Bhagya

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 10 Visualising Solid Shapes Ex 10.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.2

1. Look at the given map of a city:

Answer the following:
(a) Colour the map as follows: Blue – Water, Red – Fire Station, Orange – Library, Yellow – Schools, Green – Park, Pink – College, Purple – Hospital, Brown – Cemetery.
(b) Mark a Green ‘X’ at the intersection of Road ‘C’ and Nehru Road, Green *Y’ at the intersection of Gandhi Road and Road A.
(c) In red, draw a short street route from Library to the bus depot.
(d) Which is further east, the city park or the market?
(e) Which is further south, the primary school or the Sr. Secondary School?
Solution:
[Note: Enjoy (a) to (c) by doing oneself. (d) city park (e) Sr. Secondary School]

2. Draw a map of your classroom using proper scale and symbols for different objects.

3. Draw a map of your school compound using proper scale and symbols for various features like playground, main building, garden, etc.

4. Draw a map giving instructions to your friend so that she reaches your house without any difficulty.

[Note: Enjoy while doing 2, 3, 4 by yourself.]

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 10 Visualising Solid Shapes Ex 10.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.1

1. For each of the given solid, the two views are given. Match for each solid the corresponding top and front views: The first one is done for you.


Solution:
(a) → (iii) → (iv)
(b) → (i) → (v), (c) → (iv) → (ii)
(d) → (v) → (iii), (e) → (ii) → (i)

2. For each of the given solid, the three views are given. Identify for each solid the corresponding top, front and side views.


Solution:
(a)

Object	(i)	(ii)	(iii)
View	Front view	Side view	Top view

(b)

Object	(i)	(ii)	(iii)
View	Side view	Front view	Top view

(c)

Object	(i)	(ii)	(iii)
View	Front view	Side view	Top view

(d)

Object	(i)	(ii)	(iii)
View	Front view	Side view	Top view

3. For each given solid, identify the top view, front view and side view:


Solution:
(a)

Object	(i)	(ii)	(iii)
View	Top view	Front view	Side view

(b)

Object	(i)	(ii)	(iii)
View	Side view	Front view	Top view

(c)

Object	(i)	(ii)	(iii)
View	Top view	Side view	Front view

(d)

Object	(i)	(ii)	(iii)
View	Side view	Front view	Top view

(e)

Object	(i)	(ii)	(iii)
View	Front view	Top view	Side view

4. Draw the front view, side view and top view of the given objects:

Solution:

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 11 Mensuration InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 11 Mensuration InText Questions

From Textbook : [Textbook Page No. 170]

Can you write an expression for the perimeter of each of the above shapes?
Solution:
1. Rectangle → a × b
2. Square → a × a
3. Triangle → \(\frac {1}{2}\) × b × h
4. Parallelogram → b × h
5. Circle → πb²

Yes, perimeter of above shapes are as follows:
1. Perimeter of rectangle = 2 (length + breadth)
2. Perimeter of square = 4 × (side)
3. Perimeter of triangle = sum of lengths of three sides
4. Perimeter of parallelogram = 2 × (sum of adjacent sides)
5. Perimeter (circumference) of circle = 2πr

Try These : [Textbook Page No. 170]

(a) Match the following figures with their respective areas in the box.

Solution:

Area of square
= l × l = 7 × 7

(b) Write the perimeter of each shape.
Solution:
Perimeter of given shapes:
1. Perimeter of this figure can’t be found as breadth is not given [7 cm is the height, not breadth].

2. Perimeter of semicircle = πr + 2r
= \(\frac {22}{7}\) × 7 + 2 × 7
= 22 + 14
= 36 cm

3. Perimeter of triangle = sum of lengths of three sides
= 14 + 11 + 19 = 34cm

4. Perimeter of rectangle = 2(l + b)
= 2 (14 + 7)
= 2 × 21
= 42cm

5. Perimeter of square = 4l
= 4 × 7 = 28cm

Try These : [Textbook Page No. 172]

1. Nazma’s sister also has a trapezium-shaped plot. Divide it into three parts as shown in figure. Show that the area of trapezium WXYZ = \(\frac {a + b}{2}\).

Solution:
Area of ∆ PWZ = \(\frac {1}{2}\) × base × height
= \(\frac {1}{2}\) × c × h
= \(\frac {1}{2}\)ch
Area of rectangle PQYZ = length × breadth
= b × h = bh
Area of ∆ QXY = \(\frac {1}{2}\) × base × height
= \(\frac {1}{2}\) × d × h
= \(\frac {1}{2}\)dh
Now, Area of trapezium WXYZ
Area of ∆ PWZ + Area of rectangle PQYZ + Area of ∆ QXY
= \(\frac {1}{2}\) ch + bh + \(\frac {1}{2}\)dh
= \(\frac {1}{2}\) ch + \(\frac {1}{2}\)dh + bh
= \(\frac {1}{2}\) (c + d)h + bh
= \(\frac {1}{2}\)(a – b)h + bh (∵ c + d = a – b)
= [\(\frac {1}{2}\)(a – b) + b]h
= [\(\frac {a – b}{2}\) + b] h
= \(\frac {a-b+2b}{2}\) h
= \(\frac {h (a + b)}{2}\)
Thus, area of trapezium WXYZ = \(\frac{h(a+b)}{2}\)

2. If h = 10 cm, c = 6 cm, b = 12 cm, d = 4 cm, find the values of each of its parts separetely and add to find the area WXYZ. Verify it by putting the values of h, a and b in the expression \(\frac{h(a+b)}{2}\).
Solution:
h = 10 cm, c = 6 cm, b = 12 cm, d = 4 cm.
Area of ∆ PWZ = \(\frac {1}{2}\) × c × h
= (\(\frac {1}{2}\) × 6 × 10)cm²
= 30 cm²
Area of ∆ QXY = \(\frac {1}{2}\) × d × h
= (\(\frac {1}{2}\) × 4 × 10)cm²
= 20 cm²
Area of rectangle PQYZ
= length × breadth
= 12 × 10
= 120 cm²
∴ Area of trapezium WXYZ = Area of ∆ PWZ + Area of ∆ QXY + Area of rectangle PQYZ
= (30 + 20 + 120) cm²
= 170 cm²
Now, Area of trapezium WXYZ
= \(\frac{h(a+b)}{2}\)
= \(\frac{10(22+12)}{2}\)
= \(\frac{10(34)}{2}\)
= 170cm²
∵ a = c + b + d
= 6 + 12 + 4
= 22 cm
Thus, area of trapezium verified.

Try These : [Textbook Page No. 173]

1. Find the area of the following trapeziums:

Solution:
(i) Area of given trapezium = \(\frac {1}{2}\) × (9 + 7) × 3
= \(\frac {1}{2}\) × 16 × 3 cm² = 24 cm²
(ii) Area of given trapezium = \(\frac {1}{2}\) × (10 + 5) × 6
= \(\frac {1}{2}\) × 15 × 6 cm² = 45 cm²

Try These : [Textbook Page No. 174]

1. We know that parallelogram is also a quadrilateral.

Let us also split such a quadrilateral into two triangles, find their areas and hence that of the parallelogram. Does this agree with the formula that you know already?
Solution:
Let XYZW be a given quadrilateral, which is a parallelogram.

Join the diagonal WY of the parallelogram XYZW. It divides the parallelogram into two triangles ∆WXY and ∆YZW.
Then, area of parallelogram WXYZ
= Area of ∆ WXY + Area of ∆ YZW
= (\(\frac {1}{2}\) × xy × h) + (\(\frac {1}{2}\) × wz × h)
= (\(\frac {1}{2}\) × b × h) + (\(\frac {1}{2}\) × b × h)
= \(\frac {1}{2}\) bh + \(\frac {1}{2}\) bh
= bh sq units
Area of parallelogram = base × height
= b × h
= bh sq units
As we have studied that, a parallelogram is a special case of a trapezium, where parallel sides are equal.
∴ Area of trapezium XYZW
= \(\frac {1}{2}\) × (b + b) × h
= (\(\frac {1}{2}\) × 2b × h) sq. units
= bh sq units
Thus, the above relation prooves the formula which already we know.

Think, Discuss and Write: [Textbook Page No. 175]

1. A parallelogram is divided into two congruent triangles by drawing a diagonal across it. Can we divide a trapezium into two congruent triangles?
Solution:
No, by dividing trapezium by diagonal two congruent triangles cannot be obtained.

Let us understand from given figures. Here, by drawing diagonals of a quadrilateral ABCD (which is a trapezium) congruent triangles are not obtained.

Try These : [Textbook Page No. 175]

1. Find the area of these quadrilaterals:

Solution:
(i) h₁ = 3 cm, h₂ = 5 cm
d = length of diagonal AC = 6 cm
∴ Area of quadrilateral ABCD
= \(\frac {1}{2}\)d(h₁ + h₂)
= \(\frac {1}{2}\) × 6 × (3 + 5)
= 3 × 8 cm²
= 24 cm²

(ii) d₁ = 7 cm, d₂ = 6 cm
∴ Area of rhombus ABCD
= \(\frac {1}{2}\) × d₁ × d₂
= \(\frac {1}{2}\) × 7 × 6
= 7 × 3 cm²
= 21 cm²

(iii) [Note : Here, given figure is a parallelogram. It’s diagonals divides it into two congruent triangles. From this figure, we can see that base of the triangle is 8 cm and height is 2 cm.]
∴ Area of parallelogram
= 2 (area of ∆ ADC)
= 2 × (\(\frac {1}{2}\) × b × h)
= 2 × (\(\frac {1}{2}\) × 8 × 2) cm²
= 2 × 8
= 16 cm²

Try These : [Textbook Page No. 176]

(i) Divide the following polygons into parts (triangles and trapezium) to find out its area.

Solution:
(a)

Let’s draw perpendicular on diagonal \(\overline{\mathrm{FI}}\).
Here, GA ⊥ FI, EB ⊥ FI and HC ⊥ FI are drawn.
Area of polygon EFGHI = Area of ∆ GFA + Area of the trapezium ACHG + Area of ∆ HCI + Area of ∆ BIE + Area of ∆ FBE
= [\(\frac {1}{2}\) × FA × GA] + [\(\frac {1}{2}\) (AG + CH) × AC] + [\(\frac {1}{2}\) × CI × HC] + [\(\frac {1}{2}\) × BI × BE] + [\(\frac {1}{2}\) × FB × BE]
[Note : We can also use Area of ∆ EFI in place of Area of ∆ BIE + Area of ∆ FBE]

(b)

Let’s draw perpendicular on diagonal \(\overline{\mathrm{NQ}}\).
\(\overline{\mathrm{OE}}\) ⊥ \(\overline{\mathrm{NQ}}\), \(\overline{\mathrm{MF}}\) ⊥ \(\overline{\mathrm{NQ}}\), \(\overline{\mathrm{PG}}\) ⊥ \(\overline{\mathrm{NQ}}\) and \(\overline{\mathrm{RH}}\) ⊥ \(\overline{\mathrm{NQ}}\) are drawn.
Area of polygon MNOPQR = Area of ∆ NEO + Area of trapezium EGPO + Area of ∆ GQP + Area of ∆ HQR + Area of trapezium MRHF + Area of ∆ NFM
= [\(\frac {22}{7}\) × NE × OE] + [\(\frac {22}{7}\) × (OE + PG) × EG] + [\(\frac {22}{7}\) × GQ × PG] + [\(\frac {22}{7}\) × HQ × HR] + [\(\frac {22}{7}\) × (FM + HR) × FH] + [\(\frac {22}{7}\) × NF × FM]

(ii) Polygon ABODE is divided into parts as shown in figure.

Find its area if AD = 8 cm, AH = 6 cm,
AG = 4 cm, AF = 3 cm and perpendiculars BF = 2 cm, CH = 3 cm, EG = 2.5 cm.
Area of polygon ABODE = area of ∆ AFB + ….
Area of ∆ AFB = \(\frac {1}{2}\) × AF × BF
= \(\frac {1}{2}\) × 3 × 2 = ………..
Area of trapezium FBCH
= FH × \(\frac{(BF + CH)}{2}\)
= 3 × \(\frac{(2+3)}{2}\) [FH = AH – AF]
Area of ∆ CHD = \(\frac {1}{2}\) × HD × CH= …………;
Area of ∆ ADE = \(\frac {1}{2}\) × AD × GE = ………….
So the area of polygon ABCDE = …………
Solution:

Here, AD = 8 cm,
AH = 6 cm,
HD = 2 cm,
AG = 4 cm,
GD = 4 cm,
AF = 3 cm and
GF = 1 cm
Area of polygon ABCDE = Area of ∆ AFB + Area of trapezium FBCH + Area of ∆ CHD + Area of ∆ ADE
Now,
Area of ∆ AFB = \(\frac {1}{2}\) × AF × BF
= \(\frac {1}{2}\) × 3 × 2 = 3 cm2 … (i)
Area of trapezium FBCH
= \(\frac {1}{2}\) × (BF × CH) × FH
= \(\frac {1}{2}\) × (2 + 3) × 3 [∵ FH = AH – AF]
= \(\frac {1}{2}\) × 5 × 3 = \(\frac {15}{2}\)
= 7.5 cm² … (ii)
Area of ∆ CHD = \(\frac {1}{2}\) × HD × CH
= \(\frac {1}{2}\) × (AD – AH) × CH
[∵ HD = AD – AH]
= \(\frac {1}{2}\) × (8 – 6) × 3
= \(\frac {1}{2}\) × 2 × 3
= 3 cm² … (iii)
Area of ∆ ADE = \(\frac {1}{2}\) × AD × GE
= \(\frac {1}{2}\) × 8 × 2.5
= 4 × 2.5
= 10 cm² ………. (iv )
∴ Area of polygon ABCDE = Area of [(i) + (ii) + (iii) + (iv)]
= (3 + 7.5 + 3 + 10) cm²
= 23.5 cm²

(iii) Find the area of polygon MNOPQR if MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm, MA = 2 cm NA, OC, QD and RB are perpendiculars to diagonal MP.

Solution:

Here,
MP = 9 cm,
MD = 7 cm
∴ DP = MP – MD
= (9 – 7) cm
= 2 cm
MC = 6 cm ∴ CP = 3 cm
MB = 4 cm ∴ BP = 5 cm
MA = 2 cm ∴ AC = MC – MA
= (6 – 2) cm = 4 cm
MB + BD = MD
∴ BD = (7 – 4) cm
= 3 cm
Area of ∆ MAN = \(\frac {1}{2}\) × MA × AN
= \(\frac {1}{2}\) × 2 × 2.5
= 2.5 cm² ……. (i)
Area of trapezium ACON
= \(\frac {1}{2}\) × (AN + OC) × AC
= \(\frac {1}{2}\) × (2.5 + 3) × 4
= \(\frac {1}{2}\) × 4 × 5.5
= 2 × 5.5
= 11 cm² … (ii)
Area of ∆ CPO = \(\frac {1}{2}\) × CP × CO
= \(\frac {1}{2}\) × 3 × 3
= \(\frac {9}{2}\) = 4.5 cm² … (iii)
Area of ∆ MBR = \(\frac {1}{2}\) × MB × BR
= \(\frac {1}{2}\) × 4 × 2.5
= 2 × 2.5
= 5 cm² … (iv )
Area of trapezium BRQD
= \(\frac {1}{2}\) × (BR + DQ) × BD
= \(\frac {1}{2}\) × (2.5 + 2) × 3
= \(\frac {1}{2}\) × 4.5 × 3
= 6.75 cm² ………. (v)
Area of ∆ DQP = \(\frac {1}{2}\) × DP × DQ
= \(\frac {1}{2}\) × 2 × 2
= 2 cm² … ( vi )
∴ Area of polygon MNOPQR = Area of [(i) + (ii) + (iii) + (iv) + (v) + (vi)]
= (2.5 + 11 + 4.5 + 5 + 6.75 + 2) cm²
= 31.75 cm²

Think, Discuss and Write : [Textbook Page No. 180]

1. Why is it incorrect to call the solid shown here a cylinder?
Solution:

It is incorrect to call given solid as a cylinder because, a cylinder has two identical circular faces, parallel to each other. The radii of both the faces are the same.

Try These :[Textbook Page No. 181]

1. Find the total surface area of the following cuboids:

Solution:
(i) Here, length (i) = 6 cm,
breadth (b) = 4 cm and
height (h) = 2 cm.
Total surface area of a cuboid = 2 (lb + bh + hl)
= 2(6 × 4 + 4 × 2 + 2 × 6)
= 2(24 + 8 + 12)
= 2 (44)
= 88 cm²

(ii) Here, length (l) = 4 cm,
breadth (b) = 4 cm and
height (h) = 10 cm.
Total surface area of a cuboid = 2 (lb + bh + hl)
= 2 (4 × 4 + 4 × 10 + 10 × 4)
= 2 (16 + 40 + 40)
= 2(96) = 192 cm²

Think, Discuss and Write :[Textbook Page No. 181]

1. Can we say that the total surface area of cuboid = lateral surface area + 2 × area of base?
Solution:
Yes, the total surface area of cuboid = lateral surface area + 2 × area of base [Note: Area of base and area of top of a cuboid, both are same.]

2. If we interchange the lengths of the base and the height of a cuboid to get another cuboid, will its lateral surface area change?

Solution:
(a) Lateral surface area of cuboid (i)
= 2(l + b) × h
(b) Lateral surface area of cuboid (ii)
= 2 (h + b) × l
Thus, it is clear that by interchanging the lengths of the base and the height of a cuboid, its lateral surface area will change.

Try These : [Textbook Page No. 182]

1. Find the surface area of cube A and lateral surface area of cube B.

Solution:
For cube A:
Side = 10 cm
Surface area of cube A = 6 × (side)²
= 6 × (10)²
= 6 × 100
= 600 cm²

For cube B:
Side = 8 cm
Lateral surface area of a cube B
= 4 × (side)²
= 4 × (8)²
= 4 × 64
= 256 cm²

Think, Discuss and Write : [Textbook Page No. 183]

Question (i)
Two cubes each with side b are joined to form a cuboid. What is the surface area of this cuboid? Is it 12b²? Is the surface area of cuboid formed by joining three such cubes, 18b²? Why?

Solution:
(a) By joining two cubes end-to-end with side b, a cuboid shape formed.
For cuboid:
length = b + b = 2b, breadth = b and height = b
Total surface area of this cuboid = 2 (lb + bh + lh)
= 2 [(2b × b) + (b × b) + (2b × b)]
= 2 (2b² + b² + 2b²)
= 2 (5b²)
= 10b²
So surface area of the cuboid formed by joining two cubes is not equal to 12b²), it is 10b²).

(b) By joining three cubes end-to-end with side b, a cuboid shape forms.
For cuboid:
length = b + b + b = 3b, breadth = b and height = b
Total surface area of this cuboid = 2 (lb + bh + lh)
= 2 [(3b × b) + (b × b) + (3b × b)]
= 2 (3b² + b² + 3b²)
= 2(7b²) = 14b²
So the surface area of cuboid formed by joining three such cubes is not equal to 18b², it is 14b².

(ii) How will you arrange 12 cubes of equal length to form a cuboid of smallest surface area?
Solution:
Let us arrange 12 cubes of equal length say b to form a cuboid in different situations.
(a)

Here, length = 12b, breadth = b and height = b
Total surface area of a cuboid = 2 (lb + bh + lh)
= 2 [(12b × b) + (b × b) + (12b × b)]
= 2 (12b² + b² + 12b²)
= 2 (25b²) – 50b²

(b)

Here, length = 3b, breadth = 2b and height = 2b
Total surface area of a cuboid
= 2 (lb+ bh+ lh)
= 2 [(3b × 2b) + (2b × 2b) + (2b × 3b)]
= 2 (6b² + 4b² + 6b²)
= 2 (16b²) = 32b²

(c)

Here, length = 6b, breadth = 2b and height = b
Total surface area of a cuboid
= 2 (lb + bh + lh)
= 2 [(6b × 2b) + (2b × b) + (b × 6b)]
= 2 (12b² + 2b² + 6b²)
= 2 (20b²)
= 40b²
Hence, from above results we can conclude that, if we arrange 12 cubes of equal length according to situation (b), we get smallest surface area.

(iii) After the surface area of a cube is painted, the cube is cut into 64 smaller cubes of same dimensions. How many have no face painted? 1 face painted? 2 faces painted? 3 faces painted?
Solution:
1. 8 cubes, which have no face painted. (∵ Middle 4 × 2)
2. 24 cubes, which have 1 face painted. (∵ on each surface 4 × 6)
3. 24 cubes, which have 2 faces painted. (∵ on each surface 4 × 6)
4. 8 cubes, which have 3 faces painted. (∵ on each surface 2 × 4)

Try These : [Textbook Page No. 184]

1. Find total surface area of the following cylinders:

Solution:
(i) Radius of cylinder r = 14 cm
Height of cylinder h = 8 cm
Total surface area of a cylinder = 2πr (r + h)
= 2 × \(\frac {22}{7}\) × 14(14 + 8)
= 2 × 22 × 2 × 22
= 44 × 44
= 1936 cm²

(ii) Radius of cylinder r = \(\frac{\text { diameter }}{2}=\frac{2}{2}\) = 14 cm
Height of cylinder h = 2 cm
Total surface area of a cylinder = 2πr (r + h)
= 2 × \(\frac {22}{7}\) × 1(1 + 2)
= 2 × \(\frac {22}{7}\) × 1 × 3
= \(\frac {132}{7}\)
= 18\(\frac {6}{7}\) m²

Think, Discuss and Write: [Textbook Page No. 184]

1. Note that lateral surface area of a cylinder is the circumference of base × height of cylinder. Can we write lateral surface area of a cuboid as perimeter of base × height of cuboid?
Solution:

Let l be the height, b be the breadth and h be the height of cuboid.
∴ Lateral surface area of a cuboid = Area of 4 walls of the cuboid
= (l × h) + (l × h) + (b × h) + (b × h)
= 2 lh + 2 bh
= 2(1 + b) × h
= Perimeter of base × height
Yes, we can write lateral surface area of a cuboid as perimeter of base × height of cuboid.

Try These : [Textbook Page No. 188]

1. Find the volume of the following cuboids:

Solution:
(i) For given cuboid:
length (l) = 8 cm, breadth (b) = 3 cm s and height (h) = 2 cm
Volume of a cuboid
= Area of base × height
= (l × b) × h
= (8 × 3) × 2
= 24 × 2 = 48 cm³
OR
Volume of cuboid = l × b × h
= (8 × 3 × 2) cm³
= 48 cm³

(ii) Area of base of cuboid = 24 m³
height (h) = 3 cm = \(\frac {3}{100}\) m
Volume of cuboid
= Area of base × height
= 24 × \(\frac {3}{100}\) = 0.72 m³

Try These: [Textbook Page No. 189]

1. Find the volume of the following cubes
(a) With a side 4 cm
Solution:
For given cube : side = 4 cm
∴ Volume of the cube = (side)³
= (4)³ = 4 × 4 × 4
= 64 cm³

(b) With a side 1.5m
Solution:
For given cube : side = 1.5 m
∴ Volume of the cube = (side)³
= (1.5)³ = 1.5 × 1.5 × 1.5
= \(\frac{15}{10} \times \frac{15}{10} \times \frac{15}{10}=\frac{3375}{1000}\)
= 3.375 m³

Think, Discuss and Write: [Textbook Page No. 189]

1. A company sells biscuits. For packing purpose they are using cuboidal boxes:
box A → 3 cm × 8 cm × 20 cm,
box B → 4 cm × 12 cm × 10 cm. What size of the box will be economical for the company? Why? Can you suggest any other size (dimensions) which has the same volume but is more economical than these?
Solution:
For box A:
Given dimension = 3 cm × 8 cm × 20 cm
length = 20 cm, breadth = 8 cm and height = 3 cm
∴ Volume of the box A = l × b × h
= 20 × 8 × 3
= 480 cm³
Total surface area of the box A
= 2 (lb + bh + lh)
= 2 [(20 × 8) + (8 × 3) + (20 × 3)]
= 2 (160 + 24 + 60 )
= 2 (244)
= 488 cm³

For box B:
Given dimension = 4 cm × 12 cm × 10 cm
length = 12 cm, breadth = 10 cm and height = 4 cm
∴ Volume of the box B = l × b × h
= 12 × 10 × 4
= 480 cm³
Total surface area of the box B
= 2 (lb + bh + lh)
= 2 [(12 × 10) + (10 × 4) + (12 × 4)]
= 2(120 + 40 + 48 )
= 2(208)
= 416 cm³
From above results, we can conclude that both boxes have same volume, but surface area of box B is less than that of box A.
∴ Box B is more economical than box A.
Now, let another box of size be 8 cm × 6 cm × 10 cm.
∴ length = 8 cm, breadth = 6 cm and height =10 cm
Volume = l × b × h
= 8 × 6 × 10 = 480 cm³
It’s surface area = 2 (lb + bh + Ih)
= 2 [(8 × 6) + (6 × 10) + (8 × 10)]
= 2 (48 + 60 + 80)
= 2 (188) = 376 cm²
Surface area of this box is less than that of box B.
∴ This box is more economical for company.

Try These: [Textbook Page No. 189]

1. Find the volume of the following cylinders:

Solution:
(i) For given cylinder :
radius (r) = 7 cm, height (h) = 10 cm
Volume of the cylinder = πr²h
= \(\frac {22}{7}\) × 7² × 10
= \(\frac {22}{7}\) × 7 × 7 × 10
= 22 × 70
= 1540cm³

(ii) For given cylinder:
base area 250 m², height (h) = 2m
Volume of the cylinder
= base area × height
= 250 × 2
= 500 m³

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 6 Squares and Square Roots Ex 6.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.4

1. Find the square root of each of the following numbers by Division method.

Question (i).
2304
Solution:

Question (ii).
4489
Solution:

Question (iii).
3481
Solution:

Question (iv).
529
Solution:

Question (v).
3249
Solution:

Question (vi).
1369
Solution:

Question (vii).
5776
Solution:

Question (viii).
7921
Solution:

Question (ix).
576
Solution:

Question (x).
1024
Solution:

Question (xi).
3136
Solution:

Question (xii).
900
Solution:

2. Find the number of digits in the square root of each of the following numbers (without any calculation).

Question (i).
64
Solution:
Here, number of digits, n = 2
(Which is an even number.)
∴ Number of digits in the square root of 64 = \(\frac{n}{2}=\frac{2}{2}\) = 1

Question (ii).
144
Solution:
Here, number of digits, n = 3
(Which is an odd number.)
∴ Number of digits in the square root of 144 = \(\frac{n+1}{2}=\frac{3+1}{2}\)
= \(\frac {4}{2}\)
= 2

Question (iii).
4489
Solution:
Here, number of digits, n = 4
(Which is an even number.)
∴ Number of digits in the square root of 4489 = \(\frac{n}{2}=\frac{4}{2}\)
= 2

Question (iv).
27225
Solution:
Here, number of digits, n = 5
(Which is an odd number.)
∴ Number of digits in the square root of 27225 = \(\frac{n+1}{2}=\frac{5+1}{2}\)
= \(\frac {6}{2}\)
= 3

Question (v).
390625
Solution:
Here, number of digits, n = 6
(Which is an even number.)
∴ Number of digits in the square root of 390625 = \(\frac{n}{2}=\frac{6}{2}\)
= 3

3. Find the square root of the following decimal numbers.

Question (i).
2.56
Solution:

Here, number of decimal places are two.
∴ The number of decimal places in square root should be one.

Question (ii).
7.29
Solution:

Here, number of decimal places are two.
∴ The number of decimal places in square root should be one.

Question (iii).
51.84
Solution:

Here, number of 51.84 decimal places are two.
∴ The number of decimal places in square root should be one.

Question (iv).
42.25
Solution:

Question (v).
31.36
Solution:

4. Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.

Question (i).
402
Solution:

Here, the remainder is 2. It shows that 20² is less than 402 by 2.
So, to get a perfect square, 2 must be subtracted from given number.
∴ Required perfect square number = 402 – 2 = 400
\(\sqrt{400}\) = 20

Question (ii).
1989
Solution:

Here, the remainder is 53. It shows that 44² is less than 1989 by 53.
So, to get a perfect square, 53 must be subtracted from the given number.
∴ Required perfect square number = 1989 – 53 = 1936
\(\sqrt{1936}\) = 44

Question (iii).
3250
Solution:

Here, the remainder is 1. It shows that 57² is less them 3250 by 1.
So, to get a perfect square, 1 must be subtracted from the given number.
∴ Required perfect square number = 3250 – 1 = 3249
\(\sqrt{3249}\) = 57

Question (iv).
825
Solution:

Here, the remainder is 41. It shows that 28² is less than 825 by 41.
So, to get a perfect square, 41 must be subtracted from the given number.
∴ Required perfect square number = 825 – 41 = 784
\(\sqrt{784}\) = 28

Question (v).
4000
Solution:

Here, the remainder is 31. It shows that 63² is less than 4000 by 31.
So, to get a perfect square, 31 must be subtracted from the given number.
∴ Required perfect square number = 4000 – 31 = 3969
\(\sqrt{3969}\) = 63

5. Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.

Question (i).
525
Solution:

Here, the remainder is 41.
525 > 22²,
and the number next to 22 is 23.
23² = 529.
∴ The required number to be added
= 23² – 525
= 529 – 525 = 4
Now, 525 + 4 = 529
\(\sqrt{529}\) = 23
Thus, 4 is the least number which must be added to 525 to get a perfect square.

Question (ii).
1750
Solution:

Here, the remainder is 69.
1750 > 41²,
and the number next to 41 is 42.
42² = 1764.
∴ The required number to be added = 42² – 1750
= 1764 – 1750
= 14
Now, 1750 + 14 = 1764
\(\sqrt{1764}\) = 42
Thus, 14 is the least number which must be added to 1750 to get a perfect square.

Question (iii).
252
Solution:

Here, the remainder is 27.
252 > 15²,
and the number next to 15 is 16.
16² = 256.
∴ The required number to be added = 16² – 252
= 256 – 252
= 4
Now, 252 + 4 = 256
\(\sqrt{256}\) = 16
Thus, 4 is the least number which must be added to 252 to get a perfect square.

Question (iv).
1825
Solution:

Here, the remainder is 61.
1825 > 42²,
and the number next to 42 is 43.
43² = 1849.
∴ The required number to be added 43² – 1825
= 1849 – 1825
= 24
Now, 1825 + 24 = 1849
\(\sqrt{1849}\) = 43
Thus, 24 is the least number which must be added to 1825 to get a perfect square.

Question (v).
6412
Solution:

Here, the remainder is 12.
6412 > 80²,
and the number next to 80 is 81.
81² = 6561.
∴ The required number to be added = 81² – 6412
= 6561 – 6412
= 149
Now, 6412 + 149 = 6561
\(\sqrt{6561}\) = 81
Thus, 149 is the least number which must be added to 6412 to get a perfect square.

6. Find the length of the side of a square whose area is 441 m².
Solution:
Let the side of a square be x m.
Area of a square = x × x = x²
Area of a square = 441 (given)

∴x² = 441
∴ x = \(\sqrt{441}\)
∴ x = 21
Thus, the length of the side of the square is 21 m.

7. In a right triangle ABC, ∠B = 90° :
(a) If AB = 6 cm, BC = 8 cm, find AC.
(b) If AC = 13 cm, BC = 5 cm, find AB.
Solution:
[Note: In a right triangle, the longest side is called the hypotenuse. The square of the hypotenuse is equal to the sum of the squares of the remaining two sides.]
Let us use this theorem here.
(a) Here, ∠B = 90°, AB = 6 cm, BC = 8 cm
In ΔABC, AC is hypotenuse.

AC² = AB² + BC²
= (6)² + (8)²
= 36 + 64
=100
∴ AC = \(\sqrt{100}\)
= 10cm

(b) Here, ∠B= 90°, AC = 13cm, BC = 5cm
In ΔABC, AC Is hypotenuse.


AC² = AB² + BC²
∴ (13)² = AB² + (5)²
∴ 169 = AB² + 25
∴ AB² = 169 – 25
= 144
∴ AB = \(\sqrt{144}\)
= 12 cm

8. A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.
Solution:
Total number of plants = 1000
The number of plants in a row = The number of plants in a column
Let the number of plants planted in a row be x.
So, the number of plants planted in a column is x.
∴ Total plants = x × x = x²
∴ x² > 1000
∴ x > \(\sqrt{1000}\)
Here, the remainder is 39.
(31)² < 1000
The next square number would be 32.
32² = 1024
∴ The number of plants required to be added = 1024 – 1000
= 24
Thus, 24 more plants needed.

9. There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement?
Solution:
Total number of children in a school = 500
The number of rows = The number of columns
Let the number of children in a row be x.
So, the number of children in a column is x.
Total number of children = x × x = x².
∴ x² < 500
∴ x < \(\sqrt{500}\)
Here, the remainder is 16.
500 > 22²
∴ 500 > 484
500 – 484 = 16
Thus, 16 children would be left out in this arrangement.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 6 Squares and Square Roots Ex 6.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.3

1. What could be the possible ‘one’s’ digits of the square root of each of the following numbers ?

Question (i).
9801
Solution:
The possible digit at one’s place of the square root of 9801 can be either 1 or 9.
(∵ 1 × 1 = 1 and 9 × 9 = 81)

Question (ii).
99856
Solution:
The possible digit at one’s place of the square root of 99856 can be either 4 or 6.
(∵ 4 × 4= 16 and 6 × 6 = 36)

Question (iii).
998001
Solution:
The possible digit at one’s place of the square root of 998001 can be either 1 or 9.
(∵ 1 × 1 = 1 and 9 × 9 = 81)

Question (iv).
657666025
Solution:
The possible digit at one’s place of the square root of 657666025 can be 5.
(∵ 5 × 5 = 25)

2. Without doing any calculation, find the numbers which are surely not perfect squares:
[Note : The ending digit of perfect square is 0, 1, 4, 5, 6 or 9.
∴ A number having end digit 2, 3, 7 or 8 can never be a perfect square.

Question (i).
153
Solution:
153
Here, the end digit is 3.
∴ 153 cannot be a perfect square.

Question (ii).
257
Solution:
257
Here, the end digit is 7.
∴ 257 cannot be a perfect square.

Question (iii).
408
Solution:
408
Here, the end digit is 8.
∴ 408 cannot be a perfect square.

Question (iv).
441
Solution:
441
Here, the end digit is 1.
∴ 441 can be a perfect square.

3. Find the square roots of 100 and 169 by the method of repeated subtraction.

Question (i).
100
Solution:
100 – 1 = 99   99 – 3 = 96
96 – 5 = 91   91 – 7 = 84
84 – 9 = 75   75 – 11 = 64
64 – 13 = 51   51 – 15 = 36
36 – 17 = 19   19 – 19 = 0
∴ 100 is a perfect square.
∴ \(\sqrt{100}\) = 10

Question (ii).
169
Solution:
169 – 1 = 168   168 – 3 = 165
165-5 = 160   160 – 7 = 153
153-9 = 144   144 – 11 = 133
133-13 = 120   120 – 15 = 105
105-17 = 88   88 – 19 = 69
69-21 = 48   48 – 23 = 25
25 – 25 = 0
∴ 169 is a perfect square.
∴ \(\sqrt{169}\) = 13

4. Find the square roots of the following numbers by the Prime Factorisation Method:

Question (i).
729
Solution:
729
\(\begin{array}{l|r}
3 & 729 \\
\hline 3 & 243 \\
\hline 3 & 81 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
729 = 3 × 3 × 3 × 3 × 3 × 3
= 3² × 3² × 3²
∴ \(\sqrt{729}\) = 3 × 3 × 3
= 27

Question (ii).
400
Solution:
400
\(\begin{array}{l|r}
2 & 400 \\
\hline 2 & 200 \\
\hline 2 & 100 \\
\hline 2 & 50 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
400 = 2 × 2 × 2 × 2 × 5 × 5
= 2² × 2² × 5²
∴ \(\sqrt{400}\) = 2 × 2 × 5
= 20

Question (iii).
1764
Solution:
1764
\(\begin{array}{l|r}
2 & 1764 \\
\hline 2 & 882 \\
\hline 3 & 441 \\
\hline 3 & 147 \\
\hline 7 & 49 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
1764 = 2 × 2 × 3 × 3 × 7 × 7
= 2² × 3² × 7²
∴ \(\sqrt{1764}\) = 2 × 3 × 7
= 42

Question (iv).
4096
Solution:
4096
\(\begin{array}{l|r}
2 & 4096 \\
\hline 2 & 2048 \\
\hline 2 & 1024 \\
\hline 2 & 512 \\
\hline 2 & 256 \\
\hline 2 & 128 \\
\hline 2 & 64 \\
\hline 2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}\)
4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2² × 2² × 2² × 2² × 2² × 2²
∴ \(\sqrt{4096}\) = 2 × 2 × 2 × 2 × 2 × 2
= 64

Question (v).
7744
Solution:
7744
\(\begin{array}{r|r}
2 & 7744 \\
\hline 2 & 3872 \\
\hline 2 & 1936 \\
\hline 2 & 968 \\
\hline 2 & 484 \\
\hline 2 & 242 \\
\hline 11 & 121 \\
\hline 11 & 11 \\
\hline & 1
\end{array}\)
7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11
= 2² × 2² × 2² × 11²
∴ \(\sqrt{7744}\) = 2 × 2 × 2 × 11
= 88

Question (vi).
9604
Solution:
9604
\(\begin{array}{l|r}
2 & 9604 \\
\hline 2 & 4802 \\
\hline 7 & 2401 \\
\hline 7 & 343 \\
\hline 7 & 49 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
9604 = 2 × 2 × 7 × 7 × 7 × 7
= 2² × 7² × 7²
∴ \(\sqrt{9604}\) =2 × 7 × 7
= 98

Question (vii).
5929
Solution:
5929
\(\begin{array}{r|r}
7 & 5929 \\
\hline 7 & 847 \\
\hline 11 & 121 \\
\hline 11 & 11 \\
\hline & 1
\end{array}\)
5929 = 7 × 7 × 11 × 11
= 7² × 11²
∴ \(\sqrt{5929}\) = 7 × 11
= 77

Question (viii).
9216
Solution:
9216
\(\begin{array}{r|r}
2 & 9216 \\
\hline 2 & 4608 \\
\hline 2 & 2304 \\
\hline 2 & 1152 \\
\hline 2 & 576 \\
\hline 2 & 288 \\
\hline 2 & 144 \\
\hline 2 & 72 \\
\hline 2 & 36 \\
\hline 2 & 18 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
9216 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
= 2² × 2² × 2² × 2² × 2² × 3²
∴ \(\sqrt{9216}\) = 2 × 2 × 2 × 2 × 2 × 3
= 96

Question (ix).
529
Solution:
529
\(\begin{array}{l|r}
23 & 529 \\
\hline 23 & 23 \\
\hline & 1
\end{array}\)
529 = 23 × 23
= 23²
∴ \(\sqrt{529}\) = 23

Question (x).
8100
Solution:
8100
\(\begin{array}{l|r}
2 & 8100 \\
\hline 2 & 4050 \\
\hline 3 & 2025 \\
\hline 3 & 675 \\
\hline 3 & 225 \\
\hline 3 & 75 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5
= 2² × 3² × 3² × 5²
∴ \(\sqrt{8100}\) = 2 × 3 × 3 × 5
= 90

5. For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained:

Question (i).
252
Solution:
252
\(\begin{array}{l|r}
2 & 252 \\
\hline 2 & 126 \\
\hline 3 & 63 \\
\hline 3 & 21 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
252 = 2 × 2 × 3 × 3 × 7
The prime factor 7 is unpaired.
∴ [252] × 7 = [2 × 2 × 3 × 3 × 7] × 7
1764 = 2 × 2 × 3 × 3 × 7 × 7
= 2² × 3² × 7²
∴ \(\sqrt{1764}\) = 2 × 3 × 7 = 42
Thus, 252 should be multiplied by smallest whole number 7 to get a perfect square.

Question (ii).
180
Solution:
180
\(\begin{array}{l|r}
2 & 180 \\
\hline 2 & 90 \\
\hline 3 & 45 \\
\hline 3 & 15 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
180 = 2 × 2 × 3 × 3 × 5
Here, the prime factor 5 is unpaired.
∴ [180] × 5 = [2 × 2 × 3 × 3 × 5] × 5
∴ 900 = 2 × 2 × 3 × 3 × 5 × 5
= 2² × 3² × 5²
∴ \(\sqrt{900}\) = 2 × 3 × 5 = 30
Thus, 180 should be multiplied by smallest whole number 5 to get a perfect square.

Question (iii).
1008
Solution:
1008
\(\begin{array}{l|r}
2 & 1008 \\
\hline 2 & 504 \\
\hline 2 & 252 \\
\hline 2 & 126 \\
\hline 3 & 63 \\
\hline 3 & 21 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7
Here, the prime factor 7 is unpaired.
∴ [1008] × 7 = [2 × 2 × 2 × 2 × 3 × 3 × 7] × 7
∴ 7056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
= 2² × 2² × 3² × 7²
∴ \(\sqrt{7056}\) = 2 × 2 × 3 × 7 = 84
Thus, 1008 should be multiplied by smallest whole number 7 to get a perfect square.

Question (iv).
2028
Solution:
2028
\(\begin{array}{r|r}
2 & 2028 \\
\hline 2 & 1014 \\
\hline 3 & 507 \\
\hline 13 & 169 \\
\hline 13 & 13 \\
\hline & 1
\end{array}\)
2028 = 2 × 2 × 3 × 13 × 13
Here, the prime factor 3 is unpaired.
∴ [2028] × 3 = [2 × 2 × 3 × 13 × 13] × 3
∴ 6084 = 2 × 2 × 3 × 3 × 13 × 13
= 2² × 3² × 13²
∴ \(\sqrt{6084}\) = 2 × 3 × 13 = 78
Thus, 2028 should be multiplied by smallest whole number 3 to get a perfect square.

Question (v).
1458
Solution:
1458
\(\begin{array}{l|r}
2 & 1458 \\
\hline 3 & 729 \\
\hline 3 & 243 \\
\hline 3 & 81 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3
Here, the prime factor 2 is unpaired.
∴ [1458] × 2 = [2 × 3 × 3 × 3 × 3 × 3 × 3] × 2
∴ 2916 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
= 2² × 3² × 3² × 3²
∴ \(\sqrt{2916}\) = 2 × 3 × 3 × 3 = 54
Thus, 1458 should be multiplied by smallest whole number 2 to get a perfect square.

Question (vi).
768
Solution:
768
\(\begin{array}{l|r}
2 & 768 \\
\hline 2 & 384 \\
\hline 2 & 192 \\
\hline 2 & 96 \\
\hline 2 & 48 \\
\hline 2 & 24 \\
\hline 2 & 12 \\
\hline 2 & 6 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
Here, the prime factor 3 is unpaired.
∴ [768] × 3 = [2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3] × 3
∴ 2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
= 2² × 2² × 2² × 2² × 3²
∴ \(\sqrt{2304}\) = 2 × 2 × 2 × 2 × 3 = 48
Thus, 768 should be multiplied by smallest whole number 3 to get a perfect square.

6. For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained:

Question (i).
252
Solution:
252
\(\begin{array}{r|r}
2 & 252 \\
\hline 2 & 126 \\
\hline 3 & 63 \\
\hline 3 & 21 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
252 = 2 × 2 × 3 × 3 × 7
Here, the prime factor 7 is unpaired. So, given number should be divided by 7.
∴ [252] ÷ 7 = [2 × 2 × 3 × 3 × 7] ÷ 7
∴ 36 = 2 × 2 × 3 × 3
= 2² × 3²
∴ \(\sqrt{36}\) = 2 × 3 = 6
Thus, 252 should be divided by smallest whole number 7 to get a perfect square number.

Question (ii).
2925
Solution:
2925
\(\begin{array}{r|r}
3 & 2925 \\
\hline 3 & 975 \\
\hline 5 & 325 \\
\hline 5 & 65 \\
\hline 13 & 13 \\
\hline & 1
\end{array}\)
2925 = 3 × 3 × 5 × 5 × 13
Here, the prime factor 13 is unpaired. So, given number should be divided by 13.
∴ [2925] ÷ 13 = [3 × 3 × 5 × 5 × 13] ÷ 13
∴ 225 = 3 × 3 × 5 × 5
= 3² × 5²
∴ \(\sqrt{225}\) = 3 × 5 = 15
Thus, 2925 should be divided by smallest whole number 13 to get a perfect square number.

Question (iii).
396
Solution:
396
\(\begin{array}{r|r}
2 & 396 \\
\hline 2 & 198 \\
\hline 3 & 99 \\
\hline 3 & 33 \\
\hline 11 & 11 \\
\hline & 1
\end{array}\)
396 = 2 × 2 × 3 × 3 × 11
Here, the prime factor 11 is unpaired. So, given number should be divided by 11.
∴ [396] ÷ 11 = [2 × 2 × 3 × 3 × 11] ÷ 11
∴ 36 = 2 × 2 × 3 × 3
= 2² × 3²
∴ \(\sqrt{36}\) = 2 × 3 = 6
Thus, 396 should be divided by smallest whole number 11 to get a perfect square number.

Question (iv).
2645
Solution:
2645
\(\begin{array}{r|r}
5 & 2645 \\
\hline 23 & 529 \\
\hline 23 & 23 \\
\hline & 1
\end{array}\)
2645 = 5 × 23 × 23
Here, the prime factor 5 is unpaired. So, given number should be divided by 5.
∴ [2645] ÷ 5 = [5 × 23 × 23] ÷ 5
∴ 529 = 23 × 23 = 23²
∴ \(\sqrt{529}\) = 23
Thus, 2645 should be divided by smallest whole number 5 to get a perfect square number.

Question (v).
2800
Solution:
2800
\(\begin{array}{l|r}
2 & 2800 \\
\hline 2 & 1400 \\
\hline 2 & 700 \\
\hline 2 & 350 \\
\hline 5 & 175 \\
\hline 5 & 35 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7
Here, the prime number 7 is unpaired. So, given number should be divided by 7.
∴ [2800] ÷ 7 = [2 × 2 × 2 × 2 × 5 × 5 × 7] ÷ 7
∴ 400 = 2 × 2 × 2 × 2 × 5 × 5
= 2² × 2² × 5²
∴ \(\sqrt{400}\) = 2 × 2 × 5 = 20
Thus, 2800 should be divided by smallest whole number 7 to get a perfect square number.

Question (vi).
1620
Solution:
1620
\(\begin{array}{l|r}
2 & 1620 \\
\hline 2 & 810 \\
\hline 3 & 405 \\
\hline 3 & 135 \\
\hline 3 & 45 \\
\hline 3 & 15 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5
Here, the prime factor 5 is unpaired. So, given number should be divided by 5.
∴ [1620] ÷ 5 = [2 × 2 × 3 × 3 × 3 × 3 × 5] ÷ 5
∴ 324 = 2 × 2 × 3 × 3 × 3 × 3
= 2² × 3² × 3²
∴ \(\sqrt{324}\) = 2 × 3 × 3 = 18
Thus, 1620 should be divided by 5 to get a perfect square number.

7. The students of Class VIII of a school donated ₹ 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.
Solution:
Let the number of students be x.
Amount each student donated = Number of students in the class.
So, amount donated by each student = ₹ x
Total amount donated by class = ₹ x × x = x²
\(\begin{array}{l|r}
7 & 2401 \\
\hline 7 & 343 \\
\hline 7 & 49 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
∴ x² = 2401
∴ \(\sqrt{x^{2}}=\sqrt{2401}\)
∴ x = \(\sqrt{7 \times 7 \times 7 \times 7}\)
= \(\sqrt{7^{2} \times 7^{2}}\)
∴ x = 7 × 7 = 49
Hence, number of students in the class is 49.

8. 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
Solution:
Let the number of rows be x.
Number of rows = Number of plants in each row
So, number of plants in a row = x
∴ Number of plants to be planted in a garden = x × x = x²
\(\begin{array}{l|r}
3 & 2025 \\
\hline 3 & 675 \\
\hline 3 & 225 \\
\hline 3 & 75 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
∴ x² = 2025
∴ \(\sqrt{x^{2}}=\sqrt{2025}\)
∴ x = \(\sqrt{3 \times 3 \times 3 \times 3 \times 5 \times 5}\)
∴ \(\sqrt{3^{2} \times 3^{2} \times 5^{2}}\)
∴ x = 3 × 3 × 5 = 45
Hence, the number of rows is 45 and the number of plants in each row is 45.

9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.
Solution :
[Note: LCM is the number, which is divided by all factors of it without leaving remainder. ]
Here, the smallest square number divisible by each one of 4, 9 and 10 is equal to some multiple of the LCM of 4, 9 and 10.
\(\begin{array}{l|ll}
2 & 4, & 9, & 10 \\
\hline 2 & 2, & 9, & 5 \\
\hline 3 & 1, & 9, & 5 \\
\hline 3 & 1, & 3, & 5 \\
\hline 5 & 1, & 1, & 5 \\
\hline & 1, & 1, & 1
\end{array}\)
LCM of 4, 9 and 10 = 2 × 2 × 3 × 3 × 5 = 180
The prime factor 5 is unpaired.
So, 180 must be multiplied by 5.
∴ [180] × 5 = [2 × 2 × 3 × 3 × 5] × 5
∴ 900 = 2² × 3² × 5²
Hence, 900 is the required perfect square number.

10. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.
Solution:
[Note: LCM is the number, which is divided by all factors of it without leaving remainder.]
Here, the smallest square number divisible by each of 8, 15 and 20 is equal to some multiple of the LCM of 8, 15 and 20.
\(\begin{array}{r|rrr}
2 & 8, & 15, & 20 \\
\hline 2 & 4, & 15, & 10 \\
\hline 2 & 2, & 15, & 5 \\
\hline 3 & 1, & 15, & 5 \\
\hline 5 & 1, & 5, & 5 \\
\hline & 1, & 1, & 1
\end{array}\)
LCM of 8, 15 and 20 = 2 × 2 × 2 × 3 × 5 = 120
The prime factors 2, 3 and 5 are unpaired.
So, 120 should be multiplied by 2 × 3 × 5 = 30.
∴ [120] × 2 × 3 × 5 = [2 × 2 × 2 × 3 × 5] × 2 × 3 × 5
∴ 3600 = 2 × 2 × 2 × 3 × 5 × 2 × 3 × 5
∴ 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
Hence, 3600 is the required perfect square number.

Punjab State Board PSEB 6th Class English Book Solutions English Grammar Noun Exercise Questions and Answers, Notes.

PSEB 6th Class English Grammar Noun

A noun is a name of a person, place or thing; as-
किसी व्यक्ति स्थान अथवा वस्तु के नाम को अंग्रेजी में noun कहते हैं।

India. Mohan, taxi, class, toy, boy, table, etc.
Look at these sentences:

1. Geeta went to Patiala.
2. The fox is looking at the grapes.
3. The boys are playing football.

The underlined words in the above sentences are all nouns because they are the names of some person, place, animal or thing.

There are four kinds of noun:

1. Common Noun
2. Proper Noun
3. Abstract Noun
4. Collective Noun.

1. Common Nouns

A Common Noun is the name of every person, place or thing of the same class; as- pen, cow. bird, man, animal, bridge.
Look at these sentences:

1. The boys are playing.
2. These mangoes are pulpy.
3. The birds build nests.

The underlined words in the given sentences are Common Nouns because they are common to every person, place or thing.

Exercises (Solved)

I. Underline the Common Nouns in the following sentences:

1. Keep the books on the table.
2. The shops are closed today.
3. The tiger lives in the forest.
4. The farmer bought a tractor.
5. This building has many offices.
6. There is a dairy near our house.
7. Ail birds do not build their nests.
8. A fish lives in water and not on land.

Hints:
1. books, table
2. shops
3. tiger, forest
4. farmer, tractor building, offices
6. dairy, house
7. birds, nests
8. fish.

II. Add five Common Nouns in each set:

1. birds : parrot, sparrow, peacock, crow, niehtineale. pigeon.
2, colours : red, white, black, green, yellow, orange
3. games : hockey, football, volleyball, cricket, basketball, baseball.
4. animals : dog, camel, cow, sheep, buffalo, goat.
5. vegetables : potato, tomato, cabbage, radish, brinial, pumpkin.
6. fruits : mango, grape, apple, banana, papaya, pear.
7. In a school : library, science room, assembly hall office, staff-room, class room
8. In a house : kitchen, bathroom, dining room, store, bed-room, guest room.

2. Proper Nouns

A Proper Noun is the name of some particular (विशेष) person.
Delhi, Kolkata, Beas, Kama!

Look at these sentences:

1. Moti loves to play.
2. My brother lives in Amritsar.
3. J.C. Bose was a great scientist.
4. The Shan-e-Puniab has left just now.

The underlined w’ords in the above sentences are proper nouns because they are the names of particular persons, places or things.

Note that-
A Proper Noun always begins with a capital letter.

Proper Nouns include (शामिल हैं) the names of people, countries, cities, villages, rivers, ships, streets, buildings, mountains, seas, months of the year, days of the week, festivals, etc.

Exercises (Solved)

I. Underline the Proper Nouns in the following sentences

1. We named the cat Silky.
2. Kabir was a great saint.
3. We visited the. Taj in Agra.
4. Delhi is the capital of India.
5. I have never been to Mumbai.
6. Misha and Monu went to Delhi.
7. Do you know Sunny and Chinkv ?
8. We visited the Golden Temple on Sunday.

Hints:
1. Silky
2. Kabir
3. Taj. Agra
4. Delhi. India
5. Mumbai
6. Misha, Manu, Delhi
7. Sunny, Chinky
8. Golden Temple.

II. Rewrite each Proper Noun correctly in these sentences:

1. Have you visited the taj mahal ?
Have you visited the Taj Mahal ?
2. lam going to ropar on Monday.
3. The amritsar mail goes to kolkata.
4. muslims go to mosques on fridays,
5. black beauty is the story of a horse.
6. Where were the last Olympics held ?
7. bill clinton was the president of ameriea.
Hints:
2. Ropar, Monday
3. Amritsar Mail, Kolkata
4. Muslims, Mosques, Fridays
5. Black Beauty
6. Olympics
7. Bill Clinton, America.

3. Abstract Nouns

An Abstract Noun is the name of a quality, feeling or state (गुण भाव या स्थिति); as-
goodness, hardness, wisdom, love, hatred, theft, boyhood, slavery, freedom.

Look at these sentences:

1. Fire gives us heat.
2. He had pain in his body.
3. He acted upon my advice.
4. Do you know the depth of this well ?

The underlined words in the given sentences are abstract nouns because they refer to some quality, feeling or state.

The following words are all Abstract Nouns:
theft
peace
poverty
kindness
hope
misery
honesty
darkness
truth
greed
courage
weakness
sleep
sorrow
sickness
childhood
death
hunger
patience
treatment

Exercises (Solved)

I. Underline the Abstract Nouns in the following sentences:

1. Please control your anger.
2. Honesty is the best policy.
3. There was silence all around.
4. We get knowledge from books.
5. There was darkness in the room.
6. What is the height of this building?
7. You should have kindness for the poor.
8. Wars always bring death and destruction.
Hints:
1. anger
2. Honesty, best policy
3. silence
4. knowledge
5. darkness
6. Height
7. kindness
8. death, destruction.

II. Form Abstract Nouns from the given words:

laugh – laughter
hate – hatred
true – truth
treat – treatment
child – childhood
soft – softness
cruel – cruelty
bright – brightness
brave – bravery
strong – strength
punctual – punctuality
dangerous – danger

III. Use any five Abstract Nouns in sentences of your own:

1. She likes the softness of her skin.
2. Always speak the truth.
3. He was rewarded for his bravery.
4. Punctuality is a great virtue.
5. Her life was in danger.

4. Collective Nouns

A Collective Noun is the name of a group of persons, animals or things of the same kind; as-
flock, cattle, class, army, family, committee.

Look at these sentences:

1. Our army won the battle.
2. I have lost my bunch of keys.
3. The cattle are grazing in the field.

The underlined words in the above sentences are collective nouns because they refer to a collection of persons or things of the same kind.
The word army is a collection of soldiers.
The word cattle is a collection of farm animals.
The word bunch is a collection of things tied together.

Learn the following Collective Nouns:
1. a shoal of fish
2. a hive of bees
3. a pride of lions
4. a herd of cattle
5. a flight of stairs
6. a bunch of keys
7. a flock of sheep
8. a crew of sailors
9. a heap of stones
10. a string of pearls
11. a suite of rooms
12. a basket of fruits
13. a gang of thieves
14. a library of books
15. a bundle of sticks
16. a bench of judges
17. a crowd of people
18. a brood of chickens
19. a band of musicians
20. a wardrobe of clothes
21. a regiment of soldiers
22. a fleet of ships or cars
23. a litter of pups / piglets
24. a pack of cards / wolves

Exercises (Solved)

I. Match the Collective Nouns with the given phrases:

1. A collection of pups	Pack
2. A collection of ships	flock
3. A collection of sheep	fleet
4. A collection of books	suite
5. A collection of rooms	litter
6. A collection of wolves	herd
7. A collection of flowers	library
8. A collection of elephants	bouquet

Hints:
1. litter
2. fleet
3. flock
4. library
5. suite
6. pack
7. bouquet
8. herd.

II. Fill in the blanks with suitable Collective Nouns:

1. A filght of stairs.
2. A ………… of fish.
3. A ………… of lions.
4. A ………… of cows.
5. A ………… of cards.
6. A ………… of fruits.
7. A ………… of pearls.
8. A ………… of judges.
9. A ………… of grapes.
10. A ………… of clothes.
11. A ………… of thieves.
12. A ………… of soldiers.

Hints:
2. shoal
3. pride
4. herd
5. pack
6. basket
7. string
8. bench
9. bunch
10. wardrobe
11. gang
12. regiment.

Miscellaneous Exercises (Solved)

I. What is a Noun ?

II. Name the different kinds of Noun.
Give two examples of each.

III. The italicized words in the following sentences are Nouns. Classify these Nouns (Common / Proper /Abstract / Collective):

1. He won much praise.
2. Nitin lives in Mumbai.
3. I saw a flock of sheep.
4. Silver is a white metal.
5. You cannot cheat God.
6. My sweater is made of wool.
7. I bought some new furniture.
8. The old woman was very happy now.
Hints:
1. praise – abstract
2. Mumbai – roper
3. sheep – common
4. silver – common, metal- common
5. God – proper
6. sweater – common, wool – common
7. furniture – collective
8. woman – common.

IV. Choose suitable Nouns to fill in the blanks:

duty, profit, courage, marriage, need, weight, freedom, childhood

1. Be careful about your weight.
2. We want to live in …………..
3. Her ………….. took place last month.
4. It is our………….. to obey our parents.
5. Seema lost her parents in her …………..
6. We helped him when he was in …………..
7. The soldier was rewarded for his …………..
8. Jatin made good ………….. from his business.
Hints:
2. freedom
3. marriage
4. duty
5. childhood
6. need
7. courage
8. profit.

V. Pick out the Nouns in the following sentences and say whether they are Common, Proper, Collective or Abstract:

1. I love music.
2. Meera studies in sixth class.
3. Ludhiana is an industrial city.
4. He bought a doll for his sister.
5. These tables are made of wood.
6. A drowning man catches at a straw.
7. His father left for London yesterday.
8. Mathematics is my favourite subject.
Hints:
1. music – abstract.
2. Meera – proper; class – collective.
3. Ludhiana – proper, city – common.
4. doll – common, sister – common.
5. tables – common, wood – common.
6. man – common, straw – common.
7. father – common, London – proper.
8. Mathematics – collective, subject – common.

VI. Choose a suitable Abstract Noun to match each phrase:

pride, silence, poverty, courage, strength, greatness, innocence, intelligence

1. A quiet room [silence]
2. A clever boy
3. A great king
4. A strong girl
5. A proud child
6. A poor beggar
7. A brave policeman
8. An innocent woman
Hints:
1. silence
2. intelligence
3. greatness
4. strength
5. pride
6. poverty
7. courage
8. innocence.

The Noun – Number

Singular and Plural Nouns
A noun that refers to one thing is said to be Singular; as-
ball, chair, book, town, cow etc.

A noun that refers to more than one thing is said to be Plural; as-
books, balls, chairs, towns, animals, etc.

Now look at these sentences:

1. Rita has three dolls.
2. Reema has a bag of books.
3. All the babies were crying.
4. Joy got a big ball on his birthday.

The underlined nouns in the above sentences are either singular or plural. They tell whether they refer to one or more than one thing.

Forming Plurals of Nouns

1. As a general rule, the plural of a noun is formed by adding -s to a singular noun.

Singular – Plural
cat – cats
cap – caps
ball – balls
flag – flags
doll – dolls
bird – birds
hare – hares
goat – goats
horse – horses
rat – rats
toy – toys
son – sons
owl – owls
lion – lions
page – pages
table – tables
sister – sisters
orange – oranges

2. Nouns ending in -s, -x, -ch, or -sh form their plurals by adding -es.

Singular – Plural
bunch – bunches
brush – brushes
dish – dishes
church – churches
match – matches
fox – foxes
bush – bushes
dress – dresses
gas – gases
class – classes
loss – losses
box – boxes
glass – glasses
tax – taxes

3. Nouns ending in -y (with a consonant before them) form their plural by changing -y to -ies.

Singular – Plural
city – cities
story – stories
fairy – fairies
lady – ladies
pony – ponies
sky – skies
dairy dairies
baby – babies
family – families
puppy – puppies
butterfly – butterflies
country – countries

4. Nouns ending in -y (with a vowel before them), form their plural by taking an -s only.

Singular – Plural
key – keys
valley – valleys
ray – rays
storey – storeys
day – days
holiday – holidays
boy – boys
journey – journeys
play – plays
monkey – monkeys

5. Nouns ending in -f or -fe form their plural by changing -f or -fe to -ves.

Singular – Plural
calf – calves
loaf – loaves
wolf – wolves
shelf – shelves
life – lives
half – halves
knife – knives
thief – thieves

6. Some nouns ending in -/ or -fe form their plural by taking -s only.

Singular – Plural
roof – roofs
safe – safes
proof – proofs
hoof – hoofs
chief – chiefs
dwarf – dwarfs

7. Nouns ending in -o (with a consonant before them), form their plural by taking -es.

Singular – Plural
echo – echoes
negro – negroes
hero – heroes
mango – mangoes
potato – potatoes
volcano – volcanoes
buffalo – buffaloes
mosquito – mosquitoes

Exceptions : The words photo and piano take -s only to form their plural.

8. Nouns ending in -o (with a vowel before them), form their plural by taking -s only.

Singular – Plural
radio – radios
cuckoo – cuckoos
bamboo – bamboos

9. Some nouns have irregular plurals.

Singular – Plural Singular – Plural
man – men
foot – feet
tooth – teeth
goose – geese
ox – oxen
louse – lice
mouse – mice
child – children

10. A compound noun generally forms its plural by adding -s to the principal word.

daughters-in-law	lookers-on	step-daughters
mothers-in-law	step-sons	maid-servants
fathers-in-law	sons-in-law	passers-by

11. The following Compound Nouns take a double plural.

man-servant – men-servants
woman-teacher – women-teachers
woman-servant – women-servants

Exercises (Solved)

I. Give the plural form of:

fly – flies
box – boxes
hero – heroes
roof – roofs
shoe – shoes
shelf – shelves
dwarf – dwarfs
potato – potatoes
pencil – pencils
mouse – mice
life – lives
fish – fishes
foot – feet
child – children
piano – pianos

II. Give the singular form of:

foxes – fox
teeth – tooth
halves – half
armies – army
watches – watch
gases – gas
ladies – lady
mosquitoes – mosquito
oxen – ox
copies – copy
knives – knife
negroes – negro
chimneys – chimney
shoes – shoe
wolves – wolf

III. Rewrite each sentence using the plural form of Nouns:

1. The monkey was in a cage.
The monkeys were in cages.
2. The knife is on the shelf.
3. He put his foot on the bench.
4. The hero in the film acted well.
5. The policeman chased the thief.
6. The woman told the child a story.
7. Sam plucked a leaf from the tree.
8. The maid washed the glass and the dish.
Answer:
2. The knives are on the shelves.
3. He put his feet on the benches.
4. The heroes in the films acted well.
5. The policemen chased the thieves.
6. The women told the children stories.
7. Sam plucked leaves from the trees.
8. The maids washed the glasses and the dishes.

IV. Rewrite each sentence using the singular form of Nouns

1. The oxen are pulling the carts The ox is pulling the cart.
2. Neha heard the cries of wolves.
3. The women rode on the ponies.
4. The loaves are kept in the boxes.
5. The mice were afraid of the geese.
6. The children were bitten by mosquitoes.
7. These stories are about witches and fairies.
8. The men told the ladies stories of Indian heroes.
Answer:
2. Neha heard the cry of a wolf.
3. The woman rode on a pony.
4. The loaf is kept in the box.
5. The mouse was afraid of the goose.
6. The child was bitten by a mosquito.
7. This story is about a witch and a fairy.
8. The man told the lady a story of an Indian hero.

Remember that-
1. Some Nouns have the same form in the plural and the singular; as-
कुछ Nouns एकवचन तथा बहवचन में एक जैसे होते हैं: जैसे-
deer, sheep, fish, dozen, score, hundred, thousand.

The following Nouns have a plural form but always take the singular verb; as-
निम्नलिखित Nouns बहुवचन में होते हैं, परन्तु उनके साथ हमेशा एकवचन क्रिया (verb) लगती है; जैसे-
news, civics, politics, physics, mathematics, means, gallows.

1. This news is true.
2. Physics is a difficult subject.

3. The following Nouns are always used in the plural form and take the plural verb; as-
निम्नलिखित Nouns का प्रयोग हमेशा बहुवचन में किया जाता है और उनके साथ बहुवचन क्रिया लगती है; जैसे-
thanks, scissors, trousers, pants, alms, wages, spectacles, socks.

1. My thanks are to you all.
2. The scissors were blunt.

4. The following Nouns are used only in the singular form and take the singular verb; as-
नीचे लिखे Nouns केवल एकवचन में प्रयोग किये जाते हैं और उनके साथ एकवचन क्रिया लगती है; जैसे-
furniture, scenery, luggage, machinery, advice, bread, hair, business, mischief.

1. This furniture is not for sale.
2. Where is my luggage ?

5. The word ‘hair’ is used in the plural when a definite number of hairs are to be mentioned.
जब बालों का निश्चित संख्या में उल्लेख किया जाए, तो hair शब्द का प्रयोग बहुवचन (hairs) में किया जाता है।
1. There were two hairs in my tea.
2. My aunt has four white hairs on her head.

Miscellaneous Exercises (Solved)

I. Give the plural of the following nouns:
1. ox
2. leaf
3. knife
4. chief
5. tooth
6. fox
7. wife
8. child
9. story.
10. mouse.
Answer:
1. oxen
2. leaves
3. knives
4. chiefs
5. teeth
6. foxes
7. wives
8. children
9. stories
10. mice.

II. Rewrite each sentence with a plural subject:

1. A cow eats grass.
2. The child is playing
3. The army was fighting.
4. A crow is sitting in the tree.
5. The ox is grazing in the field.
6. This road is closed for repairs.
Answer:
1. Cows eat grass.
2. The children are playing.
3. The armies were fighting.
4. Crows are sitting in the tree.
5. The oxen are grazing in the field.
6. These roads are closed for repairs.

III. Fill in the blanks with the correct form of the given words:

1. She has white …………..
2. I have lost my …………..
3. This ………….. is not true.
4. The ………….. were crying.
5. The house has two …………..
6. Your ………….. were not new.
Hints:
1. teeth
2. shoes
3. news
4. babies
5. storeys
6. trousers

IV. Correct the following sentences:

1. Her hairs are black.
2. Your scissor is blunt.
3. Where is my trouser ?
4. Please accept my thank.
5. These furnitures are for sale.
6. We saw many wolfs in the zoo.
Answer:
1. Her hair is black
2. Your scissors are blunt.
3. Where are my trousers ?
4. Please accept my thanks.
5. This furniture is for sale.
6. We saw many wolves in the zoo.

The Noun – Gender

Gender means being a male (नर) or a female (मादा).
On the basis of gender, we can classify nouns into four kinds:

1. Masculine Gender
2. Feminine Gender
3. Common Gender
4. Neuter Gender

1. A noun that refers to a male is said to be of the Masculine (पुरुषवाचक) Gender; as-
horse, boy, man, lion, king, dog.

2. A noun that refers to a female is said to be of the Feminine (स्तरीवाचक) Gender; as-
mare, girl, woman, lioness, queen, bitch.

3. A noun that refers to both a male and a female, is said to be of the Common (सामान्य) Gender; as-
child, baby, parent, cousin, friend, student, thief.

4. A noun that refers to neither a male nor a female, is said to be of the Neuter (नपुंसक) Gender; as-
book, pen, toy, house, table, knife, etc.

Genders

1. Masculine (male)
2. Feminine (female)
3. Common (either sex)
4. Neuter (neither sex)

Change of Gender

We can change the gender of a Noun in different ways; as-
1. By using a different word:

Masculine – Feminine
monk – nun
fox – vixen
father – mother
uncle – aunt
boy – girl
son – daughter
man – woman
nephew – niece
bull – cow
cock – hen
king – queen
brother – sister
husband – wife
sir – madam
gentleman – lady
dog – bitch
horse – mare
bachelor – maid

2. By adding ‘-ess’, ‘-ss’ to the masculine and making some change in it:

Masculine – Feminine
god – goddess
prince – princess
lion – lioness
master – mistress
tiger – tigress
emperor – empress

3. By changing a part of the word:

Masculine – Feminine
bride – bridegroom
granduncle – grandaunt
peacock – peahen
he-goat – she-goat
landlord – landlady
headmaster – headmistress
milkman – milkwoman
father-in-law – mother-in-law
grandfather – grandmother
brother-in-law – sister-in-law

Exercises (Solved)

I. Put each word in the column it belongs to:

van
duke
horse
milkmaid
bull
child
flower
governess
box
book
parent
gentleman
nun
baby
servant
hairdresser
aunt
table
duchess
shopkeeper
road
monk
daughter
policeman
duchess
Answer:
Table

II. Change the Gender of the following:

sir – madam
lion lioness
bull – cow
cock – hen
mare – horse
uncle – aunt
tigress – tiger
peacock – peahen
gentleman – lady
grandfather – grandmother

Miscellaneous Exercises (Solved)

I. Give the opposite Gender of the following:

1. sir
2. aunt
3. mare
4. king
5. lady
6. cock
7. horse
8. tiger
9. wife
10. male
11. lioness
12. mother
Answer:
1. madam
2. uncle
3. horse
4. queen
5. gentleman
6. hen
7. mare
8. tigress
9. husband
10. female
11. lion
12. father.

II. Rewrite each sentence, changing the Gender of Nouns and Pronouns:

1. A cruel man killed the fox.
2. Mr. Sharma is a businessman.
3. The Emperor welcomed the Duke.
4. The dog is barking at the servant.
5. Madam, my aunt wants to see you.
6. His nephew went to Shimla with his son.
7. The headmaster punished the naughty boys.
8. The bride touched the feet of her mother-in-law.
Answer:
1. A cruel woman killed the vixen.
2. Mrs. Sharma is a businesswoman.
3. The Empress welcomed the Duchess.
4. The bitch is barking at the maid.
5. Sir, my uncle wants to see you.
6. Her niece went to Shimla with her daughter.
7. The headmistress punished the naughty girls.
8. The bridegroom touched the feet of his father-in-law.

III. Fill in the blanks with the Feminine gender of the words in italics:

1. We pray to gods and …………….
2. The hotel has a waiter and a …………….
3. The actor married an ……………. in Mumbai.
4. The lion and the ……………. are in their den.
5. The witch changed the prince into a …………….
6. The tiger and the ……………. look after their cubs.
7. The emperor and the ……………. of Japan live in Tokyo.
8. The guests were received by the host and the …………….
Hints:
1. goddesses
2. waitress
3. actress
4. lioness
5. wizard, princess
6. tigress
7. empress
8. hostess.

Punjab State Board PSEB 6th Class English Book Solutions English Grammar Tenses Exercise Questions and Answers, Notes.

PSEB 6th Class English Grammar Tenses

The form of verb that shows the Time or State of an action is called the Tense.
Verb का जो रुप क्रिया के समय अथवा स्थिति के बारे में बताती है उसे Tense कहते है।

We have three main Tenses in English.
They are-

1. The Present Tense वर्तमानकाल
2. The Past Tense भूतकाल
3. The Future Tense. भविष्यत्काल

Look at the following sentences:

1. Tanu is here today.
2. The girls are happy.
3. Tanu was here yesterday.
4. The girls were happy.
5. Tanu will be here tomorrow.
6. The girls will be happy.

Sentences 1 and 2 show the present state.
We can say they are in the Present Tense.

Sentences 3 and 4 show the past state.
We can say they are in the Past Tense.

Sentences 5 and 6 show the future state.
We can say they are in the Future Tense.

Please Note: 1. We use V₁ for Present Tense.
2. We use V₂ for Past Tense.
3. We use will / shall + V₁ for Future Tense.

Forms of Tenses काल के रूप

Each Tense in English has four different forms:

1. Indefinite
2. Continuous
3. Perfect
4. Perfect Continuous

Thus, there are twelve different Tenses.
They are-

1. Present Indefinite	I write poems.
2. Past Indefinite	I wrote poems.
3. Future Indefinite	I shall write poems
4. Present Continuous	I am writing poems.
5. Past Continuous	I was writing poems.
6. Future Continuous	I shall be writing poems.
7. Present Perfect	I have written poems.
8. Past Perfect	I had written poems.
9. Future Perfect	I shall have written poems.
10. Present Perfect Continuous	I have been writing poems.
11. Past Perfect Continuous	I had been writing poems.
12. Future Perfect Continuous	I shall have been writing poems.

1. Present Indefinite Tense

The Present Indefinite Tense is used to express a universal truth (सर्वमान्य सत्य) or habitual action (आदत की क्रिया)”; as-

1. The baby likes bread.
2. We do our duty.
3. Aman speaks the truth.
4. The earth moves round the sun.

The underlined verbs are all in the Present Indefinite Tense.
For Positive Statements in this tense:
We use V₁ for I, You and a Plural subject; as,

1. You learn your lessons.
2. We pray to God daily
3. I buy milk from this dairy.

We use V₁ + s/es, for third person Singular subject; as,

1. Meena tells lies.
2. My sister cooks delicious food.
3. Mr. Singh teaches us English.

Exercises (Solved)

I. Put each sentence into the plural:

1. A cat eats meat.
2. A dog hates a cat.
3. A writer writes a book.
4. An apple grows on a tree.
Answer:
Cats eat meat.
2. Dogs hate cats.
3. Writers write books
4. Apples grow on trees.

II. Put each sentence into the singular:

1. Houses have roofs.
2. Postmen wear caps.
3. They drink tea out of cups.
4. Classrooms have blackboards.
Answer:
1. A house has a roof.
2. A postman wears a cap.
3. He/She drinks tea out of a cup.
4. A classroom has a blackboard.

Framing Negative Questions

We use do not + V₁ for I, you and a Plural subject; as-
1. They do not learn their lessons.
2. You do not do your homework daily.
3. We do not buy milk from this dairy.

We use does not + V₁ for third person Singular subject; as
1. She does not waste her time.
2. Rita does not cook delicious food.
3. Mr. Singh does not teach us English.

Exercise (Solved)

Rewrite each sentence as a Negative:
1. Ram goes home for lunch.
2. I like coffee.
3. She looks beautiful.
4. These boys run fast.
5. We go for a walk daily.
6. You obey your teachers.
7. He takes care of his health.
8. “They take tea in the evening.
Answer:
1. Ram does not go home for lunch.
2. I do not like coffee.
3. She does not look beautiful.
4. These boys do not run fast.
5. We do not go for a walk daily.
6. You do not obey your teachers.
7. He does not take care of his health.
8. They do not take tea in the evening.

Framing Questions

We use the following sentence pattern:
Do + Plural subject + V₁ + Complement ?
Does + third person Singular subject + V₁ + Complement ?

Exercise (Solved)

Rewrite each sentence as a Question:

1. Owls hoot at night.
2. They work on Sundays.
3. Children play on the road.
4. A postman delivers letters.
5. Farmers grow crops for us.
6. Mosquitoes spread Malaria.
7. She helps her mother in the kitchen.
8. Your brother knows many people in this town.
Answer:
1. Do owls hoot at night ?
2. Do they work on Sundays ?
3. Do children play on the road ?
4. Does a postman, deliver letters ?
5. Do farmers grow crops for us ?
6. Do mosquitoes spread Malaria ?
7. Does she help her mother in the kitchen ?
8. Does your brother know many people in this town?

Framing Negative Questions

We can put not before the main verb or in short form after the helping verb; as-

1. Does Rani not tell lies ?
= Doesn’t Rani tell lies ?

2. Do you not take a bath daily ?
= Don’t you take a bath daily ?

Exercise (Solved)

Rewrite each sentence as a Negative Question:

1. Do cows live on grass ?
Do cows not live on grass ?
(or)
Don’t cows live on grass ?

Note : Do not = Don’t
Does not = Doesn’t

2. She does not like coffee.
3. The sun rises in the east.
4. Do they come here daily ?
5. Does Kusha bring flowers ?
6. We do not pluck the flowers.
7. Does Nitin obey his parents ?
8. He does not drive his car very fast.
9. These boys do not respect their teachers.
Answer:
2. Does she not like coffee ?
Or
Doesn’t she like coffee ?
3. Does the sun not rise in the east ?
4. Do they not come here daily ?
Or
Don’t they come here daily ?
5. Does Kusha not bring flowers ?
6. Do we not pluck the flowers ?
7. Does Nitin not obey his parents ?
8. Does he not drive his car very fast ?
9. Do these boys not respect their teachers ?

2. Past Indefinite Tense

Past Indefinite Tense is used to express an action which took place in the past or was completed before the time of speaking; as-
Past Indefinite Tense का प्रयोग बीते समय (भूतकाल) में हुई क्रिया के लिए किया जाता है जो उसके बारे में बात करने से पहले ही पूरी हो गई हो।

1. Simi liked ice cream.
2. Rohan went to the market.

For Positive Statements in this tense, we use V₁ with all subjects (singular or plural); as-

1. He worked hard.
2. We took milk in the morning today.
3. I bought this book last week.

Exercise (Solved)

Rewrite each sentence using the Past form of the given verbs:

1. Rahul (want) a shirt.
2. Deepa (eat) an ice cream.
3. Nancy (wear) simple clothes.
4. Raj (come) to India in March.
5. They (build) a house in Delhi.
6. The boys (laugh) at the beggar.
7. I (go) to the market with my friend.
8. My mother (buy) a new dress for me.
Answer:
1. Rahul wanted a shirt.
2. Deepa ate an ice cream.
3. Nancy wore simple clothes.
4. Raj came to India in March.
5. They built a house in Delhi.
6. The boys laughed at the beggar.
7. I went to the market with my friend.
8. My mother bought a new dress for me.

Negative Statements

We use did not + V₁ for all subjects (singular or plural); as-

1. He did not work hard.
2. We did not take milk in the morning today.
3. I did not buy this book last week.

Please note that did always take V₁ form with it; it never takes V₂ form.

Exercise (Solved)

Rewrite each sentence as a Negative:

1. Misha told the truth.
Misha did not tell the truth.
2. He took my pen.
3. Tony polished his shoes.
4. She cooked food for me.
5. Rohan respected his teachers.
6. They finished their work in time.
7. The naughty boys broke the glass.
8. Ranjan and his friends went for a picnic.
Answer:
2. He did not take my pen.
3. Tony did not polish his shoes.
4. She did not cook food for me.
5. Rohan did not respect his teachers.
6. They did not finish their work in time.
7. The naughty boys did not break the glass.
8. Ranjan and his friends did not go for a picnic.

For Questions

We use the following sentence pattern:

Exercise (Solved)

Rewrite each sentence as a Question:

1. Nancy danced at the party.
Did Nancy dance at the party ?
2. He invited us to dinner.
3. My uncle sent me a gift.
4. Our team won the match.
5. You paid your fees yesterday.
6. We spent our holidays at Shimla.
7. They plucked flowers in the garden.
8. Sonu broke his leg in the accident.
Answer:
2. Did he invite us to dinner ?
3. Did my uncle send me a gift ?
4. Did our team win the match ?
5. Did you pay your fees yesterday ?
6. Did we spend our holidays at Shimla ?
7. Did they pluck flowers in the garden ?
8. Did Sonu break his leg in the accident ?

For Negative Questions

We can put not before the main verb or in short form after the helping verb; as
1. Did she not tell lies ?
= Didn’t she tell lies ?

2. Did you not apply for the post ?
= Didn’t you apply for the post ?

Exercise (Solved)

Rewrite each sentence as a Negative Question:

1. Did the peon ring the bell ?
Did the peon not ring the bell ?
(or)
Didn’t the peon ring the bell ?
2. Did he tell a lie ?
3. We called him a fool.
4. Reeta ate all biscuits.
5. Did she reply your letter ?
6. Your sister painted this picture.
7. Did she finish her work in time ?
8. Did they congratulate you on your success ?
Answer:
2. Did he not tell a lie ?
(or)
Didn’t he tell a lie ?
3. Did we not call him a fool ?
4. Did Reeta not eat all biscuits ?
5. Did she not reply your letter ?
6. Did your sister not paint this picture ?
7. Did she not finish her work in time ?
8. Did they not congratulate you on your success ?

3. Present Continuous Tense

Present Continuous Tense is used to express an action that is going on at the time of speaking; as-

1. Radha is doing her homework.
2. Mona is cooking food in the kitchen.

The italicised words express an action which is going on at present. So we can say these sentences are in the Present Continuous Tense.

Positive Statements (is/am/are + V₁ -ing)

1. I am flying a kite.
2. She is doing her work.
3. They are reading a new lesson.

Negative Statements (is/am/are + not + V₁ -ing)

1. I am not flying a kite.
2. She is not doing her work.
3. They are not reading a new lesson.

Questions
(Helping verb before the Subject)

1. Am I flying a kite ?
2. Is she doing her work ?
3. Are they reading a new lesson ?

Negative Questions

(‘not is used before the main verb or in short form after the helping verb)

1. Am I not flying a kite ?
= An’t I flying a kite ?
(We use an’t in spoken language only.)

2. Is she not doing her work ?
= Isn’t she doing her work ?

3. Are they not reading a new lesson ?
= Aren’t they reading a new lesson ?

Exercises (Solved)

I. Use the Present Continuous Tense to complete each sentence:

1. Mona …………… a test. (take)
2. I ………. my breakfast. (have)
3. The hunter …………. the lion. (kill)
4. The trees …………… their leaves. (shed)
5. The farmers ……….. their fields. (water)
6. The pain in my arm ………… worse. (get)
7. The tailors………… the uniforms. (not make)
Answer:
1. Mona is taking a test.
2. I am having my breakfast.
3. The hunter is killing the lion.
4. The trees are shedding their leaves.
5. The farmers are watering their fields.
6. The pain in my arm is getting worse.
7. The tailors are not making the uniforms.

II. Rewrite each sentence as a question:

1. I am reading a book.
2. She is not doing her work.
3. They are watching a movie.
4. You are not listening to me.
5. We are going for a picnic today.
6. The girls are playing in the park.
7. The boys are not teasing the animals.
Answer:
1. Am I reading a book ?
2. Is she not doing her work ?
3. Are they watching a movie ?
4. Are you not listening to me ?
5. Aren’t we going for a picnic today?
6. Are the girls playing in the park ?
7. Are the boys not teasing the animals ?

4. Past Continuous Tense

The Past Continuous tense is used to express an action which was actually taking place at some particular moment (विशेष क्षण/समय पर) in the past.

Positive Statements
We use was/were + V₁ -ing; as-

1. Harjit was reading a book.
2. They were going to the fair.

Negative Statements
We use was/were + not + V₁ -ing; as-

1. Harjit was not reading a book.
2. They were not going to the fair.

Questions
We put the helping verb before the subject; as-

1. Was Harjit reading a book ?
2. Were they going to the fair ?

Negative Questions
We can put ‘not before the main verb or in short form after the helping verb; as-
1. Was Harjit not reading a book ?
= Wasn’t Harjit reading a book ?

2. Were they not going to the fair ?
= Weren’t they going to the fair ?

Exercises (Solved)

I. Complete each sentence using the Past Continuous Tense:

1. Children ………….. in the bushes. (hide)
2. They ………. through the zoo. (walk).
3. The waiter ……….. the people. (serve)
4. Meera …………. with her friends. (not play)
5. The baby ………. all the morning. (not cry)
6. The dancers ………… on the stage. (not perform)
Answer:
1. Children were hiding in the bushes.
2. They were walking through the zoo.
3. The waiter was serving the people.
4. Meera was not playing with her friends.
5. The baby was not crying all the morning.
6. The dancers were not performing on the stage.

II. Rewrite each sentence as a question:

1. The peon was ringing the bell.
2. We were not going to our village.
3. The boys were wearing red turbans.
4. Hema was not working at that time.
5. The children were playing in the street.
6. The teacher was writing on the blackboard.
7. The little girl was not playing with her doll.
8. Anu and Rosy were not talking to each other.
Answer:
1. Was the peon ringing the bell ?
2. Were we not going to our village ?
3. Were the boys wearing red turbans ?
4. Was Hema not working at that time?
5. Were the children playing in the street ?
6. Was the teacher writing on the blackboard ?
7. Was the little girl not playing with her doll ?
8. Were Anu and Rosy not talking to each other?

Miscellaneous Exercises

I. Use Simple Past form of the given verb to complete each sentence:

1. Did you ………….. this film ? (enjoy)
2. Did Roma …………. this picture ? (paint)
3. Columbus ……….. America in 1492. (discover)
4. She ………. to her village last month. (go)
5. The peon ……….. (not) the bell in time. (ring)
6. The fool didn’t ………. from experience. (learn)
Hints.
1. enjoy
2. paint
3. discovered
4. went
5. did not ring
6. learn.

II. Use Simple Present form of the given verb to complete each sentence:

1. I …….. for a walk daily. (go)
2. The sun …………… in the east. (rise)
3. They ……………. (not) bad workers. (like)
4. Kusha …………. (not) her parents. (obey)
5. Teachers …………… good students. (love)
6. We ………… milk and eggs for breakast.
Hints:
1. go
2. rises
3. do not like
4. does not obey
5. love
6. take

III. Rewrite each sentence in Past Indefinite Tense:

1. The bird flies to its nest.
2. They drink coffee every day.
3. Does he pay his fees regularly ?
4. Do you have milk for breakfast ?
5. Do we not fall ill by over-eating ?
6. You do not finish your work in time.
7. Kusha does not wear simple clothes.
8. Does he not help his friends in need ?
Answer:
1. The bird flew to its nest.
2. They drank coffee every day.
3. Did he pay his fees regularly ?
4. Did you have milk for breakfast ?
5. Did we not fall ill by over-eating ?
6. You did not finish your work in time.
7. Kusha did not wear simple clothes.
8. Did he not help his friends in need ?

IV. Rewrite each sentence in Past Continuous Tense:

1. Isn’t it raining heavily ?
2. We are waiting for the bus.
3. The teacher is teaching the children.
4. I am not living with my aunt these days.
5. They are not going home in the evening.
6. Is the lady knitting a sweater for her son ?
7. Aren’t Anu and Manu playing in the street ?
8. Am I wasting my time in watching Discovery Channel ?
Answer:
1. Wasn’t it raining heavily?
2. We were waiting for the bus.
3. The teacher was teaching the children.
4. I was not living with my aunt these days.
5. They were not going home in the evening.
6. Was the lady knitting a sweater for her son ?
7. Weren’t Anu and Manu playing in the street ?
8. Was I wasting my time in watching Discovery Channel ?

V. Rewrite each sentence in Present Continuous Tense:

1. Do you not speak the truth?
2. The students ask many questions.
3. I sit on the front bench in my class.
4. Does Kamla teach dance and music ?
5. These boys do not respect their elders.
6. She does not play with the poor children.
Answer:
1. Are you not speaking the truth?
2. The students are asking many questions.
3. I am sitting on the front bench in my class.
4. Is Kamla teaching dance and music ?
5. These boys are not respecting their elders.
6. She is not playing with the poor children.

VI. Rewrite each sentence in Past Continuous Tense:

1. The girls did not pluck flowers.
2. Did the peon not ring the bell ?
3. Did Ram break the windowpanes ?
4. My friends talked to me in English.
5. The watchman did not open the gate.
6. He spent all his money in good deeds.
Answer:
1. The girls were not plucking flowers.
2. Was the peon not ringing the bell ?
3. Was Ram breaking the windowpanes.
4. My friends were talking to me in English.
5. The watchman was not opening the gate.
6. He was spending all his money in good deeds.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 6 Squares and Square Roots Ex 6.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.2

1. Find the square of the following numbers:

Question (i).
32
Solution:
Let us use: (a + b)² = a² + 2ab + b²
(32)² = (30 + 2)²
= (30)² + 2 (30) (2) + (2)²
= 900 + 120 + 4
= 1024

Question (ii).
35
Solution:
(35)² = (30 + 5)²
= (30)² + 2 (30) (5) + (5)²
= 900 + 300 + 25
= 1200 + 25
= 1225
Second method :
[Note : The unit digit of 35 is 5.]
(35)² = 3 × (3 + 1) × 100 + 25
= 3 × 4 × 100 + 25
= 1200 + 25
= 1225

Question (iii).
86
Solution:
(86)² = (80 + 6)²
= (80)² + 2 (80) (6) + (6)²
= 6400 + 960 + 36
= 7396

Question (iv).
93
Solution:
(93)² = (90 + 3)²
= (90)² + 2 (90) (3) + (3)²
= 8100 + 540 + 9
= 8649

Question (v).
71
Solution:
(71 )² = (70 + 1)²
= (70)² + 2 (70) (1) + (1)²
= 4900 + 140 + 1
= 5041

Question (vi).
46
Solution:
(46)² = (40 + 6)²
= (40)² + 2 (40) (6) + (6)²
= 1600 + 480 + 36
= 2116

2. Write a Pythagorean triplet whose one member is:

Question (i).
6
Solution:
Here, 2n = 6
∴ n = 3

Now, n² – 1 = 3² – 1
= 9 – 1
= 8
and n² + 1 = 3² + 1
= 9 + 1 = 10
Thus, the required Pythagorean triplet is 6, 8, 10.

Question (ii).
14
Solution:
Here, 2n = 14
∴ n = 7
Now, n² – 1 = 7² – 1 = 49 – 1 = 48
and n² + 1 = 7² + 1 = 49 + 1 = 50
Thus, the required Pythagorean triplet is 14, 48, 50.

Question (iii).
16
Solution:
Here, 2n = 16
∴ n = 8
Now, n² – 1 = 8² – 1 = 64 – 1 = 63
and n² + 1 = 8² + 1 = 64 + 1 = 65
Thus, the required Pythagorean triplet is 16, 63, 65.

Question (iv).
18
Solution:
Here, 2n = 18
∴ n = 9
Now n² – 1 = 9² – 1 = 81 – 1 = 80
and n² + 1 = 92 + 1 = 81 + 1 = 82
Thus, the required Pythagorean triplet is 18, 80, 82.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 6 Squares and Square Roots Ex 6.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1

1. What will be the unit digit of the squares of the following numbers?

Question (i).
81
Solution:
81
Here, the ending digit is 1.
1 × 1 = 1
∴ The unit digit of (81)² will be 1.

Question (ii).
272
Solution:
272
Here, the ending digit is 2.
2 × 2 = 4
∴ The unit digit of (272)² will be 4.

Question (iii).
799
Solution:
799
Here, the ending digit is 9.
9 × 9 = 81
∴ The unit digit of (799)² will be 1.

Question (iv).
3853
Solution:
3853
Here, the ending digit is 3.
3 × 3 = 9
∴ The unit digit of (3853)² will be 9.

Question (v).
1234
Solution:
1234
Here, the ending digit is 4.
4 × 4 = 16
∴ The unit digit of (1234)² will be 6.

Question (vi).
26387
Solution:
26387
Here, the ending digit is 7.
7 × 7 = 49
∴ The unit digit of (26387)² will be 9.

Question (vii).
52698
Solution:
52698
Here, the ending digit is 8.
8 × 8 = 64
∴ The unit digit of (52698)² will be 4.

Question (viii).
99880
Solution:
99880
Here, the ending digit is 0.
0 × 0 = 0
∴ The unit digit of (99880)² will be 0.

Question (ix).
12796
Solution:
12796
Here, the ending digit is 6.
6 × 6 = 36
∴ The unit digit of (12796)² will be 6.

Question (x).
55555
Solution:
55555
Here, the ending digit is 5.
5 × 5 = 25
∴ The unit digit of (55555)² will be 5.

2. The following numbers are obviously not perfect squares. Give reason:

Question (i).
1057
Solution:
1057
Here, the ending digit is 7, which is not one of the digits 0, 1, 4, 5, 6 or 9.
∴ 1057 is not a perfect square.

Question (ii).
23453
Solution:
23453
Here, the ending digit is 3, which is not one of the digits 0, 1, 4, 5, 6 or 9.
∴ 23453 is not a perfect square.

Question (iii).
7928
Solution:
7928
Here, the ending digit is 8, which is not one of the digits 0, 1, 4, 5, 6 or 9.
∴ 7928 is not a perfect square.

Question (iv).
222222
Solution:
222222
Here, the ending digit is 2, which is not one of the digits 0, 1, 4, 5, 6 or 9.
∴ 222222 is not a perfect square.

Question (v).
64000
Solution:
64000
Here, the number of zeros at the end is odd.
∴ 64000 is not a perfect square.

Question (vi).
89722
Solution:
89722
Here, the ending digit is 2, which is not one of the digits 0, 1, 4, 5, 6 or 9.
∴ 89722 is not a perfect square.

Question (vii).
222000
Solution:
222000
Here, the number of zeros at the end is odd.
∴ 222000 is not a perfect square.

Question (viii).
505050
Solution:
505050
Here, the ending digit is 0. (Number of odd zero.)
∴ 505050 is not a perfect square.

3. The squares of which of the following would be odd numbers ?

(i) 431
(ii) 2826
(iii) 7779
(iv) 82004
Solution:
[Note: The square of an odd natural number is odd and that of an even number is an even number.]
(i) 431
This is an odd number.
∴ (431)² is an odd number.

(iii) 7779
This is an odd number.
∴ (7779)² is an odd number.

4. Observe the following pattern and find the missing digits:

Solution :
Observing the above pattern, we can find the missing digits as follows :

5. Observe the following pattern and supply the missing numbers :

Solution:
Observing the above pattern, we can find the missing numbers as follows :

6. Using the given pattern, find the missing numbers:
1² + 2² + 2² = 3²
2² + 3² + 6² = 7²
3² + 4² + 12² = 13²
4² + 5² + …….² = 21²
5² + ……² + 30² = 31²
6² + 7² + ……² = ……²
[To find pattern:
Third number is related to first and second number. How?
Fourth number is related to third number. How?]
Solution:
(a)² + (a + 1)² + {a (a + 1 )}²
= {a (a + 1) + 1}²
The missing numbers are :
1² + 2² + 2² = 3²
2² + 3² + 6² = 7²
3² + 4² + 12² = 13²
4² + 5² + 20² = 21²
5² + 6² + 30² = 31²
6² + 7² + 42² = 43²

7. Without adding, find the sum:

Question (i).
1 + 3 + 5 + 7 + 9
Solution:
The sum of first five odd numbers = 5²
= 5 × 5
= 25

Question (ii).
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
Solution:
The sum of first ten odd numbers = 10²
= 10 × 10
= 100

Question (iii).
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Solution:
The sum of first twelve odd numbers = 12²
= 12 × 12
= 144

8.

Question (i).
Express 49 as the sum of 7 odd numbers.
Solution:
49 = 7² = Sum of first seven odd numbers
∴ 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13

Question (ii).
Express 121 as the sum of 11 odd numbers.
Solution :
121 = 11² = Sum of first eleven odd numbers
∴ 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

9. How many numbers lie between squares of the following numbers?
[Note : Between n² and (n + 1)², there are 2n non-square numbers.]

Question (i).
12 and 13
Solution:
Natural numbers between 12² and 13²
= 2 × 12
= 24. (2n, n = 12)

Question (ii).
25 and 26
Solution:
Natural numbers between 25² and 26²
= 2 × 25
= 50. (2n, n = 25)

Question (iii).
99 and 100
Solution:
Natural numbers between 99² and 100²
= 2 × 99
= 198. (2n, n = 99)

Punjab State Board PSEB 6th Class English Book Solutions English Grammar Pronouns Exercise Questions and Answers, Notes.

PSEB 6th Class English Grammar Pronouns

A pronoun is a word used in place of a noun.
संज्ञा (Noun) के स्थान पर प्रयोग किए जाने वाले शब्द को अंग्रेज़ी में Pronoun कहते है।

The words they, she, he and it are used in place of Nouns. They are Pronouns.
There are three main kinds of Pronoun:

1. Personal Pronouns. (पुरुषवचक सर्वनाम)
2. Demonstrative Pronouns (संकेतवाचक सर्वनाम)
3. Interrogative Pronouns (प्रश्नवाचक सर्वनाम)

I. Personal Pronouns Pronouns used for persons are called Personal Pronouns; as-
1, we, you, he, they, me, our, etc.
There are three kinds of Personal Pronouns.

1. Pronouns of the First Person:

Singular	Plural
I	we
my, mine	our, ours
me	us
myself	ourselves

2. Pronouns of the Second Person:

Singular	Plural
you	you
your, yours	your, yours
yourself	yourselves

3. Pronouns of the Third Person:

Singular	Plural
he, she, it	they
him, her, it	them
his, her, its	their, theirs
himself, herself, itself	themselves

Points to Remember:
याद रखने योग्य बातें:

1. He, his, him, himself are Pronouns of Masculine Gender. (पुरुषवचक लिंग)
2. She, her, hers, herself are Pronouns of Feminine Gender. (स्त्रीरीवाचक लिंग)
3. It, its, itself are Pronouns of Neuter Gender. (नपुंसक लिंग).
   We can use “It for little babies and lifeless things.
4. All Plural Pronouns are Pronouns of Common Gender (सामान्य लिंग). They can be used for both masculine and feminine genders.

Use of Personal Pronouns

1. I, we, he, she and they are used as Subject.

2. Me, us, him, her and them are used as Object.

Subject	Verb	Object
I	him.	know
We	them.	know
He	her.	knows
She	me.	knows
They	us.	know

3. My, mine, our, ours, your, yours, his, her, hers, its and theirs are used to express ownership.

This is my book.	This book is mine.
That is her pen.	That pen is hers.
This is our school.	This school is ours.

4. We never use an apostrophe (‘) with the personal pronouns.

Incorrect	Correct
Your’s sincerely/faithfully	Yours sincerely/faithfully
This school of our’s.	This school of ours.
That picture of her’s.	That picture of hers.

5. Personal Pronouns used to express emphasis are called Emphatic Pronouns; as-
1. I did it myself.
2. We did it ourselves.
3. You did it yourself.
4. He did it himself.
5. She did it herself.
6. They did it themselves.

6. Emphatic pronouns are never used as a Subject.

Incorrect	Correct
Myself did it.	I myself did it.
Yourself did it.	You yourself did it.
Himself did it.	He himself did it.

Exercises (Solved)

I. Rewrite each sentence using a suitable pronoun in place of the Noun in bold type:

1. Neha is not here.
Neha has gone to see her mother.

2. “The Panchtantra’ has many stories.
‘The Panchtantra’ is a good book.

3. Where is Anu’s school?
How does Anu go to school?

4. Karan has high fever.
Karan will not go to school today.

5. I have invited Micky and Joy.
Micky and Joy are my friends.

6. Yash ans Rahul are very happy.
Yash and Rahul are going on a picnic.

7. My aunt bought me a computer.
The computer cost my aunt a lot.

8. Simran gave her parents a gift.
The gift was liked by her parents.
Hints:
1. She has gone to see her mother.
2. It ia a good book.
3. How does she go to school ?
4. He will not go to school today.
5. They are my friends.
6. They are going on a picnic.
7. It cost her a lot.
8. The gift was liked by them.

II. Choose the correct Pronoun to fill in each blank:

1. That horse is ………… (our / ours)
2. This is ……….. pen, not mine. (your / yours)
3. The horse fell and broke ………. leg. (its / it)
4. Here is your book; take ………….. away. (it / its)
5. The girls were tired; ………. are resting. (they / them)
Hints:
1. ours
2. your
3. its
4. it
5. they.

III. Choose a suitable Pronoun for each blank:

(myself, yourself, himself, itself, themselves)
1. He …………. did all this.
2. I will do this …………
3. You should take care of …………..
4. They ………… admitted their fault.
5. The town …………… is not very large.
Hints:
1. himself
2. myself
3. yourself
4. themselves
5. itself

2. Demonstrative Pronouns

Pronouns used to point to some object or objects are called Demonstrative Pronouns; as-

The words this, these, that, those are used to point to some object or objects. We call them Demonstrative Pronouns.

Exercise (Solved)

Choose the correct Demonstrative Pronoun to fill in each blank:

1. ……… is not my fault. (This / These)
2. Are ……….. your books ? (that / those)
3. ……….. are very tasty sweets. (This / These)
4. Was …………. a costly hotel ? (that / those)
Hints:
1. This
2. those
3. These
4. that

3. Interrogative Pronouns

Pronouns used to ask questions are called Interrogative Pronouns; as-

1. Who is he?
2. What is your name?
3. Which is your school ?
4. Whose books are these ?
5. Whom did you tell the story?
The pronouns who, what, which, whose, whom have been used to ask questions. These are Interrogative Pronouns.

Exercise (Solved)

Fill in each blank with a suitable Interrogative Pronoun:

1. What are you doing?
2. ………… did they invite ?
3. ………… did the teacher say ?
4. …………. of these is your pen ?
5. …………. visited you yesterday ?
6. …………… is the price of this table ?
7. ………… does this book belong to ?
8. ………. is better, honour or riches ?
Hints:
2. Whom
3. What
4. Which
5. Who
6. What
7. Who
8. Which.

Miscellaneous Exercises

I. Rewrite each sentence, changing the Nouns and Pronouns into their plural form. Make other necessary changes also:

1. I love my sister.
2. That is her doll.
3. This is my book.
4. He is flying a kite.
5. She is a good girl.
6. He gave me his book.
7. I did this work myself.
8. You are my dear friend.
9. She was playing with her doll.
Hints:
1. We love our sisters.
2. Those are their dolls.
3. These are our books.
4. They are flying kites.
5. They are good girls.
6. They gave us their books.
7. We did this work ourselves.
8. You are our dear friends.
9. They were playing with their dolls.

II. Fill in the blanks with suitable Pronouns:

1. ………… is Reema’s doll.
2. ………… is a girl from Goa.
3. ………… are ripe mangoes.
4. …………. has taken my ball ?
5. ………… have done our best.
6. ……….. do you want to eat ?
7. …………. gave you that knife ?
8. …………. of your cows was ill ?
9. She will do this work …………..
Hints:
1. This/That
2. She
3. These/Those
4. Who
5. We
6. What
7. Who
8. Which
9. herself.

III. Say what kind of Pronoun each of the underlined words is:

1. That is my book. (Demonstrative)
2. This is their house. (Personal)
3. Who teaches you English ? (Interrogative)
4. What are you doing here ? (Interrogative)
5. She is not like her sister. (Personal)
6. You have been very kind. (Personal)
7. It is a book åbout animals. (Personal)
8. These are all fresh flowers. (Demonstrative)
9. Which of these is your bike ? (Interrogative)
10. We are students of class six. (Personal)

IV. Choose the correct Pronoun to fill in each blank:

1. This is ………… pen. (she, her, hers)
2. This pen is ………….. (she, her, hers)
3. I am ………… sincerely. (you, your, yours)
4. Anita has hurt ………… (she, her, herself)
5. He did this work ………. (myself, himself)
6. I looked at ………….. in the mirror. (my, mine, myself)
7. ………. shall finish this work today. (We, Us, our)
8. The teacher asked ……. a question. (he, him, his)
Hints:
1. her
2. hers
3. yours
4. herself
5. himself
6. myself
7. We
8. him.

V. Choose the correct Pronoun for each blank:

1. We love ……….. motherland. (our / his)
2. She is as wise as ………… am. (me/I)
3. He has gifted ………….. a watch. (me / mine)
4. This watch is better than …………. (yours / your)
5. This book of stories is for ……….. (she / her)
6. This house belongs to ………… father. (my / me)
7. Lalit is as gentle as ………….. brother is. (he / his)
8. This family is not as poor as ……….. are. (us / we)
Hints:
1. our
2. I
3. me
4. yours
5. her
6. my
7. his
8. we