-\frac{35}{20}\\ \text{}=\frac{44-35}{20}\\ \text{}=\frac{9}{20}\text{min}\end{array}$

Q.188 In a Mathematics test, the following marks were obtained by 40 students. Arrange these marks in a table using tally marks.

8 1 3 7 6 5 5 4 4 2

4 9 5 3 7 1 6 5 2 7

7 3 8 4 2 8 9 5 8 6

7 4 5 6 9 6 4 4 6 6

(a) Find how many students obtained marks equal to or more than 7.

(b) How many students obtained marks below 4?

Ans

Arranged data in a table as given below:

 Marks Tally Marks Number of Students 1 $||$ 2 2 $|||$ 3 3 $|||$ 3 4 $\overline{)||||}||$ 7 5 $\overline{)||||}|$ 6 6 $\overline{)||||}||$ 7 7 $\overline{)||||}$ 5 8 $||||$ 4 9 $|||$ 3

(a) 12 students obtained marks equal to or more than 7.

(b) 8 students obtained marks below 4.

Q.189 Following is the choice of sweets of 30 students of Class VI.

Rasgulla, Jalebi, Ladoo, Barfi, Rasgulla,
Ladoo, Barfi, Rasgulla, Rasgulla, Jalebi,

(a) Arrange the names of sweets in a table using tally marks.

(b) Which sweet is preferred by most of the students?

Ans

(a)

 Sweets Tally Marks Number of Students Ladoo $\overline{)||||}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\overline{)||||}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|$ 11 Barfi $|||$ 3 Jalebi $\overline{)||||}\text{\hspace{0.17em}}||$ 7 Rasgulla $\overline{)||||}\text{\hspace{0.17em}}||||$ 9 30

(b) Ladoo is preferred by most of the students.

Q.190 Catherine threw a dice 40 times and noted the number appearing each time as shown below:
1 3 5 6 6 3 5 4 1 6 2 5 3 4 6 1 5 5 6 1 1 2 2 3
5 2 4 5 5 6 5 1 6 2 3 5 2 4 1 5
Make a table and enter the data using tally marks. Find the number that appeared.
(a) The minimum number of times
(b) The maximum number of times
(c) Find those numbers that appear an equal number of times.

Ans

 Numbers Tally Marks Number of times 1 $\overline{)||||}||$ 7 2 $\overline{)||||}|$ 6 3 $\overline{)||||}$ 5 4 $||||$ 4 5 $\overline{)||||}\overline{)||||}|$ 11 6 $\overline{)||||}||$ 7

(a) The number that appeared the minimum number of times is 4.
(b) The number that appeared the maximum number of times is 5.
(c) 1 and 6 both appeared 7 times.

Q.191 Following pictograph shows the number of tractors in five villages.

 Villages Number of tractors = 1 Tractor Village A Village B Village C Village D Village E

Observe the pictograph and answer the following questions.

(i) Which village has the minimum number of tractors?

(ii) Which village has the maximum number of tractors?

(iii) How many more tractors village C has as compared to village B?

(iv) What is the total number of tractors in all the five villages?

Ans

(i) Village D has minimum number of tractors.

(ii) Village C has maximum number of tractors.

(iii) Village C has 8 tractors and village B has 5 tractors. So, village C has 3 more tractors as compared to village B.

(iv) Total number of tractors in 5 villages
= 6 + 5 + 8 + 3 + 6
= 28

Q.192 The number of girl students in each class of a co-educational middle school is depicted by the pictograph:

 Classes Number of girl students -4 Girls I II III IV V VI VII VIII

Observe this pictograph and answer the following questions:
(a) Which class has the minimum number of girl students?
(b) Is the number of girls in Class VI less than the number of girls in Class V?
(c) How many girls are there in Class VII?

Ans

(a) Class VIII has minimum number of girl students i.e., 6 girls.
(b) No, the number of girls in class VI is not less than the number of girls in class V.
(c) Number of girls in class VII is 12.

Q.193 The sale of electric bulbs on different days of a week is shown below:

 Days Number of electric bulbs = 2 bulbs Monday Tuesday Wednesday Thursday Friday Saturday Sunday

Observe the pictograph and answer the following questions:

(a) How many bulbs were sold on Friday?

(b) On which day were the maximum number of bulbs sold?

(c) On which of the days same number of bulbs were sold?

(d) On which of the days minimum number of bulbs were sold?

(e) If one big carton can hold 9 bulbs. How many cartons were needed in the given week?

Ans

From the given pictograph, we can conclude the following:
(a) Number of bulbs sold on Friday is 14.

(b) Maximum number of bulbs was sold on Sunday.

(c) Same number of bulbs were sold on Wednesday and Saturday.

(d) Minimum number of bulbs was sold on Wednesday and Saturday.

(e) Number of bulbs sold on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday are 12, 16, 8, 10, 14, 8 and 18 respectively. The total number of bulbs sold in the given week is 86.

Number of cartons needed in the given week =

$\frac{86}{9}=9.5$

Therefore, 9 cartons were needed in the given week.

Q.194 In a village six fruit merchants sold the following number of fruit baskets in a particular season:

 Name of fruit merchants Number of fruit baskets = 100 fruit baskets Rahim Lakhanpal Anwar Martin Ranjit Singh Joseph

Observe this pictograph and answer the following questions :
(a) Which merchant sold the maximum number of baskets?

(b) How many fruit baskets were sold by Anwar?

(c) The merchants who have sold 600 or more number of baskets are planning to buy a godown for the next season. Can you name them?

Ans

(a) Martin sold maximum number of baskets
i.e., 950.

(b) Anwar sold baskets = 7

$×$

100
= 700
(c) The names of the merchants are Anwar, Martin and Ranjit Singh.

Q.195 Total number of animals in five villages are as follows:
Village A: 80 Village B: 120
Village C: 90 Village D: 40
Village E: 60
Prepare a pictograph of these animals using one symbol

$\otimes$

to represent 10 animals and answer the following questions :
(a) How many symbols represent animals of village E?
(b) Which village has the maximum number of animals?
(c) Which village has more animals: village A or village C?

Ans

Pictograph is as follows:

 Villages = 10 animals Village A Village B Village C Village D Village E

(a) Six symbols represent animals of village E.
(b) Village B has the maximum number of animals.
(c) Village C has more animals than village A.

Q.196 Total number of students of a school in different years is shown in the following table.

 Years Number of students 1996 400 1998 535 2000 472 2002 600 2004 623

A. Prepare a pictograph of students using one symbol to represent 100 students and answer the following questions:

(a) How many symbols represent total number of students in the year 2002?

(b) How many symbols represent total number of students for the year 1998?

B. Prepare another pictograph of students using any other symbol each representing 50 students. Which pictograph do you find more informative?

Ans

A. The required pictograph of students is given below:

 Years = 100 students 1996 1998 2000 2002 2004

(a) 6 symbols represent total number of students in the year 2002.

(b) 5 complete and 1 incomplete symbols represent total number of students for the
year 1998.

B. The required pictograph of students is given below:

 Years = 50 students 1996 1998 2000 2002 2004

Q.197 A survey of 120 school students was done to find which activity they prefer to do in their free time.

 Preferred activity Number of students Playing 45 Reading story books 30 Watching TV 20 Listening to music 10 Painting 15

Draw a bar graph to illustrate the above data taking scale of 1 unit length = 5 students.
Which activity is preferred by most of the students other than playing?

Ans

Reading story books is preferred by most of the students other than playing.

Q.198 The number of Mathematics books sold by a shopkeeper on six consecutive days is shown below:

 Days Number of books sold Sunday 65 Monday 40 Tuesday 30 Wednesday 50 Thursday 20 Friday 70

Draw a bar graph to represent the above information choosing the scale of your choice.

Ans

The required bar graph is given below:

Q.199 Following table shows the number of bicycles manufactured in a factory during the years 1998 to 2002. Illustrate this data using a bar graph. Choose a scale of your choice.

 Years Number of bicycles manufactured 1998 800 1999 600 2000 900 2001 1100 2002 1200

(a) In which year were the maximum number of bicycles manufactured?
(b) In which year were the minimum number of bicycles manufactured?

Ans

(a) In 2002, the maximum number of bicycles were manufactured.
(b) In 1999, the minimum number of bicycles were manufactured.

Q.200 Number of persons in various age groups in a town is given in the following table.

 Age-group (in years) Number of persons 1 – 14 2 lakhs 15 – 29 1 lakh 60 thousands 30 – 44 1 lakh 20 thousands 45 – 59 1 lakh 20 thousands 60 – 74 80 thousands 75 and above 40 thousands

Draw a bar graph to represent the above
information and answer the following
questions.
(take 1 unit length = 20 thousands)

(a) Which two age groups have same
population?

(b) All persons in the age group of 60 and
above are called senior citizens. How
many senior citizens are there in the
town?

Ans

(a) Two age groups which have same population are 30 – 44 and 45 – 59.

(b) 1, 20,000 senior citizens are there in the town.

##### FAQs (Frequently Asked Questions)
1. Will it be enough if I study from the NCERT books only?

Absolutely, it’s enough for students if they only want to refer to the NCERT books. However, if students want to improve their scores, or they need more practice, Extramarks NCERT Solutions will definitely come in handy.

2. Do I need to solve each and every question in every chapter?

Students must know how to solve every question in every chapter so that there is no scope of missing out on any question in the examination.

3. How many chapters are there in the class 6 NCERT Mathematics book?

There are a total of 14 chapters in NCERT Mathematics.

4. What if I complete the NCERT book and still have time left for the exam?

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