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Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 12 Perimeter and Area Ex 12.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 12 Perimeter and Area Ex 12.1

1. Find the perimeter of the following shapes:

Question (i)

Solution:
Perimeter = AB + BC + CD + DA
= 8 cm + 7 cm + 12 cm + 9 cm
= 36 cm

Question (ii)

Solution:
Perimeter = XY + YZ + ZX
= 10m + 10m + 8m
= 28m

Question (iii)

Solution:
Perimeter = PQ + QR + RS + SP
= 15 cm + 12 cm + 15 cm + 12 cm
= 54 cm

Question (iv)

Solution:
Perimeter = MN + NO + OP + PL + LM
= 8 cm + 7 cm + 5 cm + 6 cm + 7 cm
= 33 cm

Question (v)

Solution:
Perimeter = AB + BC + CD + DE + EF + FG + GM + MA
=8m + 2m + 6m + 4m + 6m + 2m + 8m + 8m
= 44 m

Question (vi)

Solution:
Perimeter = LM + MN + NO + OP + PQ + QR + RS + SL
= 3 cm + 3 cm + 3 cm + 3 cm + 3 cm + 3 cm + 3 cm + 3 cm
= 24 cm

2. Find the perimeter of the triangle with sides:

Question (i)
5 cm, 6 cm and 7 cm
Solution:
Perimeter of a triangle
= Sum of lengths of its sides
= 5 cm + 6 cm + 7 cm = 18 cm

Question (ii)
10 m, 12 m, 18 m
Solution:
Sides of triangle
= 10 m, 12 m, 18 m
∴ Perimeter of a triangle
= Sum of lengths of its sides
= 10 m + 12 m + 18 m = 40 m

Question (iii)
4.6 cm, 3.2 cm and 5.8 cm.
Solution:
Sides of triangle
= 4.6 cm, 3.2 cm and 5.8 cm
∴ Perimeter of triangle
= Sum of lengths of its sides
= 4.6 cm + 3.2 cm + 5.8 cm
= 13.6 cm

3. Find the perimeter of an isosceles triangle with 15 cm as length of equal side and 18 cm as base.
Solution:
Sides of isosceles triangle = 15 cm, 15 cm, 18 cm
Area of isosceles triangle = Sum of lengths of its sides
= (15 + 15 + 18) cm = 48 cm

4. Find the perimeter of a square with side:

Question (i)
16 cm
Solution:
Side of square = 16 cm
∴ Perimeter of square = 4 × side
= 4 × 16 cm
= 64 cm

Question (ii)
4.8 mm
Solution:
Side of square = 4.8 mm
∴ Perimeter of square = 4 × side
= 4 × 4.8 mm
= 19.2 mm

Question (iii)
125 cm
Solution:
Side of square = 125 cm
∴ Perimeter of square = 4 × side
= 4 × 125 cm
= 500 cm

Question (iv)
45 m
Solution:
Side of square = 45 m
∴ Perimeter of square = 4 × side
= 4 × 45 m
= 180 m

Question (v)
39 cm.
Solution:
Side of square = 39 cm
∴ Perimeter of square = 4 × side
= 4 × 39 cm
= 156 cm

5. Find the perimeter of a rectangle with:

Question (i)
Length 20 m and breadth 15 m
Solution:
Length of rectangle = 20 m
Breadth of rectangle = 15 m
∴ Perimeter of rectangle = 2 × (length + breadth)
= 2 × (20 + 15) m
= 2 × 35 m
= 70 m

Question (ii)
Length 25 m and breadth 35 m
Solution:
Length of rectangle = 25 m
Breadth of rectangle = 35 m
∴ Perimeter of rectangle = 2 × (length + breadth)
= 2 × (25 + 35) m
= 2 × 60 m
= 120 m

Question (iii)
Length 40 cm and breadth 28 cm
Solution:
Length of rectangle = 40 cm Breadth of rectangle = 28 cm
∴ Perimeter of rectangle = 2 × (length + breadth)
= 2 × (40 + 28) cm
= 2 × 68 cm
= 136 cm

Question (iv)
Length 18.3 cm and breadth 6.8 cm
Solution:
Length of rectangle = 18.3 cm
Breadth of rectangle = 6.8 cm
∴ Perimeter of rectangle
= 2 × (length + breadth)
= 2 × (18.3 + 6.8) cm
= 2 × 25.1 cm = 50.2 cm

Question (v)
Length 0.125 m and breadth 15 cm.
Solution:
Length of rectangle
= 0.125 m = 12.5 cm
Breadth of rectangle = 15 cm
∴ Perimeter of rectangle = 2 × (length + breadth)
= 2 × (12.5 + 15) cm
= 2 × 27.5 cm
= 55 cm

6. Find the perimeter of a regular hexagon with side:

Question (i)
5 cm
Solution:
Side of a regular hexagon = 5 cm
Perimeter of a regular hexagon = 6 × side
= 6 × 5 cm
= 30 cm

Question (ii)
12 cm
Solution:
Side of a regular hexagon = 12 cm
Perimeter of a regular hexagon = 6 × side
= 6 × 12 cm
= 72 cm

Question (iii)
7.2 cm.
Solution:
Side of a regular hexagon = 7.2 cm
Perimeter of a regular hexagon = 6 × side
= 6 × 7.2 cm
= 43.2 cm

7. Find the perimeter of an equilateral triangle with side:

Question (i)
10 cm
Solution:
Side of an equilateral triangle
= 10 cm
∴ Perimeter of an equilateral triangle = 3 × side
= 3 × 10 cm
= 30 cm

Question (ii)
8 m
Solution:
Side of an equilateral triangle = 8 m
∴ Perimeter of an equilateral triangle = 3 × side
= 3 × 8 m
= 24 m

Question (iii)
24 m
Solution:
Side of an equilateral triangle = 24 m
∴ Perimeter of an equilateral triangle = 3 × side
= 3 × 24 m
= 72 m

Question (iv)
5.6 m
Solution:
Side of an equilateral triangle = 5.6 m
∴ Perimeter of an equilateral triangle = 3 × side
= 3 × 5.6 m
= 16.8 m

Question (v)
12.1 cm.
Solution:
Side of an equilateral triangle = 12.1 cm
∴ Perimeter of an equilateral triangle = 3 × side
= 3 × 12.1 cm
= 36.3 cm

8. If the perimeter of a triangle is 48 cm and two sides are 12 cm and 17 cm. Find the third side.
Solution:
Perimeter of a triangle = 48 cm Sum of length of two sides = (12 + 17) cm = 29 cm
∴ Third side = 48 cm – 29 cm
= 19 cm

9. Find the side of an equilateral triangle, if the perimeter is:

Question (i)
45 cm
Solution:
Given perimeter of an equilateral triangle = 45 cm
Perimeter of an equilateral triangle = 3 × (side of the triangle)
⇒ 45 cm = 3 × side
⇒ Side of the triangle
= 15cm

Question (ii)
69 mm
Solution:
Given perimeter of an equilateral triangle = 69 mm
Perimeter of an equilateral triangle = 3 × (side of the triangle)
⇒ 69 = 3 × (side of the triangle)
⇒ Side of the triangle = \(\frac{69 \mathrm{~mm}}{3}\)
= 23 mm

Question (iii)
117 cm.
Solution:
Given perimeter of an equilateral triangle = 117 cm
Perimeter of an equilateral triangle = 3 × (side of the triangle)
⇒ 117 = 3 × (side of the triangle)
⇒ Side of the triangle = \(\frac{117 \mathrm{~cm}}{3}\)
= 39 cm

10. Find the side of a square if the perimeter is:

Question (i)
52 cm
Solution:
Given Perimeter of a square = 52 cm
Perimeter of a square = 4 × (side of square)
⇒ Side of square
= \(\frac{\text { perimeter of a square }}{\text { 4 }}\)
= \(\frac{52 \mathrm{~cm}}{4}\)
= 13 cm

Question (ii)
60 cm
Solution:
Given perimeter of a square = 60 cm
Side of a square
= \(\frac{\text { perimeter of a square }}{\text { 4 }}\)
= \(\frac{60 \mathrm{~cm}}{4}\)
= 15 cm

Question (iii)
112 cm.
Solution:
Given perimeter of a square = 112 cm
Side of a square
= \(\frac{\text { perimeter of a square }}{\text { 4 }}\)
= \(\frac{112 \mathrm{~cm}}{4}\)
= 28 cm

11.

Question (i)
The perimeter of rectangular field is 260 m. If its length is 80 m then find its breadth.
Solution:
Given perimeter of rectangular field = 260 m
and Length of the rectangular field = 80 m
∴ Perimeter of rectangular field = 2 × (length + breadth)
⇒ 260 = 2 × (80 + breadth)
⇒ \(\frac {260}{2}\) = 80 + breadth
⇒ 80 + breadth = 130
⇒ breadth = 130 – 80 = 50 m
Hence breadth of rectangular field = 50 m

Question (ii)
The perimeter of a rectangular garden is 140 m. If its breadth is 45 m then find its length.
Solution:
Given perimeter of rectangular garden = 140 m
and breadth of rectangular garden = 45 m
∴ Perimeter of rectangular garden = 2 × (length + breadth)
⇒ 140 = 2 × (length + 45)
⇒ \(\frac {260}{2}\) = length + 45
⇒ length = 70 – 45 = 25 m
Hence length of rectangular garden = 25 m

Question (iii)
The perimeter of a rectangle is 114 cm. If its length is 32 cm then find its breadth.
Solution:
Given perimeter of rectangle = 114 cm
and length of rectangle = 32 cm
∴ Perimeter of rectangle
= 2 × (length + breadth)
⇒ 114 = 2 × (32 + breadth)
⇒ \(\frac {114}{2}\) = 32 + breadth
⇒ breadth = 57 – 32 = 25 cm
Hence breadth of rectangle = 25 cm

12. The side of a triangular field are 15 m, 20 m and 18 m. Find the total distance travelled by a boy in taking 2 complete rounds of this field.
Solution:
Sides of a triangular fields = 15 m, 20 m and 18 m
Distance covered in one round of a triangular field = Perimeter of rectangular field = Sum of the length of the sides of a rectangular field
= 15 m + 20 m + 18 m
= 53 m
∴ Distance covered in taking 2 complete rounds of this field
= 2 × 53 m
= 106 m

13. Find the cost of fencing a square field of side 26 m at the rate of ₹ 3 per metre.
Solution:
Given side of the square field = 26 m
∴ Perimeter of the square field = 4 × side
= 4 × 26 m = 104 m
Perimeter of fencing = 104 m
Cost of 1 m of fencing = ₹ 3
Cost of 104 m of fencing
= 104 × ₹ 3
= ₹ 312

14. Mani runs around a square park of side 75 m. Kush runs around a rectangular park of length 60 m and breadth 45 m. Who covers less distance?
Solution:
Side of a square park = 75 m
Perimeter of square park = 4 × (side)
= 4 × (75 m) = 300 m
∴ Distance covered by Mani = 300 m
Length of rectangular park = 60 m
Breadth of rectangular park = 45 m
Perimeter of rectangular park = 2 (length + breadth)
= 2 × (60 + 45) m
= 2 × (105) m
= 210 m
∴ Distance covered by Kush = 210 m
Kush covers less distance.

15. Find the cost of framing a rectangular whiteboard with length 240 cm and breadth 150 cm at the rate of ₹ 6 per cm.
Solution:
Length of rectangular white board = 240 cm
Breadth of rectangular white board = 150 cm
Perimeter of rectangular white board = 2 × (length + breadth)
= 2 × (240 + 150) cm
= 2 × (390) cm
= 780 cm
Cost of fencing 1 cm = ₹ 6 × 780
= ₹ 4680

16. If length of a rectangle is ‘a’ units and breadth is 5 units. Find the perimeter of the rectangle.
Solution:
Given length of rectangle = ‘a’ units,
and Breadth of rectangle = 5 units
Perimeter of rectangle = 2 × (length + breadth)
= 2 × (a + 5) units
= 2 (a + 5) units

17. Fill in the blanks:

Question (i)
The sum of lengths of all sides of a polygon is called ……………. .
Solution:
perimeter

Question (ii)
Perimeter of Square = ……………. × side.
Solution:
4

Question (iii)
Perimeter of Rectangle = 2 × (………. +………) .
Solution:
length, breadth

Question (iv)
Side of a square = (……………) ÷ 4.
Solution:
perimeter

Question (v)
Perimeter of an equilateral triangle = …………….. × side.
Solution:
3.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 10 Circles Ex 10.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1

Question 1.
How many tangents can a circle have?
Solution:
Since at any point on a circle, there can be one and only one tangeni. But circle is a collection of infinite points, so we can draw infinite number of tangents to a circle.

Question 2.
Fill in the blanks:
(i) A tangent to a circle intersects it in ………………. point(s).
Solution:
one

(ii) A line intersecting a circle in two points is called a ………………..
Solution:
secant.

(iii) A circle can have ……………. parallel tangents at the most.
Solution:
two

(iv) The common point of a lingent h, circle and the circit is called ………………
Solution:
point of contact.

Question 3.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is:
(A) 12 cm
(B) 13 cm
(C) 8.5 cm
(D) \(\sqrt{119}\) cm.
Solution:
According to given information we draw the figure such that,

OP = 5 cm and OQ = 12 cm
∵ PQ is a tangent and OP is the radius
∵ ∠OPQ = 90°
Now, In right angled ∆OPQ.
By Pythagoras Theorem,
OQ² = OP² + QP²
Or (12)² = (5)² + QP²
Or QP² = (12)² – (5)²
Or QP² = 144 – 25 = 119
Or QP = \(\sqrt{119}\) cm.
Hence, option (D) is correct.

Question 4.
Draw a circle and two lines parallel to given line such that one is a tangent and other a secant to the circle.
Solution:
According to thc given information we draw a circle having O as centre and l is the given line.

Now, m and n be two lines parallel to a given line l such that m is tangent as well as parallel to l and n is secant to the circle as well as parallel to l.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 9 Some Applications of Trigonometry Ex 9.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1

Question 1.
A circus artist is climbing 220 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, ¡f the angle made by the rope with the ground level is 30° (see fig.).

Solution:
Let AB be the heignt of pole;
AC = 20 m be the length of rope.
The angle of elevation in this situation is 30°.
Various arrangements are as shown in figure.

In right angled ∆ABC,
\(\frac{\mathrm{AB}}{\mathrm{AC}}\) = sin 30°

or \(\frac{\mathrm{AB}}{20}=\frac{1}{2}\)

or AB = \(\frac{1}{2}\) × 20 = 10
Hence, height of pole is 10 m.

Question 2.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 300 with it. The distance between the foot of the tree to the point where the top touches the ground is 8 rn Find the height of the tree.
Solution:
Let BD be length of tree before storm.
After storm AD = AC = length of broken part of tree.
The angle of elevation in this situation is 30°.
Various arrangements are as shown in the figure.

In right angled ∆ABC,

\(\frac{\mathrm{AB}}{\mathrm{AC}}\) = tan 30°

or \(\frac{h_{1}}{8}=\frac{1}{\sqrt{3}}\)
or h₁ = \(\frac{8}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{8}{3} \sqrt{3}\) m ……….(1)

\(\frac{\mathrm{BC}}{\mathrm{AC}}\) = cos 30°

or \(\frac{8}{h_{2}}=\frac{\sqrt{3}}{2}\)

or \(h_{2}=\frac{8 \times 2}{\sqrt{3}}=\frac{16}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

h₂ = \(\frac{16}{3}\) √3 …………..(2)

Total height of the tree = h₁ + h₂
= \(\frac{8}{3}\) √3 + \(\frac{16}{3}\) √3 [Using (1) & (2)]

= \(\left(\frac{8+16}{3}\right) \sqrt{3}=\frac{24}{3} \sqrt{3}\) = 8√3 m.
Hence, height of the tree is 8√3 m.

Question 3.
A contractor plants to install two slides for the children to play in a park. For the children below the age of 5 years, she
prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Solution:
Case I:
For children below 5 years.
Let AC = l₁ m denote the length of slide and BC = 1.5 m be the height of slide. The angle of elevation is 30°.
Various arrangements are shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{BC}}{\mathrm{AC}}\) = sin 30°

or \(\frac{1 \cdot 5}{l_{1}}=\frac{1}{2}\)

or l₁ = 1.5 × 2 = 3 m.

Case II:
For Elder children
Let AC = 12 m represent the length of slide and BC = 3 m be the height of slide. The angle of elevation is 60°. Various arrangements are as shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{BC}}{\mathrm{AC}}\) = sin 60°

or \(\frac{3}{l_{2}}=\frac{\sqrt{3}}{2}\)

or l₂ = \(\frac{3 \times 2}{\sqrt{3}}=\frac{6}{\sqrt{3}}\)

= \(\frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{6 \sqrt{3}}{3}\)

= 2√3 m.

Hence, length of slides for children below 5 years and elder children are 3 m and 2 m.

Question 4.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
Solution:
Let BC = h m be the height of tower and AB = 30 m be the distance at ground level. Various arrangements are as shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{BC}}{\mathrm{AB}}\) = tan 30°

or \(\frac{h}{30}=\frac{1}{\sqrt{3}}\)

or h = \(\frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{30 \sqrt{3}}{3}\)

= 10√3 = 10 × 1.732
h = 17.32 (approx).
Hence, height of tower is 17.32 m.

Question 5.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Solution:
Let us suppose position of the kite is at point CAC = l m be length of string with which kite is attached. The angle of elevation for this situation be 60°. Various arrangements are as shown in the figure.

In right angled ∆ABC,

\(\frac{\mathrm{CB}}{\mathrm{AB}}\) = sin 60°

or \(\frac{60}{l}=\frac{\sqrt{3}}{2}\)

or l = \(\frac{60 \times 2}{\sqrt{3}}=\frac{120}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

= \(\frac{120 \sqrt{3}}{3}\) = 40√3 m.
Hence, length of the string be 40√3 m.

Question 6.
A 15 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution. Let ED = 30 m be the height of building and EC = l5 m be the height of boy.
The angle of elevation at different situation are 30° and 60° respectively.
Various arrangements are as shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{DC}}{\mathrm{AC}}\) = tan 30°

or \(\frac{28 \cdot 5}{x+y}=\frac{1}{\sqrt{3}}\)

or x + y = 28.5 × √3 m ………………(1)

Now, in right angled ∆BCD,

\(\frac{\mathrm{DC}}{\mathrm{BC}}\) = tan 60°

or \(\frac{28 \cdot 5}{y}=\sqrt{3}\)

or y = \(\frac{28 \cdot 5}{\sqrt{3}}\)

or y = \(\frac{28 \cdot 5}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{28 \cdot 5 \times \sqrt{3}}{3}\) ……….(2)

Distance covered towards building = x = (x + y) – y
= (28.5 × √3) – (\(\frac{28.5}{3}\) × √3) m [sing (1) and (2)]

= 28.5 (1 – \(\frac{1}{3}\)) √3 m

= 28.5 (\(\frac{3-1}{4}\)) √3 m

= [28.5 × \(\frac{2}{3}\)]√3 m = 19√3 m.

Hence, distance covered by boy towards the building is 19√3 m.

Question 7.
From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Solution:
Let BC = 20 m be the height of building and DC = h m be the height of transmission tower. The angle of elevation of
the bottom and top of a transmission tower are 45° and 60° respectively.
Various arrangements are as shown in the figure.

In right angled ∆ABC,
\(\frac{A B}{B C}\) = cot 45°

or \(\frac{\mathrm{AB}}{20}\) = 1
or AB = 20 m ………………..(1)
Also, in right angled ∆ABD,
\(\frac{A B}{B C}\) = cot 60°

or \(\frac{\mathrm{AB}}{20+h}=\frac{1}{\sqrt{3}}\)

AB = \(\frac{(20+h)}{\sqrt{3}}\) ………….(2)

From (1) and (2), we get

20 = \(\frac{(20+h)}{\sqrt{3}}\)
or 20√3 = 20 + h
or h = 20√3 – 20
or h = 20 (√3 – 1) m
= 20 (1.732 – 1) m
= 20 × 0.732 = 14.64 m.

Hence, height of the tower is 14.64 m.

Question 8.
A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Solution. Let BC = h m be the height of Pedestal and CD = 1.6 m be the height of statue.
The angle of elevation of top of statue and top of pedestal are 60° and 45° respectively. Various arrangements are as shown in the figure.

In right angled ∆ABC,
\(\frac{A B}{B C}\) = cot 45°

or \(\frac{A B}{h}\) = 1

or AB = h m ………….(1)

In right angled ∆ABC,
\(\frac{\mathrm{AB}}{\mathrm{BD}}\) = cot 60°

or \(\frac{\mathrm{AB}}{h+1.6}=\frac{1}{\sqrt{3}}\)

or AB = \(\frac{h+1.6}{\sqrt{3}}\) ……….(2)

From (1) and (2), we get
h = \(\frac{h+1.6}{\sqrt{3}}\)
or √3h = h + 1.6
or (√3 – 1) h = 1.6
or (1.732 – 1) h = 16
or (0.732) h = 1.6
or h = \(\frac{1.6}{0.732}\) = 2.1857923
= 2.20 m (approx.)
Hence, height of pedestal is 2.20 m.

Question 9.
The angle of elevation of the top of a building from the foot of the tower is 300 and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the
building.
Solution:
Let BC = 50 m be height of tower and AD = h m be height of building. The angle of elevation of the top of a building from the foot of tower and top of tower from foot of the building are 30° and 60° respectively. Various arrangement are as shown in figure.

In right angled ∆ABC,
\(\frac{A B}{B C}\) = cot 60°

or \(\frac{\mathrm{AB}}{50}=\frac{1}{\sqrt{3}}\)

or AB = \(\frac{50}{\sqrt{3}}\) …………(1)

Also, in right angled ∆DAB,
\(\frac{\mathrm{AB}}{\mathrm{DA}}\) = cot 30°

or \(\frac{A B}{h}\) = √3
or AB = h√3 ……………(2)

From (1) and (2), we get
\(\frac{50}{\sqrt{3}}\) = h√3

or \(\frac{50}{\sqrt{3}} \times \frac{1}{\sqrt{3}}\) = h

or h = \(\frac{50}{3}\) = 16.6666

or h = 16.70 m (approx).
Hence, height of building is 16.70 m.

Question 10.
Two poles of equal heights are tanding opposite each other on either side of he road, which is 80 m wide. From a point
between them on the road the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.
Solution:
Let BC = DE = h m he height of two equal poles and point A be the required position where the angle of elevations of top of two poles are 30° and 60° respectively. Various arrangement are as shown in the figure.

In right angled ∆ADE,

\(\frac{E D}{D A}\) = tan 30°

or \(\frac{h}{x}=\frac{1}{\sqrt{3}}\)

or h = \(\frac{x}{\sqrt{3}}\) ……………(1)

In right angled ∆ABC,

\(\frac{\mathrm{BC}}{\mathrm{AB}}\) = tan 60°

or \(\frac{h}{80-x}\) = √3

or h = (80 – x) √3 …………(2)

From (1) and (2), we get
\(\frac{x}{\sqrt{3}}\) = (80 – x)
or x = (80 – x) √3 × √3
or x = (80 – x) 3
or x = 240 – 3x
or 4x = 240
or x = \(\frac{240}{4}\) = 60
Substitute this value of x in (I), we get
h = \(\frac{60}{\sqrt{3}}=\frac{60^{\circ}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

= \(\frac{60 \sqrt{3}}{3}=20 \sqrt{3}\)

= (20 × 1.732) m = 34.64 m
DA = x = 60 m
and AB = 80 – x = (80 – 60) m = 20 m.
Hence, heigth of the poles are 3464 m and the distances of the point from the poles are 20 m and 60 m respectively.

Question 11.
A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From a point 20 m away from this point on the same bank, the angle of elevation of the top of the tower is 30° (see fig.). Find the height of the tower and the width of the canal.

Solution:
Let BC = x m be the width of canal and CD = h m be height of TV tower. The angles of elevation of top of tower at different position are 30° and 60° respectively. Various arrangements are as shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{AB}}{\mathrm{BC}}\) = tan 60°

or \(\frac{h}{x}\) = √3
or h = √3x …………..(1)

Also, in right angled ∆ABD,
\(\frac{\mathrm{AB}}{\mathrm{BD}}\) = tan 30°

or \(\frac{h}{20+x}=\frac{1}{\sqrt{3}}\)

or h = \(\frac{20+x}{\sqrt{3}}\) ……………….(2)

From (1) and (2), we get

√3x = \(\frac{20+x}{\sqrt{3}}\)
or √3(√3x) = 20 + x
or 3x = 20 + x
or 2x = 20
or x = \(\frac{20}{2}\) = 10

Substitute this value of x in (1), we get
h = 10(√3)
= 10 × 1.732
h = 17.32 m
Hence, height of TV tower is 17.32 m and. width of the canal is 10 m.

Question 12.
From the top of a 7m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Solution:
Let BD = hm be the height of cable tower and AE = 7 m be the height of building. The angle of elevation of the top of a cable tower and angle of depression of its foot from top of a building are 60° and 45° respectively.
Various arrangements are as shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{AB}}{\mathrm{AE}}\) = cot 45°

or \(\frac{\mathrm{AB}}{7}\) = 1

or AB = 7 m. ……………..(1)

Also, in right angled ∆DCE,

\(\) = cot 60°
or \(\frac{\mathrm{EC}}{h-7}=\frac{1}{\sqrt{3}}\)

or EC = \(\frac{h-7}{\sqrt{3}}\) ……………..(2)

But AB = EC ………….(Given)
7 = \(\frac{h-7}{\sqrt{3}}\) [Using (1) and (2)]
or 7√3 = h – 7
h = 7√3 + 7 = 7 (√3 + 1)
or h = 7 (1.732 + 1) = 7(2.732)
or h = 19.124
or h = 19.20 m (approx.)
Hence, height of the tower is 19.20 m.

Question 13.
As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Solution:

Let CD = 75 m be the height of light house and point D be top of light house from w’here angles of depression of two ships are 30° and 45° respectively. Various arrangements are as shown in the figure.

In right angled ∆BCD,
\(\frac{\mathrm{BC}}{\mathrm{CD}}\) = cot 45°

or \(\frac{y}{75}\) = 1
or y = 75 m ……………(1)

Also, in right angled ∆ACD
\(\frac{\mathrm{AC}}{\mathrm{CD}}\) = cot 30°

or \(\frac{x+y}{75}\) = √3
or x + y = 75√3
or x + 75 = 75√3 [using (1)]
or x = 75√3 – 75
= 75 (√3 – 1)
= 75( 1.732 – 1)
= 75 (0.732)
or x = 54.90
Hence, distance between the two ships is 54.90 m.

Question 14.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant ¡s 60°. After some time, the angle of elevation reduces to 30° (see fig.). Find the distance travelled by the balloon during the interval.

Solution:
Let ‘AB’ be the position of 1.2 m tall girl, at the point of the angles of elevation of balloon at
different distances are 30° and 60° respectively. Various arrangements are as shwon in th figure.
According to question,
FG = ED = CE – CD
= 88.2 m – 1.2 m
= 87 m

In right angled ∆AGF,
\(\frac{A G}{G F}\) = cot 60°

or \(\frac{x}{87}=\frac{1}{\sqrt{3}}\)

or x = \(\frac{87}{\sqrt{3}}\) m.

Also, in right angled ∆ADE,
\(\frac{A D}{E D}\) = cot 30°

or \(\frac{x+y}{87}\) = √3

or x + y = 87√3
or \(\frac{87}{\sqrt{3}}\) + y = 87√3
or y = 87√3 – \(\frac{87}{\sqrt{3}}\)

or y = 87√3 – \(\frac{1}{\sqrt{3}}\)

or y = 87 \(\frac{3-1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

or y = \(\frac{87 \times 2 \times \sqrt{3}}{3}\)

or y = 58√3
or y = 58(1.732) = 100.456
or y = 100.456 m.
Hence, distance travelled by the balloon during the interval is 100.46 m.

Question 15.
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the further t(me taken by the car to reach the foot of the tower.
Solution:
Let CD = h m. be the tower of height.
Let A be initial position of the car and after six seconds the car be at 13. The angles of depression at A and B are 30° and 60° respectively. Various arrangements are as shown in figure.

Let speed of the car be υ metre per second using formula, Distance = Speed x Time
AB = Distance covered by car in 6 seconds
AB = 6υ metre
Also, time taken by car to reach the tower be ‘n’ seconds.
∴ BC = nυ metre
In right angled ∆ACD.
\(\frac{\mathrm{CD}}{\mathrm{AC}}\) = tan 30°

or \(\frac{h}{6 v+n v}=\frac{1}{\sqrt{3}}\)

or h = \(\frac{6 v+n v}{\sqrt{3}}\) ……………….(1)

Also, in right angled ∆BCD,
\(\frac{C D}{B C}\) = tan 60°

or \(\frac{h}{n v}\) = √3
h = nv (√3) ……….(2)

From (1) and (2), we get
\(\frac{6 v+n v}{\sqrt{3}}\) = nυ(√3)
or 6υ + nυ = nυ(√3)
or 6υ + nυ = 3nυ
or 6υ = 2nυ
or n = \(\frac{6 v}{2 v}\) = 3
Hence, time taken by car to reach the foot of tower is 3 seconds.

Question 16.
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Solution:
Let CD = h m be the height of tower and B ; A be the required points which are at a distance of 4 m and 9 m from the tower respectively. Various arrangements are as shown in the figure.

In right angled ∆BCD
\(\frac{\mathrm{CD}}{\mathrm{BC}}\) = tan θ

or \(\frac{h}{4}\) = tan θ ………….(1)

Also, in right angled ∆ACD,
\(\frac{C D}{A C}\) = tan (90 – θ)

or \(\frac{h}{9}\) = cot θ

Multiplying (1) and (2), we get
\(\frac{h}{4} \times \frac{h}{9}\) = tan θ cot θ

or \(\frac{h^{2}}{36}=\tan \theta \times \frac{1}{\tan \theta}\)

or h² = 36 = (6)²
or h = 6
Hence, height of the tower is 6 m.

Punjab State Board PSEB 8th Class English Book Solutions English Grammar Tenses Exercise Questions and Answers, Notes.

PSEB 8th Class English Grammar Tenses

Tense शब्द लातीनी भाषा के शब्द tempus से बना है जिसका अर्थ है समय (Time) । अर्थात् Tense हमें किसी कार्य या घटना के समय या काल का ज्ञान कराता है। नीचे दिए गए वाक्य पढ़ें:

1. I play some game daily.
2. I played cricket yesterday.
3. I shall play hockey tomorrow.

पहले वाक्य के verb (play) से Present Time का बोध होता है। दूसरे वाक्य के verb (played) से Past. Time का बोध होता है। तीसरे वाक्य के verb (shall play) से Future Time का बोध होता है। इस तरह-
(1) जिस verb से Present Time का बोध होता है, वह Present Tense कहलाता है-
I take bath in the morning.
I go for a walk in the evening.

(2) जिस verb से Past Time का बोध होता है, वह Past Tense कहलाता है
I took bath in the,morning.
I went for a walk in the evening.

(3) जिस verb से Future Time का बोध होता है, वह Future Tense कहलाता है
I shall take bath in the morning.
I shall go for a walk in the evening.
इस प्रकार तीन मुख्य Tenses हैं-

1. The Present Tense
2. The Past Tense
3. The Future Tense.

अब यह जान लेना भी आवश्यक है कि प्रत्येक काल में कार्य अथवा क्रिया (verb) की स्थिति भिन्न-भिन्न होती है। हो सकता है कि काम चल रहा हो। यह भी सम्भव है कि कार्य पूरा हो चुका हो अथवा किसी अनिश्चित (indefinite) स्थिति में हो। इस तरह कार्य की मुख्य रूप से चार अवस्थाएं होती हैं। हम यूं भी कह सकते हैं कि प्रत्येक मुख्य Tense के चार रूप होते हैं और कुल मिला कर 12 Tenses होते हैं। यहां हम इन सभी Tenses का अलग-अलग अध्ययन करेंगे।

1. Simple Present Tense
OR
Present Indefinite Tense
Present Indefinite or Simple Present Tense का प्रयोग होता है-
(i) किसी आदत का वर्णन करने के लिए; जैसे,
He takes bath daily.
He goes for a walk everyday.

(ii) किसी सर्वमान्य सत्य को व्यक्त करने के लिए; जैसे,
The sun rises in the east.

(iii) किसी कहानी में किसी बीती हुई घटना को बताने में; जैसे,
The brave dog now kills the snake and waits for his master to come.

(iv) भविष्य में होने वाली किसी घटना को व्यक्त करने के लिए जो किसी योजना अथवा व्यवस्था का भाग हो; जैसे,
Our examination begins on Monday.

(v) Time तथा Condition की Clauses में साधारण Future Tense के स्थान पर; जैसे,
If it rains, we shall not go out for a walk.

(vi) कथनों को व्यक्त करने के लिए ; जैसे,
They say, “To err is human.

(vii) किसी ऐसी स्थिति को दर्शाने के लिए जो न बदलने वाली हो; जैसे,
Our house faces the east.

2. Present Continuous Tense
Present Continuous Tense का प्रयोग होता है-
(i) किसी ऐसे विशेष कार्य को प्रकट करने के लिए जो अभी पूरा न हुआ हो अथवा जारी हो; जैसे,
My brother is singing a song.

3. Present Perfect Tense
Present Perfect Tense का प्रयोग किया जाता है-
(i) ऐसे कार्य को प्रकट करने के लिए जो भूतकाल से अब तक जारी हो; जैसे,
I have never tasted tea. (I still do not drink it.)

(ii) किसी ऐसे पूर्ण कार्य अथवा घटना को प्रकट करने के लिए जो भूतकाल में विशेष समय को लेकर वर्तमान काल तक किया जाए; जैसे,
There have been two accidents on the road during 1984.

4. Present Perfect Continuous Tense.
यह Tense उस कार्य को व्यक्त करने के लिए प्रयोग होता है जो अतीत में किसी समय आरम्भ हुआ हो और अब भी चल रहा हो। समय को व्यक्त करने के लिए for (अनिश्चित समय) और since (निश्चित समय) का प्रयोग करते हैं; जैसे-
The match has been going on for an hour.
The man has been waiting for a reply.
You have been wasting your time since morning.

5. Past Indefinite Tense
OR
Simple Past Tense
Past Indefinite Tense का प्रयोग किया जाता है-
(i) भूतकाल की किसी आदत, लोकप्रिय कार्य अथवा सर्वमान्य तथ्य को प्रकट करने के लिए; जैसे,
People then believed that the sun moved round the earth.

(ii) ऐसे कार्य को व्यक्त करने के लिए जिस में भूतकाल में काफी समय लगा हो परन्तु जो अब समाप्त हो चुका हो; जैसे-
He lived in Delhi for ten years. (but he does not live there now)

(iii) भूतकाल में पूरे किए गए किसी कार्य को व्यक्त करने के लिए। इस प्रकार प्रायः बीते हुए समय को व्यक्त . करने के लिए किसी Adverb या Adverb phrase का प्रयोग किया जाता है: जैसे,
My father left for Delhi yesterday.

Note 1. कभी-कभी समय की अभिव्यक्ति Adverb की बजाये भाव (समय का) से भी हो सकती है; जैसे
I bought this watch in Mumbai.

Note 2. क्रमबद्ध घटनाओं में भी Time के Adverb की आवश्यकता नहीं पड़ती; जैसे,
He came. He saw. He conquered.

(iv) किसी प्रश्न का उत्तर देने में; जैसे,
How did he go to school ?
Answer:
He went on foot.

6. Past Continuous Tense
Past Continuous (Progressive) Tense का प्रयोग किया जाता है-
(i) किसी ऐसे कार्य को व्यक्त करने के लिए जो भूतकाल में किसी विशेष समय पर किया जा रहा हो, भले। ही कार्य करने का समय बताया गया हो, या न बताया गया हो; जैसे,-
At 7 a.m. this morning I was reading the newspaper.

7. Past Perfect Tense
Past Perfect Tense का प्रयोग किया जाता है-
(1) भूतकाल में एक साथ घटित होने वाले कार्यों में से पहले पूरा होने वाले कार्य के लिए; जैसे,
The patient had died before the doctor came.

8. Past Perfect Continuous Tense इस Tense का प्रयोग किसी ऐसे कार्य या घटना का वर्णन करने के लिए किया जाता है जो अतीत में किसी एक निश्चित समय (Point of Time) तक या किसी अवधि (Period of Time) में जारी रहे; जैसे,-
The phone had been ringing for a minute before Ram lifted it. वाक्य में स्पष्ट है कि जब राम ने फोन उठाया तब उससे पहले एक मिनट तक फोन की घण्टी बजती रही थी।

9. Future Indefinite Tense Simple Future Tense का प्रयोग ऐसे कार्य के लिए किया जाता है जो भविष्य में अभी किया जाना है; जैसे,
I shall finish the work tomorrow.
Tomorrow will be Monday.

10. Future Continuous Tense:
Future Continuous Tense का प्रयोग उस कार्य को करने के लिए किया जाता है जो भविष्य में किसी समय चल रहा हो; जैसे,
I shall be writing the letters then.
When I reach the station, the train will be moving.

11. Future Perfect Tense

• They will have heard the news by the time you reach.
• The teacher will have taken the roll-call before you enter the class.

12. Future Perfect Continuous Tense Future Perfect Continuous Tense का प्रयोग ऐसे कार्य का उल्लेख करने के लिए होता है जो भविष्य में किसी निश्चित समय के बाद भी जारी रहने का बोध कराता है; जैसे-

• The students will have been studying History since morning.
• He will have been studying Law for two years by next April.

Exercises From Board’s Grammar (Solved)

I. Fill in the blanks with the Simple Present or Present Continuous forms of the verbs given in the brackets:

1. The population of India ……………. very fast. (increase)
2. Water ……………. at 0° Celsius. (freeze)
3. The sun ……………. in the West. (not rise)
4. ……………. you ……………. Mr. Jain? (know)
5. ……………. he ever ……………… cricket? (play)
6. The Ganges …………… into the Bay of Bengal. (flow)
7. Why …………….. you ……………… this ? (eat)
8. She …………….. a bath. (have)
9. I …………… cricket everyday, but today I …………… tennis.. (play)
10. She usually ……………. a skirt but today she ……………… trousers. (wear)
Hints:
1. is increasing
2. freezes
3. does not rise
4. Do, know
5. Does, play
6. flows
7. are, eating
8. is having
9. play, am playing
10. wears, is wearing.

II. Fill in the blanks with the Present Perfect or Present Perfect Continuous forms of the verbs given in brackets:

1. Someone ……………. the window. (break)
2. Rita …………….. her pen. (lose)
3. The train ……………… just ………….. at the platform. (arrive)
4. We …………….. many medals. (win)
5. I ………. for a house for two months. (search)
6. …………. he ……………. a beard ? (grow)
7. ……………. you ……………. the Bible ? (read)
8. ……….. my uncle for months. (not visit)
9. She ……………… to China twice. (be)
10. We ……………. already ……………… our breakfast. (have)
Hints:
1. has broken
2. has lost
3. has, arrived
4. have won
5. have been searching
6. Has, grown
7. Have, read
8. have not visited
9. has been
10. have, had.

III. Fill in the blanks with the Simple Past Tense forms of the verbs given in the brackets:

Sher Singh smiled. He tossed his revolver in the air and ………… (catch) it by the handle. He ………… (take) careful aim at an empty sardine tin and …….3….. (fire) another six shots. The bullets ……4….. (go) through into the earth kicking up whiffs of dust. His Alsatian dog …….5….. (begin) to bark with excitement. He ………… (leap) up with a growl and ………… (run) down the canal embankment. He …….8….. (sniff) at the tin and …….9….. (take) it up in his mouth and …….10….. (run) back with it and ……. 11…. (lay) it at his master’s feet.
Hints:
1. caught
2. took
3. fired
4. went
5. began
6. leapt
7. ran
8. sniffed
9. took
10. ran
11. laid.

IV. Fill in the blanks with the Simple Past or Past Perfect forms of the verbs given in the brackets:

1. The plane ……………. When we reached the airport. (leave)
2. Ramesh …………… home when I phoned him. (return)
3. …………… he ……………. his old car before he bought a new one ? (sell)
4. The children ……………. before I came home. (sleep)
5. The film had already begun when we …………….. the theatre. (reach)
6. The teacher ……………. the book before the examination began. (finish)
7. The robber had run away before the police. (come)
8. Tom …………. sleepy after having a good lunch. (feel)
9. I ……………. the message before you came. (receive)
10. He ……………… for India last year. (play)
Hints:
1. had left
2. had returned
3. Had, sold
4. had slept
5. reached
6. had finished
7. came
8. felt
9. had received
10. played.

V. Correct the following sentences:

1. The rain has stopped yesterday.
2. He had been born in 1950.
3. He is suffering from fever since last night.
4. Stephenson has invented the steam engine.
5. He will reach home before the storm will come.
6. I left Bihar before the earthquake occurred.
7. She will reach the station before the train will go.
8. The great reformer had died in 1977.
9. I waited at home for her since 9 o’clock.
10. She finished her dinner when I saw her.
Hints:
1. The rain stopped yesterday.
2. He was born in 1950.
3. He has been suffering from fever since last night.
4. Stephenson invented the steam engine.
5. He will have reached home before the storm comes.
6. I had left Bihar before the earthquake occurred.
7. She will have reached the station before the train goes.
8. The great reformer died in 1977.
9. I had been waiting at home for her since 9 o’clock.
10. She had finished her dinner when I saw her.

Errors in the Use of Tenses

The Simple Past is often used wrongly for the Present Perfect Tense; as,

Incorrect : He did not write the letter yet.
Correct : He has not written the letter yet.
Incorrect : We did not hear from him for a week.
Correct : We have not heard from him for a week.
Incorrect : I lived in Ambala since 1990.
Correct : I have lived in Ambala since 1990.

The Present Perfect is often used wrongly for the Simple Past; as,

Incorrect – Columbus has discovered America.
Correct – Columbus discovered America.
Incorrect – Babar has won the First Battle of Panipat.
Correct – Babar won the First Battle of Panipat.
Incorrect – The servant has not answered when called.
Correct – The servant did not answer when called.

The Present Perfect Tense में Past Time को व्यक्त करने वाला adverb या कोई अन्य शब्द प्रयोग नहीं किया जा सकता जैसे-

Incorrect – I have made a call to him yesterday.
Correct – I made a call to him yesterday.
Incorrect – A new bookshop has been opened last Monday.
Correct – A new bookshop was opened last Monday.
Incorrect – I have finished my work last night.
Correct – I finished my work last night.

The Past Perfect is often used wrongly for the Simple Past; as,

Incorrect – I had visited her yesterday.
Correct – I visited her yesterday.
Incorrect – He had gone to Kolkata last year.
Correct – He went to Kolkata last year.
Incorrect – We had gone for a picnic last Sunday.
Correct – We went for a picnic last Sunday.
Incorrect – Nehru had died in 1964.
Correct – Nehru died in 1964.

The Simple Past is often used wrongly for the Past Perfect; as,

Incorrect – The patient died before the doctor came.
Correct – The patient had died before the doctor came.
Incorrect – The train left before we bought the tickets.
Correct – The train had left before we bought the tickets.
Incorrect – I finished my work before my father came.
Correct – I had finished my work before my father came.

The Past Perfect or Perfect Continuous, and not the Simple Past or Past Continuous, is used to express something that continued up to a past time after beginning at a still earlier time; as,

Incorrect – He told me that he was ill for four days.
Correct – He told me that he had been ill for four days.
Incorrect – She was writing a novel for six weeks when I visited her.
Correct – She had been writing a novel for six weeks when I visited her.

The Simple Future is often used wrongly for the Future Perfect; as;

Incorrect – We shall reach home before the sun will set.
Correct – We shall have reached home before the sun sets.
Incorrect – I shall leave for Ludhiana by the time he will come.
Correct – I shall have left for Ludhiana by the time he comes.

Punjab State Board PSEB 8th Class English Book Solutions English Vocabulary Idioms and Phrases Exercise Questions and Answers, Notes.

PSEB 8th Class English Vocabulary Idioms and Phrases

1. Act up to (के अनुसार काम करना)- He always acted up to his promise.
2. Bear away (जितना)- She won the race and bore away the prize.
3. Break in (सिधाना)- The trainee is breaking in the horse.
4. Break into (सेंध लगाना)- Last night three thieves broke into his house.
5. Break off (सम्बन्ध विछेद करना)- He has broken off with his religion.
6. Break out (भड़काना फूटना भड़क उठना)- The Second World War broke out in 1939. Cholera has broken out in London.
7. Break up (समाप्त होना)- The meeting broke up at 5 p.m.
8. Bring up (पालना)- He was brought up by his uncle.

9. Bring round (होरा में लाना)- The doctor soon brought the patient round.
10. Bring to book (दण्ड देना)- The police brought the pick-pocket to book.
11. Call on (किसी व्यक्ति से भेंट करना)- I called on him last evening.
12. Call in (बुला भेजना)- Call in the doctor at once.
13. Carry on (जारी रखना)- Carry on with your work.
14. Carry out (आज्ञा पालन करना)- We should carry out the orders of our teacher.
15. Come of (सम्बन्ध रखना)- He comes of a noble family.
16. Come off (होना)- My brother’s marriage comes off next Monday.
17. Do away with (मार देना समाप्त करना)- The woman did away with her step-son. We should do away with old customs.
18. Draw near (समीप आना)- My examination is drawing near.
19. Fall back upon (आश्रय होना)- He has nothing to fall back upon in old age.
20. (a) Fall off (झड़ जाना)- Tree leaves fall off in autumn.
(b) Fall upon (झपटना)- The lion fell upon the mouse.
21. Fall out (झगड़ना)- He always falls out over trifles.
22. Get into (पूरी तरह फिट आना)- I can’t get into this shirt. It is very tight.
23. Get on (चल निकलना)- How are you getting on in your school ?
24. Get through (उत्तीर्ण होना)- He got through the examination.
25. Get rid of (छुटकारा पाना)- I want to get rid of my servant.
26. Give away (बांटना)- The president gave away the prizes.
27. Give in (पराजित होना)- The soldiers fought bravely but gave in at last.
28. Give up (छोड़ना)- Give up your bad habits.
29. Go through (पढ लोना)- She has gone through the book.
30. Keep away (अनुपस्थित रहना)- Do not keep away from the school.
31. Keep on (जारी रखना)- He kept on laughing.
32. Keep one’s word (वचन पूरा करना)- He always keeps his word.
33. Lay by (बचाना)- Always lay by something for a rainy day.
34. Look after (देखभाल करना)- The mother looks after the children.
35. Look for (खोजना)- She is looking for her missing ring.
36. Look down upon (घृणा करना)- Don’t look down upon the poor.
37. Look upon (समझना)- I always look upon her as my sister.
38. Look into (जाँच करना)- The police is looking into the matter.

39. Make away with (मार देना)- The robbers made away with the rich men.
40. Make out (समझना)- I cannot make out the meaning of this sentence.
41. Make up (कमी पूरी करना)- Try to make up your deficiency in English.
42. Pass away (स्वर्गवास होना)- His father passed away last night.
43. Put off (स्थगित करना)-The meeting was put off to some later date.
44. Put out (बुझाना)- Put out the fire.
45. Put up with (सहन करना)- I cannot put up with this insult.
46. Run down (स्वास्थ्य गिरना)- He has run down on account of overwork.
47. Run over (कुचला जाना)- The child was run over by a car.
48. See off (विदा करना)- I went to the railway station to see my mother off.
49. Send for (बुला भेजना)- His father sent for the doctor.
50. Set off (चल पड़ना)- He set off for Delhi in no time.
51. Set out (चल पड़ना) – His father set out on a long journey.
52. Stand by (साथ देना)- Always stand by your friends in need.
53. Take after (राक्ल मिलना)- The child takes after his mother.
54. Take for (भूल से किसी को और कोई समझना)- I took the rope for a snake.
55. Turn down (रह करना)- He turned down my request.
56. At first sight (पहली नज़र में)- They fell in love at first sight.
57. At sea (कोरा)- He is at sea in English.
58. At home in (निपुण)- He is at home in Mathematics.
59. In course of (समय पाकर)- You will know everything in course of time.
60. In order to (क़े लिए)- He went there in order to see his brother.
61. In spite of (के बावजूद)- In spite of his hard work, he failed in the examination.
62. In search of (की खोज में)- He wandered here and there in search of water.
63. At arm’s length (परे)- Keep that man at arm’s length.
64. To and fro (इधर-उधर)-The people were moving to and fro on the platform.
65. At any rate (चाहे कुछ भी हो)- You must finish your work today at any rate.
66. To the backbone (पूर्णतया)- He is a patriot to the backbone.
67. To live from hand to mouth (जो कमाया सो खा लेना) He lives from hand to mouth.
68. To make up one’s mind (निशचय करना)- I have made up my mind to work hard.
69. To make good the loss (क्षतिपूर्ति करना)- He is trying to make good the loss he has suffered.
70. All in all (कर्ता धर्ता)- The principal is all in all in the school.
71. Fair and square (ईमानदार)- I am fair and square in my dealings with everybody.
72. Ups and downs (उतार चढ़ाव)- The old man has seen many ups and downs in his life.
73. In full swing (पूरे योवन पर)- The fair was in full swing.

74. Out of pocket (धन की कमी धन हानि खर्च किया हुआ धन)- I can’t go out; I am out of pocket right now.
75. At sixes and sevens (बिखरी हुई)- Everything in the house was at sixes and sevens.
76. Part and parcel (आवश्यक अंग)- Air is a part and parcel of our life.
77. Every inch (पूर्णतया)- He is every inch a patriot.
78. Hard up (हाथ तंग होना)- I am hard up these days.
79. In a fix (दुविधा में)- I am in a fix to know what to do.

Punjab State Board PSEB 8th Class English Book Solutions English Grammar Voice (Active and Passive Voice) Exercise Questions and Answers, Notes.

PSEB 8th Class English Grammar Voice (Active and Passive Voice)

अंग्रेज़ी भाषा में Voice (वाच्य) दो प्रकार की होती है-
1. Active Voice
2. Passive Voice

1. Active Voice : जब किसी वाक्य में Subject (कर्ता) कार्य करता है तो वाक्य का Verb, Active Voice में होता है।
2. Passive Voice : जब किसी वाक्य में Subject (कर्ता) कार्य नहीं करता है बल्कि उस पर कार्य किया जाता है तो वाक्य का Verb, Passive Voice में होता है।
Example : The hunter killed the lion. शिकारी ने शेर को मारा।
The lion was killed by the hunter. शेर शिकारी द्वारा मारा गया।

नोट-विद्यार्थियों को Verb की तीसरी फ़ार्म लगाना कभी नहीं भूलना चाहिए। Let, hit, cut आदि शब्दों की फ़ार्म लगाते समय विशेष सावधानी बरतनी चाहिए।
Imperative तथा To + 1st form वाले वाक्य में Be का प्रयोग होता है।

“Active से Passive बनाने की विधि

1. Verb ‘Be’ के सैंप

2. Verb की तीसरी फ़ार्म
Note-पुस्तक में Verb की forms दी गई हैं। विद्यार्थियों को चाहिए कि वे इन्हें अच्छी तरह याद कर लें।

3. Subject और Object का आपसी परिवर्तन

(B) नोट : “You’ और ‘It’ subject और object दोनों रूपों में नहीं बदलते।
2. Possessive Case of Pronoun में भी कोई परिवर्तन नही होता जैसे-
1. My brother = by my brother
2. Our teacher = by our teacher

Be के रूप का प्रयोग

की फ़र्म रूप का चुनाव करने से पूर्व हमें दिया गए वाक्य में verb को देखना चाहिए verb की फ़र्म या tense के आधार पर ही हम be की फ़र्म लागयोंगे (i) यदि दिए गए वाक्य में verb की पहली फ़र्म (present Indefinite tense) हो तो Passive Voice बनाने के लिए Be की पहली फ़ार्म अर्थात् is, am, are में से किसी एक का प्रयोग किया जाएगा; जैसे,-

(1) Auxiliary Verbs
Rule. Can, could, may, might, shall, should आदि वाले वाक्यों को passive voice में बदलने के लिए can, could, may, might, shall, should आदि के पश्चात् ‘be’ तथा क्रिया (Verb) की तीसरी फ़ार्म लगाई जाती है।

Active Voice	Passive Voice
1. I can solve this sum.	This sum can be solved by me.
2. You may take this pen.	This pen may be taken by you.
3. She may catch the train.	The train may be caught by her.
4. They might miss the bus.	The bus might be missed by them.
5. You should obey the rules.	The rules should be obeyed by you.
6. They must take exercise.	Exercise must be taken by them.
7. I cannot do it.	It cannot be done by me.
8. He might not win the race.	The race might not be won by him.
9. We must not tell a lie.	A lie must not be told by us.
10. One should do one’s duty.	Duty should be done.

(2) Verbs having Prepositions
Note. कुछ क्रियाओं के बाद at, on, in आदि Prepositions दी होती हैं। Passive Voice बदलते समय Prepositions of का स्थान नहीं बदलता है। उन्हें verbs के साथ ही रखा जाना चाहिए।

Active Voice	Passive Voice
1. He knocked at the door.	The door was knocked at by him.
2. He does not care for you.	You are not cared for by him.
3. They laughed at the poor man.	The poor man was laughed at by them.
4. She sent for me.	I was sent for by her.
5. The dog barked at them.	They were barked at by the dog.
6. I was waiting for him.	He was being waited for by me.
7. She did not listen to me.	I was not listened to by her.
8. They aimed at the lion.	The lion was aimed at by them.

(3) Prepositions other than ‘By Rule. कुछ वाक्यों में क्रिया (Verb) की तीसरी फ़ार्म के पश्चात् by’ के स्थान पर कोई अन्य Preposition लगती है जैसे to, at, in, with आदि।

Active Voice	Passive Voice
1. He knows me.	I am known to him.
2. The news shocked him.	He was shocked at the news.
3. His work satisfied me.	I was satisfied with his work.
4. Your habits worry me.	I am worried at your habits.
5. This jug contains milk.	Milk is contained in this jug.
6. You cannot please him.	He cannot be pleased with you.
7. Your answer does not satisfy me.	I am not satisfied with your answer.
8. The news surprised me.	I was surprised at the news.

(4) It is time से आरम्भ होने वाले वाक्य

Active Voice	Passive Voice
1. It is time to open the shop.	It is time for the shop to be opened.
2. It is time to write our letters.	It is time for our letters to be written.
3. It is time to read the newspaper.	It is time for the newspaper to be read.
4. It is time to take tea.	It is time for tea to be taken.
5. It is time to close the shop.	It is time for the shop to be closed.
6. It is time to pray to God.	It is time for God to be prayed to.
7. It is time to ring the bell.	It is time for the bell to be rung.

(5) Interrogative Sentences

Active Voice	Passive Voice
1. Does she tell a story?	Is a story told by her ?
2. Do you play football ?	Is football played by you ?
3. Did you read the letter ?	Passive Voice Was the letter read by you ?
4. Will you help me ?	Shall I be helped by you ?
5. Is she singing a song ?	Is a song being sung by her ?
6. Have you finished your work ?	Has your work been finished by you?
7. When does he take tea ?	When is tea taken by him ?
8. What is he doing ?	What is being done by him ?
9. Why were you making a noise ?	Why was a noise being made by you ?
10. Who teaches you English ?	By whom are you taught English ?

(6) Imperative Sentences

Active Voice	Passive Voice
1. Shut the windows.	Let the windows be shut.
2. Post this letter.	Let this letter be posted.
3. Show me your book.	Let your book be shown to me.
4. Change your clothes.	You are advised to change your clothes. Or Let your clothes be changed.
5. Write it clearly.	Let it be written clearly.
6. Always speak the truth.	You are advised always to speak the truth.
7. Work hard.	You are advised to work hard.
8. Never tell a lie.	You are advised never to tell a lie.
9. Let me finish it.	Let it be finished by me.
10. Tell him to keep quiet.	Let him be told to keep quiet.
11. Do not run fast.	You are advised not to run fast. Or You are forbidden to run fast.
12. Do not steal things.	You are advised not to steal things. Or Let things be not stolen.

(7) Typical Sentences

Active Voice	Passive Voice
1. The rose smells sweet.	The rose is sweet when (it is) smelt.
2. Sit down.	Be seated.
3. You have to do it.	It has to be done by you.
4. They say that honesty is the best policy.	It is said that honesty is the best policy.
5. God helps those who help themselves.	Those who help themselves are helped by God.
6. We elected him President.	He was elected President by us.
7. One should do one’s duty.	Duty should be done.
8. Someone has picked his pocket.	His pocket has been picked.
9. He has to pay the fine.	The fine has to be paid by him.
10. I hope to stand first.	It is hoped that I shall stand first.

Exercises (Solved) Change the voice:

I. 1. Harish plays cricket.
2. She likes singing.
3. We fly kites.
4. Meera helps the poor.
5. I open an account in the bank.
6. Hamid does his homework.
7. The boys watch television.
8. The cobbler mends my shoes.
9. She hates liars.
10. Children like sweets.
Answer:
1. Cricket is played by Harish.
2. Singing is liked by her.
3. Kites are flown by us.
4. The poor are helped by Meera.
5. An account is opened in the bank by me.
6. His homework is done by Hamid.
7. Television is watched by the boys.
8. My shoes are mended by the cobbler.
9. Liars are hated by her.
10. Sweets are liked by children.

II. 1. Rama lost his book.
2. They welcomed me.
3. Mina wrote a letter.
4. Mohan did not sing a song.
5. Radha did not drink coffee.
6. Harish did not paint a picture.
7. Did the boys fly kites ?
8. Did you close the door ?
9. Did Kavita help you?
10. The Prime Minister honoured Kapil Dev.
Answer:
1. His book was lost by Rama.
2. I was welcomed by them.
3. A letter was written by Mina.
4. A song was not sung by Mohan.
5. Coffee was not drunk by Radha.
6. A picture was not painted by Harish.
7. Were kites flown by the boys ?
8. Was the door closed by you ?
9. Were you helped by Kavita ?
10. Kapil Dev was honoured by the Prime Minister.

III. 1. Manohar will solve the sums.
2. You will miss your bus.
3. The teacher will punish the boys.
4. The doctor will examine the patient.
5. The children will enjoy this game.
6. You will join the party.
7. He will not cook the food.
8. She will not wash the clothes.
9. Will they elect the President ?
10. Will she speak the truth?
Answer:
1. The sums will be solved by Manohar.
2. Your bus will be missed by you.
3. The boys will be punished by the teacher.
4. The patient will be examined by the doctor.
5. This game will be enjoyed by the children.
6. The party will be joined by you.
7. The food will not be cooked by him.
8. The clothes will not be washed by her.
9. Will the President be elected by them?
10. Will the truth be spoken by her?

IV. 1. The leader is making a speech.
2. They are playing hockey.
3. Șudhir is telling a story.
4. Pakistan is making an atom bomb.
5. The girls are making chairs.
6. The workers are not repairing the road.
7. We are not taking exercise.
8. I am not favouring you.
9. Are they knocking at the door?
10. Is he serving his country?
Answer:
1. A speech is being made by the leader.
2. Hockey is being played by them.
3. A story is being told by Sudhir.
4. An atom bomb is being made by Pakistan.
5. Chairs are being made by the girls.
6. The road is not being repaired by the workers.
7. Exercise is not being taken by us.
8. You are not being favoured by me.
9. Is the door being knocked at by them?
10. Is his country being served by him ?

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 2 Linear Equations in One Variable Ex 2.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3

Solve the following equations and check your results :

Question 1.
3x = 2x + 18
Solution:
3x = 2x + 18
∴ 3x – 2x = 18 (Transposing 2x to LHS)
∴ x = 18

Check:
LHS = 3x = 3 × 18 = 54
RHS = 2x + 18
= 2(18) + 18
= 36 + 18 = 54
LHS = RHS
Thus, the answer is correct.

Question 2.
5t – 3 = 3t – 5
Solution:
5t – 3 = 3t – 5
∴ 5t – 3t – 3 = – 5 (Transposing 3t to LHS)
∴ 2t – 3 = -5
∴ 2t = – 5 + 3 (Transposing -3 to RHS)
∴ 2t = – 2
∴ \(\frac{2 t}{2}=\frac{-2}{2}\) (Dividing both the sides by 2)
∴ t = – 1

Check:
LHS = 5t – 3
= 5 (- 1) – 3
= – 5 – 3 = -8
RHS = 3t – 5
= 3 (- 1) – 5
= – 3 – 5 = – 8
∴ LHS = RHS
Thus, the answer is correct.

Question 3.
5x + 9 = 5 + 3x
Solution:
5x + 9 = 5 + 3x
∴ 5x + 9 – 3x = 5 (Transposing 3x to LHS)
∴ 2x + 9 = 5
∴ 2x = 5 – 9 (Transposing 9 to RHS)
∴ 2x = -4
∴ \(\frac{2 x}{2}=\frac{-4}{2}\) (Dividing both the sides by 2)
x = – 2

Check:
LHS = 5x + 9
= 5 (-2) + 9
= – 10 + 9 = -1
RHS = 5 + 3x
= 5 + 3 (-2)
= 5 – 6 = -1
∴ LHS = RHS
Thus, the answer is correct.

Question 4.
4z + 3 = 6 + 2z
Solution:
4z + 3 = 6 + 2z
∴ 4z + 3 – 2z = 6 (Transposing 2z to LHS)
∴ 2z + 3 = 6
∴ 2z = 6 – 3 (Transposing 3 to RHS)
∴ 2z = 3
∴ \(\frac{2 z}{2}=\frac{3}{2}\) (Dividing both the sides by 2)
∴ z = \(\frac {3}{2}\)

Check:
LHS = 4z + 3
=4 (\(\frac {3}{2}\)) + 3
= 6 + 3 = 9
RHS = 6 + 2z
= 6 + 2(\(\frac {3}{2}\))
= 6 + 3 = 9
LHS = RHS
Thus, the answer is correct.

Question 5.
2x – 1 = 14 – x
Solution:
2x – 1 = 14-x
∴ 2x – 1 + x = 14 (Transposing -x to LHS)
∴ 3x – 1 = 14
∴ 3x = 14 + 1 (Transposing -1 to RHS)
∴ 3x = 15
∴ \(\frac{3 x}{3}=\frac{15}{3}\) (Dividing both the sides by 3)
x = 5

Check:
LHS = 2x – 1
= 2 (5) – 1
= 10 – 1 = 9
RHS = 14 – x
= 14 – 5 = 9
LHS = RHS
Thus, the answer is correct.

Question 6.
8x + 4 = 3(x – 1) + 7
Solution:
8x + 4 = 3(x – 1) + 7
∴ 8x + 4 = 3x – 3 + 7
∴ 8x + 4 = 3x + 4
∴ 8x + 4 – 3x = 4 (Transposing 3x to LHS)
∴ 5x + 4 = 4
∴ 5x = 4 – 4 (Transposing 4 to RHS)
∴ 5x = 0
∴ \(\frac{5 x}{5}=\frac{0}{5}\) (Dividing both the sides by 5)
∴ x = 0

Check:
LHS = 8x + 4
= 8 (0) + 4
= 0 + 4 = 4
RHS = 3 (x – 1) + 7
= 3(0 – 1) + 7
= 3 (-1) + 7
= – 3 + 7 = 4
∴ LHS = RHS
Thus, the answer is correct.

Question 7.
x = \(\frac {4}{5}\) (x + 10)
Solution:
x = \(\frac {4}{5}\) (x + 10)
∴ x = \(\frac{4 x}{5}+10 \times \frac{4}{5}\)
∴ x = \(\frac{4 x}{5}+8\)
∴ x – \(\frac{4 x}{5}\) = 8 (Transposing \(\frac{4 x}{5}\) to LHS)
∴ \(\frac{5 x-4 x}{5}\) = 8 (LCM = 5)
∴ \(\frac{x}{5}\) = 8
∴ \(\frac{x}{5}\) × 5 = 8 × 5 (Multiplying both the sides by 5)
∴ x = 40

Check:
LHS = x = 40
RHS = \(\frac {4}{5}\) (x + 10)
= \(\frac {4}{5}\) (40 + 10)
= \(\frac {4}{5}\) (50)
= 4 × 10 = 40
∴ LHS = RHS
Thus, the answer is correct.

Question 8.
\(\frac{2 x}{3}+1=\frac{7 x}{15}+3\)
Solution:
\(\frac{2 x}{3}+1=\frac{7 x}{15}+3\)
∴ \(\frac{2 x}{3}\) = \(\frac{7 x}{15}\) + 3 – 1 (Transposing 1 to RHS)
∴ \(\frac{2 x}{3}\) = \(\frac{7 x}{15}\) + 2
∴ \(\frac{2 x}{3}-\frac{7 x}{15}\) = 2 (Transposing \(\frac{7 x}{15}\) to LHS)
∴ \(\frac{2 x \times 5-7 x}{15}\) (LCM = 15)
∴ \(\frac{10 x-7 x}{15}\) = 2
∴ \(\frac{3 x}{15}\) = 2
∴ \(\frac{x}{5}\) = 2
∴ \(\frac{x}{5}\) × 5 = 2 × 5 (Multiplying both the sides by 5)
∴ x = 10

Check:
LHS = \(\frac{2 x}{3}\) + 1
= \(\frac{2(10)}{3}\) + 1
= \(\frac{20}{3}\) + 1 = \(\frac{23}{3}\)
RHS = \(\frac{7 x}{15}\) + 3
= \(\frac{7(10)}{15}\) + 3
= \(\frac {14}{3}\) + 3
= \(\frac {23}{3}\)
∴ LHS = RHS
Thus, the answer is correct.

Question 9.
2y + \(\frac {5}{3}\) = \(\frac {26}{3}\) – y
Solution:
2y + \(\frac {5}{3}\) = \(\frac {26}{3}\) – y
∴ 2y + y + \(\frac {5}{3}\) = \(\frac {26}{3}\) (Transposing -y to LHS)
∴ 3y + \(\frac {5}{3}\) = \(\frac {26}{3}\)
∴ 3y = \(\frac{26}{3}-\frac{5}{3}\) (Transposing \(\frac {5}{3}\) to RHS)
∴ 3y = \(\frac{26-5}{3}\)
∴ 3y = \(\frac {21}{3}\)
∴ 3y = 7
∴ \(\frac{3 y}{3}=\frac{7}{3}\) (Dividing both the sides by 3)
∴ y = \(\frac {7}{3}\)

Check:
LHS = 2y + \(\frac {5}{3}\)
= 2(\(\frac {7}{3}\) ) + \(\frac {5}{3}\)
= \(\frac{14}{3}+\frac{5}{3}\)
= \(\frac{14+5}{3}\)
= \(\frac {19}{3}\)
RHS = \(\frac {26}{3}\) – y
= \(\frac{26}{3}-\frac{7}{3}\)
= \(\frac{26-7}{3}\)
= \(\frac {19}{3}\)

Question 10.
3m = 5m – \(\frac {8}{5}\)
Solution:
3m = 5m – \(\frac {8}{5}\)
∴ 3m – 5m = –\(\frac {8}{5}\) (Transposing 5m to LHS)
∴ -2m = –\(\frac {8}{5}\)
∴ 2m = \(\frac {8}{5}\) [Multiplying both the sides by (-1)]
∴ \(\frac{2 m}{2}=\frac{8}{5} \times \frac{1}{2}\) (Dividing both the sides by 2)
∴ m = \(\frac {4}{5}\)

Check:
LHS = 3m
= 3(\(\frac {4}{5}\))
= \(\frac {12}{5}\)
RHS = 5m – \(\frac {8}{5}\)
= 5(\(\frac {4}{5}\)) – \(\frac {8}{5}\)
= 4 – \(\frac {8}{5}\)
= \(\frac{20-8}{5}\)
= \(\frac {12}{5}\)
∴ LHS = RHS
Thus, the answer is correct.

Punjab State Board PSEB 8th Class English Book Solutions English Grammar Finite and Non-Finite Verbs Exercise Questions and Answers, Notes.

PSEB 8th Class English Grammar Finite and Non-Finite Verbs

निम्नलिखित वाक्यों को पढ़ो और इन वाक्यों में दिये गए italicised (तिरछे) शब्दों के प्रयोग पर विचार करो-

1. (a) (i) I saw the girls jumping.
(ii). I did not see the dancing girl.

(b) (i) He got his shoes mended.
(ii) He is a worried man now.

2. (i) I went to see the match.
(ii) We go home to take rest.

3. (i) Dancing is an art.
(ii) She enjoys dancing.

सभी italicised शब्द ऐसे शब्द हैं जो अपने आप में किसी वाक्य का Predicate नहीं बन सकते। हम ऊपर दिए गए किसी भी वाक्य का italicised शब्द के साथ predicate नहीं बना सकते। अर्थात् ‘I jumping’, ‘I dancing’ आदि predicate नहीं बना सकते, इस प्रकार के Verbs को Non-Finite Verbs कहते हैं। इसके विपरीत वे Verbs या Verb Phrases जो किसी वाक्य के Predicate बन सकते हैं, Finite Verbs कहलाते हैं।

एक अन्य परिभाषा

Italicised शब्दों पर Tense, Person अथवा Number का कोई प्रभाव दिखाई नहीं देता। अर्थात् जिन Verbs पर Tense, Person के Number का प्रभाव नहीं होता, Non-Finite Verbs कहलाते हैं। Tense अथवा Subject बदलने के पश्चात् भी इन Verbs का रूप नहीं बदलता। इसके विपरीत Finite Verbs का रूप Tense तथा Person के अनुसार बदल सकता है। आओ वाक्यों पर पुनः विचार करें-
1. (a) I saw the train moving.
I see the train moving.
He sees the train moving.

(b) He got his watch repaired.
He gets his watch repaired.
They will get their watches repaired.

2. (i) I want to see the match.
(ii) He wants to see the match.
(iii) We wanted to see the match.

3. (i) She enjoys dancing.
(ii) She will enjoy dancing.
(iii) They enjoyed dancing.

अतः स्पष्ट है कि Finite Verbs (underlined) का परिवर्तन होने पर भी Non-finite Verbs में कोई परिवर्तन नहीं होता।

पूर्ण स्पष्टीकरण
अब दाईं तथा बाईं ओर दिए गए शब्द-समूहों का अध्ययन करो। आप देखेंगे कि Non-finites किस प्रकार Predicate का रूप धारण नहीं कर सकते।

Finite Verbs:
He takes tea.
He can drive well.
They have gone home.
The dog was beaten by the boys

Non-Finite Verbs:
He taking tea.
He to drive well.
They going home.
The dog beaten by the boys.

बाईं ओर के सभी शब्द समूह वाक्य हैं। ऐसा इसलिए है क्योंकि इनके Verbs Predicate का काम करते हैं परन्तु दाईं ओर के Verbs Non-Finites हैं। क्योंकि Non-Finites स्वयं Predicate का निर्माण नहीं करते, इसलिए दाईं ओर के शब्द-समूह वाक्य नहीं हो सकते।

Non-Finites का वर्गीकरण-
(1) Present Participle
(2) Past Participle

1. (a) (i). I saw the girl jumping. (Present Participle)
(ii) I did not see the dancing girl. (Present Participle)

(b) (i) He got his shoes mended. (Past Participle)
(ii) He is a ‘worried man now. (Past Participle)

2. (i) I want to see the match. (Infinitive)
(ii) We go home to take rest. (Infinitive)

3. (i) Dancing is an art. (Gerund)
(ii) She enjoys dancing (Gerund)

I. The Infinitive

I. Infinitive का प्रयोग Noun के रूप में हो सकता है।
1. Verb के Subject के रूप में:

• To forgive is divine.
• To drive a car requires skill.
• To err is human.

2. Object के रूप में:

• She wishes to rise higher in life.
• No one likes to die.
• I want to learn music.

3. Complement के रूप में:

• This house is to let.
• He seems to act well.
• Her desire was to do good.

4. Preposition के Object के रूप में:

• He was about to speak.
• The match was going to start.
• She was about to die.

5. Noun या Pronoun के Apposition के रूप में:

• It is easy to advise others.
• It is bad to find faults with others.
• It is good to help the poor.

II. Infinitive का प्रयोग adjective के रूप में भी हो सकता है।

1. Bere 2017 बताने के लिए

• He got up to ask a question.
• I went to see the Principal.
• He studied to become a doctor.

2. Noun या Pronoun की विशेषता बताने के लिए

• My decision to go is final.
• I have no friends to talk to.
• The topics to be written are known to all.

3. Preposition या Object की विशेषता बताने के लिए

• He is too old to walk.
• She is too young to understand.
• They are too busy to attend the function.

4. verb या complement की विशेषता बताने के लिए

• To tell the truth, I hate shirkers.
• To sum up, he is the best of friends.
• To say in a few words, Mohan achieved the object of his life.

Bare Infinitive या बिना to के Infinitive

इसका प्रयोग होता है:
1. bid, feel, hear, know, let, make, notice, observe, see, watch if Verbs

• I made him give up smoking.
• He bade me open the window.
• I let the boy go.

2. shall, will, would, should, do, have may, must, can, could if Auxiliaries as:

• You may leave now.
• I do not like him.
• You must not disobey your parents.

3. ‘had better’, ‘had rather’, ‘would rather’, ‘had sooner:

• You had better leave this place.
• I would rather starve than beg.
• He would rather solve the problem better.

4. but’, ‘than’s are:

• We could not but laugh.
• He did more than help his friend.

II. The Gerund

The Gerund का निर्माण Verb की पहली फार्म + ing से होता है। Gerund का प्रयोग निम्नलिखित ढंग से हो सकता है।

1. Verb के Subject के रूप में :

• Swimming is a good exercise.
• Speaking is easier than writing.
• Dancing is an art.

2. Verb के Object के रूप में:

• I hate waiting at bus stops.
• He likes reading novels.
• She stopped playing.

3. Preposition के Object के रूप में:

• I am tired of thinking.
• He is thinking of leaving this place.
• He started his journey after resting for an hour.

4. Verb at Complement के रूप में:

• Thinking is doing.
• Talking to him is wasting time.
• Seeing is believing

5. Absolute construction के रूप में:

• Speaking the truth being his habit, we like him.
• Reading the books being his hobby, we appreciate him.

Note : यदि Gerund से पहले कोई noun या pronoun आये तो उसका Possessive रूप ही प्रयोग करना सकता है।

• He stopped my going there.
• He likes my doing this job.
• I do not like Ram’s coming here.

6. Noun Compounds as party के रूप में:

• He bought a new dining table.
• The dancing girl was full of thrill.
• She wastes hours before her looking glass.

Note : निचे कुछ विशेष verbs दिए गए है जिनके साथ Gerund का प्रयोग होता है

• He avoided seeing the Principal.
• He admitted telling a lie.
• She denied using force.
• He dislikes deceiving people.
• I enjoy playing with children.
• He cannot help laughing.
• I don’t mind waiting for an hour.
• I missed seeing that film.
• He postponed his going to Delhi.
• She stopped going there.
• I suggest going for a walk.

III. Participle (Present and Past)

Present Participle : Present Participle का निर्माण verb की पहली फार्म तथा ing से होता है।
Note : Present Participle तथा Gerund दोनों का निर्माण ‘ing’ से होता है; प्रतनु दोनों में अत्नर है (i) Participle adjective के रूप में प्रयोग होता है
उदाहरण:
I like new coat.
I like shining-coat.
यहाँ ‘shining new की तरह adjective का काम कर रहा है। इसलिए यह Participle है।

(ii) Gerund noun के रूप में प्रयुक्त होता है। इसलिए यह वाक्य में वे सभी स्थान ले सकता है जो Noun के होते है; जैसे
Swimming is an exercise. (Subject के रूप में)
I like swimming. (Object के रूप में)

Present Participle का प्रयोग
1. Present Participle का प्रयोग subject के बाद आने वाले Noun के Adjective के रूप में होता है:

• His speech was expressing.
• Her lectures were interesting.
• The results were encouraging.

2. जब दो कार्य एक ही Subject द्वारा एक के बाद एक किए जायें, तो पहले कार्य को व्यक्त करने के लिए Present Participle का प्रयोग किया जाता है, जैसे,

• Seeing his father, the boy ran away.
• She entered, closing the door behind her.
• Crying, she went to qazi.

3. जब दो साथ-साथ हों तो उनमें से एक को Present Participle दुरा यक्त किया जाता है:

• He went into the room singing.
• He came to me running.
• The birds flew away chirping.

4. Present Participle ‘Object complement’ के रूप में भी कार्य कर सकता है:

• We found him studying in his room.
• The doctor found the patient sitting up in bed.
• I saw him watering the plants in his garden.

5. Present Participle का प्रयोग absolutely’ भी होता है; जैसे,

• The weather being fine (having been fine), we decided to go out for a walk.
• The dinner being over, the guests started leaving.
• The song being over, the dancers stopped dancing.

6. कभी- कभी Present Participle का प्रयोग Perfect Participle के रूप में होता है। ऐसा तब किया जाता है जब यह व्यक्त करना हो कि दूसरा कार्य आरम्भ होने से पूर्व पहला कार्य पूरा हो चुका था; जैसे,

• Having seen my sister off, I came home.
• Having done her homework, she went out to play.
• Having seen the film, they went out to a restaurant.

7. Perfect Participle का प्रयोग Passive constructions में भी होता है; जैसे,

• Having been betrayed once, he did not fall into the trap again.
• Having been defeated several times, the army finally surrendered.
• Having been insulted twice, I never went to see him again.

Past Participle : Past Participle verb की तीसरी फार्म होती है।
Past Participle का प्रयोग
Past Participle का प्रयोग निम्नलिखित प्रकार से होता है

1. Adjective के रूप में; जैसे

• His spoken English is much better than his written English.
• The written words have much weight.
• The planned object was achieved.

2. Passive भावना को यकत करने के लिए; जैसे

• The Chief Minister arrived, accompanied by the Minister for Education.
• Shot by an arrow, the bird fell to the ground.
• Disgusted, he left his home.

3. Subject complement के रूप में; जैसे

• They grew tired.
• We were left bored.
• Don’t be disappointed.

4. Object complement के रूप में; जैसे

• I got a new shirt made.
• We got a new home built.
• He had his hair cut.

Combination of Sentences

(Using Non-Finites)
Participle, Infinitive या Gerund (Non-finites) की सहायता से दो वाक्यों को जोड़ कर एक वाक्य भी बनाया जा सकता है।

1. Infinitive के प्रयोग द्वारा
(1) Separate : We go to a cinema. We see a movie.
Combined : We go to a cinema to see a movie.

(2) Separate : The principal called Mrs. Sharma. She would teach English.
Combined : The principal called Mrs. Sharma to teach English.

(3) Separate : I shall go to the market. I shall buy rice.
Combined : I shall go to the market to buy rice.

(4) Separate : She is very poor. She cannot pay her fee.
Combined : She is too poor to pay her fee.

(5) Separate : I want to go to my brother. I shall assist him.
Combined :: I want to go to my brother to assist him.

(6) Separate : I go to the playground. I play there.
Combined : I go to the playground to play.

(7) Separate : I bent. I picked the ball.
Combined : I bent to pick the ball.

2. Participle के प्रयोग द्वारा

(1) Separate : He picked up his umbrella. He went out.
Combined : Picking up his umbrella, he went out.

(2) Separate : The thieves ran away. They saw the policeman.
Combined : Seeing the policeman, the thieves ran away.

(3) Separate : The students stopped talking. They saw the headmaster.
Combined : Seeing the headmaster, the students stopped talking.

(4) Separate : He lost his book. He began to cry.
Combined : Having lost his book, he began to cry.

(5) Separate : He picked the pocket. He ran away.
Combined : Having picked the pocket, he ran away.

(6) Separate : The old lady was helped by the little boy. She was able to cross the road.
Combined : Helped by the little boy, the old lady was able to cross the road.

(7) Separate : I saw some monkeys. They were jumping from branch to branch.
Combined : I saw some monkeys jumping from branch to branch.

(8) Separate : We watched a cricket match. It was being played in our school.
Combined : We watched a cricket match being played in our school.

(9) Separate : I met a girl. She was weeping in the street.
Combined : I met a weeping girl in the street.

(10) Separate : We heard a noise. It was coming from a nearby house.
Combined : We heard a noise coming from a nearby house.

3. Gerund के प्रयोग द्वारा

(1) Separate : Mohan waits for the bus everyday. He can’t bear it.
Combined : Mohan can’t bear waiting for the bus everyday.

(2) Separate : Gopal watches hockey matches. He likes it.
Combined : Gopal likes watching hockey matches.

(3) Separate : Kamla writes stories. She is very fond of it.
Combined : Kamla is very fond of writing stories.

(4) Separate : He helped my brother. I appreciate it.
Combined : I appreciate his helping my brother.

(5) Separate : I avoided Ram. I did not meet him.
Combined : I avoided meeting Ram.

(6) Separate : He was seeing the match. He saw it for some time.
Combined : He went on seeing the match for some time.

(7) Separate : The bird spread the wings. It flew away.
Combined : The bird flew away by spreading the wings.

(8) Separate : You go there. I do not approve of it.
Combined : I do not approve of your going there.

Exercises (Solved) (With Hints) Set-I

Combine the following sentences in each pair using participle:

1. He took aim. He shot the tiger.
2. He hurt his foot. He stopped.
3. He was unwilling to go any further. He returned home.
4. They saw the uselessness of punishment. They changed their way.
5. He was tired of failure. He went to another city.
6. I received no answer. I knocked it second time.
7. He felt tired. He laid his work aside.
8. I went to Delhi last year. I wished to see a doctor.
9. He lost money. He gave up gambling.
10. He gave up the job. He was not satisfied with the salary.
11. He went straight on. He met Ram on the path.
12. A dog stole a piece of meat. He went outside the city to enjoy it.
13. The magician took pity on the mouse. He turned it into a cat.
14. My sister liked the book. She bought it at once.
15. The letter was badly written. I had great difficulty in reading it.
16. The hungry fox saw some grapes. They were hanging from a vine.
17. I was walking along the bank. I saw a dead snake.
18. He ran at top speed. He got out of breath.
19. He jumped up. He ran away.
20. He was tired. He sat down to rest.
21. He finished his dinner. He went out for a walk.
22. He felt sleepy. He went to bed.
23. He aimed at the bird. He shot an arrow.
24. He failed in the examinations. He gave up studies.
25. He ran after the thief. He caught him.
Hints:
1. Taking aim
2. Having hurt
3. Unwilling
4. Seeing the
5. Tired of
6. Having received
7. Feeling tired
8. Wishing to
9. Having lost
10. Dissatisfied with
11. Going
12. Having stolen, the dog went
13. Taking pity, the magician turned
14. Having liked, my sister bought it
15. The letter being badly written.
16. The hungry fox saw some grapes hanging
17. Walking along the bank
18. Running at
19. Jumping up
20. Being tired
21. Having finished
22. Feeling sleepy
23. Aiming at the bird
24. Having failed
25. Running after.

Set-II
Combine the following sentences by using infinitives:

1. I went to the playground. I wanted to see the match.
2. I worked very hard. I wanted to assist him.
3. I want to go to my brother. I want to assist him.
4. I won a scholarship. I had to work very hard for it.
5. She is very poor. She cannot pay her fee.
6. He is very selfish. He will not help you.
7. I shall go to the market. I shall buy sugar.
8. I was trying to lift the box. He helped me.
9. I speak the truth. I am not afraid of it.
10. Everyone should do his duty. The country expects this of everyone.
11. He must apologise to me. This is the only way to escape punishment.
12. I shall succeed. I am sure of it.
13. He will stand first. He is hopeful of it.
14. You will catch the train. You need not run for it.
15. She visits the poor. It is in this way that she can help.
16. He took out the knife. His object was to stab the passer-by.
17. I am very tired. I cannot work.
18. The hunter took up his gun. He wanted to shoot the tiger.
19. He bought a box. He needed it for keeping ornaments in it.
20. The king was very pleased. He heard of the success of his army.
21. This load is very heavy. I cannot lift it.
22. He heard the happy news. He was overjoyed.
23. The problem was difficult. It could not be solved.
24. My friend has gone to Delhi. He will attend a wedding there.
25. This book is very expensive. I cannot buy it.
26. We go to a cinema. We see a movie there.
27. We telephoned the airport. We wanted to ask for some information.
28. The school appointed Miss Sheela. She would teach English.
29. I wanted to meet my parents. I returned home.
30. He wanted to learn the art of bowling. The coach taught him.
31. We bow before our teacher. We respect him.
32. She bought a car. She would travel fast.
33. They use kerosene. They would/will cook their food.
Hints:
1. playground to see
2. hard to assist
3. brother to assist him
4. very hard to win a scholarship
5. too poor to pay her fees
6. too selfish to help
7. market to buy
8. He helped me lift
9. afraid to speak the truth
10. expects everyone of us to do our duty
11. to escape punishment
12. sure to succeed
13. he hopes to stand
14. run to catch
15. She visits the poor to help them
16. knife to stab
17. too-to
18. his gun to shoot
19. a box to keep
20. pleased to hear
21. too heavy for me to
22. overjoyed to hear
23. too difficult to be
24. gone to Delhi to attend
25. too expensive for me to
26. to a cinema to see
27. the airport to ask
28. appointed Miss Sheela to teach
29. home to meet
30. taught him to learn
31. our teacher to respect
32. a car to travel fast
33. use kerosene to cook

Exercises From Board’s Grammar (Solved)

1. Pick out Infinitives in the following sentences:
1. To lie is a sin.
2. I saw him enter.
3. She let me watch the film.
4. He promised to come.
5. To forgive is divine.
6. He is too weak to walk.
7. I don’t know where to go.
8. It is shameful to cheat your friend.
9. I watched her dance.
10. Straw is used to make paperboard.
Answer:
1. To lie
2. enter
3. watch
4. to come
5. To forgive
6. to walk
7. to go
8. to cheat
9. dance
10. to make.

II. Complete the following sentences by filling in the blank spaces with appropriate non-finite forms:

1. (Err) is human, (forgive) is divine.
2. You ought (get) up earlier.
3. It is easy (make) mistakes.
4. Why not (take) the day off?
5. He made me (repeat) the lessons.
6. You needn’t (say) anything.
7. I am sorry (disappoint) you.
8. He heard a cock (crow) in the neighbouring village.
9. Would you (like) (come) in my car?
10. He will be able (swim) very soon.
Answer:
1. To err, to forgive.
2. to get
3. to make
4. take.
5. repeat
6. say
7. to disappoint
8. crow.
9. like, to come.
10. to swim.

III. Combine the following pairs of sentences into one sentence each using too / enough + infinitive:

1. You are very young. You can’t have a gun.
2. He is very ill. He can’t eat anything.
3. The coffee is strong. It won’t keep us awake.
4. Tom was very foolish. He told lies to the police.
5. He was furious. He couldn’t speak.
6. You are quite thin. You could slip between the bars.
7. It is very cold. We can’t bathe.
8. It is very cold. We can’t go out.
9. The fire isn’t very hot. It won’t boil water in a kettle.
10. I am rather old. I can’t walk that far.
Answer:
1. You are too young to have a gun.
2. He is too ill to eat anything.
3. The coffee is not strong enough to keep us awake.
4. Tom was foolish enough to tell lies to the police.
5. He was too furious to speak.
6. You are thin enough to slip between the bars.
7. It is too cold for us to bathe.
8. It is too cold for us to go out.
9. The fire isn’t hot enough to boil water in a kettle.
10. I am too old to walk that far.

IV. Pick out gerunds in the following sentences:

1. Gambling is a bad habit.
2. She enjoys sleeping.
3. Old men enjoy gossiping.
4. I hate waiting.
5. Stealing is a crime.
6. He is fond of walking.
7. I am good at spelling.
8. We took part in boating.
9. My sister does not like cooking.
10. She is fond of dancing.
Answer:
1. Gambling.
2. sleeping.
3. gossiping.
4. waiting.
5. stealing
6. walking
7. spelling
8. boating.
9. cooking
10. dancing.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.2

Question 1.
If you subtract \(\frac {1}{2}\) from a number and multiply the result by \(\frac {1}{2}\), you get \(\frac {1}{8}\). What is the number?
Solution:
Let the required number be x.
By subtracting \(\frac {1}{2}\) from x, we get x-\(\frac {1}{2}\) and by multiplying this result by \(\frac {1}{2}\),
we get \(\frac {1}{2}\)(x – \(\frac {1}{2}\))
But, the result is \(\frac {1}{8}\)
\(\frac {1}{2}\)(x – \(\frac {1}{2}\)) = \(\frac {1}{8}\)
∴ \(\frac {1}{2}\)(x – \(\frac {1}{2}\)) × 2 = \(\frac {1}{8}\) × 2 (Multiplying both the sides by 2)
∴ x – \(\frac {1}{2}\) = \(\frac {1}{4}\)
∴ x = \(\frac{1}{4}+\frac{1}{2}\) (Transposing –\(\frac {1}{2}\) to RHS)
∴ x = \(\frac{1+2}{4}\) (LCM = 4)
∴ x = \(\frac {3}{4}\)
Thus, the required number = \(\frac {3}{4}\)

Question 2.
The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?
Solution:
Perimeter of the pool = 154 m
Let breadth = x metres
Length is 2 m more than twice its breadth.
Length = 2 (breadth) + 2
= (2x + 2) metres
Perimeter of a rectangle = 2 (length + breadth)
2 (length + breadth) = Perimeter
∴ 2[(2x + 2) + x] = 154
∴ 2 [2x + 2 + x] = 154
∴ 2 (3x + 2) = 154
∴ \(\frac{2(3 x+2)}{2}=\frac{154}{2}\) (Dividing both the sides by 2)
∴ 3x + 2 = 77
∴ 3x = 77 – 2 (Transposing 2 to RHS)
∴ 3x = 75
∴ \(\frac{3 x}{3}=\frac{75}{3}\) (Dividing both the sides by 3) x — 25
Breadth = 25 m
Length = 2x + 2
= 2 (25) + 2
= 50 + 2
= 52 m
Thus, the length of the pool is 52 m and its breadth is 25 m.

Question 3.
The base of an isosceles triangle is \(\frac {4}{3}\) cm. The perimeter of the triangle is 4\(\frac {2}{15}\) cm. What is the length of either of the remaining equal sides?
Solution:
Base of an isosceles triangle = \(\frac {4}{3}\) cm
Let the length of each of the equal sides = x cm
Perimeter of the triangle = \(\frac {4}{3}\) + x + x
= \(\frac {4}{3}\) + 2x
Perimeter of the triangle = 4\(\frac {2}{15}\) cm (Given)
∴ \(\frac {4}{3}\) + 2x = 4\(\frac {2}{15}\)
∴ \(\frac {4}{3}\) + 2x = \(\frac {62}{15}\)
∴ 2x = \(\frac{62}{15}-\frac{4}{3}\)(Transposing to RHS)
∴ 2x = \(\frac{62-20}{15}\) (LCM = 15)
∴ 2x = \(\frac {42}{15}\)
∴ \(\frac{2 x}{2}=\frac{42}{15} \times \frac{1}{2}\) (Dividing both the sides by 2)
∴ x = \(\frac {21}{15}\)
∴ x = \(\frac{7 \times 3}{5 \times 3}\)
∴ x = \(\frac {7}{5}\)
∴ x = 1\(\frac {2}{5}\)
Thus, the required length of either of the remaining equal sides is 1\(\frac {2}{5}\) cm.

Question 4.
Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Solution:
Let the smaller number be x
∴ The greater number = x + 15
Their sum is 95.
∴ x + (x + 15) = 95
∴ x + x + 15 = 95
∴ 2x + 15 = 95
∴ 2x = 95 – 15 (Transposing 15 to RHS)
∴ 2x = 80
∴ \(\frac{2 x}{2}=\frac{80}{2}\) (Dividing both the sides by 2)
∴ x = 40
The smaller number = x = 40
The greater number = x + 15 = 40 + 15 = 55
Thus, 40 and 55 are the required numbers.

Question 5.
Two numbers are in the ratio 5 : 3. If they differ by 18, what are the numbers ?
Solution:
Ratio of the two numbers = 5 : 3
Let the two numbers be 5x and 3x.
Difference = 18
∴ 5x – 3x = 18
∴ 2x = 18
∴ \(\frac{2 x}{2}=\frac{18}{2}\) (Dividing both the sides by 2)
∴ x = 9
∴ Greater number = 5x = 5 × 9 = 45
Smaller number = 3x = 3 × 9 = 27
Thus, 45 and 27 are the numbers.

Question 6.
Three consecutive integers add up to 51. What are these integers ?
Solution:
Let the consecutive integers be, x, x + 1 and x + 2.
Their sum is 51.
∴ x + (x + 1) + (x + 2) = 51
∴ x + x + 1 + x + 2 = 51
∴ 3x + 3 = 51
∴ 3x = 51 – 3 (Transposing 3 to RHS)
∴ 3x = 48
∴ \(\frac{3 x}{3}=\frac{48}{3}\) (Dividing both the sides by 3)
∴ x = 16
∴ First number = x = 16
Second number = x + 1 = 16 + 1 = 17
Third number = x + 2 = 16 + 2 = 18
Thus, the required consecutive integers are 16, 17 and 18.

Question 7.
The sum of three consecutive multiples of 8 is 888. Find the multiples.
Solution:
Let the three multiples of 8 be x, x + 8, and x + 8 + 8 = x + 16.
Their sum is 888.
∴ x + (x + 8) + (x + 16) = 888
∴ x + x + 8 + x + 16 = 888
∴ 3x + 24 = 888
∴ 3x = 888 – 24 (Transposing 24 to RHS)
∴ 3x = 864
∴ \(\frac{3 x}{3}=\frac{864}{3}\) (Dividing both the sides by 3)
∴ x = 288
∴ First number = x = 288
∴ Second number = x + 8 = 288 + 8 = 296
Third number = x + 16 = 288 + 16 = 304
Thus, the required three consecutives multiples of 8 are 288, 296 and 304.

Question 8.
Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.
Solution:
Let the three consecutive integers be, x, (x + 1) and (x + 2).
According to the condition,
∴ 2 × (x) + 3 × (x + 1) + 4 × (x + 2) = 74
∴ 2x + 3x + 3 + 4x + 8 = 74
∴ 9x + 11 = 74
∴ 9x = 74 – 11 (Transposing 11 to RHS)
∴ 9x = 63
∴ \(\frac{9 x}{9}=\frac{63}{9}\) (Dividing both the sides by 9)
∴ x = 7
∴ First integer = x – 7
Second integer = x + 1 = 7 + 1 = 8
Third integer = x + 2 = 7 + 2 = 9
Thus, the required integers are 7, 8 and 9.

Question 9.
The ages of Rahul and Haroon are in the ratio 5 : 7. Four years later the sum of their ages will be 56 years. What are their present ages?
Solution:
Ages of Rahul and Haroon are in the ratio of 5 : 7.
Let their present ages be 5x and 7x years.
∴ 4 years later,
the age of Rahul will be 5x + 4 years and the age of Haroon will be 7x + 4 years.
According to the condition,
(5x + 4) + (7x + 4) = 56
∴ 5x + 4 + 7x + 4 = 56
∴ 12x + 8 = 56
∴ 12x = 56 – 8 (Transposing 8 to RHS)
∴ 12x = 48
∴ \(\frac{12 x}{12}=\frac{48}{12}\) (dividing both the sides by 12)
∴ x = 4
Present age of Rahul = 5x = 5 × 4
= 20 years
Present age of Haroon = 7x = 7 × 4
= 28 years
Thus, present age of Rahul is 20 years and that of Haroon is 28 years.

Question 10.
The number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the number of girls. What is the total class strength?
Solution:
Number of boys : Number of girls = 7 : 5
Let the number of boys be 7x, and the number of girls be 5x.
According to the condition
7x = 5x + 8
∴ 7x – 5x = 8 (Transposing 5x to LHS)
∴ 2x = 8
∴ \(\frac{2 x}{2}=\frac{8}{2}\) (Dividing both the sides by 2)
∴ x = 4
Number of boys = 7x = 7 × 4 = 28
Number of girls = 5x = 5 × 4 = 20
Total class strength = 28 + 20 = 48
Thus, total class strength is 48.

Question 11.
Bharat’s father is 26 years younger than Bharat’s grandfather and 29 years older than Bharat. The sum of the ages of all the three is 135 years. What is the age of each one of them ?
Solution:
Let the age of Bharat be x years,
His father’s age = (x + 29) years
His grandfather’s age = x + 29 + 26
= (x + 55) years
Sum of their ages is 135 years.
∴ x + (x + 29) + (x + 55) = 135
∴ x + x + 29 + x + 55 = 135
∴ 3x + 84 = 135
∴ 3x = 135 – 84 (Transposing 84 to RHS)
∴ 3x = 51
∴ \(\frac{3 x}{3}=\frac{51}{3}\) (Dividing both the sides by 3)
∴ x = 17
Bharat’s age = x = 17 years
His father’s age = x + 29
= 17 + 29
= 46 years
His grandfather’s age = x + 55
= 17 + 55
= 72 years
Thus, Bharat’s age is 17 years, his father’s age is 46 years and his grandfather’s age is 72 years.

Question 12.
Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?
Solution:
Let Ravi’s present age be x years.
4 times his present age be 4x years.
15 years from now his age be x + 15 years.
According to the condition, x + 15 = 4x
∴ 4x = x + 15 (Interchanging both the sides)
∴ 4x – x = 15 (Transposing x to LHS)
∴ 3x = 15
∴ \(\frac{3 x}{3}=\frac{15}{3}\) (Dividing both the sides by 3)
∴ x = 5
Thus, Ravi’s present age is 5 years.

Question 13.
A rational number is such that when you multiply it by \(\frac {5}{2}\) an\(\frac {2}{3}\) add g to the product, you get –\(\frac {7}{12}\). What is the number ?
Solution:
Let the required rational number be x.
We get \(\frac{5 x}{2}\) by multiplying x with \(\frac {5}{2}\)
By adding \(\frac {2}{3}\) to it we get \(\frac{5 x}{2}+\frac{2}{3}\)
But, the result is \(\frac {-7}{12}\)
∴ \(\frac{5 x}{2}+\frac{2}{3}=\frac{-7}{12}\)
\(\frac{5 x}{2}=-\frac{7}{12}-\frac{2}{3}\) (Transposing \(\frac {2}{3}\) to RHS)
∴ \(\frac{5 x}{2}=\frac{-7-8}{12}\) (LCM = 12)
∴ \(\frac{5 x}{2}=\frac{-15}{12}\)
∴ \(\frac{5 x}{2} \times \frac{2}{5}=\frac{-15}{12} \times \frac{2}{5}\) (Multiplying both the sides by \(\frac {2}{5}\))
∴ x = –\(\frac {1}{2}\)
Thus, the required rational number is –\(\frac {1}{2}\).

Question 14.
Lakshmi is a cashier in a bank. She has currency notes of denominations ₹ 100, ₹ 50 and ₹ 10, respectively. The ratio of the number of these notes is 2 : 3 : 5. The total cash with Lakshmi is ₹ 4,00,000. How many notes of each denomination does she have?
Solution:
Let the number of
₹ 100 notes be 2x,
₹ 50 notes be 3x,
₹ 10 notes be 5x.
Value of ₹ 100 notes = 2x × 100 = ₹ 200x
Value of ₹ 50 notes = 3x × 50 = ₹ 150x
Value of ₹ 10 notes = 5x × 10 = ₹ 50x
According to the condition, value of
₹ 200x + ₹ 150x + ₹ 50x = ₹ 4,00,000
∴ 200x + 150x + 50x = 4,00,000
∴ 400x = 400000
∴ \(\frac{400 x}{400}=\frac{400000}{400}\)
∴ x = 1000
Number of ₹ 100 notes = 2x
= 2 × 1000
= 2000
Number of ₹ 50 notes = 3x
= 3 × 1000
= 3000
Number of ₹ 10 notes = 5x
= 5 × 1000
– 5000
Thus, Lakshmi has 2000 notes of ₹ 100, 3000 notes of ₹ 50 and 5000 notes of ₹ 10.

Question 15.
I have a total of ₹ 300 in coins of denomination ₹ 1, ₹ 2 and ₹ 5. The number of ₹ 2 coins is 3 times the number of ₹ 5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Solution:
Let the number of ₹ 5 coins be x.
Then, the number of ₹ 2 coins = 3x
Total number of coins = 160
Number of ₹ 1 coins = 160 – 3x – x
= 160 – 4x
Now, value of
₹ 5 coins = ₹ 5 × x = ₹ 5x
₹ 2 coins = ₹ 2 × 3x = ₹ 6x
₹ 1 coins = ₹ 1 × (160 – 4x)
= ₹ (160 – 4x)
According to the condition,
5x + 6x + (160 – 4x) = 300
∴ 5x + 6x + 160 – 4x = 300
∴ 11x – 4x + 160 = 300
∴ 7x + 160 = 300
∴ 7x = 300 – 160 (Transposing 160 to RHS)
7x = 140
∴ \(\frac{7 x}{7}=\frac{140}{7}\) (Dividing both the sides by 7)
∴ x = 20
Number of
₹ 5 coins = x = 20
₹ 2 coins = 3x = 3 × 20 = 60
₹ 1 coins = 160 – 4x
= 160 – 4 × 20
= 160 – 80
= 80
Thus, I have 20 coins of ₹ 5, 60 coins of ₹ 2 and 80 coins of ₹ 1.

Question 16.
The organisers of an essay competition decide that a winner in the competition gets a prize of ₹ 100 and a participant who does not win gets a prize of ₹ 25. The total prize money distributed is ₹ 3000. Find the number of winners, if the total number of participants is 63.
Solution:
Let the number of winners be x
∴ Number of participants who are not winners = (63 – x)
Prize money given to winners = x × ₹ 100 = ₹ 100x
Prize money given to non-winner
participants = ₹ 25 × (63 -x)
= ₹ 25 × 63 – ₹ 25x
= ₹ 1575 – ₹ 25x
According to the condition
100x + 1575 – 25x = 3000
∴ 75x + 1575 = 3000
∴ 75x = 3000 – 1575 (Transposing 1575 to RHS)
∴ 75x = 1425
\(\frac{75 x}{75}=\frac{1425}{75}\) (Dividing both the sides by 75)
∴ x = 19
Thus, the number of winners are 19.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4

Question 1.
Express the trigonometric ratios of sin A, sec A and tan A in terms of cot A.
Solution:
By using Identity,
cosec² A – cot² A = 1
⇒ cosec² A = 1 + cot² A
⇒ (cosec A)² = cot² A + 1
⇒ \(\left(\frac{1}{\sin A}\right)^{2}\) = cot² A + 1
⇒ (sin A)² = \(\frac{1}{\cot ^{2} \mathrm{~A}+1}\)
⇒ sin A = ± \(\frac{1}{\sqrt{\cot ^{2} \mathrm{~A}+1}}\)
We reject negative values of sin A for acute angle A.
Therefore, sin A = \(\frac{1}{\sqrt{\cot ^{2} A+1}}\)
By using identity,
sec² A – tan² A = 1
⇒ sec² A = 1 + tan² A
= 1 + \(\frac{1}{\cot ^{2} A}\)
= \(\frac{\cot ^{2} A+1}{\cot ^{2} A}\)

⇒ sec A = \(\sqrt{\frac{\cot ^{2} A+1}{\cot ^{2} A}}\)

tan A = \(\frac{1}{\cot A}\).

Question 2.
Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution:
By using Identity
sin² A + cos² A = 1
⇒ sin² A = 1 – cos² A
= 1 – \(\frac{1}{\sec ^{2} \cdot A}\) = \(\frac{\sec ^{2} A-1}{\sec ^{2} A}\)

⇒ (sin A)² = \(\frac{\sec ^{2} A-1}{\sec ^{2} A}\)

⇒ sin A = ± \(\sqrt{\frac{\sec ^{2} A-1}{\sec ^{2} A}}\)

[Reject – ve sign for acute angle A]
⇒ sin A = ± \(\sqrt{\frac{\sec ^{2} A-1}{\sec ^{2} A}}\)
cos A = \(\frac{1}{\sec A}\)
1 + tan² A = sec² A
tan² A = sec² A – 1
(tan A)² = sec² A – 1
⇒ tan A = ± \(\sqrt{\sec ^{2} A-1}\)
[Reject – ve sign for acute angle A]
i.e., tan A = \(\sqrt{\sec ^{2} A-1}\)
cosec A = \(\frac{1}{\sin A}=\frac{1}{\sqrt{\sec ^{2} A-1}}\)

= \(\frac{\sec A}{\sqrt{\sec ^{2} A-1}}\)

cot A = \(\frac{1}{\tan A}=\frac{1}{\sqrt{\sec ^{2} A-1}}\).

Question 3.
Evaluate:
(i) \(\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}\)

(ii) sin 25° cos 65° + cos 25° sin 65°.
Solution:
(i) \(\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}\)

= \(\frac{\left\{\sin \left(90^{\circ}-27^{\circ}\right)+\sin ^{2} 27^{\circ}\right\}}{\cos ^{2} 17^{\circ}+\left\{\cos \left(90^{\circ}-17^{\circ}\right)\right\}^{2}}\)
[∵ sin(90 – θ) = cos θ and cos (90 – θ) = sin θ]

= \(\frac{\left\{\cos 27^{\circ}\right\}^{2}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\left\{\sin 17^{\circ}\right\}^{2}}\)

= \(\frac{\cos ^{2} 27^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\sin ^{2} 17^{\circ}}\)
= \(\frac{1}{1}\) = 1.

(ii) sin 25° cos 65° + cos 25° sin 65°
= sin 25° × cos (90° – 25°)
+ cos 25° × sin (90° – 25°)
[∵ cos (90° – θ) = sin θ
sin(90° – θ) = cos θ].
= sin 25° × sin 25° + cos 25° × cos 25°
= sin² 25° + cos² 25° = 1.

Question 4.
Choose the correct option. Justify your choice:
(i) 9 sec² A – 9 tan² A =
(A) 1
(B) 9
(C) 8
(D) 0.

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =
(A) θ
(B) 1
(C) 2
(D) – 1.

(iii) (sec A + tan A) (1 – sin A) =
(A) sec A
(B) sin A
(C) cosec A
(D) cos A.

(iv) \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\) =
(A) sec²A
(B) – 1
(C) cot² A
(D) tan² A.

Solution:
(i) Consider, 9 sec² A – 9 tan² A
= 9 (sec² A – tan² A)
= 9 × 1 = 9.
Option (B) is correct.

(ii) Consider, (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
= \(\left\{1+\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right\} \times\left\{1+\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta}\right\}\)

= \(\left\{\frac{\cos \theta+\sin \theta+1}{\cos \theta}\right\} \times\left\{\frac{\sin \theta+\cos \theta+1}{\sin \theta}\right\}\)

= \(\begin{array}{r}
\{(\cos \theta+\sin \theta)+1\} \\
\times\{(\cos \theta+\sin \theta)-1\} \\
\hline \cos \theta \times \sin \theta
\end{array}\)

= \(\frac{(\cos \theta+\sin \theta)^{2}-(1)^{2}}{\cos \theta \times \sin \theta}\)

[∵ (a + b) (a – b) = a² – b²]

= \(\frac{\cos ^{2} \theta+\sin ^{2} \theta+2 \cos \theta \sin \theta-1}{\cos \theta \times \sin \theta}\)

= \(\frac{1+2 \cos \theta \sin \theta-1}{\cos \theta \sin \theta}\) = 2.

Option (C) is correct.

(iii) Consider, (sec A + tan A) (1 – sin A)
= \(\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)\) × (1 – sin A)

= \(\frac{(1+\sin A)}{\cos A}\) × (1 – sin A)

= \(\frac{(1+\sin A)(1-\sin A)}{\cos A}\)

= \(\frac{(1)^{2}-(\sin A)^{2}}{\cos A}=\frac{1-\sin ^{2} A}{\cos A}=\frac{\cos ^{2} A}{\cos A}\)
[∵ cos² A = 1 – sin² A]
= cos A.
Option (D) is correct.

(iv) Consider, \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\)

= tan² A.
Option (D) is correct.

Question 5.
Prove the following Identities, where the angles involved are acute angles for which the expressions are defined.
(i) (cosec θ – cot θ) = \(\frac{1-\cos \theta}{1+\cos \theta}\)

(ii) \(\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}\) = 2 sec A.

(iii) \(\frac{\tan \theta}{1+\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\) = 1 + sec θ cosec θ
[Hint : Write the expression in terms of sin θ and cos θ]

(iv) \(\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}\)
[Hint: Simplify L.H.S. and R.H.S. separately]

(v) \(\frac{\cos A-\sin A+1}{\cos A+\sin A-1}\) using the identity cosec² A = 1 + cot² A.

(vi) \(\sqrt{\frac{1+\sin A}{1-\sin A}}\) = sec A + tan A

(vii) \(\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}\) = tan θ

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A.

(ix) (cosec A – sin A) (sec A – cos A) = \(\frac{1}{\tan A+\cot A}\)
[Hint : Simplify L.H.S. and R.H.S. separately]

(x) \(\left(\frac{1+\tan ^{2} A}{1+\cot A^{2}}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}\) = tan² A.

Solution:
(i) L.H.S. = (cosec θ – cot θ)²
= \(\left\{\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right\}^{2}\)

= \(\left(\frac{1-\cos \theta}{\sin \theta}\right)^{2}=\frac{(1-\cos \theta)^{2}}{\sin ^{2} \theta}\)
Using identity, sin² θ + cos² θ = 1
⇒ sin² θ = 1 – cos² θ
= \(\frac{(1-\cos \theta)^{2}}{1-\cos ^{2} \theta}\)
= \(\)
[∵ a² – b² = (a + b) (a – b)]

= \(\)

∴ L.H.S. = R.H.S.
Hence, (cosec θ – cot θ)² = \(\frac{1-\cos \theta}{1+\cos \theta}\)

(ii) L.H.S. = \(\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}\)

= \(\frac{2}{\cos A}\) = cos A
∴L.H.S. = R.H.S.
Hence, \(\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}\) = 2 sec A.

(iii) L.H.S. = \(\frac{\tan \theta}{1+\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\) = 1 + sec θ cosec θ

= \(\frac{\left(\frac{\sin \theta}{\cos \theta}\right)}{\left(1-\frac{\cos \theta}{\sin \theta}\right)}+\frac{\left(\frac{\cos \theta}{\sin \theta}\right)}{\left(1-\frac{\sin \theta}{\cos \theta}\right)}\)

= \(\frac{1+\sin \theta \cos \theta}{\cos \theta \sin \theta}=\frac{1}{\cos \theta \sin \theta}+1\)

= 1 + \(\left(\frac{1}{\cos \theta}\right)\left(\frac{1}{\sin \theta}\right)\) = 1 + sec θ cosec θ
∴L.H.S. = R.H.S.
Hence, \(\frac{\tan \theta}{1+\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\) = 1 + sec θ cosec θ

(iv) L.H.S. = \(\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}\)
= \(\frac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}\)
= 1 + cos A …………….(1)
R.H.S = \(\frac{\sin ^{2} A}{1-\cos A}\)
(∵ 1 – cos² A = sin² A.)
= \(\frac{1-\cos ^{2} A}{1-\cos A}\)

= \(\frac{(1+\cos A)(1-\cos A)}{(1-\cos A)}\)

= 1 + cos A. …………….(2)
From (1) and (2) it is clear that
∴ L.H.S. = R.H.S.
Hence, \(\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}\).

(v) L.H.S. = \(\frac{\cos A-\sin A+1}{\cos A+\sin A-1}\) using the identity cosec² A

= cosec A + cot A
= R.H.S
∴ L.H.S. = R.H.S.
Hence, \(\frac{\cos A-\sin A+1}{\cos A+\sin A-1}\) using the identity cosec² A = 1 + cot² A.

(vi) L.H.S. = \(\sqrt{\frac{1+\sin A}{1-\sin A}}\)

= \(\sqrt{\frac{(1+\sin A)(1+\sin A)}{(1-\sin A)(1+\sin A)}}\)

= \(\sqrt{\frac{(1+\sin A)^{2}}{(1)^{2}-(\sin A)^{2}}}\)

= \(\sqrt{\frac{(1+\sin A)^{2}}{1-\sin ^{2} A}}=\sqrt{\frac{(1+\sin A)^{2}}{\cos ^{2} A}}\)

= \(\frac{1+\sin A}{\cos A}=\frac{1}{\cos A}+\frac{\sin A}{\cos A}\)
= sec A + tan A
∴ L.H.S. = R.H.S.
Hence, \(\sqrt{\frac{1+\sin A}{1-\sin A}}\) = sec A + tan A.

(vii) L.H.S. = \(\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}\)

∴ L.H.S. = R.H.S.
Hence, \(\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}\) = tan θ

(viii) L.H.S. = (sin A + cosec A)² + (cos A + sec A)²
= (sin² A + cosec² A + 2 sin A × cosec A) + {cos² A + sec² A
+ 2 cos A × sec A)
= [sin² A + co² A + 2sin A × \(\frac{1}{\sin A}\)] + [cos² A + sec² A + 2 cosA × \(\frac{1}{\cos A}\)]
= (sin² A + cosec² A + 2) + (cos² A + sec² A + 2)
= 2 + 2 + (sin² A + cos² A) + sec² A + cosec² A
= 2 + 2 + 1 + 1 + tan² A + 1 + cot² A
= 7 tan² A + cot² A
∴ L.H.S. = R.H.S.
Hence, (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A.

(ix) L.H.S. = (cosec A – sin A) (sec A – cos A)

From (1) and (2), it is clear that
L.H.S. = R.H.S.
Hence, (cosec A – sin A) (sec A – cos A) = \(\frac{1}{\tan A+\cot A}\)

(x) \(\left(\frac{1+\tan ^{2} A}{1+\cot A^{2}}\right)\)
(∵ 1 + tan² A = sec² A
and 1 + cot² A = cosec² A)

From (1) and (2), it is clear that
LH.S. = R.H.S.
Hence, \(\left(\frac{1+\tan ^{2} A}{1+\cot A^{2}}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}\) = tan² A.