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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:

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.

(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:

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.

(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:

Then, we plot these points on the Cartesian plane and draw the line passing through them. This line is the graph of y = 3x.

(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:

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.

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.

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:
Then, 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.

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

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

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.

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.

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.

(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.

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:

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.

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:

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.

(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

(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.

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 \(\frac {9}{-33}\)

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

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

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

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 \(\frac{1}{-3}\)

Question (iv).
\(\frac{-12}{-36}\)
Solution:
\(\frac{-12}{-36}\)
∵ H.C.F. of 12 and 36 is 12.

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.

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

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

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 \(\frac{-3}{2}, \frac{-2}{4}, \frac{-3}{8}\)

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

\(<\frac{4}{7}<\frac{13}{21}<\frac{2}{3}<\frac{5}{7}\)
Hence, five rational numbers between \(\frac {1}{3}\) and \(\frac {5}{7}\) are
\(\frac{8}{21}, \frac{3}{7}, \frac{10}{21}, \frac{4}{7}, \frac{13}{21}\).

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 :
\(\frac{5}{-35}, \frac{6}{-42}, \frac{7}{-49}, \frac{8}{-56}\)

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.

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.

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.

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.

9. Multiple Choice Questions :

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

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) \(\frac{-4}{5}<\frac{-3}{10}\)

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:

∴ Prime factorisation of 96 = 2 × 2 × 2 × 2 × 2 × 3

Question (ii)
120
Solution:

∴ Prime factorisation of 120 = 2 × 2 × 2 × 3 × 5

Question (iii)
180.
Solution:

∴ Prime factorisation of 180 = 2 × 2 × 3 × 3 × 5

2. Complete each factor tree:

Question (i)

Solution:

Question (ii)

Solution:

Question (iii)

Solution:

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

Question (i)
420
Solution:

∴ 420 = 2 × 2 × 3 × 5 × 7

Question (ii)
980
Solution:

∴ 980 = 2 × 2 × 5 × 7 × 7

Question (iii)
225
Solution:

∴ 225 = 3 × 3 × 5 × 5

Question (iv)
150
Solution:

∴ 150 = 2 × 3 × 5 × 5

Question (v)
324
Solution:

∴ 324 = 2 × 2 × 3 × 3 × 3 × 3

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.1

Question 1.
Check whether the following are quadratic equations:
(i) (x + 1)² = 2(x – 3)
(ii) x² – 2x = (-2) (3 – x)
(iii) (x – 2) (x + 1) = (x – 1) (x + 3)
(iv) (x – 3)(2x + 1) = x (x + 5)
(v) (2x – 1) (x – 3) = (x + 5) (x – 1)
(vi) x² + 3x + 1 = (x – 2)
(vii) (x + 2)³ = 2x(x² – 1)
(viii) x³ – 4x² – x + 1 = (x – 2)³

Solution:
(i) Given that
(x + 1)² = 2(x – 3)
Or x² + 1 + 2x = 2x – 6
Or x² + 1 + 2x – 2x + 6 = 0
Or x² + 7 = 0
Or x² + 0x + 7 = 0
which is in the formof ax² + bx + c = 0;
∴ It is a quadratic equation.

(ii) Given that
x² – 2x = (-2) (3 – x)
Or x² – 2x = -6 + 2x
Or x² – 2x + 6 – 2x = 0
Or x² – 4x + 6 = 0
which is the form of ax² + bx + c = 0; a ≠ 0
∴ It is the quadratic equation.

(iii) Given that ,
(x – 2) (x + 1) = (x – 1) (x + 3)
Or x² + x – 2x – 2 = x² + 3x – x – 3
Or x² – x – 2 = x² + 2x – 3
Or x² – x – 2 – x² -2x + 3 = 0
Or -3x + 1 = 0 which have no term of x².
So it is not a quadratic equation.

(iv) Given that
(x – 3)(2x + 1) = x(x + 5)
Or 2x² + x – 6x – 3 = x² + 5x
Or 2x² – 5x – 3 – x² – 5x = 0
Or x² – 10x – 3 = 0
which is a form of ax² + bx + c = 0; a ≠ 0
∴ It is a quadratic equation.

(v) Given that ,
(2x – 1) (x – 3) = (x + 5) (x – 1)
0r2x² – 6x – x + 3 = x² – x + 5x – 5
Or 2x² – 7x + 3 = x² + 4x – 5
Or 2x² – 7x + 3 – x² – 4x + 5 = 0
Or x² – 11x + 8 = 0
which is a form of ax² + bx + c = 0; a ≠ 0
∴ It is a quadratic equation.

(vi) Given that
x²+3x+1 = (x – 2)²
Or x² + 3x + 1 = x² + 4 – 4x
Or x² + 3x + 1 – x² – 4 + 4x = 0
Or 7x – 3 = 0
which have no term of x².
So it is not a quadratic equation.

(vii) Given that
(x + 2)³ = 2x(x² – 1)
Or x³ + (2)³ + 3 (x)² 2 + 3(x)(2)² = 2x³ – 2x
Or x³ + 8 + 6x² + 12x = 2x³ – 2x
Or x³ + 8 + 6x² + 12x – 2x³ + 2x = 0
Or -x³ + 6x² + 14x + 8 = 0
Here the highest degree of x is 3. which is a cubic equation.
∴ It is not a quadratic equation.

(viii) Given that
x³ – 4x² – x+ 1= (x – 2)³
Or x³ – 4x² – x + 1 = x³ – (2)³ + 3(x)² (-2) + 3 (x) (-2)²
Or x³ – 4x² – x + 1 = x³ – 8 – 6x² + 12x
Or x³ – 4x² – x + 1 – x³ + 8 + 6x² – 12x = 0
Or 2x² – 13x + 9 = 0
which is in the form of ax² + bx +c = 0; a ≠ 0
∴ It is a quadratic equation.

Question 2.
Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Solution:
(i) Let Breadth of rectangular plot = x m
Length of rectangular plot= (2x + 1) m
∴ Area of rectangular plot = [x (2x + 1)] m² = (2x² + x) m²
According to question,
2x² + x = 528
S = 1
P = -528 × 2 = -1056
0r 2x² + x – 528 = 0
Or 2x² – 32x + 33x – 528 = 0
Or 2x(x – 16) + 33(x – 16) = 0
Or (x – 16) (2x + 33) = 0
Either x – 16 = 0 Or 2x + 33 = 0
x = 16 Or x = 2
∵ breadth of any rectangle cannot be negative, so we reject x = \(\frac{-33}{2}\), x = 16
Hence, breadth of rectangular plot = 16 m
Length of rectangular plot = (2 ×16 + 1)m = 33m
and given problem in the form of Quadratic Equation are 2x² + x – 528 = 0.

(ii) Let two consecutive positive integers are x and x + 1.
Product of Integers = x (x + 1) = x² + x
According to question,
Or x² + x – 306 = 0
S = 1, P = – 306
Or x² + 18x – 17x – 306 = 0
Or x(x + 18) -17 (x + 18) = 0
Or (x + 18) (x – 17) = 0
Either x + 18 = 0 Or x – 17 = 0
x = -18 Or x = 17
∵ We are to study about the positive integers, so we reject x = – 18.
x = 17
Hence, two consecutive positive integers are 17, 17 + 1 = 18
and given problem in the form of Quadratic Equation is x² + x – 306 = 0.

(iii) Let present age of Rohan = x years
Rohan’s mother’s age = (x + 26) years
After 3 years, Rohan’s age = (x + 3) years
Rohan’s mother’s age = (x + 26 + 3) years = (x + 29) years
∴ Their product = (x + 3) (x + 29)
= x² + 29x + 3x + 87
= x² + 32x + 87
According to question,
x² + 32x + 87 = 360
Or x² + 32x + 87 – 360 = 0
Or x² + 32x – 273 = 0
Or x² + 39x – 7x – 273 = 0
S = 32, P = – 273
Or x(x + 39) – 7(x + 39) = 0
Or (x + 39) (x – 7) =
Either x + 39 = Or x – 7 = 0
x = -39 Or x = 7
∵ age of any person cannot be negative so, we reject x = -39
∴ x = 7
Hence, Rohans present age = 7 years
and given problem in the form of Quadratic Equation is x² + 32x – 273 = 0.

(iv) Let u km/hour be the speed of train.
Distance covered by train = 480 km
Time taken by train = \(\frac{480}{u}\) hour
[ Using, Speed = \(\frac{\text { Distance }}{\text { Time }}\)
or Time = \(=\frac{\text { Distance }}{\text { Speed }}\) ]

If speed of train be decreased 8km/hr.
∴ New speed of train = (u – 8) km/hr.
and time taken by train = \(\frac{480}{u-8}\) hour
According to question.

or 3840 = 3 (u² – 8u)
or u² – 8u = 1280
or u² – 8u – 1280=0
or u² – 40u + 32u – 1280 = 0
S = -8, P = – 1280
or u(u – 40) + 32 (u – 40) = 0
or (u – 40)(u + 32) = 0
Either u – 40 = 0
or u + 32 = 0
u = 40 or u = -32
But, speed cannot be negative so we reject
u = – 32
∴ u = 40.
Hence speed of train is 40 km/hr Ans.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 4 Linear Equations in Two Variables Ex 4.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2

Question 1.
Which one of the following options is true, and why?
y = 3x + 5 has
(i) a unique solution,
(ii) only two solutions,
(iii) infinitely many solutions.
Answer:
Option (iii) is true. Since y = 3x + 5 is a linear equation in two variables, it has infinitely many solutions. e.g., (1, 8), (2, 11), (3, 14), (4, 17), (0, 5), (- 1, 2) are all solutions of the given equation y = 3x + 5.

Question 2.
Write four solutions for each of the following equations:
(i) 2x + y = 7
Answer:
2x + y = 7
∴ y = 7 – 2x
Taking x = 0, 1, 2, 3, we get the values of y as 7, 5, 3 and 1 respectively. Thus, (0, 7), (1, 5), (2, 3) and (3, 1) are four solutions of the given equation 2x + y = 7. We can give other answers as well because the given linear equation in two variables has infinitely many solutions.

(ii) πx + y = 9
Answer:
πx + y = 9
∴ y = 9 – πx
For x = 0, y = 9.
For x = 1, y = 9 – π.
For x = – 1, y = 9 + π.
For x = , y = 8.
Thus, (0, 9), (1, 9 – π), (- 1, 9 + π) and (\(\frac{1}{\pi}\), 8) are four of the infinitely many solutions of the given equation πx + y = 9.

(iii) x = 4y
Answer:
x = 4y
For y = 0, x = 0.
For y = 1, x = 4.
For y = – 1, x = – 4.
For y = 2, x = 8.
Thus, (0, 0), (4, 1), (- 4, – 1) and (8, 2) are four of the infinitely many solutions of the given equation x = 4y.

Question 3.
Check which of the following are solutions of the equation x – 2y = 4 and which are not:
(i) (0, 2)
Answer:
Substituting x = 0 and y = 2, we get x – 2y = 0 – 2(2) = – 4, which is not equal to 4. Hence, (0, 2) is not a solution of x – 2y = 4.

(ii) (2, 0)
Answer:
Substituting x = 2 and y = 0, we get x – 2y = 2 – 2 (0) = 2, which is not equal to 4. Hence, (2, 0) is not a solution of x – 2y = 4.

(iii) (4, 0)
Answer:
Substituting x = 4 and y = 0, we get x – 2y = 4 – 2 (0) = 4. Hence, (4, 0) is a solution of x – 2y = 4.

(iv) (√2, 4√2)
Answer:
Substituting x = √2 and y = 4√2, we get x – 2y = √2 – 2(4√2) = – 7√2, which is not equal to 4. Hence, (√2, 4√2) is not a solution of x – 2y = 4.

(v) (1, 1)
Answer:
Substituting x = 1 and y = 1, we get x – 2y = 1 -2(1) = – 1, whIch is not equal to 4. Hence, (1, 1) is not a solution of x – 2y = 4.

Question 4.
Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Answer:
x = 2, y = 1 is a solution of equation 2x + 3y = k. Hence, x = 2 and y = 1 satisfy the equation.
∴ 2(2) + 3(1) = k
∴ 4 + 3 = k
∴ 7 = k
∴ k = 7

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 4 Linear Equations in Two Variables Ex 4.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 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.
(Take the cost of a notebook to be and that of a pen to be ₹ y.)
Answer:
Let the cost of a notebook be ₹ x and the cost of a pen be ₹ y.
Then, according to the given data, the cost of a notebook is twice the cost of a pen.
∴ x = 2y
Thus, x = 2y, i.e., x – 2y = 0 is the required linear equation in two variables.

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:
(i) 2x + 3y = 9.3\(\overline{5}\)
Answer:
2x + 3y = 9.3\(\overline{5}\)
∴ 2x + 3y – 9.3\(\overline{5}\) = 0
Here, a = 2, b = 3 and c = – 9.3\(\overline{5}\).

(ii) x – \(\frac{y}{5}\) – 10 = 0
Answer:
x – \(\frac{y}{5}\) – 10 = 0
∴x – \(\frac{1}{5}\)y – 10 = 0
Here, a = 1, b = – \(\frac{1}{5}\) and c = – 10.

(iii) – 2x + 3y = 6
Answer:
– 2x + 3y = 6
∴ – 2x + 3y – 6 = 0
Here, a = 1, b = – 3 and c = – 6.

(iv) x = 3y
Answer:
x = 3y
∴ x – 3y + 0 = 0
Here, a = 1, b = – 3 and c = 0.

(v) 2x = – 5y
Answer:
2x = – 5y
∴ 2x + 5y + 0 = 0
Here, a = 2, b = 5 and c = 0

(vi) 3x + 2 = 0
Answer:
3x + 2 = 0
∴ 3x + oy – 2 = 0
Here, a = 3, b = 1 and c = – 2

(vii) y – 2 = 0
Answer:
y – 2 = 0
∴ 0x + y – 2= 0
Here a = 0, b = 1 and c = – 2

(viii) 5 = 2x
Answer:
5 = 2x
∴ 5 – 2x = 0
∴ – 2x + 0y + 5 = 0
Here a = – 2, b = 0, c = 5

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 3 Playing with Numbers Ex 3.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.2

1. Find the common factors of the followings:

Question (i)
16 and 24
Solution:
The factors of 16
= 1, 2, 4, 8, 16
The factors of 24
= 1, 2, 3, 4, 6, 8, 12, 24
Common factors of 16 and 24
= 1, 2, 4, 8

Question (ii)
25 and 40
Solution:
The factors of 25
= 1, 5, 25
The factors of 40
= 1, 2, 4, 5, 8, 10, 20, 40
Common factors of 25 and 40
= 1, 5

Question (iii)
24 and 36
Solution:
The factors of 24
= 1, 2, 3, 4, 6, 8, 12, 24
The factors of 36
= 1, 2, 3, 4, 6, 12, 18, 36
Common factors of 24 and 36
= 1, 2, 3, 4, 6, 12

Question (iv)
14, 35 and 42
Solution:
The factors of 14
= 1, 2, 7, 14
The factors of 35
= 1, 5, 7, 35
The factors of 42
= 1,2,3, 6, 7, 21, 42
Common factors of 14, 35 and 42
= 1, 7

Question (v)
15, 24 and 35.
Solution:
The factors of 15
= 1, 3, 5, 15
The factors of 24
= 1, 2, 3, 4, 6, 8, 12, 24
The factors of 35
= 1, 5, 7, 35
Common factors of 15, 24 and 35.
= 1

2. Find first three common multiples of the followings:

Question (i)
3 and 5
Solution:
The multiples of 3
= 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45
The multiples of 5
= 5, 10, 15, 20, 25, 30, 35, 40,45
First three common multiples of 3 and 5
= 15, 30 and 45

Question (ii)
6 and 8
Solution:
The multiples of 6
= 6, 12, 18, 24, 30, 36, 42, 48 54, 60, 66, 72
The multiples of 8
= 8, 16, 24, 32, 40, 48, 56, 64, 72
First three common multiples of 6 and 8
= 24, 48 and 72

Question (iii)
2, 3 and 4.
Solution:
The multiples of 2
= 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36
The multiples of 3
= 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36
The multiples of 4
= 4, 8, 12, 16, 20, 24, 28, 32, 36
First three common multiples of 2, 3 and 4
= 12, 24 and 36

3. Which of the following numbers are divisible by 2 or 4?

Question (i)
52314
Solution:
52314 is divisible by 2 as it is even number.
52314 is not divisible by 4 because the last two digits i.e. 14 which is not divisible by 4

Question (ii)
678913
Solution:
678913 is not divisible by 2. As it is an odd number.
678913 is not divisible by 4 because the last two digits i.e. 13 is not divisible by 4.

Question (iii)
4056784
Solution:
4056784 is divisible by 2. As it is an even number.
4056784 is also divisible by 4 because the last two digits i.e. 84 which is divisible by 4.

Question (iv)
21536
Solution:
21536 is divisible by 2. As it is an even number.
21536 is divisible by 4. As number formed by their last two digits is divisible by 4.

Question (v)
412318.
Solution:
412318 is divisible by 2. As it is an even number.
412318. is not divisible by 4. As number formed by their last two digits is not divisible by 4.

4. Which of the following numbers are divisible by 3 or 9?

Question (i)
654312
Solution:
654312 is divisible by 3.
As sum of its digits = 6 + 5 + 4 + 3 + 1 + 2 = 21, which is divisible by 3.
654312 is not divisible by 9.
As sum of its digits = 21, which is not divisible by 9.

Question (ii)
516735
Solution:
516735 is divisible by 3.
As sum of its digits = 5 + 1 + 6 + 7 + 3 + 5 = 27, which is divisible by 3.
516735 is also divisible by 9.
As sum of its digits = 27, which is divisible by 3.

Question (iii)
423152
Solution:
423152 is divisible by 3.
As sum of its digits = 4 + 2 + 3 + 1 + 5 + 2=17, which is not divisible by 3.
423152 is also not divisible by 9.
As sum of its digits = 17, which is not divisible by 9.

Question (iv)
704355
Solution:
704355 is divisible by 3.
As sum of its digits = 7 + 0 + 4 + 3 + 5 + 5 = 24, which is divisible by 3.
704355 is not divisible by 9.
As sum of its digits = 24, which is not divisible by 9.

Question (v)
215478.
Solution:
215478 is divisible by 3.
As sum of its digits = 2 + 1 + 5 + 4 + 7 + 8 = 27, which is divisible by 3.
215478 is divisible by 9.
As sum of its digits = 27, which is divisible by 9.

5. Which of the following numbers are divisible by 5 or 10?

Question (i)
456803
Solution:
456803 is not divisible by 5
As its last digit is not 0 or 5.
456803 is not divisible by 10
As its last digit is not 0.

Question (ii)
654130
Solution:
654130 is divisible by both 5 and 10
As its last digit is 0.

Question (iii)
256785
Solution:
256785 is divisible by 5
As its last digit is 5.
256785 is not divisible by 10
As its last digit is not 0.

Question (iv)
412508
Solution:
412508 is not divisible by 5
As its last digit is not 0 or 5.
412508 is not divisible by 10
As its last digit is not 0.

Question (v)
872565.
Solution:
872565 is divisible by 5
As its last digit is 5.
872565 is not divisible by 10
As its last digit is not 0.

6. Which of the following numbers are divisible by 8?

Question (i)
457432
Solution:
457432 is divisible by 8, because its last three digits are 432, which is divisible by 8.

Question (ii)
5134214
Solution:
5134214 is not divisible by 8, because its last three digits are 214, which is not divisible by 8.

Question (iii)
7232000
Solution:
7232000 is divisible by 8, because its last three digits are 000, which is divisible by 8.

Question (iv)
5124328
Solution:
5124328 is divisible by 8, because its last three digits are 328, which is divisible by 8.

Question (v)
642516.
Solution:
642516 is not divisible by 8, because its last three digits are 516, which is not divisible by 8.

7. Which of the following numbers are divisible by 6?

Question (i)
425424
Solution:
425424 is divisible by 2 because, it has 4 in its units place.
Sum of digits = 4 + 2 + 5+4 + 2 + 4 = 21
Sum of digits of 425424 is divisible by 3.
∴ 425424 is divisible by 2 as well as 3
Hence, 425424 is divisible by 6.

Question (ii)
617415
Solution:
617415 is not divisible by 2 because, it has 5 in its units place.
∴ 617415 is not divisible by 6.

Question (iii)
3415026
Solution:
3415026 is divisible by 2 because, it has 6 in its units place.
Sum of digits = 3 + 4 + 1 + 5 + 0 + 2 + 6 = 21
Sum of digits of 3415026 is divisible by 3
So, 3415026 is divisible by 3
∴ 3415026 is divisible by 2 as well as 3
Hence, 3415026 is divisible by 6.

Question (iv)
4065842
Solution:
4065842 is divisible by 2 because, it has 2 in its units place.
Sum of digits = 4 + 0 + 6 + 5 + 8 + 4 + 2 = 29
Sum of digits of 4065842 is not divisible by 3.
So, 4065842 is not divisible by 3.
∴ 4065842 is divisible by 2 but not by 3.
Hence, 4065842 is not divisible by 6.

Question (v)
725436.
Solution:
725436 is divisible by 2 because, it has 6 in its units place.
Sum of digits = 7 + 2 + 5 + 4 + 3 + 6 = 27
Sum of digits of 725436 is divisible by 3.
So, 725436 is divisible by 3.
∴ 725436 is divisible by 2 as well as 3
Hence, 725436 is divisible by 6.

8. Which of the following numbers are divisible by 11?

Question (i)
4281970
Solution:
4281970 is divisible by 11.

Since sum of its digits in odd places = 4 + 8 + 9 + 0 = 21 and
sum of its digits in even places = 2 + 1 + 7 = 10
Their difference = 21 – 10=11, which is odd places digits divisible by 11.

Question (ii)
8049536
Solution:
8049536 is divisible by 11.

Since sum of its digits in odd places = 8 + 4 + 5 + 6 = 23
and sum of its digits in even places = 0 + 9 + 3 = 12
Difference = 23 – 12 = 11, which is divisible by 11.

Question (iii)
1234321
Solution:
1234321 is divisible by 11.

Since sum of its digits in odd places = 1 + 3 + 3 + 1 = 8
and sum of its digits in even places = 2 + 4 + 2 = 8
Difference = 8 – 8 = 0.

Question (iv)
6450828
Solution:
6450828 is not divisible by 11.

Since sum of its digits in odd places = 6 + 5 + 8 + 8 = 27
and sum of its digits in even places = 4 + 0 + 2 = 6
Difference = 27 – 6 = 21, which is not divisible by 11.

Question (v)
5648346.
Solution:
5648346 is divisible by 11.

Since sum of its digits in odd places = 5 + 4 + 3 + 6 = 18 and
sum of its digits in even places = 6 + 8 + 4 = 18.
Difference = 18 – 18 = 0.

9. State True or False:

Question (i)
If a number is divisible by 24, then it is also divisible by 3 and 8.
Solution:
True

Question (ii)
60 and 90 both are divisible by 10 then their sum is not divisible by 10.
Solution:
False

Question (iii)
If a number is divisible by 8 then it is also divisible by 16.
Solution:
False

Question (iv)
If a number is divisible by 15 then it is also divisible by 3.
Solution:
True

Question (v)
144 and 72 are divisible by 12 then their difference is also divisible by 12.
Solution:
True

10. If a number is divisible by 5 and 9 then by which other numbers will that number be always divisible?
Solution:
If a number is divisible by 5 and 9. Then the number is also divisible by their product i.e. 5 × 9 = 45.

11. Which of the following pairs are co-prime?

Question (i)
25, 35
Solution:
Two numbers are said to be co-prime if they do not have a common factor other than 1.
Given numbers are 25 and 35 Factors of 25 = 1, 5, 25
Factors of 35 = 1, 5, 7, 35
Since 25 and 35 have 1 and 5 two common factors
∴ 25 and 33 are not co-prime.

Question (ii)
16,21
Solution:
Given numbers are 16 and 21
Factors of 16 = 1, 2, 4, 8, 16
Factors of 21 = 3, 7, 21
There is only 1 common factors 16 and 21 are co-prime
∴ 16 and 21 are co-prime.

Question (iii)
24, 41
Solution:
Given numbers are 24 and 41
Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Factors of 41 = 1, 41
There is only one (1) common factors.
∴ 24 and 41 are co-prime.

Question (iv)
48,33
Solution:
Given numbers are 48 and 33
Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Factors of 33 = 1, 3, 11
There are two common factors 1 and 3.
∴ 48 and 33 are not co-prime.

Question (v)
20, 57.
Solution:
Given numbers are 20 and 57
Factors of 20 = 1, 2, 4, 5, 10, 20
Factors of 57 = 1, 3, 19, 57
There is only only one (1) common factors.
∴ 20 and 57 are co-prime.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 3 Coordinate Geometry MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 3 Coordinate Geometry 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.
For x = 2, y = 3, u = – 2 and v = – 3, point (x + y, u + v) lies in the ……………. quadrant.
A. first
B. second
C. third
D. fourth
Answer:
D. fourth

Question 2.
For x = 4, y = – 5, u = – 6 and v = 8, point (x + y, u + v) lies In the quadrant.
A. first
B. second
C. third
D. fourth
Answer:
B. second

Question 3.
If (x, y) and (y, x) represent the same point in the coordinate plane, then is possible.
A. x = 5, y = 2
B. x = 2, y = 5
C. x = – 5. y = – 2
D. x = 5, y = 5
Answer:
D. x = 5, y = 5

Question 4.
The line Joining P(3, -2) and Q(3, 4)
A. is parallel to the x-axis
B. is parallel to the y-axis
C. is perpendicular to the y-axis
D. intersects both the axes
Answer:
B. is parallel to the y-axis

Question 5.
The line joining A(- 2, 5) and B(- 2, – 8)
A. is parallel to the x-axis
B. is perpendicular to the x-axis
C. intersects the y-axis
D. intersects both the axes
Answer:
B. is perpendicular to the x-axis

Question 6.
The line joining A (- 2, 5) and B (3, 5) intersects ……………….. .
A. the x-axis at (- 2, 0)
B. the x-axis at (3, 0)
C. the y-axis at (0, 5)
D. the x-axis at (5, 0)
Answer:
C. the y-axis at (0, 5)

Question 7.
The line joining A (3, 2) and B (3, – 2) intersects …………………….. .
A. the x-axis at (0, 3)
B. the x-axis at (3, 0)
C. the y-axis at (0, 2)
D. the y-axis at (0, – 2)
Answer:
B. the x-axis at (3, 0)