PSEB 9th Class Maths MCQ Chapter 4 Linear Equations in Two Variables

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 4 Linear Equations in Two Variables MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 4 Linear Equations in Two Variables MCQ Questions

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
If (2, – 2) is a root of 5x – 2y = k, then k = ………………. .
A. – 40
B. 6
C. 14
D. 10
Answer:
C. 14

PSEB 9th Class Maths MCQ Chapter 4 Linear Equations in Two Variables

Question 2.
If x = 2 and y = 1 is one of the solutions of 4x + ky = 11, then k = ……………… .
A. 2
B. 3
C. 5
D. 6
Answer:
B. 3

Question 3.
If (3, – 2) is one of the solutions of kx – 3y = 21, then k = ……………………. .
A. 3
B. – 3
C. 2
D. 5
Answer:
D. 5

Question 4.
The graph of 2x – 3y = 6 passes through points ……………… .
A. (2, – 3) and (- 2, 3)
B. (2, 3) and (3, 2)
C. (0, 2) and (- 3, 0)
D. (0, – 2) and (3, 0)
Answer:
D. (0, – 2) and (3, 0)

PSEB 9th Class Maths MCQ Chapter 4 Linear Equations in Two Variables

Question 5.
Expressing 4x = 2y – 7 in the y-form, we get y = ………….. .
A. 4x + 7
B. 4x + \(\frac{7}{2}\)
C. 2x + \(\frac{7}{2}\)
D. 2x – \(\frac{7}{2}\)
Answer:
C. 2x + \(\frac{7}{2}\)

Question 6.
If F = \(\left(\frac{9}{5}\right)\)C + 32, then C = ……………….. .
A. 5F – 160
B. \(\frac{1}{9}\) (5F – 32)
C. \(\frac{5}{9}\) F – 32
D. \(\frac{5}{9}\)(F – 32)
Answer:
D. \(\frac{5}{9}\)(F – 32)

PSEB 9th Class Maths MCQ Chapter 4 Linear Equations in Two Variables

Question 7.
For the equation F = \(\left(\frac{9}{5}\right)\)C + 32, F and C are numerically equal when ……………….. .
A. C = 45
B. C = – 40
C. C = 40
D. C = 32
Answer:
B. C = – 40

PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.4

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 4 Linear Equations in Two Variables Ex 4.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.4

Question 1.
Give the geometric representation of y = 3 as an equation
(i) in one variable
Answer:
If the equation y = 3 is treated as an equation in one variable, its graphical representation is a point on the number line as shown below:

PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.4 1

PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.4

(ii) in two variables.
Answer:
If the equation y = 3 is treated as an equation in two variables, it can be written as 0x + y = 3. Here, for any value of x, the value of y remains 3. Hence, we can easily take (0, 3), (2, 3) and (4, 3) as three solutions of the equation 0x + y = 3. Then, we plot these points in the Cartesian plane and draw the line passing through them. This line is the graph of equation y = 3 as an equation in two variables. This graph is perpendicular to the y-axis and parallel to the x-axis.

PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.4 2

Question 2.
Give the geometric representations of 2x + 9 = 0 as an equation
(i) In one variable
Answer:
If the equation 2x + 9 = 0, i.e., x = – \(\frac{9}{2}\) is treated as an equation in one variable. its graphical representation is a point on the number line as shown below:

PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.4 3

PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.4

(ii) In two variables.
Answer:
If the equation 2x + 9 = 0 is treated as an equation in two variables, it can be written as 2x + 0y + 9 = 0. Here, for any value of y, the value of x remains – \(\frac{9}{2}\). Hence, we can easily take \(\left(-\frac{9}{2}, 0\right)\), \(\left(-\frac{9}{2}, 2\right)\), and \(\left(-\frac{9}{2}, 4\right)\) as three solutions of the equation 2x + 9 = 0. Then, we plot these points in the Cartesian plane and draw the line passing through them. This line is the graph of the equation 2x + 9 = 0 as an equation in two variables. This graph is perpendicular to the x-axis and parallel to the y-axis.

PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.4 4

PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2

Question 1.
Find the roots of the following quadratic equations by factorisation:
(i) x2 – 3x – 10 = 0
(ii) 2x2 + x – 6 = 0
(iii) √2x2 + 7x + 5√2 = 0
(iv) 2 x2 – x + \(\frac{1}{8}\) = 0
(v) 100x2 – 20x + 1 = 0
Solution:
(i) Given quadratic
x2 – 3x – 10 = 0
Or x2 – 5x + 2x – 10 = 0
S = -3, p = -10
Or x (x – 5) + 2 (x – 5) = 0
Or (x – 5) (x + 2) = 0
Either x – 5 = 0 Or x + 2 = 0
x = 5 Or x = -2
Hence, 5 and -2 are roots of given Quadratic Equation.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

(ii) Given quadratic equation
2x2 + x – 6 = 0 =1
0r 2x2 + 4x – 3x – 6 = 0
S = 1 P = -6 × 2 = -12
Or 2x (x + 2) -3 (x + 2) = 0
Or (x + 2) (2x – 3) = 0
Either x + 2 = 0 Or 2x – 3 = 0
x = -2 Or x = –\(\frac{3}{2}\)
Hence, – 2 and \(\frac{3}{2}\) are roots of given quadratic equation.

(iii) Given Quadratic Equation,
√2x2 + 7x + 5√2 = 0
Or √2x2 + 2x + 5x + 5√2 = 0
S = 7, P = √2 × 5√2 = 10
Or √2x (x + √2) + 5 (x + √2) = 0
Or (x + √2) (√2x + 5) = 0
Either x + √2 = 0 Or √2x + 5 = 0
x = -√2 Or x = –\(\frac{-5}{\sqrt{2}}\)
Hence, -√2 and \(\frac{-5}{\sqrt{2}}\) are roots of given quadratic equation.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

(iv) Given quadratic equation
2x2 – x + \(\frac{1}{8}\) = 0
Or \(\frac{16 x^{2}-8 x+1}{8}\) = 0
Or 16x2 – 8x + 1 = 0
S = -8, P = 16 × 1 = 16
Or 16x2 – 8x + 1 = 0
Or 16x2 – 4x – 4x + 1 = 0
Or 4x(4x – 1) -1(4x – 1) = 0
Or (4x – 1) (4x – 1) = 0
Either 4x – 1 = 0
Or 4x – 1 = 0
x = \(\frac{1}{4}\) Or x = \(\frac{1}{4}\)
Hence, \(\frac{1}{4}\) and \(\frac{1}{4}\) are roots of given quadratic equation.

(v) Given quadratic equation,
100x2 – 20x + 1 = 0
Or 100x2 – 10x – 10x + 1 = 0
S = -20, P = 100 × 1 = 100
Or 10x(10x – 1) – 1 (10x – 1) = 0
Or (10x – 1)(10x – 1) = 0
Either 10x – 1 = 0 Or 10x – 1 = 0
x = \(\frac{1}{10}\) Or x = \(\frac{1}{10}\)
Hence, \(\frac{1}{10}\) and \(\frac{1}{10}\) are roots of given quadratic equation.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

Question 2.
Solve the problems given in Example I. Statements of these problems are given below:
(i) John and Jivanti together have 45 marbles. Both of them lost S marbles each, and the product of the number of marbles they now have is 124. We would lfke to find out how many marbles they had to start with.

(ii) A cottage Industry produces a certain number of toys in a day. The cost of production of each toy (In rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was 750. We would like to find out the number of toys produced on that day.

Solution:
(i) Let the number of marbles John had be x.
Then the number of marbles Jivanti had = 45 – x
The number of marbles Íeft withJohn, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5 = 40 – x
Therefore, their product = (x – 5) (40 – x)
= 40x – x2 – 200 + 5x
= -x2 + 45x – 200
According to question,
-x2 + 45x – 200 = 124
Or -x2 + 45x – 324 = 0
Or x2 – 45x + 324 =0
Or x2 – 36x – 9x + 324 = 0
S = -45, P = 324
Or x(x – 36) – 9(x – 36) = 0
Or (x – 36)(x – 9) = 0
Either x – 36 = 0, Or x – 9 = 0
x = 36 Or x = 9
∴ x = 36, 9
Hence, number of marbles they had to start with were 36 and 9 or 9 and 36.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

(ii) Let the number of toys produced on that day be x.
Therefore, the cost of production (in rupees) of each toy that day = 55 – x
So, the total cost of production (in rupees) that day = x (55 – x)
According to question.
x(55 – x) = 750
Or 55x – x2 = 750
Or -x2 + 55x – 750 = 0
Or x2 – 55x – 750 = 0
Or x2 – 30x – 25x + 750=0
S = -33, P = 750
Or x(x – 30) – 25(x – 30) = 0
Or (x – 30)(x – 25) = 0
Either x – 30 = 0 Or x – 25 = 0
x = 30 Or x = 25
∴ x = 30, 25
Hence, number of toys produced on that day were 30 and 25 or 25 and 30.

Question 3.
Find two numbers whose sum is 27 and product is 182.
Solution:
Let one number = x
2nd number = 27 – x
Their product = x (27 – x) = 27x – x2
According to question,
27x – x2 = 182
Or – x2 + 27x – 182 = 0
Or x2 – 27x + 182 = 0
S = -27, P = 182
Or x2 – 13x – 14x + 182 = 0
Or x(x – 13) – 14(x – 13) = 0
Or (x – 13) (x – 14) = 0
Either x – 13 = 0 Or x – 14 = 0
x = 13 Or x = 14
x = 13, 14
Hence, two numbers are 13 and 14 Or 14 and 13.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

Question 4.
Find two consecutive positive integers, sum of whose squares is 365.
Solution:
Let one positive integer = x
2nd positive integer = x + 1
According to question,
(x)2 + (x + 1)2 = 365
Or x2 + x2 + 1 + 2x = 365
Or 2x2 + 2x + 365 = 0
Or 2x2 + 2x – 364 = 0
Or x2 + x – 182 = 0
Or x2 + 14x – 13x – 182 = 0
S = 1, P = -182
Or x(x + 14) – 13(x + 14) = 0
(x + 14)(x— 13) = O
Either x + 14 = 0
Or x = -14
Or
x – 13 = 0
x = 13
∵ We have positive integers.
So, we reject x = – 14.
∴ x = 13
∴ One positive integer = 13
and 2nd positive integer = 13 + 1 = 14
Hence, required consecutive positive integers are 13 and 14.

Question 5.
The altitude of a right triangle is 7 cm less than its base. 1f the hypotenuse is 13 cm, find the other two sides.
Solution:
Let base of right triangle = x cm
Altitude of right triangle = (x – 7) cm
and hypotenuse of right triangle = 13 cm (Given)
According to Pythagoras Theorem,
(Base)2 + (Altitude)2 = (Hypotenuse)2
(x)2 ÷ (x – 7)2 = (13)2
Or x2 + x2 + 49 – 14x = 169
Or 2x2 – 14x + 49 – 169 = 0
Or 2x2 – 14x – 120 = 0
Or 2[x2 – 7x – 60] = 0
Or x2 – 7x – 60 = 0
Or x2 – 12x + 5x – 60 = 0
S = – 7 P = – 60
Or x(x – 12) + 5(x – 12) = 0
Or (x – 12) (x + 5) = 0
Either x – 12 = 0 Or x + 5 = 0
x = 12 Or x= – 5
∵ Length of any triangle cannot be negative.
So, we reject x = – 5
∴ x = 12
Hence, base of right triangle = 12 cm
Altitude of right triangle = (12 – 7) cm = 5 cm.

Question 6.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was 90, find the number of articles produced and the cost of each article.
Solution:
Let, number of pottery articles produced by industry in one day = x
Cost of production of each article = ₹ (2x + 3)
∴ Total cost of production in panicular day = ₹ [x(2x + 3)] = ₹ (2x2 + 3x)
According to question,
2x2 + 3x = 90
2x2 + 3x – 90 = 0
S = 3, P = 2 × -90 = -180
Or 2x2 – 12x + 15x – 90 = 0
Or 2x (x – 6) + 15 (x – 6) = 0
Or (x – 6) (2x + 15) = 0
Either x – 6 = 0 Or 2x + 15 = 0
x = 6 Or x = \(\frac{-15}{2}\)
∵ number of articles cannot be negative
So, we reject x = 2
∴ x = 6
Hence, number of articles produced on certain day = 6
and cost of production of each article = ₹ [2 × 6 + 3] = ₹ 15.

PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.3

Question 1.
Draw the graph of each of the following linear equations in two variables:
(i) x + y = 4
Answer:
To draw the graph of x + y = 4, we need at least two solutions of x + y = 4. And to be on the safer side, we get three solutions of x + y = 4.
For x = 0, we get 0 + y = 4, i.e., y = 4.
For x = 2, we get 2 + y = 4. i.e., y = 2.
For x = 4. we get 4 + y = 4, i.e., y = 0.
We can represent these solutions In the tabular form as below:
PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3 1
Then, we plot these points on the Cartesian plane and draw the line passing through them.
This line is the graph of x + y = 4.
PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3 2

PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3

(ii) x – y = 2
Answer:
x – y = 2
To draw the graph of x – y = 2, we
find three solutions of x – y = 2. For convenience, we express the equation in y-form as y = x – 2.
For x = 0, y = – 2.
For x = 2, y = 0.
For x = 4, y = 2.
We represent these solutions in the tabular form as below:
PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3 3
Then, we plot these points on the Cartesian plane and draw the line passing through them. This line is the graph of x – y = 2.
PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3 4

PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3

(iii) y = 3x
Answer:
y = 3x
To draw the graph of y = 3x, we find three solutions of y = 3x.
For x = 0, y = 3 × 0 = 0.
For x = 1, y = 3 × 1 = 3.
For x = 2, y = 3 × 2 = 6.
We represent these solutions in the tabular form as below:
PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3 5
Then, we plot these points on the Cartesian plane and draw the line passing through them. This line is the graph of y = 3x.
PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3 6

PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3

(iv) 3 = 2x + y
Answer:
To draw the graph of 3 = 2x + y. we find three solutions of 3 = 2x + y by expressing it as y = 3 – 2x.
For x = 0, y = 3 – 2 × 0 = 3.
For x = 1, y = 3 – 2 × 1 = 1.
For x = 2, y = 3 – 2 × 2 = – 1.
We represent these solutions in the tabular form as below:
PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3 7
Then, we plot these points on the Cartesian plane and draw the line passing through them. This line is the graph of 3 = 2x + y.
PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3 8

Question 2.
Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
Answer:
Equations y = 7x and x + y = 16 are two equations of lines passing through point (2, 14) as the coordinates of the point satisfy both the equations.
There are infinitely many equations which are satisfied by the coordinates of the point. Few examples of such equations are y – x = 12, y = 6x + 2, x – y = – 12, etc. This happens so because infinitely many lines pass through a point given in a plane.

PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3

Question 3.
If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
Answer:
As the point (3, 4) lies on the graph of the equation 3y = ax + 7, its coordinates, i.e., x = 3 and y = 4 must satisfy the equation.
Hence, we get
3(4) = a(3) + 7
∴ 12 = 3a + 7
∴ 12 – 7 = 3a
∴ 5 = 3a
∴ 3a = 5
∴ a = \(\frac{5}{3}\)

Question 4.
The taxi fare in a city is as follows: For the first kilometre, the fare is ₹ 8 and for the subsequent distance it is ₹ 5 per km. Taking the distance covered a x km and total fare as ₹ y, write a linear equation for this information, and draw its graph.
Answer:
Let the total distance covered be x km and the total fare be ₹ y. Now, the fare for the first km is ₹ 8 and for the remaining (x – 1) km, it is ₹ 5 per km. Hence, the total fare will turn out to be ₹ [(8 + 5 (x – 1)]. Hence, we get the equation as
8 + 5(x – 1) = y
∴ 8 + 5x – 5 = y
∴5x – y + 3 = 0
To draw the graph of this equation, we find three solutions of the equation by expressing the equation in the form y = 5x + 3.

Note: Distance travelled cannot be zero or negative. Hence, we choose only positive values of x.
For x = 1, y = 5(1) + 3 = 8.
For x = 2, y = 5(2) + 3 = 13.
For x = 3, y = 5(3) + 3 = 18.
We represent these solutions in the tabular form as below:
PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3 9Then, we plot these points on the Cartesian plane and draw the line passing through them. This line is the graph of the equation 5x – y + 3 = o derived above.
PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3 10

PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3

Question 5.
From the choices given below, choose the equation whose graphs are given in figure (1) and figure (2):
For figure (1)
(i) y = x
(ii) x + y = 0
(iii) y = 2x
(iv) 2 + 3y = 7x
PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3 11
Answer:
For figure (1): The graph of equation (ii) x + y = 0 is given in figure (1) as all the three points represented on the line, i.e., (- 1, 1), (0, 0) and (1, – 1) satisfy equation x + y = 0. For other equations, (1) y = x and (iii) y = 2x are satisfied by the point (0, 0), but not by the other two points. Equation (iv) 2 + 3y = 7x is not satisfied by any point.

For figure (2)
(i) y = x + 2
(ii) y = x – 2
(iii) y = – x + 2
(iv) x + 2y = 6
PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3 12
Answer:
For figure (2): The graph of equation (iii) y = – x + 2 is given in figure (2) as all the three points represented on the line, i.e., (- 1, 3), (0, 2) and (2, 0) satisfy equation y = – x + 2. Equation (i) y = x + 2 is satisfied by only one point (0, 2). Equation (ii) y = x – 2 is satisfied by only one point (2, 0). Equation (iv) x + 2y = 6 is not satisfied by any point.

PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3

Question 6.
If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is (i) 2 units (ii) 0 unit.
Answer:
We know well that
Work done = Force × Distance travelled.
Let the work done be y, the distance travelled be x and here the constant force applied is 5 units.
Then, the relation reduces to y = 5x which is a linear equation in two variables.
To draw the graph of y = 5x. we find three solutions of the equation and represent them in the tabular form.
For x = 1, y = 5 × 1 = 5.
For x = 3, y = 5 × 3 = 15.
For x = 4, y = 5 × 4 = 20.
PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3 13
Then, we plot these points on the Cartesian plane and draw the line passing through them. This line is the graph of the equation derived above. The coordinates of any point on the line will satisfy the derived equation.
PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3 14
(i) From the graph, we observe that when the distance travelled (x) is 2 units, the work done (y) is 10 units.
(ii) From the graph, we observe that when the distance travelled (x) is 0 unit, the work done (y) is 0 unit.

PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3

Question 7.
Yamini and Fatima, two students of class IX of a school, together contributed ₹ 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as ₹ x and ₹ y). Draw the graph of the same.
Answer:
1et the contribution of Yamini be ₹ x and the contribution of Fatima be ₹ y. Then, their total contribution is ₹ (x + y). Their total contribution is given to be ₹ 100. Hence, we get the linear equation x + y = 100.
Now, to draw the graph, we find three solutions.
For x = 0, y = 100. For x = 50, y = 50, For x = 100, y = 0.
We represent these solution in the tabular form as below:
PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3 15
We plot these three point in the Cartesian plane and draw the line passing through them. This line is the graph of the mathematical representation of the information given in the data.
PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3 16

PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3

Question 8.
In countries like USA and Canada, temperature is measured In Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius:
F = \(\left(\frac{9}{5}\right)\)C + 32
(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.
Answer:
F = \(\left(\frac{9}{5}\right)\)C + 32
For C = – 15,
F = \(\frac{9}{5}\)(- 15) + 32 = – 27 + 32 = 5
For C = 10,
F = \(\frac{9}{5}\)(10) + 32 = 18 + 32 = 50
For C = 60,
F = \(\frac{9}{5}\)(60) + 32 = 108 + 32 = 140
Hence, three solutions of F = \(\left(\frac{9}{5}\right)\)c + 32 can be given in the tabular form as below:
PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3 17
To draw the graph of F = \(\left(\frac{9}{5}\right)\)c + 32.
we take Celsius (°C) on the x-axis and Fahrenheit (°F) on the y-axis with scale 1 cm = 10 units on both the axes.
Then, we plot the points (- 15, 5), (10, 50) and (60, 140) and draw the line passing through them which is the graph of the equation F = \(\left(\frac{9}{5}\right)\)C + 32.

PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3 18

(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?
Answer:
If the temperature is 30 °c means the x-coordinate of the point is 30. Now, the point on the graph of F = \(\left(\frac{9}{5}\right)\)C + 32 having x-coordinate 30 has y-coordinate 86.
From the equation for C = 30, we get
F = \(\left(\frac{9}{5}\right)\)3o + 32 = 54 + 32 = 86.
Hence, if the temperature is 30°C, in the Fahrenheit scale it is 86°F.

(iii) If the temperature is 95 °E what is the temperature in Celsius?
Answer:
If the temperature is 95°F means the y-coordinate of the point is 95. Now, the point on the graph of F = \(\left(\frac{9}{5}\right)\)C + 32 having y-coordinate 95 has x-coordinate 35.
From the equation for F = 95, we get
95 = \(\left(\frac{9}{5}\right)\)C + 32
∴ 63 = \(\left(\frac{9}{5}\right)\)C
∴ C = 63 × \(\frac{5}{9}\)
∴ C = 35

PSEB 9th Class Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3

(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0 °F, what is the temperature in Celsius?
Answer:
As explained in (ii) and (iii) above, if the temperature is 0 °C, in the Fahrenheit scale it is 32°F. And if the temperature is 0 °F, in the Celsius scale it is – 18 °C approximately as observed from the graph. From the equation, for C = 0, we get
F = \(\left(\frac{9}{5}\right)\)0 + 32 = 0 + 32 = 32, and for
F = 0, we get
0 = \(\left(\frac{9}{5}\right)\)C + 32
∴ – 32 = \(\left(\frac{9}{5}\right)\)C
∴ C = – 32 × \(\frac{5}{9}\)
∴ C = – \(\frac{160}{9}\)
∴ C = – 17 \(\frac{7}{9}\)

(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
Answer:
Yes, there is a temperature which is numerically the same in both Fahrenheit and Celsius.
As we see, the graph of F = \(\left(\frac{9}{5}\right)\)C + 32 passes through the point (- 40, – 40), i.e., – 40°C = – 40°F.
From the equation for F = C, we get
C = \(\left(\frac{9}{5}\right)\)C + 32
∴ – 32 = \(\frac{9}{5}\)C – C
∴ – 32 = \(\frac{4}{5}\)C
∴ C = – 32 × \(\frac{5}{4}\)
∴ C = – 40
Hence, – 40 is the temperature which is numerically the same in both Fahrenheit and Celsius.

PSEB 7th Class Maths Solutions Chapter 9 Rational Numbers Ex 9.1

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 9 Rational Numbers Ex 9.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.1

1. Write two equivalent rational numbers of the following :

Question (i).
\(\frac {4}{5}\)
Solution:
\(\frac {4}{5}\) = \(\frac {4}{5}\) × \(\frac {2}{2}\)
= \(\frac {8}{10}\)
\(\frac {4}{5}\) = \(\frac {4}{5}\) × \(\frac {3}{3}\)
= \(\frac {12}{15}\)
∴ Equivalent rational numbers of \(\frac {4}{5}\) are \(\frac {8}{10}\) and \(\frac {12}{15}\)

Question (ii).
\(\frac {-5}{9}\)
Solution:
\(\frac {-5}{9}\) = \(\frac {-5}{9}\) × \(\frac {2}{2}\)
= \(\frac {-10}{18}\)
\(\frac {-5}{9}\) = \(\frac {-5}{9}\) × \(\frac {3}{3}\)
= \(\frac {-15}{27}\)
∴ Equivalent rational numbers of \(\frac {-5}{9}\) are \(\frac {-10}{18}\) and \(\frac {-15}{27}\)

Question (iii).
\(\frac {3}{-11}\)
Solution:
\(\frac {3}{-11}\) = \(\frac {3}{-11}\) × \(\frac {2}{2}\)
= \(\frac {6}{-22}\)
\(\frac {3}{-11}\) = \(\frac {3}{-11}\) × \(\frac {3}{3}\)
= \(\frac {9}{-33}\)
∴ Equivalent rational numbers of \(\frac {3}{-11}\) are \(\frac {6}{-22}\) and \(\frac {9}{-33}\)

PSEB 7th Class Maths Solutions Chapter 9 Rational Numbers Ex 9.1

2. Find the standard form of the following rational numbers :

Question (i).
\(\frac {35}{49}\)
Solution:
\(\frac {35}{49}\)
∵ H.C.F. of 35 and 49 is 7
PSEB 7th Class Maths Solutions Chapter 9 Rational Numbers Ex 9.1 1
So dividing both the numerator and denominator by 7 we get.
\(\frac {35}{49}\) = \(\frac{35 \div 7}{49 \div 7}\) = \(\frac {5}{7}\)
∴ Standard form of \(\frac {35}{49}\) is \(\frac {5}{7}\)

Question (ii).
\(\frac {-42}{56}\)
Solution:
\(\frac {-42}{56}\)
∵ H.C.F. of -42 and 56 is 14
PSEB 7th Class Maths Solutions Chapter 9 Rational Numbers Ex 9.1 2
So dividing both the numerator and denominator by 14 we get.
\(\frac {-42}{56}\) = \(\frac{-42 \div 14}{56 \div 14}\) = \(\frac{-3}{4}\)
∴ Standard form of \(\frac {-42}{56}\) is \(\frac{-3}{4}\)

Question (iii).
\(\frac {19}{-57}\)
Solution:
\(\frac {19}{-57}\)
∵ H.C.F. of 59 and 57 is 19
PSEB 7th Class Maths Solutions Chapter 9 Rational Numbers Ex 9.1 3
So dividing both the numerator and denominator by 19 we get.
\(\frac {19}{-57}\) = \(\frac{-19 \div 19}{-57 \div 19}\) = \(\frac{1}{-3}\)
∴ Standard form of \(\frac {19}{-57}\) is \(\frac{1}{-3}\)

PSEB 7th Class Maths Solutions Chapter 9 Rational Numbers Ex 9.1

Question (iv).
\(\frac{-12}{-36}\)
Solution:
\(\frac{-12}{-36}\)
∵ H.C.F. of 12 and 36 is 12.
PSEB 7th Class Maths Solutions Chapter 9 Rational Numbers Ex 9.1 4
So dividing both the numerator and denominator by 12 we get.
\(\frac{-12}{-36}\) = \(\frac{-12 \div 12}{-36 \div 12}\) = \(\frac{1}{3}\)
Standard form of \(\frac{-12}{-36}\) is \(\frac{1}{3}\)

3. Which of the following pairs represent same rational number ?

Question (i).
\(\frac{-15}{25}\) and \(\frac{18}{-30}\)
Solution:
\(\frac{-15}{25}\) = \(\frac{-15 \div 5}{25 \div 5}\)
= \(\frac{-3}{5}\)
\(\frac{18}{-30}\) = \(\frac{18 \div-6}{-30 \div-6}\)
= \(\frac{-3}{5}\)
∴ \(\frac{-15}{25}\) and \(\frac{18}{-30}\) represents the same number.

Question (ii).
\(\frac{2}{3}\) and \(\frac{-4}{6}\)
Solution:
\(\frac{2}{3}\) = \(\frac{2}{3}\) × \(\frac{1}{1}\)
= \(\frac{2}{3}\)
\(\frac{-4}{6}\) = \(\frac{-4 \div 2}{6 \div 2}\)
= \(\frac{-2}{3}\)
∴ \(\frac{-2}{3}\) and \(\frac{-4}{6}\) doesnot represents the same rational numbers.

Question (iii).
\(\frac{-3}{4}\) and \(\frac{-12}{16}\)
Solution:
\(\frac{-3}{4}\) = \(\frac{-3}{4}\) × \(\frac{4}{4}\)
= \(\frac{-12}{16}\)
\(\frac{-12}{16}\) = \(\frac{-12}{16}\)
∴ \(\frac{-3}{4}\) and \(\frac{-12}{16}\) represents the same rational number.

Question (iv).
\(\frac{-3}{-7}\) and \(\frac{3}{7}\)
Solution:
\(\frac{-3}{4}\) = \(\frac{-3 \div-1}{-7 \div-1}\)
= \(\frac{-3}{4}\)
∴ \(\frac{-3}{-7}\) and \(\frac{3}{7}\) represents the same rational number.

PSEB 7th Class Maths Solutions Chapter 9 Rational Numbers Ex 9.1

4. Which is greater in each of the following ?

Question (i).
\(\frac{3}{7}\), \(\frac{4}{5}\)
Solution:
Given rational nrnnbere are \(\frac{3}{7}\) and \(\frac{4}{5}\)
L.C.M. of 7 and 5 is 35
∴ \(\frac{3}{7}\) = \(\frac{3 \times 5}{7 \times 5}\)
= \(\frac{15}{35}\)
and \(\frac{4}{5}\) = \(\frac{4 \times 7}{5 \times 7}\)
= \(\frac{28}{35}\)
∵ Numerator of second is greater than first i.e. 28 > 15
So \(\frac{4}{5}\) > \(\frac{3}{7}\)

Question (ii).
\(\frac{-4}{12}\), \(\frac{-8}{12}\)
Solution:
Given rational numbere are \(\frac{-4}{12}\) and \(\frac{-8}{12}\)
∵ Numerator of first is greater than second i.e. -4 > – 8
∴ \(\frac{-4}{12}\) > \(\frac{-8}{12}\)

Question (iii).
\(\frac{-3}{9}\), \(\frac{4}{-18}\)
Solution:
Given rational numbers are \(\frac{-3}{9}\), \(\frac{4}{-18}\)
\(\frac{-3}{9}\) = \(\frac{-3 \times 2}{9 \times 2}\)
= \(\frac{-6}{18}\)
\(\frac{4}{-18}\) = \(\frac{4 \times-1}{-18 \times-1}\)
\(\frac{-4}{18}\)
Since -4 > – 6.
\(\frac{4}{-18}\) > \(\frac{-3}{9}\)

Question (iv).
-2\(\frac{3}{5}\), -3\(\frac{5}{8}\)
Solution:
-2\(\frac{3}{5}\) = \(\frac{-13}{5} \times \frac{8}{8}\)
= \(\frac{-104}{40}\)
-3\(\frac{5}{8}\) = \(\frac{-29}{8} \times \frac{5}{5}\)
= \(\frac{-135}{40}\)
∵ -104 > -135
∴ \(\frac{-13}{5}\) > \(\frac{-29}{8}\)
Thus, -2\(\frac{3}{5}\) > -3\(\frac{5}{8}\)

PSEB 7th Class Maths Solutions Chapter 9 Rational Numbers Ex 9.1

5. Write the following rational numbers in ascending order.

Question (i).
\(\frac{-5}{7}, \frac{-3}{7}, \frac{-1}{7}\)
Solution:
\(\frac{-5}{7}, \frac{-3}{7}, \frac{-1}{7}\)
Here -5 < -3 < -1
i.e. \(\frac{-5}{7}, \frac{-3}{7}, \frac{-1}{7}\)
Therefore, the ascending order is:
\(\frac{-5}{7}, \frac{-3}{7}, \frac{-1}{7}\)

Question (ii).
\(\frac{-1}{5}, \frac{-2}{15}, \frac{-4}{5}\)
Solution:
\(\frac{-1}{5}, \frac{-2}{15}, \frac{-4}{5}\)
L.C.M of 5, 15, 5 is 15
PSEB 7th Class Maths Solutions Chapter 9 Rational Numbers Ex 9.1 5

Question (iii).
\(\frac{-3}{8}, \frac{-2}{4}, \frac{-3}{2}\)
Solution:
\(\frac{-3}{8}, \frac{-2}{4}, \frac{-3}{2}\)
L.C.M of 8, 4, 2 is 8
∴ \(\frac{-3}{8}=\frac{-3}{8} \times \frac{1}{1}=\frac{-3}{8}\)
\(\frac{-2}{4}=\frac{-2 \times 2}{4 \times 2}=\frac{-4}{8}\)
\(\frac{-3}{2}=\frac{-3 \times 4}{2 \times 4}=\frac{-12}{8}\)
∴ -12 < -4 < -3
or \(\frac {-12}{8}\) < \(\frac {-4}{8}\) < \(\frac {-3}{8}\)
Hence assending order is \(\frac{-3}{2}, \frac{-2}{4}, \frac{-3}{8}\)

PSEB 7th Class Maths Solutions Chapter 9 Rational Numbers Ex 9.1

6. Write five rational numbers between following rational numbers.

Question (i).
-2 and -1
Solution:
Given rational numbers are -2 and -1
Let us write -2 and -1 as rational numbers with 5 + 1 = 6 as denominator.
We have -2 = -2 × \(\frac {6}{6}\)
= \(\frac {-6}{6}\)
\(\frac {-12}{6}\) < \(\frac {-11}{6}\) < \(\frac {-10}{6}\) < \(\frac {-9}{6}\) < \(\frac {-8}{6}\) < \(\frac {-7}{6}\) < \(\frac {-6}{6}\)
Hence five rational numbers between -2 and -1 are :
\(\frac {-11}{6}\),\(\frac {-10}{6}\),\(\frac {-9}{6}\),\(\frac {-8}{6}\),\(\frac {-7}{6}\)
i.e. \(\frac {-11}{6}\),\(\frac {-5}{3}\),\(\frac {-3}{2}\),\(\frac {-4}{3}\),\(\frac {-7}{6}\)

Question (ii).
\(\frac {-4}{5}\) and \(\frac {-2}{3}\)
Solution:
Given rational numbers are \(\frac {-4}{5}\) and \(\frac {-2}{3}\)
First we find equivalent rational numbers having same denominator
Thus \(\frac {-4}{5}\) = \(\frac{-4 \times 9}{5 \times 9}\)
= \(\frac {-36}{45}\)
and \(\frac {-2}{3}\) = \(\frac{-2 \times 15}{3 \times 15}\)
= \(\frac {-30}{45}\)
Now, we choose any five integers -35, -34, -33, -32, -31 between the numerators -36 and -30
Then the five rational numbers between \(\frac {-36}{45}\) and \(\frac {-30}{45}\) are:
\(\frac{-35}{45}, \frac{-34}{45}, \frac{-33}{45}, \frac{-32}{45}, \frac{-31}{45}\)
Hence, five rational numbers between \(\frac {-4}{5}\) and \(\frac {-2}{3}\) are
\(\frac{-35}{45}, \frac{-34}{45}, \frac{-33}{45}, \frac{-32}{45}, \frac{-31}{45}\)
i.e. \(\frac{-7}{9}, \frac{-34}{45}, \frac{-11}{15}, \frac{-32}{45}, \frac{-31}{45}\)

Question (iii).
\(\frac {1}{3}\) and \(\frac {5}{7}\)
Solution:
Given rational numbers are \(\frac {1}{3}\) and \(\frac {5}{7}\)
First we find equivalent rational numbers having same denominator
PSEB 7th Class Maths Solutions Chapter 9 Rational Numbers Ex 9.1 6
\(<\frac{4}{7}<\frac{13}{21}<\frac{2}{3}<\frac{5}{7}\)
Hence, five rational numbers between \(\frac {1}{3}\) and \(\frac {5}{7}\) are
\(\frac{8}{21}, \frac{3}{7}, \frac{10}{21}, \frac{4}{7}, \frac{13}{21}\).

PSEB 7th Class Maths Solutions Chapter 9 Rational Numbers Ex 9.1

7. Write four more rational numbers in each of the following.

Question (i).
\(\frac{-1}{5}, \frac{-2}{10}, \frac{-3}{15}, \frac{-4}{20}, \ldots\)
Solution:
The given rational numbers are :
\(\frac{-1}{5}, \frac{-2}{10}, \frac{-3}{15}, \frac{-4}{20}, \ldots\)
\(\frac {-1}{5}\) is the rational number in its lowest form
Now, we can write
\(\frac{-2}{10}=\frac{-1}{-5} \times \frac{2}{2}\),
\(\frac{-3}{15}=\frac{-1}{5} \times \frac{3}{3}\) and \(\frac{-1}{5}=\frac{-1}{5} \times \frac{4}{4}\)
Thus, we observe a pattern in these numbers.
The next four rational numbers would be
\(\frac{-1}{5} \times \frac{5}{5}=\frac{-5}{25}\),
\(\frac{-1}{5} \times \frac{6}{6}=\frac{-6}{30}\),
\(\frac{-1}{5} \times \frac{7}{7}=\frac{-7}{35}\)
\(\frac{-1}{5} \times \frac{8}{8}=\frac{-8}{40}\)
Hence required four more rational numbers are :
\(\frac{-5}{25}, \frac{-6}{30}, \frac{-7}{35}, \frac{-8}{40}\)

Question (ii).
\(\frac{-1}{7}, \frac{2}{-14}, \frac{3}{-21}, \frac{4}{-28}, \ldots\)
Solution:
The given rational numbers are
\(\frac{-1}{7}, \frac{2}{-14}, \frac{3}{-21}, \frac{4}{-28}, \ldots\)
\(\frac {-1}{7}\) is the rational number in its lowest form
Now, we can write
\(\frac{2}{-14}=\frac{-1}{7} \times \frac{-2}{-2}=\frac{2}{-14}, \frac{3}{-21}\)
= \(\frac{-1}{7} \times \frac{-3}{-3}\) and \(\frac{4}{-28}=\frac{-1}{7} \times \frac{-4}{-4}\)
Thus, we observe a pattern in these numbers.
The next four rational numbers would be :
\(-\frac{1}{7} \times \frac{-5}{-5}=\frac{5}{-35}\), \(\frac{-1}{7} \times \frac{-6}{-6}=\frac{6}{-42}\),
\(\frac{-1}{7} \times \frac{-7}{-7}=\frac{7}{-49}\), \(\frac{-1}{7} \times \frac{-8}{-8}=\frac{8}{-56}\)
Hence required four more rational numbers are :
\(\frac{5}{-35}, \frac{6}{-42}, \frac{7}{-49}, \frac{8}{-56}\)

PSEB 7th Class Maths Solutions Chapter 9 Rational Numbers Ex 9.1

8. Draw a number line and represent the following rational number on it.

Question (i).
\(\frac {2}{4}\)
Solution:
Draw a line and choose a point O on it to represent the rational number zero (0). We choose a point A to the right of 0 to represent 1.
Divide the segment OA into four equal parts. Second part from O to the right represents the rational number \(\frac {2}{4}\) as shown in the figure.
PSEB 7th Class Maths Solutions Chapter 9 Rational Numbers Ex 9.1 7

Question (ii).
\(\frac {-3}{4}\)
Solution:
Draw a line and choose a point O on it to represent the rational number zero (0). We choose a point A to the right of 0 to represent -1.
Divide the segment OA into four equal parts. Third part from O to the left represents the rational number \(\frac {-3}{4}\) as shown in the figure.
PSEB 7th Class Maths Solutions Chapter 9 Rational Numbers Ex 9.1 8

Question (iii).
\(\frac {5}{8}\)
Solution:
Draw a line and choose a point O on it to represent the rational number zero (0).
We choose a point A to the right of 0 to represent 1.
Divide the segment OA into eight equal parts. Fifth part from O to the right represents the rational number as shown in the figure.
PSEB 7th Class Maths Solutions Chapter 9 Rational Numbers Ex 9.1 9

Question (iv).
\(\frac {-6}{4}\)
Solution:
Draw a line and choose a point O on it to represent the rational number zero (0).
We choose a point A to the left of O to represent -2.
Divide the segment OA into eight equal parts. Sixth part from O to the left represents the rational number \(\frac {-6}{4}\) as shown in the figure.
PSEB 7th Class Maths Solutions Chapter 9 Rational Numbers Ex 9.1 10

9. Multiple Choice Questions :

Question (i).
\(\frac{3}{4}=\frac{?}{12}\) then ? =
(a) 3
(b) 6
(c) 9
(d) 12.
Answer:
(c) 9

PSEB 7th Class Maths Solutions Chapter 9 Rational Numbers Ex 9.1

Question (ii).
\(\frac{-4}{7}=\frac{?}{14}\) then ? =
(a) -4
(b) -8
(c) 4
(d) 8
Answer:
(b) -8

Question (iii).
The standard form of rational number \(\frac {-21}{28}\) is
(a) \(\frac {-3}{4}\)
(b) \(\frac {3}{4}\)
(c) \(\frac {3}{7}\)
(d) \(\frac {-3}{7}\)
Answer:
(a) \(\frac {-3}{4}\)

Question (iv).
Which of the following rational number is not equal to \(\frac {7}{-4}\) ?
(a) \(\frac {14}{-8}\)
(b) \(\frac {21}{-12}\)
(c) \(\frac {28}{-16}\)
(d) \(\frac {7}{-8}\)
Answer:
(d) \(\frac {7}{-8}\)

Question (v).
Which of the following is correct ?
(a) 0 > \(\frac {-4}{9}\)
(b) 0 < \(\frac {-4}{9}\)
(c) 0 = \(\frac {4}{9}\)
(d) None
Answer:
(a) 0 > \(\frac {-4}{9}\)

Question (vi).
Which of the following is correct ?
(a) \(\frac{-4}{5}<\frac{-3}{10}\)
(b) \(\frac{-4}{5}>\frac{3}{-10}\)
(c) \(\frac{-4}{5}=\frac{3}{-10}\)
(d) None
Answer:
(a) \(\frac{-4}{5}<\frac{-3}{10}\)

PSEB 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.3

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 3 Playing with Numbers Ex 3.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.3

1. Find prime factors of the following numbers by factor tree method:

Question (i)
96
Solution:
PSEB 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.3 1
∴ Prime factorisation of 96 = 2 × 2 × 2 × 2 × 2 × 3

PSEB 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.3

Question (ii)
120
Solution:
PSEB 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.3 2
∴ Prime factorisation of 120 = 2 × 2 × 2 × 3 × 5

Question (iii)
180.
Solution:
PSEB 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.3 3
∴ Prime factorisation of 180 = 2 × 2 × 3 × 3 × 5

2. Complete each factor tree:

Question (i)
PSEB 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.3 4
Solution:
PSEB 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.3 5

PSEB 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.3

Question (ii)
PSEB 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.3 6
Solution:
PSEB 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.3 7

Question (iii)
PSEB 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.3 8
Solution:
PSEB 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.3 9

PSEB 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.3

3. Find the prime factors of the following numbers by division method:

Question (i)
420
Solution:
PSEB 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.3 10
∴ 420 = 2 × 2 × 3 × 5 × 7

Question (ii)
980
Solution:
PSEB 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.3 11
∴ 980 = 2 × 2 × 5 × 7 × 7

Question (iii)
225
Solution:
PSEB 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.3 12
∴ 225 = 3 × 3 × 5 × 5

PSEB 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.3

Question (iv)
150
Solution:
PSEB 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.3 13
∴ 150 = 2 × 3 × 5 × 5

Question (v)
324
Solution:
PSEB 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.3 14
∴ 324 = 2 × 2 × 3 × 3 × 3 × 3

PSEB 6th Class Social Science Notes Chapter 22 Public Property and its Protection

This PSEB 6th Class Social Science Notes Chapter 22 Public Property and its Protection will help you in revision during exams.

PSEB 6th Class Social Science Notes Chapter 22 Public Property and its Protection

→ Personal Property: Personal property is the property owned by individuals or a group of individuals for private enjoyment and use.

→ Public Property: Public property is owned by the public or a whole community.

→ Public Services: Basic facilities provided by the government to the public are called public services.

PSEB 6th Class Social Science Notes Chapter 22 Public Property and its Protection

→ Family Property: The family property is the property that belongs to the family.

→ Historical Monuments: Historical monuments are evidence of historical past and ancient glory in the shape of forts, palaces, temples, mosques, gurdwaras, and churches.

→ The Ancient Monuments and Archaeological Sites and Remains Act, 1958: The Ancient Monuments and Archaeological Sites and Remains Act was passed by the Government of India in 1958.

→ Under this Act, the persons who damage historical monuments can be prosecuted and punished by law.

PSEB 6th Class Social Science Notes Chapter 21 Urban Development – Local Self Government

This PSEB 6th Class Social Science Notes Chapter 21 Urban Development – Local Self Government will help you in revision during exams.

PSEB 6th Class Social Science Notes Chapter 21 Urban Development – Local Self Government

→ Institutions of local self-government in urban areas: Nagar Panchayat, Municipal Council or Municipal Committee, and Municipal Corporation are the three institutions of local self-government in urban areas.

→ Nagar Panchayat: A Nagar Panchayat is formed in a town with a population of not more than 20,000.

→ Municipal Councilor Municipal Committee: A city with a population from 20,000 to one lac, has a Municipal Councilor Municipal Committee.

→ Municipal Corporation: A Municipal Corporation is a local body for a big city.

PSEB 6th Class Social Science Notes Chapter 21 Urban Development - Local Self Government

→ It is the highest institution of urban local self-government. It is established in a city with a population of lakhs.

→ Councilors: Members of a Municipal Council (Committee) and a Municipal Corporation are called councilors.

→ Mayor: Mayor is the head of a Municipal Corporation. He is elected by the members of the Municipal Corporation.

→ District Administration: The administration at the district level is called District Administration. It is headed by Deputy Commissioner.

→ Judicial Administration: There are two types of courts in a district. Civil courts deal with matters related to property and money, while criminal courts deal with cases like theft, murder, and criminal assault.

PSEB 6th Class Social Science Notes Chapter 20 Rural Development and Local Self Government

This PSEB 6th Class Social Science Notes Chapter 20 Rural Development and Local Self Government will help you in revision during exams.

PSEB 6th Class Social Science Notes Chapter 20 Rural Development and Local Self Government

→ Local Self-Government: A local Self-Government is a form of government at the local level. It is formed to solve local problems.

→ India-A Land of Villages: India is a land of villages. There are about six lac villages. Nearly 75% population of India lives in villages.

→ Panchayati Raj: The structure of rural local self-government in India is known as Panchayati Raj.

→ Institutions of Panchayati Raj: Village Panchayat, Panchayat (Block) Samiti, and Zila Parishad are the three institutions of Panchayati Raj.

PSEB 6th Class Social Science Notes Chapter 20 Rural Development and Local Self Government

→ Village Panchayat: Village Panchayat is the lowest unit of Panchayati Raj. It is established in a village with a population of 500 or more.

→ Panchayat(Block) Samiti: Panchayat (Blpgk) Samiti is a committee formed of members from Village Panchayats and others. It works at the block level. A block is a group of villages in a part of a district.

→ Zila Parishad: Zila Parishad is the highest unit of Panchayati Raj. It is a council of members drawn from Panchayat Samiti, the State Legislative Assembly and Legislative Council, and also the members of Parliament.

→ Tenure of the Institutions of Panchayati Raj: Tenure of all the three institutions of Panchayati Raj (Village Panchayat, Panchayat Samiti, and Zila Parishad) is five years.

PSEB 6th Class Social Science Notes Chapter 19 Community Meet Human Needs

This PSEB 6th Class Social Science Notes Chapter 19 Community Meet Human Needs will help you in revision during exams.

PSEB 6th Class Social Science Notes Chapter 19 Community Meet Human Needs

→ Society: Society is a system where people live together in organised communities and interact with each other.

→ Community: A community is a group of people who have definite characteristics and who live in a place, district or country.

PSEB 6th Class Social Science Notes Chapter 19 Community Meet Human Needs

→ Family: Family is a system in which parents, children, or grandchildren live together. It is the first unit of society.

→ Man as a social animal: Man is a social animal. He cannot live alone. He always lives with his fellow beings.

→ Our basic needs: Food, clothes, and home are our basic needs. We need food to eat, clothes to wear, and a home to live in.

→ Civic life: Civic life is the sharing of joys and sorrows with one another and cooperating with each other in family, neighbourhood, school, or society.

→ Duty: Doing work according to the rules and dictates of one’s organization, society or country is called duty.

→ Man in primitive times: Man in primitive times was a nomad. He had to wander from place to place in search of food and shelter.

→ But the invention of agriculture made him live in one place permanently.

→ India as a country with unity in diversity: India is known as a country with unity in diversity.

→ Here, large-scale physical, social, economic, religious, regional, and political diversities are found. Still, there is emotional unity among people.

PSEB 6th Class Social Science Notes Chapter 19 Community Meet Human Needs

→ Life and co-operation: Life depends upon co-operation. Every man has to depend upon others for the fulfillment of his needs.