• 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)

• Divisibility Rule:
  The process of checking whether a number is divisible by a given number or not with actual division is called ‘Divisibility rule’ for that number.
• A number is completely divisible by another number when it leaves zero as remainder.
• A number is divisible by 2 if it has the digits 0, 2, 4, 6 or 8 in its ones place.
• If the sum of the digits in a number is a mul- tiple of 3 then the number is divisible by 3.
• If a number is divisible by both 2 and 3 then it is also divisible by 6.
• A number is divisible by 9, if the sum of the digits of the number is divisible by 9.
• A number is divisible by 5, if and only if its last digit is either 5 or 0.
• A number is divisible by 10, if and only if its last digit is 0.
• If the number formed by the last two digits of the given number is divisible by 4, then the given number is divisible by 4.
• If the number formed by the last three digits of the given number is divisible by 8, then the given number is divisible by 8.
• A number is divisible by 11 if and only if the difference of the numbers obtained on adding the alternate digits of the number is either O or divisible by 11.
  Factors:
• A number which divides the other number without remainder is called a ‘factor’.
• 1 is factor of every number and is the smallest of all factors.
• Every number is a factor of itself and is the greatest of its factors.
• Every factor is less than or equal to the given number.
• Number of factors of a given number are countable.
• Every number is a multiple of itself.
• Every number is multiple of each of its factors.
• Every multiple of a given number is greater than or equal to that number.
• The number other than 1, with only factors namely 1 and the number itself is called a “Prime number”.
• Numbers that have more than two factors are called “Composite numbers”.
• ‘1’ is neither prime nor composite.
• Finding prime numbers, easy method was given by the Greek Mathematician “Eratosthenes”.
• The smallest even prime number is 2.
• Every prime number other than 2 is odd.
• The numbers which have only 1 as the common factor are called “Co-primes” or “relatively prime”.
• Only two primes are co-primes but all the co-primes need not be primes.
• “Twin primes” are prime numbers that differ from each other by two.
• A number divisible by two coprime numbers is divisible by their product also.
• If a number is divisible by another number then it is divisible by each of the factors of that number.
• When a number is expressed as a product of its factors we say that the number has been factorized. The process of finding the factors is called ‘factorization’.
• Methods of prime factorization :
  • Division Method
  • Factor Tree Method.
• “Common factors” are those numbers which are factors of all the given numbers.
• HCF – Highest Common Factor
• GCD – Greatest Common Divisor
• The Highest Common Factor (HCF) of two or more given numbers is the highest (or greatest) of their common factors. It is also called Greatest Common Divisor (GCD).
• Methods for finding HCF:
  • Prime Factorization Method
  • Continued Division Method.
• The method of division was invented by the famous Greek Mathematician “Euclid”.
• Continued division method is useful to find the HCF of large numbers.
• LCM – Lowest Common Multiple
• The Lowest Common Multiple (LCM) of two or more given numbers is the lowest of their common multiples.
• Methods for finding LCM:
  • Prime Factorization Method
  • Division Method
• If one of the two given numbers is a multiple of the other, then the greater number will be their LCM.
• Relation between LCM and HCF : Product of LCM and HCF of the two numbers Product of the two numbers.
• 1221 is a “Palindrome number”.
• If two numbers are divisible by a number then their sum and difference are also divisible by that number.
• A factor of a number is an exact divisor of that number.
• The number of multiples of a given number are infinite.
• Every number is a multiple of itself.
• A number for which sum of all its factors is equal to twice the number is called a perfect number.
• The least perfect number is 6.
• The numbers 2, 4, 6, 8, 10, ………… are called even numbers.
• The numbers 1, 3, 5, 7, numbers ……. are called odd

Find the possible factors of 45, 30 and 36.

Solution. Factors of 45 are 1, 3, 5, 9, 15, 45.

Factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30.

Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36.

Haryana Board Class 6 Maths Solutions For  Chapter 3 Playing With Numbers Exercise-3.1

Question 1. Write all the factors of the following numbers:

Solution. a) 24

24 = 1 × 24

24 = 2 × 12

24 = 3 x 8

24 = 4 × 6

Factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24.

Haryana Board Class 6 Maths Playing With Numbers solutions

b) 15

15 = 1 × 15

15 = 3 × 5

Factors of 15 are 1, 3, 5, 15.

Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18

c) 21

21 = 1 × 21

21 = 3 × 7

Factors of 21 are 1, 3, 7 and 21.

d) 27

27 = 1 × 27

27 = 3 × 9

Factors of 27 are 1, 3, 9 and 27.

e) 12

12 = 1 × 12

12 = 2 × 6

12 = 3 × 4

Factors of 12 are 1, 2, 3, 4, 6 and 12.

f) 20

20 = 1 × 20

20 = 2 x 10

20 = 4 x 5

Factors of 20 are 1, 2, 4, 5,10 and 20.

g) 18

18 = 1 x 18

18 = 2 x 9

18 = 3 x 6

Factors of 18 are 1, 2, 3, 6, 9 and 18.

h) 23

23 = 1x 23

Factors of 23 are 1 and 23.

i) 36

36 = 1 x 36

36 = 2 x 18

36 = 3 x 12

36 = 4 x 9

36 = 6 x 6

Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.

Question 2. Write first five multiples of:

a) 5

Solution. The first five multiples of 5 are 5, 10, 15, 20, 25.

b) 8

Solution. The first five multiples of 8 are 8, 16, 24, 32 and 40.

c) 9

Solution. The first five multiples of 9 are 9, 18, 27, 36 and 45.

Question 3. Match the items in column – 1 with the items in column – 2

Solution. 1-b, 2-d, 3-a, 4-f, 5-e.

Question 4. Find all the multiples of 9 upto 100.

Solution. The multiples of 9 upto 100 are, 9, 18, 27, 36, 45,54, 63, 72, 81, 90 and 99.

Question 5. Observe that 2 x 3 + 1 = 7 is a prime number. Here, 1 has been added to a multiple of 2 to get a prime number. Can you find some more numbers of this type?

Solution. 2 × 5 + 1 = 10 + 1 = 11, Prime number

2 x 6 + 1 = 12 + 1 = 13, Prime number

2 x 8 + 1 = 16 + 1 = 17, Prime number

2 × 9 + 1 = 18 + 1 = 19, Prime number

2 x 11 + 1 = 22 + 1 = 23, Prime number

Haryana Board Class 6 Maths Solutions For  Chapter 3 Playing With Numbers Exercise 3.2

Question 1. What is the Sum of any two

a) Odd numbers

Solution. The sum of any two odd numbers is an even number.

b) Even numbers

Solution. The sum of any two even numbers is an even number.

Question 2. State whether the following statements are True (T) or False (F).

a) The sum of three odd numbers is even.

Solution. False. [For eg: 3 +5 +9 = 17 is an odd number.]

b) The sum of two odd numbers and one even number is even..

Solution. True.

c) The product of three odd numbers is odd.

Solution. True.

d) If an even number is divided by 2, the quotient is always odd.

Solution. False. [Even numbers are those which are divided by 2. The quotient need not always be odd.]

e) All prime numers are odd.

Solution. False [The prime number 2 is an even number]

f) Prime numbers do not have any factors.

Solution. False.

HCF and LCM Class 6 HBSE Maths

g) Sum of two prime numbers is always even.

Solution. False. [For eg. The total of two prime numbers 2 and 3 is not an even number.]

h) 2 is the only even prime number.

Solution. True.

i) All even numbers are composite numbers.

Solution. False.

j) The porduct of two even numbers is always even.

Solution. True.

Question 3. The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers upto 100.

Solution. 17, 71; 37,73; 79,97.

Question 4. Write down separately the prime and composite numbers less than 20.

Solution.

Question 5. What is the greatest prime number between 1 and 10?

Solution. The greatest prime number between 1 and 10 is 7.

Question 6. Express the following as the sum of two odd primes.

a) 44

Solution. 44 = 7 + 37

b) 36

Solution. 36 = 7 + 29

c) 24

Solution. 24 = 5 + 19

d) 18

Solution. 18 = 5 + 13 and 7 + 11

Question 7. Give three pairs of prime numbers whose difference is 2. (Remark: Two prime numbers whose difference is 2 are called twin primes)

Solution. The three pairs of prime numbers whose difference is 2 are. 3, 5; 5, 7:11, 13.

Question 8. Which of the following numbers are prime?

a) 23

b) 51

c) 37

d) 26

Solution. 23, 37 are prime numbers.

Question 9. Write seven consecutive composite numbers less than 100 so that there is no prime number between them.

Solution. 90, 91, 92, 93, 94, 95, 96.

Question 10. Express each of the following numbers as the sum of three odd primes:

a) 21

b) 31

c) 53

d) 61

Solution.

a) 21 = 3 + 7 + 11

b) 31 = 7 + 11 + 13

c) 53 = 11 + 13 + 29

d) 61 = 11 + 13 + 37

Question 11. Write five pairs of prime numbers less than 20 whose sum is divisible by 5.

(Hint: 3 + 7 = 10)

Solution. 2. 3; 2, 13: 3, 7:3, 17; 11, 19.

Question 12. Fill in the blanks:

a) A number which has only two factors is called a …

Answer. Prime Number

b) A number which has more than two factors is called a

Answer. Composite Number

c) 1 is neither………nor………….

Answer. Prime, Composite

d) The smallest prime number is….

Answer. 2

e) The smallest composite number is

Answer. 4

f) The smallest even number is…

Answer. 2

Haryana Board Class 6 Maths Solutions For  Chapter 3 Playing With Numbers Exercise-3.3

Question 1. Using divisibility tests, determine which of the following numbers are divisible by 2, by 3, by 4, by 5, by 6, by 8, by 9, by 10, by 11 (say: yes or no)

Solution.

Question 2. Using divisibility tests, determine which of the following numbers are divisible by 4; by 8:

a) 572

b) 726352

c) 5500

d) 6000

e) 12159

f) 14560

g) 21084

h) 31795072

i) 1700

j) 2150

Solution. a) 572

1) Test for divisibility by 4:

Number with last two digits = 72

∴ 72 is divisible by 4.

∴ 572 is divisible by 4.

Last two digits are divisible by 4, givne number is divisible by 4.

2) Test for divisibility by 8

Number with last three digits = 572

Remainder is not zero.

∴ 572 is not divisible by 8.

Last three digits are divisible by 8. So given number is divisible by 8.

b) 726352

1) Test for divisibility by 4:

Number with last two digits = 52

The remainder is zero.

∴ 52 is divisible by 4

∴ 726352 is divisible by 4.

2) Test for divisibility by 8:

Number with last three digits = 352

The remainder is 0, 352 is divisible by 8.

∴ 726352 is divisible by 8.

Prime and composite numbers Class 6 HBSE

c) 5500

1) Test for divisibility by 4:

As the last two digits are ‘0’ and.

‘0’ is divisible by 4

∴ 5500 is divisible by 4.

2) Test for divisibility by 8:

Number with last three digits = 500

The remainder is not zero.

∴ 500 is not divisible by 8

∴ 5500 is not divisible by 8.

d) 6000

1) Test for divisibility by 4:

The number with last two digits is 00, always divisible by 4.

∴ 6000 is divisible by 4.

2) Test for divisibility by 8:

The number with last three digits is 000, always divisible by 8.

∴ 6000 is divisible by 8.

e) 12159

1) Test for divisibility by 4:

The number with last two digits = 59

The remainder is not zero.

∴ 59 is not divisible by 4.

∴ 12159 is not divisible by 4.

2) Test for divisibility by 8:

The number with last three digits = 159

Remainder is zero.

∴ 159 is divisible by 8.

∴ 12159 is divisible by 8.

f) 14560

1) Test for divisibility by 4:

The number with last two digits = 60

The remainder is zero.

∴ 60 is divisible by 4.

∴ 14560 is divisible by 4.

2) Test for divisibility by 8:

The number with last three digits = 560

Remainder is not zero.

∴ 560 is not divisible by 8.

∴ 14560 is not divisible by 8.

g) 21084

1) Test for divisibility by 4:

The number with last two digits = 84

The remainder is zero.

∴ 84 is divisible by 4.1

∴ 21084 is divisible by 4.

2) Test for divisibility by 8:

The number with last two digits = 084.

Remainder is not zero.

∴ 84 is not divisible by 8.

∴ 21084 is not divisible by 8.

h) 31795072

1) Test for divisibility by 4:

The number with last two digits = 50

Remainder is zero.

∴ 72 is divisible by 4.

∴ 31795072 is divisible by 4.

2) Test for divisibility by 8:

The number with last three digits = 072

Remainder is zero.

∴ 72 is divisible by 8

∴ 31795072 is divisible by 8.

i) 1700

1) Test for divisibility by 4:

The number formed with last two digits = 00, which is always divisible by 4.

∴ 1700 is divisible by 4.

2) Test for divisibility by 8:

The number with last three digits = 700

Remainder is not zero.

∴ 700 is not divisible by 8

∴ 1700 is not divisible by 8.

j) 2150

1) Test for divisibility by 4:

The number with last two digits = 50

Remainder is not zero.

∴ 50 is not divisible by 4.

∴ 2150 is not divisible by 4.

2) Test for divisibility by 8:

The number with last three digits = 150

Remainder is not zero.

∴ 150 is not divisible by 8

∴ 2150 is not divisible by 8.

Divisibility rules Class 6 Haryana Board

Question 3. Using divisiblity tests, determine. which of the following numbers are divisible by 6:

Solution. 6 is divisible by 2, 3

a) 297144

1) Test for divisibility by 2: Units digit = 4 is divisible by 2.

∴ 297144 is divisible by 2.

2) Test for divisibility by 3:

Sum of digits = 2 + 9 + 7 + 1 + 4 + 4 = 27

27 is divisible by 3

∴ 297144 is divisible by 3.

Since 297144 is divisible by both 2 and 3, it is divisible by 6.

b) 1258

1) Test for divisibility by 2:

Units digit = 8 is divisible by 2.

∴ 1258 is divisible by 2.

2) Test for divisibility by 3:

Sum of digits = 1 + 2 + 5 + 8 = 16

16 is not divisible by 3

∴ 1258 is not divisible by 3.

∴ 1258 is not divisible by 3 but divisible by 2, So it is not divisible by 6.

c) 4335

1) Test for divisibility by 2:

Units digit = 5 is not divisible by 2.

∴ 4335 is not divisible by 2.

∴ 4335 is not divisible by 6.

d) 61233

1) Test for divisibility by 2:

Units digit = 3 is not divisible by 2.

∴ 61233 is not divisible by 2.

∴ 61233 is not divisible by 6.

e) 901352

1) Test for divisibility by 2:

Units digit = 2.

∴ 901352 is divisible by 2.

2) Test for divisibility by 3: Sum of the digits = 9 + 0 + 1 + 3 + 5 + 2 = 20.

20 is not divisible by 3

∴ 901352 is not divisible by 3.

∴ 901352 is not divisible by 6.

f) 438750

1) Test for divisibility by 2:

Units digit = 0 is always divisible by 2

∴ 438750 is divisible by 2.

2) Test for divisibility by 3:

Sum of digits = 4 + 3 + 8 + 7 + 5 + 0 = 27.

27 is divisible by 3

∴ 438750 is divisible by 3.

∴ 438750 is divisible by 2 and 3.

∴ 438750 is divisible by 6.

g) 1790184

1) Test for divisibility by 2:

Units digit = 4 is divisible by 2.

∴ 1790184 is divisible by 2.

2) Test for divisibility by 3:

Sum of the digits = 1 + 7 + 9 + 0 + 1 + 8 + 4 = 30

30 is divisible by 3

∴ 1790184 is divisible by 3.

1790184 is divisible by both 2 and 3.

∴ 1790184 is divisible by 6.

h) 12583

1) Test for divisibility by 2:

Units digit = 3 is not divisible by 2.

∴ 12583 is not divisible by 2.

∴ 12583 is not divisible by 6.

i) 639210

1) Test for divisibility by 2:

Units digit = 0

∴ 639210 is divisible by 2.

2) Test for divisibility by 3:

Sum of the digits = 6 + 3 + 9 + 2 + 1 + 0 = 21.

21 is divisible by 3

∴ 639210 is divisible by 3.

∴ 639210 is divisible by 2 and 3.

∴ 639210 is divisible by 6.

j) 17852

1) Test for divisibility by 2:

Units digit = 2

∴ 17852 is divisible by 2.

2) Test for divisible by 3:

Sum of the digits = 1 + 7 + 8 + 5 + 2 = 23

23 is not divisible by 3

∴ 17852 is divisible by 3.

∴ 17852 is not divisible by 6.

Question 4. Using divisibility tests, determine which of the following numbers are divisible by 11:

a) 5445

b) 10824

c) 7138965

d) 70169308

e) 10000001

f) 901153

Solution. a) 5445

Sum of the digits at odd places from the right = 5 + 4 = 9

Sum of the digits at even places from the right = 4 + 5 = 9

Difference = 9 – 9 = 0

‘0’ is divisible by 11.

∴ 5445 is divisible by 11.

b) 10824

Sum of the digits at odd places from the right = 4 + 8 + 1 = 13

Sum of the digits at even places from the right = 2 + 0 = 2

Difference = 13 – 2 = 11 divisible by 11

∴ 10824 is divisble by 11.

c) 7138965

Sum of the digits at odd places from the right = 5 + 9 + 3 + 7 = 24

Sum of the digits at even places from the right = 6 + 8 + 1 = 15

Difference = 24 – 15 = 9, not divisible by 11

∴ 7138965 is not divisible by 11.

d) 70169308

Sum of the digits at odd places from the right = 8 + 3 + 6 + 0 = 17

Sum of the digits at even places from the right = 0 + 9 + 1 + 7 = 17

Difference 17 – 17 = 0

‘0’ is divisible by 11

∴ 70169308 is divisible by 11.

e) 10000001

Sum of the digits at odd places from the right = 1 + 0 + 0 + 0 = 1

Sum of the digits at even places = 0 + 0 + 0 + 1 = 1

Difference = 1 – 1 = 0

‘0’ is divisible by 11.

∴ 10000001 is divisible by 11.

f) 901153

Sum of the digits at odd places from the right = 3 + 1 + 0 = 4

Sum of the digits at even places from the right = 5 + 1 + 9 = 15

Difference = 15 – 4 = 11

∴ 11 is divisible by 11

∴ 901153 is divisible by 11.

Question 5. Write the smallest digit and the greatest digit in the blank space of each of the following numbers so that the number formed is divisible by 3.

Solution. a) _6724

1) Smallest digit

Sum of the digits = 6 + 7 + 2 + 4 = 19

19 is not divisible by 3.

21 is the nearest number to 19 which is divisible by 3.

So, 2 is the smallest number that fills the blank.

∴ Smallest number is 2 i.e. 26724.

2) Largest digit

The largest digit is 8 (19 + 8 = 27)

b) 4765_2

Sum of the given digits

= 4 + 7 + 6 + 5 + 2 = 24

24 is divisible by 3

∴ Smallest digit is 0

Largest digit is 9.

Question 6. Write the digit in the blank space of each of the following numbers so that the number formed is divisible by 11:

a) 92_389

b) 8_9484

Solution. a) 92_389

The sum of digits at odd places from the right = 9 + 3 + 2 = 14

Sum of the digits at even places from the right = 8 + 9 = 17

Difference = 17 – 14 = 3

To make the difference divisible by 11, we have to add 8 to 3.

∴ Blank space must be 8.

∴ Required number-928389.

b) 8_9484

Sum of the digits at odd places from the right = 4 + 4 = 8

Sum of the digits at even places from the right = 8 + 9 + 8 = 25

Difference 25 – 8 = 17

To make the difference divisible by. 11, we have to deduct 6 from 17.

∴ Digit in the blank space = 6

∴ Required number = 869484.

Question 7. Find the common factors of

a) 8, 20

b) 9, 15

Solution. a) Factors of 8 are 1, 2, 4, 8.

Factors of 20 are 1, 2, 4, 5, 10, 20.

Common factors are 1, 2, 4.

b) Factors of 9 are 1, 3, 9.

Factors of 15 are 1, 3, 5, 15.

Common factors are 1, 3.

Chapter 3 Playing With Numbers Exercise 3.4

Question 1. Find the common factors of:

a) 20 and 28

b) 15 and 25

c) 35 and 50

d) 56 and 120.

Solution. a) 20 and 28.

Factors of 20 are 1, 2, 4, 5, 10 and 20.

Factors of 28 are 1, 2, 4, 7, 14 and 28.

Common factors are 1, 2, 4.

b) 15 and 25

Factors of 15 are 1, 3, 5 and 15.

Factors of 25 are 1, 5 and 25:

Common factors are 1,5.

c) 35 and 50

Factors of 35 are 1, 5, 7 and 35.

Factors of 50 are 1, 2, 5, 10, 25 and 50.

Common factors are 1, 5.

d) 56 and 120

Factors of 56 are 1, 2, 4, 7, 8, 14, 28, 56.

Factors of 120 are 1, 2, 3, 4, 5, 6, 10, 12, 20, 24, 30, 40, 60, 120.

Common factors are 1, 2, 4.

Question 2. Find the common factors of:

a) 4, 8 and 12

b) 5, 15 and 25

Solution. a) 4, 8 and 12

Factors of 4 are 1, 2, 4.

Factors of 8 are 1, 2, 4 and 8.

Factors of 12 are 1, 2, 3, 4, 6, 12.

Common factors are 1, 2 and 4.

b) 5, 15 and 25

Factors of 5 are 1 and 5.

Factors of 15 are 1, 3, 5 and 15.

Factors of 25 are 1,5 and 25.

Common factors are 1, 5.

Question 3. Find first three common multiples of:

a) 6 and 8

b) 12 and 18

Solution. a) 6 and 8

Multiples of 6 are 6, 12, 18, 24, 30, 30, 36, 42, 48, 54, 60, 66, 72, 78, ……….

Multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72,…….

First three common multiples are 24, 48 and 72.

b) 12 and 18

Multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, ………

Multiples of 18 are 18, 36, 54, 72, 90, 108, 126 …….

First three common multiples are 36, 72, 108.

Question 4. Write all the numbers less than 100 which are common multiples of 3 and 4.

Solution. The multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105…..

Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100…..

Common multipies of 3 and 4 are 12, 24, 36, 48, 60,72, 84, 96, 108……

∴ Common multiples less than 100 are 12, 24, 36, 48, 60, 72, 84, 96

Question 5. Which of the following numbers are co-prime?

a) 18 and 35

b) 15 and 37

c) 30 and 415

d) 17 and 68

e) 216 and 215

f) 81 and 16

Solution. Two numbers having 1 as a common factor are called co-prime numbers.

a) 18 and 35

Factors of 18 are 1, 2, 3, 6, 9 and 18.

Factors of 35 are 1, 5, 7 and 35.

Common factor is 1.

∴ 18 and 35 are co-prime numbers.

b) 15 and 37

Factors of 15 are 1, 3, 5 and 15.

Factors of 37 are 1 and 37.

Common factor is 1.

∴ 15 and 37 are co-prime numbers.

c) 30 and 415

Factors of 30 are 1, 2, 3, 5, 6, 10, 15. and 30.

Factors of 415 are 1, 5, 83 and 415

Common factors are 1 and 5

∴ 30 and 415 are not co-primes.

d) 17 and 68

Factors of 17 are 1 and 17.

Factors of 68 are 1, 2, 4, 17, 34 and 68.

Common factors are 1 and 17.

∴ 17 and 68 are not co-prime numbers.

e) 216 and 215

Factors of 216 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108 and 216.

Factors of 215 are 1, 5, 43, and 215.

Common factors is 1 only.

∴ 216 and 215 are co-prime numbers.

f) 81 and 16

Factors of 81 are 1, 3, 9, 27 and 81

Factors of 16 are 1, 2, 4, 8 and 16.

Common factor is 1.

∴ 81 and 16 are co-prime numbers.

Question 6. A number is divisible by both 5 and 12. By which other number will that number be always divisible?

Solution. The number divisible by both 5 and 12 is 60(12 × 5). The other number by which 60 is always divisible is 15. (15 × 4 = 60)

Question 7. A number is divisible by 12. By what other numbers will that number be divisible?

Solution. Factors of 12 are 1, 2, 3, 4, 6 and 12.

∴ The number is also divisible by 1, 2, 3, 4 and 6 also.

Question 8. Write the prime factorisation of 16, 28, 38.

Solution. 16 = 2 x 2 x 2 x 2

28 = 2 × 2 × 7

38 = 2 × 19

Haryana Board Class 6 Maths Solutions For  Chapter 3 Playing With Numbers Exercise 3.5

Question 1. Here are two different factor trees for 60. Write the missing numbers.

a)

Solution.

b)

Solution.

Question 2. Which factors are not included in the prime factorisation of a composite number?

Solution. 1 and composite factors are not included in the prime factorisation of a composite number.

Question 3. Write the greatest four digit number and express it in terms of its prime factors.

Solution. The greatest four digit number is 9999.

∴ 99993 x 3 x 11 x 101 x 1

Factors and multiples Class 6 HBSE

Question 4. Write the smallest five digit number and express it into the form of prime factors.

Solution. The smallest five digit number is 10000

10000 = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5 x 1.

Question 5. Find all the prime factors of 1729 and arrange them in ascending order. Now state the relation, if any; between two consecutive prime factors.

Solution.

1729 = 7 x 13 × 19 × 1

All the prime factors of 1729 are 7, 13 and 19.

The ascending order is 7, 13, 19.

We, observe that the difference is the same between two consecutive prime factors.

13 – 7 = 6

19 – 13 = 6

Question 6. The product of three consecutive number is always divisible by 6. Verify this statement with the help of some examples.

Solution. Example 1:

Let us take three consecutive numbers be 21, 22 and 23.

21 is divisible by 3

22 is divisible by 2

∴ 21 x 22 is divisible by 6.

∴ 21 x 22 x 23 is divisible by 6.

Example 2:

Let the three consecutive numbers be 47, 48 and 49.

48 is divisible by both 2 and 3.

(2 x 3 = 6)

∴ 47 x 48 x 49 is divisible by 6.

Question 7. The sum of two consecutive odd num- ber is divisible by 4. Verify this statement with the help of some examples.

Solution. Example 1:

Let us take two consecutive odd numbers 13 and 15

Sum of the number = 13 + 15 = 28.

∴ 28 is divisible by 4.

Example 2:

Let the two consecutive odd numbers be 17 and 19

Sum of the numbers = 17 + 19 = 36

∴ 36 is divisible by 4.

Example 3:

Let the two consecutive odd numbers be 25 and 27.

Sum of the numbers = 25 + 27 = 52

∴ 52 is divisible by 4.

Question 8. In which of the following expressions, prime factorisation has been done?

a) 24 = 2 x 3 x 4

b) 56 = 7 x 2 x 2 x 2

c) 70 = 2 x 5 x 7

d) 54 = 2 × 3 × 9

Solution. a) Prime factorisation has been done.

b) Prime factorisation has not been done.

c) Prime factorisation has been done.

d) Prime factorisation has not been done.

Question 9. 18 is divisible by both 2 and 3. It is also divisible by 2 × 3 = 6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4 × 6 = 24? If not, give an example to justify your answer.

Solution. No! we cannot conclude that the number will be divisible by 4 × 6 = 24 if it is divisible by both 4 and 6 because 4 and 6 are not co-prime numbers they have two common factors 1 and 2.

Example: 36 is divisible by 4 and 6 but 36 is not divisible by 24.

Question 10. I am the smallest number, having four different prime factors. Can you find me?

Solution. The four different prime factors = 2, 3, 5,7.

To find out the smallest number, we have to write them in their ascending order and multiply 2 × 3 × 5 × 7 = 210.

210 is the smallest number, having four different prime factors 2,3,5 and 7.

Question 11. Find the HCF of the following:

1) 24 and 36

2) 15, 25 and 30

3) 8 and 12

4) 12, 16 and 28.

Solution. 1) 24 and 36

Factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.

Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.

Highest common factor is 12.

∴ HCF of 24 and 36 is 12.

2) 15, 25 and 30

Factors of 15 are 1, 3, 5 and 15

Factors of 25 are 1,5 and 25

Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30.

∴ HCF of 15, 25 and 30 is 5.

3) 8 and 12

Factors of 8 are 1, 2, 4 and 8

Factors of 12 are 1, 2, 3, 4, 6 and 12

HCF of 8 and 12 is 4.

4) 12, 16 and 28

Factors of 12 are 1, 2, 3, 4, 6 and 12

Factors of 16 are 1, 2, 4, 8 and 16

Factors of 28 are 1, 2, 4, 7, 14 and 28.

HCF of 12, 16 and 28 is 4.

Haryana Board Class 6 Maths Solutions For  Chapter 3 Playing With Numbers Exercise 3.6

Question 1. Find the HCF of the following numbers:

a) 18, 48

b) 30,42

c) 18, 60

d) 27,63

e) 36,84

f) 34, 102

g) 70, 105, 175

h) 91, 112, 49

j) 12, 45, 75

i) 18, 54, 81

Solution. a) 18, 48

Factors of 18 are 1, 2, 3, 6, 9 and 18.

Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.

Common factors are 1, 2, 3 and 6.

∴ HCF of 18 and 48 is 6.

b) 30, 42

Factors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30.

Factors of 42 are 1, 2, 3, 6, 7, 14, 21 and 42.

Common factors are 1, 2, 3 and 6.

∴ HCF of 30 and 42 is 6.

c) 18, 60

Factors of 18 are 1, 2, 3, 6, 9 and 18.

Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.

Common factors are 1, 2, 3, and 6.

∴ HCF of 18 and 60 is 6.

d) 27, 63

Factors of 27 are 1, 3, 9 and 27.

Factors of 63 are 1,3,7,9, 21 and 63.

Common factors are 1, 3 and 9.

∴ HCF of 27 and 63 is 9.

e) 36, 84

Factors of 36 are 1,2,3,4,6,9, 12, 18 and 36.

Factors of 84 are 1, 2, 3, 4, 7, 12, 14, 21, 42 and 84.

Common factors are 1, 2, 3, 4 and 12.

∴ HCF of 36 and 84 is 12.

f) 34, 102

Factors of 34 are 1, 2, 17 and 34.

Factors of 102 are 1, 2, 3, 6, 17, 34, 51 and 102.

Common factors are 1, 2, 17 and 34.

∴ HCF of 34 and 102 is 34.

g) 70, 105, 175

Factors of 70 are 1, 2, 5, 7, 10, 14, 35 and 70.

Factors of 105 are 1, 3, 5, 7, 15, 21, 35 and 105:

Factors of 175 are 1, 5, 7, 25, 35 and 175.

Common factors are 1, 5 and 35.

∴ HCF of 70, 105 and 175 is 35.

h) 91, 112, 49

Factors of 91 are 1, 7, 13 and 91.

Factors of 112 are 1, 2, 4, 7, 8, 14, 16, 28, 56 and 112.

Factors of 49 are 1, 7 and 49.

Common factors are 1 and 7.

∴ HCF of 91, 112, 49 is 7.

i) 18, 54, 81

Factors of 18 are 1, 2, 3, 6, 9 and 18.

Factors of 54 are 1, 2, 3, 6, 9, 18, 27 and 54.

Factors of 81 are 1, 3, 9, 27 and 81.

Common factors are 1, 3, 9.

∴ HCF of 18, 54, 81 is 9.

j) 12, 45, 75

Factors of 12 are 1, 2, 3, 4, 6 and 12.

Factors of 45 are 1,3,5,9, 15 and 45.

Factors of 75 are 1, 3, 5, 15, 25 and 75.

Common factors are 1 and 3.

∴ HCF of 12, 45 and 75 is 3.

Question 2. What is the HCF of two consecutive.

a) Numbers ?

b) Even numbers?

c) Odd numbers?

Solution. a) Numbers

HCF of two consecutive numbers is 1.

b) Even numbers

HCF of two consecutive even number is 2.

c) Odd numbers

HCF of two consecutive odd numbers is 1.

Question 3. HCF of co-prime numbers 4 and 15 was found as follows by factorisation : 4 = 2 × 2 and 15 = 3 x 5 since there is no common prime factor, so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?

Solution. No, the answer is not correct.

The HCF of 4 and 15 is 1.

Haryana Board Class 6 Maths Solutions For  Chapter 3 Playing With Numbers Exercise – 3.7

Question 1. Renu purchases two bags of fertiliser of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertiliser exact number of times.

Solution. Factors of 75 are 1, 3, 5, 15, 25 and 75

Factors of 69 are 1, 3, 23 and 69.

Common factors are 1 and 3.

HCF of 75 and 69 is 3.

The maximum capacity of weight which can measure the weight of the fertiliser exact number of times is 3 kg.

Question 2. Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?

Solution.

LCM of 63, 70 and 77

= 2 x 3 x 3 x 5 x 7 x 11 = 6930

The minimum distance each should cover so that all cover the distance in complete steps is 6930 cm. or 69 m 30 cm.

Word problems on HCF and LCM for Class 6 HBSE

Question 3. The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimen- sions of the room exactly.

Solution. Factors of 825 are 1, 3, 5, 11, 15, 25, 33, 55, 75, 165, 275 and 825.

Factors of 450 are 1, 2, 3, 5, 6, 9, 10, 15, 18,25,30,45,50, 75, 90, 150, 225 and 450.

Common factors are 1, 3, 5, 15, 25 and 75.

HCF = 75

The length of largest tape which can measure the three dimensions of the room exactly is 75 cm.

Question 4. Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.

Solution.

L.C.M = 2 x 2 x 3 x 1 x 2 x 1 = 24

Multiples of 24 are 24, 48, 72, 96, 120, 144……

The smallest 3-digit number which is exactly divisible by 6, 8 and 12 is 120.

Question 5. Determine the greatest 3-digit num- ber exactly divisible by 8, 10 and 12.

Solution.

LCM = 2 × 2 × 2 × 3 × 5 = 120

Multiples of 120 are 120, 240, 360, 480, 600, 720, 840, 960, 1080 ……….

∴ The largest 3-digit number exactly divisible by 8, 10 and 12 is 960.

Question 6. The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m. at what time will they change simultaneously again?

Solution.

L.C.M. = 2 x 2 x 2 x 2 x 3 x 3 x 3 = 432

432 seconds = \(\frac{432}{60}\) = 7 min. 12 seconds.

(1 minute = 60 secs)

∴ They will change simultaneously again at 7 min. 12 seconds after 7 am:.

Question 7. Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times.

Solution. Factors of 403 are 1, 13, 31 and 403.

Factors of 465 are 1, 3, 5, 15, 31, 93, 155 and 465.

Factors of 434 are 1, 2, 7, 14, 31, 62, 217 and 434.

Common factors are 1 and 31. HCF = 31 The maximum capacity of the container that can measure the diesel of the three containers exact number of times is 31 litres.

Question 8. Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.

Solution.

LCM = 2 x 3 x 3 x 5 = 90

∴ Required number = 90 + 5 = 95.

Question 9. Find the smallest four digit number which is divisible by 18, 24 and 32.

Solution.

LCM = 2 × 2 × 2 x 2 x 3 x 3 x 2 = 288

Multiples of 288 are 288 × 1

288 × 1 = 288

288 x 2 = 576

288 × 3 = 864

288 x 4 = 1152

∴ The smallest 4-digit number which is divisible by 18, 24 & 32 is 1152.

Question 10. Find the LCM of the following numbers:

a) 9 and 4

b) 12 and 5

c) 6 and 5

d) 15 and 4.

Solution. a) 9 and 4

LCM of 9 and 4 is 9 x 4 = 36

b) 12 and 5

LCM of 12 and 5 is 12 x 5 = 60

c) 6 and 5

LCM of 6 and 5 is 6 x 5 = 30

d) 15 and 4

LCM of 15 and 4 is 15 x 4 = 60

We observe a common property in the obtained LCM’s that the LCM is the product of two numbers in each case and also the LCM is always a multiple of 3.

Question 11. Find the LCM of the following numbers in which one number is the factor of the other.

a) 5,20

b) 6, 18

c) 12,48

d) 9,45

What do you observe in the results obtained?

Solution. a) 5, 20

5 = 5 x 1

20 = 2 × 2 × 5

LCM of 5 and 20 = 2 x 2 x 5 = 20

b) 6,18

6 = 2 × 3

18 = 2 x 3 x 3

LCM of 6 and 18 is 2 × 3 × 3 = 18

c) 12, 48

12 = 2 × 2 × 3

48 = 2 x 2 x 2 × 2 × 3

LCM of 12 and 48

= 2 × 2 × 2 × 2 × 3 = 48.

d) 9, 45

9 = 3 × 3

45 = 3 × 3 × 5

LCM of 9 and 45 = 3 x 3 x 5 = 45

In the results obtained, we observe that the LCM of the two numbers in which one number is a factor of the other is the greater number.

Haryana Board Class 6 Maths Solutions For  Chapter 3 Playing With Numbers Very Short Answer Questions

Question 1. Are the numbers 452, 673, 259, 356 divisible by 2? Verify.

Solution. A number is divisible by 2, if it has any of the digits 0, 2, 4, 6 and 8 in its ones place.

∴ The numbers 452 and 356 have 2 and 6 in their ones place and the other num- bers does not have.

∴ 452 and 356 are divisible by 2.

Question 2. Is 8430 divisible by 6? Why?

Solution. If a number is divisible by both 2 and 3 then it is also divisible by 6.

The ones place of 8430 is 0, so it is divisible by 2.

The sum of digits of 8430 is 8 + 4 + 3 + 0 = 15.

15 is multiple of 3. So 8430 is divisible by 3.

∴ The number 8430 is divisible by both 2 and 3. It is divisible by ‘6’.

Question 3. Test whether 6669 is divisible by 9.

Solution. The sum of digits of 6669

= 6 + 6 + 6 + 9 = 27

It is a multiple of 9. So the number 6669 is divisible by 9.

Question 4. Without actual division, find whether 8989794 is divisible by 9.

Solution. The sum of digits of 8989794

= 8 + 9 + 8 + 9 + 7 + 9 + 4 = 54

It is a multiple of 9. So the number 8989794 is divisible by 9.

Question 5. Find the smallest number that must be added to 128, so that it becomes exactly divisible by 5.

Solution. Given number is 128

If the number is exactly divisible by 5, its ones place should have the digits 5 or 0.

∴ The nearest number to 128 which is having 0 or 5 in one’s place is 130.

∴ The smallest number that must be added is 130 – 128 = 2

Question 6. What is the smallest prime number?

Solution. 2 is the smallest prime number.

Question 7. What is the smallest composite number?

Solution. 4 is the smallest composite number.

Question 8. What is the smallest odd prime number?

Solution. 3 is the smallest odd prime number.

Question 9. What is the smallest odd composite number?

Solution. 9 is the smallest odd composite number.

Question 10. Write 10 odd and 10 even composite numbers.

Solution. 10 odd composite numbers = 9, 15, 21, 25, 27, 33, 35, 39, 45, 49.

10 even composite numbers = 4, 6, 8, 10, 12, 14, 16, 18, 20, 22.

Haryana Board Class 6 Maths Solutions For  Chapter 3 Playing With Numbers Short Answer Questions

Question 11. Are the numbers 28570, 90875 divisible by 5? Verify by actual division also.

Solution. The number with zero or five at ones place is divisible by 5.

In the numbers 28570 and 90875; the digits at ones place are 0 and 5. So these numbers are divisible by 5.

Verifying by actual division:

Question 12. Find the smallest number that has to be subtracted from 276 so that it becomes exactly divisible by 10.

Solution. Given number is 276

The digit in ones place is 6.

For the number exactly divisible by 10, the digit in ones place must be ‘0’

∴ The smallest number that has to be subtracted = 276 – 6 = 270

∴ The smallest number that has to be subtracted is 6.

Question 13. Write the greatest four digit number which is divisible by 9. Is it divisible by 3? What do you notice? 

Solution. The greatest four digit number which is divisible by 9 is 9999.

∴ The sum of digits = 9 + 9 + 9 + 9 = 36

36 is divisible by 3.

∴ 9999 is divisible by 3.

∴ Yes, the greatest four digit number which is divisible by 9 is also divisible by 3.

∴ We noticed that the numbers which are divisible by 9 are also divisible by 3.

Question 14. Write the nearest number to 12345 which is divisible by 4.

Solution. A number is divisible by 4 if the number formed by the last two digits is divisible by 4.

Given number is 12345.

Number formed by last two digits is 45.

If 1 is subtracted from it.

45 – 1 = 44 is divisible by 4.

So the nearest number to 12345 divisible by 4 is 12344.

Question 15. Find the factors of 60.

Solution. Factors of 60 = 1 × 60

= 2 x 30

= 3 x 20

= 4 x 15

= 5 x 12

= 6 x 10.

= 10 x 6

= 12 x 5

= 15 x 4

= 20 x 3

= 30 x 2

= 60 × 1

∴ The factors of 60 are = 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.

Chapter 3 Playing With Numbers Long Answer Questions

Question 16. Take any three 4 digit numbers and check whether they are divisible by 6.

Solution. Let us take three four digits numbers.

1) 3420 2) 7942 3) 6036

1) 3420

The ones place of 3420 is 0. So it is divisible by 2.

The sum of digits = 3 + 4 + 2 + 0 = 9.

It is divisible by 3.

∴ The number 3420 is divisible by both 2 and 3. So it is divisible by 6 also.

2) 7942

The ones place of 7942 is 2. So it is divisible by 2.

The sum of digits = 7 + 9 + 4 + 2 = 22.

It is not divisible by 3.

∴ The number 7942 is divisible by 2 but not divisible by 3. So it is not divisible by 6.

3) 6036

The ones place of 6036 is 6. So it is divisible by 2.

The sum of digits = 6 + 0 + 3 + 6 = 15.

It is divisible by 3.

∴ The number 6036 is divisible by both 2 and 3. So it is divisible by 6 also.

Question 17. Check whether the numbers 598, 864, 4782 and 8976 are divisible by 4. Use divisibility rule and verify by ac- tual division.

Solution. Divisibility rule of 4:

A number is divisible by 4 if the number formed by its last two digits. (i.e., ones and tens) is divisible by 4.

1) 598

The number formed by last two digits is 98.

98 is not divisible by 4.

∴ 598 is not divisible by 4.

2) 864

The number formed by last two digits is 64.

64 is divisible by 4.

∴ 864 is divisible by 4.

3) 4782

The number formed by last two digits is 82.

82 is not divisible by 4.

∴ 4782 is not divisible by 4.

4) 8976

The number formed by last two digits is 76.

76 is divisible by 4.

∴ 8976 is divisible by 4.

Question 18. Determine which of the following numbers are divisible by 5 and by 10.

25, 125, 250, 1250, 10205, 70985, 45880.

Check whether the numbers that are divisible by 10 are divisible by 2 and 5.

Solution. 25, 125, 10205, 70985, 250, 1250, 45880

These numbers are having ‘5’ or ‘0’ in the ones place.

∴ These numbers are divisible by 5

250, 1250, 45880

These numbers having ‘0’ in the ones place

∴ These numbers are divisible by 10

These numbers are divisible by ‘5’

these numbers are having ‘0’ in the ones place,

These numbers are also divisible by 2

∴ The numbers which are divisible by 10 are divisible by 2 and 5 also.

Difference between factors and multiples Class 6

Question 19. Write all the numbers between 100 and 200 which are divisible by 6.

Solution. The numbers divisible by 2 and 3 are divisible by 6.

The numbers divisible by 2 between 100 and 200 are = 102, 104, 106, 108, 110, 112, 114, 116, 118, 120, 122, 124 ………… 196, 198

The numbers divisible by 3 between 100 and 200 are = 102, 105, 108, 111, 114, 117, 120, 123, 126 …….. 195,198

The numbers which are divisible by both 2 and 3 are = 102, 108, 114, 120, 126, 132, 138, 144, 150, 156, 162, 168, 174, 180, 186, 192, 198

So all these numbers are divisible by 6.

Question 20. Using divisibility rules, determine which of the following numbers are divisible by 11.

1) 6446

2) 10934

3) 7138965

4) 726352

Solution. A number is divisible by 11, if the difference between the sum of the digits at odd places and the sum of the digits at even places (from the right) is either ‘0’ or multiple of 11.

1) 6446

The sum of digits at odd places = 6 + 4 = 10

The sum of digits at even places = 4 + 6 = 10

∴ Their difference = 10 – 10 = 0

So 6446 is divisible by 11.

2) 10934

The sum of digits at odd places = 4 + 9 + 1 = 14

The sum of digits at even places = 3 + 0 = 3

∴ Their difference = 14 3 = 11

It is a multiple of 11.

So 10934 is divisible by 11.

3) 7138965

The sum of digits at odd places = 5 + 9 + 3 + 7 = 24

The sum of digits at even places = 6 + 8 + 1 = 15

∴ Their difference = 24 – 15 = 9

It is not a multiple of 11.

So 7138965 is not divisible by 11.

4) 726352

The sum of digits at odd places = 2 + 3 + 2 = 7

The sum of digits at even places = 6 + 8 + 1 = 15

∴ Their difference = 18 – 7 = 1

It is a multiple of 11.

So 726352 is divisible by 11.

Haryana Board Class 6 Maths Solutions For  Chapter 3 Playing With Numbers Objective Type Questions

Choose the correct answer :

Question 1. In the following which is a composite number?

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

Answer. 4. 4

Question 2. Is 107 a prime number?

1. yes
2. no
3. can’t say
4. infinite

Answer. 1. yes

Question 3. Divide 29 by 4 it gives quotient

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

Answer. 2. 7

Question 4. A number is completely divisible by another number it leaves a remainder

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

Answer. 4. 0

Question 5. Which of the following numbers is divisible by 9?

1. 197235
2. 726352
3. 988975
4. 345880

Answer. 1. 197235

Question 6. Factors of 12 are

1. 2,3
2. 1,2,3
3. 1,2,3,4,6
4. 1, 2, 3, 4, 6, 12

Answer. 4. 1, 2, 3, 4, 6, 12

Question 7. The HCF of two consecutive numbers is

1. 1
2. The smallest number
3. The greatest number
4. 0

Answer. 1. 1

Question 8. The greatest two digit prime number.

1. 77
2. 79
3. 97
4. 99

Answer. 3. 97

Question 9. The number of factors of 80

1. 8
2. 10
3. 11
4. 12

Answer. 2. 10

Question 10. The another name for GCD

1. LCM
2. HCF
3. GCM
4. LCF

Answer. 2. HCF

Question 11. The number divisible by 4

1. 200094
2. 249874
3. 249884
4. 249854

Answer. 3. 249884

Question 12. The least two digit prime number

1. 11
2. 10
3. 12
4. 13

Answer. 1. 11

Question 13. 512215 is divisible by …..

1. 2
2. 4
3. 11
4. 6

Answer. 3. 11

Question 14. Example for twin prime numbers.

1. (41,43)
2. (42, 44)
3. (43,45)
4. (44, 46)

Answer. 1. (41,43)

Question 15. Number of common factors of 12 and 18 is

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

Answer. 2. 4

Question 16. The least prime number is

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

Answer. 1. 2

Question 17. 51225 is divisible by.

1. 51
2. 2
3. 4
4. 5

Answer. 4. 5

Question 18. HCF of 3, 4 is.

1. 0.5
2. 1
3. 2
4. 3

Answer. 2. 1

Question 19. Choose the correct matching.

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

Answer. 4. a-2, b-4, c-1, d-3

Question 20. Which of the following is not divisible by 2?

1. 11150
2. 63874
3. 81111
4. 1496

Answer. 3. 81111

Question 21. Greatest factor of 120 is………….

1. 8
2. 15
3. 120
4. 12

Answer. 3. 120

Question 22………… is neither prime nor composite.

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

Answer. 3. 1

Question 23. LCM of 8 and 12 is…………….

1. 11
2. 13
3. 12
4. 24

Answer. 4. 24

Question 24. HCF of 40, 56 and 60 is…………….

1. 8
2. 4
3. 60
4. 16

Answer. 2. 4

Question 25. How many prime numbers less than 100 are there?

1. 15
2. 25
3. 30
4. 26

Answer. 2. 25

Question 26. 42 = ……………..

1. 7 × 3 × 2
2. 7 × 7 x 9
3. 2 × 21 × 3
4. 7 × 6 × 4

Answer. 1. 7 x 3 x 2

Question 27. Identify a palindrome number.

1. 1225
2. 1331
3. 11452
4. 31119

Answer. 2. 1331

Question 28. 2 x 2 × 3 × 7 = ……….

1. 84
2. 62
3. 81
4. 22

Answer. 1. 84

Question 29. Number of factors of 36 is ………

1. 6
2. 11
3. 10
4. 9

Answer. 9

Question 30. The smallest odd composite number is …..

1. 17
2. 15
3. 9
4. 16

Answer. 3. 9

Question 31. Identify the correct statement from the following:

1. 733 is divisible by 3
2. The greatest prime number after 20 is 23
3. 4, 6, 8, 10, 12 are even composite numbers
4. 7221 is a factor of 11

Answer. 3. 4, 6, 8, 10, 12 are even composite numbers

Question 32. Factors of 115 are.

1. 5
2. 23
3. 115
4. All the above

Answer. All the above

Question 33. How many prime numbers are there in between 10 and 30?

1. 9
2. 6
3. 8
4. 19

Answer. 2. 6

Question 34. 9846 is not divisible by …….

1. 9
2. 3 only
3. 2 only
4. 5

Answer. 5

Question 35. Sum of the digits of 36129 is

1. 16
2. 3
3. 21
4. 8

Answer. 3. 21

Question 36. How many twin primes are below 20?

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

Answer. 4. 4

Question 37. 53 = …………..

1. 13 + 17 + 20
2. 13 + 17 + 23
3. 18 + 10 + 16
4. 17 + 13 + 33

Answer. 2. 13 + 17 + 23

Question 38. HCF of 28, 35 is ………

1. 6
2. 9
3. 7
4. 8

Answer. 7

Question 39. LCM of 10 and 11 is ……..

1. 119
2. 122
3. 111
4. 110

Answer. 4. 110

Question 40. LCM of a and b if HCF of a, b = 1.

1. \(\frac{a}{b}\)
2. a+b
3. ab
4. a-b

Answer. 3. ab

Question 41. Two prime numbers are said to be twin primes, if they differ each other by

1. 0
2. 1
3. 2
4. None

Answer. 3. 2

Question 42. Choose the correct matching.

1. a-1, b-2, c-3
2. a-3, b-2, c-1
3. a-1, b-3, c-2
4. a-2, b-3, c-1

Answer. 2. a-3, b-2, c-1

Question 43. The numbers which have 1 as the common factor are called…….

1. co-prime
2. relatively prime
3. Both A & B
4. None

Answer. 3. Both A & B

Question 44. The H.C.F. of any two twin primes is………….

1. 0
2. 1
3. 2
4. None

Answer. 2. 1

Question 45. The product of L.C.M and H.C.F of two numbers =

1. Sum of two numbers
2. Difference of two numbers
3. Both A & B
4. Product of two numbers

Answer. 4. Product of two numbers

Question 46. A number which divides the other number with out remainder is called

1. factor
2. multiple
3. prime number
4. composite number

Answer. 1. factor

Question 47. The number which is divisible by 1 and itself is called

1. factor
2. multiple
3. prime number
4. composite number

Answer. 3. Prime number

Question 48. If the L.C.M of 8 and 12 is 24 then their H.C.F. is

1. 3
2. 4
3. 16
4. 96

Answer. 2. 4

Question 49. If the H.C.F. of 18 and 27 is 9 then their L.C.M is

1. 9
2. 486
3. 54
4. None

Answer. 3. 54

Question 50. ……………. is a factor of 12.

1. 7
2. 9
3. 4
4. 18

Answer. 3. 4

Question 51. ……………… is a factor of every number.

1. 3
2. 12
3. 0
4. 1

Answer. 4. 1

Question 52. The fifth multiple of 6 is.

1. 30
2. 35
3. 40
4. 45

Answer. 1. 30

Question 53. Every prime number except ………. is odd.

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

Answer. 4. 2

Question 54. 128 is divisible by

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

Answer. 1. 4

Question 55. 2 x 2 x 2 x 2 = ……….

1. 18
2. 8
3. 12
4. 16

Answer. 4. 16

Question 56. HCF of 24 and 36 is ….

1. 14
2. 36
3. 12
4. 24

Answer. 3. 12

Question 57. LCM of 24 and 90 is …..

1. 360
2. 300
3. 460
4. 306

Answer. 1. 360

Question 58. Which of the following number is divisible by 2?

1. 133
2. 124
3. 135
4. 111

Answer. 2. 124

Question 59. Two numbers having only 1 as a common factor are called …………..numbers.

1. co-prime
2. prime
3. odd
4. even

Answer. 1. co-prime

Question 60. Which of the following number is not divisible by 5?

1. 140
2. 222
3. 1000
4. 110
5. Answer. 2. 222

Haryana Board Class 6 Maths Solutions For  Chapter 3 Playing With Numbers Fill In The Blanks:

Question 61. HCF of 2021 and 2022 is

Answer. 1

Question 62. The least composite number is ………..

Answer. 4

Question 63. Number of primes below 98 is ………..

Answer. 25

Question 64. LCM of 2, 4, 6 ………………..

Answer. 12

Question 65. HCF of 3, 5, 19 is …………

Answer. 1

Question 66.

In the above factor tree of 112 the value of x = ………..

Answer. 7

Question 67. HCF of 30 and 48 is ……………

Answer. 6

Question 68. HCF of two consecutive even numbers is ………….

Answer. 2

Question 69. Express 44 as sum of two odd primes ………….

Answer. 13 + 31

Question 70. Number of factors of 80 is …………..

Answer. 10

Question 71. 2, 4, 6, 8 are examples of ………….numbers.

Answer. even

Question 72. 2 × 3 + 1 = …………

Answer. 7

Question 73. …………….. is neither a prime nor a composite number

Answer. 1

Question 74. 4 × 17 = ……………..

Answer. 68

Question 75. Number of factors of a given number are.

Answer. finite

Question 76. Every number is a factor of ………..

Answer. itself

Question 77. 1, 2, 4 are called …………………..

Answer. factors

Question 78. Sum of the digits of a number is divisible by 3, then it is divisible by

Answer. 3

Question 79. LCM of 8 and 12 is. ……….

Answer. 24

Question 80. HCF of any two consecutive numbers is ………..

Answer. 1

Question 81. LCM means ………….

Answer. Least Common Multiple

• Perimeter is the distance covered along the boundary forming a closed figure when you go round the figure once.
• The length of the boundary of a closed figure is called its ‘Perimeter’ (Or) Sum of all the lengths of a polygon is called its ‘Perimeter’.
• The perimeter of a rectangle = 2 [length + breadth]
• Perimeter of a square = 4 x [length of its side]
• Perimeter of a equilateral triangle = 3 x length of the side.
• Circumference [Perimeter] of a circle = πd (Or) 2πr
• The place occupied by an object is called area of the object. Area is measured in square units.
• Area of a rectangle = length x breadth
• Area of a square = side x side

Question 1. Meera went to a park 150 m long and 80m wide. She took one complete round on its boundary. What is the distance covered by her?

Solution.

Distance covered by Meera

= AB + BC + DC + DA

= (150 + 80 + 150 + 80) m

= 460 m

Haryana Board Class 6 Maths Mensuration solutions

Question 2. Find the perimeter of the following figures:

Solution. a)

Perimeter

= AB + BC + CD + DA

= 40 m + 10 m + 40 m + 10m

= 100 m

b)

Perimeter

= AB + BC + CD + AD

= 5m + 5m + 5m + 5m

= 20 m

c)

Perimeter

= AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA

= 1 cm + 3 cm + 3 cm + 1 cm + 3 cm + 3cm + 1 cm + 3 cm + 3 cm + 1 cm + 3 cm + 3 cm

= 28 cm

Perimeter and area formulas for Class 6 HBSE Maths

d)

Perimeter

= AB + BC + CD + DE + EF + FA

= 100 m + 120 m + 90m + 45m + 60 m + 80 m

= 495 m

Question 3. Find the perimeter of the following rectangles:

Solution. 1) Perimeter = 2(1 + b) = 2 (0.5 m + 0.25 m)

= 2(0.75) m = 1.5 m

Perimeter by adding all the sides

= 0.5 + 0.25 + 0.5 + 0.25

= 1.5 m

2) Perimeter = 2(1 + b)= 2(18 cm + 15 cm) = 66 cm.

Perimeter by adding all the sides

= 18 + 15 + 18 + 15 = 66 cm.

3) Perimeter = 2(1 + b)

= 2 (10.5 cm + 8.5 cm)

= 2 (19 cm) = 38 cm.

Perimeter by adding all the sides

= 10.5 + 8.5 + 10.5 + 8.5

= 38 cm.

Volume of cube and cuboid Class 6 HBSE Maths

Question 4. Find various objects from your surroundings which have regular shapes and find their perimeters.

1) Square:

Solution. 1) A square has four equal sides.

2) Let’s assume the length of one side is 5 cm.

3) The perimeter of a square is calculated by multiplying the length of one side by 4.

4) Therefore, the perimeter of the square would be 5 cm x 4 = 20 cm.

2) Rectangle:

Solution. 1) A rectangle has two pairs of equal sides (opposite sides)

2) Let’s assume the length of one side is 6 cm and the length of the adjacent side is 8 cm.

3) The perimeter of a rectangle is calculated by adding the lengths of all four sides.

4) Therefore, the perimeter of the rectangle would be (6 cm + 8 cm) x 2 = 28 cm..

3) Circle:

Solution. 1) A circle is a curved shape with a constant radius.

2) Let’s assume the radius of a circle is 3 cm.

3) The perimeter of a circle is called its circumference and is calculated using the formula: 2 × π × radius.

4) Therefore, the perimeter (circumference) of the circle would be

2 x 3.14 x 3 cm = 18.84 cm.

Please note that these are just examples, and the actual objects around you may have different dimensions and perimeters.

Haryana Board Class 6 Maths Solutions For Chapter 10 Mensuration Exercise 10.1

Question 1. Find the perimeter of each of the following figures:

Solution. a) Perimeter

5 cm + 1 cm + 2 cm + 4 cm

= 12 cm

b) Perimeter

= 40 cm + 35 cm + 23 cm + 35 cm

= 133 cm

c) Perimeter =

= 15 cm + 15 cm + 15 cm + 15 cm

= 60 cm

d) Perimeter = 4 cm + 4 cm + 4 cm + 4 cm + 4 cm = 20 cm

e) Perimeter = 1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5cm + 4cm = 15cm

f) Perimeter = 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm, + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm = 52 cm

Circumference and area of a circle Class 6 Haryana Board

Question 2. The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?

Solution. Length of rectangular box = 40 cm

Breadth of rectangular box = 10 cm

Length of the tape required

= 2 (length + breadth)

= 2(40 cm + 10 cm)

= 2 × 50 cm

= 100 cm

= 1 m

Question 3. A table top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?

Solution. Length = 2 m 25 cm = 2.25 m

Breadth = 1 m 50 cm = 1.5 m

Perimeter of top of the table

= 2(length + breadth)

= 2(2.25 + 1.5) m.

= 2(3.75 m)

= 7.5 m.

Question 4. What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?

Solution. Length = 32 cm

Breadth = 21 cm

Length of wooden strip required

= 2(length + breadth)

= 2(32 + 21) cm

= 2 x 53 cm

= 106 cm.

Question 5. A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?

Solution. Length = 0.7 km

Breadth = 0.5 km

Perimeter of a rectangle

= 2 (length + breadth)

= 2 (0.7 km + 0.5 km)

= 2.4 km.

Length of wire needed

= 4 × perimeter

= 4 × 2.4 km

= 9.6 km.

Mensuration Class 6 HBSE important questions

Question 6. Find the perimeter of each of the following shapes.

a) A triangle of sides 3 cm, 4 cm, and 5 cm.

Solution. Perimeter of the triangle

= 3 cm + 4 cm + 5 cm

= 12 cm

b) An equilateral triangle of side 9 cm.

Solution. Perimeter = 9 cm + 9 cm + 9 cm

= 27 cm

c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.

Solution. Perimeter of an isosceles triangle

= 8 cm + 8 cm + 6 cm

= 22 cm

Question 7. Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.

Solution. Perimeter of the triangle

= Sum of the lengths of sides

= 10 cm + 14 cm + 15 cm

= 39 cm.

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

Solution. Perimeter of regular hexagon

= 6 x length of side

= 6 × 8 m

= 48 m.

Question 9. Find the side of the square whose perimeter is 20 m.

Solution. Perimeter of square = 20 m

4 x length of the side = 20 m

Length of the side = \(\frac{20 \mathrm{~m}}{4}=5 \mathrm{~m}\)

Question 10. The perimeter of a regular pentagon is 100 cm. How long is its each side?

Solution. Perimeter of regular pentagon = 100 cm

5 x length of the side = 100 cm

Length of the side = \(\frac{100 \mathrm{~cm}}{5}=20 \mathrm{~cm} .\)

Question 11. A piece of string is 30 cm long. What will be the length of each side if the string is used to form:

a) a square?

b) an equilateral triangle ?

c) a regular hexagon?

Solution. a) Perimeter of a square = 30 cm

4 × length of side = 30 cm

length of side = \(\frac{30 \mathrm{~cm}}{4}=7.5 \mathrm{~cm}\)

b) Perimeter of equilateral triangle = 30 cm.

3 x length of side = 30 cm

Length of side = \(\frac{30 \mathrm{~cm}^{\prime}}{3}=10 \mathrm{~cm} .\)

c) Perimeter of regular hexagon = 30 cm

6 x length of side = 30 cm

Length of side = \(\frac{30 \mathrm{~cm}}{6}=5 \mathrm{~cm} .\)

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

Solution. Perimeter of a triangle = Sum of the lengths of 3 sides

36 cm = 12 cm + 14 cm + length of third side

36cm = 26 cm + length of third side

∴ length of third side

= 36 cm – 26 cm = 10 cm

Question 13. Find the cost of fencing a square park of side 250 m at the rate of ₹ 20 per metre.

Solution. Perimeter of square park

= 4 x length of one side.

= 4 x 250 m

= 1000 m

∴ Cost of fencing at the rate of ₹ 20 per metre.

= ₹ 1000 x ₹ 20

= ₹ 20,000

Question 14. Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of ₹ 12 per metre.

Solution. Perimeter of a rectangular park

= 2(length + breadth)

= 2(175 + 125)m

= 2(300)m

= 600 m

∴ Cost of fencing the rectangular park at the rate of ₹ 12 per metre.

= ₹ 12 × ₹ 600

= ₹ 7200

Question 15. Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length 60 m and breadth 45m. Who covers less distance?

Solution. Perimeter of a square park

= 4 x length of one side

= 4 x 75 m.

= 300 m.

Perimeter of a rectangular park

= 2(1 + b)

= 2 (60 m + 45 m)

= 2 (105 m)

= 210 m

here 300 m > 210 m

∴ Bulbul covers less distance.

Question 16. What is the perimeter of each of the following figure? What do you infer from the answers?

Solution. a) Perimeter = 25 cm + 25 cm + 25 cm +25 cm 100 cm.

b) Perimeter = 30 cm + 20 cm + 30 cm + 20 cm = 100 cm

c) Perimeter = 40 cm + 10 cm + 40 cm +10 cm = 100 cm

d) Perimeter = 40 cm + 30 cm + 30 cm = 100 cm

Inference: The perimeter of all the fig- ures is 100 cm.

Question 17. Avneet buys 9 square paving slabs, each with a side of \(\frac{1}{2}\) m. He lays them in the form of a square.

a) What is the perimeter of his arrangement in the above figure (1)?

b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement in the above figure (2)?

c) Which has greater perimeter?

d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges i.e. they can- not be broken.)

Solution. a) Perimeter of the figure(1) = 4 x side

Length of the side of the square

= 3 x square

Slabs x length of each side

= 3 x 0.5 m = 1.5 m(each side = \(\frac{1}{2}\)m)

Perimeter of Avneet’s arrangement

= 4 x 1.5m = 6m.

b) Peimeter of the figure (2)

= 5 x length of each side

= 5 x 0.5 = 2.5 m (each side = \(\frac{1}{2}\)m)

Perimeter of the cross = 4 x figure

= 4 x 2.5 = 10 m

c) Shari’s arrangement has greater perimeter.

d) No, but we can achieve the same perimeter as shari’s arrangement by arranging all the squares in a row.

Haryana Board Class 6 Maths Solutions For Chapter 10 Mensuration Exercise 10.2

Question 1. Find the areas of the following figures by counting squares.

Solution. a)

b)

∴ Total area of the figure = 5 sq.cm

c) Fully-filled squares = 2

Half-filled squares = 4

Area covered by fully squares = 2 × 1 = 2 sq. cm

Area covered by half squared = 4 x \(\frac{1}{2}\) = 2 sq. cm

Total area = 4 sq.cm.

d) Fully-filled squares = 8

∴ Total area = Area covered by full squares

=8 × 1 sq.cm

= 8 sq. cm

e) Fully filled squares = 10

∴ Total area = Area covered by full squares

= 10 × 1 sq cm.

= 10 sq.cm.

f) Fully-filled squares = 2

Half-filled squares = 4

Area covered by fully squares = 2 x 1 sq.cm

= 2 sq.cm

Area covered by half squares = 4 x \(\frac{1}{2}\) sq.cm

= 2 sq. cm.

∴ Total area = 2 sq.cm + 2 sq.cm

= 4 sq.cm

g) Fully-filled squares = 4

Half-filled squares = 4

Area covered by full squares

= 4 x 1 sq. cm = 4 sq. cm

Area covered by half squares

= 4 x \(\frac{1}{2}\) sq. cm = 2 sq. cm

∴ Total area 4 sq. cm + 2 sq. cm

= 6 sq. cm.

h) Fully-filled squares = 5

∴ Total area = Area covered by full squares

= 8 x 1 sq. cm = 5 sq. cm

i) Fully-filled squares = 9

∴ Total area = Area covered by full squares

= 9 × 1 sq. cm

= 9 sq. cm

j) Fully-filled squares = 2

Half filled squares = 4

Area covered by full squares

= 2 x 1 sq. cm

= 2 sq. cm

Area covered by half squares

= 4 x \(\frac{1}{2}\) sq.cm = 2 sq.cm

∴ Total area 2 sq. cm + 2 sq. cm

= 4 sq. cm

k) Fully filled squares = 4

Half filled squares = 2

Area covered by full squares

= 4 × 1 sq. cm

= 4 sq. cm

Area covered by half squares

= 2 x \(\frac{1}{2}\) sq.cm

= 1 sq. cm

∴ Total area = 4 sq. cm + 1 sq. cm

= 5 sq. cm

1) Fully-filled squares = 4 = 4 sq.cm

Half-filled squares

= 2 = 2 x \(\frac{1}{2}\) = 1 sq.cm

More than half filled squares

= 3 = 3 x 1 sq cm = 3 sq.cm

∴ Total area

= 4 sq. cm + 3 sq. cm

= 7 sq.cm

m) Fully filled squares = 7 x 1 sq.cm

= 7 sq.cm

Half-filled squares = More than half filled squares

= 7 x 1 sq.cm = 7 sq.cm

Less than half filled squares = 4

Total area = 7 sq. cm + 7sq.cm

= 14 cm.

n) Fully filled squares = 9 x 1 = 9 sq.cm

Half-filled squares

More than half filled squares

= 9 x 1 = 9 sq.cm

Total area = 9 sq. cm + 9 sq.cm

= 18 sq.cm

Question 2. Find the area of the floor of your classroom.

Solution. Let’s assume the length of the classroom is 10 meters and the width is 8 meters.

Area = 10 meters x 8 meters = 80 sq.m.

So, the area of the floor of the classroom would be 80 sq m.

Question 3. Find the area of any one door in your house.

Solution. The most common standard door size is 80 inches (6.67) feet) in height and 36 inches (3 feet) in width area of door

= Height x Width

Area of door = 80 inches x 36 inches = 2880 sq. inches.

To convert square inches to square feet, divide the area by 144 (since there are 144 square inches in a square foot):

Area of door (in square feet) = 2880 square inches/144 = 20 square feet.

There fore, the approximate area of standard -size door in a house is around 20 square feet.

Haryana Board Class 6 Maths Solutions For Chapter 10 Mensuration Exercise 10.3

Question 1. Find the areas of the rectangles whose sides are:

a) 3 cm and 4 cm

b) 12 m and 21 m

c) 2 km and 3 km

d) 2 m and 70 cm

Solution. a) 1 = 4 cm, b = 3 cm

Area of rectangle = 1 x b

= 4 x 3

= 12 sq. cm.

b) 1 = 21 m, b = 12 m

Area of rectangle = 1 x b

= 21 × 12

= 252 sq.m.

c) 1 = 3 km, b = 2 km

Area of rectangle = 1x b

= 3 × 2

= 16 sq.km.

d) 1 = 2 m, b = 70 cm = 0.7 m

Area of rectangle = 1 x b

= 2 × 0.7

= 1.4 sq.m.

Question 2. Find the areas of the squares whose sides are:

a) 10 cm

b) 14 cm

c) 5 m

Solution. a) a = 10 cm

Area of the square a² sq. cm

= 10² sq. cm

= 100 sq. cm

b) a = 14 cm

Area of the square = a² sq. cm

= 14² sq. cm

= 196 sq. cm

c) a = 5 m

Area of the square = a²

= 5² sq.cm

= 25 sq.cm

Question 3. The length and breadth of three rectangles are as given below:

a)9 m and 6 m

b) 17 m and 3 m

c) 4 m and 14 m

Which one has the largest area and which one has the smallest ?

Solution. a) 1 = 9 m, b = 6m

Area of rectangle = l x b

= 9 x 6

= 54 sq.m.

b) 1 = 17 m, b = 3 m

Area of rectangle = 1 x b

= 17 × 3

= 51 sq. m.

c) 1 = 14 m, b = 4m

Area of rectangle

= 1 x b

= 14 × 4

= 56 sq.m

The rectangle(c) has the largest area and the rectangle (b) has the smallest area.

Word problems on Mensuration for Class 6 HBSE

Question 4. The area of a rectangular garden 50 m long is 300 sq.m. Find the width of the garden.

Solution. Area of a rectangular garden = 300 sq.m.

length = 50 m.

1 x b = 300

50 x b = 300

b = \(\frac{300}{50}\) m

b = 6m

∴ Width of the garden = 6m

Question 5. What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of 8 per hundred sq. m.?

Solution. Length of a rectangular piece of land = 500 m

Breadth of a rectangular piece of land = 200 m

Area = length x breadth

= 500 m × 200 m

= 1,00,000 sq.m.

Cost of tiling per 100sq. m = ₹ 8

Cost of tiling 1,00,000 sq. m

= \(₹ \frac{8}{100} \times 100000\)

= ₹ 8000

Question 6. A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?

Solution. Length of the table = 2 m

Breadth of the table = 1 m 50 cm = 1.5 m

Area of the table = length x breadth

= 2 × 15 sq.m

= 3 sq.m.

Question 7. A room is 4m long and 3 m 50 cm wide. How many square metres of carpet is needed to cover the floor of the room?

Solution. Length of the room = 4 m = 4 m

Breadth of the room = 3 m 50 cm

= 3.50 m

∴ Area of the room

= length x breadth

= 4m x 3.50 m

= 14 sq.m.

∴ 14 sq. m of carpet is needed to cover the floor.

Question 8. A floor is 5m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.

Solution. Length of the floor = 5 m

Breadth of the floor = 4 m

Area of the floor = length x breadth

= 5 x 4 sq. m

= 20 sq.m

Area of the square carpet = a2

= (3)2 sq. m

= 9 sq. m.

Area of the floor that is not carpeted

= 20 sq.m. – 9 sq.m

= 11 sq.m.

Question 9. Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of land?

Solution. Area of square flower bed = a² sq. m

=(1)²sq. m

= 1 sq. m

∴ Area of 5 square flower beds

= 1 x 5 sq. m

= 5 sq. m

Length of the piece of land = 5 m

Breadth of the piece of land

= length x breadth

= 4 x 5 sq. cm

= 20 sq. m

Area of remaining part of land

= 20 sq. m – 5 sq. m

= 15 sq. m

Question 10. By splitting the following figures into rectangles find their areas.

(The measures are given in centimetres)

Solution. a) Area of the figure

= (3 x 3) + (1 x 2) + (3 x 3) + (4 × 2) sq. cm

= 9 + 2 + 9 + 8 sq.cm

= 28 sq. cm

b) Area of the figure

= (3 x 1 + 3 x 1 + 3 x 1) sq. m

= 3 + 3 + 3 sq. cm

= 9 sq. cm

Question 11. Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)

Solution. a)

Area of the shape

= (8 x 2 + 12 x 2) sq. cm

= 16 + 24 sq. cm

= 40 sq. cm

b)

Area of the shape

= (7×7) + (7×7) + (7×7) + (7×7) + (7×7) sq.m

=(49 + 49 + 49 + 49 + 49) sq. m

= 245 sq. cm

c)

Area of the shape

= (5 x 1)+(4 x 1) sq. cm

= (5 + 4) sq. cm

= 9 sq. cm

Question 12. How many tiles whose length and breadth are 12 cm and 15 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively.

a) 100 cm and 144 cm

Solution. 1 = 144 cm, b = 100 cm

Area of rectangle = 1 x b

= 144 × 100

= 14400 sq.cm

Length of the tile = 5 cm

Breadth of the tile = 12 cm

Area of rectangle = 1 x b

= 5 x 12 sq.cm

= 60 sq.cm

∴ Number of tiles needed to fit the

region = \(\frac{\text { Area of theregion }}{\text { Area of tile }}\)

= \(\frac{14400}{60}\) = 240.

b) 70 cm and 36 cm

Solution. 1 = 70 cm, b = 36 cm

Area of the region = 70 × 36 sq.cm = 2520 sq.cm

Area of the tile = 60 sq.cm

Number of tiles needed = \(\frac{2520}{60}\)

= 42 sq.cm

A challenge!

On a centimetre squared paper, make as many rectangles as you can, such that the area of the rectangle is 16 sq cm (consider only natural number lengths).

a) Which rectangle has the greatest perimeter?

b) Which rectangle has the least perimeter? If you take a rectangle of area 24 sq cm, what will be your answers? Given any area, is it possible to predict the shape of the rectangle with thegreatest perimeter? With the least perimeter? Give example and reason.

Solution. Three rectangles can be made as follows:

1) Sides 16 cm and 1 cm

Perimeter = 2(1 + b)

= 2(16 + 1) = 34 cm

2) Sides 8 cm and 2 cm

Perimeter = 2(l + b)

= 2(8 + 2) = 20 cm

3) Sides 4 cm and 4 cm

Perimeter = 2(l + b)

= 2(4 + 4) = 16 cm

a) The rectangle (1) has the greatest perimeter.

b) The rectangle (3) has the least perimeter. If the area is 24 sq. cm. Four rectangles can be made as follows.

1) Sides 24 cm and 1 cm

Perimeter = 2(1 + b)

= 2(24 + 1) = 50 cm

2) Sides 4 cm and 6 cm

Perimeter = 2(1 + b)

= 2(4 + 6) = 20 cm

3) Sides 8 cm and 3 cm

Perimeter = 2 (1 + b)

= 2(8 + 3) = 22 cm

4) Sides 12 cm and 2 cm

Perimeter = 2(l + b)

= 2(12 + 2) = 28 cm

Yes! it is possible to predict the shape of the rectangle with the (1) greatest perimeter. 2) least perimeter.

Difference between perimeter and area Class 6

Haryana Board Class 6 Maths Solutions For Chapter 10 Mensuration Very Short Answer Questions

Question 1. Find the perimeters of the given figures.

In the above figures (1) and (2), find the perimeter of ΔKLM, ΔKMN and ▢KLMN.

Solution. 1) Perimeter of ΔKLM

= KL + LM + KM

= 2 + 2.6 + 3.8 = 8.4 cm

Perimeter of ΔKMN

= KM + MN + KN

= 3.8 + 2 + 2.6 = 8.4 cm

2) Perimeter of ▢KLMN

= KL + LM + MN + KN

= 2 + 2.6 + 2 + 2.6 = 9.2 cm

Question 2. Find the perimeters of the following figures.

Solution. Perimeter of AXYZ

= XY + YZ + XZ

= 2 + 2 + 2 = 6 cm

2) Perimeter of a square ABCD

= AB + BC + CD + AD

= 3 + 3 + 3 + 3 = 12 cm

3) Perimeter of ▢PQRS

= PQ + QR + RS + PS

= 2 + 2 + 2 + 2 = 8 cm

Question 3. Find the area of a square with side 16 cm.

Solution.

Length of the side of a square = 16 cm

Area of a square = side x side

= 16 x 16 = 256 cm²

Question 4. Length and breadth of a rectangle are 16 cm and 12 cm respectively. Find its

Solution.

Length of a rectangle l = 16 cm

Its breadth = b = 12 cm

Area of a rectangle

= 1 x b

= 16 × 12

= 192 cm²

Question 5. Find the area of the rectangle of measurements 15cm and 8cm as length and breadth respectively.

Solution.

Length of a rectangle = 1 = 15 cm

Its breadth = b = 8 cm

Area of a rectangle = l x b

= 15 x 8

= 120 cm²

Question 6. Find the area of a square whose perimeter is 64m.

Solution. Perimeter of a square = 64 m

Perimeter of a square

= 4 x side(s)

= 4 x s = 64

s = \(\frac{64}{4}\) = 16

Side of a square(s) = 16 m

∴ Area of a square = s x s

= 16 x 16 = 256 m²

Question 7. Ritu went to a park 130 m long and 90 m wide. She took one complete round of it. What distance did she cover?

Solution.

Total distance covered by Ritu = Perimeter of the park ABCD

= AB + BC + CD + DA

= 130 + 90 + 130 + 90

= 440 m.

Question 8. Find the perimeter of given shape.

Solution.

IJ = DC = 3 cm; EF = HG = 2 cm

AB = LK = 4 cm; FG = KJ = CB = 1 cm

AL = BC + DE + FG + HI + JK

= 1 + 2 + 1 + 2 + 1 = 7 cm.

Perimeter

= AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA

= 4 + 1 + 3 + 2 + 2 + 1 + 2 + 2 + 3 + 1 + 4 + 7 = 32 cm.

Question 9. Length and breadth of a rectangle are 15 cm. and 10 cm. Find the perimeter.

Solution. Length of the rectangle = 15 cm.

Breadth = 10 cm.

Perimeter of the rectangle

= 2 (length + breadth)

= 2 (15 + 10)

= 2 x 25 = 50 cm.

Question 10. Find the perimeter of a rectangular field which is 36 m. long and 24 m. wide.

Solution. Length of the rectangle field (1) = 36 m.

Breadth of the field (b) = 24 m.

∴ Perimeter of the field = 2(1 + b)

= 2(36 + 24)

= 2 x 60 = 120 m.

Haryana Board Class 6 Maths Solutions For Chapter 10 Mensuration Short Answer Questions

Question 11. Find the perimeter of

1) An equilateral triangle whose side is 3.5 cm.

Solution.

The length of the side of an equilateral triangle = 3.5 cm

Perimeter of an equilateral triangle

= 3 x length of the side

= 3 x 3.5 – 10.5 cm

2) A square whose side is 4.8 cm.

Solution.

The length of the side of a square = s = 4.8 cm

Perimeter of square

= 4 x length of the side

= 4 x 4.8 = 19.2 cm

Question 12. Length and breadth of top of one table is 160cm and 90cm respectively. Find how much length of beading is required for each table.

Solution.

A table is in the shape of a rectangle.

Length of a table = l = 160 cm

Its breadth = b = 90 cm

∴ The length of beading is required for each table

= Perimeter of the rectangular table.

= 2[1 + b]

= 2[160 + 90] = 2 × 250 = 500 cm

Question 13. Find the perimeter and area of a square of side 4 cm. Are these same? Give some examples to support your answer.

Solution. Length of the side of a square = 4 cm Perimeter of a square

= 4 x side = 4 x 4 = 16 cm

Area of a square = s² = 4² = 16 cm²

In this condition perimeter and area of a square are equal.

For example length of the side be 6 cm.

Perimeter of a square

= 4 x s = 4 x 6 = 24 cm

Area of a square = s x s

= 6 × 6 = 36 cm

∴ Perimeter and area of a square need not be equal for all conditions.

Question 14. Find the area of the square whose perimeter is 48cm.

Solution. Perimeter of a square = 48 cm

Perimeter of a square = 4 × side

∴ 4 × side = 48

Side = \(\frac{48}{4}\) = 12

∴ Side of the square (s) = 12 cm

∴ Area of the square = s x s

= 12 x 12 = 144cm²

Haryana Board Class 6 Maths Solutions For Chapter 10 Mensuration Long Answer Questions

Question 15. Measurements of two rectangular fields are 50m x 30m and 60m x 40m. Find their perimeters. Check whether the perimeters are 2 x length + 2 x breadth.

Solution. Measurements of rectangle ABCD:

Length (l) Breadth (b)

Breadth (b)= 30 m

Perimeter of ▢ABCD

= AB + BC + CD + AD

= 50 + 30 + 50 + 30

= 160 m

Perimeter of ▢ABCD

=2 x length + 2 x breadth

= 2 (50) + 2 (30)

= 100 + 60

= 160 m

Yes, it is true.

Length of rectangle PQRS = l = 60 m

Breadth (b) = 40 m

Perimeter of ▢PQRS

= PQ + QR + RS + PS

= 60 + 40 + 60 + 40

= 200 m

Perimeter of ▢PQRS

= 2 x length + 2 x breadth

=2 x 60 + 2 x 40

= 120 + 80.

= 200 m

Yes, it is true.

Question 16. Manasa has 24 cm of metallic wire with her. She wanted to make some polygons with equal sides whose sides are integral without milling into pieces values. Find how many such polygons she can make with the metallic wire?

Solution. The length of the wire that Manasa has = 24 cm

She has to make some polygons with equal sides i.e., [squares] whose sides are integral without milling into pieces values.

∴ The smallest length of the side of a polygon [square],

she wants to make is (s) = 1 cm

Perimeter of one polygon [square]

= 4 x side

= 4 x 1 = 4 cm

∴ Number of polygons [square] that she can make with the length of 24 cm metallic wire

= \(\frac{\text { Length of the wire }}{\text { Perimeter of } 1 \text { square }}=\frac{24}{4}=6\)

Question 17. Find the perimeter of the following figures (1) and (2).

Solution. The perimeter of figure (1) ABCDEFGHIJKL

= AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + AL

= 5 + 3 + 1 + 2 + 1 + 1 + 1 + 1 + 1 + 2 + 1 + 3 = 22 cm

The perimeter of figure (2) MNOPQRSTUVWX

= MN + NO + OP + PQ + QR + RS + ST + TU + UV + VW + WX + MX

= 1 + 2 + 1 + 1 + 1 + 1 + 1 + 2 + 1 + 1 + 5 + 1 = 18 cm

Question 18. If the length of a rectangle is 14cm and its perimeter is 3 times of its length. Find its area.

Solution. Length of a rectangle = 1 = 14cm

Its perimeter = 3 x length of a rectangle

= 3 x 14 = 42 cm

∴ Perimeter of a rectangle = 2 [1 + b]

= 2[14 + b] cm.

∴ By the problem 2[14 + b] = 42

14 + b = \(\frac{42}{2}\)

14+ b = 21

b = 21 – 14 = 7

∴ Breadth of a rectangle = b = 7 cm

∴ Area of a rectangle = 1 x b

= 14 x 7 = 98 cm²

Question 19. 14cm and 12cm are the length and breadth of a rectangle. If the breadth is increased by 6cm and length is decreased by 6cm, find the difference in areas.

Solution. Length of a rectangle = l = 14cm

Its breadth = b = 12 cm

Area of a rectangle = l x b

= 14 x 12 = 168 cm²

If the breadth is increased by 6 cm and length is decreased by 6 cm then so formed.

Length = l = (l – 6) cm;

breadth = b = (b + 6) cm = 14 – 6 = 8 cm

∴ Area of so formed rectangle

= l x b

12 + 6 = 18 cm

Area of new rectangle = 1 x b

= 8 × 18

= 144 cm²

∴ Difference in areas

= 168 cm² – 144 cm²

= 24 cm²

Haryana Board Class 6 Maths Solutions For Chapter 10 Mensuration Objective Type Questions

Choose the correct answer:

Question 1. Perimeter of an equilateral triangle of side ‘s’ is

1. 4s
2. 2s
3. 3s
4. 6s

Answer. 3. 3s

Question 2. Perimeter of a regular hexagon of side 2 cm is

1. 12 cm
2. 10 cm
3. 8 cm
4. 12 sq.cm.

Answer. 1. 12 cm

Question 3. The space occupied by a closed figure is called its

1. area
2. perimeter
3. segment
4. point

Answer. 1. area

Question 4. The perimeter of triangle with sides \(\overline{A B}\), \(\overline{B C}\), \(\overline{C A}\)

1. \(\overline{A B}\) + \(\overline{B C}\)
2. \(\overline{A B}\) + \(\overline{B C}\) + \(\overline{C A}\)
3. \(\overline{B C}\) + \(\overline{C A}\)
4. A + B + C

Answer. 2. \(\overline{A B}\) + \(\overline{B C}\) + \(\overline{C A}\)

Question 5. The number of sides of a octagon is

1. 6
2. 7
3. 10
4. 8

Answer. 4. 8

Question 6. If length and breadth of a rectangle is 12 cm and 8 cm respectively, then its perimeter is ….

1. 20 cm
2. 96 cm
3. 40 sq.m.
4. 40 cm

Answer. 4. 40 cm

Question 7. Area of a rectangle

1. l x b
2. 2(l + b)
3. 4s
4. s²

Answer. 1. l x b

Question 8. A = s² is the formula for

1. Area of a rectangle
2. Area of a square
3. Perimeter of a square
4. Perimeter of a rectangle

Answer. 2. Area of a square

Question 9. The sum of all the lengths of a polygon is called ………..

1. area
2. perimeter
3. volume
4. none

Answer. 2. perimeter

Question 10. How many sides are there in a pentagon?

1. 2
2. 9
3. 6
4. 5

Answer. 4. 5

Question 11. In a rectangle l = 3 cm, b = 0.9 cm then A = …… cm²

1. 9
2. 27
3. 2.7
4. 3

Answer. 3. 2.7

Question 12. Side of an equilateral triangle is 3 cm then its perimeter is ………. cm.

1. 3.5
2. 11
3. 12
4. 9

Answer. 4. 9

Question 13. Perimeter of rectangle is …..

1. \(\frac{2}{3}\)(l + b)
2. \(\frac{l+b}{2}\)
3. 2(l + b)
4. \(l+\frac{\mathrm{b}}{2}\)

Answer. 3. 2(l + b)

Question 14. Perimeter of a square is 20 cm then its side is ………. cm.

1. 10
2. 5
3. 4
4. 7

Answer. 2. 5

Question 15. Identify the odd one from the following

1. length
2. breadth
3. area
4. perimeter

Answer. 3. area

Question 16. P = 2(l + b) then l = ……

1. \(\frac{\mathrm{P}}{2}+\mathrm{b}\)
2. \(\frac{P}{2}-l\)
3. \(\frac{P}{2}-b\)
4. \(\frac{P}{2}-2b\)

Answer. 3. \(\frac{P}{2}-b\)

Question 17. Formula to find area of square of side ‘a’ units is ………unit².

1. a
2. 2a
3. \(\frac{a}{2}\)
4. a²

Answer. 4. a²

Question 18. The perimeter of regular hexagon of side 6 cm is ……….. cm.

1. 20
2. 36
3. 15
4. 16

Answer. 2. 36

Question 19. 1 m = ………. cm.

1. 50
2. 60
3. 70
4. 100

Answer. 4. 100

Question 20. The cost of fencing a square park of side 150 m at the rate of ₹ 10 per meter is

1. 6000
2. 5000
3. 1100
4. 1700

Answer. 1. 6000

Question 21. In a rectangle l = 36 m, b = 24 m, then P = …..m.

1. 114
2. 103
3. 120
4. 110

Answer. 3. 120

Question 22. Side of a square is 17 cm then its area is ……… cm².

1. 423
2. 324
3. 189
4. 289

Answer. 4. 2889

Question 23. The area of a rectangle is 1,125 cm2. If its breadth is 25 cm then length = ….. cm.

1. 25
2. 40
3. 15
4. 45

Answer. 4. 45

Question 24. Perimeter of a regular pentagon of side 8 cm is ….. cm.

1. 60
2. 40
3. 80
4. 70

Answer. 2. 40

Question 25. ……… are the simple closed plane figures bounded by line segments.

1. Area
2. Perimeter
3. Square
4. Polygons

Answer. 4. Polygons

Question 26. The perimeter of a triangle is 30 cm. Sum of its two sides is 21 cm, then the length of 3rd side is …….. cm.

1. 13
2. 12
3. 9
4. 11

Answer. 3. 9

Question 27. Area measured in ……….units.

1. square
2. cubic
3. Both A & B
4. none

Answer. 1. square

Question 28. If the perimeter of a square is 48 cm then its side

1. 12 cm
2. 24 cm
3. 6 cm
4. 144 cm

Answer. 1. 12 cm

Question 29. If the area of a square is 144 cm2 then its perimeter is

1. 12 cm
2. 48 cm
3. 24 cm
4. 36 cm

Answer. 2. 48 cm

Question 30. If the side of a square is doubled then its area will be increased by

1. doubled
2. 3 times
3. 4 times
4. None

Answer. 3. 4 times

Question 31. If the side of a square is halved then its area will be

1. same as original area
2. halved
3. 4 times increases
4. \(\frac{1}{4}\) of original area

Answer. 4. \(\frac{1}{4}\) of original area

Question 32. If the length and breadth of rectangle are doubled then its area will be

1. same as original area
2. halved
3. doubled
4. increased by 4 times

Answer. 4. increased by 4 times

Question 33. If the length is doubled and breadth is tripled then its area will be

1. doubled
2. increased by 3 times
3. increased by 6 times
4. same as original area

Answer. 3. increased by 6 times

Question 34. A rectangular plot of land is 240 m by 200 m. The cost of fencing per meter is ₹ 10. What is the cost of fencing the entire field?

1. ₹ 8800
2. ₹ 8000
3. ₹ 4800
4. ₹ 8880

Answer. 1. ₹ 8800

Question 35. A piece of wire is 60 cm long. What will be the length of each side if the string is used to form and equilateral triangle

1. 10 cm
2. 15 cm
3. 20 cm
4. 30 cm

Answer. 3. 20 cm

Question 36. The perimeter of a rectangle whose lengths is ‘l’ meters and breadth is ‘b’ meters is ………… meters.

1. l + b
2. lb
3. 2(l + b)
4. 2l + b

Answer. 3. 2(l + b)

Question 37. The perimeter of a rectangle whose sides are 18 cm and 15 cm is ….. cm.

1. 38
2. 60
3. 66
4. 36

Answer. 3. 66

Question 38. A farmer has to cover three times the perimeter of field. The length is 10 m and the breadth is 8 m then total distance travelled by him is …… m.

1. 108
2. 100
3. 18
4. 36

Answer. 1. 108

Question 39. Perimeter of a square whose side is ‘a’ meters is ………… meters.

1. a⁴
2. 4 + a
3. 2a
4. 4a

Answer. 4. 4a

Question 40. The perimeter of a regular pentagon whose side is 5 cm is …… cm.

1. 28
2. 25
3. 20
4. 29

Answer. 2. 25

Question 41. The area of rectangle whose sides are 3 m and 4 m is …….. sq.m.

1. 12
2. 13
3. 34
4. 7

Answer. 1. 12

Question 42. The area of a square whose side is 4 cm is …….. sq.cm.

1. 4
2. 16
3. 24
4. 8

Answer. 2. 16

Question 43. The area of rectangle is 20 sq.cm and the length is 4 cm then the breadth is ….. cm.

1. 10
2. 4
3. 3
4. 5

Answer. 4. 5

Question 44. The area of rectangle whose measurements are given by length = 3 m and breadth = 40 cm is ……. sq.m.

1. 340
2. 120
3. 1.2
4. 12

Answer. 3. 1.2

Chapter 10 Mensuration Fill in the blanks:

Question 45. The perimeter of a square whose length is 4.5 cm ………..cm

Answer. 18

Question 46. The perimeter of an equilateral triangle is 30 cm then its side is ……..cm

Answer. 10

Question 47. The area of square is 12544 cm² then its side is……cm

Answer. 112

Question 48. The perimeter of a rectangle is 48 cm, its length is 20 cm then its breadth = …………..cm.

Answer. 4

Question 49. The area of adjacent rectangle PQRS is ……………cm².

Answer. 36

Question 50. The cost of fencing a square park of side 250 m. at the rate of 20 per meter is.

Answer. 20,000

Question 51. The area of a square is 144 cm2 then its perimeter is ………..cm

Answer. 48

Question 52. The perimeter of given figure is………cm

Answer. 8.4

Question 53. The perimeter of a square is 40 cm., then its side is.

Answer. 10 cm.

Question 54. The length of the boundary of closed figure is called its

Answer. perimeter

Question 55. Area of a rectangle is 100 cm2 its length and breadth are equal then its perimeter is………..

Answer. 40 cm.

Question 56. The perimeter of a square is 24 cm., then its area is …….

Answer. 36 cm².

Question 57. The perimeter of a regular hexagon of side 8 cm., is……..

Answer. 48 cm.

Question 58. Which is different among 15 cm., 16 cm., 19 cm.2, 18 cm.

Answer. 19 cm².

Question 59. The side of a square is 6 cm. The length and breadth of rectangle are 7 cm. and 5 cm. respectively. Which has less area?………..

Answer. rectangle

Question 60. Which is different among “length, breadth, side, perimeter, area”………

Answer. area

Question 61. 2(1+b) P is the formula for …………….

Answer. perimeter of a rectangle

Question 62. Which is different among 16 cm², 17 cm², 18 cm², 20 cm² ……

Answer. 18 cm.

Question 63. The sum of the lengths of line segments of the polygon is ………..

Answer. perimeter of the polygon

Question 64. If the area of a rectangular field is 1350 m2 and its length is 50 m then its breadth is ………

Answer. 27 m.

Question 65. If the length of the side of a square is 10 m. Then 100 square mts is its …….

Answer. area

Question 66.

What is the relationship between these shapes?

Answer. perimeters are equal.

Question 67. The amount of surface enclosed by a closed figure is called its ……

Answer. area

Question 68. The perimeter of the regular hexagon is 18 cm then side is ……. cm.

Answer. 3

Question 69. The area of square whose side is 4 cm. is…………………….cm.

Answer. 16

Question 70. The area in square metre of a piece of cloth 1 m 25 cm wide and 2.m long is …….cm².

Answer. 25

Question 71. The perimeter of an equilateral triangle whose side is 4 cm is…….cm.

Answer. 12

Question 72. The perimeter of a triangle whose sides are 3 cm, 4 m and 6 cm is ………….. cm.

Answer. 13

Question 73. The perimeter of the rectangle given in the figure is ………. cm

Answer. 12

Haryana Board Class 6 Maths Solutions For Chapter 10 Mensuration Match The Following:

Question 74.

Answer. 1 – D, 2 – A, 3 – B, 4 – F, 5 – E, 6 – C.

• ‘Data’ is a collection of numbers gathered to give some information.
• ‘Information’ is either in the form of numbers or words.
• To get a particular information from the given data quickly, the data can be arranged in a tabular form using tally marks.
• Data that has been organised and presented in ‘frequency distribution tables’ can also be represented using ‘pictographs’ and ‘bar graphs’.
• Representation of the data in the form of pictures, objects or parts of objects is called ‘pictograph’ or ‘pictogram’.
• A pictograph uses pictures or symbols to represent the frequency of the data.
• In pictograph the size of all pictures should be the same and should represent the same number of items of the data.
• We use tally marks for recording and organisation of data.
• Representation of data can be done in the following ways (1) Pictographs (2) Bar graphs.
• The pictographs represents the representation of the data through pictures.
• “P.C. Mahalanobis” is known as father of Indian Statistics.

Haryana Board Class 6 Maths Solutions For Chapter 9 Data Handling Exercise 9.1

Question 1. In a Mathematics test, the following marks were obtained by 40 students. Arrange these marks in a table using tally marks.

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

b) How many students obtained marks below 4?

Solution.

a) 5 + 4 + 3 = 12 students obtained marks equal to or more than

7 (7 marks – 5 students, 8 marks – 4 students; 9 marks – 3 students)

b) 3 + 3 + 2 = 8 students obtained marks below 4.

Haryana Board Class 6 Maths Data Handling solutions

Question 2. Following is the choice of sweets of 30 students of Class VI: Ladoo, Barfi, Ladoo, Jalebi, Ladoo, Rasgulla, Jalebi, Ladoo, Barfi, Rasgulla, Ladoo, Jalebi, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo, Rasgulla, Ladoo, Ladoo, Barfi, Rasgulla, Rasgulla, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo.

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

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

Solution. a)

b) Ladoo is preferred by most of the students.

Question 3. Catherine threw a dice 40 times and noted the number appearing each time as shown below:

Make a table and enter the data using tally marks. Find the numbers that appeared:

a) The minimum number of times.

b) The maximum number of times.

c) Find those numbers that appear an equal number of times.

Solution.

a) Number 4 appeared minimum number of times.

b) Number 5 appeared maximum number of times.

c) Number 1.and 6 appeared equal number of times.

Question 4. Following pictograph shows the number of tractors in five villages.

Observe the pictograph and answer the following questions.

1) Which village has the minimum number of tractors?

Solution. Village D has the minimum number of tractors – 3.

2) Which village has the maximum number of tractors?

Solution. Village C has the maximum number of tractors – 8.

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

Solution. Village C has 8 – 5 = 3 more tractors as compared to village B.

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

Solution. Total number of tractors in all the five villages = 6 + 5 + 8 + 3 + 6 = 28.

Question 5. The number of girl students in each class of a co-educational middle school is depicted by the pictograph.

Observe this pictograph and answer the following questions:

a) Which class has the minimum number of girl students?

Solution. Class VIII has the minimum no. of girl students (6).

b) Is the number of girls in class VI less than the number of girls in Class V ?

Solution. No, the number of girls (16) in class VI is not less than the number of girls (10) in class V.

c) How many girls are there in class VII?

Solution. Number of girls in class VII = 3 x 4 = 12.

How to draw a pictograph Class 6 HBSE

Question 6. The sale of electric bulbs on different days of a week is shown below:

Observe the pictograph and answer the following questions:

a) How many bulbs were sold on Friday.

Solution. 14 bulbs were sold on Friday.

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

Solution. Maximum bulbs were sold on Sunday.

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

Solution. Equal no. of bulbs were sold on Wednesday and Saturday.

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

Solution. Minimum bulbs were sold on Wednesday and Saturday.

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

Solution. No. of bulbs can hold in one big carton = 9

No.of cartons were need = 86 ÷ 9

= 10 (Approx)

Question 7. In a village six fruit merchants sold the following number of fruit baskets in a particular season:

Observe this pictograph and answer the following questions:

a) Which merchant sold the maximum number of baskets?

Solution. Martin sold the maximum number of baskets – 950.

b) How many fruit baskets were sold by Anwar?

Solution. 7 x 100 = 700 fruit baskets were sold by Anwar.

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

Solution. Anwar, Martin and Ranjit Singh are planning to buy a godown for the next season.

Haryana Board Class 6 Maths Solutions For Chapter 9 Data Handling Very Short Answer Questions

Question 1. Give two examples of data in numerical figures.

Solution. Example: 1

The number of students in each class in a school:

Example: 2

The details of number of cricket bats sold in a week:

Question 2. Give two examples of data in words.

Solution. Example: 1

Days of a week

MON, TUE, WED, THU, FRI, SAT, SUN.

Example: 2

Important cities in Andhra Pradesh.

Vijayawada, Guntur, Tirupathi, Visakhapatnam, Machilipatnam, Nellore, Kakinada.

Question 3. Vehicles that crossed a checkpost between 10 AM and 11 AM are as follows:

car, lorry, bus, lorry, auto, lorry, lorry, bus, auto, bike, bus, lorry, lorry, zeep, lorry, bus, zeep, car, bike, bus, car, lorry, bus, lorry, bus, bike, car, zeep, bus, lorry, lorry, bus, car, car, bike, auto.

Represent the data in a frequency distribution table using tally marks.

Solution.

Frequency table and data interpretation Class 6

Question 4. The following pictograph shows the number of students use cycles, in five classes of a school.

Answer the following questions based on the pictograph given above –

1) Which class students have the maximum number of cycles?

2) Which class students have the minimum number of cycles?

3) Which class students have 9 cycles?

4) What is the total number of cycles in all the five classes?

Solution. 1) IX class students have the maximum number of cycles.

2) VI class students have the minimum number of cycles.

3) VIII class students have 9 cycles.

4) Total number of cycles in all the five classes:

VI      5

VII     10

VIII     9

IX       12

X         7

Total   43

Question 5. A book-shelf has books of different subjects. The number of books of each subject is represented as a pictograph given below. Observe them.

Answer the following questions:

1) Which books are more in number?

Solution. Maths

2) Which books are least in number?

Solution. English

3) How many total books are there ?

Solution.

Telugu      4

English     3

Hindi        4

Maths       6

Science     5

Social        4

Total         26

Question 6. 26 students in a class got the following marks in an assignment – 5, 6, 7, 5, 4, 2, 2, 9, 10, 2, 4, 7, 4, 6, 9, 5, 5, 4, 3, 7, 9, 5, 2, 4, 5, 7. The assignment was for 10 marks.

1) Organise the data and represent in the form of a frequency distribution table using tally marks.

2) Find out the marks obtained by maximum number of students.

3) Find out how many students received least marks.

4) How many students got 8 marks?

Solution. 1)

2) Maximum number of students (6) got 5 marks.,

3) Least mark (2) was obtained by 4 students.

4) No student in the class got 8 marks.

How to create a bar graph Class 6 Haryana Board

Question 7. In a class of 25, students like various games. The details are shown in the following pictograph. (No student plays more than one game).

1) How many students play badminton?

2) Which game is played by most number of students ?

3) What is the game in which least number of students are interested?

4) How many students do not play any game?

Solution. 1) 5 students play badminton.

2) Kabaddi is played by most number of students i.e., 7.

3) Tennikoit is played by least number of students i.e., 4.

4) Total number of players = 7 + 4 + 5 + 6 = 22

Number of students in the classroom = 25.

Thus, the number of students who do not play any game = 25 – 22 = 3

Question 8. Give two examples of data in numerical figures.

Solution. 1) The population of Andhra Pradesh in (2011) – 4,96,34,314

2) The cost of I phone 11 pro-1,21,889.

Question 9. Give two examples of data in words.

Solution. 1) The cost of Maruti Brezza [Z x 1] price on road in India – Nine lakh ten thousand rupees only.

2) The population of India according to census 2011 – One hundred and twenty one crore eight lakhs fifty four thousand nine hundred and seventy seven.

Haryana Board Class 6 Maths Solutions For Chapter 9 Data Handling Short Answer Questions

Question 10. A child’s Kiddy bank is opened and the coins collected are in the following denomination.

Represent the data in a frequency distribution table using tally marks.

Solution.

Question 11. The favourite colours of 25 students in a class are given below:

Blue, Red, Green, White, Blue, Green, White, Red, Orange, Green, Blue, White, Blue, Orange, Blue, Blue, White, Red, White, White, Red, Green, Blue, Blue, White.

Write a frequency distribution table using tally marks for the data. Which is the least favourite colour for the students ?

Solution.

colour for the students = 2 (Orange)

Question 12. A TV channel invited a SMS poll on ‘Ban of Liquor’ giving options:-

A – Complete ban

B – Partial ban

C – Continue sales

They received the following SMS, in the first hour-

Represent the data in a frequency distribution table using tally marks.

Solution.

Question 13. The number of wrist watches manufactured by a factory in a week are as follows:

Represent the data using a pictograph. Choose a suitable scale.

Solution. = 50 watches.

Question 14. Votes polled for various candidates in a sarpanch election are shown below, against their symbols in the following table.

Represent the data using a pictograph. Choose a suitable scale.

Answer the following questions:

1) Which symbol got least votes?

2) Which symbol candidate won in the election?

Solution. Each represents 50 votes

1) Watch symbol got least votes.

2) Pot symbol candidate won in the election.

Question 15. The sale of television sets of different companies on a day is shown in the pictograph given below. Scale: = 5 televisions

Answer the following questions:

1) How many TVs of company A were sold?

2) Which company’s TVs are people more crazy about?

3) Which company sold 15 TV sets?

4) Which company had the least sale?

Solution. 1) Number of TV’s sold by company A = 5 x 5 = 25

2) People more crazy about the TV company ‘C’.

3) Company ‘E’ sold 15 TV sets.

4) Company ‘D’ sold least sale.

Haryana Board Class 6 Maths Solutions For Chapter 9 Data Handling Long Answer Questions

Question 16. Given below are the ages of 20 Students of Class VI in a School.

13, 10, 11, 12, 10, 11, 11, 13, 12, 11, 10, 11, 12, 11, 13, 11, 10, 13, 10, 12

1) Organise the data and represent in the form of a frequency distribution table using tally marks.

Solution.

2) Find out the age having more number of students

Solution. More number of students having the age is 11 years.

3) How many students are there in 10 Years age?

Solution. 5 students

4) Find out number of students who are having more age?

Solution. Number of students who are having more age 13 years is 4.

Question 17. A die was thrown 30 times and following scores were obtained

5 3 4 6 2 3

6 2 2 3 15

2 5 4 6 2 1

4 5 1 6 2 1

3 1 3 3 4 6

Word problems on pictographs and bar graphs Class 6

1) Prepare a frequency table of the scores.

Solution.

2) Which number obtained more times?

Solution. The numbers 2 and 3 obtained more times ie., 6.

3) How many times was a score greater than 4 obtained?

Solution. The number of times a score greater than 4 obtained was 4 + 5 = 9

4) Find the total number of times an odd number obtained.

Solution. Total number of times an odd number obtained is = 5 + 6 + 4

= 15 [1, 3 and 5 are faces of a dice]

Haryana Board Class 6 Maths Solutions For Chapter 9 Data Handling Objective Type Questions

Choose the correct answer:

Question 1. The information collected in terms of numbers

theorem

data

mean

range

Answer. 2. data

Question 2. Representing the data in terms of pictures is

Bar graphs

Bar charts

Pictographs

Pie charts

Answer. 3. Pictographs

Question 3. While drawing pictures in a pictograph all ………… must be equal.

lengths

sizes

breadths

bases

Answer. 2. sizes

Question 4. || indicates ….

7

6

3

5

Answer. 1. 7

Question 5. If a * represents 3 cars, how many * are required to indicate 18 cars?

3

6

8

12

Answer. 2. 6

Question 6. If 6 cm represent 9 lakh people then the scale is ………..

1 cm = 1,50,000

2 cm = 1,00,000

1 cm = 2,00,000

1 cm = 1,20,000

Answer. 1. 1 cm = 1,50,000

Question 7. In a data maximum value = 30 and the minimum value is 10. Then the range is

31

12

20

16

Answer. 3. 20

Question 8. 13 can be represented by tally marks as

|||

||

|

Answer. 2. |||

Question 9. According to the scale \(\frac{1}{2}\) cm = 15 students then 45 students = …… cm

8

4

7

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

Answer. 4. \(\frac{3}{2}\)

Question 10. ………. are used in preparing frequency table

Symbols

Tally marks

Data

Scale

Answer. 2. Tally marks

Data Handling examples for Class 6 HBSE Maths

Question 11. Which of the following indicates ‘5’

|||||

|||

Answer. 2. Diagram

Question 12. Average of 1, 2, 3 is

1. 2
2. 3
3. 8
4. 3.5

Answer. 1. 2

Question 13. Statement – a: A table showing the frequency or count of various items is called a frequency distribution.

Statement – b: To get a particular information from the given data quickly, the data can be arranged in a tabular form using tally marks.

1. both a & b are true
2. a is true b is false
3. a is false, b is true
4. both a & b are false

Answer. 1. both a & b are true

Question 14. Father of Indian Statistics

1. Ramanujan
2. Aryabhata
3. P.C.Mahalanobis
4. Bhaskaracharya

Answer. 3. P.C.Mahalanobis

Question 15. If represents = 10 books then represent = …….books.

1. 10
2. 20
3. 30
4. 40

Answer. 4. 40

Answer the questions [16-19] from the information given below.

Question 16. How many marks obtained by maximum number of students?

1. 5
2. 6
3. 2
4. 10

Answer. 1. 5

Question 17. How many students received least marks?

1. 1
2. 2
3. 4
4. 10

Answer. 3. 4

Question 18. How many students got 8 marks?

1. 1
2. 4
3. 21
4. 0

Answer. 4. 0

Question 19. How many students got more than 8 marks?

1. 3
2. 4
3. 8
4. 21

Answer. 2. 4

Question 20. If represents 5 eggs, then how many eggs do represent?

1. 20
2. 25
3. 100
4. 30

Answer. 2. 25

Question 21. If represents 5 balloons, then the number of symbols to be drawn to represent 70 balloons is

1. 12
2. 10
3. 15
4. 14

Answer. 4. 14

Question 22. The number of members in 20 families in a village is given as 6, 8, 6, 3, 2, 5, 7, 8, 6, 5, 5, 7, 7, 8, 6, 6, 7, 7, 6, 5.

How many families are of the largest size?

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

Answer. 2. 3

Question 23. The number of times an observation occurs is called

1. frequency
2. data
3. tallymarks
4. none

Answer. 1. frequency

Haryana Board Class 6 Maths Solutions For Chapter 9 Data Handling Fill in the blanks:

Question 24. Diagram represents ……

Answer. 5

Question 25. 1 + 3 + 4 + 5 + 6 + 7 + 8 + 10 = ……….

Answer. 44

Question 26. ……………… collection of numbers gathered to give some information.

Answer. Data

Question 27. Average of -2, -1, 0, 1, 2, … ………….

Answer. 0

Question 28. According to a scale \(\frac{1}{2}\) cm = 20 students then 60 students

Answer. 3/2

Question 29. Range of first 10 whole numbers is ………

Answer. 9

Question 30. Range of first 100 even natural numbers is = …….

Answer. 98

Question 31. Show 12 as tally mark…………….

Answer. ||

Question 32. Collection of certain information is called …….

Answer. data

Question 33. Representing the data in the form of pictures is called ………

Answer. pictograph

Question 34. If Diagram represents 150 then Diagram represent …….

Answer. 750

Question 35. If 100 books = Diagram then 150 books are represented by ……….

Answer. ▢▢ ▢

Question 36. The tally marks ||| represent ………..

Answer. 13

Question 37. Which are used in preparing frequency table are ……..

Answer. Tally marks

Question 38. The tally marks to represent 16……………..

Answer. |

• The collection of all positive numbers, zero and negative numbers together are called ‘Integers’.
  The set of Integers is denoted by the symbol Z. Z = {………….. -4, -3, -2, -1, 0, 1, 2, 3, 4 …}
• The number without any sign is considered positive number only.
  Eg: 3 is considered + 3
• Brahmagupta (598 AD-670 AD) first used a special sign (-) for negative numbers and stated the rules for dealing with positive and negative quantities.
• The letter “Z” was first used by the Germans because the word for Integers in the Ger- man language is “Zehlen”, which means “Number”.
• Natural numbers, (1,2,3,4,…) are also called positive integers and Whole numbers. {0,1,2,3,4,…) are also called non-negative integers.
  Negative numbers are important to represent just opposite numbers for positive numbers.
• Every integer can be represented on a number line horizontally or vertically.
• Addition and subtraction of integers can be done on number line.
• Every negative integer is always lesser than zero.
• The smallest and the biggest positive integers are +1 and positive infinite (+∞).
• The smallest and the biggest negative integers are negative infinite (-∞) and -1
• The sum of two negative integers is always a negative integer. Eg: -2 + (-3) = -5
• The sum of two positive integers is always a positive integer. Eg: 5 + 7 = 12
• The sum of two integers, one of which is positive and the other is negative, then the sum may be either positive, negative or zero.
  Eg: 12 + (-9) = 3;-12 + 9 = -3;-3 + 3 = 0
• For any integer a, -(-a) = a
• We add the opposite of an integer (additive inverse) to an integer in subtraction.
• Integers are widely used in our daily life i.e., business, engineering, games, temperatures, medicine etc……
• Integers are closed under addition and sub- traction.
• Integers are commutative and associative under addition.
• Integers are not commutative and associative under subtraction.
• We use negative numbers to represent debit, temperature below the 0°C, past periods of time depth below sea level.
• Number Line:

• The numbers on the left side of zero (i.e., less than zero) are called ‘negative numbers’.
• The numbers on the right side of zero (i.e., greater than zero) are called ‘positive numbers’.
• We can show the addition and subtraction of integers on the number line.
• ‘O’ is neither positive nor negative.
• Any two distinct numbers that give zero when added to each other are additive inverse of each other.
  Eg. : 3 and -3
  3 + (-3) = 0
• The subtraction of integers is the same as the addition of their additive inverse.
  Eg. The additive inverse of 7 is -7
  The additive inverse of -8 is 8
• Numbers with a negtative sign are called negative numbers. They are less than zero.
• If we add ‘1’ (one) to a given number we get successor.
• If we subtract ‘one’ from a given number we get predecessor.
• If we move right side of a given number on a number line the value of the number increases and to the left the value decreases.
• If we add a positive number and a negative number we must do subtraction and put the sign of the bigger number.
• The addition of two integers is zero then each one is called additive inverse of the other.

Suppose David and Mohan have started walking from zero position in opposite directions. Let the steps to the right of zero be represented by ‘+’ sign and to the left of zero represented by “” sign. If Mohan moves 5 steps to the right of zero it can be represented as +5 and if David moves 5 steps to the left of zero it can be represented as -5. Now represent the following positions with +or – sign:

a) 8 steps to the left of zero.

Answer. -8

b) 7 steps to the right of zero.

Answer. +7

c) 11 steps to the right of zero.

Answer. +11

d) 6 steps to the left of zero.

Answer. -6

Write the succeeding number of the following:

Write the preceeding number of the following:

Write the following numbers with appropriate signs:

a. 100 m below sea level.

Solution. -100 m

b. 25°C above 0°C temperature.

Solution. + 25°C

c. 15°C below 0°C temperature.

Solution. -15°C.

d. Five numbers less than 0.

Solution. -1, -2, -3, -4, -5

Mark 3, 7,-4,-8,- 1 and 3 on the number line.

Solution.

Compare the following pairs of numbers using > or <.

0 _ 8

-1 _ -15

5 _ -5

11 _ 15

0 _ 6

– 20 _ 2

Solution.

0 > 8

-1 > -15

5 > -5

11 < 15

0 < 6

-20 < 2

Haryana Board Class 6 Maths Solutions For  Chapter 6 Integers Exercise – 6.1

Question 1. Write the opposites of the following:

a) Increase in weight

b) 30 km north

c) 80 m east

d) Loss of Rs. 700

e) 100 m above sea level

Solution. a) decrease in weight

b) 30 km south

c) 80 m west

d) Profit of Rs. 700

e) 100 m below sea level

Haryana Board Class 6 Maths Integers solutions

Question 2. Represent the following numbers as integers with appropriate signs.

a) An aeroplane is flying at a height, two thousand metre above the ground.

Solution. +2000 metre

b) A submarine is moving at a depth, eight hundred metre below the sea level.

Solution. -800 metere

c) A deposit of rupees two hundred.

Solution. +Rs. 200

d) Withdrawal of rupees seven hundred.

Solution. -Rs. 700

Question 3. Represent the following numbers on a number line.

a) +5

b) -10

c) +8

d) -1

e) -6

Addition and subtraction of integers Class 6 HBSE

Question 4. Adjacent figure is a vertical number line, representing integers. Observe it and locate the following points:

a. If the point D is + 8, then which point is -8?

Solution. Point F.

b. Is point G is a negative integer or a positive integer?

Solution. It is a negative integer.

c. Write integers for points B and E.

Solution. Integer for point B = +4

Integer for point E = -10.

d. Which point marked on this number line has the least value?

Solution. Point E.

e. Arrange all the points in decreasing order of value.

Solution. D, C, B, A, O, H, G, F, E.

Question 5. Following is the list of temperatures of five places in India, on a particular day of the

a) Write the temperatures of these places in the form of integers in blank column.

Solution.

b) Following is the number line representing the temperature in degree celsius.

Plot the name of the city against its temperature.

c) Which is the coolest place?

Solution. Siachin is the coolest place.

d) Write the names of the places where temperatures are above 10° C.

Solution. Delhi and Ahmedabad.

Question 6. In each of the following pairs, which number is to the right of the other on the number line?

a) 2,9

Solution. The number 9 is to the right of the number 2.

b) -3, -8

Solution. The number -3 is to the right of the number -8.

c) 0,-1

Solution. The number ‘0’(zero) is to the right of the number -1.

d) -11, 10

Solution. The number 10 is to the right of the number -11.

e) -6,6

Solution. The number 6 is to the right of the number -6.

f) 1, -100

Solution. The number 1 is to the right of the

Multiplication and division of integers Class 6

Question 7. Write all the integers between the given pairs (write them in the increasing or

a. 0 and -7.

Solution. -6,-5,-4,-3,-2 and -1.

b. -4 and 4.

Solution. -3, -2, -1, 0, 1, 2 and 3.

C. -8 and -15.

Solution. -14,13,12,-11,-10 and -9.

d. 30 and -23.

Solution. -29,-28,-27,-26,-25 and -24.

Question 8. a. Write four negative integers greater than 20.

Solution. 19, -18,-17 and – 16..

b. Write four negative integers less than -10.

Solution. 11, 12, 13, and – 14.

Question 9. For the following statements, write True (T) or False (F). If the statement is false, correct the statement.

a. -8 is to the right of -10 on a number line.

Solution. True (T)

b. -100 is to the right of -50 on a number line.

Solution. False (F)

c. Smallest negative integer is -1.

Solution. False (F)

d. -26 is greater than -25.

Solution. False (F)

Question 10. Draw a number line and answer the following:

a. Which number will we reach if we move 4 numbers to the right of – 2?

Solution. We reach the number ‘2’.

b. Which number will we reach if we move 5 numbers to the left of 1?

Solution. We reach the number – 4.

c. If we are at 8 on the number line, in which direction should we move to reach – 13?

Solution. We should move in the left direction.

d. If we are at 6 on the number line, in which direction should we move to reach – 1?

Solution. We should move in the right direction.

Question 11. Draw a figure on the ground in the form of a horizontal number line as shown below. Frame questions as given in the said example and ask your friends.

1) Which is greater (-2) or (-1)?

Solution. 1.

2) Which is smaller among – 6 and 3?

Solution. -6.

3) Which is the nearest positive integer to zero ?

Solution. 1.

4) Which integer is neither positive nor negative ?

Solution. ‘O’ is neither positive nor negative.

5) How many negative numbers you will find on left side of zero?

Solution. Infinite

6) Write three integers greater than -2.

Solution. -1,0,1.

7) Write five integers lesser than -1.

Solution. -2,-3,-4,-5, -6.

8) Which is the nearest positive integer to -1?

Solution. +1.

Integer number line Class 6 Haryana Board

Question 12. Take two different coloured buttons like white and black. Let us denote one white button by (+1) and one black button by (-1). A pair of one white button (+1) and black button (-1) will denoted zero i.e. [1 + (-1) = 0]

In the following table, integers are shown with the help of coloured buttons.

Let us perform additions with the help of the coloured buttons. Observe the following table and complete it.

Question 13. Find the answers of the following additions:

a. (-11)+(-12)

Solution. (-11) + (-12)

= -11 – 12 = -23.

b. (+10) + (+4)

Solution. 10 + 4 = 14.

c. (32) + (-25)

Solution. -32 + (-25)

= -32 – 25 = -57.

d. (+23) + (+40)

Solution. 23 + 40 = 63.

Question 14. Find the solution of the following:

a. (-7) + (+8)

Solution. -7 + (+8)

= -7 + 7 + 1 = 1.

b. (-9) + (+13)

Solution. (-9) + (+13)

= -9 + 9 + 4 = 4.

c. (+7) + (-10)

Solution. +7 + (-10) = +7 – 10

[∴ -10 = -7 – 3]

= 7 – 7 – 3 = -3.

d. (+12) + (-7)

Solution. (+12) + (-7) = 12 – 7

[∴ 12 = 5 + 7]

= 5 + 7 – 7 = 5.

Question 15. Find the solution of the following additions using a number line:

a) (-2) + 6

Solution.

∴ 2 + 6 = 4

b. (-6) + 2

Solution.

∴ -6 + 2 = -4

Question 16. Make two such questions and solve them using the number line.

Solution. 1) (-5) + 3

∴ 5 + 3 = -2

2) -3 + 5

Question 17. Find the solution of the following without using number line.

a) (+7) + (-11)

Solution. 7 – 11 = -4

b) (-13) + (+10)

Solution. 13 + 10 = -3.

c) (-7)+(+9)

Solution. -7 + 9 = 2.

d) (+10)+(-5)

Solution. 10 – 5 = 5.

Make five such questions and solve them.

Question 18. Find the solution of (-7) + (+3) without using the number line.

Solution. (-7) + (+3) = (-4)

Question 2. Find the solution of (+15) + (-9) without using the number line.

Solution. (+15) + (-9) = (+6)

Question 3. Find the solution of (-20) + (+20) without using the number line.

Solution. (-20) + (+20) = 0

Question 4. Find the solution of (+11) + (-11) without using the number line.

Solution. (11) + (-11) = 0

Question 5. Find the solution of (-5) + (+7) without using the number line.

Solution. (-5) + (+7) = (+2)

Haryana Board Class 6 Maths Solutions For  Chapter 6 Integers Exercise 6.2

Question 1. Using the number line write the integer which is:

a) 3 more than 5.

Solution.

∴ 3 more than 5 is 8.

Word problems on integers for Class 6 HBSE

b) 5 more than -5.

Solution.

∴ 5 more than -5 is ‘0’.

c) 6 less than 2.

Solution.

∴ 6 less than 2 is -4.

d) 3 less than -2.

Solution.

∴ 3 less than 2 is -5.

Question 2. Use number line and add the following integers :

a) 9 + (-6)

Solution.

∴ 9 – 6 = 3

b) 5 + (-11)

Solution.

∴ 5 – 11 = -6.

c) (-1) + (-7)

Solution.

∴ -1 – 7 = -8.

d) -5 + 10

Solution.

∴ -5 + 10 =5 .

e) (-1) + (-2) + (-3)

Solution.

∴ -1 – 2 – 3 = -6.

f) (-2) + 8 + (-4)

Solution.

∴ – 2 + 8 – 4 = 8 – 6 = 2.

Question 3. Add without using number line:

a) 11 + (-7).

Solution. 11 + (-7)

= 11 – 7

= 4 + 7 – 7 [∴ 11 = 4 + 7]

= 4 + 0 = 4

b) (-13) + (+18)

Solution. (-13) + (+18)

= -13 + 18

= -13 + 13 + 5

[∴ 18 = 13 + 5]

= 0 + 5

= 5

c) (-10) + (+19)

Solution. (-10) + (+19) = -10 + 19

= -10 + 10 + 9

= 0 + 9

= 9

d) (-250) + (+150)

Solution. -250 + 150 = -150 – 100 + 150

[∴ – 250 = -150 – 100]

= -150 + 150 – 100

= 0 – 100

= -100

e) (-380) + (-270)

Solution. (-380) + (-270)

= -380 – 270

= -650

f) (-217)+(-100)

Solution. (-217) + (-100)

= -217 – 100

= -317

Question 4. Find the sum of:

a) 137 and -354

Solution. 137 + (-354) = 137 – 137 – 217

[∴ -354 = -137 – 127]

= 0 – 217

= – 217

b) -52 and 52

Solution. -52 + 52 = 0

c) -312, 39 and 192.

Solution. -312, 39 and 192.

= -300 – 12 + 39 + 192

[∴ -312 – 300 – 12]

= -300 – 12 + 39 + 180 + 12

= -300 + 219 – 81.

d) -50 – 200 + 300

Solution. -50 – 200 + 300

= -250 + 300

= -250 + 250 + 50

[∴ 300 = 250 + 50]

= +50

= 50

Question 5. Find the sum?

a) (-7) + (-9) + 4 + 16

Solution. (-7) + (-9) + 4 + 16

= -16 + 20

= -16 + 16 + 4 (20 = 16 + 4)

= 0 + 4 – 4

b) (37) + (-2) + (-65) + (-8)

Solution. 37 + (-2) + (-65) + (-8) = 37 – 75

= 37 – 75

= -38

Chapter 6 Integers Exercise 6.3

Question 1. Find:

a) 35-(20)

Solution. 35 – 20 = 15 + 20 – 20 (35 = 15 + 20)

= 15 + 0

= 15

b) 72 – (90)

Solution. 72 – 90

= 72 – 72 – 18 [∴ -90 = -72 – 18]

=0 – 18

= -18

c) (-15) – (-18)

Solution. -15 – (-18)

= -15 + 18

= -15 + 15 + 3 [∴ 18 = 15 + 3]

= 0 + 3

= 3

d) (-20)-(13)

Solution. -20 – 13

= -33

e) 23 (12)

Solution. 23 – (-12)

= 23 + 12 = 35 [∴ – x – = +]

f)(-32) – (-40)

Solution. (-32) – (-40)

= -32 + 40

= -32 + 32 + 8

[∴ 40 = 32 + 8]

= 0 + 8

= 8.

Question 2. Fill in the blanks with >, < or = sign.

a) (-3)+(-6) ____ (-3)-(-6)

Solution. L.H.S. = -3 – 6 = -9

R.H.S. = -3 – (-6)

= -3 + 6 = 3

∴ (-3) + (-6) < (-3) – (-6)

b) (-21) – (-10) ____ (31) + (-11)

Solution. L.H.S. = -21 – (-10) = -21 + 10 = -11

R.H.S. = -31 + (-11) = -31 – 11 = -42

= (-21) – (-10) > (-31) + (-11)

c) (45)-(-11) ____ 57 + (-4).

Solution. L.H.S. = 45 – (-11) = 45 + 11 = 56

R.H.S. = 57 + (-4) = 57 – 4 = 53

= 45 – (-11) > 57 + (-4)

d) (-25)-(-42) ____ (-42)-(-25)

Solution. L.H.S. = -25 – (-42) = -25 + 42.

= -25 + 25 + 17

= 0 + 17 = 17

R.H.S. = (-42)-(-25) = -42 + 25

= -17 – 25 + 25

= -17 – 0 = -17

∴ (-25) – (-42) > (-42) – (-25)

Question 3. Fill in the blanks:

a) (-8) + ____ = 0

Solution. (-8) + 8= 0

b) 13 + ____ = 0

Solution. 13 + (-13) = 0

c) 12 + (-12) = ____

Solution. 12 + (-12) = 0

d) (-4) + ____ = -12

Solution. (-4) + (-8) = -12.

e) ____ -15 = -10

Solution. (+5) – 15 – 10

Question 4. Find:

a) (-7) – 8 – (-25)

Solution. (-7) -8 -(-25) = 7 – 8 + 25

= -7 – 8 + 7 + 18 [∴ 25 = 7 + 18]

= -8 + 18

= -8 + 8 + 10

= 0 + 10 = 10

b) (-13) + 32 – 8 – 1

Solution. (-13) + 32 – 8 – 1

= -13 + 32 – 9

= -13 + 23 + 9 – 9 [∴ 32 = 23 + 9]

= -13 + 23

= -13 + 13 + 10

= 0 + 10

= 10

c) (-7)+(-8)+(-90)

Solution. (-7) + (-8) + (-90)

= -7 – 8 – 90

= -15 – 90

= -105

d) 50 (-40)-(-2).

Solution. 50 – (-40) – (-2)

50 + 40 + 2 = 90 + 2

= 92

Haryana Board Class 6 Maths Solutions For  Chapter 6 Integers Very Short Answer Questions

Question 1. Put appropriate symbol > or < in the boxes given between the two integers.

1) – 1………. 0

Solution. -1 < 0

2) -3 …….. -7

Solution. -3 > -7

3)-10 ……. +10

Solution. -10 < +10

Question 2. The temperature recorded in Shimla is -4° c and in Kufri is -6°c on the same day. Which place is colder on that day? Why?

Solution. Temperature recorded in Shimla = -4°C;

Kufri = -6°C

Comparing these: -6 < -4

∴ Kufri is colder than Shimla.

Question 3. Find the sum of:

1) 120 and -274

Solution. The sum of 120 and -274 is

= 120 + (-274)

= -154

2) -68 and 28

Solution. The sum of -68 and 28 is

= -68 + 28

= -40

Question 4. Simplify

1) (-6) + (-10) + 5 + 17

Solution. -6 + (-10) + 5 + 17 = -16 + 22

= 6

2) 30 + (-30) = (-60) + (-18)

Solution. 30 + (-30) + (-60) + (-18)

= [30+(-30)] + [(-60)+(-18)]

= 0 + (-78)

= -78

Question 5. Fill in boxes with <, > or = sign.

1) (-4) + (-5) ____ (-5) – (-4)

Solution.-9 < -1

2) (-16)-(-23) ____ (-6) + (-12)

Solution. 7 > -18

3) 44-(-10) ____ 47+ (-3)

Solution. 54 > 44

Question 6. Fill in the blanks:

1) (-13) + …… = 0

Solution. (-13) + 13 = 0

2) (-16) + 16 =

Solution. (-16) + 16 = 0

3) (-5) + …………. = -14

Solution. (-5) + (-9) = -14

4) ……..+ (2 – 16) = -22

Solution. -8 + (2 – 16) = -22 [∴ 2 – 16 = -14]

Question 7. Arrange the following integers in ascending order and descending order. -1000, 10,-1,-100, 0, 1000, 1, -10

Solution. Ascending order: -1000, -100, -10, -1, 0, 10, 1000 (From smallest to greatest)

Descending order: 1000, 10, 0, 1-, -10, – 100,-1000 (From greatest to smallest)

Question 8. The temperature in Nainital is -5° C and in Shimla is -10° C. In which place is the temperature higher ?

Solution.

The temperature in Nainital is -5°C and temperature in Shimla is -10°C – 5°C is greaterthan -10°C.

∴ The temperature in Nainital is higher than the temperature in Shimla.

Question 9. Simplify (-7)+(-12)+5+12

Solution. (-7) + (-12) + 5 + 12 = [-7 – 12] + [5 + 12]

= -19 + 17

= -2

Question 10. Find 25 + (-21) + (-20) + 17 + (-1).

Solution. 25 + (-21) + (-20) + 17 + (-1)

= 25 + 17 + [-21 – 20 – 1]

= 42 + [-42]

= 42 – 42 = 0

Haryana Board Class 6 Maths Solutions For  Chapter 6 Integers Short Answer Questions

Question 11. Represent the integers on a number line as given below.

1) Integers lie between -7 and -2.

2) Integers lie between -2 and 5.

Solution. 1) The integers lie between -7 and -2 are -6, -5, -4, -3

2) The integers lie between -2 and 5 are -1, 0, 1, 2, 3, 4

Question 12. Write the following integers in increasing and decreasing order:

1) -7,5,-3

Solution. Increasing order: -7,-3,5

Decreasing order: 5, -3, -7

2) -1,3,0

Solution. Increasing order: -1,0,3

Decreasing order: 3, 0, -1

3) 1,3, -6

Solution. Increasing order: -6, 1, 3

Decreasing order: 3, 1, -6

4) -5, -3, -1

Solution. Increasing order: -5, -3, -1

Decreasing order: -1, -3, -5

Question 13. Write True or False, Correct those that are false:

1) Zero is on the right of -3

Solution. True

2) -12 and +12 represent the same integer on the number line.

Solution. False; Correct answer : -12 is a negative integer and +12 is a positive integer.

3) Every positive integer is greater than zero.

Solution. True

4) (-100) > (+100)

Solution. False; Correct answer: -100 < +100.

Question 14. Find the value of the following using a numberline.

1) (-3) + 5

2) (-5) + 3

Make your own two new questions and solve them using the number line.

Solution. 1) (-3) + 5

First move 3 steps to the left of 0 reach – 3 and then from that point we move 5 steps to the right. We reach the number.

∴ (-3) + 5 = 2

Solution. 2) (-5) + 3

First move 5 steps to the left of 0 reach – 5 and then from that point we move 3 steps to the right. We reach the number – 2.

∴ (-5) + 3 = -2.

Question 15. Add without using number line.

1) 10 + (-3)

2) (-10) + (+16)

3) (-8) + (+8)

Solution.

1) 10+ (-3)

= (7 + 3) + (-3)

= 7 + [3+(-3)]

= 7 + 0 = 7

2) (-10) + (+16)

= (-10) + 10 + 6

= [-10 + 10] + 6

= 0 + 6 = 6

3) (-8) + (+8) = 0

Chapter 6 Integers Long Answer Questions

Question 16. Find all integers which lie between the given two integers. Represent them on number line.

1) -1 and 1

2) -5 and 0

3) -6 and -8

4) 0 and -3

1) -1 and 1

Solution. The integer between -1 and 1 is 0.

2) -5 and 0

Solution. The integers between -5 and 0 are -4, -3, -2, -1.

3) -6 and -8

Solution. The integer between -6 and -8 is -7.

4) 0 and -3

Solution. The integers between 0 and -3 are -1, -2

Properties of integers Class 6 HBSE Maths

Question 17. Add the following integers using number line.

(1) 7 + (-6)

(2) (-8) + (-2)

(3) (-6)+(-5)+(+2)

1) 7 + (-6)

Solution.

First move to the right of 0 by 7 steps to each 7, then move 6 steps to the left of 7 we reach 1.

∴ 7 + (-6) = 1

2) (-8) + (-2)

Solution.

First move to the left of 0 by 8 steps to reach -8, then again move 2 steps to the left of -8 we reach -10.

3) (-6)+(-5) + (+2)

Solution.

First move to the left of 0 by 6 steps to reach – 6.

Again move 5 steps to the left of -6 we reach -11.

Now move 2 steps to the right of -11 we reach -9.

∴ (-6) + (-5) + (+2) = -9.

Question 18. In a quiz competition,where negative score for wrong answer is taken,Team A scored +10, -10, 0, -10, 10,-10 and Team B scored 10,10, -10,0,0,10 in 6 rounds successively. Which team wins the competition? How?

Solution. In a quiz, the score made by team A are +10, -10, 0, -10, 10, -10-

The total score made by team A

= 10 + (-10) + 0 + (-10) + 10 + (-10) = -10

The score made by team B are 10, 10, -10, 0, 0, 10

The total score made by team B

= 10 + 10 + (-10) + 0 + 0 + 10 = 20
Team B scored more than team A.

∴ Team B wins the quiz competition.

Question 19. Find the values of (1) (-3)+(-8)+(-5); (2) (-1)+7+ (-3) using number line.

1) (-3) + (-8) + (-5)

Solution.

First move to the left of 0 by 3 steps to reach – 3.

Again move 8 steps to the left of -3 we reach -11.

Now move 5 steps to the left of -11 we reach -16.

∴ (-3) + (-8) + (-5) = -16

2) (-1) + (7) + (-3)

Solution.

First move to the left of 0 by 1 steps to reach – 1.

Again move 7 steps to the right of -1 we reach 6.

Now move 3 steps to the left of -6 we reach 3.

∴ (-1) + 7 + (-3) = 3

Question 20. Represent the following statements using signs of integers :

1) An aeroplane is flying at a height of 3000 meters.

Solution. +3000 meters (height)

2) The fish is 10 meters below the water surface.

Solution. -10 meters (depth)

3) The temperature in Vijayawada is 35°C above 0°C.

Solution. +35° C (+ more),

4) Water freezes at 0°C temperature.

Solution. 0° C (neither ‘+’ nor ‘-‘)

5) The average temperature at the mount Everest in January is 36°C below zero degree.

Solution. -36° C (less)

6) The submarine is 500 meters below the surface of the sea.

Solution. -500 meters (below)

7) The average temperature at Darjeeling in July is 19°C below zero degree.

Solution. -19° C (less)

8) The average low temperature in Vishakapatnam during January is 18°C.

Solution. -18° C (more)

Haryana Board Class 6 Maths Solutions For  Chapter 6 Integers Objective Type Questions

Choose the correct answer :

Question 1. Set of integers is denoted by

1. N
2. Z
3. W
4. Q

Answer. 2. Z

Question 2. Write all the integers lie between 0 and -3

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

Answer. 3. -1, -2

Question 3. The sum of two negative integers is always

1. negative
2. positive
3. 0
4. None of these

Answer. 1. negative

Question 4. The sum of two integers one of which is positive and the other is negative then the sum may be

1. positive
2. negative
3. zero
4. any one of the above

Answer. 4. any one of the above

Question 5. Subtract +5 from (-5)

1. 0
2. 10
3. -10
4. 5

Answer. 3. -10

Question 6. 18 – 18 + 1 = …..

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

Answer. 4. 1

Question 7. -2 … 1

1. =
2. ⊥
3. <
4. >

Answer. 3. <

Question 8. Sum of 10 and -16 is …….

1. -6
2. 6
3. 12
4. -26

Answer. 1. -6

Question 9. Choose the correct matching:

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

Answer. 2. 1 – b, 2 – d, 3 – a, 4 – c

Question 10. The lest negative integer is

1. 1
2. -1
3. -7
4. does not exist

Answer. 4. does not exist

Question 11. Subract -8 from 8

1. 19
2. 0
3. 11
4. 16

Answer. 4. 16

Question 12. -4 – 4 + (-4) + (-4) =

1. -16
2. -12
3. 13
4. 84

Answer. 1. -16

Question 13. Additive inver of 7 – 3 is

1. 4
2. -4
3. 8
4. -7

Answer. 2. -4

Question 14. -20 – 82 = …

1. 111
2. -200
3. 100
4. -102

Answer. 4. -102

Question 15. Integer between -1 and + 1

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

Answer. 3. 0

Question 16. -131 + ……… = 0

1. -13
2. 131
3. 117
4. 814

Answer. 2. 131

Question 17. 38 – (-40) = …….

1. 31
2. 2
3. 16
4. 78

Answer. 4. 78

Question 18. -3 – (-3) = ………

1. -9
2. 0
3. 1
4. -2

Answer. 2. 0

Question 19. -8 + (-9) + (+17) =

1. -3
2. 11
3. -1
4. 0

Answer. 4. 0

Question 20. The smallest among -7, 0, 1, 5 is

1. 1
2. 5
3. 0
4. -7

Answer. 4. -7

Question 21. The greater negative integer in the following is

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

Answer. 1. -1

Question 22. …… is neither positive nor negative.

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

Answer. 4. 0

Question 23. Identify true statement from the following

1. -7 is to the left of 0
2. -3 is to the right of 1
3. -7 – 3 = -6
4. -1, -2, -3, -4 ….. are positive integers.

Answer. 1. -7 is to the left of 0

Question 24. The fish is 20 meters below the water surface can be denoted as

1. 21 m
2. -10 m
3. -20 m
4. 20 m

Answer. 3. -20 m

Question 25. Number of integers between 2017 and 2018 is

1. 0
2. 1
3. 3
4. does not exist

Answer. 1. 0

Question 26. -6 – 13 + 19 = …….

1. 10
2. -11
3. 0
4. 8

Answer. 3. 0

Question 27. ….. + (-16) = 20

1. 36
2. 20
3. 10
4. 12

Answer. 1. 36

Question 28. The profit of Rs 200/- is represented as

1. +200
2. -200
3. Both A & B
4. None

Answer. 1. +200

Question 29. 4° C below 0° is represented as

1. +4° C
2. -4° C
3. Both A & B
4. None

Answer. 2. -4° C

Question 30. Positive integers are

1. Whole numbers
2. Natural numbers
3. Both A & B
4. None

Answer. 2. Natural numbers

Question 31. Non negative integers are

1. Natural numbers
2. Either natural numbers or whole numbers
3. Whole numbers
4. Both B & C

Answer. 2. Either natural numbers or whole numbers

Question 32. Number of integers between -5 and 5

1. 10
2. 8
3. 9
4. 11

Answer. 3. 9

Question 33. Identify the false statement from the following

1. Zero is the right side of -3
2. Every positive integer is greater than zero
3. There are infinite integers in the number system
4. All integers are whole numbers

Answer. 4. All integers are whole numbers.

Question 34. The sum of two positive numbers is always

1. Positive
2. Negative
3. Either positive nor negative
4. Neither positive nor negative

Answer. 1. Positive

Question 35. The sum of two negative numbers is always

1. Positive
2. Negative
3. Either positive or negative
4. Neither positive nor negative

Answer. 1. Positive

Question 36. The sum of a positive and a negative number is

1. Positive
2. Negative
3. Either positive or negative
4. All above

Answer. 2. Negative

Question 37. a – (-a) =

1. a
2. 2a
3. 0
4. \(a^2\)

Answer. 2. 2a

Question 38. The biggest among -6, 0, 7, 1, -7, -9 is

1. -9
2. -7
3. 7
4. 0

Answer. 3. 7

Question 39. Add (-5) + (-5) + (-5) + 0

1. 15
2. -5
3. 0
4. -15

Answer. 4. -15

Question 40. Add the number 5 to additive inverse of -8 is

1. 11
2. -1
3. 1
4. 0

Answer. 2. -1

Question 41. Subtract 7 from additive inverse of -8 is

1. -1
2. 15
3. 1
4. -15

Answer. 3. 1

Question 42. Add additive inverse of 6 and additive inverse of -5

1. 1
2. -1
3. 11
4. -11

Answer. 2. -1

Question 43. Simplify 30 + (-30) + (-70) + 70 + 0

1. 100
2. 200
3. 0
4. 10