Class 8

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.2

1. Which of the following are in inverse proportion?

Question (i)
The number of workers on a job and the time to complete the job.
Solution:
If the number of workers on a job increases, then time to complete the job decreases. So, it is the case of inverse proportion.

Question (ii)
The time taken for a journey and the distance travelled in a uniform speed.
Solution:
Here, the speed is uniform.
∴ The time taken for a journey is directly proportional to the speed. So, it is not the case of inverse proportion.

Question (iii)
Area of cultivated land and the crop harvested.
Solution:
For more area of cultivated land, more crops would be harvested. So, it is not the case of inverse proportion.

Question (iv)
The time taken for a fixed journey and the speed of the vehicle.
Solution:
If speed of vehicle is more, then time to cover the fixed journey would be less. So, it is the case of inverse proportion.

Question (v)
The population of a country and the area of land per person.
Solution:
For more population, less area per person would be in the country. So, it is a case of inverse proportion.

2. In a Television game show, the prize money of ₹ 1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners?

Number of winners	1	2	4	5	8	10	20
Prize for each winner (in ₹)	1,00,000	50,000	…	…	…	…	…

Solution:
Here, more the number of winners, less is the prize money for each winner.
∴ This is a case of inverse proportion.

Now, the table is as follows:

Number of winners	1	2	4	5	8	10	20
Prize for each winner (in ₹)	1,00,000	50,000	25,000	20,000	12,500	10,000	5,000

3. Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. Help him by completing the following table:

Number of spokes	4	6	8	10	12
Angle between a pair of consecutive spokes	90°	60°	…	…	…

Question (i)
Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion?
Solution:
Here, more the number of spokes, less the measure of angle between a pair of consecutive spokes.
∴ This is a case of inverse proportion.
Here,

Now, the table is as follows:

Number of spokes	4	6	8	10	12
Angle between a pair of consecutive spokes
	90°	60°	45°	36°	30°

Question (ii)
Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.
Solution:
Let the measure of angle be x°.
∴ 15 × x° = 4 × 90°
∴ x° = \(\frac{4 \times 90^{\circ}}{15}\) = 24°
Hence, this angle should be 24°.

Question (iii)
How many spokes would be needed, if the angle between a pair of consecutive spokes is 40°?
Solution:
Let the number of spokes be n.
∴ n × 40° = 4 × 90°
∴ n = \(\frac{4 \times 90^{\circ}}{40^{\circ}}\) = 9
Thus, 9 spokes would be needed.

4. If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of the children is reduced by 4?
Solution:

Number of children x	Number of sweets y
x1 = 24	y1 = 5
x2 = 24 – 4 = 20	y2 = (?)

Here, if the number of children decreases, then the number of sweets received by each child will increase.
∴ This is a case of inverse proportion.
x₁ × y₁ = x₂ × y₂
∴ 24 × 5 = 20 × y₂
∴ y₂ = \(\frac{24 \times 5}{20}\) = 6
Thus, each child will get 6 sweets.

5. A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle?
Solution:

Number of animals x	Number of days y
x1 = 20	y1 = 6
x2 = 20 + 10 = 20	y2 = (?)

Here, the number of animals increases, so the number of days to feed them will decrease.
∴ This is a case of inverse proportion.
∴ x₁ × y₁ = x₂ × y₂
∴ 20 × 6 = 30 × y₂
∴ y₂ = \(\frac{20 \times 6}{30}\) = 4
Thus, the food will last for 4 days.

6. A contractor estimates that 3 persons could rewire Jasminder’s house in 4 days. If, he uses 4 persons instead of three, how long should they take to complete the job?
Solution:

Number of persons x	Number of days y
x1 = 42	y1 = 63
x2 = (?)	y1 = 63

Here, more the number of persons, less will be the time required to complete the job.
∴ This is a case of inverse proportion.
∴ x₁ × y₁ = x₂ × y₂
∴ 3 × 4 = 4 × y₂
∴ y₂ = \(\frac{3 \times 4}{4}\) = 3
Thus, 3 days will be required to complete the job.

7. A batch of bottles were packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled?

Solution:

Number of bottles in a box x	Number of boxes y
x1 = 12	y1 = 25
x2 = 20	y2 = (?)

Here, more the number of bottles in a box, less would be the number of boxes.
∴ This is a case of inverse proportion.
∴ x₁ × y₁ = x₂ × y₂
∴ 3 × 4 = 4 × y₂
∴ y₂ = \(\frac{3 \times 4}{4}\) = 3
Thus, 15 boxes would be filled.

8. A factory requires 42 machines to produce a given number of articles in 63 days. How many machines would be required to produce the same number of articles in 54 days?
Solution:

Number of machines x	Number of days y
x1 = 42	y1 = 63
x2 = (?)	y2 = 54

Here, if the number of days will be less, the number of machines required will be more.
∴ This is a case of inverse proportion.
∴ x₁ × y₁ = x₂ × y₂
∴ 42 × 63 = x₂ × 54
∴ x₂ = \(\frac{42 \times 63}{54}\) = 49
Thus, 49 machines would be required.

9. A car takes 2 hours to reach a destination by travelling at the speed of 60 km/h. How long will it take when the car travels at the speed of 80 km/h?
Solution:

Speed (km/h) x	Number of hours y
x1 = 60	y1 = 2
x2 = 80	y2 = (?)

Here, if the speed of car increases, then the time taken to cover the same distance will decrease.
∴ This is a case of inverse proportion.
∴ x₁ × y₁ = x₂ × y₂
∴ 60 × 2 = 80 × y₂
∴ y₂ = \(\frac{60 \times 2}{80}=\frac{3}{2}=1 \frac{1}{2}\) h
Thus, car would take 1\(\frac {1}{2}\) hours.

10. Two persons could fit new windows in a house in 3 days.

Question (i)
One of the persons fell ill before the work started. How long would the job take now?
Solution:

Number of persons x	Number of days y
x1 = 2	y1 = 3
x2 = 2 – 1 = 1	y2 = (?)

Here, less the number of persons, more would be the number of days to complete the job.
∴ This is a case of inverse proportion.
∴ x₁ × y₁ = x₂ × y₂
∴ 2 × 2 = 1 × y₂
∴ y₂ = \(\frac{2 \times 3}{1}\) = 6
Thus, it would take 6 days to complete the job.

Question (ii)
How many persons would be needed to fit the windows in one day?
Solution:

Number of days x	Number of persons y
x1 = 3	y1 = 2
x2 = 1	y2 = (?)

Here, less the number of days, more will be the number of persons needed.
∴ This is a case of inverse proportion.
∴ y₂ = ? and x₂ = 1
∴ x₁ × y₁ = x₂ × y₂
∴ 3 × 2 = 1 × y₂
∴ y₂ = \(\frac{3 \times 2}{1}\) = 6
Thus, 6 persons would be needed.

11. A school has 8 periods a day each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the number of school hours to be the same?
Solution:

Number of periods x	Length of each period (in minute) y
x1 = 8	y1 = 45
x2 = 9	y2 = (?)

Here, the number of periods is more, then the length of each period will be less.
∴ This is a case of inverse proportion.
∴ x₁ × y₁ = x₂ × y₂
∴ 8 × 45 = 9 × y₂
∴ y₂ = \(\frac{8 \times 45}{9}\) = 40
Thus, each period would be of 40 minutes.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 13 Direct and Inverse Proportions Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1

1. Following are the car parking charges near a railway station upto:

Check if the parking charges are in direct proportion to the parking time.
Solution:
Here, ratio of parking charges and parking time are as follow:

Parking time	Parking charges	Parking charge / Parking time
4 hours	₹ 60	\(\frac{60}{4}=\frac{15}{1}\)
8 hours	₹ 100	\(\frac{100}{8}=\frac{25}{2}\)
12 hours	₹ 140	\(\frac{140}{12}=\frac{35}{3}\)
24 hours	₹ 180	\(\frac{180}{24}=\frac{15}{2}\)

Here, \(\frac {15}{1}\) ≠ \(\frac {25}{2}\) ≠ \(\frac {35}{3}\) ≠ \(\frac {15}{2}\)
Thus, the parking charges are not in direct proportion to the parking time.

2. A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added:
Solution:

Parts of red pigment	1	4	7	12	20
Parts of base	8	…	…	…	…

If parts of red pigment are x₁, x₂, x₃, x₄ and x₅ respectively, then parts of base are y₁, y₂, y₃, y₄ and y₅ respectively. Here, it is clear that mixture preparation is in direct proportion.

Thus, the table is

Parts of red pigment	1	4	7	12	20
Parts of base	8	32	56	96	160

3. In Question 2 above, if 1 part of a red pigment requires 75 ml of base, how much red pigment should we mix with 1800 ml of base?
Solution:
See as per question 2 –
x₁ = 1, y₁ = 75, x₂ = ? and yx₂ = 1800
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
∴ \(\frac{1}{75}=\frac{x_{2}}{1800}\)
∴ x₂ = \(\frac{1 \times 1800}{75}\)
∴ x₂ = 24
Thus, 24 ml of red pigment should be mixed with 1800 ml of base.

4. A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Solution:
Let the number of bottles filled by machine in 5 h be x.

Number of hours (x)	6	5
Number of bottles filled (y)	840	?

Here, as the number of hours decreases, the number of bottles filled will also decrease.
∴ It is case of direct proportion.
Here, x₁ = 6, y₁ = 840, x₂ = 5 and y₂ = ?
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
∴ \(\frac{6}{840}=\frac{5}{y_{2}}\)
∴ y₂ = \(\frac{5 \times 840}{6}\)
∴ y₂ = 700
Thus, 700 bottles will be filled in 5 hours.

5. A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?
Solution:

Enlargement in picture of bacteria	Length (cm)
50,000 times enlarged (x1)	5 (y1)
1 (x2)	? (y2)

Here, length of bacteria increases as picture of bacteria enlarges.
∴ It is case of, direct proportion.

Hence, the actual length of bacteria is 10^-4 cm.
Now, the photograph is enlarged 20,000 times.

Enlargement in picture of bacteria	Length (cm)
50,000 times enlarged (x1)	5 (y1)
20,000 times enlarged (x1)	? (y2)

∴ \(\frac{50,000}{5}=\frac{20,000}{y_{2}}\)
∴ y₂ = \(\frac{20,000 \times 5}{50000}\)
∴ y₂ = 2
Thus, its enlarged length would be 2 cm.

6. In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship?
Solution:

–	Actual ship	Model ship
Length of the ship x	28 m	?
Height of mast y	12m	9 cm

This is a case of direct proportion.
x₁ = 28, y₁ = 12, x₂ = ?, y₂ = 9
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
∴ \(\frac{28}{12}=\frac{x_{2}}{9}\)
∴ x₂ = \(\frac{28 \times 9}{12}\)
∴ x₂ = 21
Thus, the length of model ship is 21 cm.

7. Suppose 2 kg of sugar contains 9 × 10⁶ crystals. How many sugar crystals are there in

Question (i)
5 kg of sugar?
Solution:

Weight of sugar (kg) x	Number of sugar crystals y
x1 = 2	y1 = 9 × 106
x2 = 5	y2 = (?)

This is a case of direct proportion.

Thus, there are 2.25 × 10⁷ crystals of sugar in 5 kg of sugar.

Question (ii)
1.2kg of sugar?
Solution:

Weight of sugar (kg) x	Number of sugar crystals y
x1 = 2	y1 = 9 × 106
x2 = 1.2	y2 = (?)

Here. x₁ = 2, x₁ = 9 × 10⁶, x² = 1.2, y² = ?
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
∴ \(\frac{2}{9 \times 10^{6}}=\frac{1.2}{y_{2}}\)
∴ y₂ = \(\frac{1.2 \times 9 \times 10^{6}}{2}\)
∴ y₂ = 0.6 × 9 × 10⁶
∴ y₂ = 5.4 × 10⁶
Thus, there are 5.4 × 10⁶ crystals of sugar in 1.2 kg of sugar.

8. Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?
Solution:

Actual distance (km) x	Distance on the map (cm) y
x1 = 18	y1 = 1
x2 = 72	y2 = (?)

This is a case of direct proportional.
Here, x₁ = 18 km, y₁ = 1 cm, x₂ = 72 km, y₂ = ?
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
∴ \(\frac{18}{1}=\frac{72}{y_{2}}\)
∴ y₂ = \(\frac{72 \times 1}{18}\)
∴ y₂ = 4
Thus, the distance covered by her on the map is 4 cm.

9. A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time

Question (i)
the length of the shadow cast by another pole 10 m 50 cm high
Solution:

Height of vertical pole x	Length of shadow y
x1 = 5 m 60 cm = 560 cm	y1 = 3 m 20 cm = 320 cm
x2 = 10 m 50 cm = 1050 cm	y2 = (?)

This is a case of direct proportionality.

Thus, the length of the shadow cast by another pole is 6 m.

Question (ii)
the height of a pole which casts a shadow 5 m long.
Solution:

Height of vertical pole x	Length of shadow y
x1 = 560 cm	y1 = 320 cm
x2 = (?)	y2 = 5 m = 500 cm

x₁ = 560, y₁ = 320, x₂ = ?, y₂ = 500
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
∴ \(\frac{560}{320}=\frac{x_{2}}{500}\)
∴ x₂ = \(\frac{560 \times 500}{320}\)
∴ x₂ = 875 cm
∴ x₂ = 8.75 cm
Thus, the height of the pole is 8,75 m.

10. A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Solution:

Distance (km) x	Time (minute) y
x1= 14	y1 = 25
x2 = (?)	y2 = 5 hours = 300

This is a case of direct proportion.
∴ Here, x₁ = 14, y₁ = 25, x₂ = ?, y₂ = 300
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
∴ \(\frac{14}{25}=\frac{x_{2}}{300}\)
∴ x₂ = \(\frac{14 \times 300}{25}\)
∴ x₂ = 168
Thus, loaded truck can travel 168 km in 5 h.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 Factorization InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 14 Factorization InText Questions

Try These : [Textbook Page No. 219]

1. Factorise:

Question (i)
12x + 36
Solution:
12x = 2 × 2 × 3 × x and
36 = 2 × 2 × 3 × 3
Common factors = 2 × 2 × 3
∴ 12x + 36 = (2 × 2 × 3 × x) + (2 × 2 × 3 × 3)
= (2 × 2 × 3) (x + 3)
= 12 (x + 3)

Question (ii)
22y – 33z
Solution:
22y = 2 × 11 × y and 33z = 3 × 11 × z
Common factor =11
∴ 22y – 33z = (2 × 11 × y) – (3 × 11 × z)
= (11) × (2 × y – 3 × z)
= 11 (2y – 3z)

Question (iii)
14pq + 35pqr
Solution:
14pq = 2 × 7 × p × q and
35pqr = 7 × 5 × p × q × r
Common factor = 7pq
∴ 14pq + 35pqr = (2 × 7 × p × q) + (7 × 5 × p × q × r)
= 7 × p × q (2 + 5 × r)
= 7pq (2 + 5r)

Try These : [Textbook Page No. 225]

1. Divide:

Question (i)
24xy²z³ by 6yz²
Solution:
= \(\frac{24 x y^{2} z^{3}}{6 y z^{2}}\)
= \(\frac{2 \times 2 \times 2 \times 3 \times x \times y \times y \times z \times z \times z}{2 \times 3 \times y \times z \times z}\)
= \(\frac{2 \times 2 \times x \times y \times z}{1}\) = 4xyz
∴ 24xy²z³ ÷ 6yz²
= 4xyz

Question (ii)
63a²b⁴c⁶ by 7a²b²c³
Solution:
= \(\frac{63 a^{2} b^{4} c^{6}}{7 a^{2} b^{2} c^{3}}\)
= \(\frac{3 \times 3 \times 7 \times a^{2} \times b^{4} \times c^{6}}{7 \times a^{2} \times b^{2} \times c^{3}}\)
= 3 × 3 × \(\frac{a^{2}}{a^{2}} \times \frac{b^{4}}{b^{2}} \times \frac{c^{6}}{c^{3}}\)
= 9 × a^2-2 × b^4-2 × c^6-3
= 9 × a⁰ × b² × c³
= 9 × 1 × b² × c³
= 9b²c³
∴ 63a²b⁴c⁶ ÷ 7a²b²c³
= 9b²c³

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 Factorization Ex 14.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.4

1. Find and correct the errors in the following mathematical statements:

Question 1.
4 (x – 5) = 4x – 5
Solution:
Error: 4 × – 5 = (- 20) and not (- 5)
Correct statement: 4 (x – 5) = 4x – 20

Question 2.
x (3x + 2) = 3x² + 2
Solution:
Error: x × 2 = 2x
Correct statement: x (3x + 2) = 3x² + 2x

Question 3.
2x + 3y = 5xy
Solution:
Error: 2x and 3y are unlike terms.
So their sum is not possible.
Correct statement: 2x + 3y = 2x + 3y

Question 4.
x + 2x + 3x = 5x
Solution:
Error: x, 2x and 3x are like terms. So sum of their coefficient =1 + 2 + 3 = 6.
Correct statement: x + 2x + 3x = 6x

Question 5.
5y + 2y + y – 7y = 0
Solution:
Error : 5y, 2y, y and – 7y all are like terms here. So sum of their coefficient = 5 + 2 + 1 – 7 = 1.
Correct statement: 5y + 2y + y – 7y = y

Question 6.
3x + 2x = 5x²
Solution:
Error: When like terms are added or subtracted their exponents do not change.
Correct statement: 3x + 2x = 5x

Question 7.
(2x)² + 4 (2x) + 7 = 2x² + 8x + 7
Solution:
Error: (2x)² = 2x × 2x = 4x²
Correct statement:
(2x)² + 4 (2x) + 7 = 4x² + 8x + 7

Question 8.
(2x)² + 5x = 4x + 5x = 9x
Solution:
Error : (2x)² = (2x × 2x) = 4x²
Correct statement: (2x)² + 5x = 4x² + 5x

Question 9.
(3x + 2)² = 3x² + 6x + 4
Solution:
Error : (3x + 2)²
= (3x)² + 2 (3x)(2) + (2)²
= 9x² + 12x + 4
Correct statement:
(3x + 2)² = 9x² + 12x + 4

10. Substituting x = – 3 in

Question (a)
x² + 5x + 4 gives (- 3)² + 5 (- 3) + 4 = 9 + 2 + 4 = 15
Solution:
Error : 5 (- 3) = – 15 and not 2
Correct statement:
Substituting x = (- 3) in, x² + 5x + 4
= (- 3)² + 5 (-3) + 4
= 9 – 15 + 4 = 9 + 4 – 15
= 13 – 15
= (-2)

Question (b)
x² – 5x + 4 gives (- 3)² – 5 (- 3) + 4 = 9 – 15 + 4 = -2
Solution:
Error: -5 (-3) = + 15 and not (-15)
Correct statement:
Substituting x = (- 3) in,
x² – 5x + 4
= (- 3)² – 5 (- 3) + 4
= 9 + 15 + 4
= 28

Question (c)
x² + 5x gives (- 3)² + 5 (- 3) = – 9 – 15 = – 24
Solution:
Error : (- 3)² = + 9
Correct statement:
Substituting x = (- 3) in, x² + 5x
= (- 3)² + 5 (- 3)
= 9 – 15 = – 6

Question 11.
(y – 3)² = y² – 9
Solution:
Error : (y – 3)²
= (y)² – 2 (y)(3) + (- 3)²
= y² – 6y + 9
Correct statement: (y – 3)² = y² – 6y + 9.

Question 12.
(z + 5)² = z² + 25
Solution:
Error : (z + 5)²
= (z)² + 2 (z)(5) + (5)²
= z² + 10 z + 25
Correct statement: (z + 5)²
= z² + 10z + 25

Question 13.
(2a + 3b) (a – b) = 2a² – 3b²
Solution:
Error : (2a + 3b) (a – b)
= 2a (a-b) + 3b (a-b)
= 2a² – 2ab + 3ab – 3b²
= 2a² + ab – 3b²
Correct statement: (2a + 3b) (a – b)
= 2a² + ab – 3b²

Question 14.
(a + 4) (a + 2) = a² + 8
Solution:
Error : (a + 4) (a + 2) = a (a + 2) + 4 (a + 2)
= a² + 2a + 4a + 8
= a² + 6a + 8
Correct statement: (a + 4) (a + 2)
= a² + 6a + 8

Question 15.
(a – 4) (a – 2) = a² – 8
Solution:
Error : (a – 4) (a – 2) = a (a – 2) – 4 (a – 2)
= a² – 2a – 4a + 8
= a² – 6a + 8
Correct statement: (a – 4) (a – 2)
= a² – 6a + 8

Question 16.
\(\frac{3 x^{2}}{3 x^{2}}\) = 0
Solution:
Error: Numerator and denominator, both are same. So their division is 1.
Correct statement:\(\frac{3 x^{2}}{3 x^{2}}\) = 1

Question 17.
\(\frac{3 x^{2}+1}{3 x^{2}}\) = 1 + 1 = 2
Solution:
Error: \(\frac{3 x^{2}+1}{3 x^{2}}=\frac{3 x^{2}}{3 x^{2}}+\frac{1}{3 x^{2}}\)
= 1 + \(\frac{1}{3 x^{2}}\)
Correct statement: \(\frac{3 x^{2}+1}{3 x^{2}}\) = 1 + \(\frac{1}{3 x^{2}}\)

Question 18.
\(\frac{3 x}{3 x+2}=\frac{1}{2}\)
Solution:
Error: Here, simplification of LHS is not possible.
Correct statement: \(\frac{3 x}{3 x+2}=\frac{3 x}{3 x+2}\)

Question 19.
\(\frac{3}{4 x+3}=\frac{1}{4 x}\)
Solution:
Error: Here, simplification of LHS is not possible.
Correct statement: \(\frac{3}{4 x+3}=\frac{3}{4 x+3}\)

Question 20.
\(\frac{4 x+5}{4 x}\) = 5
Solution:
Error: \(\frac{4 x+5}{4 x}\)
= \(\frac{4 x}{4 x}+\frac{5}{4 x}\)
= 1 + \(\frac{5}{4 x}\)
Correct statement: \(\frac{4 x+5}{4 x}\) = 1 + \(\frac{5}{4 x}\)

Question 21.
\(\frac{7 x+5}{5 x}\) = 7x
Error: \(\frac{7 x+5}{5 x}\)
= \(\frac{7 x}{5}+\frac{5}{5}\)
= \(\frac{7 x}{5}\) + 1
Correct statement: \(\frac{7 x+5}{5}=\frac{7 x}{5}\) + 1

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 Factorization Ex 14.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.3

1. Carry out the following divisions:

Question (i)
28x⁴ ÷ 56x
Solution:

Question (ii)
– 36y³ ÷ 9y²
Solution:

Question (iii)
66pq²r³ ÷ 11qr²
Solution:

Question (iv)
34x³y³z³ ÷ 51 xy²z³
Solution:

Question (v)
12a⁸b⁸ ÷ (- 6a⁶b⁴)
Solution:

2. Divide the given polynomial by the given monomial:

Question (i)
(5x² – 6x) ÷ 3x
Solution:

Question (ii)
(3y⁸ – 4y⁶ + 5y⁴) ÷ y⁴
Solution:

Question (iii)
8 (x³y²z² + x²y³z² ÷ 4 x²y²z²)
Solution:

Question (iv)
(x³ + 2x² + 3x) ÷ 2x
Solution:

Question (v)
(P³ q⁶ – p⁶ q³) ÷ p³ q³
Solution:

3. Work out the following divisions:

Question (i)
(10x – 25) ÷ 5
Solution:
= \(\frac{10 x-25}{5}\)
= \(\frac{5(2 x-5)}{5}\)
= 2x – 5

Question (ii)
(10x-25) ÷ (2x – 5)
Solution:
= \(\frac{10 x-25}{2 x-5}\)
= \(\frac{5(2 x-5)}{(2 x-5)}\)
= 5

Question (iii)
10y (6y + 21) ÷ 5 (2y + 7)
Solution:
= \(\frac{10 y(6 y+21)}{5(2 y+7)}\)
= \(\frac{2 \times 5 \times y \times 3 \times(2 y+7)}{5(2 y+7)}\)
= 2 × y × 3
= 6y

Question (iv)
9x²y²(3z – 24) ÷ 27xy (z – 8)
Solution:

Question (v)
96 abc (3a – 12) (5b – 30) ÷ 144 (a – 4)(b – 6)
Solution:
= \(\frac{96 a b c(3 a-12)(5 b-30)}{144(a-4)(b-6)}\)
= \(\frac{2 \times 2 \times 2 \times 2 \times 2 \times 3 \times a \times b \times c \times 3 \times(a-4) \times 5 \times(b-6)}{2 \times 2 \times 2 \times 2 \times 3 \times 3 \times(a-4) \times(b-6)}\)
= 2 × 5 × a × b × c
= 10 abc

4. Divide as directed:

Question (i)
5 (2x + 1) (3x + 5) ÷ (2x + 1)
Solution:
= \(\frac{5(2 x+1)(3 x+5)}{(2 x+1)}\)
= \(\frac{5 \times(3 x+5)}{1}\)
= 5(3x + 5)

Question (ii)
26xy (x + 5) (y – 4) ÷ 13x (y – 4)
Solution:
= \(\frac{26 x y(x+5)(y-4)}{13 x(y-4)}\)
= \(\frac{2 \times 13 \times x \times y(x+5)(y-4)}{13 x(y-4)}\)
= 2y(x + 5)

Question (iii)
52pqr (p + q) (q + r) (r + p) ÷ 104pq (q + r) (r + p)
Solution:
= \(\frac{52 p q r(p+q)(q+r)(r+p)}{104 p q(q+r)(r+p)}\)
= \(\frac{52 \times p \times q \times r \times(p+q)(q+r)(r+p)}{2 \times 52 \times p \times q \times(q+r)(r+p)}\)
= \(\frac{r \times(p+q)}{2}\)
= \(\frac {1}{2}\)r (p + q)

Question (iv)
20 (y + 4) (y² + 5y + 3) ÷ 5 (y + 4)
Solution:
= \(\frac{20(y+4)\left(y^{2}+5 y+3\right)}{5(y+4)}\)
= \(\frac{2 \times 2 \times 5 \times(y+4)\left(y^{2}+5 y+3\right)}{5(y+4)}\)
= 2 × 2 × (y² + 5y + 3)
= 4(y² + 5y + 3)

Question (v)
x (x + 1) (x + 2) (x + 3) ÷ x (x + 1)
Solution:
= \(\frac{x(x+1)(x+2)(x+3)}{x(x+1)}\)
= \(\frac{(x+2)(x+3)}{1}\)
= (x + 2)(x + 3)

5. Factorise the expressions and divide them as directed:

Question (i)
(y² + 7y + 10) ÷ (y + 5)
Solution:
First we factorise
y² + 7y + 10
= y² + 5y + 2y + 10
= y (y + 5) + 2 (y + 5)
= (y + 5) (y + 2)
∴ (y² + 7y + 10) ÷ (y + 5)
= \(\frac{(y+5)(y+2)}{(y+5)}\)
= y + 2

Question (ii)
(m² – 14m – 32) ÷ (m + 2)
Solution:
First we factorise
m² – 14m – 32
= m² – 16m + 2m – 32
= m (m – 16) + 2 (m – 16)
= (m – 16) (m + 2)
∴ (m² – 14m – 32) ÷ (m + 2)
= \(\frac{(m-16)(m+2)}{m+2}\)
= m – 16

Question (iii)
(5p² – 25p + 20) ÷ (p – 1)
Solution:
First we factorise
5p² – 25p + 20
= 5 (p² – 5p + 4)
= 5 (p² – 4p – p + 4)
= 5 [p (p – 4) – 1 (p – 4)]
= 5 (p – 4) (p – 1)
∴ (5p² – 25p + 20) ÷ (p – 1)
= \(\frac{5(p-4)(p-1)}{p-1}\)
= 5 (p – 4)

Question (iv)
4yz (z² + 6z – 16) ÷ 2y (z + 8)
Solution:
= \(\frac{4 y z\left(z^{2}+6 z-16\right)}{2 y(z+8)}\)
= \(\frac{2 z\left(z^{2}+6 z-16\right)}{z+8}\)
Now, factorise
z² + 6z – 16
= z² + 8z – 2z – 16
= z (z + 8) – 2 (z + 8)
= (z + 8) (z – 2)
∴ 4yz (z² + 6z – 16) ÷ 2y (z + 8)
= \(\frac{2 z(z+8)(z-2)}{z+8}\)
= 2z (z – 2)

Question (v)
5pq (p² – q²) ÷ 2p(p + q)
Solution:

Question (vi)
12xy (9x² – 16y²) ÷ 4xy (3x + 4y)
Solution:
= \(\frac{12 x y\left(9 x^{2}-16 y^{2}\right)}{4 x y(3 x+4 y)}\)
= \(\frac{12 x y(3 x-4 y)(3 x+4 y)}{4 x y(3 x+4 y)}\)
= 3 (3x – 4y)

Question (vii)
39y³ (50y² – 98) ÷ 26y² (5y + 7)
Solution:

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 Factorization Ex 14.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

1. Factorise the following expressions:

Question (i)
a² + 8a + 16
Solution:
= (a)² + 2 (a)(4) + (4)²
= (a + 4)²

Question (ii)
p² – 10p + 25
Solution:
= (p)² – 2 (p)(5) + (5)²
= (P – 5)²

Question (iii)
25m² + 30m + 9
Solution:
= (5m)² + 2 (5m) (3) + (3)²
= (5m + 3)²

Question (iv)
49y² + 84yz + 36z²
Solution:
= (7y)² + 2 (7y)(6z) + (6z)²
= (7y + 6z)²

Question (v)
4x² – 8x + 4
Solution:
= 4(x² – 2x + 1)
= 4 [(x)² – 2 (x)(1) + (1)²]
= 4 (x – 1)²

Question (vi)
121b² – 88bc + 16c²
Solution:
= (11b)² – 2 (11b)(4c) + (4c)²
= (11b – 4c)²

Question (vii)
(l + m)² – 4lm [Hint: Expand (1 + m)² first]
Solution:
= l² + 2lm + m² – 4lm
= l² + 2lm – 4lm + m²
= l² – 2lm + m²
= (l)² – 2 (l) (m) + (m)²
= (l – m)²

Question (viii)
a⁴ + 2a²b² + b⁴
Solution:
= (a²)² + 2 (a²)(b²) + (b²)²
= (a² + b²)²

2. Factorise:

Question (i)
4p² – 9q²
Solution:
= (2p)² – (3q)²
= (2p – 3q) (2p + 3q)

Question (ii)
63a² – 112b²
Solution:
= 7 (9a² – 16b²)
= 7 [(3a)² -(4b)²]
= 7 (3a – 4b) (3a + 4b)

Question (iii)
49x² – 36
Solution:
= (7x)² – (6)²
= (7x – 6) (7x + 6)

Question (iv)
16x⁵ – 144x³
Solution:
= 16x³(x² – 9)
= 16x³ [(x)² – (3)²]
= 16x³ (x-3) (x + 3)

Question (v)
(l + m)² – (l – m)²
Solution:
=[(l + m) + (l – m)] [(l + m) – (l – m)]
= [l + m + l – m] [l + m – l + m]
= (2l) (2m)
= 4lm

Question (vi)
9x²y² – 16
Solution:
= (3xy)² – (4)²
= (3xy – 4) (3xy + 4)

Question (vii)
(x² – 2xy + y²) – z²
Solution:
= (x – y)² – (z)²
= [(x – y) – z] [(x – y) + z]
= (x – y – z) (x – y + z)

Question (viii)
25a² – 4b² + 28bc – 49c²
Solution:
= (25a²) – (4b² – 28bc + 49c²)
= (5a)² – (2b – 7c)²
= [(5a) – (2b – 7c)] [(5a) + (2b – 7c)]
= (5a – 2b + 7c) (5a + 2b – 7c)

3. Factorise the expressions:

Question (i)
ax² + bx
Solution:
= x (ax + b)

Question (ii)
7p² + 21q²
Solution:
= 7 (p² + 3q²)

Question (iii)
2x³ + 2xy² + 2xz²
Solution:
= 2x(x² + y² + z²)

Question (iv)
am² + bm² + bn² + an²
Solution:
= am² + bm² + an² + bn²
= m² (a + b) + n²(a + b)
= (a + b) (m² + n²)

Question (v)
(lm + l) + m + 1
Solution:
= l (m + 1) + 1 (m + 1)
= (m + 1) (l + 1)

Question (vi)
y(y + z) + 9(y + z)
Solution:
= (y + z)(y + 9)

Question (vii)
5y² – 20y – 8z + 2yz
Solution:
= 5y² – 20y + 2yz – 8z
= 5y (y – 4) + 2z (y – 4)
= (y- 4) (5y + 2z)

Question (viii)
10ab + 4a + 5b + 2
Solution:
= 2a (5b + 2) + 1 (5b + 2)
= (5b + 2) (2a + 1)

Question (ix)
6xy – 4y + 6 – 9x
Solution:
= 6xy – 4y – 9x + 6
= 2y (3x-2)-3(3x-2)
= (3x-2) (2y – 3)

4. Factorise:

Question (i)
a⁴ – b⁴
Solution:
= (a²)² – (b²)²
= (a² – b²) (a² + b²)
= ((a)² – (b²)] (a² + b²)
= (a – b) (a + b) (a² + b²)

Question (ii)
p⁴ – 81
Solution:
= (p²)² – (9)²
= (p² – 9) (p² + 9)
= ((p)² – (3)²] (p² + 9)
= (p – 3)(p + 3)(p² + 9)

Question (iii)
x⁴ – (y + z)⁴
Solution:
= (x²)² – (a²)² (∵ y + z = a)
= (x² – a²) (x² + a²)
= (x – a) (x + a) (x² + a²)
= [x – (y + z)] [x + (y + z)] [x² + (y + z)²] (∵ a = y + z)
= (x – y – z) (x + y + z) [x² + (y + z)²]

Question (iv)
x⁴ – (x – z)⁴
Solution:
= (x²)² – [(x – z)²]²
= [x² – (x – z)²] [x² + (x – z)²]
= [x² – (x² – 2xz + z²)] [x² + (x² – 2xz + z²)]
= (x² – x² + 2xz – z²) (x² + x² – 2xz + z²)
= (2xz – z²) (2x² – 2xz + z²)
= z (2x – z) (2x² – 2xz + z²)

Question (v)
a⁴ – 2a²b² + b⁴
Solution:
= (a²)² – 2(a²)(b²) + (b²)²
= (a² – b²)²
= (a² – b²) (a² – b²)
= (a – b) (a + b) (a – b) (a + b)

5. Factorise the following expressions:

Question (i)
p² + 6p + 8
Solution:
= p² + 6p + 9 – 1
= (p² + 6p + 9) – (1)
= (p + 3)² – (1)²
= (p + 3 + 1) (p + 3 – 1)
= (P + 4) (p + 2)
Here, last term is 8.
∴ 9 – 1 = 8.

OR
p² + 6p + 8
Here, ab = 8 and a + b = 6
On solving equations, a = 4, b = 2
Now, p² + 6p + 8
= p² + 4p + 2p + 8
= p (p + 4) + 2 (p + 4)
= (p + 4) (p + 2)

Question (ii)
q² – 10q + 21
Solution:
= q² – 10q + 25 – 4
= (q² – 10q + 25) – (4)
= (q – 5)² – (2)²
= (q – 5 + 2) (q – 5 – 2)
= (q – 3) (q – 7)
Here, last term is 21.
∴ 25 – 4 = 21.

OR
q² – 10q + 21
Here, ab = 21 and a + b = (- 10)
Possible values of a = 7 or (-7)
b = 3 or (- 3)
Let us check, 7 + 3 = 10 ≠ (- 10)
∴ a = – 7, b = – 3
Now, q² – 10q + 21
= q² – 7q – 3q + 21
= q (q – 7) – 3 (q – 7)
= (q – 7) (q – 3)

Question (iii)
p² + 6p – 16
Solution:
= p² + 6p + 9 – 25
= (P² + 6p + 9) – (25)
= (p + 3)² – (5)²
= (p + 3 – 5) (p + 3 + 5)
= (p – 2) (p + 8)
Here, last term is (-16).
∴ (-25) + 9 = (-16)

OR

p² + 6p – 16
Here, ab = – 16 and a + b = 6
Possible values of a = 8 or (-8) b = 2 or (-2)
Let us check, 8 + 2 = 10 ≠ 6
(- 8) + 2 = (-6) ≠ 6
8 + (-2) = 8 – 2 = 6
Now, p² + 6p – 16
= p² + 8p – 2p – 16
= P (P + 8) – 2 (p + 8)
= (p + 8) (p – 2)

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 Factorization Ex 14.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.1

1. Find the common factors of the given terms:

Question (i)
12x, 36
Solution:
12x = 2 × 2 × 3 × x
36 = 2 × 2 × 3 × 3
∴ Common factors of 12x and 36 = 2 × 2 × 3
= 12

Question (ii)
2y, 22xy
Solution:
2y = 2 xy
22xy = 2 × 11 × x × y
∴ Common factors of 2y and 22xy
= 2 × y = 2y

Question (iii)
14pq, 28p²q²
Solution:
14pq = 2 × 7 × p × q
28p²q² = 2 × 2 × 7 × p × p × q × q
∴ Common factors of 14pq and 28p²q²
= 2 × 7 × p × q = 14pq

Question (iv)
2x, 3x², 4
Solution:
2x = 2 × x
3x² = 3 × x × x
4 = 2 × 2
∴ Common factors of 2x, 3x² and 4 = 1 [Note: 1 is a factor of each term.]

Question (v)
6abc, 24ab², 12a²b
Solution:
6abc = 2 × 3 × a × b × c
24ab² = 2 × 2 × 2 × 3 × a × b × b
12a²b = 2 × 2 × 3 × a × a × b
∴ Common factors of 6abc, 24ab² and 12a²b
= 2 × 3 × a × b
= 6ab

Question (vi)
16x³, – 4x², 32x
Solution:
16x³ = 2 × 2 × 2 × 2 × x × x × x
– 4x² = (-1) × 2 × 2 × x × x
32x = 2 × 2 × 2 × 2 × 2 × x
∴ Common factors of 16x³, – 4x² and 32x = 2 × 2 × x = 4x

Question (vii)
10pq, 20qr, 30rp
Solution:
10pq = 2 × 5 × p × q
20qr = 2 × 2 × 5 × q × r
30rp = 2 × 3 × 5 × r × p
∴ Common factors of 10pq, 20qr and 30rp = 2 × 5 = 10

Question (viii)
3x²y³, 10x³y², 6x²y²z
Solution:
3x²y³ = 3 × x × x × y × y × y
10x³y² = 2 × 5 × x × x × x × y × y
6x²y²z = 2 × 3 × x × x × y × y × z
∴ Common factors of 3x²y³, 10x³y² and 6x²y²z
= x × x × y × y = x²y²

2. Factorise the following expressions:

Question (i)
7x – 42
Solution:
7x = 7 × x
42 = 2 × 3 × 7
∴ 7 is common in both terms.
7x – 42 = 7 (x – 6)

Question (ii)
6p – 12q
Solution:
6p = 2 × 3 × p
12q = 2 × 2 × 3 × q
∴ 2 × 3 = 6 is common in both terms.
6p – 12q = 6 (p – 2q)

Question (iii)
7a² + 14a
Solution:
7a² = 7 × a × a
14a = 2 × 7 × a
∴ 7 × a = 7a is common in both terms.
∴ 7a² + 14a = 7a (a + 2)

Question (iv)
– 16z + 20z³
Solution:
– 16z = -2 × 2 × 2 × 2 × z
20z³ = 2 × 2 × 5 × z × z × z
∴ 2 × 2 × z = 4z is common in both terms.
∴ – 16z + 20z³ = 4z (- 4 + 5z²)

Question (v)
20l²m + 30alm
Solution:
= 2 × 2 × 5 × l × l × m + 2 × 3 × 5 × a × l × m
= 10lm (2l + 3a)

Question (vi)
5x²y – 15xy²
Solution:
= 5xy (x – 3y)

Question (vii)
10a² – 15b² + 20c²
Solution:
= 5 (2a² – 3b² + 4c²)

Question (viii)
– 4a² + 4ab – 4ca
Solution:
= 4a (- a + b – c)

Question (ix)
x²yz + xy²z + xyz²
Solution:
= xyz (x + y + z)

Question (x)
ax²y + bxy² + cxyz
Solution:
= xy (ax + by + cz)

3. Factorise:

Question (i)
x² + xy + 8x + 8y
Solution:
= x (x + y) + 8 (x + y)
= (x + y) (x + 8)

Question (ii)
15xy – 6x + 5y – 2
Solution:
= 3x (5y – 2) + 1 (5y – 2)
= (5y – 2) (3x + 1)

Question (iii)
ax + bx- ay – by
Solution:
= x (a + b) – y (a + b)
= (a+ b) (x- y)

Question (iv)
15pq + 15 + 9q + 25p
Solution:
= 15pq + 9q + 25p + 15
= 3q (5p + 3) + 5 (5p + 3)
= (5p + 3) (3q + 5)

Question (v)
z – 7 + 7xy – xyz
Solution:
= z – 7 – xyz + 7xy
= 1 (z – 7) – xy (z – 7)
= (z – 7) (1 – xy)

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 15 Introduction to Graphs InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 15 Introduction to Graphs InText Questions

Try These : [Textbook Page No. 244]

1. In the above example, use the graph to find how much petrol can be purchased for ₹ 800.

Solution:
We can find the quantity of petrol to be got for ₹ 800. For this take a point on the Y-axis (0, 800). Now, draw a line parallel to X-axis to meet the graph at the point B. Now, from the point B, draw a line parallel to Y-axis, which intersect X-axis in the point C. Coordinate of the point C : (16, 0).
Hence, 16 litres of petrol can be purchased for ₹ 800.

Think, Discuss and Write : [Textbook Page No. 243]

1. The number of litres of petrol you buy to fill a car’s petrol tank will decide the amount you have to pay. Which is the independent variable here? Think about it.
Solution:
Here, we clearly understand that graph of quantity of petrol (litre) and amount to pay (₹) should be a line.
Both quantities are in direct proportion. If we fill more litres of petrol, we have to pay more amount and vice versa.
∴ Petrol is the independent variable.

Try These : [Textbook Page No. 245]

1. Is Example 7, a case of direct variation?
Solution:
Yes, Example 7 given on page 245 (Textbook), is a case of direct variation. As the principal increases, the simple interest on it also increases proportionately.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 15 Introduction to Graphs Ex 15.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.3

1. Draw the graphs for the following tables of values, with suitable scales on the axes.

Question (a)
Cost of apples

Number of apples	1	2	3	4	5
Cost (in ₹)	5	10	15	20	25

Solution:

1. Draw 2 lines perpendicular to each other as X-axis and Y-axis on graph paper.
2. Take scale :
On X-axis : 1 cm = 1 apple
On Y-axis 1 cm = ₹ 5
3. Plot the points (1, 5), (2, 10), (3, 15), (4, 20) and (5, 25) on graph paper.
4. Join these points and extend line.

(b) Distance travelled by a car

Time (in hours)	6 a.m.	7 a.m.	8 a.m.	9 a.m.
Distance (in km)	40	80	120	160

Solution:
1. Draw two lines perpendicular to each other as X-axis and Y-axis on graph paper.
2. Take scale:
On X-axis : 2 cm = 1 unit (Time in hours.)
On Y-axis : 1 cm = 10 km
3. Plot the points (6, 40), (7, 80), (8, 120) and (9, 160) on graph paper.
4. Join these points and extend line.

Question (i)
How much distance did the car cover during the period 7:30 a.m. to 8:00 a.m.?
Solution:
In the graph, draw a perpendicular at the point indicating 7:30 a.m. on the X-axis such that it meets the graph at P.
From P draw a line parallel to X-axis to meet Y-axis at 100 km.
∴ Distance travelled between 7:30 am and 8:00 am.
= (120 – 100) km
= 20 km

Question (ii)
What was the time when the car had covered a distance of 100 km since it’s start?
Solution:
When the car had covered a distance of 100 km, the time was 7 : 30 am.

(c) Interest on deposits for a year.

Deposit (in ₹)	1000	2000	3000	4000	5000
Simple Interest (in ₹)	80	160	240	320	400

Solution :
1. Draw two lines perpendicular to each other as X-axis and Y-axis on graph paper.
2. Take scale:
On X-axis : 2 cm = ₹ 1000 (deposit)
On Y-axis : 1 cm = ₹ 40 (simple interest)
3. Plot the points (1000, 80), (2000, 160), (3000, 240), (4000, 320) and (5000, 400) on graph paper.
4. Join these points and extend line.

Question (i)
Does the graph pass through the origin?
Solution:
Yes, it passes through the origin.

Question (ii)
Use the graph to find the interest on ₹ 2500 for a year.
Solution:
From the graph, the interest on ₹ 2500 for a year is ₹ 200.

Question (iii)
To get an interest of ₹ 280 per year, how much money should be deposited?
Solution:
From the graph an interest of ₹ 280 can be got by depositing ₹ 3500.

2. Draw a graph for the following:

Question (i)

Side of square (in cm)	2	3	3.5	5	6
Perimeter (in cm)	8	12	14	20	24

Is it a linear graph?
Solution:
1. Draw two lines perpendicular to each other as X-axis and Y-axis on graph paper.
2. Take scale:
On X-axis : 1 cm = 1 cm (Side of a square)
On Y-axis : 1 cm = 4 cm (Perimeter of a square) )
3. Plot the points (2, 8), (3, 12), (3.5, 14), (5, 20) and (6, 24) on graph paper.
4. Join these points and extend line.

Yes, it is a linear graph.

Question (ii)

Side of square (in cm)	2	3	4	5	6
Area (in cm2)	4	9	16	25	36

Is it a linear graph?
Solution:
1. Draw two lines perpendicular to each other as X-axis and Y-axis on graph paper.
2. Take scale:
On X-axis : 1 cm = 1 cm (Side of a square)
On Y-axis : 1 cm 5 cm (Area of a square)
3. Plot the points (2, 4), (3, 9), (4, 16), (5, 25) and (6, 36) on graph paper.
4. Join these points.

No, this graph is not a straight line. So it is not a linear graph.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 4 Practical Geometry Ex 4.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.2

1. Construct the following quadrilaterals:

Question (i).
Quadrilateral LIFT.
LI = 4 cm,
IF = 3 cm,
TL = 2.5 cm,
LF = 4.5 cm,
IT = 4 cm.
Solution:

Steps of construction:

• Draw a line segment LI = 4 cm.
• With L as centre and radius = 2.5 cm, draw an arc.
• With I as centre and radius = 4 cm, draw an arc to intersect the previous arc at T.
• With L as centre and radius = 4.5 cm draw an arc.
• With I as centre and radius 3 cm, draw an arc to intersect the previous, arc at F.
• Draw \(\overline{\mathrm{LT}}, \overline{\mathrm{IF}}, \overline{\mathrm{FT}}, \overline{\mathrm{LF}}\) and \(\overline{\mathrm{IT}}\).

Thus, LIFT is the required quadrilateral.

Question (ii).
Quadrilateral GOLD
OL = 7.5 cm,
GL = 6 cm,
GD = 6 cm,
LD = 5 cm,
OD = 10 cm.
Solution:

Steps of construction:

• Draw a line segment LD = 5 cm.
• With L as centre and radius = 7.5 cm, draw an arc.
• With D as centre and radius = 10 cm, draw another arc to intersect the previous arc at O.
• With L as centre and radius = 6 cm, draw an arc.
• With D as centre and radius = 6 cm, draw another arc to intersect previous arc at G.
• Draw \(\overline{\mathrm{LO}}, \overline{\mathrm{GO}}, \overline{\mathrm{DG}}, \overline{\mathrm{LG}}\) and \(\overline{\mathrm{DO}}\).

Thus, GOLD is the required quadrilateral.

Question (iii).
Rhombus BEND
BN = 5.6 cm,
DE = 6.5 cm.
Solution:

[Note: Diagonals of a rhombus are perpendicular bisectors of each another. Here, diagonals of □ BEND \(\overline{\mathrm{DE}}\) and \(\overline{\mathrm{BN}}\) intersect at A. So AN = 2.8 cm and AB = 2.8 cm (BN = 5.6 cm, \(\frac {1}{2}\) BN = AN and AB)]
Steps of construction:

• Draw a line segment DE = 6.5 cm.
• Draw perpendicular bisector \(\overleftrightarrow{X Y}\) of \(\overline{\mathrm{DE}}\), which intersects \(\overline{\mathrm{DE}}\) at A.
• With centre at A and radius = 5.6 × \(\frac {1}{2}\) = 2.8 cm, draw two arcs intersecting \(\overleftrightarrow{X Y}\) in points B and N.
• Draw \(\overline{\mathrm{DN}}, \overline{\mathrm{EN}}, \overline{\mathrm{EB}}\) and \(\overline{\mathrm{DB}}\).

Thus, BEND is the required quadrilateral.