PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.5

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 4 Practical Geometry Ex 4.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.5

Draw the following.

Question 1.
The square READ with RE = 5.1 cm.
Solution:
PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.5 1
Steps of construction :

  • Draw a line segment RE = 5.1 cm.
  • At E, draw \(\overrightarrow{\mathrm{EM}}\), such that ∠REM = 90°.
  • With E as centre and radius =5.1 cm, draw an arc intersecting \(\overrightarrow{\mathrm{EM}}\) at A.
  • With R as centre and radius = 5.1 cm, draw an arc.
  • With A as centre and radius = 5.1 cm, draw an arc which intersect the previous arc at D.
  • Draw \(\overline{\mathrm{RD}}\) and \(\overline{\mathrm{AD}}\).

Thus, READ is the required square.

PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.5

Question 2.
A rhombus whose diagonals are 5.2 cm and 6.4 cm long.
Solution:
PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.5 2
[Note : The diagonals of a rhombus bisect each other at right angle.]
Here, in rhombus XYZW, YW 6.4 cm.
∴ OW = OY = 3.2 cm

Steps of construction:

  • Draw a line segment XZ = 5.2 cm.
  • Draw \(\overleftrightarrow{\mathrm{AB}}\), the perpendicular bisector of \(\overline{\mathrm{XZ}}\), which intersect \(\overline{\mathrm{XZ}}\) at O.
  • With O as centre and radius = 3.2 cm, draw an arc intersecting \(\overleftrightarrow{\mathrm{AB}}\) above \(\overline{\mathrm{XZ}}\) at W.
  • With O as centre and radius = 3.2 cm, draw another arc which intersects \(\overleftrightarrow{\mathrm{AB}}\) below \(\overline{\mathrm{XZ}}\) at Y.
  • Draw \(\overline{\mathrm{XY}}\), \(\overline{\mathrm{YZ}}\), \(\overline{\mathrm{ZW}}\) and \(\overline{\mathrm{XW}}\).

Thus, XYZW is the required rhombus.

Question 3.
A rectangle with adjacent sides of lengths 5 cm and 4 cm.
Solution:
PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.5 3
[Note: Each angle of a rectangle is a right angle.]
Steps of construction :

  • Draw a line segment AB = 5 cm.
  • At A, draw \(\overrightarrow{\mathrm{AX}}\), such that ∠XAB = 90°.
  • With A as centre and radius = 4 cm, draw an arc intersecting \(\overrightarrow{\mathrm{AX}}\) at D.
  • With B as centre and radius = 4 cm, draw an arc.
  • With D as centre and radius = 5 cm, draw an arc which intersects the previous arc at C.
  • Draw \(\overline{\mathrm{BC}}\) and \(\overline{\mathrm{CD}}\).

Thus, ABCD is the required rectangle.

PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.5

Question 4.
A parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm. Is it unique ?
Solution:
PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.5 4
A parallelogram cannot be constructed as sufficient measurements are not given. It is not unique as angles may vary in parallelogram drawn by given measurements.

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

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:
PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.1 1
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.

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

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.
PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.1 2
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.

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

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.
PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.1 2.1
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.

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

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.
PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.1 3
Thus, there are 2.25 × 107 crystals of sugar in 5 kg of sugar.

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

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.

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

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.
PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.1 4
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.

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

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.

PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.4

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 4 Practical Geometry Ex 4.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.4

1. Construct the following quadrilaterals:

Question (i).
Quadrilateral DEAR
DE = 4 cm
EA = 5 cm
AR = 4.5 cm
∠E = 60°
∠A = 90°
Solution:
PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.4 1
Steps of construction:

  • Draw a line segment DE = 4 cm.
  • At E, draw \(\overrightarrow{\mathrm{EM}}\) such that ∠DEM = 60°.
  • With E as centre and radius = 5 cm, draw an arc intersecting \(\overrightarrow{\mathrm{EM}}\) at A.
  • At A, draw \(\overrightarrow{\mathrm{AN}}\) such that ∠EAN = 90°.
  • With A as centre and radius = 4.5 cm, draw an arc intersecting \(\overrightarrow{\mathrm{AN}}\) at R.
  • Draw \(\overline{\mathrm{DR}}\).

Thus, DEAR is the required quadrilateral.

PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.4

Question (ii).
Quadrilateral TRUE
TR = 3.5 cm
RU = 3 cm
UE = 4 cm
∠R = 75°
∠U = 120°
Solution:
PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.4 2
Steps of construction :

  • Draw a line segment TR = 3.5 cm.
  • At R, draw a \(\overrightarrow{\mathrm{RM}}\) such that ∠TRM = 75°.
  • With R as centre and radius = 3 cm, draw an arc intersecting \(\overrightarrow{\mathrm{RM}}\) at U.
  • At U, draw \(\overrightarrow{\mathrm{UN}}\) such that ∠RUN = 120°.
  • With U as centre and radius = 4 cm, draw an arc intersecting \(\overrightarrow{\mathrm{UN}}\) at E.
  • Draw \(\overline{\mathrm{ET}}\).

Thus, TRUE is the required quadrilateral.

PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.3

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 4 Practical Geometry Ex 4.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3

1. Construct the following quadrilaterals:

Question (i).
Quadrilateral MORE
MO = 6 cm
OR = 4.5 cm
∠M = 60°
∠O = 105°
∠R = 105°
Solution:
PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.3 1

  • Draw a line segment MO = 6 cm.
  • At M, draw \(\overrightarrow{\mathrm{MA}}\), such that ∠OMA = 60°
  • At O, draw \(\overrightarrow{\mathrm{OB}}\) such that ∠MOB = 105°.
  • With O as centre and radius = 4.5 cm, draw an arc intersecting \(\overrightarrow{\mathrm{OB}}\) at R.
  • At R, draw \(\overrightarrow{\mathrm{RC}}\) such that ∠ORC = 105°.
  • Locate E at intersection of \(\overrightarrow{\mathrm{RC}}\) and \(\overrightarrow{\mathrm{MA}}\).

Thus, MORE is the required quadrilateral.

PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.3

Question (ii).
Quadrilateral PLAN
PL = 4 cm
LA = 6.5 cm
∠P = 90°
∠A = 110°
∠N = 85°
Solution:
PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.3 2
[Note : In □ PLAN, m∠P = 90°, m∠A = 110° and m∠N = 85°)
∴ m∠L = 360°- (m∠P + m∠A + m∠N)
= 360° – (90° + 110° + 85°)
= 360° – 285°
= 75°

Steps of construction:

  • Draw a line segment AL = 6.5 cm.
  • At A, draw \(\overrightarrow{\mathrm{AX}}\) such that ∠XAL = 110°. (Use protractor)
  • At L, draw \(\overrightarrow{\mathrm{LY}}\) such that ∠YLA = 75°. (Use protractor)
  • With L as centre and radius = 4 cm, draw an arc intersecting \(\overrightarrow{\mathrm{LY}}\) at P.
  • At P, draw \(\overrightarrow{\mathrm{PZ}}\) such that ∠ZPL = 90°. (∵ ∠ ZPY = 90°)
  • Locate N at intersection of \(\overrightarrow{\mathrm{AX}}\) and \(\overrightarrow{\mathrm{PZ}}\).

Thus, PLAN is the required quadrilateral.

PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.3

Question (iii).
Parallelogram HEAR
HE = 5 cm
EA = 6 cm
∠R = 85°
Solution:
PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.3 3
[Note : □ HEAR is a parallelogram.
Opposite sides of parallelogram are of equal lengths.]
HE = 5 cm, ∴ AR = 5 cm, EA = 6 cm, ∴ HR = 6 cm
Adjacent angles of a parallelogram arc supplementary.
m∠R = 85° (given)
∴ m∠H = 180° – 85° = 95°
Opposite angles of a parallelogram are of equal measures,
m ∠ R = 85°
∴ m ∠ E = 85°

Steps of construction:

  • Draw a line segment HE = 5 cm.
  • At H, draw \(\overrightarrow{\mathrm{HX}}\), such that ∠ XHE = 95°. (Use protractor)
  • With H as centre and radius = 6 cm, draw an arc intersecting \(\overrightarrow{\mathrm{HX}}\) at R.
  • At E, draw \(\overrightarrow{\mathrm{EY}}\) such that ∠ HEY = 85°. (Use protractor)
  • With E as centre and radius = 6 cm, draw an arc intersecting \(\overrightarrow{\mathrm{EY}}\) at A.
  • Draw \(\overline{\mathrm{AR}}\).

Thus, HEAR is the required parallelogram.

PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.3

Question (iv).
Rectangle OKAY
OK = 7 cm
KA = 5 cm
Solution:
PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.3 4
[Note: Here, OKAY is a rectangle. Opposite sides of a rectangle are of equal lengths.]
OK = 7 cm, ∴ AY = 7 cm and KA = 5 cm, ∴ OY = 5 cm
Moreover, all angles of a rectangle are right angles.

Steps of construction:

  • Draw a line segment OK = 7 cm.
  • At O, draw \(\overrightarrow{\mathrm{OM}}\) such that ∠ MOK = 90°.
  • With O as centre and radius = 5 cm, draw an arc intersecting \(\overrightarrow{\mathrm{OM}}\) at Y.
  • At K, draw \(\overrightarrow{\mathrm{KN}}\) such that ∠ NKO = 90°.
  • With K as centre and radius = 5 cm, draw an arc intersecting \(\overrightarrow{\mathrm{KN}}\) at A.
  • Draw \(\overline{\mathrm{AY}}\).

Thus, OKAY is the required rectangle.

PSEB 8th Class Maths Solutions Chapter 5 Data Handling Ex 5.1

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 5 Data Handling Ex 5.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.1

Question 1.
For which of these would you use a histogram to show the data?
(a) The number of letters for different areas in a postman’s bag.
(b) The height of competitors in an athletics meet.
(c) The number of cassettes produced by 5 companies.
(d) The number of passengers boarding trains from 7:00 a.m. to 7:00 p.m. at a station.
Give reasons for each.
Solution:
We represent those data by a histogram which can be grouped into class intervals. Obviously, for (b) and (d), the data can be represented by histograms.

Question 2.
The shoppers who come to a departmental store are marked as: man (M), woman (W), boy (B) or girl (G). The following list gives the shoppers who came during the first hour in the morning:
W W W G B W W M G G M M W W W W G B M W B G G M W W M M W W W M W B W G M W W W W G W M M W W M W G W M G W M M B G G W
Make a frequency distribution table using tally marks. Draw a bar graph to illustrate it.
Solution:
The frequency distribution table for the above data can be done as follows :
PSEB 8th Class Maths Solutions Chapter 5 Data Handling Ex 5.1 1
We can represent the above data by a bar graph as given below:
PSEB 8th Class Maths Solutions Chapter 5 Data Handling Ex 5.1 2

PSEB 8th Class Maths Solutions Chapter 5 Data Handling Ex 5.1

Question 3.
The weekly wages (in ₹) of 30 workers in a factory are:
830, 835, 890, 810, 835, 836, 869, 845, 898, 890, 820, 860, 832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 835, 836, 878, 840, 868, 890, 806, 840
Using tally marks make a frequency table with intervals as 800-810, 810-820 and so on.
Solution:
The lowest observation = 804
The highest observation = 898
The classes are 800-810, 810-820, etc.
∴ The frequency distribution table is
PSEB 8th Class Maths Solutions Chapter 5 Data Handling Ex 5.1 3

Question 4.
Draw a histogram for the frequency table made for the data in Question 3, and answer the following questions:
(i) Which group has the maximum number of workers?
(ii) How many workers earn ₹ 850 and more ?
(iii) How many workers earn less than ₹ 850 ?
Solution:
The histogram from the above frequency table is given below.
Here we have represented the class intervals along X-axis and frequencies of the class intervals along the Y-axis.
PSEB 8th Class Maths Solutions Chapter 5 Data Handling Ex 5.1 4
(i) The group 830 – 840 has the maximum number of workers.
(ii) Number of workers earning ₹ 850 or more = 1 + 3 + 1 + 1 – 1 – 4 = 10
(iii) Number of workers earning less than ₹ 850 = 3 + 2 + 1 + 9 + 5 = 20

PSEB 8th Class Maths Solutions Chapter 5 Data Handling Ex 5.1

Question 5.
The number of hours for which students of a particular class watched television during holidays is shown through the given graph.
Answer the following:
(i ) For how many hours did the maximum number of students watch TV ?
(ii) How many students watched TV for less than 4 hours ?
(iii) How many students spent more than 5 hours in watching TV ?
PSEB 8th Class Maths Solutions Chapter 5 Data Handling Ex 5.1 5
Solution:
(i) The maximum number of students watched TV for 4 to 5 hours.
(ii) 34 students watched TV for less than 4 hours (4 + 8 + 22 = 34).
(iii) 14 students spent more than 5 hours in watching TV (8 + 6 = 14).

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions InText Questions

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 13 Direct and Inverse Proportions InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions InText Questions

Try These : [Textbook Page No. 204]

1. Observe the following tables and find if x and y are directly proportional.

Question (i)

X 20 17 14 11 8 5 2
y 40 34 28 22 16 10 4

Solution:
\(\begin{aligned}
&\frac{20}{40}=\frac{1}{2}, \frac{17}{34}=\frac{1}{2}, \frac{14}{28}=\frac{1}{2}, \frac{11}{22}=\frac{1}{2}, \frac{8}{16}=\frac{1}{2} \\
&\frac{5}{10}=\frac{1}{2}, \frac{2}{4}=\frac{1}{2}
\end{aligned}\)
The value of \(\frac{\text {x}}{\text {y}}\) is same for different values of x and y. So these values x and y are directly proportional.

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions InText Questions

Question (ii)

X 6 10 14 18 22 26 30
y 4 8 12 16 20 24 28

Solution:
\(\begin{aligned}
&\frac{6}{4}=\frac{3}{2}, \frac{10}{8}=\frac{5}{4}, \frac{14}{12}=\frac{7}{6}, \frac{18}{16}=\frac{9}{8}, \frac{22}{20}=\frac{11}{10}, \\
&\frac{26}{24}=\frac{13}{12}, \frac{30}{28}=\frac{15}{14}
\end{aligned}\)
The values of \(\frac{\text {x}}{\text {y}}\) are different for different values of x and y respectively. So these values of x and y are not directly proportional.

Question (iii)

X 5 8 12 15 18 20
y 15 24 36 60 72 100

Solution:
The values of \(\frac{\text {x}}{\text {y}}\) are different for different values of x and y respectively.
So these values of x and y are not directly proportional.

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions InText Questions

2. Principal = ₹ 1000, Rate = 8 % per annum. Fill in the following table and find which type of interest (simple or compound) changes in direct proportion with time period.

Time Period 1 year 2 years 3 years
Simple Interest (in ₹)
Compound Interest (in ₹)

Solution:
Simple interest : SI = \(\frac{P \times R \times T}{100}\)
For calculation:
P = ₹ 1000, R = 8 %, T = …………….

Time (T) → 1 year: T = 1 2 years : T = 2 3 years : T = 3
Simple interest SI = \(\frac{P \times R \times T}{100}\) ₹ \(\frac{1000 \times 8 \times 1}{100}\)
= ₹ 80
₹ \(\frac{1000 \times 8 \times 2}{100}\)
= ₹ 160
₹ \(\frac{1000 \times 8 \times 3}{100}\)
= ₹ 240
\(\frac{\text { SI }}{\text { T }}\) \(\frac {80}{1}\) = 80 \(\frac {160}{2}\) = 80 \(\frac {240}{3}\) = 80

Here, the ratio of simple interest with time period is same for every year.
Hence, simple interest changes in direct proportion with time period.
Compound interest:
For calculation:
P = ₹ 1000, R = 8 %, T = ……………

Time → 1 year : n = 1
A = P(1 + \(\frac {R}{100}\))n
CI = A – P
A = 1000(1 + \(\frac {8}{100}\))1
= 1000 × \(\frac {108}{100}\) = 1080
∴ CI = 1080 – 1000 = ₹ 80
\(\frac{\text { CI }}{\text { T }}\) \(\frac {80}{1}\)
Time → 2 years : n = 2
A = P(1 + \(\frac {R}{100}\))n
CI = A – P
A = 1000(1 + \(\frac {8}{100}\))2
= 1000 × \(\frac {108}{100}\) × \(\frac {108}{100}\) = ₹ 1166.40
∴ CI = ₹ 1166.40 – 1000 = ₹ 166.40
\(\frac{\text { CI }}{\text { T }}\) \(\frac {166.40}{2}\)
Time → 3 years : n = 3
A = P(1 + \(\frac {R}{100}\))n
CI = A – P
A = 1000(1 + \(\frac {8}{100}\))3
= 1000 × \(\frac {108}{100}\) × \(\frac {108}{100}\) × \(\frac {108}{100}\) = ₹ 1259.712
∴ CI = ₹ 1259.712 – ₹ 1000 = ₹ 259.712
\(\frac{\text { CI }}{\text { T }}\) \(\frac {259.712}{3}\)

Here, the ratio of CI and T is not same.
Thus, compound interest is not proportional with time period.

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions InText Questions

Think, Discuss and Write: [Textbook Page No. 204]

1. If we fix time period and the rate of interest, simple interest changes proportionally with principal. Would there be a similar relationship for compound interest? Why?
Solution:
Time period (T) and rate of interest (R) are fixed, then
Simple interest = \(\frac {PRT}{100}\) = P × Constant
So simple interest depends on principal. Simple interest changes proportionally with principal.
Now, compound interest = P(1 + \(\frac {R}{100}\))T – P
i.e., A – P
= P [(1 + \(\frac {R}{100}\)T – 1]
= P × Constant
So compound interest depends on principal.
If principal increases or decreases, then compound interest will also increases or decreases.
Thus, compound interest changes with principal.

Think, Discuss and Write : [Textbook Page No. 209]

1. Take a few problems discussed so far under ‘direct variation’. Do you think that they can be solved by ‘unitary method’?
Solution:
Yes, each problem can be solved by unitary method.
e.g. Question 4 of Exercise: 13.1
Number of bottles filled in 6 hours = 840
∴ The number of bottles filled in 1 hour = \(\frac {840}{6}\) = 140
The number of bottles filled in 5 hours = 140 × 5 = 700
Thus, 700 bottles will it fill in five hours.

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions InText Questions

Try These : [Textbook Page No. 211]

1. Observe the following tables and find which pair of variables (here x and y) are in inverse proportion.

Question (i)

X 50 40 30 20
y 5 6 7 8

Solution:
x1 = 50 and y1 = 5
∴ x1y1 = 50 × 5
∴ x1y1 = 250

x2 = 40 and y2 = 6
∴ x2y2 = 40 × 6
∴ x2y2 = 240

x3 = 30 and y3 = 7
∴ x3y3 = 30 × 7
∴ x3y3 = 210

x4 = 20 and y4 = 8
∴ x4y4 = 20 × 8
∴ x4y4 = 160
Now 250 ≠ 240 ≠ 210 ≠ 160
∴ x1y1 ≠ x2y2 ≠ x3y3 ≠ x4y4
∴ x and y are not in inverse proportion.

Question (ii)

X 100 200 300 400
y 60 30 20 15

Solution:
x1 = 100 and y1 = 60
∴ x1y1 = 100 × 60
∴ x1y1 = 6000

x2 = 200 and y2 = 30
∴ x2y2 = 200 × 30
∴ x2y2 = 6000

x3 = 300 and y3 = 20
∴ x3y3 = 300 × 20
∴ x3y3 = 6000

x4 = 400 and y4 = 15
∴ x4y4 = 400 × 15
∴ x4y4 = 6000
Now x1y1 = x2y2 = x3y3 = x4y4
∴ x and y are in inverse proportion.

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions InText Questions

Question (iii)

X 90 60 45 30 20 5
y 10 15 20 25 30 35

Solution:
x1 = 90 and y1 = 10
∴ x1y1 = 90 × 10
∴ x1y1 = 900

x2 = 60 and y2 = 15
∴ x2y2 = 60 × 15
∴ x2y2 = 900

x3 = 45 and y3 = 20
∴ x3y3 = 45 × 20
∴ x3y3 = 900

x4 = 30 and y4 = 25
∴ x4y4 = 30 × 25
∴ x4y4 = 750

x5 = 20 and y5 = 30
∴ x5y5 = 20 × 30
∴ x5y5 = 600

x6 = 30 and y6 = 35
∴ x6y6 = 5 × 35
∴ x6y6 = 175

Now x1y1 = x2y2 = x3y3 ≠ x4y4 ≠ x5y5 ≠ x6y6
∴ x and y are in inverse proportion.

PSEB 8th Class Maths Solutions Chapter 14 Factorization InText Questions

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)

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

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)

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

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

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

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 × a2-2 × b4-2 × c6-3
= 9 × a0 × b2 × c3
= 9 × 1 × b2 × c3
= 9b2c3
∴ 63a2b4c6 ÷ 7a2b2c3
= 9b2c3

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

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.

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

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.

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

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.
PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2 1
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:
PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2 2

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

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

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,
PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2 3.1
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.

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

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.

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

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?
PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2 4.1
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.

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

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.

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

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.

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.4

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

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.4

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

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.4

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

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.4

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)

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.4

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

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.4

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

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.4

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

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.4

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: \(\frac{3}{4 x+3}=\frac{3}{4 x+3}\)

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.4

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

PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3

Question 1.
Given a parallelogram ABCD. Complete each statement along with the definition or property used:
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 1
(i) AD = ……………
(ii) ∠DCB = ……………
(iii) OC = ……………
(iv) m∠DAB + m∠CDA = ……………
Solution:
(i) AD = BC
(∵ Opposite sides are equal)

(ii) ∠DCB = ∠DAB
(∵ Opposite angles are equal)

(iii) OC = OA
(∵ Diagonals bisect each other)

(iv) m∠DAB + m∠CDA = 180°
(∵ Adjacent angles are supplementary)

PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3

Question 2.
Consider the following parallelograms. Find the values of the unknowns x, y, z.
(i)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 2
Solution:
□ ABCD is a parallelogram.
∴ ZB = ZD
∴ y = 100°
(∵ Opposite angles)
Now, y + z = 180° (∵ Adjacent singles are supplementary)
∴ 100° + z = 180°
∴ z = 180° – 100°
∴ z = 80°
Now, x = z (∵ Opposite angles)
∴ x = 80°
Thus, x = 80°, y = 100° and z = 80°.

(ii)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 3
Solution:
It is a parallelogram.
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 4
∴ m∠P + m∠S = 180°
(∵ Adjacent angles are supplementary)
∴ x + 50° = 180°
∴ x = 180° – 50°
∴ x = 130°
x = y (∵ Opposite angles)
∴ y = 130°
Now, m∠Q = 50° (∵ ∠S and ∠Q are opposite angles)
m∠Q + z = 180° (∵ Linear pair of angles)
∴ 50° + z = 180°
∴ z = 180° – 50°
∴ z = 130°
Thus, x = 130°, y = 130° and z = 130°.

(iii)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 5
Solution:
Vertically opposite angles are equal.
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 6
∴ m∠AMD = m∠BMC = 90°
∴ x = 90°
In ΔBMC
y + 90° + 30° = 180°
(Angle sum property of a triangle)
∴ y + 120° = 180°
∴ y = 180° – 120° = 60°
Here, ABCD is a parallelogram
∴ \(\overline{\mathrm{AD}} \| \overline{\mathrm{BC}}, \overline{\mathrm{BD}}\) is a transversal.
∴ y = z (∵ Alternate angles)
∴ z = 60° (∵ y = 60°)
Thus, x = 90°, y = 60° and z = 60°

(iv)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 7
Solution:
□ ABCD is a Dparallelogram.
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 8
∠D = ∠B (∵ Opposite angles)
∴ y = 80°
m∠A + m∠D = 180° (∵ Adjacent angles are supplementary.)
∴ x + y = 180°
∴ x + 80° = 180°
∴ x = 180° – 80°
∴ x = 100°
m∠A = m∠BCD (∵ Opposite angles are equal)
∴ 100° = m∠BCD
Now, z + m∠BCD = 180° (∵ Linear pair of angles)
∴ z + 100° = 180°
∴ z = 180° – 100°
∴ z = 80°
Thus, x = 100°, y = 80° and z = 80°

(v)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 9
Solution:
□ ABCD is a parallelogram.
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 10
∴ m∠B = m∠D (∵ Opposite angles)
∴ y = 112°
m∠A + m∠B = 180°
(∵ Adjacent angles are supplementary)
∴ (40° + z) + 112° = 180°
∴ 40° + z + 112° = 180°
∴ z + 152° = 180°
∴ z = 180° – 152°
∴ z = 28°
Now, \(\overline{\mathrm{DC}} \| \overline{\mathrm{AB}}\), \(\overleftrightarrow{\mathrm{AC}}\) is their transversal.
∴ z = x
∴ x = 28° (∵ z = 28°)
Thus, x = 28°, y = 112° and z = 28°

PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3

Question 3.
Can a quadrilateral ABCD be a parallelogram if
(i) ∠D + ∠B = 180°?
Solution:
In a quadrilateral ABCD,
∠D + ∠B = 180° (Given)
In parallelogram, the sum of measures of adjacent angles is 180°.
But here, the sum of measures of opposite angles is 180°.
∴ The quadrilateral may be a parallelogram.

(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
Solution:
In a quadrilateral ABDC,
AB = DC = 8 cm
AD = 4 cm
BC = 4.4 cm
∵ Opposite sides AD and BC are not equal.
∴ It cannot be a parallelogram.

(iii) ∠A = 70° and ∠C = 65° ?
Solution:
In a quadrilateral ABCD, m∠A = 70° and m∠C = 65°
Opposite angles ∠A ≠ ∠C
∴ It cannot be a parallelogram.

Question 4.
Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.
Solution:
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 11
Here, ∠B = ∠D (see figure)
Yet, it is not a parallelogram.

PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3

Question 5.
The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram.
Solution:
Let ABCD be a parallelogram in which adjacent angles ∠A and ∠B are 3x and 2x respectively.
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 12
Adjacent angles are supplementary.
∴ m∠A + m∠B = 180°
∴ 3x + 2x – 180°
∴ 5x = 180°
∴ \(\frac{180^{\circ}}{5}\)
∴ m∠A = 3x = 3 × 36° = 108°
m∠B = 2x = 2 × 36° = 72°
Now, m∠A = m∠C and m∠B = m∠D (∵ Opposite angles)
∴ m∠C = 108° and m∠D = 72°
Thus, ∠A = 108°, ∠B = 72°, ∠C = 108° and ∠D = 72°.

Question 6.
Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Solution:
Let ABCD be a parallelogram such that adjacent angles ∠A = ∠B.
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 13
∴ m∠A + m∠B = 180°
(∵ Adjacent angles are supplementary)
∴ m∠A + m∠A = 180° (m∠B = m∠A)
∴ 2m∠A = 180°
∴ m∠A = \(\frac{180^{\circ}}{2}\) = 90°
∴ m∠B = 90°
Now, m∠A = m∠C and m∠B = m∠D (∵ Opposite angles)
∴ m∠C = 90° and m∠D = 90°
Thus, ∠A = 90°, ∠B = 90°, ∠C = 90° and ∠D = 90°.

PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3

Question 7.
The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 14
Solution:
∠POA is the exterior angle of ΔHOE
∴ y + z = 70° (Exterior)
∴ m∠HOP = 180° – 70° = 110°
∠HEP = ∠HOP = 110° (Opposite angles of a parallelogram)
∴ x = 110°
∵ \(\overline{\mathrm{EH}} \| \overline{\mathrm{PO}}\), \(\overleftrightarrow{\mathrm{PH}}\) is a transversal.
∴ y = 40° (∵ Alternate angles are equal)
Now, y + z = 70°
∴ 40° + z = 70°
∴ z = 70° – 40° = 30°
Thus, x = 110°, y = 40° and z = 30°

Question 8.
The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)
(i)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 15
Solution :
□ GUNS is a parallelogram.
∴ GS = NU and SN = GU (Y Opposite sides)
∴ 3x = 18 and 26 = 3y – 1
∴ x = \(\frac {18}{3}\)
∴ x = 6

∴ 3y – 1 = 26
∴ 3y = 26 + 1
∴ 3y = 27
∴ y = \(\frac {27}{3}\) = 9
Thus, x = 6 cm and y = 9 cm

(ii)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 16
Solution:
□ RUNS is a parallelogram.
∴ Its diagonals bisect each other.
x + y = 16 ……(1)
and y + 7 = 20
y = 20 – 7 = 13 …..(2)
Substituting value of y in (1)
x + y = 16
x + 13 = 16
∴ x = 16 – 13
= 3
Thus, x = 3 cm and y = 13 cm

PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3

Question 9.
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 17
In the above figure both RISK and CLUE are parallelograms. Find the value of x.
Solution:
□ RISK is a parallelogram.
∴ m∠R + m∠K = 180°
(∵ Adjacent angles are supplementary)
∴ m∠R + 120° = 180°
∴ m∠R = 180° – 120°
∴ m∠R = 60°
∠R and ∠S are opposite angles of parallelogram.
∴ m∠S = 60°
□ CLUE is a parallelogram.
∴ m∠E = m∠L = 70° (∵ Opposite angles)
In ΔESQ
∴ m∠E + m∠S + x = 180° (∵ Angle sum property of triangle)
∴ 70° + 60° + x = 180°
∴ 130° + x = 180°
∴ x = 180° – 130°
∴ x = 50°
Thus, x = 50°

Question 10.
Explain how this figure is a trapezium. Which of its two sides are parallel ? (Fig3.32)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 18
Solution:
Since, 100° + 80° = 180°
i. e., ∠M and ∠L are supplementary.
∴ \(\overline{\mathrm{NM}} \| \overline{\mathrm{KL}}\)
(∵ Interior angles along on the same side of the transversal are supplementary)
One pair of opposite side of □ LMNK is parallel.
∴ □ LMNK is a trapezium.

PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3

Question 11.
Find m∠C in figure 3.33 if \(\overline{\mathbf{A B}} \| \overline{\mathbf{D C}}\).
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 19
Solution:
In □ ABCD, \(\overline{\mathrm{AB}} \| \overline{\mathrm{DC}}\) and BC is their transversal,
∴ m∠B + m∠C = 180°
(∵ Interior angles on the same side of transversal are supplementary)
∴ 120° + m∠C = 180°
∴ m∠C = 180° – 120° = 60°
Thus, m∠C = 60°

Question 12.
Find the measure of ∠P and ∠S if \(\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}\) in figure 3.34. (If you find m∠R, is there more than one method to find m∠P ?)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 20
Solution:
In □ PQRS \(\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}\)
∴ □ PQRS is a trapezium.
\(\overleftrightarrow{\mathrm{PQ}}\) is a transversal of \(\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}\)
m∠P + m∠Q = 180°
(∵ Interior angles on the same side of transversal are supplementary)
∴ mZP + 130° = 180°
∴ mZP = 180° – 130°
∴ mZP = 50°
In □ PQRS, ∠R = 90°
\(\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}\), \(\overleftrightarrow{\mathrm{RS}}\) is their transversal.
∴ m∠S + m∠R = 180°
∴ m∠S + 90° = 180°
∴ m∠S= 180°-90°
∴ m∠S = 90°
Yes, the sum of the measures of all angles of quadrilateral is 360°. ∠P and ∠R can be found.
m∠P + m∠Q + m∠R + m∠S = 360°
∴ m∠P + 130° + 90° + 90° = 360°
∴ m∠P + 310° = 360°
∴ m∠P = 360° – 310°
∴ m∠P = 50°