Haryana Board Class 7 Maths Solutions For Chapter 13 Visualising Solid Shapes

Key Concepts

• Introduction:
  In our day-to-day life, we see several objects around us which have different shapes. One thing common about most of these items is that they all have some length, breadth, and height or depth.
  That is they all occupy space and have three dimensions. Hence they are called three-dimensional shapes.
• Two-dimensional figures:
  The figures drawn on a paper which have only length and breadth are called two-dimensional figures.
• Faces, Edges, and Vertices:
  The comers of a solid shape are called its vertices. The line segment of its skeleton are its edges. Its flat surfaces are its faces. The 8 corners of the cube are its vertices.
  The 12 line segments that form the skeleton of the cube are its edges.
  The 6 flat square surfaces that are the skin of the cube are its faces.
• 2 – dimensional figures

• 3- dimensional figures –

HBSE Class 7 Visualising Solid Shapes Solutions Ex 13.1

Hint: The sum of number of faces and number of vertices – number of edges = 2

i.e, F+V- E = 2

Net: A net is a sort of skeleton outline of a solid that can be folded to make it

• The same solid can have several types of nets.
• Solid shapes can be drawn on a flat surface like paper realistically.
• We call this “2-D representation of a 3-D solid”
• Drawing solids on a flat surface:
  • Visual illusion: Our drawing surface is a paper, which is flat. When we draw a. solid shape, the images are somewhat distorted, to make them appear three-dimensional. It is a visual illusion.
• There are two ways of drawing solids on a flat surface. They are
  • oblique sketches,
  • Isometric sketches.
• Oblique sketches:
  
  Here is a picture of a cube. It gives a clear idea of how the cube looks like. When seen from the front, We do not see certain faces. In the drawn picture, the lengths are not equal, as they should be in a cube. Such a sketch of a solid is called an “oblique sketch’
• In an oblique sketch, it is clear that.
  1. The sizes of the front faces and its opposite faces are same.
  2. The edges which are all equal in a cube, appear so in the sketch, though the actual measures of edges are not taken so.
• Isometric sketches:
  Isometric dot sheet is such a sheet which divides the paper into small equilateral triangles made up of dots or lines. To draw sketches, in which measurements also agree with those ofthe solid, we can use isometric dot sheets
• Isometric means equal measurements.
• In an isometric sketch, The vertical lines denotes height of the solid and the horizontal lines generally drawn at 30° to the baseline to denote length and width.
• A shadowplay:

1. 1. Another way is by observing a 2-D shadow of a 3-D shape.
   2. A third way is to look at the shape from different angles. The ‘front’ view, the side view’ and the ‘top view’ can provide a lot of information about the shape observed.

Haryana Board Class 7 Maths Visualising Solid Shapes solutions

Solutions To Try These

Match the shape with the name:

Solution. (i)- b (11)- d (iii)- a (iv) – c (v) – f (vi) – e

Solutions To Try These

Match the 2-dimensional figures with the names:

Solution:

(1) – b (2) – a (3) – e (4)- c (5)- d

Solutions For Pratice

Complete the following table:

Solution:

Solutions To Try These

Here youfind four nets. There are two correct nets among them to make a tetrahedron. See if you can work out which nets will make a tetrahedron.

Solution:

(1) and (2) are the correct nets for making a tetrahedron

Haryana Board Class 7 Maths Solutions For Chapter 13 Exercise 13.1

1. Identify make cubes the nets(cut out which copies can of be the used nets to a and try it):

Solution: Nets (2), (3),(4) and (5) can be used to make cubes.

2) Dice are cubes with dots on each face. Opposite faces of a die always have a total of neven dots on them.

Here are two nets to make dice (cubes). the numbers inserted in each square indicate the number of dots in that box.

Insert suitable numbers in the blanks, remembering that the number on the opposite faces should total to 7.

Solution:

HBSE 7th Class Visualising Solid Shapes Real-Life Applications

3. Can this be a net for a die?

Explain your answer.

Solution:

No, we cannot use the given net as a die Because one pair at opposite face will have 1 and 4 on them whose total is not equal to 7 and another pair of opposite faces will have 3 and 6 on them whose total is also not equal to 7.

HBSE Class 7 Visualising Solid Shapes Chapter 13 Definitions Faces Edges Vertices

4. Here is an incomplete net to making a cube. Complete it in at least two different ways. Remember that a cube has six faces. How many are there in the net here? (Give two separate diagrams. If you like, you may use a squared sheet for easy manipulation )

Solution:

To complete the net for making a cube in at latest two different ways Such that a cube has six faces as follows:

It is clear that there are three faces in the net.

How to draw 3D shapes on paper Class 7 HBSE

5. Match the nets with appropriate Solids

Solution: a -(2); b -(3) ;c – (4); d)- (1)

Haryana Board Class 7 Maths Solutions For Chapter 13 Exercise-13.2

1. Use isometric dot paper and make an isometric sketch for each one of the given shapes:

(1)

Top Front Side Views of Solids Class 7 Haryana Board

(2)

(3)

(4)

Solution:

(1)

(2)

(3)

Faces, edges, and vertices of 3D shapes Class 7 HBSE

(4)

2. The dimensions of a cuboid are 5 cm, 3cm and 2 cm. Draw three different isometric sketches of this cuboid.

Solution:

(1)

(2)

(3)

The above are the three different isometric sketches of cuboid of dimensions 5 cm, 3 cm and 2 cm. By changing length, breadth andheight we get the above sketches.

Drawing Oblique Sketches Class 7 HBSE Solutions

3. Three cubes each with 2 cm edge are placed side by side to form a cuboid. Sketch an oblique or isometric sketch of this cuboid.

Solution: Three cubes each with 2 cm edge are placed side by side to form a cuboid then

length = 2 + 2 + 2 = 6cm

Breadth = 2 cm

Height = 2 cm

Oblique sketch of the cuboid:

Isometric sketch of the cuboid :

4. Make an oblique sketch for each one of the given isometric shapes.

Solution: Oblique sketch:

5. Give (1) an oblique sketch and (2) an isometric sketch for each of the following:

1) A cuboid of dimensions 5 cm, 3 cm and 2 cm. (Is your sketch unique?)

2) A cube with an edge 4 cm long.

Solution:

(1) Oblique sketches:

1)

2)

(2) Isometric sketches:

1)

2)

Nets of solid shapes examples Class 7 HBSE

Solutions To Try These

Try to guess the number of cubes in the. following arrangements:

Solution:

The number of cubes in the arrangements are :

(1) 24 cubes (2) 8 ctibes (3) 9 cubes

Solutions To Try These

1. Two dice are placed side by side as shown: Can you say what the total would be on the face opposite to

1) 5 + 6

2) 4 + 3

(Remember that in a die sum of the numbers on opposite faces is 7).

Solution:

Numbers on the opposite face

1) 5 + 6 is 2 + 1

2) 4 + 3 is 3 + 4

2. Three cubes each with 2 cm edge are $ § § placed side by side to form a cuboid. Try to make an oblique sketch and say what could be its length, breadth and height.

Solution:

Length = 2 cm + 2 cm + 2 cm = 6 cm

Breadth = 2 cm

Height =2 cm

Haryana Board Class 7 Maths Solutions For Chapter 13 Exercise-13.3

1. What cross-sections do you get when you give a
(1) vertical cut (2) horizontal cut to the following solids?

1) A brick; 2) A round apple 3) A die 4) A circular pipe 5) An ice cream cone

Solution:

Haryana Board Class 7 Maths Solutions For Chapter 13 Exercise-13.4

1. A bulb is kept binning just above the following solids. Name the shape of the shadows obtainedin each case. Attempt to give a rough sketch of the shadow. (You may try to experiment first and then answer these questions)

(1) A ball (2) A cylindrical pipe (3) A book

Solution:

1)

2)

3)

2. Here are shadows of some 3-D objects, when seen under the lamp of an overhead projector. Identify the solid(s) that match each shadow. (There may be multiple answers for these !)

3. Examine if the following are true statements:

(1) The cube can cast a shadow in the shape of a rectangle.
(2) The cube can cast a shadow in the shape of a hexagon

Solution: (1) True (2) False

Solutions To Try These

1. For each solid, the three views (1), (2), (3) are given. Identify for each solid the corresponding top, front, and side views.

Solution:

1) 1 -> Front 2 -> Side 3 -> Top

2) 1 -> Top 2 -> Side 3 -> Front

3) 1 -> Side 2 -> Front3 -> Top

4) 1 -> Side 2 -> Top 3 -> Front

HBSE Class 7 Maths Chapter 13 Guide Visualising Solid Shapes

2. Draw a view of each solid as seen from the direction indicated by the arrow.

Solution:

View of solid as seen from the direction indicated by the arrow:

Haryana Board Class 7 Maths Solutions For Chapter 13 Very Short Answer Questions!

1. Give examples of ‘plane figures’

Solution:

The circle, the square, the rectangle, the quadrilateral, and the triangle.

2. Give examples of ‘solid shapes’.

Solution:

The cube, the cuboid, the sphere, the cylinder, the cone, and the pyramid.

3. How many types of sketches of a solids are possible? What are they?

Solution:

There are-two types of sketches of a solid that are possible. They are (1) Oblique sketches (2) Isometric sketches.

Key Questions in Visualising Solid Shapes for Class 7 HBSE

4. What shape is (1) A brick (2) A road roller (3) A sweet laddu

Solution:

(1) A brick- Cuboid
(2) A road roller- Cylinder
(3) A sweet laddu – Sphere

Different views of solid shapes Class 7 Haryana Board

5. Give two examples of the following

(1) Cone (2) Cylinder (3) Cuboid

Solution:

(1) Cone- Ice cream cone; birthday cap
(2) Cylinder – Pillar, road roller
(3) Cuboid- Book, matchbox

6. Write the difference between an oblique sketch and an isometric sketch.

Solution:

An oblique sketch:

It does not have proportional lengths. Still it conveys all important aspects of the appearance of the solid.

An isometric sketch:

It is drawn on an isometric dot paper. In anisometric sketch of the solid the measurements are of exact size.

Haryana Board Class 7 Maths Solutions For Chapter 13 Short Answer Questions

7. Write names of at least 2 objects from day-to-day life, which arc in the shape of the basic 3D shapes given below:

Haryana Board Class 7 Maths Solutions For Chapter 13 Long Answer Questions

8. Given below are the pictures of some objects. Categorise and write their names according to the shape and fill the table with name of it.

Solution:

9. Use an isometric dot paper and make an isometric sketch for each one of the given shapes.

1)

2)

3)

Solution:

1) Isometric sketch for the given shape

2) Isometric sketches for the figure given

3) Isometric sketches for the figure given

Important Concepts Visualising Solid Shapes Class 7 HBSE

10. The dimensions of a cuboid are 5 cm, 3 cm and 2 cm. Draw two different isometric sketches of this cuboid.

Solution: The two different isometric sketches of the cuboid whose dimensions are 5 cm, 3 cm and 2 cm. are
1)

2)

11. Three cubes each with 2 cm edge are placed side by side to form a cuboid. Draw an oblique or isometric sketch of this cuboid.

Solution:

Three cubes each with 2 cm edge are placed side by side to form a cuboid;

1) The oblique sketch of thus formed cuboid

2) Isometric sketch of cuboid.

Haryana Board Class 7 Maths Solutions For Chapter 13 Multiple Choice Answer Questions

1. Match the following.

1. 1- a,2 – d,3 – e, 4-b
2. 1 – b,2- c,3 – d,4- e
3. 1 – a,2- c,3 – d,4 -b
4. 1 – c,2- d,3 – a,4-b

Answer: 1

Practice Problems Visualising Solid Shapes Class 7 Haryana Board with Nets of Solids

2. What is the number on the face opposite to 4 on a die?

1. 5
2. 1
3. 3
4. 2

Answer: 3

3. What is the horizontal cross-section of cone?

1. Circle
2. Right triangle
3. Rectangle
4. Square

Answer: 1

4. What is the vertical cross-section of cone?

1. Circle
2. Right triangle
3. Rectangle
4. Square

Answer: 2

5. When we rotate a right triangle we get a

1. Cube
2. Cuboid
3. Cone
4. Cylinder

Answer: 3

6. The solid with one curved surface and one flat surface is……..

1. Cuboid
2. Cylinder
3. Cone
4. Sphere

Answer: 3

7. When we cut a brick horizontally then the shape of the cutting is……..

1. Square
2. Rectangle
3. Triangle
4. Circle

Answer: 2

You can draw sketches in which measurements also agree with those of a given solid. To do this we need anisometric sheet.

Read the above para and answer the following (8-9).

8. What is the difference between oblique sketch and isometric sketch?

1. Shape
2. Faces
3. Measurements
4. None

Answer: 3

9. Ravi wants to draw 6 cm, 3 cm, and 2 cm cuboid exactly with these measurements. Which method is suitable?

1. Oblique sketch
2. Isometric sketch
3. Kitchen play
4. Shadow play

Answer: 2

10. Nani cuts the carrot as shown in figure. What is the shape of cross-section?

Answer: 1

11. Pyramid is a ……dimensional object.

1. 1
2. 2
3. 3
4. Infinite

Answer: 3

12. A point has…….. dimensions.

1. 1
2. 2
3. 3
4. ‘0’ (zero)

Answer: 4

13. A two-dimensional figure in the following is

1. Ball
2. Square
3. Cylinder
4. Matchbox

Answer: 2

14. A three-dimensional figure in the following is

1. Ball
2. Square
3. Rectangle
4. Triangle

Answer: 1

15. Can this be a net for a cube?

1. Yes
2. No
3. Sometimes
4. None

Answer: 2

16. Which of the following is not the net of a cube?

Answer: A

17. Identify the correct statement

1. A cube has 8 vertices
2. A cuboid has 10 faces
3. A cone has 2 vertices
4. A cylinder has vertex

Answer: 1

18. One face of a dice has 3 dots then its opposite face contains……..number of dots.

1. 3
2. 4
3. 5
4. 1

Answer: 2

19. Which solid has four triangular faces and one square face?

Answer: C

20. is the net diagram of the following

1. cylinder
2. triangular pyramid
3. cube
4. square pyramid

Answer: 2

21. The top view of the cylinder is…………..

Answer: A

22. Rishi made a cube of side 3 cm with a wire. What is the length of the wire? 

1. 32 cm
2. 42 cm
3. 48 cm
4. 36 cm

Answer: 4

23. The following is the shadow of which 3D object?

1. sphere
2. cylinder
3. cuboid
4. cone

Answer: 3

24. The diagram represents the front view of…………..

1. die
2. book
3. ball
4. pyramid

Answer: 4

25. The solid which has square shaped from front, top view and side view is

1. cuboid
2. cylinder
3. cube
4. cone

Answer: 3

26. The following has only one flat surface

Answer: A

27. What is the shape Of resulting figure of the combination of two cubes?

1. rectangle
2. cuboid
3. cylinder
4. cone

Answer: 2

28. The number of vertices of the figure

1. 4
2. 5
3. 6
4. 7

Answer: 2

29. The net of cylinder is …………

Answer: B

30. How many cubes are there in the figure?

1. 6
2. 8
3. 10
4. 12

Answer: 2

31.  What are the measurements of resulting figure?

1. l = 6 cm, b = 3 cm, h = 3 cm
2. l = 3 cm, b = 3 cm, h = 6 cm
3. l = 3 cm, b = 6 cm, h = 3 cm
4. l = 6 cm, b = 6 cm, h = 6 cm

Answer: 1

32. Relavent 3D shape of the net

Answer: A

Haryana Board Class 7 Maths Solutions For Chapter 13 Fill in the blanks:

33. Plane figures are of……………

Answer: two dimension

34. ……….shapes are of three dimensions.

Answer: Solid

35. The corners of a solid shape are called……..

Answer: vertices

36. The line segments that forms the skeleton of the cube are its……..

Answer: edges

37. The flat surfaces that are the skin of the cube are its………

Answer: faces

38. A …….is a skeleton outline of a solid that can be folded to make it.

Answer: net

39…………is a very useful skill.

Answer: Visualising solid shapes

40. A cylinder has …….faces.

Answer: 2

41. The vertical cut of a circular pipe is a ………..

Answer: circle

42. Ball and laddu are examples for a ……….shape.

Answer: sphere

Shape No. of faces or edges

43. Match the following:

Shape                                                                             No Of Faces Or Edges

1. Cube                                                                              (     )  A) 4

2. Triangular pyramid                                                      (     )  B) 12

3. Triangular prism in the shape of a kaleidoscope      (     )  C) 6

4. Number of edges in a cuboid                                     (     )  D) 8

5. Number of verticesin a cuboid                                   (     )  E) 5

Answer:

1. C 2. A 3. E 4. B 5. D

Haryana Board Class 7 Maths Solutions For Chapter 10 Algebraic Expressions

Key Concepts

1. Introduction:
   Expressions are a central concept in algebra. x + 3, y- 5, 4x + 5,10y- 5 are some simple algebraic expressions.
2. Variable:
   A variable can take various values. Its value is not fixed. We use the letters x, y, l, m ………..etc. to denote variables.
3.  Constant:
   A constant has a fixed value.
   Examples: 4, 100, -17 etc.
   
4. Algebraic Expression:
   A number of combination of numbers using the signs of fundamental operations is called an expression.
   We combine variables and constants to make algebraic expressions. For this we use the operations of addition, subtraction, multiplication, and division.
   Examples: 4x + 5, 10y – 20.
   Look at how the following expressions are obtained:
   1. x²,
   2. 2y²,
   3. 3x²- 5,
   4. xy,
   5. 4xy+7
      1. The expression x2 is obtained by multiplying the variable x by itself.
         x × x = x². It is commonly read as squared
         x × x × x=x³. It is read as ‘x’ cubed.
         x, x², x³……..are all algebraic expressions obtainedfrom x.
      2. The expression 2y² is obtained from y = 2y²= 2 × y × y. Here by multiplying y with y, we obtain y² and then we multiply y² by the constant 2.
      3. In 3x² – 5, we first obtain x² and multiply it by 3 to get 3x². From 3x², we subtract 5 to finally arrive at 3x²- 5.
      4. In xy, we multiply the vanable x with another variable y. Thus, x xy = xy.
      5. In,4xy + 7, we first obtain xy, and multiply it by 4 to get 4xy one add 7 to 4xy to get the expression.
5. Terms of an expression:
   The expression is separated by ‘+’ or ‘ – ‘ into several parts each part along with its sign is known as the term of the expression.
   Example: 8x²- 6xy. The terms in this expression are 8x² and- 6xy.

Solutions To Try These

Describe how the following expressions are obtained.

7xy + 5,x²y, 4X²- 5x 

Solutions:

7xy + 5: We multiply the variable x with another variable y to obtain xy and then multiply by the constant 7 to get 7xy. Adding 5 to 7xy we obtain 7xy + 5.

x²y: Multiply the variable x with itself to obtain x² and then multiply with y to get x²y.

4x²- 5x: Multiply the variable x with itself to obtain x² and. then multiply with 4 to get 4x².

Multiply the variable x with a constant 5 to get 5x. Then subtract 5x from 4x² to get 4x²-5x

Solutions To Try These

1. What are the terms in the following expressions? Show how the terms are formed. Draw a tree diagram for each expression :

1. 8y + 3x²,
2. 7mn- 4,
3. 2x²y.

1) 8y + 3x²

Solution: Terms: 8y, 3x²

The term 8y is formed by multiplying the variable y by 8.

The term 3x² is formed by multiplying 3, x, and x.

Tree diagram

Expression: 8y+3x²

2) 7mn- 4

Solutions: Terms: 7mn,- 4

The term 7mn is formed by multiplying 7, m, and n. The term – 4 is a constant.

Tree diagram

Expression: 7mn- 4

HBSE Class 7 Algebraic Expressions Solutions Ex 10.1 Solved

Solution: Terms: 2x²y

The term 2x²y is formed by multiplying

2, x, x and y

Tree diagram

Expression: 2x²y

Haryana Board Class 7 Maths Algebraic Expressions solutions

2. Write three expressions each having 4 terms.

Solution:

(1) 4x²- 3xy + 4x + 13

(2) 3x²- 5y²-+ 7 xy + 8

(3) 5x³- 5x²- 5x- 5

Solutions To Try These

Identify the coefficients of the terms of the following expressions :

(1) 4x-3y

Solutions: 4 is the coefficient of x; – 3 is the coefficient of. y.

2) a + b + 5

Solution: The coefficient of a and b is 1.

3) 2y + 5

Solution: The coefficient of y is 2.

4) 2xy

Solution:

The coefficient of xy is 2.

The coefficient of x is 2y.

The coefficient of y is 2x.

Solutions To Try These

Group the like terms together from the following: 12x, 12, -25x, -25, -25y, 1, x, 12y, y

Solution:

Like terms are 12x, -25x, x

-25y, 12y,y

12, -25, 1

Solutions To Try These

Classify the following expressions as a monomial, a binomial or a trinomial: a, a + b, ab + a + b, ab + a +b- 5, xy, xy + 5, 5x²- x + 2, 4pq – 3q + 5p, 7, 4m- 7n + 10, 4mn + 7.

Solution:

Monomials: a, xy, 7

Binomials: a + b, xy + 5, 4mn + 7

Trinomials : ab + a + b, 5x²- x + 2,

4m- 7n + 10, 4pq- 3p + 5p

Polynomial: ab+a+b-5

Haryana Board Class 7 Maths Solutions For Chapter 10 Exercise-10.1 :

1. Get the algebraic expression in the following cases using variables, constants, and arithmetic operations.

1. Subtraction of z from y. → y- z
2. One-half of the sum of numbers, x and \( y \rightarrow \frac{1}{2}(x+y) \)
3. The number z multiplied by itself. → z²
4. One-fourth of the product of numbers p and q.→\( \frac{1}{4} \mathrm{pq}\)
5. Numbers x and y both squared and added. x²+, y²
6. Number 5 added to three times the product of numbers m and n. → 3mn + 5
7. Product of numbers y and z subtracted from 10. → 10- yz
8. Sum of numbers a and b subtracted from their product→ ab-(a + b)

2. (1) Identify the terms and their factors in the following expressions. Show the terms and factors by tree diagrams.

1) x- 3

Solution:

Expression: x-3

2) 1+x + x²

Solution: Expression: 1 + x + x²

3) y – y³

Solution:

Expression: y – y³

HBSE Class 7 Maths Chapter 10 Get Algebraic Expressions

3) 2x²y

4) 5xy² + 7x²y

Solution:

Expression: 5xy²+ 7x²y

5) -ab + 2b²-3a²

Solution:

Expression: -ab + 2b²-3a²

(2) Identify terms and factors in the expressions given below:

1) – 4x + 5

Solution:

Terms: – 4x, 5

Factors : – 4, x; 5

2) – 4x + 5y

Solution:

Terms: – 4 x; 5y

Factors: – 4, x; 5, y

3) 5y + 3y²

Solution:

Terms: 5y, 3y²

Factors: 5, y; 3, y, y

4) xy +2x²y²

Solution:

Terms: xy; 2x²y²

Factors: x, y; 2, x, x, y, y

5) pq + q

Solution:

Terms: pq, q

Factors : p, q; q

6) 1.2ab – 2.4b + 3.6a

Solution:

Terms: 1.2ab; -2.4b; 3.6a

Factors: 1.2, a, b; -2.4, b; 3.6, a

7) \( \frac{3}{4} x+\frac{1}{4} \)

Solution:

Terms: \( \frac{3}{4} x ; \frac{1}{4} \)

Factors: \( \frac{3}{4}, x ; \frac{1}{4} \)

8) 0.1p² + 0.2q²

Solution:

Terms: 0.1p², 0.2q²

Factors: 0.1, p, p; 0.2, q, q

3. Identify the numerical coefficients of terms (other than constants) in the following expressions:

1) 5 – 3t²

Solution:

Term which Is not constant is -3t²

Numerical coefficient is -3

2) 1 + t + t²+ t³

Solution:

Terms which are not constant are t, t²,t³

Numerical coefficients: 1, 1,1

3) x + 2xy + 3y

Solution:

Terms which are not constant are x, 2xy,3y

Numerical coefficients: 1, 2, 3

HBSE 7th Class Algebraic Expressions Word Problems Solutions

4)100m + 1000n

Solution:

Terms which are not constant are 100m 1000n.

Numerical coefficients: 100, 1000

5) – p²q² + 7pq

Solution:

Terms which are not constant are -p²q²;7pq

Numerical coefficients: -1, 7

6) 1.2a + 0.8b

Solution:

Terms which are not constant are 1.2a; 0.8b

Numerical coefficients: 1.2, 0.8

7) 3.14r²

Solution:

Term: 3.14r²

Numerical coefficient is 3.14

How to simplify algebraic expressions Class 7 HBSE

8) 2(l+ b)

Solution:

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

Terms: 2l; 2b

Numerical coefficients: 2, 2

9) 0.1y + 0.01y²

Solution:

Terms: 0.1y + 0.01y²

Numerical coefficients: 0.1; 0.01

4. a) Identify terms which contain x and give the coefficient of x.

1) y³x + y 

Solution:

Term which contain x is y²x

Coefficient of x is y²

2) 13y² – 8yx

Solution:

Term which contain x is -8 yx

Coefficient of x is- 8y

3) x+y + 2

Solution:

Term which contain x is x

Coefficient ‘of x is 1

4) 5 + z + zx

Solution:

Term which contain x is zx

Coefficient of x is z

5) 1 + x + xy

Solution:

Term which contain x are x, xy

Coefficients of x are 1, y

6) 12xy² + 25

Term which contain x is 12xy²

Coefficient of x is 12y².

7) 7x + xy²

Solution:

Terms which contain x are 7x and xy²

Coefficients of x are 7 and y²

2) Identify terms which contain y² and give the coefficient of y².

1) 8-xy²

Solution:

Term which contain y² is – xy²

Coefficient of y² is – x.

2) 5y² + 7x

Solution:

Term which contain y² is 5y²

Coefficient of y² is 5

3) 2x²y – 15xy² + 7y²

Solution:

Terms which contains y² is -15xy²; 7y²

Coefficients of y² are -15x; 7

5. Classify into monomials, binomials and trinomials.

1. 4y-7z
2.  y²
3. x + y- xy
4. 100
5. ab – a – b
6. 5 – 3t
7. 4p²q-4pq²
8. 7mn
9. z²- 3z + 8
10.  a² + b²
11. z² + z
12. 1 + x + x²

Solution:

Monomials: (2) y², (4)100, (8) 7 mn

Binomials: (1) 4y- 7z,(6) 5 -3t, (7) 4p²q -4pq², (10) a² + b², (11) z² + z

Trinomials : (3) x + y- xy, (5) ab- a-b, (9) z²- 3z + 8, (12)1+ x + x²

6. State whether a given pair of terms is of like or unlike terms.

1) 1,100

Solution: 1, 100 are like terms.

2) \( -7 x, \frac{5}{2} x \)

Solution: \( -7 x, \frac{5}{2} x \) are like terms.

3) -29x, -29y

Solution: -29x, -29y are unlike terms.

4) 14xy, 42yx

Solution: 14xy, 42yx are like terms.

5) 4m²p,4mp²

Solution: 4m²p, 4mp² are unlike terms.

7) 12xz, 12x²z²

Solution: 12xz, 12x²z² are unlike terms.

7. Identify like terms in the following:

1) -xy², – 4yx², 8x², 2xy², 7y, -11x², -100x,-11yx, 20x²y,- 6x², y, 2xy, 3x

Solution:

Like terms are:-

xy², 2xy²;- 4yx², 20x²y;- 8x², -6x², – 11x²; 7y, y; -100x, 3x; -11yx, 2xy

2)10pq, 7p, 8q, -p²q², -7qp, – 100q, -23,12q²p², -5p², 41, 2405p, 78qp,13p²q, qp², 701p²

Solution:

Like terms are :

10pq, -7qp, 78qp; 7p, 2405p; 8q, -100q; -p²q²,12qy;- 23, 41; -5p², 701p²; 13p²q, qp²

Haryana Board Class 7 Maths Solutions For Chapter 10 Exercise-10.2

1. If m = 2, find the value of:

1) m-2

Solution: m-2

Putting m = 2 in m- 2

we get m-2=2-2=0

2) 3m -5

Solution: Putting m = 2 in 3m- 5

we get 3m-5 = 3 x 2-5

=6-5=1

3) 9 -5m

Solution: 9 -5m

Putting m = 2 in 9- 5m

we get 9- 5m = 9- (5 x 2) ,

= 9- 10 = -1

4) 3m² -2m -7

Solution:

Putting m = 2 in 3m²- 2m- 7 we get

3m²- 2m- 7 = 3(2)² -2(2) -7

=3 x 4-2 x 2-7

=12-4-7

=12-11 =1

5) \( \frac{5 m}{2}-4 \)

Solution:

Putting m = 2 in \( \frac{5 m}{2}-4 \)

we get

=5-4=1

2. If p- – 2 find the value of:

(1) 4p + 7

Solution:

Putting p = -2 in 4p + 7 we get

4p + 7 = 4(-2) + 7

=-8+7

=-1

2) – 3p² + 4p + 7 

Solution:

Putting p =- 2 in -3p² + 4p + 7 we get

– 3p² + 4p + 7 = – 3(-2)²+ 4 (-2) + 7

= (-3 x 4) + (-4 x 2) + 7

=-12 – 8 + 7

= -20 + 7 = -13

Simplifying Algebraic Expressions Class 7 HBSE Solved Examples

3) -2p³ – 3p² + 4p + 7

Solution:

Putting p = -2 in -2p³- 3p² + 4p + 7 we get

get

-2p³- 3p² + 4p + 7 =

-2 (-2)³- 3 (-2)² + 4 (-2) + 7

=-2 x (-8) -3 x 4-4 x 2 + 7

=16-12-8 + 7

= 23 – 20 = 3

Practice Problems Algebraic Expressions Class 7 Haryana Board with Substitution

3. Find the value of the following expressions, when x = -1:

1) 2x-7

Solution:

Putting x = -1 in 2x- 7 we get

2x- 7 = 2(-1) -7

= -2-7

= -9

2) -x + 2

Solution:

Putting x = -1 in -x + 2 we get

= x + 2 = – (-1) + 2

=1 + 2

= 3

3) x² + 2x +1

Solution:

Putting x = -1 in x² + 2x +1 we get

x²+ 2x +1 = (-1)² + 2 (-1) +1

=1-2+1 =2-2=0

4) 2x²- x- 2

Solution:

Putting x = -1 in 2x²- x- 2 we get

2x²- x- 2 = 2(-1)²- (-1)- 2

=2+1-2 =3-2=1

4. If a = 2 b = -2, find the values of:

1) a² + b²

Solution:

Putting a = 2;b = -2 in a² + b² we get a² + b² = (2)² + (-2)²

-4+4=8

2) a² + ab + b²

Solution: Putting a = 2; b = -2in a² + ab + b² we get

a²+ ab + b²

= (2)² + 2(-2) + (-2)²

=4-4+4

=8-4=4

3) a²- b²

Solution:

Putting a = 2; b = -2 in a²- b² we get

a²-b²=(2)²-(-2)²

=4-4=0

5. When a = 0, b = -1, find the value of the given expressions:

1) 2a + 2b

Solution:

Putting a =0; b = -1 in 2a + 2b we get

2a + 2b = 2(0) +2(-1)

= 0- 2 = -2

Addition and subtraction of algebraic expressions Class 7

2) 2a² + b²+1

Solution:

Putting a = 0; b= -1 in 2a²+ b² +1 we get

2a² + b² +1 = 2(0)² + (-1)² +1

=0+1 +1 =2

3) 2a²b + 2ab² + ab

Solution:

Putting a = 0;b = -1 in 2a²b + 2ab² + ab we get

2a²b + 2ab² + ab =

2(0)² (-1) + 2(0) (-1)² +0(-1)

0+0+0=0

4) a² + ab + 2

Solution:

Putting a = 0;b =-1 in a² + ab+ 2 we get

a² + ab + 2 = (0)² + 0 (-1) + 2

=0+0+2=2

6. Simplify the expressions and find the value if x is equal to 2

1) x + 7 + 4 (x- 5)

Solution:

x + 7 + 4 (x- 5)

= x + 7 + 4x-20

= x + 4x + 7- 20

= 5 x- 13

Putting x = 2 in 5x- 13 we get

5x- 13 = 5(2) -13

=10-13

=- 3

2) 3(x + 2) + 5x- 7

Solution:

3(x + 2) + 5x- 7

= 3x + 6 + 5x- 7

= 3x + 5x + 6 – 7

=8x-1

Putting x = 2 in 8x-1 we get

8x-1 = 8(2) -1

=16-1=15

3) 6x + 5 (x- 2)

Solution:

6x + 5(x- 2)

= 6x + 5x- 10

=11x -10

Putting x = 2 in 11x- 10 we get

11x- 10 = 11(2) – 10

= 22-10 =12

Important Concepts Algebraic Expressions Class 7 HBSE NCERT Based

4) 4(2x- 1) + 3x +11

Solution: 4(2x- 1) + 3x + 11

= 8x- 4 + 3x +11

= 8x + 3x +11 – 4

=11x + 7

Putting x = 2inllx + 7 we get

11x + 7 = 11(2) + 7= 22 + 7= 29

7. Simplify these expressions and find their values if x = 3, a = -1, b = -2.

1) 3x- 5 – x + 9 ‘

Solution: 3x-5-x + 9

=3x-x-5 + 9 = 2x + 4

Putting x = 3 in 2x + 4 we get

2x + 4 = 2(3) + 4

= 6 + 4 = 10

2) 2- 8x + 4x + 4

Solution: 2-8x + 4x + 4

= -8x + 4x + 4 + 2

= -4x + 6

Putting x = 3 in- 4x + 6 we get

-4x + 6 = -4 (3) + 6

= -12 + 6 = -6

3) 3a + 5 – 8a +1

Solution: 3a + 5- 8a +1

= 3a- 8a + 5 +1

= -5a + 6

Putting a = -1 in -5a + 6 we get

-5a + 6- -5 (-1) + 6

= 5 + 6 =11

4) 10- 3b – 4 – 5b

Solution: 10-3b-4-5b

= -3b- 5b + 10- 4

= -8b + 6

Putting b = -2 in – 8b + 6 we get

– 8b + 6 = -8 (-2) + 6

= 16 + 6 = 22

5) 2a -2b -4 -5 + a

Solution:

2a-2b-4-5 + a

= 2a + a -2b- 4- 5

= 3a- 2b- 9

Putting a = -1; b = -2 in 3a – 2b- 9 we get

3a -2b- 9 = (3 (-1) -2 (-2) -9)

= -3 + 4- 9 = – 12 + 4

= -8

8. (1) If z = 10 find the value of z³ – 3(z- 10)

Solution:

z³ – 3 (z- 10)

Putting z = 10 in z³- 3 (z- 10) we get

z³- 3(z- 10) = (10)3- 3(10- 10)

= 1000- 3 x 0

=1000- 0 = 1000

Substituting Values in Algebraic Expressions Class 7 Haryana Board Solutions Ex 10.3

2) If p = -10 find the value of p²-2p -100

Solution:

p²- 2p -100

Putting p = -10 in p²- 2p -100 we get

p²- 2p -100 = (-10)² – 2(-10) – 100

= 100 + 20-100

=120-100 = 20

9. What should be the value of ‘a’ if the value of 2x² + x – a equals to 5, when x = 0 ?

Solution:

2x²+ x-a = 5

Putting x = 0 in 2 x²+ x- a = 5 we get ,

2(0)² + 0-a = 5

-a = 5 1

a =- 5

10. Simplify the expression and find its value when a = 5 and b =- 3.

2(a² + ab) + 3 – ab

Solution:

2(a² + ab) + 3- ab = 2a²+ 2ab + 3- ab

= 2a² + 2ab- ab + 3

= 2a² + ab + 3

Putting a = 5 and b = -3 in 2a²+ ab + 3

2a² + ab + 3 = 2(5)² + (5) (-3) + 3

= 2×25-15 + 3

= 50-15 + 3

= 53-15 = 38

Haryana Board Class 7 Maths Solutions For Chapter 10 Very Short Answer Questions

1. Define

1. Monomial
2. Binomial
3. Trinomial

Solution:

1. An expression with only one term is called a monomial.
2. An expression which contain two unlike terms is called a binomial.
3. An expression which contain three terms is called a trinomial.

2. Define polynomial.

Solution: In general an expression with one or more terms is called a polynomial.

3. Give examples to each

1. Monomial
2. Binomial
3. Trinomial.

Solution:

1. Monomial: – 3x,- 5m
2. Binomial : -2x + y,z-3
3. Trinomial: -a+b+3,x+y+z

4. Find the value of the expression a³- b³ for a = 3;b = 2

Solution:

Substituting a =3;b = 2 in a³- b³ we get

(3)³- (2)³ = 3 x 3 x 3 – 2 x 2 x 2

= 27-8 =19

5. Simplify the expression 4 (2x-1)+3x +11 

Solution:

4 (2x-1)+ 3x + 11 =- 8

= 4 x 2x- 4 x1 + 3x + 11

= 8x-4 + 3x +11

= 8x + 3x + 11 – 4

=11x + 7

Important formulas for algebraic expressions Class 7

6. Write 3 algebraic expressions with 3 terms each.

Solution:

2x² + 3x + 5

px² + q x + r

ax² + bx + c

7. Find the value of the expression – 9x if x = -3.

Solution:

Given x = -3

=  -9x

= -9(-3)

= 27

8. Write the expression whose value is equal to -9 when x = -3.

Solution: -9

= -3×3

= (-3)3

= (x)3 [∵Given x = -3]

= 3x

The required expression is 3x.

Haryana Board Class 7 Maths Solutions For Chapter 10 Short Answer Questions

9. Identify the expressions given below as monomial, binomial, trinomial, and polynomial.

1) 5x² + y + 6

Solution:

In this expression three unlike terms.

So, it is Trinomial.

2) 3xy

Solution: In this expression, only one term

So, it is Monomial.

3) 5x²y + 6x 

Solution: In this expression, two unlike terms

So, it is Binomial.

4) a +4x-xy + xyz

Solution:

In this expression, more than one, unlike terms. So, it is polynomial.

10. Identify and write the like terms in each of following groups.

1) a², b², – 2a², c², 4a

Solution:

a², b², – 2a², c2, 4a are unlike terms

Here, a²,- 2a², are like terms

2) 3a, 4xy, – yz, 2zy •

Solution:

3a, 4xy, – yz, 2zy are unlike terms.

Here, – yz, 2zy are like terms.

3) -2xy², x²y, 5y²x, x²z

Solution:

–2xy², and 5y²x, are like terms.

4) 7p, 8pq, -5pq, -2p, 3p

Solution:

7p, 8pq, -5pq, -2p, 3p are unlike terms.

8pq, -5pq, are like terms

7p, -2p, 3p are like terms.

11. Find the value of the following monomials, if x =1.

Given x = 1.

1) -x 

Solution: Consider -x

= -(1)

= -1

2) 4x.

Solution: Consider 4x

= 4(1)

= 4

3) -2X²

Solution: Consider

-2x²

– -2(1)²

= -2(1)

= -2

12. Simplify and find the value of 4x + x- 2x² + x-1 when x = -1.

Solution:

Consider

4x + x – 2x² + x-1

=-2x²+ (4 +1 + 1)x-1

= -2x² + 6x- 1

But given x = -1

= -2(-1)² + 6(-1) -1

= -2(1) -6-1

= -9

13. Write the expression

5x² – 4 – 3x² + 6x + 8 + 5x – 13 in its simplified form. Find its value when x = -2.

Solution:

5x²- 4- 3x² + 6x + 8 + 5x- 13

= (5x²- 3x²) + (6x + 5x) + (8- 4- 13)

= (5 – 3)x² + (6 + 5)x + (8 – 17)

= 2x² +11x – 9

But given x = -2

= 2(-2)² + 11(-2) – 9

= 2(4) – 22- 9

= 8-22-9

= 8-31

= -23

Key Questions in Algebraic Expressions Ex 10.3 for Class 7 HBSE

14. If x = 1, y = 2 find the values of the following expressions.

Given x =1,y = 2

1) 4x-3y + 5

Solution:

4x- 3y + 5

= 4(1) -3(2) +5

= 4-6+5

=’ 9-6 = 3

2) x² + y²

Solution:

Consider

x² + y²

= (1)² +(2)²

= 1+ 4

= 5

3) xy + 3y-9

Solution: Consider

xy + 3y- 9

= (1) (2) +3(2) -9

= 2+6-9

= 8-9 = -1

15. Group the like terms together 12x, 12, 25x, – 25, 25y, 1, x, 12y, y, 25xy, 5x²y, 7xy²,2xy, 3xy², 4x²y.

Solution:

Group A                 12x, 25x, x

Group B                  25y,12y,y

Group C                  25,xy, 2xy

Group D                 5x²y, 4x²y

Group E                 7xy², 3xy²

Group F                 12,1,-25

Haryana Board Class 7 Maths Solutions For Chapter 10 Long Answer Questions

16. State true or false and give reasons for your answer.

1) 7x² and 2x are unlike terms.

Solution: It is True.

Both terms contain the same variable x.

However, their exponents are not same.

In the first term, the exponent of x is

2 and in the second term it is 1.

2) pq² and -4pq² are like terms

Solution: It is True.

Both terms contain the same variables p and q. However, the exponent of p is 1, and exponent of q is 2.

Multiplication of polynomials Class 7 Haryana Board

3) xy, -12x²y and 5xy² are like terms.

Solution: It is false.

The above terms contain the same variables x and y. However, their exponents are not the same.

In the first term, the exponent of x is 1, and second it is 1.

In the second term, the exponent of x is 2, and second it is 1.

In the third term, the exponent of x is ,1 and second it is 2.

17. State whether the algebraic expression given below is monomial, binomial, trinomial or polynomial.

1) y²

Solution: Monomial.

2) 4y-7z

Solution: Binomial

3) 1+x + x²

Solution: Trinomial

4) 7mn

Solution: Monomial

5) a² + b²

Solution: Binomial

6) 100 xyz

Solution: Monomial,

7) ax + 9

Solution: Binomial

8) p²- 3pq +r

Solution: Trinomial

9) 3y² – x²y² + 4x

Solution: Trinomial

10) 7x²-2xy + 9y²-11

Solution: Polynomial

Haryana Board Class 7 Maths Solutions For Chapter 10 Multiple Choice Answer Questions

Choose the correct answers:

1. In the expression 4x + 5 variable is

1. 4
2. 5
3. x
4. x+5

Answer: 3

2. The coefficient of y in 2y + 5 is

1. 2
2. 2y
3. 5
4. y + 5

Answer: 1

3. The nth term of the number pattern 11, 21, 31, 41. is

1. 10n
2. n + 10
3. 10 (n + 1)
4. 10n +1

Answer: 4

4. What is the coefficient of xin 6xy² + 7y is

1. 6
2. y2
3. 6y²
4. 7

Answer: 3

5. The value of 4z +1 for z = 2 is

1. 5
2. 1
3. 8
4. 9

Answer: 4

6. Which type of expression is 2x² + 3x +1 ?

1. monomial
2. binomial
3. trinomial
4. multinomial

Answer: 3

7. How many terms are there in this expression 4x² y⁴ z?

1. 1
2. 2
3. 3
4. 4

Answer: 1

8. What is the value of 3P² -5P + 6 at P =1?

1. 6
2. 4
3. 5
4. 7

Answer: 2

9. Find the value of 6b – 3a for a = 2, b = 1.

1. 3
2. 4
3. 0
4. 7

Answer: 3

10. Choose the correct matching.

1) \(\frac{2 m}{5} \) at m = 5                  (  ) a) 20

2) x² + 8x at x = 2                                                     (  ) b) 6

3) a² + b at a = 0, b =1                                             (  )(c) 1

4) pq at p = 2, q = 3                                                  (  ) d) 2

1. i – d,ii – a,iii – c, iv – b
2. i -b,ii – c,iii – a, iv – d
3. i – b,ii – a,iii – c, iv – d
4. i – a,ii – b,iii – d, iv-c

Answer: 1

11. What is the value of the expression 2x²y + xy² + xy at x = ( -1) and y = 2?

1. -4
2. -2
3. -6
4. -8

Answer: 2

12. Value of x²-y + 2 at x = 0, y = -1 is

1. -3
2. 2
3. 3
4. 1

Answer: 3

13. What is the coefficient of ‘P²’in 4P²y- 5P

1. 4
2. 4y
3. A or B
4. None

Answer: 2

14. What do we call the terms with same algebraic factors?

1. unlike terms
2. like terms
3. constants
4. variables

Answer: 2

15. a², b², c² are called

1. Like terms
2. Unlike terms
3. Numerical terms
4. None, of these

Answer: 2

16. Which of the following is a trinomial?

1. 2x
2. x²
3. 2a-3b + c
4. 2x + y

Answer: 3

17. If A = 2x- 4 then- 3A =

1. -6x +12
2. -4 + 2x
3. 8x – 12
4. 0

Answer: 1

18. If ‘n’ denotes the natural number, then formula for even number is

1. n +1
2. 2n +1
3. 2n
4. n-1

Answer: 3

19. “The cost of 5 pencils and 7 pens is 50” Expression algebraic form.

1. 5x + 7y = 90
2. 5x + 7y = 50
3. 5y + 7x = 50
4. 5y + 7x = 90

Answer: 2

20. Value of x²- 5y + 2 at x = 0, y = 3 is

1. -3
2.  2
3. -13
4. 1

Answer: 3

Haryana Board Class 7 Maths Solutions For Chapter 10 Fill in the blanks:

21…….. are formed from variables and constants

Answer: Algebraic expressions

22. Expressions are made up of……….

Answer: terms

23. A term is a ………..

Answer: product of factors

24. Terms which have the same algebraic factors are called………

Answer: like terms

25. The…..is the numerical or alphabetical factor of the term.

Answer: coefficient

26. Factors containing variables are said to be……….

Answer: algebraic factors

27. Tire value of 7x – 3 at x = 5 is…….

Answer: 32

28. The nth term of tire number pattern 6, 11,16, 21…..is………

Answer: 5n +1

29. Match the following:

1. The perimeter of the equilateral triangle whose side l is                                 (  ) A) 4l

2. The perimeter of a square whose side l is                                                         (  ) B) 2n 

3. The perimeter of a regular pentagon whose sidel is                                        (  ) 3) 3l

4. If a natural number is denoted by n, general form of a even number is        (  ) D) 2n+1

5. If a natural number is denoted by n, the general form of an odd number is (  ) E) 5l

Answer:

1. C 2. A 3. E 4. B 5. D

Haryana Board Class 9 Maths Solutions For Chapter 4 Linear Equations In Two Variables

• Equation: An equation is a statement in which one expression equals to another expression.
  Eg: 2x + 3 = 7y
  5m – 9 = 0
  3 + 6 = 9
  3\(x^2\)-5x+6=0
• Linear equation in one variable: An equation with only one variable of degree one is called linear equation in one variable. (or) An equation of the form ax + b = 0, where a, and b are real numbers such that a ≠ 0, is called a linear equation in one variable.
  Eg: 5x + 6 = 7; 3p = -7
• Standard form: ax + b = 0, where a and b ∈ R and a ≠ 0.
• Solution: A linear equation in one variable has a number that can satisfy the equation. This numbers are called the solution of the linear equation in one ” variable.
  • Linear equation has unique solution.

• Linear equation in two variables
  • An equation with two variables both of degree one is called linear equation in two variables. (or) An equation of the form ax + by + c = 0, where a, b and c are real numbers such that a ≠ 0 and b ≠ 0, is called a linear equation in two variables.
    Eg: 7a + 3b = 12
    2x = 3y – 5
  • Standard form: ax + by + c = 0, where a, b, c ∈ R and a, b ≠ 0.
• Solution of linear equation in two variables:
  A linear equation in two variables has a pair of numbers that can satisfy the equations. This pair of numbers is called the solution of the linear equation in two variables.
  • There are infinitely many solutions for a single linear equation in two variables.
  • The process of finding solution(s) is called solving an equation.
  • The solution of a linear equation is not affected when
    • the same number is added to (subtracted from) both sides of the equation.
    • both sides of the equation are mutiplied or divided by the same non-zero number.
• Graphical representation of linear equations:
  • Any linear equation in the standard form ax + by + c = 0 has a pair of solutions (x,y), that can be represented in the coordinate plane.
  • The graph of every linear equation in two variables (ax + by + c = 0) is a straight line.
  • Every point on the graph of a linear equation in two variables is a solution of the linear equation.
  • Every solution of the linear equation is a point on the graph of the linear equation.
  • The linear equation with constant value zero (in ax + by + c = 0, c = 0) passes through origin.
  • An equation of the type y = mx represents a straight line passing through the origin. Certain linear equations exist such that their solution is (0,0).
• Steps to draw the graph of linear equations in two variables:
  • Step 1: Let the given equation be ax + by + c = 0.
  • Step 2: Make the y as subject. i.e., y = \(-\left(\frac{a x+c}{b}\right)\)
  • Step 3: Take any 2 values (Most probably integrals) to x and calculate the values of y to obtain solutions (ordered pairs).
  • Step 4: Plot the ordered pairs on the graph paper on a suitable
  • Step 5: Draw the line passing through plotted points.
    Now the obtained line represents the equation: ax + by + c = 0.
    Note: We can take more than 2 values to x to get more solutions to check the correctness of the graph.
• Equations of the lines parallel to the coordinate axes:
  • The equation of the X-axis: y = 0.
  • The equation of the Y-axis: x = 0.
  • The equation of the straight line parallel to X-axis: y = k. It lies at k units from X-axis and passes through (0,k).

• • The equation of straight line parallel to Y-axis: x = k. It lies at k units from Y-axis and passes through (k, 0).

Haryana Board Class 9 Maths Chapter 4 Solutions

Haryana Board Class 9 Maths Solutions For Chapter 4 Linear Equations In Two Variables Exercise – 4.1

Question 1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.

Solution. Let the cost of a notebook = ₹ x

Let the cost of a pen = ₹ y

The cost of a notebook is twice the cost of a pen.

Cost of a notebook = 2 × cost of a pen

⇒ x = 2xy

⇒ x = 2y

⇒ x – 2y = 0

∴ Required equation

Question 2. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:

Solution. (1) 2x + 3y = 9.35

Given equation: 2x + 3y = 9.35

⇒ 2x + 3y – 9.35 = 0

⇒ 2x + 3y + (-9.35) = 0

On comparing with ax + by + c = 0,

We get, a = 2,b = 3,c = -9.35

(2) Given equation: x – \(\frac{y}{5}\) – 10 = 0

⇒ \(x+\left(-\frac{1}{5}\right) y+(-10)=0\)

On comparing with ax + by + c = 0,

We get, a = 1, b = [latex-\frac{1}{5}[/latex], c = -10

(3) Given equation: -2x + 3y = 6

⇒ -2x + 3y – 6 = 0

On comparing with ax + by + c = 0,

We get, a = -2, b = 3, c = -6

(4) Given equation: x = 3y

⇒ x – 3y = 0

⇒ x + (-3)y + 0 = 0

On Comparing with ax + by + c = 0,

We get, a = 1, b = -3, c = 0

(5) Given equation: 2x = -5y

⇒ 2x + 5y = 0

⇒ 2x + 5y + 0 = 0

On comparing with ax + by + c = 0,

We get, a = 2, b = 5, c = 2.

(6) Given equation: x = 3y

⇒ x – 3y = 0

⇒ x + (-3)y + 0 = 0

On comparing with ax + by + c = 0,

We get, a = 1, b = -3, c = 0.

(5) Given equation : 2x = -5y

⇒ 2x + 5y = 0

⇒ 2x + 5y + 0 = 0

On comparing with ax + by + c = 0,

We get, a = 2, b = 5, c = 0.

(6) Given equation: 3x + 2 = 0

⇒ 3x + (0)y + 2 = 0

On comparing with ax + by + c = 0,

We get, a = 3, b = 0, c = 2.

(7) Given equation: y – 2 = 0

⇒ 0(x) – 1(y) – 2 = 0

On comparing with ax + by + c = 0,

We get, a = 0,b = 1, c = -2.

(8) Given equation: 5 = 2x

⇒ 2x – 5 = 0

⇒ 2x + (0)y + (-5) = 0

On comparing with ax + by + c = 0,

We get, a = 2, b = 0, c = -5.

Class 9 Maths Chapter 4 Important Questions Haryana Board

Haryana Board Class 9 Maths Solutions For Chapter 4 Linear Equations In Two Variables Exercise – 4.2

Question 1. Which one of the following options is true, and why?

(1) A unique solution

(2) Only two solutions

(3) Infinitely many solutions

Solution. A linear equation in two variables has infinitely many solutions.

So, y = 3x + 5 has infinitely many solutions. So, option (3) is correct.

Question 2. Write four solutions for each of the following equations:

(1) 2x + y = 7

Solution. Given equation: 2x + y = 7

⇒ y = 7 – 2x

(a) Let x = 0 ⇒ y = 7 – 2(0) = 7 – 0 = 7.

Here solution = (0,7).

(b) Let x = 1 ⇒ y = 7 – 2(1) = 7 – 2 = 5.

Here solution = (1,5)

(c) Let x = -1 ⇒ y = 7 – 2(-1) = 7 + 2 = 9.

Here solution = (-1,9).

(d) Let x = 2 ⇒ y = 7 – 2(2) = 7 – 4 = 3.

Here solution = (2, 3).

∴ The solutions are (0,7), (1, 5), (-1, 9), (2, 3).

(2) πx + y = 9

Solution. Given equation: πx + y = 9

⇒ y = 9 – πx

(a) Let x = 0 ⇒ y = 9 – π(0) = 9 – 0 = 9.

Here solution = (0,9)

(b) Let x = 1 ⇒ y = 9 – π(1) = 9 – π

Here solution = (1, 9 – π)

(c) Let x = -1 ⇒ y = 9 – π(-1) = 9 + π

Here solution = (-1, 9 + π)

(d) Let x = 2 ⇒ y = 9 – π(2) = 9 – 2π

Here solution = (2, 9 – 2π)

∴ The solutions are (0,9), (1, 9 – π), (-1, 9 + π), (2, 9 – 2π)

(3) x = 4y

Solution. Given equation: x = 4y

⇒ y = \(\frac{x}{4}\)

(a) Let x = 0 ⇒ y = \(\frac{0}{4}\) = 0

Here solution = (0,0).

(b) Let x = 1 ⇒ y = \(\frac{1}{4}\)

Here solution = (1,7).

(c) Let x = 4 ⇒ y = \(\frac{4}{4}\) = 1

Here solution (4, 1).

(d) Let x = 2 ⇒ y = \(\frac{2}{4}\) = \(\frac{1}{2}\)

Here solution = (2, \(\frac{1}{2}\))

∴ The solutions are (0,0), (1, \(\frac{1}{4}\)), (4,1), (2,\(\frac{1}{2}\)).

Step-by-step Solutions for Class 9 Maths Chapter 4 Haryana Board

Question 3. Check which of the following are solutions of the equation x – 2y = 4 and which are not:

(1)(0, 2)

(2) (2,0)

(3) (4,0)

(4) (√2,4√2)

(5) (1,1)

Given equation: x – 2y = 4

Solution. (1) (0,2)

Substitute x = 0 and y = 2

LHS = x – 2y = 0 – 2(2) = 0 – 4 = -4 ≠ RHS

∴ (0, 2) is not a solution of the given equation.

(2) (2, 0)

Substitute x = 2 and y = 0

LHS = x – 2y = 2 – 2(0) = 2 – 0 = 2 ≠ RHS

∴ (2, 0) is not a solution of the given equation.

(3) (4,0)

Substitute x = 4 and y = 0

LHS = x – 2y = 4 – 2(0) = 4 – 0 = 4 = RHS

∴ (4,0) is a solution of the given equation.

(4) (√2,4√2)

Substitute x = √2 and y = 4√2

LHS = x – 2y = √2 – 2(4√2)

= √2 – 8√2 = -7√2 ≠ RHS

∴ (√2, 4√2) is not a solution of the given equation.

(5) (1,1)

Substitute x = 1 and y = 1

LHS = x – 2y = 1 – 2(1) = 1 – 2 = -1 ≠ RHS

∴ (1, 1) is not a solution of the given equation.

Question 4. Find the value of K, if x = 2, y = 1 is a solution of the equation 2x + 3y = K.

Solution. Given equation 2x + 3y = K.

x = 2, y = 1 is a solution.

on substituting x = 2 and y = 1.

⇒ 2(2) + 3(1) = K

4 + 3 = K

7 = K

∴ K = 7

Haryana Board Class 9 Maths Solutions For Chapter 4 Linear Equations In Two Variables Very Short Answer Type Questions

Question 1. Define linear equation in two variables.

Solution. An equation of the form ax + by + c = 0, where a, b and c are real numbers such that a ≠ 0 and b ≠ 0, is called a linear equation in two variables.

Question 2. Panth and Pandya scored 125 runs together. Express the information in the form of an equation.

Solution. Let the runs scored by Panth = x

Let the runs scored by Pandya = y

Total runs = 125

⇒ x + y = 125

Graphical Method of Solving Linear Equations Class 9 Haryana Board

Question 3. Express each of the equation y = 3 in the form of ax + by + c = 0 and write the values of a, b and c.

Solution. Given equation : y = 3

⇒ 0.x + y + (-3) = 0

On comparing with ax + by + c = 0,

We get, a = 0, b = 1, c = -3.

Question 4. Express each of the equation x – 5 = √3y in the form of ax + by + c = 0 and write the values of a, b and c.

Solution. Given equation: x – 5 = √3y

⇒ x – √3y – 5 = 0

⇒ x +(-√3)y + (-5) = 0

On comparing with ax + by + c = 0,

We get, a = 1, b = -√3,c = -5.

Question 5. How many solutions does a linear equation in two variables have?

Solution. A linear equation in two variables has infinitely many solutions.

Question 6. How many linear equations in two variables exist for which (2, 4) is a solution?

Solution. There are infinitely many linear equations in two variables exist for which (2, 4) is a solution.

Example: x + y = 6; x – y = -2; y = 2x.

Question 7. Write any two linear equations in two variables whose solution is (6,2).

Solution. Given solution: (6,2)

Sum of x-coordinate and y coordinate = 8 ⇒ x + y = 8

Difference of x-coordinate and y coordinate ⇒ 4x – y = 4

Question 8. Check whether (2, -5) is a solution of equation 2x + 5y = 2 or not.

Solution. Given equation : 2x + 5y = 2

Substitute x = 2 and y = -5

LHS = 2x + 5y = 2(2) + 5(-5)

= 4 – 25 = -21 ≠ RHS

∴ (2,-5) is not a solution of the given equation.

Question 9. Check whether (5, 0) is a solution of equation x + 3y = 5 or not.

Solution. Given equation: x + 3y = 5

Substitute x = 5 and y = 0

LHS = x + 3y = 5 + 3(0)

= 5 + 0 = 5 = RHS

∴ (5, 0) is a solution of the given

Question 10. Find 2 different solutions of x + y = 9

Solution. Given equation: x + y = 9

⇒ y = 9 – x

(a) Let x = 0 ⇒ y = 9 – 0 = 9.

Here solution = (0,9).

(b) Let x = 2 ⇒ y = 9 – 2 = 7.

Here solution = (2,7).

Question 11. Find two different solutions of 4x + y = 3.

Solution. Given equation: 4x + y = 3

⇒ y = 3 – 4x

(a) Let x = 0 ⇒ y = 3 – 4(0) = 3-0 = 3.

Here solution = (0,3).

(b) Let x = 2 ⇒ y = 3 – 4(2) = 3 – 8 = -5

Here solution = (2,-5).

Question 12. Which type of graph of a linear equation ax + by + c = 0 (Here a, b and c ∈ R & a ≠ 0, b ≠ 0) represents?

Solution. The graph of every linear equation in two variables (ax + by + c = 0) is a straight line.

Question 13. If (2, 0) is a solution of the linear equation 5x – 4y = k, then find the value of k.

Solution. Given equation: 5x – 4y = k

(2, 0) is a solution.

On substituting, x = 2 and y = 0.

⇒ 5(2) – 4(0) = k

⇒ 10 – 0 = k

⇒ 10 = k

∴ k = 10

Question 14. Write the equations of coordinate axes x and y.

Solution. The equation of the X-axis: x = 0

The equation of the Y-axis: y = 0

Question 15. Linear equation x – 2 = 0 is parallel to which axis?

Solution. Given equation: x – 2 = 0 ⇒ x = 2.

It is in the form of x = k.

∴ It is parallel to Y-axis.

Question 16. Write the equation of the line parallel to y-axis and passing through the point (-7,3).

Solution. The equation of the straight line parallel to Y-axis: x = k

Required equation: x = -7

⇒ x + 7 = 0

Question 17. Write the equation of the line parallel to X-axis and passing through the point (-2,-4)

Solution. The equation of the straight line parallel to X-axis: y = k

∴ Required equation: y = -4

⇒ y + 4 = 0

Question 18. Write the equation of three lines that are parallel to X-axis

Solution. The equation of three lines that are parallel to X-axis:

(1) y = 2

(2) y + 9 = 0

(3) 2y = 3

Question 19. Write the equation of three lines that are parallel to Y-axis

Solution. The equation of three lines that are parallel to X-axis:

(1) x = -2

(2)x – 9 = 0

(3) 5x = -3

Question 20. Find the distance between the graph of x – 5 = 0 and the Y-axis.

Solution. Given equation: x – 5 = 0 ⇒ x = 5

The graph of x = k is a straight line parallel to Y-axis. It lies at k units from Y-axis.

∴ The distance between the graph of x – 5 = 0 and the Y-axis is 5 units.

Question 21. Find the distance between the graph of 2y – 5 = 0 and the X-axis.

Solution. Given equation: 2y – 5 = 0 ⇒ x = 2.5

The graph of y k is a straight line parallel to X-axis. It lies at k units from X-axis.

∴ The distance between the graph of 2y – 5 = 0 and the Y-axis is 2.5 units.

Haryana Board Class 9 Maths Solutions For Chapter 4 Linear Equations In Two Variables Short Answer Type Questions

Question 22. Bhargavi got 10 more marks than double of the marks of Sindhu. Express the information in the form of an equation.

Solution. Let marks got by Bhargavi = x

Let marks got by Sindhu = y

Given that Bhargavi got 10 more marks than double of the marks Sindhu

∴ x = 10 + 2y

⇒ x – 2y – 10 = 0

Question 23. A number is 27 more than the number obtained by reversing its digits. If its units and tens digits are x and y respectively, write the linear equation representing the above statement.

Solution. Let the digit in units place = x

Let the digit in tens place = y

The number = 10y + x

If we reverse the digits, then the new number = 10x + y

From problem,

(Two-digit number) – (number formed by reversing the digits) = 27.

i.e., 10y + x – (10x + y) = 27

⇒ 10y + x – 10x – y – 27 = 0

⇒ 9y – 9x – 27 = 0

⇒ y – x – 3 = 0

⇒ x – y + 3 = 0

Haryana Board 9th Class Maths Chapter 4 Exercise Solutions

Question 24. The cost of a ball pen is 5 less than half the cost of a fountain pen. Write a linear equation in two variables to rep- resent this statement.

Solution. Let the cost of fountain pen = ₹ x

Let the cost of ball pen = ₹ y

Given that the cost of a ball pen is 5 less than half the cost of a fountain pen.

⇒ y = \(\frac{x}{2}\) – 5

⇒ 2y = x – 10

⇒ x – 2y = 10

Question 25. If x = 2k + 1 and y = k is a solution of the equation 5x + 3y – 7 = 0, find the value of k.

Solution. Given equation: 5x + 3y – 7 = 0

x = 2k + 1 and y = k is a solution of the equation.

On substituting, x = 2k + 1 and y = k.

⇒ 5(2k+1) – 3(k) – 7 = 0

⇒ 10k + 5 – 3k – 7 = 0

⇒ 7k – 2 = 0

⇒ 7k = 2

⇒ k = \(\frac{2}{7}\)

Question 26. Find 4 different solutions of 5x + y = 3

Solution. Given equation: 5x + y = 3

⇒ y = 3 – 5x

(a) Let x = 0 ⇒ y = 3 – 5(0) = 3 – 0 = 3.

Here solution = (0,3).

(b) Let x = 1 ⇒ y = 3 – 5(1) = 3 – 5 = -2.

Here solution = (1,-2).

(c) Let x = -1 ⇒ y = 3 – 5(-1) = 3 + 5 = 8.

Here solution = (-1, 8).

(d) Let x = 2 ⇒ y = 3 – 5(2) = 3 – 10 = -7

Here solution (2, -7).

(e) ∴ The solutions are

(0,3), (1, -2), (-1, 8), (2, -7).

Question 27. At which point the graph of the linear equation 2x – 3y = 6 cuts the Y-axis.

Solution. Given equation: 2x – 3y = 6

The x-coordinate of any point on the y-axis is zero.

Let the point of the line cuts the y-axis is (0, a).

On substituting, x = 0 and y = a.

⇒ 2(0) – 3(a) = 6

⇒ 0 – 3a = 6

⇒ -3a = 6

⇒ a = \(-\frac{6}{3}\)

⇒ a = -2

∴ The required point = (0, -2).

Haryana Board Class 9 Maths Solutions For Chapter 4 Linear Equations In Two Variables Long Answer Type Questions

Question 28. If (0, a) and (b, 0) are the solutions of the following linear equation 2x – 3y = 6. Find ‘a’ and ‘b’.

Solution. Given equation: 2x – 3y = 6.

(0, a) is one of the solutions of equation.

On substituting, x=0 and y = a.

⇒ 2(0) – 3(a) = 6

⇒ 0 – 3n = 6

⇒ -3a = 6

⇒ a = \(-\frac{6}{3}\)

⇒ a = -2

(b, 0) is another solution of equation.

On substituting, x = b and y = 0.

⇒ 2(b) – 3(0) = 6

⇒ 2b – 0 = 6

⇒ 2b = 6

⇒ b = \(\frac{6}{2}\)

⇒ b = 3

∴ a = -2 and b = 3

Haryana Board Class 9 Maths Solutions For Chapter 4 Linear Equations In Two Variables Objective Type Questions

Multiple Choice Questions:

Question 1. Which of the following is an equation?

1. 2x + 3 ≠ 4
2. 3x + y – 3
3. 7x – 7
4. 3m²+ m = 1

Answer. 4. 3m² + m = 1

Question 2. General form of linear equation in two variables is where a, b, c ∈ R and a and b ≠ 0.

1. ax + b = c
2. ax + by = cz
3. ax + by + c = 0
4. ay + bx = c

Answer. 3. ax + by + c = 0

Question 3. Sankar and Sanvi collected 100 together. Suitable equation for this information is

1. x + y + 100 = 0
2. x + y = 100
3. x – y = 100
4. x – y = -100

Answer. 2. x + y = 100

MCQ Questions on Linear Equations in Two Variables Class 9 Haryana Board

Question 4. On comparing 4x – y = 0 with ax + by + c = 0, we get c =

1. 4
2. -1
3. 0
4. -4

Answer. 3. 0

Question 5. On comparing -y = 0 with ax + by + c = 0, we get a + b + c =.

1. -1
2. 0
3. -2
4. 3

Answer. 1. -1

Question 6. The linear equation 2x+5 has _______ solution(s).

1. Unique
2. two
3. No
4. Infinitely many

Answer. 1. Unique

Question 7. The linear equation 2x + y = 5 has _______ solution(s).

1. Unique
2. two
3. No
4. Infinitely many

Answer. 4. Infinitely many

Question 8. The linear equation x + y = 5 has _______ natural solution(s).

1. Unique
2. two
3. No
4. Infinitely many

Answer. 1. Unique

Question 9. The solution of equation x – 2y = 4 are

(1)(0-2)

(2) (8.0)

(3)(6, 1)

(4) (-2,3)

1. (1) only
2. (1) & (3) only
3. (2) & (4) only
4. (4) only

Answer. 2. (1) & (3) only

Question 10. Number of linear equations in x and y can be satisfied by x=1 and y = 2 is

1. one
2. two
3. zero
4. Infinitely many

Answer. 4. Infinitely many

Question 11. If (2, 0) is a solution of the linear equation 2x + 3y = k, then the value of k is:

1. 4
2. -4
3. 3
4. 5

Answer. 1. 4

Question 12. x = 9, y = 4 is a solution of the linear equation

1. x + y + 13 = 0
2. x – y = 5
3. x – y + 5 = 0
4. x + y = -5

Answer. 2. x – y = 5

Question 13. (3,1) is the solution of the equation

1. 2x – y = 7
2. x – 2y = -7
3. 2x + y = 7
4. x – 2y = 7

Answer. 3. 2x + y = 7

Question 14. (2,-1) is the solution of the equation

(1) 3x + y = 5

(2) 2x – y = 5

(3) x + y + 1 = 0

(4) x + 2y = 1

1. (1) & (2) only
2. (1) & (3) only
3. (3) & (4) only
4. (4) only

Answer. 1. (1) & (2) only

Question 15. If x = -2 and y = 3 is the solution of equation x + 2y = k, then k=

1. 4
2. -4
3. 3
4. 5

Answer. 1. 4

Question 16. The geometrical representation of linear equation in two variables is

1. Straight line
2. Circle
3. Parabola
4. Triangle

Answer. 1. Straight line

Class 9 Maths Chapter 4 Theorems and Formulas Haryana Board

Question 17. For the equation 5x – 7y = 35, if y = 5, then the value of ‘x’ is

1. -14
2. 0
3. 14
4. 10

Answer. 3. 14

Question 18. The straight line passing through the points (0,0), (-1, 1) and (1, -1) has the equation

1. x – y = 0
2. x + y = 0
3. x + y = -1
4. x – y = 1

Answer. 2. x + y = 0

Question 19. Any point of the form (a, a) always lies on the graph of the equation

1. x – y = 0
2. x + y = 0
3. x + y = -1
4. x – y = 1

Answer. 1. x – y = 0

Question 20. Which of the following is not a solution of 2x – y + 3 = 0?

1. (3,9)
2. (0,3)
3. (-1,1)
4. (-1,-2)

Answer. 4. (-1,-2)

Question 21. The graph of the equation 2x + 3y = 6 cuts the X-axis at the point

1. (3,0)
2. (0,-3)
3. (-2,0)
4. (0,2)

Answer. 1. (3,0)

Question 22. The graph of linear equation x + 2y = 2, cuts the Y-axis at the point

1. (1,0)
2. (0,-3)
3. (-2,0)
4. (0,1)

Answer. 4. (0,1)

Question 23. Which of the following is true?

1. The line y = 2 parallel to Y-axis.
2. The line y – 3 = 0 parallel to X-axis.
3. The line y = 2 passes through (2,0)
4. The line y – 3 = 0 passes through (-3,0)

Answer. 2. The line y – 3 = 0 parallel to X-axis.

Question 24. Which of the following is not a solution of 3x – y = 6?

1. (0, -6)
2. (2, 0)
3. (-1,9)
4. (1,-3)

Answer. 3. (-1,9)

Question 25. The value of k if x = 2, y = 1 is a solution of equation 2x – k = -3y is

1. 7
2. 6
3. -6
4. -7

Answer. 1. 7

Question 26. Which of the following lines pass through origin?

(1) √3x + 3y = 0

(2) 4y = 3

(3) 5x = 2

(4) 4y = 3y

1. (1) & (2) only
2. (2) & (3) only
3. (3) & (4) only
4. (1) & (4) only

Answer. 4. (1) & (4) only

Question 27. The equation 2x+5y= 7 has a unique solution, if x, y are _______ numbers.

1. natural
2. integers
3. rational
4. real

Answer. 1. natural

Linear Equations in Two Variables Word Problems Class 9 Haryana Board

Question 28. A linear equation in two variables is of the form ax + by + c = 0 where

(1) a ≠ 0

(2) b ≠ 0

(3) c ≠ 0

(4) c = 0

1. (1) & (2) only
2. (2) & (3) only
3. (3) & (4) only
4. (4) only

Answer. 1. (1) & (2) only

Question 29. A linear equation in two variables is of the form ax + by + c = 0 passes through origin if

(1) a = 0

(2) b ≠ 0

(3) c ≠ 0

(4) c = 0

1. (1) & (2) only
2. (2) & (3) only
3. (3) & (4) only
4. (4) only

Answer. 4. (4) only

Question 30. The number of linear equations in 2 variables passes through 2 distinct points is

1. one
2. two
3. zero
4. Infinitely many

Answer. 1. one

Question 31. The positive solutions of the equation ax + by + c = 0 always lie in

1. 1st Quadrant
2. 2nd Quadrant
3. 3rd Quadrant
4. 4th Quadrant

Answer. 1. 1st Quadrant

Question 32. If we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation

1. Changes
2. Remains the same
3. Changes in case of multiplication only
4. Changes in case of division only

Answer. 2. Remains the same

Question 33. Which of the following equation has graph parallel to Y-axis?

1. x = 3
2. y = 5
3. 2x + 3y = 0
4. 2x = 3y

Answer. 1. x = 3

Question 34. Which of the following equation has graph parallel to X-axis ?

1. 4x = 3
2. y + 6 = 0
3. 2x – y = 0
4. 2x + 4y = 8

Answer. 2. y + 6 = 0

Question 35. The equation of the line parallel to X-axis and passing through the point

1. y – 3 = 0
2. y + 3 = 0
3. y + 2 = 0
4. y – 2 = 0

Answer. 2. y + 3 = 0

36-40: Direction: There are two statements are given in each question. Select the options as the following.

1) Both statements are true

2) Statement A is true, statement B is false.

3) Both statements are false

4) Statement A is false, statement B is true.

Question 36. Statement A: The linear equation 2x – 5y = 7 has infinitely many solutions.

Statement B: Only one linear equation in x and y can be satisfied by x = 1 and y = 2.

Answer. 2. Statement A is true, statement B is false.

Question 37. Statement A: (1, -4) is one of the solution of equation x – 2y = 9.

Statement B: The equation x + y = 5 has only one pair of natural number solutions.

Answer. 1. Both statements are true

Question 38. Statement A: Equation of Y-axis is : y = 0.

Statement B: y = 2 line parallel to y axis.

Answer. 3. Both statements are false

Question 39. Statement A: The line 3x + 5y = 0 passes through origin.

Statement B: The y = x passes through (4,4).

Answer. 1. Both statements are true

Question 40. Statement A: The line parallel to the Y-axis at a distance 4 units to the left of Y-axis is x = -4.

Statement B: The equation of X-axis is of the form y = 0.

Answer. 4. Statement A is false, statement B is true.

41-50: Assertion and Reasoning questions

Direction: In each of the following questions, a statement of Assertion is given followed by a corresponding statement of Reason just below it. Of the statements, mark the correct answer as

1) Both assertion and reason are true and reason is the correct explanation of assertion.

2) Both assertion and reason are true but reason is not the correct explanation of assertion.

3) Assertion is true but reason is false.

4) Assertion is false but reason is true.

Question 41. Assertion: There are infinite number of lines which passes through (-3,5)

Reason: A linear equation in two variables has infinitely many solutions.

Answer. 1. Both assertion and reason are true and reason is the correct explanation of assertion.

Question 42. Assertion: The graph of the equation 3x + y = 0 is a line passing through the origin.

Reason: An equation of the form ax + by + c = 0, where a, b, c R and a0, b0 is a linear equation in x and y.

Answer. 1. Both assertion and reason are ture and reason is the correct explanation of assertion.

Question 43. Assertion: The point (0, 3) lies on the graph of the linear equation 3x + 4y = 12.

Reason: (0, 3) satisfies the equation 3x + 4y = 12.

Answer. 2. Both assertion and reason are true but reason is not the correct explanation of assertion.

Question 44. Assertion The graph of the linear equation x – 2y = 1 passes through the point (-1, -1)

Reason: The linear equation x – 2y = 1 has unique solution.

Answer. 3. Assertion is true but reason is false.

Question 45. Assertion: The point (2, 2) lies on the line y = x

Reason: Any point on the line y = x is of the form (a, a)

Answer. 1. Both assertion and reason are true and reason is the correct explanation of assertion.

Question 46. Assertion: A linear equation 2x + 3y = 5 has a unique solution.

Reason: There are infinitely many solutions for a linear equation in two variables.

Answer. 4. Assertion is false but reason is true.

Question 47. Assertion: The graph of a line 3x – 4y + 12 = 0 intersects Y-axis at (0,3).

Reason: The line ax + by + c = 0 intesects X-axis at (\(-\frac{c}{a}\),0).

Answer. 2. Both assertion and reason are true but reason is not the correct explanation of assertion.

Question 48. Assertion: All the points (0, 0), (0,5), (0,3) and (0, 6) lie on the Y-axis.

Reason: Equation of the Y-axis is x = 0.

Answer. 1. Both assertion and reason are true and reason is the correct explanation of assertion.

Question 49. Assertion: The line y = 5x passes through origin..

Reason: The linear equation y = mx + c (c+0) passes through origin.

Answer. 3. Assertion is true but reason is false.

Question 50. Assertion: The geometric representation of x = -2 meets the X-axis at (0, -2).

Reason: The line y = k is parallel to X-axis and passes through the point (0, k).

Answer. 4. Assertion is false but reason is true.

Haryana Board Class 9 Maths Solutions For Chapter 4 Linear Equations In Two Variables Match the following:

Question 51. Match the following linear equations and points on those lines.

Answer. 1 – C, 2 – E, 3 – B, 4 – A, 5 – G.

Question 52. Match the following with suitable equations

Answer. 1 – G, 2 – C,3 – B, 4 – D, 5 – A.

Question 53. Match the following general form of points with their equations. (Here a ≠ 0)

Answer. 1 – C, 2 – D, 3 – A, 4 – E, 5 – F.

Haryana Board Class 9 Maths Solutions For Chapter 4 Linear Equations In Two VariablesTrue or False Questions

Question 54. Any equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero is called a linear equation in two variables

Answer. True

Question 55. 5x – 3 = 2x is a linear equation in one variable

Answer. True.

Question 56. 3x – y² = 5 is a linear equation in two variables

Answer. False

Question 57. ax² + by + c = 0 where a, b and c are real numbers, is a linear equation in two variables.

Answer. False

Question 58. There are infinitely many solutions for a given linear equation in two variables

Answer. True

Question 59. 3x + 5 = 0 has a unique solution

Answer. True

Question 60. The linear equation 2x + 5y = 7 has unique natural solutions.

Answer. True

Question 61. The linear equation 2x – 3y = 4 has unique solution.

Answer. False

Question 62. The point of the form (a,a) always lies on X-axis

Answer. False

Question 63. The graph of every linear equation in two variables need not be a line

Answer. False

Question 64. (0,2) is a solution for x – 2y = 4

Answer. False

Question 65. The point (0,-3) lies on the graph of the linear equation 3x + 4y = 12

Answer. True

Question 66. The graph of the linear equation x + 2y = 7 passes through the point (0,7)

Answer. False

Question 67. The graph of linear equation 2x – 3y = 0 is a line parallel to X-axis

Answer. False

Question 68. The line x – y = 4 intersects X-axis at (4,0)

Answer. True

Question 69. The line ax + by + c = 0 intersects Y-axis at (0, \(-\frac{c}{a}\))

Answer. False

Question 70. The line ax + by + c = 0 passes through origin if c = 0.

Answer. True

Question 71. (3, -3) lies on x + y = 3.

Answer. False

Question 72. The line passes through (4,7) and parallel to X-axis is y = 7.

Answer. True

Question 73. The line passes through (-3,7) and parallel to Y-axis is x – 3 = 0

Answer. False

Question 74. The line y = -3 intersects Y-axis at (0,-3)

Answer. True

Question 75. The line x + 2 = 0 intersects Y-axis at (0,-2)

Answer. False

Question 76. The line y = mx passes through origin.

Answer. True

Question 77. x = 0 is the equation of the X-axis.

Answer. False

Question 78. The linear equation 3x + 2 = 0 represents a line parallel to Y-axis.

Answer. True

Question 79. The graph of y = 6 is a line parallel to X-axis at a distance 6 units from the X-axis

Answer. True

Haryana Board Class 9 Maths Solutions For Chapter 4 Linear Equations In Two Variables Fill in the Blanks:

Question 80. A statement in which one expression equals to another expression is _______

Answer. equation

Question 81. An equation with only one variable of degree one is called as _______ equation in one variable.

Answer. linear

Question 82. General form of linear equation in one variable is _______

Answer. ax + b = 0

Question 83. Solution of linear equation ax + b = 0 is _______

Answer. –\(\frac{b}{a}\)

Question 84. An equation with two variables both of degree one is called as _______ equation in _______ variables.

Answer. linear, 2

Question 85. General form of linear equation in two variables is _______

Answer. ax + by + c = 0

Question 86. The number of solutions of linear equations in two variables is _______

Answer. infinite

Question 87. The graph of every linear equation in two variables (ax + bt + c = 0) is a _______

Answer. line

Question 88. If x = 1 and y = 1 is one of the solutions of x – y = k then k = _______

Answer. 0

Question 89. All the points (2,0), (-3,0), and (5,0) lie on the _______ axis.

Answer. x

Question 90. All the points (0,3), (0,0), (0,-4) and (0,7) lie on the _______ axis.

Answer. y

Question 91. Abscissa of all points on the Y-axis is _______

Answer. 0

Question 92. The negative solutions of the equation ax + by + c = 0 always lie in the _______ quadrant.

Answer. 3rd

Question 93. The positive solutions of the equation ax + by + c = 0 always lie in the _______ quadrant.

Answer. 1st

Question 94. If the point (3,4) lies on the graph of 3y = ax + 7, then the value of a = _______

Answer. \(\frac{5}{3}\)

Question 95. The line ax + by + c = 0 intersects X-axis at _______

Answer. (-\(\frac{c}{a}\), 0)

Question 96. The graph of the linear equation 2x + 3y = 6 is a line which meets the X-axis at _______

Answer. (3,0)

Question 97. The graph of the linear equation 3x – 4y – 12 = 0 is a line which meets the Y-axis at _______

Answer. (0,-3)

Question 98. The value of y if x = 2 in the linear equation 3x – 4y – 12 = 0 is _______

Answer. 3

Question 99. The value of x for which y = -4 is a solution of the linear equation 5x – 8y = 40 is _______

Answer. \(\frac{8}{5}\)

Question 100. An ordered pair that satisfy an equation in two variables is called its _______

Answer. solution

Question 101. If x = 1 and y = 0 is the solution of equation 2x + y = 3a, then the value of a _______

Answer. \(\frac{2}{3}\)

Question 102. If (3,-2) is a solution of the equation 3x – py – 7 = 0, then the value of p is _______

Answer. -1.

Question 103. If 7x – 3y = k passes through origin, then k = _______

Answer. 0

Question 104. The point of the form (a,a) always lies on _______

Answer. x = y

Question 105. The equation x = 7, in two variables, can be written as _______

Answer. 1.x + 0.y – 7 = 0

Question 106. If (a,1) lies on the graph of 3x – 2y + 4 = 0, then a = _______

Answer. –\(\frac{2}{3}\)

Question 107. The area of a triangle formed by coordinate axes and line x + y = 4 is _______

Answer. 8 sq. units

Question 108. The area of a rectangle formed by coordinate axes, line x = 2 and y = 6 is _______

Answer. 12 sq. units

Question 109. The line passes through (0,p) and parallel to X-axis is _______

Answer. y = p

Question 110. The linear equation such that each point on its graph has an ordinate 3 times its abscissa _______

Answer. y = 3x

Question 111. The line y = mx passes through _______

Answer. origin

Question 112. The equation of the X-axis is _______

Answer. y = 0

Question 113. The line parallel to the Y-axis at a distance 4 units to the left of Y-axis is given by the equation _______

Answer. x + 4 = 0

Question 114. The graph of y = 6 is a line parallel to _______ axis.

Answer. x

Question 115. The line x + 3 = 0 passes through _______ and _______ quadrants.

Answer. 2nd, 3rd

Question 116. The line passes through (-6,-5) and parallel to Y-axis is _______

Answer. x + 6 = 0

Question 117. The line passes through (-2,-3) and parallel to X-axis is _______

Answer. y + 3 = 0

Question 118. The line x + 2 = 0 intersects X-axis at _______

Answer. (-2,0)

Question 119. The line y = 2 intersects Y-axis at _______

Answer. (0,2)

Haryana Board Class 7 Maths Solutions For Chapter 9 Perimeter And Area

Key Concepts

1. Perimeter: Perimeter is the distance around a closed figure.
2. Area: Area is the region occupied by a closed figure.
3. Remember:
   1. Perimeter of a regular polygon = number of sides x length of one side
   2. Perimeter of square = 4 x side
   3. Perimeter of a rectangle = 2x(l+b)
   4. Area of a rectangle = l x b
   5. Area of a square = side x side
   6. Increase of perimeter does not necessarily imply that area also increases.

• A quadrilateral is a closed figure with four sides, four angles and four vertices.
• Quadrilateral ABCD is said to be a convex quadrilateral if all line segments joining points in the interior of the quadrilateral also tie in interior of the quadrilateral.
• Quadrilateral PQRS is said to be a concave quadrilateral if all line segments joining points in the interior.of the quadrilateral do not
  necessarily lie in the interior of the quadrilateral.
• Trapezium is a quadrilateral with one pair of parallel sides.
• The diagonals ofa parallelogram bisect each other.
• The diagonals of a rhombus are perpendicular bisectors of one another.
• Area of parallelogram = base x height
  Area of rectangle = length x breadth (length = base; breadth = height)
• If all the sides of a parallelogram are equal,it is called a Rhombus’.
• The area of a rhombus is equal to half of the product of its diagonals i.e., A = \( \frac{1}{2} d1:d2 \)
• The approximate value of the ratio of the circumference to the diameter of a circle is \( \frac{22}{7} \) or 3.14. It is a constant and is denoted by π (Pi).
• \( \frac{c}{d} \) = where ‘c’ is the circumference of the circle and ‘d’ is its diameter

Since, \( \frac{c}{d} \) =
= n, where ‘c’ is the circumference of the circle and W is its diameter.
= n, c= nd

Since, diameter of a circle is twice the radius i.e. d = 2r, c=π x 2r or c=2πr.

The area of a triangle is equal to half of the product of its base (b) and height (h) i.e. A =\( \frac{1}{2} \) bh

Solutions To Try These

Find the area of the following parallelograms:

1)

Solution:

Base = 8 cm

Height = 3.5 cm

Area of the parallelogram= base x height = 8 X 3.5 = 28 cm²

2)

Solution:

Base = 8 cm

Height = 2.5 cm

Area of the parallelogram=base X height = 8 X 2.5 = 20 cm²

Haryana Board Class 7 Maths Perimeter and Area Solutions

3) In a parallelogram ABCD, AB = 7.2 cm, and the perpendicular from C on AB is 4.5 cm.

Solution:

Base = 7.2 cm

Height = 4.5 cm

Area of the parallelogram = base X height = 7,2 X 4.5 = 32. 40 cm²

Solutions To Try These

1. Try the above activity with different types of triangles.

Solution:

Try yourself

2. Take different parallelograms. Divide each of the parallelograms into two triangles by cutting along any of its diagonals. Are the triangles congruent?

Solution:

Try yourself.

Hint: All the congruent triangles are equal in area but the triangles equal in area need not be congruent.

Haryana Board Class 7 Maths Solutions For Chapter 9Exercise-9.1 :

1. Find the area of each of the following parallelogram:

1)

Solution:

Base = 7 cm 7cm

Height = 4 cm

Area of parallelogram= base X height

= 7 x 4 = 28 cm²

2)

Solution:

Base = 5 cm

Height = 3 cm

Area of parallelogram = b x h

= 5×3 = 15 cm²

3)

Solution:

Base = 2.5 cm

Height = 3.5 cm

Area of parallelogram =b x h

= 2.5 x 3.5 = 8.75 cm²

Area of parallelogram =b x h

= 2.5 x 3.5 = 8.75 cm²

Class 7 Maths Chapter 9 Perimeter and Area Haryana Board

4)

Solution:

Base = 5 cm

Height = 4.8 cm

Area of parallelogram = b x h

= 5 X 4.8 = 24 cm²

HBSE Class 7 Rational Numbers Solutions Ex 9.1

5)

Solution:

Base = 2 cm

Height = 4.4 cm

Area of parallelogram = base x height

= 2×4.4 = 8.8 cm²

2. Find the area of each of the following triangles:

Solution:

1)

Base = 4 cm

Height = 3 cm

Area of triangle = \( \frac{1}{2} b x h \)x b x h

2)

Base = 5 cm

Height= 3.2 cm

Area of triangle = \( \frac{1}{2} \times b \times h \)

3)

Solution:

Base = 3 cm

Height =4 cm

Area of triangle =\( \frac{1}{2} \times b \times h \)

HBSE Class 7 Rational Numbers Solutions Ex 9.2

4)

Solution:

Base = 3 cm

Height = 2 cm

Area of triangle = \( \frac{1}{2} \times b \times h \)

Haryana Board 7th Class Maths Perimeter and Area Questions and Answers

3. Find the missing values:

Solution:

Area of parallelogram = bh

4. Find the missing values:

Solution:

Area of triangle = \( \frac{1}{2} \mathrm{bh} \)

5. PQRS is a parallelogram. QM is the height from Q to SR and QN is the height from Q to PS.If SR = 12 cm and QM = 7.6 cm.

Find:

1) The area of the parallelogram PQRS

2) QN,if PS = 8 cm

Solution:

1)

PQRS is a parallelogram.

Its base SR = 12 cm

Height QM = 7.6 cm

Area of the parallelogram PQRS

= base x height

=12×7.6 = 91.2 cm2

2) Base PS = 8 cm

Corresponding height QN =?

Area of parallelogram PQRS

= 91.2 cm²

= 8 X QN = 91.2

QN = \( \frac{91.2}{8}=11.4 \mathrm{~cm} \)

Key Questions in Rational Numbers Ex 9.1 for Class 7 HBSE

6. DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD. If the area of the parallelogram is 1470cm², AB = 35cm, and AD = 49 cm, find the length of BM and DL.

Solution:

ABCD is a parallelogram;

AD = 49 cm

Base AB = 35 cm

DL and BM are the heights.

Area of parallelogram ABCD =1470 cm²

AD X BM = 1470

49 X BM = 1470

The length of BM = 30 cm

Area of parallelogram ABCD = 1470 cm²

AB x DL = 1470

35 X DL = 1470

The length of DL 42 cm

Chapter 9 Perimeter and Area Class 7 Solutions in Hindi Haryana Board

7. AABC is right-angled at A. AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm, and AC = 12 cm. Find the Area of ΔABC. Also, find the length of AD.

Solution:

AABC is right-angled at A.

When we take base AC = 12 cm and height AB = 5 cm

Area of the triangle ABC

AD is perpendicular to BC.

When we take base BC = 13 cm

Area of the triangle ABC = 30 cm2

Its height AD =?

8. ΔABCis isosceles with AB=AC=7.5 cm and BC = 9 cm. The height AD from A to BC, is 6 cm. Find the area of AABC. What will be height from C to AB i.e CE?

Solution:

AABC is an isosceles triangle.

AB = AC = 7.5.cm

Base BC = 9 cm

Height AD = 6 cm

Area of triangle ABC

Area of the AABC = 27 cm²

When we take base AB = 7.5 cm

Its corresponding height CE =?

Area of triangle ΔABC

The height from C to AB is 7.2 cm

Practice Problems Rational Numbers HBSE Class 7

Solutions To Try These

From the figure,

Which square has the larger perimeter?

Solution: The outer square has the larger perimeter.

Which is larger, perimeter of smaller square or the circumference of the circle?

Solution:

The circumference of the circleis larger.

Solutions To Try These

Draw circles of different radii on a graph paper. Find the area by counting the number of squares. Also find the area by using formula. Compare the two answers.

Solution:

Try yourself with the help your teacher.

Haryana Board Class 7 Maths Solutions For Chapter 9 Exercise-9.2

1. Find the circumference of the circles with the following radius:

1) 14 cm

Solution:

Radius of the circle (r) = 14 cm

Circumference of the circle = 2Πr

= 88 cm

2) 28 mm

Solution:

Radius of the circle (r) = 28 mm

Circumference of the circle = 2Πr

Haryana Board Class 7 Maths Exercise 9.1 Solutions

3) 21 cm

Solution:

Radius of the circle (r) = 21 cm

Circumference of the circle = 2Πr

=176 mm

= 132 cm

2. Find the area of the following circles, given that :

1) radius = 14 mm

Solution:

Radius of the circle (r) = 14 mm

Area of the circle = Πr²

2)diameter = 49 m

Solution:

Diameter = 49 m

Radius = \( \frac{49}{2} \mathrm{~m} \) \( r=\frac{d}{2} \)

Area of the circle = Πr²

= 1886.5 m2

3) radius = 5 cm

Solution:

Radius ( r) 5 cm

Area of the circle = Πr²

3. if the circumference of a circular sheet is 154 m, find its radius. Also, find the area of the sheet.

Solution:

Circumference of the circular sheet 154 m

2Πr =154

Area of the circular sheet = Πr²

4. A gardener wants to fence a circular garden of diameter 21m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the cost of the rope, if it costs Rs. 4 per meter. \( \text { (Take } \pi=\frac{22}{7} \text { ) } \)

Solution:

Diameter of the circular garden = 21 m

Circumference of the garden = Πd

Length of the rope to make one round of fence = 66 m

Length of the rope to make 2 rounds of fence = 66 x 2 = 132 m

Cost of 1 m rope = Rs. 4

Cost of 132 m rope = Rs, 4 x 132= Rs. 528

HBSE 7th Class Rational Number Word Problems

5. From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet (Take π = 3.14)

Solution:

Radius of a circular sheet = 4 cm

Area of the outer circle = Πr³

=> 3.14 x 4 x 4 = 3.14 X 16 = 50.24 cm³

Radius of the inner circle =3 cm

Area of the inner circle = Πr²

= 3.14 x 3 x 3 = 28.26 cm³

Area of the remaining sheet

= Area of outer circle – Area of inner circle

= 50.24 – 28.26 = 21.98 cm²

Important Questions for Class 7 Maths Chapter 9 Haryana Board

6. Saima wants to put a lace on the edge of a circular tabic cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace costs Rs. 15. (Take π = 3.14)

Solution:

Diameter of the circular table cover = 1.5 m

Circumference of the table cover =πd

=3.14 x 1.5 = 4.71 m

Length of the lace required

Circumference of the circular table cover = 4.71 m

Cost of lm of lace = Rs. 15

Cost of 4.71 m of lace = Rs. 15 x 4.71 = Rs. 70.65

7. Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

Solution:

Diameter of the semicircle ‘d’= 10 cm

Circumference of the semicircle

Perimeter of semicircle

=Circumference of semicircle + diameter

= 15.7 + 10 = 25.7 cm

8. Find the cost of polishing a circular table – top of diameter 1.6 m, if the rate of polishing is Rs. 15/mJ. (Take Π = 3.14)

Solution:

Diameter of the circular table-top = 1.6m

Radius = \( \frac{1.6}{2} \) = 0.8m

Area of the table-top

= Πr²

= 3.14 x (0.8)²

= 3.14 x 0.8 x 0.8

= 3.14 x 0.64 = 20096 m²

Cost of l m² area = Rs. 15

Cost of 2.0096 m²area = Rs 15 x 2.0096

= Rs 30 .144

= Rs 30. 14 (approx)

9. Shazli took a wire of length 44 cm and bentit into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides ? Which figure encloses more area, the circle or the square? \( \text { (Take } \pi=\frac{22}{7} \text { ) } \)

Solution:

Length of the wire = 44 cm

Itis bent into the shape of a circle.

Circumference of the circle = 44 cm

27 πr = 44

Radius of that circle = 7 cm

Area of the circle = Πr²

The wire is bent into the shape of a square.

Perimeter of the square = Length of the wire 4 X side = 44

Side =\( \frac{44}{4} \) =11 cm

Area of the square = side x side = 11 X 11 = 121 cm²

The circle encloses more area than the square.

Important Concepts Rational Numbers Class 7 HBSE Chapter 9

10. From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm, and a rectangle of length 3 cm and breadth 1 cm are removed (as shown in the adjoining figure), find the area of the 22 remaining sheet.

Solution:

Radius of the circular card sheet=14 cm

Area of the circular card sheet

=πr²

=22 x 2 x 14 =616 cm²

Area of two circles of radius = 3.5 cms.

= 2 X 38.5 = 77 cm²

Length of the rectangle = 3 cm

Breadth =1 cm

Area of this rectangle = l x b

=3 x 1=3 cm²

Area of the sheet = Area of the circular card sheet- Area of the two circles Area of the rectangle.

= 616-77-3

= 616-80

= 536 cm²

Step-by-Step Solutions for Perimeter and Area Class 7 Haryana Board

11. A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the leftover aluminium sheet? (Take π= 3.14)

Solution: Side of the square piece of an aluminium sheet = 6 cm

Area of this sheet = side x side = 6 x 6 = 36 cm²

Radius of the circle = 2 cm

Area of the circle

= Πr²

= 3.14 x(2)² = 3.14x2x2

= 3.14×4 = 12.56 cm²

Area of the aluminium sheet left over

= Area of the square- Area of the circle

= 36 -12.56 = 23.44 cm²

12. The circumference of a circleis 31.4 cm. Find the radius and area of the circle? (Take Π = 3.14)

Solution:

Circumference of a circle = 31.4 cm

2Πr= 31.4

2 X3.14 x r = 31.4

Area of the circle = Πr²

= 3.14 X (5)²

= 3.14x5x5

= 3.14 X 25 = 78.5 cm²

13. A circular flowerbed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (Take π= 3.14)

Solution:

Diameter of the flower bed = 66 m

Radius of the flower bed = \( \frac{66}{2}=33 \mathrm{~m} \)

Area of the flower bed

= πr²

= 3.14 x (33)²

= 3.14 X 33 X 33

= 3.14 X 1089 = 3419.46 m²

Width of the path = 4 m

Radius of the flower bed with path = 33 + 4 = 37 m

Area of the flower bed with path

= Π²

= 3.14 x(37)²

= 3.14x37x37

= 3.14×1369 = 4298.66 m²

Area of the path = Area of flowerbed with path- Area of the flower bed = 4298.66- 3419. 46 = 879.20 m²

14. A circular flower garden has an area of 314 m2. A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden? (Take Π – 3.14)

Solution:

Area of the circular flower garden = 314 m²

Πr²= 314

3.14 xr² = 314

r² = 100

r = 100 = 10×10 = 10 m

Given that the sprinkler can cover the area that has a radius 12m.

12 m > 10 m

The sprinkler will water the entire garden.

HBSE Class 7 Maths Chapter 9 Guide Rational Numbers

5. Find the circumference of the inner and outer circles, shown in the adjoining figure. (Take π= 3.14)

Solution:

Radius of the outer circle = 19m

Circumference of the outer circle

= 2Πr

= 2×3.14×19

= 38×3.14 = 119.32 m

Radius of the inner circle =19m-10m = 9m

Circumference of the inner circle

= 2Πr

= 2×3.14×9

= 18X3.14 = 56.52 m

16. How many times a wheel of radius 28 cm must rotate to go 352 m?

Solution:

Radius of the wheel = 28 cm

Circumference of the wheel = 27Πr

Distance covered by the wheel in one rotation = Circumference of the wheel 1 m = 100 cm

352 m = 352 x 100 = 35200 cm

= 2 X 22 x 4 = 176 cm

Number of times the wheel must rotate to go 352 m.

17. The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in1 horn? (Take Π = 3.14)

Solution:

Length of the minute hand of a circular clock =15 cm

Radius of the circular clock ‘r’=15 cm

Circumference of the circle = 27Πr

=2X3.14X15 = 94.2 cm

In one hour i.e., in 60 minutes; the minute hand of the clock completes 1 rotation.

The tip of the minute hand moves 94.2 cm in 1 hour.

Haryana Board Class 7 Maths Solutions For Chapter 9 Very Short Answer Questions

1. Define ‘Area’.

Solution:

The amount of surface enclosed by a closed figure is called its ‘area’.

2. Define ‘Perimeter’.

Solution:

Perimeter is the distance around a closed figure.

3. The area of a rectangular sheet is 500 cm². If the length of the sheet is 25 cm. What is its width?

Solution:

Length of the rectangular sheet (Z)=25 cm

Area of the rectangular sheet = 500 cm²

l x b = 500 => 25 x b = 500

4. The perimeter of a rectangle is 180 cm. If the breadth of the rectangle is 40 cm. Find its length.

Solution:

The breadth of a rectangle = 40 cm

Perimeter of the rectangle = 180 cm

2 ( l + b ) = 180

2 ( l + 40) = 180

l = 90 – 40 = 50 cm

5. Find the area of the parallelogram whose base is 10 cm and the height is 4 cm.

Solution:

Base of the parallelogram = 10 cm

Height = 4 cm

Area of the parallelogram = Base x Height = 10 X 4 = 40 cm²

Rational Numbers on the Number Line Class 7 Haryana Board

6. Find the area of the triangle whose base is 6 cm and height 3 cm:

Solution:

Area of triangle

7. What is the circumference of a circular 22 disc of radius 14 cm?\( \)

\( \text { (Take } \pi=\frac{22}{7} \text { ) } \)

Solution:

Radius of circular disc = 14 cm

Circumference of disc = 2Πr

8. Diameter of a circular garden is 9.8 m. Find its area.

Solution:

Diameter d = 9.8 m

Radius (r) = \(\frac{9.8}{2} \) = 4.9 m

Area of the cirle = Πr²

9. Find the base of a triangle whose area is 220 cm2 and height is 11 cm.

Solution:

Given area of triangle = 220 cm²

10. Find the circumference of a circle whose radius is (1) 35 cm (2) 4.2 cm (3) 15.4 cm

Solution:

Circumference of a circle = 2Πr

1) r = 35 cm; circumference

= 2 x \( \frac{22}{7} \) x 35 cm = 220 cm

2) r = 4.2 cm; circumference

= 2 x \( \frac{22}{7} \) x 4.2 = 26.4 cm

3) r = 15.4 cm; circumference

= 2 x \( \frac{22}{7} \) x 15.4 = 26.4 cm

11. If the circumference of a circle is 264 cm, find its radius. \( \text { Take } \pi=\frac{22}{7} \) .

Solution:

Circumference of a circle = 2rcr = 264cm

Given

r = \( \frac{264 x 7}{2 x 22} \) = 42 cm

12. If the circumference of a circle is 33 cm, find its diameter.

Solution:

Given

Circumference of a circle = Πd = 33 cm

i.e, \( \frac{22}{7} \times d=33 \)

d = \( \frac{33×7}{22} [latex] = [latex] \frac{21}{2} \) = 10.5 cm

Haryana Board Class 7 Maths Solutions For Chapter 9 Short Answer Questions

13. Find the area of each of the following triangles.

1)

Solution:

1) Area of triangle = \( \frac{1}{2} \mathrm{bh} =

= [latex] \frac{1}{2} \) x 5 x 8 = 20 cm²

2)

Solution:

2) Area of triangle = \( \frac{1}{2} \mathrm{bh}

= [latex] \frac{1}{2} \) x 6 x 4 = 12 cm²

3)

Solution:

3) Area of triangle = \( \frac{1}{2} \) x 5.4 x 7.5 = 20.25 cm²

4)

Solution:

4) Area of triangle = \( \frac{1}{2} \) x 6 x 4 = 12 cm²

14. ΔABC is right-angled at A. AD is perpendicular to BC AB = 5 cm, BC = 13 cm, and AC =12 cm. Find the area of ΔABC. Also, find the length of AD.

Solution:

ABC is a right-angled triangle, either AB or AC can be considered as base or height

Take AC (Base 12 cm); height (AB) = 5 cm

Area of triangle = \( \frac{1}{2} \mathrm{bh}=\frac{1}{2} \times 12 \times 5=30 \mathrm{~cm}^2 \) …….(1)

Now take BC (base) = 13 cm and AD =h

1/2 bh = area; substituting, 13 cm for base

15. APQRis isosceles withPQ = PR = 7.5 cm and QR = 9 cm. The height PS from P to QR is 6 cm. Find the area of ΔPQR. What will be the height from R to PQ i.e. RT?

Solution:

By Question, base QR = 9 cm; height PS = 6 cm

Area of triangle \( =\frac{1}{2} \mathrm{bh} \)

= \(\frac{1}{2}\)x9x6 = 27cm² (1)………

Again, take PQ as base = PR = 7.5 cm (It is a triangle) & Height (TR) =h cm

Height = \( \frac{27×2}{75} \) x 10 (decimal removed)

16. Find the area of the following rhombuses.

Solution:

Area of rhombus = \( \frac{1}{2} \mathrm{~d}_1 \mathrm{~d}_2 \)

Solution:

Area of rhombus = \( \frac{1}{2} \mathrm{~d}_1 \mathrm{~d}_2 \)

= 24 cm²

Haryana Board Class 7 Maths Solutions For Chapter 9 Long Answer Questions

17. Find the circumference of circle whose diameter is

(1) 17.5 cm (2) 5.6 cm (3) 4.9 cm

Note: take \( \pi=\frac{22}{7}\) in the above two questions.

Solution:

Circumference of circle = Πd

1) Circumference of circle = \( \frac{22}{7} \times 17.5=55.0 \mathrm{~cm} \)

2) Circumference of circle =\( \frac{22}{7} \times 5.6=17.6 \mathrm{~cm} \)

3) Circumference of circle =\( \frac{22}{7} \times 4.9=15.4 \mathrm{~cm} \)

18. 1) Taking Π = 3.14, find the circumference of a circle whose radius is

(1) 8 cm (2) 15 cm (3) 20 cm

2) Calculate the radius of a circle whose circumference is 44 cm.

Solution:

1) Circumference of circle = 2Πr

1) given r = 8 cm and Π= 3.14, circumference = 2 x 3.14 x 8 = 50.24 cm

2) given r = 15 cm and Π = 3.14, circumference = 2 x 3.14 x 15 = 94.20 cm

3) given r=20 cm and Π= 3.14, circumference = 2 x 3.14 x 20 = 125.60 cm

2) Given circumference = 44 cm. To find radius, 27Πr = 44

i.e \( 2 \times \frac{22}{7} \times r=44 \)

r = \( \frac{44×7}{2×22} \)

= 7

Finding Rational Numbers Between Two Numbers Class 7 HBSE

19. Arectangle ABCD with AB = 8 cm, BC = 16 cm and AE = 4 cm.Find the area of ΔBCE. Is the area of ΔBEC equal to the sum of the area of ΔBAE and ΔCDE. Why?

Solution:

Area of a rectangle =l xb = 16 x 8 cm² = 128 cm²

Area of a triangle (BEC) = \( \frac{1}{2} \mathrm{bh} \) = \( \frac{1}{2} \) x 16 x 8 = 64 cm²

Area of a triangle (BAE) =\( \frac{1}{2} \mathrm{bh} \) = \( \frac{1}{2} \) x 4 x 8 = 16 cm²

Area of a triangle (CDE) =\( \frac{1}{2} \mathrm{bh} \) = \( \frac{1}{2} \) x 8 x 12 = 48 cm²

Observation:

1) Area of triangle BEC = \( \frac{1}{2} \) (area of rectangle)

2)Area of A BAE +A CDE = Area of A BEC because area of A BEC = Area of A BEC = \( \frac{1}{2} \) (Area of rectangle) the remaining portion of rectangle containing two triangles

BAE and CDE = BEC = 64 cm²

So, the area of ΔBEC is equal to the sum of the area of ΔBAE and ΔCDE.

Haryana Board Class 7 Maths Solutions For Chapter 9 Multiple Choice Question and Answers

1. 1 Hectare =

1. 100 m²
2. 1000 m²
3. 1000 mm²
4. 10 m²

Answer: 3

2. Side of a square is 50 cm find its area

1. 2500 cm²
2. 2000 cm²
3. 2500 cm
4. 200 cm

Answer: 1

3. The perpendicular dropped on that side from the opposite vertex is known as

1. median
2. height
3. side
4. length

Answer: 2

4. If the area of a parallelogram is 24 cm2 and the base is 4 cm then its height is

1. 4 cm
2. 6 cm
3. 48 cm
4. 96 cm

Answer: 2

5. Diameter of a circle is 10 cm. Its radius is

1. 20 cm
2. 10 cm
3. 5 cm
4. 40 cm

Answer: 3

6. If the area of a circle is 2464 m² find its diameter.

1. 14 m
2. 28 m
3. 56m
4. 45 m

Answer: 3

7. The circumference of a circle is 44 m. What is its area?

1. 44 cm²
2. 154 cm²
3. 164 cm²
4. 144 cm²

Answer: 2

8. Two sides of a right-angled triangle are 100 cm and 8.6 cm find its area.

1. 340 cm²
2. 530 cm²
3. 430 cm²
4. 240 cm²

Answer: 3

9. Find the altitude of a triangle whose base is 24 m and area 672 cm².

1. 56 cm
2. 4 cm
3. 26 cm
4. 36 cm

Answer: 1

10. Find the area of the triangle whose base is 14 cm and height is 650 cm

1. 3550 cm²
2. 4550 cm²
3. 2550 cm²
4. 5550 cm²

Answer: 2

11. If the perimeter of a semi-circle is 144 cm. What is its area?

1. 1132 cm²
2. 1432 cm²
3. 1232 cm²
4. 1332 cm²

Answer: 3

12. If each side of a square is 1 m which of the following is its area?

1. 100 cm²
2. 1000 cm²
3. 10000 cm²
4. 100000 cm²

Answer: 3

13. The sides of a triangle are 3 cm, 4 cm, and 5 cm respectively then the perimeter is

1. 10 cm
2. 12 cm
3. 20 cm
4. 15 cm

Answer: 2

14. If the side of an equilateral triangle is 6 cm then its perimeter is

1. 15 cm
2. 12 cm
3. 18 cm
4. 24 cm

Answer: 3

15. If the side of a right-angled isosceles triangle is 2 m then its area is

1. 4 m²
2. 6 m²
3. 5 m²
4. 2 m²

Answer: 4

16. The diagonals of a rhombus are 8 cm and 12 cm then its area is

1. 64 sq.cm
2. 40 sq.cm
3. 48 sq.cm
4. 70 sq.cm

Answer: 3

17. If the base of an isosceles right triangle is 30 cm then its area is

1. 300 sq.cm
2. 400 sq.cm
3. 450 sq.cm
4. 500 sq.cm

Answer: 3

18. If the diameter of the circle is 52 cm then its radius is

1. 26 cm
2. 27 cm
3. 28 cm
4. 29 cm

Answer: 1

19. If the radius of a circle is 12 m then the circumference of a circle is

1. 20πm
2. 24πm
3. 48πm
4. 42πm

Answer: 2

20. Choose the correct matching.

1) Circumference of circle              (  )    1/2 bh

2) Area of circle                              (  )      l x b

3) Area of triangle                          (  )     2Πr

4) Area of rectangle                       (  )     Πr2

                                                        (  )  \( \frac{\pi r^2}{2} \)

1. 1 – c,2 – d,3 – a,4- b
2. 1 – e,2 – a,3 – b,4- d
3. 1 – a,2 – c,3 – b,4- d
4. 1 – e,2 – b,3 – c,4- a

Answer: 1

21. The area of a rhombus is 60cm2 and one of its diagonals is 8 cm find the other diagonal.

1. 10 cm
2. 15 cm
3. 20 cm
4. 25 cm

Answer: 2

22. If the area of a rectangle is 144 m2 then its length and breadth are

1. 18 m, 8 m
2. 16 m, 9m
3. 24 m, 6m
4. All of them

Answer: 4

23. The perimeter of a square is 1 meter then its side is…..

1. 10 cm
2. 25 cm
3. 20 cm
4. 50 cm

Answer: 2

24. The product of length of two diagonals of a rhombus is 90 cm2 then its area is……

1. 90 cm²
2. 45 cm²
3. 180 cm²
4. 135 cm²

Answer: 2

25. Choose the correct matching.

1. i-c,ii-b,iii-a
2. i-a,ii-c,iii-b
3. i-a,ii-b,iii-c
4. i-c,ii-a,iii-b

Answer: 2

26. ABCD is a rectangle. The ratio of the areas of ABCD and AED is………..

1. √2:1
2. 2:1
3. 3:1
4. 3:√2

Answer: 2

27. If the outer radius of a circular path is R and its width is W, then its area in sq. units is

1. Π(2R-W) W
2. Π(R-W)W
3. Π(R+W)W
4. Π(R+W)(R-W)

Answer: 1

28. If the radii of two circles are in the ratio 9:16 then the ratio of their areas is

1. 81:256
2. 265:84
3. 3:4
4. 4:3

Answer: 1

29. If the perimeter of a circle is 16 times to the perimeter of square then the ratio of radius to the side is……….

1. 112:11
2. 11:112
3. 11: 128
4. 256: 11

Answer: 1

30. Statement – 1: If r is a radius of a circle then 2itr is the circumference of the cird.

Statement -II: If the side of the square is 14cm then the ratio of the square and circle perimeter is 14: 11

1. Both statements are true.
2. Both statements are false.
3. Statement – 1 is true statement – II is false.
4. Statement – 1 is false statement -II Is true.

Answer: 1

31. Statement – 1: If the inner and outer radii of circular rings arc 3.5 m and 7 m then area of the ring is 1185 m².

Statement -II: Area of the ring whose inner and outer radii are r, R is Π(R + r)(R – r) sq. units.

1. Both statements are true. Statement – II is the correct explanation of statement -I.
2. Both statements are true. Statement -II is not correct explanation statement-I.
3. Statement – 1 is true, statement – II is false.
4. Statement – I Is false, statement II is true.

Answer: 4

32 Base and height of a parallelogram are 6 cm and 13 cm respectively. Its area is

1. 39 sq.m
2. 78 sq.m
3. 82 sq.m
4. 38 sq.m

Answer: 2

33. Area of the parallelogram is 1470 cm2. Its base is 49 cm then the corresponding height is

1. 42 cm
2. 21cm
3. 30 cm
4. 32 m

Answer: 3

34. If the inner and outer radii of circular rings are 4m and 8m then the area of the ring is……..m².

1. 148
2. 150.86
3. 160.81
4. 140.72

Answer: 2

35. In ΔABC, ∠A = 90. AB=5 cm; AC= 12 cm; BC=13 cm, then area of ΔABC is

1. 30 cm²
2. 31.5 cm²
3. 78 cm²
4. 60 cm²

Answer: 1

36. The diagonals of a rhombus are li2 cm and 16 cm then its area is……square cms.

1. 192
2. 96
3. 126
4. 108

Answer: 2

37. The circumference of a circle whose radius is 4.2 cm

1. 27.2 cm
2. 26.4 cm
3. 18.6 cm
4. 13.2 cm

Answer: 2

38. If the circumference of a circle is 264 cm, then its radius is

1. 36 cm
2. 40 cm
3. 42 cm
4. 38 cm

Answer: 3

39. A road roller makes 200 rotations in covering 2200m then the radius of the road roller is

1. 1.75 m
2. 1.25 m
3. 2.15 m
4. 2.25 m

Answer: 1

40. In a rhombus ABCD, ∠A = 60°, AB = 6 cm then the length of the diagonal BD in cm is

1. 12
2. 9
3. 6
4. 3

Answer: 3

41. In the adjacent figure side of a square is 7cm. A circle is inscribed in the square. The perimeter of the circle is

1. 11 cm
2. 44 cm.
3. 22 cm
4. 28 cm

Answer: 3

Observe the table and answer the following questions. (42 – 44)

42. What is the formula for area of a rectangle?

1. 2(l +b)
2. l x b
3. l ÷ b
4. l + b

Answer: 2

43. What are the dimensions here?

1. length
2. breadth
3. A or B
4. A and B

Answer: 4

44. What information does this table give us?

1. It gives formula for area of rectangle
2. It gives formula for perimeter of rectangle
3. A and B
4. None

Answer: 3

45, What In the length of QS?

1. 4 cm
2. 3 cm
3. 2 cm
4. 6 cm

Answer: 2

46. Find the bane of n triangle whose area is 220 cm2 and height is 11 cm.

1. 40 cm
2. 50 cm
3. 60 cm
4. 70 cm

Answer: 1

Read the above table and answer the following questions (47 – 49)

47. About which area this table tells us

1. Parallelogram
2. Rectangle
3. Square
4. Rhombus

Answer: 4

48. What are the values of a and x?

1. 96 cm², 6 cm
2. 6 cm, 96 cm²
3. 30 cm², 6 cm
4. 36 cm², 6 cm

Answer: 1

49. What is the formula for area of a rhombus?

1. bh
2. \( \frac{1}{2} \mathrm{bh} \)
3. \( \frac{1}{2} \mathrm{~d}_1 \mathrm{~d}_2 \)
4. l x b

Answer: 3

50. If the circumference is 30cm more than the diameter of the circle find the radius of the circle.

1. 7 cm
2. 8 cm
3. 9 cm
4. 10 cm

Answer: 1

51. Area of a semi-circle is 77cm². Its perimeter is equal to

1. 35 cm
2. 44 cm
3. 42 cm
4. 36 cm

Answer: 4

52. Area of the parallelogram is 1470 cm².Its base is 30 cm then the corresponding height

1. 42 cm
2. 21cm
3. 49 cm
4. 32 cm

Answer: 3

53. What is the length of DL if AB = 13 cm and area of parallelogram is 156 cm² ?

1. 13 cm
2. 12 cm
3. 14 cm
4. 15 cm

Answer: 2

54. What is the area of quadrilateral?

1. 25 m²
2. 35 m²
3. 45 m²
4. 55 m²

Answer: 3

55. What is the difference of circumferences of the circles shown ?

1. 22 cm
2. 33 cm
3. 44 cm
4. 66 cm

Answer: 1

Haryana Board Class 7 Maths Solutions For Chapter 9 Fill in the blanks:

56. Perimeter of a regular polygon is……………..

Answer: Number of sides x Length of one side

57. All congruent triangles are equal in…………

Answer: area

58. The distance around a circular region is known as its ………

Answer: circumference

59. The perimeter of a parallelogram whose base is 5 units and height 3 units is………

Answer: 16 units

60. The area of a triangle is 36 cm² and the height is 3 cm. Its base is………..

Answer: 24 cm

Haryana Board Class 7 Maths Solutions For Chapter 9 Match the following:

61. Figure                             Area

1. Square                (   ) A) Length x Breadth

2. Rectangle           (   ) B) Base x Height

3. Triangle              (   ) C)r²

4. Parallelogram    (   ) D) (Side)²

5. Circle                  (   ) E) 1/2 x base X height

Answer: 1. D 2. A 3. E . 4. B 5. C

62.

1. Perimeter of a square                                                                                    (  ) A) π x diameter

2. Perimeter of a rectangle                                                                               (  ) B) 20 cm

3. Circumference of a circle                                                                              (  ) C) 4 X Side

4. Area of a rectangular sheet is 500 cm2, length is 25 cm its breadth is    (  ) D) 6m²

5. Base of a triangle is 3 cm, height is 4 cm its area is                                   (  ) E) 2 (Length + Breadth)

Answer: 1. C 2. E 3. A 4. B 5. D

Haryana Board Class 7 Maths Solutions For Chapter 7 Comparing Quantities

1. Introduction: In our daily life, there are many occasions when we compare two quantities.

Example:

1. Heena is 150 cm toll; Amir is 75 cm tall.
   1. Heena is two times taller than Amir.
   2. Amir’s height is 1/2 of Heena’s.
2. Speed of cheetah is 120 km per hour. Speed of man is 20 km per hour.
   1. The speed of cheetah is 6 times the speed of man.
   2. The speed of a man is 1/6 of the speed of a cheetah.

In example: 1

1. We write the ratio of their heights as Heena’s height: Amir’s height – 150: 75 or 2: 1

In example: 2

2. We write the ratio of their speeds as

Speed of man: Speed of cheetah : 20: 120 or 1: 6

To compare two quantities, the units must be the same.

2. Ratio :

We compare two quantities of some kind by division. We use Y this symbol to express the ratio.

For any non-zero numbers a and b, a is to b is a ratio.

a/b is written as a: b.

‘a’ is known as the first term or antecedent.

‘b’ is known as the second term or consequent.

3. Equivalent ratio:

A ratio does not change if its first and second terms are multiplied or divided by the same non-zero number.

Example: 15: 25 (Multiplying by 2)

15X2: 25×2=30: 50

15:25 (Dividing by 5)

15 +5: 25 +5 = 3:5

HBSE Class 7 Comparing Quantities Solutions

4. Ratio in the Simplest Form:

A ratio a: b is said to be in the simplest form if its antecedent ‘a’ and consequent ‘b’ have no common factors except 1. A ratio in the simplest
form is also called the ratio in the lowest terms.

Example: 24: 72

24: 72=\frac{24}{72}=\frac{1}{3}=1: 3

5. Comparison of ratios:

Steps :

1. Write each of the ratios in the form of a fraction in the simplest form.
2. Find the LCM ofdenominators of the two fractions.
3. Make each of the fractions its respective equivalent fractions in such way that the denominators should be equal to the LCM.
4. Compare the numerators of each of the equivalent fractions. The fraction having the larger numerator will be larger than the other.

Example:

In 5: 6; 6:7 which is bigger ?

Solution:

LCM of 6 and 7 is 6×7=42

The ratio 6: 7 is bigger than 5: 6.

6. Proportion:

The ratios which are equivalent are said to be in proportion.

Four numbers a, b, c, d are in proportion ad=bc

Product ofmeans = Product of extremes;

Proportions are also used in the making of National flags

7. Unitary method:

The method offinding the value ofone article first from the value of the given number of articles and then the value of the required number of articles is called the unitary method.

Example: 6 bowls cost Rs. 90/-. What would be the cost of 10 such bowls ?

Solution:

Cost of 6 bowls = Rs. 90

Cost of 1 bowl = Rs \( \frac{90}{6} \)

Hence the cost of 10 bowls = Rs.\( \frac{90}{6} \times 10\)

= Rs. 15X10 = Rs. 150

Ratios also appear in the form of percentages.

We use percentages to express profit, loss,discount and interest.

Expressing them in percentages makes comparisons easy.

The money borrowed or lent outfor a certain period is called the Principal. This money would be used by the borrowerfor some time before it is returned. For keeping this money for some time the borrower has to pay some extra money to the bank. This is known as Interest.

The amount that is to be repayed back is equal to the sum of the borrowed principal and the interest

That is Amount = Principal + Interest.

Interest is generally expressed as percent of the principal for a period of one year. It is written as say 10% peryear orper annum or in short as 10% p.a

Haryana Board Class 7 Maths Comparing Quantities solutions

1. Find the percentage of children of different heights for the following data.

Solution:

2. A shop has the following number of shoe pairs of different sizes.
Size 2: 20; Size 3: 30; Size 4: 28; Size 5: 14; Size 6: 8

Write this information in tabular form as done earlier and find the Percentage of each shoe size available in the shop

Solution:

1. A collection of 10 chips with different colours is given.

Fill the table and find the percentage of chips of each colour

Solution:

Key Questions in Comparing Quantities for Class 7 HBSE

2. Mala has a collection of bangles. She has 20 gold bangles and10 silver bangles. What is the percentage of bangles of each type? Can you put it in the tabular form as donein the above example?

Solution:

1. Look at the examples below and in each of them, discuss which is better for comparison. In the atmosphere,1 g of air contains:

Solution: In the atmosphere the quantity of air contained in percent is better for comparison at a glance..

The second one is better for comparison.

2. A shirt has:

Solution:

Quantity of blending of cotton and polysterin a shirtis easy to understand in terms of percentage.

The second oneisbetter for comparison

1) Can you eat 50% of a cake ? Can you eat 100% of a cake ? Can you eat 150% of a cake ?

Solution:

Yes, we can eat 50% of a cake.

Yes, we can eat 100% of a cake.

No, we cannot eat 150% of a cake.

2) Can a price of an item go up by 50% ?

Can a price of an item go up by 100% ?

Can a price of an item go up by 150% ?

Solution:

Yes, theprice of anitemcan goupby 50%.

Yes, the price of an item can go up by 100%.

Yes, the price of an item can go up by 150%.

1. Convert the following to percents:

\( \frac{12}{16} \)

3.5

\( \frac{49}{50} \)

\( \frac{2}{2} \)

0.05

Solution:

1) \( \frac{12}{16}=\frac{12}{16} \times 100=75 \% \)

2) 3.5 = 3.5×100

3) \( \frac{49}{50}=\frac{49}{50} \times 100=98 \% \)

4) \( \frac{2}{2}=\frac{2}{2} \times 100=1 \times 100=100 \% \)

5) 0.05 = 0.05 x 100

2.1) Out of 32 students, 8 are absent. What present of the students are absent?

Solution:

Out of 32 students 8 are absent. Writing

this as a fraction we get \( \frac{8}{32} \).

25% students are absent.

How to calculate percentage Class 7 HBSE

2) There are 25 radios, 16 of them are put of order. What percent of radios are out of order?

Solution:

Total number of radios = 25

Number of radios which are out of order = 16

Writing this as a fraction we get

64% of radios are out of order.

3) A shop has 500 items, out of which 5 are defective. What percent are defective ?

Solution:

Total number of items = 500

Number of items defective = 5

Writing this as a fraction = \( \frac{5}{500} \)

l% of items are defective

Practice Problems Comparing Quantities Class 7 Haryana Board

4) There are 120 voters, 90ofthem voted yes. What percent voted yes ?

Solution:

Total number of voters = 120

Numbers of voters voted yes = 90

Writing this as a fraction = \( \frac{90}{120} \)

75% voters voted yes.

Look at the table, observe, and complete it:

1. 35% +________% = 100%;
64% + 20% +____________% = 100%
45% = 100% -____________%;
70% = _____________%-30%

Solution:

35% + 65% = 100%;
64% + 20% + 16% = 100%
45% = 100% -55%;
70% =100% -30%

2. If 65% of students in a class have a bicycle, what percent of the students do not have bicycles ?

Solution:

Number of students having bicycle in a class = 65%

Out of 100 students 65 of them have a bicycle.

Number of students do not have a bicycle = 100% – 65%. = 35%.

35% of students do not have a bicycle.

3. We have a basket full of apples, oranges and mangoes.If 50% are apples, 30% are oranges, then whatpercent are mangoes?

Solution:

Out of 1,00 given fruits, apples are 50,oranges are 30 and the remaining are mangoes.

Percentage of mangoes

= 100% -50% -30%
= 100%’-80% =20%

Consider the expenditure made on a dress. 20% on embroidery, 50%pn cloth, 30% on stitching. Can you think of more such examples ?

Solution:

Yes, some more examples are as follows:

1) Maths examination was conducted for 100 marks and was observed that 35% of the students,got marks below 50. 29% of the students gotmarks between 60
and 75. 46% of the students gotmarks above 80%.

2) An alloy is made of following compositions :

Copper: 35%; Nickel: 40%; Zinc: 25%

What percent of these figures are shaded?

You can make some more figures yourself and ask your friends to estimate the shaded parts.

Solution:

1) Fractions which are,shaded

Shaded triangular partin the fraction

Percentage of shaded triangular part

2) Fraction of tangram which is shaded

Percentage of total shaded parts

1. Find

1) 50% of 164

Solution: 50% of 164

2) 75% of 12

Solution: 75% of 12

3) \( 12 \frac{1}{2} \% \text { of } 64 \)

Solution:

2. 8% children of a class of 25 like getting wet in the rain. How many children like getting wet in the rain ?

Solution.

Children who like getting wet in the rain = 8% of 25

1. 9 is 25% of what number ?

Solution:

Let the required number be P

25% of P = 9

P= 9 x 4 = 36

Required number is 36

Important Concepts Comparing Quantities Class 7 HBSE

2. 75% at what number is 15?

Salution:

let the number be P

75% of P=15

Required number is 20

Haryana Board Class 7 Maths Solutions For Chapter 7 Exercise-7.1

1. Convert the given fractional numbers to percents.

1) \( \frac{1}{8} \)

Solution:

2. \( \frac{5}{4} \)

Solution:

= 125%

Profit and loss formula Class 7 Haryana Board

3. \( \frac{3}{40} \)

Solution:

4. \( \frac{2}{7} \)

Solution:

2. Convert the given decimal fractions to percents

1) 0.65

Solution:

2) 2.1

Solution:

3) 0.02

Solution:

4) 12.35

Solution:

HBSE Class 7 Maths Chapter 7 Guide Comparing Quantities

3. Estimate what part of the figures are coloured and hence find the percent which is coloured.

Solution:

1) \( \begin{aligned}
&\frac{1}{4} \text { part is coloured }\\
&\frac{1}{4}=\frac{1}{4} \times 100 \%=25 \%
\end{aligned} \)

2) \( \begin{aligned}
&\frac{3}{5} \text { part is coloured }\\
&\frac{3}{5}=\frac{3}{5} \times 100 \%=60 \%
\end{aligned} \)

3)

4. Find:

1) 15% of 250

Solution: 15% of 250

2) 1% of 1 hour

Solution:

1% of 1 hour

3) 20% of Rs. 2500

Solution:

20% of 2500 \( =\frac{20}{100} \times \text { Rs. } 2500 \)

= Rs. 500

4) 75% of 1 kg

Solution:

75% of 1 kg.

5. Find the whole quantity if

1) 5% of it is 600

Solution:

Let the whole quantity be P

5% of P is 600

P = 600 x 20 = 12000

The whole quantity is 12000.

2) 12% of it is Rs. 1080.

Solution:

Let the whole quantity he Rs. P

12% of Rs. P = Rs. 1060

3P = 1060 x 25

The whole quantity is Rs. 9000.

3) 40% of it is 500 km.

Solution:

Let the whole quantity be P km 40% of P km = 500 km

2P = 500 x 5

The whole quantity is 1230 km

4) 70% of it is 14 minutes

Solution:

Let the whole quantity be P minutes

70% of P minutes = 14 minutes

7P =14 x 10

The whole quantity is 20 minutes

3) 8% of it is 40 litres.

Solution:

let the whole quantity be P litres.

8% of P litres = 40 litres

The whole quantity is 500 litres

6. Convert given percents to decimal fractions and also to fractions in simplest forms:

1) 25%

Solution:

2) 150%

Solution:

3) 20%

Solution:

4) 5%

Solution:

Ratio and Proportion Class 7 Haryana Board

7. In a city, 30% are females, 40% are males and remaining are children. What percent are children?
Solution:

Total population of the city = 100%

Females – 30%

Males = 40%

Children = 100- (30 + 40)

=100-70 = 30%

8. Out of 15,000 voters in a constituency,60% voted. Find the percentage of voters who didnot vote. Can you now find how many actually didnot vote ?

Solution:

Number of voters in the constituency = 15,000

Voters who voted = 60%

Voters who did not vote = 100% -60% = 40%.

Actual number of voters who did not vote = 40% of 15,000

9. Meeta saves Rs. 4000 from her salary.If this is 10% of her salary. What is her salary?

Solution: Let the salary of Meeta be Rs. P

Given Rs 4000 is 10% of her salary

10% of P = 4000

P = 4000x 10 = 40000

The salary of Meetais Rs. 4000.

10. A local cricket team played 20 matches in one season.It won 25% of them. How many matches did they win?

Solution:

Let the total number of matches won be P.

Given 25% of matches were won.

25% of 20 = P

Number of matches won by the team = 5

Solutions To Try These

1. Divide 15 sweets between Manu and Sonu so that they get 20% and 80% of them respectively.

Solution: Total number of sweets = 15

No. of sweets Manu gets = 20% of 15

No. of sweets Sonu gets = 80% of15

2. If angles of a triangle are in the ratio.2:3:4. Find the value of each angle.

Solution:

Angles of a triangle are in the ratio 2:3:4

Total of the parts = 2 + 3 + 4 = 9

Sum of the measures of the three angles of a triangle = 180°

The value of each angle = 40°, 60°, 80°

Solutions To Try These

1. Find percentage of increase or decrease:

1) Price of shirt decreased from Rs. 280 to Rs. 210.

Solution:

Decreased price of shirt- Rs. 280-Rs. 210

=Rs. 70

Percent decrease

2) Marks in a test increased from 20 to 30.

Solution:

Increased marks- 30 – 20 = 10

Percent increase

HBSE 7th Class Comparing Quantities Word Problems

2. My mother says, in her childhood petrol was Rs.1 a litre.It is Rs. 52 per litre today.By what percentage has the prife gone up?

Solution:

Cost of 1 litre petrol in her childhood = Rs.1

Cost of 1 litre petrol today is = Rs. 52

Increase,in the price of 1 litre petrol = Rs. 52 – Rs. 1 = Rs. 51

Percentage of price increase

Solutions To Try These

1. A shopkeeper bought a chair for Rs. 375 and sold it for Rs 400. Find the gain percentage

Solution: C.P. of a chair = Rs. 375

S.P. of a chair = Rs. 400

SP>CP

Gain = S.P. – C.P

= Rs. 400 – Rs. 375- Rs. 25

= \( \begin{aligned}
& \frac{\text { Gain }}{\text { C.P. }} \times 100 \% \\
& \quad=\frac{25}{375} \times 100 \% \\
& =\frac{20}{3} \%=6 \frac{2}{3} \%
\end{aligned} \)

2. Cost of an item is Rs. 50.It was sold with a profit of 12%. Find the selling price.

Solution:

Cost price of an item Rs. 50

Gain = 12% of Rs. 50

S.P.= C.P. + Gain = Rs. 50 + Rs. 6 = Rs. 56

The selling price of the item Rs. 56

3. An article was sold for Rs 250 with a profit of 5%. What was its cost price?

Solution:

Selling price of an article = Rs. 250

Gain = 5% of C.P.

S.P. = C.P. + Gain

The cost price of the article

4. An item was sold for Rs. 540 at a loss of 5%. What was its cost price?

Solution:

Selling price of an item = Rs. 540

Loss = 5%

Actual loss =5% of C.P

S.P. = CP- Loss

The cost price of the item

Solutions To Try These

1. Rs. 10,000 is invested at 5% interest rate p.a. Find the interest at the end of one year.

Solution:

Principal (P) = Rs. 10,000

Rate of interest (R) = 5%

Interest at the end of one year \( I=\frac{P R T}{100} \)

Interest at the end of one year = Rs.500

Sample Problems Comparing Quantities Haryana Board Class 7

2. Rs. 3,500 is given at 7%. p.a. rate of interest. Find the interest which will be received at the end of two years.

Solution:

Principal (P) = Rs. 3,500

Rate of interest (R) = 7%

Time (T) = 2 years

Interest at the end of two years

Interest at the end of two years is Rs. 490

3. Rs. 6,050 is borrowed at 6.5% rate of interest p.a. Find the interest and the amount to be paid at the end of 3 years.

Solution:

Principal (P) = Rs, 6,050

Rate of interest (R) =6.5%

Time (T) =3 years

Interest at the end of 3 years

= Rs. 1179.75

Amount (A) = Principal + Interest

Amount to be paid at tire end of 3 yearsis

= Rs. 6050 + Rs. 1179.75

= Rs. 7229.75

4. Rs. 7,000 is borrowed at 3.5% rate of interest p.a. borrowed for 2 years. Find the amount to be paid at the end of the second year.

Solution:

Principal (P) = Rs. 7,000

Rate of Interest (R) = 3.5%

Time (T) = 2 years

Interest to be paid at the end of second year

Amount (A) = Principal + Interest

Amount to be paid at the end of the second year is = Rs. 7000 + Rs. 490 = Rs. 7490

Solutions To Try These

1. You have Rs. 2400 in your account and the interest rate is 5% After how many years would you earn Rs. 240 as interest.

Solution:

Principal (P) = Rs. 2,400

Rate of interest (R) = 5%

Interest (I) = Rs. 240

Time (T) =?

After 2 years you would earn Rs. 240 as interest.

2. On a certain sum the interest paid after 3 years is Rs. 450 at 5% rate of interest per annum. Find the sum

Solution:

Let the sum be Rs. P

Rate of interest (R) = 5%

Time (T) = 3 years

Interest (I) = Rs. 450

The required sum = Rs. 3000

Haryana Board Class 7 Maths Solutions For Chapter 7 Exercise-7.2

1. Tell what is the profit or loss in the following transactions. Also find profit percent (or) loss per centin each case.

1) Gardening shears bought forRs. 250 and sold for Rs. 325.

Solution:

CP of gardening shears = Rs. 250

SP of gardening shears = Rs. 325

SP>CP

Profit = SP- CP = Rs. 325- Rs. 250

= Rs.75

2) A refrigerator bought for Rs. 12,000 and sold at Rs. 13,500

Solution: CP of refrigerator = Rs. 12,000.

SP of refrigerator = Rs. 13,500

SP>CP

Profit = SP- CP

= Rs. 13,500- Rs. 12,000 = Rs. 1,500

3) A cupboard bought for Rs. 2500 and sold at Rs. 3,000

Solution:

CP of cupboard = Rs. 2,500

SP of cupboard = Rs. 3,000

SP>CP

Profit = SP- CP

= Rs. 3000- Rs. 2500

= Rs. 500

Word problems on profit and loss Class 7 HBSE Maths

4) A skirt bought for Rs. 250 and sold at Rs. 150.
Solution:

CP of skirt = Rs. 250.

SP of a skirt = Rs. 150

CP>SP

Loss = CP- SP

= Rs. 250- Rs. 150

= Rs. 100

2. Convert each part of the ratio to percentage:

1) 3:1

Solution: Given ratio 3:1

Total parts =3 +1=4

Percentage of first part

Percentage of second part

2) 2:3:5

Solution:

Given ratio = 2:3:5

Total parts = 2 + 3 + 5 = 10

Percentage of first part

Percentage of second part

Percentage of third part

3) 1:4

Solution:

Given ratio 1:4

Total parts =1+4 = 5

Percentage of first part

Percentage of second part

4) 1:2:5

Solution: Given ratio 1:2:5

Total parts =l+2+5=8

Percentage of firstpart

Percentage of second part

Percentage of third part

= 62.5%

3. The population of a city decreased from 25,000 to 24,500. Find the percentage of decrease.

Solution:

Population at the begining = 25,000

Population after = 24,500

Actual decrease = 500

Percent decrease =

4. Arun bought a car for Rs. 3,50,000 The next year, theprice wentupto Rs. 3,70,000. What was the percentage of price increase?

Solution:

Price of the carin the first year = Rs. 3,50,000

Price of the carin the next year = Rs. 3,70,000

Increased price of the car
= Rs. 3,70,000 -Rs. 3,50,000
= Rs. 20,000

Percent increase

5. I buy a TV for Rs. 10,000 and sell it at a profit of 20%. How much money do I get for it ?

Solution:

C.P. of a TV= Rs. 10,000

Profit = 20% of C.P

SP of TV = CP +’Profit
= Rs. 10,000 + Rs. 2,000
= Rs. 12,000
I will get Rs. 12,000 for it.

6. Juhi sells a washing machine for Rs. 13,500. She loses 20% in the bargain. What was the price at which she bought it?

Solution: SP of washing machine = Rs. 13,500

Loss =20% of CP
Now SP = CP- Loss
13,500 = CP -20% of CP

CP =3375 X 5 = Rs. 16,875

Juhi bought the washing machine at Rs. 16,875

7.

1) Chalk contains calcium, carbon and oxygen in the ratio10:3: 12. Find the percentage of carbon in chalk.

Solution:

Given ratio =10:3:12

Total parts = 10 + 3 + 12 = 25

Percent of carbon = \( \frac{3}{25} \times 100 \%=12 \% \)

2) If in a stick of chalk,carbon is 3 g, what is the weight of the chalk stick?

Solution:

Let the weight of the chalk stick be

P grams.

12% of P = 3g

The weight of the chalk stick = 25 g

8. Amina buys a book for Rs. 275 and sells it at a loss of 15%. How much does she sell it for ?

Solution:

CP of book = Rs. 275

Loss = 15% of CP

SP of the book = CP – Loss

= Rs. 275 -Rs. 41.25

= Rs. 233.75

Amina sells the book for Rs. 233.75

Percentage Problems Class 7 HBSE

9. Find the amount to be paid at the end of 3 years in each case:

1) Principal = Rs. 1,200 at 12% p.a.

Solution:

Principal (P) = Rs. 1,200

Rate of interest (R) =12% pa.

Time (T) = 3 years

Amount (A) = P +1

= Rs. 1200 + Rs. 432

= Rs. 1632

Amount to be paid at the end of 3 years, is Rs. 1632.

2) Principal = Rs. 7,500 at 5%pa.

Solution:

Principal (P) = Rs. 7500

Rate of interest (R) = 5% p.a.

Time (T) = 3 years

= Rs.1125

Amount(A)= P+I

= Rs. 7300 + Rs 1125

= Rs, 8625

Amount to be paid at the end of 3 years is Rs. 8625.

10. What rate gives Rs. 280 as interest on a sum of Rs. 56,000in 2 years ?

Solution:

Principal (P) = Ps. 56,000

Interest (T) = Rs. 280

Time (T) =2 years

Rate of interest (R) =?

The rate of interest is 0.25 percent per annum.

11. If Meena gives an interest of Rs. 45 for one year at 9% rate pa. What is the sum she has borrowed?

Solution:

Let the sum Meena borrowed be Rs. P.

Rate of interest (R) = 9% per annum

Time (T) =1 year

Interest (I) = Rs. 45

= 5 x 100 = Rs. 500

Meena has borrowed a sum of Rs.500

Haryana Board Class 7 Maths Solutions For Chapter 7 Very Short Answer Questions

1. What is meant by ‘ratio’?

Solution:

The comparison of two quantities is known as ratio.

2. Find the ratio of 3 km to 300 m.

Solution:

3 km = 3 X 1000 m = 3000 m

The required ratio = 3 km: 300 m

= 3000:300 =10:1.

3. Write the equivalent ratio of 2: 3

Solution:

4. What is meant by ‘Unitary method’?

Solution:

The method in which we find the value of one unit first and then the value of the required number of units is known as unitary method.

5. What is a ‘proportion’?

Solution:

If two ratios are equivalent then the four quantities are said to be in proportion.

The ratios 8: 2 and 16: 4 are equivalent.

8, 2, 16, 4 arein proportion

6. Find 25% of 40

Solution:

7. If Manohar pays an.interest of Rs 750 for 2 years on a sum of Rs 4,500 find the rate of interest.

Solution:

Principal (P)=Rs 4500

Interest (I) = Rs 750

Time (T) = 2 years

Rate (R) =?

8. Last year the cost of 1000 articles was 5000 /-. This year, it went down to 4000/-. Find the percentage decrease in price.

Solution:

Original cost = 5000/-

Present cost = 4000/-

decreasein cost = 1000

% decrease in price = \( \frac{1000}{5000} \) x 100

= 20 %

9. Out,of 12,000 voters in a constituency, 60% voted. Find the No. of people voted in the constituency?

Solution:

By Question, 60% voted means \( \frac{60}{100} \)

Persons voted =\( \frac{12000×60}{100} \) = 7200

Simple Interest formula examples Class 7 HBSE

10. 40% of a number is 800 then find the number.

Solution:

40% of the number is 800

1 % of the number is \( \frac{800}{40} \)

and 100% (Actual No) = \( \frac{800}{40} \) x 100

= 2000

11. Tamarind was soldlast year at? 75/-perkg. This year it is sold at 125 per kg. Find the percentage increase in price?

Solution:

Percentage increase = \( \frac{\text { dif.in price }}{\text { original price }} \times 100 \)

The increase in tamarind price

= \( \frac{125-75}{75} \) x 100

= \( \frac{50}{75} \) x 100

12. Suppose a person buys an article for 650/- and gains 6% on selling it. Find the selling price?

Solution:

CP = 650

Profit % = 6%

Thus, profit = 6% of 650

= \( \frac{6}{100} \) x 650 = 39

We know that SP = CP + Profit

=650+39=689

Thus, the SP = 689

13. Ajay bought a TV for 15,000 and sold it for 14,100. Find the loss percent.

Solution:

By Question, loss =? (15,000.- 14,100) = 900

% loss = \( \frac{900 \times 100}{C P} \)

= 6%

14. The marked Price of a book is 225. The publisher allows a discount of 10% onit. Find the selling price of it.

Solution:

MP = 225; discount = 10%

SP (100 down) = \( \frac{100 \text {-discount }}{100} \times \text { M.P } \)

= 202.50

15. A dealer allows a discount of 10% and still gains by 10%. Find the marked price of an article which cost him 900.

Solution:

Given CP = 900

gain =10%

SP (100 down) =

\(\frac{900}{100} \times { 110 } \) = 990

again, discount = 10%

MP (100 up) = \( \frac{\mathrm{SP} \times 100}{100-\text { discount }} \)

\(\frac{990×100}{90} \) = 1100

16. Find the interest on a sum of? 8250 for 3 years at the rate of 8% per annum.

Solution:

Principal (P) = 8250

Rate of interest (R) = 8%

Interest per 3 years (I)

17. 3000 is lent out at 9% rate of interest. Find the interest which will be received at the end of 2½ years.

Solution:

Amount (P) = 3000,

Rate of Interest (R) = 9%.

= 675

18. A child-friendly bank annnounces a , savings scheme for school children. Theywill give kiddybanks to children. Children have to keep their savings in it and the bank collects all the money once in a year. To encourage children savings, they give 6% interest if the amount exceeds 10,000, and otherwise 5%. Find the interest received by the school if they deposit 9000 for one year.

Solution:

Since the deposit is only 9,000 the children get 5% interest only.

Interest = \( \frac{\text { PTR }}{100}=\frac{9000 \times 1 \times 5}{100} \)

= 450

Haryana Board Class 7 Maths Solutions For Chapter 7 Short Answer Questions

19. Shyam’smonthlyincomeis?10,000.He spends 60% of it on family expenses 10% onmedical expenses, 5% on donations and saves by 25%. Find the
amount he spends on each item.

Solution:

Amount spent on family expenses

= 60% of total income

= 60% of 10000

\( =\frac{60}{100} \times 10000 \) = 6000

Similarly, amount spent on medical expenses

\( =\frac{10}{100} \times 10000 \) = 1000

Amount spent on donations

\( =\frac{5}{100} \times 10000 \) = 500

Amount saved = \( =\frac{25}{100} \times 10000 \) = 2500

20. Ramu sold a plot of land for 2,40,000 gaining 20%. Find the costprice ofplot.

Solution:

Method -1:

SP = 2, 40,000 gain 20%

CP (100 up) = \( \frac{2,40,000 x 100}{120} \)

= 2,00,000

Method – 2: Since Ramu gained 20% his CP : SP = 100 : 120

By question 100: 120 = x : 2,40,000

x = \( \frac{2,40,000 x 100}{120} \)

= 2,00,000

21. A shopkeeper gives successive discounts of10% and5% onhis articles. Find the net discount on the whole.

Solution:

Let his MP be 100

by two successive discounts,his SP (100

down) = 100x \( \frac{90}{100} \) x \( \frac{95}{100} \)= \( \frac{171}{2} \)

= 85.50

Net discountis (100- 85.5) = 14.5,%

22. In what time will? 6880 amount to 7224, if simple interest is calculated at 10% per annum?

Solution:

Amount = ? 7224

Principle = ? 6880

S.I = Amount – Principle

= 7224-6880 = 344

R% = 10%

344 x 100 = 6880 x 10 x T

Therefore, T = \( \frac{344×100}{6880×10} \)

= \( \frac{1}{2} \) years

= 6 months

23. What sum will yield an interest of 3927 in 2 years and 4 months at 8% per annum?

Solution:

S.I =3927,

R% = 8%

T = 2 year + 4 months

3927 x 100 x 3 = P x 8 x 7

Therefore,P =\( \frac{3927x100x3}{8×7} /latex]

Thus, P = 21037.50

Therefore, Principal. = 21037.50

Haryana Board Class 7 Maths Solutions For Chapter 7 Long Answer Questions

24. A shopkeeper bought a TV for 9000 and he sold it for 10,000. Find the profit or loss? calculate percentage.

Solution:

Method-1:

Cost price (CP) of the TV = 9000

Selling price (SP) of the TV = 10,000

As SPis greater than CP, the shopkeeper makes a profit:

Profit (P) = 10000 -9000 = 1000

Thus, when the CP is 9000, the shopkeeper makes a profit of 1000

The ratio of profit and cost price is [latex] \frac{1000}{9000} \)

To find the profit percentage we multiply this ratio with 100%

i.e, \( \frac{1000}{9000} \) x 100%

Method -II:

When the CP is 9000, the profit is  1000 Now, when CP is? 100, let the profit be x.

We know that the CP and profit are directly proportional thus, ratio of profit and the ratio of cost price (CP) will be same in both cases.

Therefore, x: 1000 = 100: 9000

9000 x x = 1000 x 100

Thus, the profit % \( =11 \frac{1}{9} \% \)

25. In 4 years ? 6,500 amounts to ? 8840 at a certain rate of interest.In what time will 1600 amount to 1816 at the same rate?

Solution:

Case (1):

P = 6500

T = 4 yrs

A = 8840 such that I = (8840- 6500)

= 2340

we have to find rate %

we know \( \frac{\mathrm{PTR}}{100}=\mathrm{I} \)

R = \( \frac{2340×100}{6500×4} \) = 9 %

Case (2):

given P = 1600 A =1816

T = ? (such that I=(1816-1600)=216)

R = 9%

Substituting these values in \( \frac{\mathrm{PTR}}{100}=\mathrm{I} \)

we get 1600 x T x \( \frac{9}{100} \) = 216

T = \( \frac{216×100}{1600×9} \) = \( \frac{3}{2} \)

Haryana Board Class 7 Maths Solutions For Chapter 7 Multiple Choice Answer Questions

1. Ratio of Rs 20 to Rs 50 is

1. 2: 5
2. 5: 2
3. 2: 3
4. 3: 5

Answer: 1

2. Equivalent ratio of 6 : 4 is

1. 2:8
2. 8:2
3. 8:12
4. 12:8

Answer: 4

3. Percentages are numerators of fractions with denominator

1. 10
2. 100
3. 1000
4. 50

Answer: 2

4. If the cost of 10 cans of juice is Rs 200 then the cost of 6 cans of juice is

1. Rs 180
2. Rs 160
3. Rs 120
4. Rs 140

Answer: 3

5. \( \mathrm{I}=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100} \) in this formula P is called

1. Interest
2. Amount
3. Principal
4. Rate of interest

Answer: 3

6.Convert \( \frac{1}{3} \) as percent.

1. \( \frac{3}{100} \% \)
2. \( \frac{100}{3} \% \)
3. \( \frac{50}{3} \% \)
4. \( \frac{3}{50} \% \)

Answer: 2

7. Choose the correct matching.

1) 0.2                     (  )           a) 75%

2) 0.09                  (  )            b) 20%

3) 0.75                  (  )            c) 9%

1. 1-a,2-b,3-c
2. 1-b,2-a,3-c
3. 1-a,2-c,3-b
4. 1-b,2-c,3-a

Answer: 4

8. If 2: 5 and 6 : x are equal then find fourth proportion.

1. 20
2. 25
3. 15
4. 30

Answer: 3

9. lire ratio of two angles is 3: 2 and the larger angle is 60. What is the smaller one ?

1. 40
2. 50
3. 60
4. 70

Answer: 1

10. Choose the correct matching

1) %                (   )         a) Rate

2) /                 (   )         b) Proportion

3) ::                 (   )         c) Percent

1. 1- a,2-b,3-c
2. 1- c,2-a,3-b
3. 1- b,2-a,3-c
4. 1- c,2-b,3-a

Answer: 2

11. 72% + 15% + X = 100% => X = ………

1. 15%
2. 20%
3. 18%
4. 13%

Answer: 4

12. If x and y arein direct proportion, then which of the following is true ?

1. \( \frac{x}{y} \text { is constant } \)
2. xy is constant
3. \( \frac{1}{x} \text { is constant } \)
4. y is constant

Answer: 1

13. If C.P of 12 mangoes is equal to the selling price of 15 mangoes, find the loss percentage.

1. 10%
2. 20%
3. 30%
4. 40%

Answer: 2

14. If 25% ofa number is 300 then the number is………..

1. 1200
2. 1250
3. 1300
4. 1350

Answer: 1

15. If the cost of 10 pens is Rs. 300/- then the cost of one pen is…………..

1. Rs. 30
2. Rs. 25
3. Rs. 50
4. Rs. 17.50

Answer: 1

16. If 3, 8, x and 16 are in proportion. The value of x is…….

1. 12
2. 4
3. 8
4. 6

Answer: 4

17. The ratio of 8 books to 20 books.

1. 2:5
2. 5:2
3. 4: 5
4. 5: 4

Answer: 1

18. What sum will yield an interest of Rs. 1000/-in 2 years at 5% (in Rs) ?

1. 10,000
2. 12,000
3. 15,000
4. 20,000

Answer: 1

19. At what rate will ?10,000 amount to ?16,000in 3 yrs ?

1. 10%
2. 15%
3. 20%
4. 25%

Answer: 3

20. An amount becomes double in 5 yrs at some SI rate. Find the rate %.

1. 25%
2. 20%
3. 15%
4. 10%

Answer: 2

21.What sum will amount to ? 880in 2 years at 5% SI (in Rs)

1. 600
2. 660
3. 800
4. 500

Answer: 3

22. Certain sum amounts to ? 7000in 4 yrs and? 6000 in 2 years, the rate % is

1. 10%
2. 5%
3. \( 12 \frac{1}{2} \% \)
4. \( 6 \frac{1}{4} \% \)

Answer: 1

23. If the compound ratio of 5: 8 and 3: 7 is 45: x, then x is

1. 128
2. 72
3. 168
4. 105

Answer: 3

24. By selling an article for 1100 which is marked at 1200, discount percent is

1. \( 6 \frac{1}{4} \% \)
2. \( 12 \frac{1}{2} \% \)
3. \( 8 \frac{1}{3} \% \)
4. 10%

Answer: 3

25. The 30% of 40% of a number is 69 then the number is

1. 557
2. 575
3. 757
4. 775

Answer: 2

26. At what rate per annum will X 6360 yield an interest of 1378 in \( 2 \frac{1}{2} \) years ?

1. \( 6 \frac{2}{3} \% \)
2. \( 8 \frac{1}{3} \% \)
3. \( 8 \frac{2}{3} \% \)
4. \( 7 \frac{1}{4} \% \)

Answer: 3

27. Ratio’s of sellingprice and costprice are 4: 5, then percentage is

1. 10%
2. 20%
3. 19%
4. 25%

Answer: 2

28. Profit is calculated on

1. selling price
2. cost price
3. marked price
4. above allAnswer: 2

29. Discount is calculated on

1. marked price
2. selling price
3. cost price
4. none of the above

Answer: 1

30. 21% of 85 =?

1. 16.85
2. 17.05
3. 18.05
4. 17.85

Answer: 4

31. If 3: 5 = 4.5: x then x – 5 = ?

1. 7.5
2. 5
3. 4.5
4. 2.5

Answer: 4

32. 8% of 400 – 4% of 800 + 1% of 500 = ?

1. 10
2. 20
3. 4
4. 5

Answer: 4

33. What is the mean proportion of 25: 20:: 20: 16 ?

1. 25
2. 20
3. 16
4. 100

Answer: 2

34. If 25% of a number is 40 then the number is

1. 100
2. 140
3. 160
4. none

Answer: 3

35. While doing the problem “whatis the cost of 9 bananasif the cost of a dozen bananas is 20 ?” Write the steps in order.

1) Cost of 9 bananas =\( \frac{20}{12} \times 9 \) = 15

2) Cost of 12 bananas = 20

3) Cost of 1 banana = \( \frac{20}{12} \)

1. 1,2,3
2. 2,1,3
3. 2,3,1
4. 1,3,2

Answer: 3

36. 40% of a numberis 800 then find the number.

1. 1000
2. 2000
3. 3000
4. 4000

Answer: 2

37. If the price of an item goes up by 100%, then the cost of article will be

1. doubled
2. tripled
3. five times
4. eight times

Answer: 1

38. When 35% is expressedin decimalit is equal to

1. 0.35
2. 0.7
3. 0.14
4. 0.21

Answer: 1

39. \( \frac{4}{16} \) is equal to……..%

1. 30%
2. 25%
3. 35%
4. 40%

Answer: 2

40. Arectangleis enlarged fromits originalposition. Which conceptis usedin this aspect?

1. Profit
2. Percentage
3. Proportion
4. Simple interest

Answer: 3

41. Two is 10% of x and 20% of y, what is x-y ?

1. 1
2. 2
3. 5
4. 10

Answer: 4

42. The markedprice of a machineis Rs. 18,000.Ifitis sold at 20% discount, there is a loss of 4%. Then its costin Rs. is

1. 15,000
2. 16,000
3. 14,000
4. 13,000

Answer: 1

43. A is 20% less than B. At what percent B is more than A?

1. 25%
2. 20%
3. 18%
4. 22%

Answer: 1

44. What percentage of numbers from1 to 70 has1 or 9in its unit’s digit ?

1. 1%
2. 14%
3. 20%
4. 21%

Answer: 3

45. Rani’s salary is increased from 630 to 700, find the increase percent.

1. \( 11 \frac{1}{9} \% \)
2. \( 8 \frac{1}{2} \% \)
3. \( 13 \frac{2}{9} \% \)
4. \( 9 \frac{1}{9} \% \)

Answer: 1

46. 12% of a certain sum of money is 43.5. Find the sum.

1. 360
2. 362.50
3. 340.50
4. 352.50

Answer: 2

47. Out of 40 students 5 are absent. What percent of the students are present ?

1. \( 12 \frac{1}{2} \% \)
2. 40%
3. \( 87 \frac{1}{2} \% \)
4. 4%

Answer: 3

48. The C.P of 25 articles is equal to the S.P of 20 articles. Whatis the gain% ?

1. 0.25%
2. 2.5%’
3. 25%
4. 0.025%

Answer: 3

49. A man buys aradio for ? 500 and sellsit at a gain of 25%. What is the S.P of the radio?

1. 600
2. 575
3. 625
4. 675

Answer: 3

50. At what rate of interest per annum will a sum double itselfin 8 years ?

1. \( 6 \frac{1}{4} \% \)
2. \( 12 \frac{1}{2} \% \)
3. \( 11 \frac{1}{2} \% \)
4. \( 11 \frac{1}{4} \% \)

Answer: 2

51. Suppose aperson buys an article for? 650 and gains 6% on sellingit. The sellingprice is

1. 700
2. 650
3. 698
4. 689

Answer: 4

52. Ramu sold a plot of land for? 24000 gaining 20%. The cost price of the plot is

1. 28,000
2. 2,800
3. 20,000
4. 19,200

Answer: 3

53. 3000 is lent out at 9% rate of interest. The interest which will be received at the end of \( 2 \frac{1}{2} \) years is

1. 675
2. 725
3. 756
4. 657

Answer: 1

54. At what rate per annum will the principal triples in 16 years ?

1. 25%
2. 24%
3. \( 12 \frac{1}{2} \% \)
4. 20%

Answer: 3

55. At what rate per annus will ? 6360 yield an interest of 1378in 2 \( [/2 \frac{1}{2}latex] years.

1. [latex] 6 \frac{2}{3} \% \)
2. \( 8 \frac{1}{3} \% \)
3. \( 8 \frac{2}{3} \% \)
4. \( 7 \frac{1}{4} \% \)

Answer: 4

56. What is the decimal form of this figure?

1. \( \frac{17}{10} \)
2. 0.17
3. 0.18
4. 1.8

Answer: 3

Fill in the blanks:

57. To compare two quantities the units must be the ………

Answer:

58. 1 Dozen =………..items.

Answer:

59. Percentage is derived from the Latin word…………

Answer:

60. The buying price of any item is known as its …………….

Answer:

61. Amount = ………+………..

Answer:

62. Match the following:

1. Out of 25 children in a class 15 are girls. The percentage of girls is items.                                         (  ) A) 20%

2. Write – as a percent                                                                                                                                  (  ) B) 50%

3. Convert 0.2 as a percent                                                                                                                           (  ) \( 33 \frac{1}{3} \% \)

4. A school team won 6 games this year against 4 games won last year. Write the percent increase. (  ) D) 0.25%

5. What rate gives Rs 280 as interest on a sum of Rs 56000 in 2 years?                                                  (  ) E) 60%

Answer:

1. E 2. C 3. A 4. B 5. D

Haryana Board Class 7 Maths Solutions For Chapter 8 Rational Numbers

Key Concepts

1. 1. Introduction :
   1. The numbers used for counting objects around us are called Counting numbers (or) Natural numbers They are 1, 2, 3, 4,……,…
   2. Including ‘0’ to natural numbers we get the whole numbers i.e. 0,1,2,3,4, ………
   3. The negatives of natural numbers were put together with whole numbers to make up integers. They are……..- 3,-2,-1, 0,1,2,3,……….
   4. The numbers of the form \( \frac{\text { numerator }}{\text { denominator }} [latex] where the numerator is either 0 or a positive integer and the denominator, a positive integer are called fractions.
      Need for rational numbers: We know that integers can be. used to denote opposite situations involving numbers.
2. Definition of a rational number:
   A rational number is defined as number that can be expressed in the form of [latex] \frac{p}{q} \) where p and q are integers and q ≠ 0.

Example :

1. Is the number \( \frac{2}{-3} \) rational ? Think about it.

Solution: Yes; \( \frac{2}{-3} \) is a rational number.

It is in the form of latex] \frac{p}{q} [/latex], where p = 2; q = -3 both are integers.

2. List ten rational numbers.

Solution:

HBSE Class 7 Rational Numbers Solutions

Fill in the boxes:

1)

Solution:

2)

Solution:

Solutions To Try These

1. Is 5 a positive rational number?

Solution:

Yes,5 is a positive rational number. It can be written as \( \frac{5}{1} \) . The numerator is 5 and denominator is 1.

2. List five more positive rational numbers.

Solution:

Solutions To Try These

1. Is – 8 a negative rational number?

Solution:

Yes,- 8 is a negative rational number. It can be written as \( \frac{-8}{1} \) . The numerator is a negative integer and the denominator is
a positive integer.

2. List five more negative rational numbers.

Solution:

Solutions To Try These

Which of these are negative rational numbers?

1. \( \frac{-2}{3} \)
2. \( \frac{5}{7} \)
3. \( \frac{3}{-5} \)
4. 0
5. \( \frac{6}{11} \)
6. \( \frac{-2}{-9} \)

Solution:

1) \( \frac{-2}{3} \) and 3. \( \frac{3}{-5} \) are negative rational numbers.

Solutions To Try These

Find the standard form of

1) \( \frac{-18}{45} \)

Solution: The HCF of 18 and 45 is 9.

2) \( \frac{-12}{18} \)

Solution: The HCF of 12 and 18 is 6.

Solutions To Try These

Find five rational numbers between

\( \frac{-5}{7}\) and \( \frac{-3}{8}\).

Solution:

\( \frac{-40}{56}<\frac{-39}{56}<\frac{-38}{56}<\frac{-29}{56} \) \( <\frac{-27}{56}<\frac{-22}{56}<\frac{-21}{56} \)

\( \frac{-5}{7}<\frac{-39}{56}<\frac{-38}{56}<\frac{-29}{56}<\frac{-27}{56}<\frac{-22}{56} \) \( <\frac{-3}{8} \)

The five rational numbers between \( <\frac{-5}{7} \) and \( <\frac{-3}{8} \) are

Haryana Board Class 7 Maths Solutions For Chapter 8  Exercise-8.1

1. List five rational numbers between: (1) -1 and 0 . (2) -2 and -1

(3) \( \frac{-4}{5} \text { and } \frac{-2}{3} \)

4. \( \frac{-1}{2} \text { and } \frac{2}{3} \)

Solution:

First we find equivalent rational numbers having same denominator.

1) \( -1=\frac{-1}{1}=\frac{-1 \times 10}{1 \times 10}=\frac{-10}{10} \)

The five rational numbers between -1 and 0 are

Haryana Board Class 7 Maths Rational Numbers Solutions

2) – 2 and -1

Solution:

The five rational numbers between -2 and -1 are

HBSE 7th Class Rational Number Word Problems

3) \( \frac{-4}{5} \text { and } \frac{-2}{3} \)

Solution:

The five rational numbers between \( \frac{-4}{5} \) and \( \frac{-2}{3} \text { are } \) are

Class 7 Maths Chapter 8 Rational Numbers Haryana Board

4) \( \frac{-1}{2} \text { and } \frac{2}{3} \)

Solution:

The five rational numbers between \( \frac{-1}{2} \text { and } \frac{2}{3} \text { are } \)

2. Write four more rational numbers in each of the following patterns :

1) \( \frac{-3}{5}, \frac{-6}{10}, \frac{-9}{15}, \frac{-12}{20} \ldots . . \)

Solution:

Thus, we observe a pattern in these numbers.

The next four numbers would be

The required four rational numbers are

2) \( \frac{-1}{4}, \frac{-2}{8}, \frac{-3}{12}, \ldots . \)

Solution:

Thus we observe a pattern in these numbers.

The next four numbers would be

The required four rational numbers are

Haryana Board 7th Class Maths Rational Numbers Questions and Answers

3) \( \frac{-1}{6}, \frac{2}{-12}, \frac{3}{-18}, \frac{4}{-24}, \ldots \ldots \)

Solution:

Thus we observe a pattem in these numbers.

The required four numbers are \( \frac{5}{-30} ; \frac{6}{-36} \)

4) \( \frac{-2}{3}, \frac{2}{-3}, \frac{4}{-6}, \frac{6}{-9} \ldots . \)

Solution:

Thus we observe a pattern in these numbers.

The next four numbers would be

The required four numbers are

3. Give four rational numbers equivalent to:

(1)\( \frac{-2}{7} \)
(2)\( \frac{5}{-3} \)
(3)\( \frac{4}{9} \)

Solution:

(1)

Thefourrational numbers equivalent to \( \frac{-2}{7} \text { are } \frac{-4}{14}, \frac{-6}{21}, \frac{-8}{28}, \frac{-10}{35} \)

(2)\( \frac{5}{-3} \)

Solution:

The four rational numbers equivalent

Sample Problems Rational Numbers Haryana Board Class 7

(3)\( \frac{4}{9} \)

Solution:

The four rational numbers equivalent to \( \frac{4}{9} \) are

4. Draw the number line and represent the following rational numbers on it:

1) \( \frac{3}{4} \)

Solution:

2) \( \frac{-5}{8} \)

Solution:

3) \( \frac{-7}{4} \)

Solution:

4) \( \frac{7}{8} \)

Solution:

5. The points P, Q, R, S, T, U, A, and B on the number line are such that, TR = RS = SU and AP= PQ = QB. Name the rational numbers represented by P, Q, R, and S.

Solution:

The points P, Q, R,S on the number line such that TR = RS = SU

The rational number represented by P

The rational number represented by Q,

The rational number represented by R,

The rational number represented by S,

6. Which of the following pairs represent the same rational number ?

1) \( \frac{-7}{21} \text { and } \frac{3}{9} \)

Solution:

\( \frac{-7}{21} \) is a negative rational number.

\( \frac{3}{9} \) is a positive rational number.

The given pair does not represent the same rational number.

2)

\( \frac{-16}{20} \text { and } \frac{20}{-25} \)

Solution:

The given pair represents the same rational number

3) \( \frac{-2}{-3} \text { and } \frac{2}{3} \)

Solution:

The given pair represents the same rational number.

4) \( \frac{-3}{5} \text { and } \frac{-12}{20} \)

Solution:

The given pair represents the same rational number.

5) \( \frac{8}{-5} \text { and } \frac{-24}{15} \)

The given pair represents the same rational number.

6) \( \frac{1}{3} \text { and } \frac{-1}{9} \)

Solution:

\( \frac{1}{3} \) is a positive rational number.

\( \frac{-1}{9} \) is a negative rational number.

The given pair does not represent the same rational number.

7) \( \frac{-5}{-9} \text { and } \frac{5}{-9} \)

Solution:

\( \frac{5}{9} \) is a positive rational number.

\( \frac{5}{-9} \) is a negative rational number.

The given pair does not represent the same rational number.

Operations on Rational Numbers Class 7 HBSE

7. Rewrite the following rational numbers in the simplest form:

1) \( \frac{-8}{6} \)

Solution:

HCF of 8 and 6 is 2.

2) \( \frac{25}{45} \)

Solution:

HCF of 25 and 45 is 5.

Chapter 8 Rational Numbers Class 7 Solutions in Hindi Haryana Board

3) \( \frac{-44}{7 \cdot 2} \)

Solution:

HCF of 44 and 72 is 4.

4) \( \frac{-8}{10} \)

Solution:

HCF of 8 and 10 is 2.

8. Fill in the boxes with the correct symbol out of >, < and =

1) \( \frac{-5}{7}\) \( \frac{2}{3}\)

Solution:

LCM of 7 and 3 is 21.

2) \( \frac{-4}{5} \) \( \frac{-5}{7} \)

Solution:

LCM of 5 and 7 is 35

3) \( \frac{-7}{8} \) \( \frac{14}{-16} \)

Solution:

4) \( \frac{-8}{5} \) \( \frac{-7}{4} \)

Solution:

LCM of 5 and 4 is 20

5) \( \frac{1}{-3} \) \( \frac{-1}{4} \)

Solution:

LCM of3 and 4 is 12

6) \( \frac{5}{-11} \) \( \frac{-5}{11} \)

Solution:

Equivalent Rational Numbers Class 7 Haryana Board

7) 0 rec \( \frac{-7}{6} \)

Solution: \begin{aligned}
& 0=\frac{0}{6} \\
& \text { Hence } 0>\frac{-7}{6}
\end{aligned}

9. Which is greater in each of the following:

1) \( \frac{2}{3}, \frac{5}{2} \)

Solution:

LCM of 3 and 2 is 6

Important Questions for Class 7 Maths Chapter 8 Haryana Board

2) \( \frac{-5}{6}, \frac{-4}{3} \)

Solution: LCM of 6 and 3 is 6.

3) \( \frac{-3}{4}, \frac{2}{-3} \)

LCM of 4 and 3 is 12.

Solution:

4) \( \frac{-1}{4}, \frac{1}{4} \)

Solution: \( \frac{1}{4}>\frac{-1}{4} \)

5) \( -3 \frac{2}{7},-3 \frac{4}{5} \)

Solution:

LCM of 7 and 5 is 35.

10. Write the following rational numbers in ascending order :

1) \( \frac{-3}{5}, \frac{-2}{5}, \frac{-1}{5} \)

Solution:

Denominators of each rational number is 5.

-3<-2<-l

2) \( \frac{-1}{3}, \frac{-2}{9}, \frac{-4}{3} \)

Solution: LCM of 3, 9, 3 is 9.

-12 < -3 < -2

HBSE Class 7 Maths Chapter 8/9 Guide Rational Numbers

3) \( \frac{-3}{7}, \frac{-3}{2}, \frac{-3}{4} \)

Solution: LCM of 7, 2, 4 is 28.

– 42 < -21 < -12

Solutions To Try These

Find (1) \( \frac{-13}{7}+\frac{6}{7} \)

Solution: \( \frac{-13}{7}+\frac{6}{7}=\frac{-13+6}{7}=\frac{-7}{7}=-1 \)

2) \( \frac{19}{5}+\left(\frac{-7}{5}\right) \)

Solution: \( \frac{19}{5}+\frac{-7}{5}=\frac{19+(-7)}{5}=\frac{19-7}{5}=\frac{12}{5} \)

Solutions To Try These

Find:

(1) \( \frac{-3}{7}+\frac{2}{3} \)

Solution: LCM of 7 and 3 is 21.

2) \( \frac{-5}{6}+\frac{-3}{11} \)

Solution: LCM of 6 and 11 is 66

Solutions To Try These

What will be the additive inverse of \( \frac{-3}{9}, \frac{-9}{11}, \frac{5}{7} ? \)

Solution:

Solutions To Try These

Find 1) \( \frac{7}{9}-\frac{2}{5} \)

Solution:

2) \( 2 \frac{1}{5}-\frac{(-1)}{3} \)

Solution:

Solutions To Try These

What will be

1) \( \frac{-3}{5} \times 7 ? \)

Solution: \( \frac{-3}{5} \times 7=\frac{(-3) \times 7}{5}=\frac{-21}{5}=-4 \frac{1}{5} \)

2) \( \frac{-6}{5} \times(-2) ? \)

Solution: \( \frac{-6}{5} \times(-2)=\frac{(-6) \times(-2)}{5}=\frac{12}{5}=2 \frac{2}{5} \)

Important Concepts Rational Numbers Class 7 HBSE

Solutions To Try These

Find:

(1) \( \frac{-3}{4} \times \frac{1}{7} \)

Solution: \( \frac{-3}{4} \times \frac{1}{7}=\frac{(-3) \times 1}{4 \times 7}=\frac{-3}{28} \)

(2) \( \frac{2}{3} \times \frac{-5}{9} \)

Solution: \( \frac{2}{3} \times \frac{-5}{9}=\frac{2 \times(-5)}{3 \times 9}=\frac{-10}{27} \)

Solutions To Try These

What will be the reciprocal of \( \frac{-6}{11} \) and \( \frac{-8}{5} \) ?

Solution:

Solutions To Try These

Find:

(1) \( \frac{2}{3} \times \frac{-7}{8} \)

Solution:

2) \( \frac{-6}{7} \times \frac{5}{7} \)

Solution: \( \begin{aligned}
&\frac{-6}{7} \times \frac{5}{7}=\frac{(-6) \times 5}{7 \times 7}\\
&=\frac{-30}{49}
\end{aligned} \)