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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 InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 6 Squares and Square Roots InText Questions

Try These (Textbook Page No. 90)

1. Find the perfect square numbers between (i) 30 and 40 (ii) 50 and 60
Solution:
(i) Since,
1 × 1 = 1,
2 × 2 = 4,
3 × 3 = 9,
4 × 4 = 16,
5 × 5 = 25,
6 × 6 = 36,
7 × 7 = 49.
Thus, 36 is the only perfect square number between 30 and 40.

(ii) Since, 7 × 7 = 49 and 8 × 8 = 64.
∴ There is no perfect square number between 49 and 64.
Thus, there is no perfect square number between 50 and 60.

Try These (Textbook Page No. 90 – 91)

1. Can we say whether the following numbers are perfect squares? How do we know ?
(i) 1057
(ii) 23453
(iii) 7928
(iv) 222222
(v) 1069
(vi) 2061
Write five numbers which you can decide by looking at their units digit that they are not square numbers.
Solution:
(i) 1057
The ending digit is 7. All square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.
∴ 1057 cannot be a perfect square.

(ii) 23453
The ending digit is 3. All square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.
∴ 23453 cannot be a perfect square.

(iii) 7928
The ending digit is 8. All square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.
∴ 7928 cannot be a perfect square.

(iv) 222222
The ending digit is 2. All square numbers end with 0, 1, 4, 5, 6 or s 9 at unit’s place.
∴ 222222 cannot be a perfect square.

(v) 1069
The ending digit is 9. All square jj numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.
∴ 1069 may or may not be a square number.
30 × 30 = 900, 31 × 31 = 961,
32 × 32 = 1024 and 33 × 33 = 1089.
e.g. No natural number between 1024 and 1089 is a perfect square.
∴ 1069 cannot be a perfect square.

(vi) 2061
The ending digit is 1. All square s numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.
∴ 2061 may or may not be a square number.
45 × 45 = 2025, 46 × 46 = 2116,
e.g. No natural number between 2025 and 2116 is a square number.
∴ 2061 is not a square number.

2. Write five numbers which you cannot decide just by looking at their unit’s digit (or units place) whether they are square numbers or not.
Solution:
Any natural number ending with 0, 1, 4, 5, 6 or 9 can be or cannot be a square number.
Five such numbers are :
719, 2431, 524, 215, 326
[Note: You can write many more numbers.]

Try These (Textbook Page No. 91)

1. Which of 1232, 772, 822, 1612, 1092 would end with digit 1 ?
Solution:
The square of those numbers end in 1 which end in either 1 or 9. Here, the square of 161 and 109 would end in 1.
OR
The number having 1 or 9 in units place has the digit 1 in units place of its square.

Try These (Textbook Page No. 91)

Which of the following numbers would have digit 6 at unit place ?
(i) 19²
(ii) 24²
(iii) 26²
(iv) 36²
(v) 34²
Solution:
The number having 4 or 6 in the units place has the digit 6 in units place of its square. Except 19² all other numbers have 6 in their units place.
OR
(i) 19²
Units place digit is 9.
∴ 19² would not have units digit as 6. (9 × 9 = 81)

(ii) 24²
Units place digit is 4.
∴ 24² would have units digit as 6. (4 × 4= 16)

(iii) 26²
Units place digit is 6.
∴ 26² would have 6 at units place. (6 × 6 = 36)

(iv) 36²
Units place digit is 6.
∴ 36² would have 6 at units place. (6 × 6 = 36)

(v) 34²
Units place digit is 4.
∴ 34² would have 6 at units place. (4 × 4 = 1²)

Try These (Textbook Page No. 92)

What will be the “one’s digit” in the square of the following numbers ?

Question (i).
1234
Solution:
1234
The ending digit is 4.
4 × 4 = 16
∴ (1234)² will have 6 as the one’s place.

Question (ii).
26387
Solution:
26387
The ending digit is 7.
7 × 7 = 49
∴ (26387)² will have 9 as the one’s place.

Question (iii).
52698
Solution:
52698
The ending digit is 8.
8 × 8 = 64
∴ (52698)² will have 4 as the one’s place.

Question (iv).
99880
Solution:
99880
The ending digit is 0.
0 × 0 = 0
∴ (99880)² will have 0 as the one’s place.

Question (v).
21222
Solution:
21222
The ending digit is 2.
2 × 2 = 4
∴ (21222)² will have 4 as the one’s place.

Question (vi).
9106
Solution:
9106
The ending digit is 6.
6 × 6 = 36
∴ (9106)² will have 6 as the one’s place.

Try These (Textbook Page No. 92)

1. The square of which of the following numbers would be an odd number/an even number? Why?

Question (i).
727
Solution:
727
Here, the ending digit is 7.
It is an odd number.
∴ Its square is also an odd number.

Question (ii).
158
Solution:
158
Here, the ending digit is 8.
It is an even number.
∴ Its square is also an even number.

Question (iii).
269
Solution:
269
Here, the ending digit is 9.
It is an odd number.
∴ Its square is also an odd number.

Question (iv).
1980
Solution:
1980
Here, the ending digit is 0.
It is an even number.
∴ Its square is also an even number.

2. What will be the number of zeros in the square of the following numbers?

Question (i).
60
Solution:
60
In 60, number of zero is 1.
∴ (60)² will have 2 zeros. (∵ 60² = 3600)

Question (ii).
400
Solution:
400
In 400, number of zeros are 2.
∴ (400)² will have 4 zeros.
(∵ 400² = 160000)

Try These (Textbook Page No. 94)

1. How many natural numbers lie between 9² and 10²? Between 11² and 12²?
Solution:
(a) We can find 2n natural numbers between two consecutive natural numbers, n² and (n + 1)².
Here, n = 9, n + 1 = 9 + 1 = 10.
∴ Natural numbers between 9² and 10² are 2n = 2 × 9 = 18.
Thus, 18 natural numbers lie between 9² and 10².

(b) We can find 2n natural numbers between two consecutive natural numbers, n² and (n + 1)².
Here, n = 11, n + 1 = 11 + 1 = 12.
∴ Natural numbers between 11² and 12² are 2n = 2 × 11 = 22.
Thus, 22 natural numbers lie between 112 and 122.

2. How many non square numbers lie between the following pairs of numbers:

Question (i).
100² and 101²
Solution:
100² and 101²
Here, n = 100 and n + 1 = 100 + 1
= 101
∴ non square numbers between 100² and 101² = 2 × n
= 2 × 100
= 200
Thus, 200 non square numbers lie between 100² and 101².

Question (ii).
90² and 91²
Solution:
90² and 91²
Here, n = 90 and n + 1 = 90 + 1 = 91.
∴ Natural numbers between 90² and 91²
= 2 × n
= 2 × 90
= 180.
Thus, 180 non square numbers lie between 90² and 91².

Question (iii).
1000² and 1001²
Solution:
1000² and 1001²
Here, n = 1000 and
n + 1 = 1000 + 1 = 1001.
∴ Natural numbers between 1000² and 1001²
= 2 × n
= 2 × 1000
= 2000.
Thus, 2000 non square numbers lie between 1000² and 1001².

Try These (Textbook Page No. 94)

Find whether each of the following numbers is a perfect square or not:

Question (i).
121
Solution:
121
121 – 1 = 120, 120 – 3 = 117
117-5 = 112, 112 – 7 = 105
105-9 = 96, 96 – 11 = 85
85 – 13 = 72, 72 – 15 = 57
57- 17 = 40, 40 – 19 = 21
21 – 21=0
e.g. 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
Thus, 121 is a perfect square.

Question (ii).
55
Solution:
55
55 – 1 = 54, 54 – 3 = 51
51 – 5 = 46, 46 – 7 = 39
39 – 9 = 30, 30 – 11 = 19
19 – 13 = 6, 6 – 15 = – 9
Thus, 55 cannot be expressed as the sum of successive odd numbers starting from 1.
∴ 55 is not a perfect square.

Question (iii).
81
Solution:
81
81 – 1 = 80, 80 – 3 = 77
77 – 5 = 72, 72 – 7 = 65
65 – 9 = 56, 56 – 11 = 45
32 – 15 = 17, 32 – 15 = 17
17 – 17 = 0
∴ 81 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17
Thus, 81 is a perfect square.

Question (iv).
49
Solution:
49
49 – 1 = 48, 48 – 3 = 45
45 – 5 = 40, 40 – 7 = 33
33 – 9 = 24, 24 – 11 = 13
13 – 13 = 0
∴ 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
Thus, 49 is a perfect square.

Question (v).
69
Solution:
69
69 – 1 = 68, 68 – 3 = 65
65 – 5 = 60, 60 – 7 = 53
53-9 = 44, 44 – 11 = 33
33- 13 = 20, 20 – 15 = 5
5 – 17 = – 12
Thus, 69 cannot be expressed as the sum of successive odd numbers starting from 1.
∴ 69 is not a perfect square.

Try These (Textbook Page No. 95)

1. Express the following as the sum of two consecutive integers:

Question (i).
21²
Solution:
Remember : n² = \(\frac{n^{2}-1}{2}+\frac{n^{2}+1}{2}\)
21²
n = 21

∴ 21² = 220 + 221 = 441

Question (ii).
13²
Solution:
13²
n = 13

∴ 13² = 84 + 85 = 169

Question (iii).
11²
Solution:
11²
n = 11

∴ 11² = 60 + 61 = 121

Question (iv).
19²
Solution:
19²
n = 19

∴ 19² = 180 + 181 = 361

2. Do you think the reverse is also true, e.g. is the sum of any two consecutive positive integers is perfect square of a number? Give example to support your answer.
Solution:
No, the reverse is not always true.
(i) 3 + 4 = 7, 7 is not a perfect square.
(ii) 10 + 11 = 21, 21 is not a perfect square.
But,
(i) 4 + 5 = 9, 9 is a perfect square.
(ii) 12 + 13 = 25, 25 is a perfect square.

Try These (Textbook Page No. 95)

Write the square, making use of the above pattern:

Question (i).
111111²
Solution:
(111111)²
The given number is a six-digit number,
∴ middle number 6

Question (ii).
1111111²
Solution:
(1111111)²
The given number is a seven-digit number.
∴ middle number 7

Try These (Textbook Page No. 95)

Can you find the square of the following numbers using the above pattern ?
(i) 6666667²
(ii) 66666667²
Solution:
(i) Yes, 6666667² = 44444448888889
(ii) Yes, 66666667² = 4444444488888889

Try These (Textbook Page No. 97)

Find the squares of the following numbers containing 5 in unit’s place:

Question (i).
15
Solution:
[Note : A number with unit digit 5, e.g. a5
(a5)² = a(a + 1) × 100 + 25]
( i ) (15)² = 1 × (1 + 1) × 100 + 25
= 1 × 2 × 100 + 25
= 200 + 25
= 225
Hint:

Question (ii).
95
Solution:
(95)² = 9 (9 + 1) × 100 + 25
= 9 × 10 × 100 + 25
= 9000 + 25
= 9025

Question (iii).
105
Solution:
(105)² = 10 × (10 + 1) × 100 + 25
= 10 × 11 × 100 + 25
= 11000 + 25
= 11025
Hint:

Question (iv).
205
Solution:
(205)² = 20 × (20 + 1) × 100 + 25
= 20 × 21 × 100 + 25
= 42000 + 25
= 42025

Try These (Textbook Page No. 99)

Question (i).
11² = 121. What is the square root of 121 ?
Solution:
Square root of 121 is 11.

Question (ii).
14² = 196. What is the square root of 196 ?
Solution:
Square root of 196 is 14.

Think, Discuss and Write (Textbook Page No. 99)

Question (i).
(-1)² = 1. Is – 1, a square root of 1 ?
Solution:
[Note: There are two (positive as well as negative) integral square roots of a perfect square number.]
(- 1) × (- 1) = 1, (- 1)² = 1
∴ Square root of 1 can also be (-1).

Question (ii).
(-2)² = 4. Is -2, a square root of 4 ?
Solution:
(- 2) × (- 2) = 4, (-2)² = 4
∴ Square root of 4 can also be (-2).

Question (iii).
(-9)² = 81. Is – 9, a square root of 81 ?
Solution:
(-9) × (-9) = 81, (- 9)² = 81
∴ Square root of 81 can also be (-9).

Try These (Textbook Page No. 100)

By repeated subtraction of odd numbers starting from 1, find whether the following numbers are perfect squares or not? If the number is a perfect square then find its square root:

Question (i).
121
Solution:
Subtracting the successive odd numbers from 121 :
121 – 1 = 120, 120 – 3 = 117
117 – 5 = 112, 112 – 7 = 105
105 – 9 = 96, 96 – 11 = 85
85 – 13 = 72, 72 – 15 = 57
57 – 17 = 40, 40 – 19 = 21
21 – 21 = 0
∴ \(\sqrt{121}\) = 11
∴ 121 is a perfect square.

Question (ii).
55
Solution:
∵ 55 – 1 = 54, 54 – 3 = 51
51 – 5 = 46, 46 – 7 = 39
39 – 9 = 30, 30 – 11 = 19
19 – 13 = 6, 6 – 15 = – 9
55 does not reduced to 0 after subtracting odd numbers starting from 1.
∴ 55 is not a perfect square.

Question (iii).
36
Solution:
∵ 36 – 1 = 35, 35 – 3 = 32
32 – 5 = 27, 27 – 7 = 20
20 – 9 = 11, 11 – 11 = 0
∴ \(\sqrt{36}\) = 6
∴ 36 is a perfect square.

Question (iv).
49
Solution:
49 – 1 = 48, 48 – 3 = 45
45 – 5 = 40, 40 – 7 = 33
33 – 9 = 24, 24 – 11 = 13
13- 13 = 0
∴ \(\sqrt{49}\) = 7
∴ 49 is a perfect square.

Question (v).
90
Solution:
90 – 1 = 89, 89 – 3 = 86
86 – 5 = 81, 81 – 7 = 74
74 – 9 = 65, 65 – 11 = 54
54 – 13 = 41, 41 – 15 = 26
26 – 17 = 9, 9 – 19 = – 10
90 does not reduced to 0 after subtracting odd numbers starting from 1.
∴ 90 is not a perfect square.

Think, Discuss and Write (Textbook Page No. 103)

Can we say that if a perfect square is of n-digits, then its square root will have \(\frac{n}{2}\) digits if n is even or \(\frac{(n+1)}{2}\) if n is odd ?
Solution:
Yes, we can say that if a perfect square is of n-digits, then its square root will have
(a) \(\frac{n}{2}\) digits, if n is even.
(b) \(\frac{(n+1)}{2}\) digits, if n is odd.

Try These (Textbook Page No. 105)

Without calculating square roots, find the number of digits in the square root of the following numbers:

Question (i).
25600
Solution:
Here, the number of digits, n = 5 (an odd number.)
∴ Number of digits in the square root of 25600 = \(\frac{(n+1)}{2}\)
= \(\frac{5+1}{2}\)
= \(\frac {6}{2}\)
= 3

Question (ii).
100000000
Solution:
Here, the number of digits, n =9 (an odd number.)
∴ Number of digits in the square root of 100000000 = \(\frac{(n+1)}{2}\)
= \(\frac{9+1}{2}\)
= \(\frac {10}{2}\)
= 5

Question (iii).
36864
Solution:
Here, the number of digits, n = 5 (an odd number.)
∴ Number of digits in the square root of 36864 = \(\frac{(n+1)}{2}\)
= \(\frac{5+1}{2}\)
= \(\frac {6}{2}\)
= 3

Try these (Textbook Page No. 107)

Estimate the value of the following to the nearest whole number:

Question (i).
\(\sqrt{80}\)
Solution:
\(\sqrt{80}\)
10² = 100, 9² = 81, 8² = 64
∴ 80 is between 64 and 81.
∴ 64 < 80 < 81
∴ 8² < 80 < 9²
∴ 8 < \(\sqrt{80}\) < 9
Thus, \(\sqrt{80}\) lies between 8 and 9.
\(\sqrt{80}\) is much closer to 81 than 64.
∴ \(\sqrt{80}\) is approximately 9.

Question (ii).
\(\sqrt{1000}\)
Solution:
\(\sqrt{1000}\)
30² = 900, 31² = 961, 32² = 1024
∴ 1000 is between 961 and 1024.
∴ 961 < 1000 < 1024
∴ 31² < 1000 < 32²
∴ 31 < \(\sqrt{1000}\) < 32
Thus, \(\sqrt{1000}\) lies between 31 and 32.
\(\sqrt{1000}\) is much closer to 32 than 31.
∴ \(\sqrt{1000}\) is approximately 32.

Question (iii).
\(\sqrt{350}\)
Solution:
\(\sqrt{350}\)
18² = 324 and 19² = 361
∴ 350 is between 324 and 361.
∴ 324 < 350 < 361
∴ 18² < 350 < 19²
∴ 18 < \(\sqrt{350}\) < 19
Thus, \(\sqrt{350}\) lies between 18 and 19.
\(\sqrt{350}\) is much closer to 19 than 18.
∴ \(\sqrt{350}\) is approximately 19.

Question (iv).
\(\sqrt{500}\)
Solution:
\(\sqrt{500}\)
22² = 484 and 23² = 529
∴ 500 is between 484 and 529.
∴ 484 < 500 < 529
∴ 22² < 500 < 23²
∴ 22 < \(\sqrt{500}\) < 23
Thus, \(\sqrt{500}\) lies between 22 and 23.
\(\sqrt{500}\) is much closer to 22 than 23.
∴ \(\sqrt{500}\) is approximately 22.

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 2 ਸੰਖਿਆਵਾਂ ਉੱਪਰ ਮੁੱਢਲੀਆਂ ਕਿਰਿਆਵਾਂ Ex 2.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 2 ਸੰਖਿਆਵਾਂ ਉੱਪਰ ਮੁੱਢਲੀਆਂ ਕਿਰਿਆਵਾਂ Ex 2.6

ਪ੍ਰਸ਼ਨ 1.
ਇੱਕ ਕਾਪੀ ਦਾ ਮੁੱਲ ਤੋਂ 15 ਹੈ ।9 ਕਾਪੀਆਂ ਦਾ ਮੁੱਲ ਕਿੰਨਾ ਹੋਵੇਗਾ ?
ਹੱਲ:
ਇੱਕ ਕਾਪੀ ਦਾ ਮੁੱਲ = ₹ 15
9 ਕਾਪੀਆਂ ਦਾ ਮੁੱਲ = ₹ 15 × 9
= ₹ 135.

ਪ੍ਰਸ਼ਨ 2.
ਇੱਕ ਪੈਕਟ ਵਿੱਚ 75 ਪੈਨਸਿਲਾਂ ਆਉਂਦੀਆਂ ਹਨ । ਇਸ ਵਰਗੇ 19 ਪੈਕਟਾਂ ਵਿੱਚ ਕਿੰਨੀਆਂ ਪੈਨਸਿਲਾਂ ਆਉਣਗੀਆਂ ?
ਹੱਲ:
ਇੱਕ ਪੈਕਟ ਵਿੱਚ ਪੈਨਸਿਲਾਂ = 75
19 ਪੈਕਟਾਂ ਵਿੱਚ ਪੈਨਸਿਲਾਂ = 75 × 19
= 1425.

ਪ੍ਰਸ਼ਨ 3.
ਇੱਕ ਮਾਲਾ ਵਿੱਚ 79 ਮੋਤੀ ਹਨ । ਇਸ ਵਰਗੀਆਂ 68 ਮਾਲਾ ਵਿੱਚ ਕਿੰਨੇ ਮੋਤੀ ਹੋਣਗੇ ?
ਹੱਲ:
ਇੱਕ ਮਾਲਾ ਵਿਚ ਮੋਤੀ = 79
68 ਮਾਲਾ ਵਿਚ ਮੋਤੀ = 79 × 68
= 5372.

ਪ੍ਰਸ਼ਨ 4.
ਇੱਕ ਛੋਟੀ ਸਾਇਕਲ ਦਾ ਮੁੱਲ ਤੋਂ 1560 ਹੈ । ਅਜਿਹੀਆਂ 6 ਸਾਇਕਲਾਂ ਦਾ ਮੁੱਲ ਕਿੰਨਾ ਹੋਵੇਗਾ ?
ਹੱਲ:
ਇੱਕ ਛੋਟੀ ਸਾਇਕਲ ਦਾ ਮੁੱਲ = ₹ 1560
ਅਜਿਹੀਆਂ 6 ਸਾਇਕਲਾਂ ਦਾ ਮੁੱਲ = ₹ 1560 × 6 = ₹ 9360.

ਪ੍ਰਸ਼ਨ 5.
ਜੇਕਰ ਕ੍ਰਿਕੇਟ ਦੀ ਇੱਕ ਟੀਮ ਵਿੱਚ 11 ਖਿਡਾਰੀ ਹੋਣ ਤਾਂ 12 ਟੀਮਾਂ ਵਿੱਚ ਕਿੰਨੇ ਖਿਡਾਰੀ ਹੋਣਗੇ ?
ਹੱਲ:
ਇੱਕ ਕ੍ਰਿਕੇਟ ਟੀਮ ਵਿਚ ਖਿਡਾਰੀ = 11
12 ਟੀਮਾਂ ਵਿੱਚ ਖਿਡਾਰੀ = 11 × 12
= 132

ਪ੍ਰਸ਼ਨ 6.
ਇੱਕ ਡੱਬੇ ਵਿਚ 1440 ਸਾਬਣ ਦੀਆਂ ਟਿੱਕੀਆਂ ਹਨ । 6 ਡੱਬਿਆਂ ਵਿੱਚ ਸਾਬਣ ਦੀਆਂ ਕਿੰਨੀਆਂ ਟਿੱਕੀਆਂ ਹੋਣਗੀਆਂ ?
ਹੱਲ:
ਇੱਕ ਡੱਬੇ ਵਿੱਚ ਸਾਬਣ ਦੀਆਂ ਟਿੱਕੀਆਂ = 1440
6 ਡੱਬਿਆਂ ਵਿੱਚ ਸਾਬਣ ਦੀਆਂ ਟਿੱਕੀਆਂ = 1440 × 6 = 8640.

ਪ੍ਰਸ਼ਨ 7.

ਤੁਹਾਡੇ ਮਾਤਾ ਜੀ ਬਜ਼ਾਰ ਗਏ-
(a) ਉਨ੍ਹਾਂ ਨੇ 2 ਕਿਲੋਗ੍ਰਾਮ ਸੇਬ ਅਤੇ 2 ਕਿਲੋਗ੍ਰਾਮ ਅਮਰੂਦ ਖ਼ਰੀਦੇ । ਉਨ੍ਹਾਂ ਨੇ ਕਿੰਨੀ ਰਾਸ਼ੀ ਦੁਕਾਨਦਾਰ ਨੂੰ ਦਿੱਤੀ ਹੈ ?
ਹੱਲ:
2 ਕਿਲੋਗ੍ਰਾਮ ਸੇਬਾਂ ਦਾ ਮੁੱਲ = 120 × 2 = ₹ 240
2 ਕਿਲੋਗ੍ਰਾਮ ਅਮਰੂਦ ਦਾ ਮੁੱਲ = ₹ 35 × 2 = ₹ 70

ਦੁਕਾਨਦਾਰ ਨੂੰ ਕੁੱਲ ਰਾਸ਼ੀ ਦਿੱਤੀ = ₹ 310.

(b) ਜੇਕਰ ਉਹ 3 ਕਿਲੋਗ੍ਰਾਮ ਸੰਤਰੇ ਅਤੇ 2 ਕਿਲੋਗ੍ਰਾਮ ਅਨਾਰ ਖਰੀਦਣ ਤਾਂ ਉਨ੍ਹਾਂ ਨੂੰ ਕਿੰਨੇ ਰੁਪਏ ਦੇਣੇ ਪੈਣਗੇ ?
ਹੱਲ:
3 ਕਿਲੋਗ੍ਰਾਮ ਸੰਤਰੇ ਦਾ ਮੁੱਲ = ₹ 45 × 3 = ₹ 135
2 ਕਿਲੋਗ੍ਰਾਮ ਅਨਾਰ ਦਾ ਮੁੱਲ ₹ 140 × 2 = ₹ 280

ਜਿੰਨੀ ਰਾਸ਼ੀ ਦੇਣੀ ਪਵੇਗੀ = ₹ 415

ਪ੍ਰਸ਼ਨ 8.
ਹੇਠ ਦਿੱਤੇ ਸਾਰੇ ਨੋਟ ਅਤੇ ਸਿੱਕੇ, ਕਰਨ ਨੂੰ ਉਸ ਦੇ ਜਨਮਦਿਨ ‘ਤੇ ਮਿਲੇ । ਤੁਸੀਂ ਦੱਸੋ ਕਿ ਕਰਨ ਕੋਲ ਕਿੰਨੇ ਰੁਪਏ ਹਨ ?

ਹੱਲ:
ਕਰਨ ਕੋਲ ਕੁੱਲ ਰੁਪਏ ਹਨ = ₹ 500 × 5 + ₹ 50 × 3 + ₹ 10 × 7 + ₹ 2 × 3
= ₹ 2500 + ₹ 150 + ₹ 70 + 6
= ₹ 2726.

ਪ੍ਰਸ਼ਨ 9.
ਇੱਕ ਕਾਰ 1 ਲੀਟਰ ਪੈਟਰੋਲ ਨਾਲ 16 ਕਿਲੋਮੀਟਰ ਦੀ ਦੂਰੀ ਤੈਅ ਕਰਦੀ ਹੈ । 28. ਲੀਟਰ ਪੈਟਰੋਲ ਨਾਲ ਉਹ ਕਾਰ ਕਿੰਨੀ ਦੂਰੀ ਤੈਅ ਕਰੇਗੀ ?
ਹੱਲ:
1 ਲੀਟਰ ਪੈਟਰੋਲ ਨਾਲ ਤੈਅ ਕੀਤੀ ਦੂਰੀ = 16 ਕਿਲੋਮੀਟਰ
28 ਲੀਟਰ ਪੈਟਰੋਲ ਨਾਲ ਤੈਅ ਕੀਤੀ ਦੂਰੀ = 16 ਕਿਲੋਮੀਟਰ × 28 = 448 ਕਿਲੋਮੀਟਰ ।

ਪ੍ਰਸ਼ਨ 10.
ਇੱਕ ਫੈਕਟਰੀ ਵਿੱਚ ਇਕ ਘੰਟੇ ਵਿੱਚ 125 ਸਾਬਣ ਦੀਆਂ ਟਿੱਕੀਆਂ ਬਣਦੀਆਂ ਹਨ ? 8 ਘੰਟਿਆਂ ਵਿੱਚ ਕਿੰਨੀਆਂ ਸਾਬਣ ਦੀਆਂ ਟਿੱਕੀਆਂ ਬਣਨਗੀਆਂ ?
ਹੱਲ:
1 ਘੰਟੇ ਵਿੱਚ ਸਾਬਣ ਦੀਆਂ ਟਿੱਕੀਆਂ ਬਣਦੀਆਂ ਹਨ = 125
8 ਘੰਟੇ ਵਿੱਚ ਸਾਬਣ ਦੀਆਂ ਟਿੱਕੀਆਂ ਬਣਦੀਆਂ ਹਨ = 8 × 125
= 1000

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 8th Class Maths Book Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1

1. Identify the terms, their coefficients for each of the following expressions:

(i) 5xyz² – 3zy
(ii) 1 + x + x²
(iii) 4x²y² – 4x²y²z² + z²
(iv) 3 – pq + qr – rp
(v) \(\frac{x}{2}+\frac{y}{2}\) – xy
(vi) 0.3a – 0.6ab + 0.5b
Solution:

Terms	Coefficient of terms
(i) 5 xyz2 -3zy	5 – 3
(ii) 1 x x2 ‘	1 1 1
(iii) 4x2y2 – 4x2y2z2 z2	4 -4 1
(iv) 3 – pq qr – rp	3 – 1 1 – 1
(v) \(\frac {x}{2}\) \(\frac {y}{2}\) – xy	\(\frac {1}{2}\) \(\frac {1}{2}\) – 1
(vi) 0.3 a – 0.6ab 0.5b	0.3 – 0.6 0.5

2. Classify the following polynomials as monomials, binomials, trinomials. which polynomials do not fit in any of these three categories?
x + y, 1000, x + x² + x³ + x⁴, 7 + y + 5x, 2y – 3y², 2y – 3y² + 4y³, 5x – 4y + 3xy, 4z – 15z², ab + bc + cd + da, pqr, p²q + pq², 2p + 2q.
Solution:

Monomials	Binomials	Trinomials
1000 pqr	x + y 2y – 3y2 4z – 15y2 p2q +pq2 2p + 2q	7 + y + 5x 2y – 3y2 + 4y3 5x – 4y + 3xy

Following polynomials do not fit in any catagories:
x + x² + x³ + x⁴ [∵ Polynomial has 4 terms] ab + bc + cd + da [∵ Polynomial has 4 terms]

3. Add the following:

Question (i)
ab – bc, bc – ca, ca – ab
Solution:
To add, let us arrange like terms one below the other.

Question (ii)
a – b + ab, b – c + bc, c – a + ac
Solution:

Question (iii)
2p²q² – 3pq + 4, 5 + 7pq – 3p²q²
Solution:

Question (iv)
l² + m², m² + n², n² + l², 2lm + 2mn + 2nl
Solution:

4.

Question (a)
Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3
Solution:

Question (b)
Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
Solution:

Question (c)
Subtract 4p²q – 3pq + 5pq² – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq² + 5p²q
Solution:

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 10th Class Maths Book Solutions Chapter 15 Probability Ex 15.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1

Question 1.
Complete the following statements:

(i) Probability of an event E + Probability of the event ‘not E’ = _________.
Solution:
Probability of an event E +
Probability of the event ‘not E’ = 1

(ii) The probability of an event that cannot happen is ___________. Such an event is called _________.
Solution:
The probability of an event that cannot happen is 0. Such an event is called impossible event.

(iii) The probability of an event that is certain to happen is _________. Such an event is called ________.
Solution:
The probability of an event that is certain to happen is 1. Such event is called sure event.

(iv) The sum of the probabilities of all the elementary events of an experiment is __________.
Solution:
The sum of the probabilities of all the elementary events of an experiment is 1.

(v) The probability of an event is greater than or equal to _________ and less than or equal to _________.
Solution:
The probability of an event is greater than or equal to 0 and less than or equal to 1.

Question 2.
Which of the following experiments have equally likely outcomes? Explain.

(i) A driver attempts to start a car. The car starts or does not start.
Solution:
When a dnver attempts to start a car the car starts normally. Only when there is some defects the car does not start. So the outcome is not equally likely.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
Solution:
When a player attempts to shoot a basketball the outcome in this situation is not equally likely because the outcome depends on many factors such as the training of the player, quality of the gun used etc.

(iii) A trial is made to answer a true – false question. The answer is right or wrong.
Solution:
Since for a question there are two possibilities either right or wrong the outcome in this trial of true-false question is either true or false i.e. one out of the two and both have equal chances to happen. Hence, the two outcomes are equally likely.

(iv) A baby is born. It is a boy or a girl.
Solution:
A new baby (i.e. who took birth at a moment) can be either a boy or a girl and both the outcome have equally likely chances.

Question 3.
Why is tossing a coin considered to he a fair way of deciding which team should get the ball at the beginning of a football game?
Solution:
When a coin is tossed there are only two possibilities i.e. Head or tail both are equally likely to happen. Result of the toss of a fair coin is completely unpredictable.

Question 4.
Which of the following cannot be the probability of an event?
(A) \(\frac{2}{3}\)
(B) – 1.5
(C) 15 %
(D) 0.7
Solution:
As we know probability of event cannot be less than O and greater than 1
i.e. 0 ≤ P ≤ 1
∴ (B) – 1.5 is not possible.

Question 5.
If P(E) = 0.05, what is the probability of not E.
Solution. As we know P (E) + P \((\overline{\mathrm{E}})\) = 1
P\((\overline{\mathrm{E}})\) = 1 – P(E)
= 1 – 0.05 = 0.95.

Question 6.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Solution:
(i) Since bag contains only lemon flavoured candies
∴ There is no orange candies
∴ It is impossible event.
∴ Probability of getting orange flavoured = 0.

(ii) Since there are only lemon flavoured candies, it is sure event
∴ Probability, of getting lemon flavoured candy = \(\frac{1}{1}\) = 1.

Question 7.
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 studenís have the same birthday?
Solution:
Let A is event that two students have same birthday
∴ \((\overline{\mathrm{A}})\) is event that 2 students not having same birthday is 0.992
∴ P \((\overline{\mathrm{A}})\) = 0.992
∴ P (A) = 1 – P (A) (P (A) + P \((\overline{\mathrm{A}})\) = 1)
= 1 – 0.992 = 0.008
∴ Probability that two students have saine birthday = 0.008.

Question 8.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probabifity that the ball drawn is
(i) red?
(ii) not red?
Solution:
Number of Red balls = 3
Number of Black balls = 5
Total number of balls = 3 + 5 = 8
One hail is drawn at random
(i) Probability of getting Red ball = \(\frac{\text { Number of favourable cases }}{\text { Total number of cases }}\)
P (Red ball) = \(\frac{3}{8}\).

(ii) Probability of getting not red ball = 1 – P (Red ball)
= 1 – \(\frac{3}{8}\) = \(\frac{3}{8}\) [P \((\overline{\mathrm{A}})\) = 1 – P(E)].

Question 9.
A box contaIns 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red ?
(ii) white?
(iii) not green?
Solution:
Number of red marbles = 5
Number of white marbles = 8
Number of green marbles = 4
Total number of marbles = 5 + 8 + 4 = 17
Since, one marble is taken out
(i) There are 5 Red marbles
Probability of drawing Red marble = \(\frac{\text { Number of favourable cases }}{\text { Total number of cases }}\)
= \(\frac{5}{17}\)

(ii) Since there are 8 white marbles
Probability of drawing white marble = \(\frac{\text { Number of favourable cases }}{\text { Total number of cases }}\)
= \(\frac{8}{17}\)

(iii) There are 4 green bails
Probability of drawing green ball = \(\frac{\text { Number of favourable cases }}{\text { Total number of cases }}\)
= \(\frac{4}{17}\)

∴ Probability of not drawing green ball = 1 – Probability of green ball
= 1 – \(\frac{4}{17}\) = \(\frac{17-4}{17}\) = \(\frac{13}{17}\).

Question 10.
A piggy bank contains hundred 50p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin
(i) will be a 50 p coin?
(ii) will not be a ₹ 5 coin?
Solution;
Number of 50 coins = 100
Number of ₹ 1 coins = 50
Number of ₹ 2 coins =20
Number of ₹ 5 coins = 10
∴ Total number of coins = 100 + 50 + 20 + 10 = 180

(i) Since there are 100 ; 50’ p coin
Probability of getting 50p coin = \(\frac{\text { Number of favourable cases }}{\text { Total number of outcomes }}\)

= \(\frac{100}{180}\)

P (50 p coins) = \(\frac{5}{9}\).

(ii) Number of ₹ 5 coins = 10
∴ Probability of getting ₹ 5 coin = \(\frac{\text { Number of favourable cases }}{\text { Total number of cases }}\)

P (₹ 5 coins) = \(\frac{10}{180}\) = \(\frac{1}{18}\)
Probability of getting not ₹ 5 coin = 1 – P (₹ 5 coins)
= 1 – \(\frac{1}{18}\)
= \(\frac{18-1}{18}\) = \(\frac{17}{18}\).

Question 11.
Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?

Solution:
Number of male fish = 5
Number of female fish = 8
Total number of fish in the tank = 5 + 8 = 13
Probability of getting a male fish = \(\frac{\text { Number of favourable cases }}{\text { Total number of cases }}\)
P(Male fish) = \(\frac{5}{13}\)

Question 12.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 and these are equally likely outcomes. What is the probability that it will point at
(i) 8?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?

Solution:
(i) Total number of outcomes = {1, 2, 3, 4, 5, 6, 7, 8)
Probability of getting ‘8’ = \(\frac{1}{8}\)

(ii)Odd numbers are = {1, 3, 5, 7)
Probability of getting odd number = \(\frac{4}{8}=\frac{1}{2}\)

(iii) Numbers greater than 2 are {3, 4, 5, 6, 7, 8)
∴ Probability of getting number greater than 2 = \(\frac{6}{8}=\frac{3}{4}\)
P (number greater than 2) = \(\frac{3}{4}\).

(iv) Numbers less than 9 are: {1, 2, 3, 4, 5, 6, 7, 8)
∴ Probability of getting number less than 9 = \(\frac{8}{8}\)
P(a numher less than 9) = 1.

Question 13.
A die is thrown once. Find the probability of getting
(i) a prime number,
(ii) a number lying between 2 and 6;
(iii) an odd number.
Solution:
When dice is thrown number of possible outcomes
S = {1, 2, 3, 4, 5, 6)
(i) Prime numbers are {2, 3, 5)
∴ Probability of getting prime number = \(\frac{3}{6}=\frac{1}{2}\)

(ii) Numbers lying between 2 and 6 = {3, 4, 5}
Probability of getting number between 2 and 6 = \(\frac{3}{6}=\frac{1}{2}\).

(iii) The odd numbers are = {1, 3, 5}
Probability of getting an odd number = \(\frac{3}{6}=\frac{1}{2}\)
P (odd number) = \(\frac{1}{2}\).

Question 14.
One card is drawn from a well. shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds.
Solution:
There are 52 cards in a pack
(i) There are two red kings i.e. king of heart and king of diamond
Probability of getting red king = \(\frac{2}{52}=\frac{1}{26}\)
P(Red king) = \(\frac{1}{26}\)

(ii) There are 12 face cards
i.e. 4 Jack, 4 Queens and 4 kings
Probability of getting face card = \(\frac{12}{52}\)
∴ P (A face card) = \(\frac{3}{13}\).

(iii) Since there are 6 Red face cards i.e 2 Jacks; 2 Queens and 2 Kings
∴ Probability of getting 6 Red face cards = \(\frac{6}{52}\)
P (Red face card) = \(\frac{3}{26}\).

(iv) There is only one Jack of Heart
∴ Probability of getting Jack of Heart = \(\frac{1}{52}\)
P (A Jack card) = \(\frac{1}{52}\)

(v) Since there are 13 spade cards
∴ Probability of getting a spade card = \(\frac{13}{52}\)
P (A spade card) = \(\frac{1}{4}\).

(vi) Since there is only one queen of diamonds
∴ Probability of getting queen of spade card = \(\frac{1}{52}\)
P (A queen of spade) = \(\frac{1}{52}\).

Question 15.
Five cards – the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) if the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Solution:
Five cards are ten, jack, queen, king and ace
(i) Probability of getting queen = \(\frac{1}{5}\)
∴ P (A queen) = \(\frac{1}{5}\).

(ii) If the queen is drawn and put aside then there are 4 cards left – Ten, a Jack, a king and an ace.
(a) Probability of getting an ace = \(\frac{1}{4}\)
P (An Ace) = \(\frac{1}{4}\).
There’s no queen left

(b) Probability of getting a queen = \(\frac{0}{4}\) = 0
P (a queen) = 0.

Question 16.
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
Number of defective pens = 12
Number of good pens = 132
∴ Total number of pens = 12 + 132 = 144
Probability of getting good pen = \(\frac{132}{144}=\frac{11}{12}\)
P (a good pen) = \(\frac{11}{12}\).

Question 17.
(i) A lot of 20 bulbs contains 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in
(i) is not defective and is not replaced. Now one bulb is defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Solution:
(i) Number of defective bulbs 4
Number of good bulbs (Not defective) = 16
Total number of bulbs = 4 + 16 = 20
Probability of getüng defective bulb = \(\frac{4}{20}=\frac{1}{5}\).

(ii) When a defective bulb drawn is not being replaced, we are left with 19 bulbs
Now probability of getting not defective bulb = \(\frac{15}{19}\)
∴ P (Not defective bulb) = \(\frac{15}{19}\)

Question 18.
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5.
Solution:
From 1 to 90 there are 90 numbers in all and 81 two – digit numbers from 10 to 90
(i) Probability of getting two digit number = \(\frac{81}{90}\)
∴ P (two digit number) = \(\frac{81}{90}=\frac{9}{10}\).

(ii) Perfect square numbers are (1, 4, 9, 16, 25, 36, 49, 64, 81 } there are 9 perfect square numbers between 1 to 90
Probability of getting perfect square = \(\frac{9}{90}=\frac{1}{10}\)
∴ P (Perfect square) = \(\frac{1}{10}\)

(iii) Numbers divisible by 5 are (5, 10, 15, 20, 25, 30, 35, 40, 45. 50, 55. 60, 65, 70, 75, 80, 85, 90}
There are 18 numbers divisible by 5
∴ Probability of number getting divisible by 5 = \(\frac{18}{90}=\frac{1}{5}\)
∴ Required probability = \(\frac{1}{5}\).

Question 19.
A child has a die whose six faces show the letters as given below:

The die is thrown. What is the probability of getting
(i) A ?
(ii) D?
Solution:
Number of faces of a die = 6
S = {A, B, C, D, E, A}
n(S) = 6
(i) Since there are two A’s
∴ Probability of getting A = \(\frac{2}{6}=\frac{1}{3}\)
P(A) = \(\frac{1}{3}\)

(ii) Since there is only one face with D
Probability of getting D = \(\frac{1}{6}\)
∴ P(D) = \(\frac{1}{6}\)

Question 20.
Suppose you drop a die at random on the rectangular region shown in Fig. What is the probability that it will land inside the circle with diameter 1 m?

Solution:
Length of rectangle (l) = 3 m
Width of rectangle (b) = 2 m
∴ Area of rectangle = 3 m × 2 m = 6m²
Diameter of circle = 1 m
Radius of circle (R) = \(\frac{1}{2}\) m
∴ Area of circle = πR² = π(\(\frac{1}{6}\))²
= \(\frac{\pi}{4}\) m².

Probability of die to land on a circle = \(\frac{\text { Area of circle }}{\text { Area of rectangle }}\)
= \(\frac{\frac{\pi}{4} \mathrm{~m}^{2}}{6 \mathrm{~m}^{2}}=\frac{\pi}{24}\)
∴ Required Probability = \(\frac{\pi}{24}\).

Question 21.
A lot consists of 144 ball pens of which 20 are défective and the others are good. Nun will buy a pen if it is good, but will not buy it if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) She will buy it?
(ii) She will not buy it?
Solution:
Total number of Pens in lot = 144
Number of defective Pens = 20
∴ Number of good Pens = 144 – 20 = 124

(i) Let ‘A’ is event showing she buy the pen
∴ Probability that she buy a Pen = \(\frac{124}{144}\)
P(A) = \(\frac{31}{36}\)

(ii) \(\bar{A}\) is event showing that she will not buy the pen
P \((\bar{A})\) = 1 – P(A)
= 1 – \(\frac{31}{36}\) = \(\frac{36-31}{36}\)
∴ P (Not buy the pen) = \(\frac{5}{36}\).

Question 22.
Two dice, one blue and one grey are thrown at the same time. Write down all the possible outcomes
(i) Complete the following table:

(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9 10, 11 and 12. Therefore, each of them has a
probability \(\frac{1}{11}\) Do you agree with this argument ? Justify your answer.
Solution:
When two dices are thrown total number of possible outcomes
{(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
S (4, 1) (4, 2) (4, 3) (4, 4) (4,5) (4,6)
(5, 1) (5, 2) (5.3) 5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)}
n(S) = 36
Let A is event of getting sum as 3
∴ A = {(1,2) (2, 1)}
n(A) = 2
∴ Probability of getting sum as 3 = \(\frac{2}{36}=\frac{1}{18}\)
P(A) = \(\frac{1}{18}\)

Let B is event of getting sum as 4 B = ((1, 3), (3, 1), (2, 2))
n(B) = 3
∴ P(B) = \(\frac{3}{36}=\frac{1}{12}\)

Let C is event of getting sum as 5.
C = {(1, 4) (4, 1) (2, 3) (3, 2)}
n(C) = 4
P(C) = \(\frac{4}{36}=\frac{1}{9}\)

Let D is event of getting sum as 6
D = {(1, 5) (5, 1)(2, 4) (4,2) (3, 3)}, n (D) = 5
∴ P(6) = \(\frac{5}{36}\)

Let E is event of getting sum as 7
E = {(1, 6) (6, 1) (2, 5) (5,2) (4, 3) (3, 4)}
∴ P (E) = P (Sum as 7) = \(\frac{6}{36}=\frac{1}{6}\)

Let F is event of getting sum as 8
F = {(2, 6) (6, 2) (3, 5) (4, 4) (5, 3)}
∴ n(F) = 5
P(F) = P(sum as 8) = \(\frac{5}{36}\)

Let G is event of getting sum as 9 when two dices are thrown
G = {(4, 5) (5, 4) (3, 6) (6, 3))
n(G) = 4
∴ P (G) P (Sum as 8) = \(\frac{4}{36}=\frac{1}{9}\)

Let H is event of getting sum as 10
H= {(6, 4) (4, 6) (5, 5)}
n(H) = 3
∴ P (H) = P (sum as 10) = \(\frac{3}{36}=\frac{1}{12}\)

Let I is event of getting sum as 11
I = ((5,6) (6, 5))
n(I) = 2
∴ P(D) = \(\frac{2}{36}=\frac{1}{18}\)

Let J is event of,getting sum as 12
J = {(6,6)}; n(J) = 1
∴ P(J) = \(\frac{1}{36}\)

(ii) No, here all 11 possible outcomes are not equally likely
∴ Three probabilites are different.

Question 23.
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e. three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Solution:
When a coin tossed three times, then possible out comes are
S = {HHH, HHT HTH, THH, HTF, THT, TTH, TTT)
n(S) = 8
Let A is event of getting all the three same results i.e., {HHH, TTT}
∴ P(A) = \(\frac{2}{8}=\frac{1}{4}\)
Probability of lossing the game = 1 – P (A)
P \((\bar{A})\) = 1 – \(\frac{1}{4}\)
= \(\frac{4-1}{4}\) = \(\frac{3}{4}\)
∴ Probability of losing the game = \(\frac{3}{4}\).

Question 24.
A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
(ii) 5 will come up at least once?
Solution:
When a die is thrown twice all possible outcomes are
S = {(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) 5) (4, 6) (5, 1) (5, 2) (5, 3) (5,4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)}
n(S) = 36
Ler A is event that 5 will come up either time
A = {(1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5) (5, 1) (5, 2) (5, 3) (5, 4) (5, 6)}
n(A) = 11
∴ \((\bar{A})\) is event that 5 will not come up either time.
n\((\bar{A})\) = 36 – 11 = 25.

(i) ∴ Probability of not getting 5 up either time = \(\frac{25}{36}\)
P \((\bar{A})\) = \(\frac{25}{36}\)
Probability that 5 will come up at least once = \(\frac{11}{36}\)
∴ P(A) = \(\frac{11}{36}\).

Question 25.
Which of the following arguments are correct ? Give reasons for your answer:
(i) 1f two coins are tossed simultaneously there are three possible outcomes – two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is \(\frac{1}{3}\):
(ii) If a die is thrown, there are two possible outcomes – an odd number or an even number. Therefore, the probability of getting an odd number is \(\frac{1}{2}\).
Solution:
(i) When two coins are tossed the possible outcomes are S = {HH, HT, TH, TT}
Probability of getting 2 Heads = \(\frac{1}{4}\)
P(HH) = \(\frac{1}{4}\)
Probability of getting two tails = \(\frac{1}{4}\)
P(TT) = \(\frac{1}{4}\)
Probability of getting one head and one tail = \(\frac{2}{4}=\frac{1}{2}\)
∴ (i) argument is incorrect.

(ii) When a die is thrown possible outcomes are S = (1, 2, 3, 4, 5, 6)
n(S) = 6
Odd numbers are 1, 3, 5
∴ Probability of getting odd number = \(\frac{3}{6}=\frac{1}{2}\)
Even numbers are 2, 4, 6
∴ Probability of getting even number = \(\frac{3}{6}=\frac{1}{2}\)
(ii) argument is correct.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 11 Mensuration Ex 11.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4

1. Given a cylindrical tank, in which situation will you find surface area and in which situation volume:

Question (a)
To find how much it can hold.
Solution:
To find how much a cylindrical tank can hold, we will find its volume.

Question (b)
Number of cement bags required to plaster it.
Solution:
To find number of cement bags required to plaster a cylindrical tank, we will find its surface area.

Question (c)
To find the number of smaller tanks that can be filled with water from it.
Solution:
To find the number of smaller tanks that can be filled with water from it, we will find volume of the tank.

2. Diameter of cylinder A is 7 cm and the height is 14 cm.

Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?
Solution:
Radius of cylinder B is double them that of cylinder A.
∴ Volume of cylinder B should be more than that of cylinder A.

For cylinder A:
radius (r) = \(\frac{diameter}{2}\) = \(\frac{7}{2}\)
height (h) = 14 cm
Volume of cylinder A = πr²h
= \(\frac{22}{7}\) × (\(\frac{22}{7}\))² × 14
= \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 14
= 11 × 7 × 7
= 593 cm³

For cylinder B:

radius (r) = \(\frac{diameter}{2}\) = \(\frac{14}{2}\) = 7cm and
height (h) = 7 cm
Volume of cylinder B = πr²h
= \(\frac{22}{7}\) × 7² × 7
= \(\frac{22}{7}\) × 7 × 7 × 7
= 22 × 7 × 7
= 1078 cm³

Total surface area:
For cylinder A:

radius (r) = \(\frac{7}{2}\) cm
height (h) = 14 cm
Total surface area of cylinder A
= 2πr (r + h)
= 2 × \(\frac{22}{7}\) × \(\frac{7}{2}\)(\(\frac{7}{2}\) +14)
= 22(\(\frac{35}{2}\))
= 11 × 35
= 385 cm²

For cylinder B :
radius (r) = 7 cm, height (h) = 7 cm
Total surface area of cylinder B
= 2πr (r + h)
= 2 × \(\frac {22}{7}\) × 7(7 + 7)
= 44(14) = 616 cm²
So the surface area of cylinder B is greater than that of cylinder A. Hence, the cylinder with greater volume also has greater surface area.

3. Find the height of a cuboid whose base area is 180 cm² and volume is 900 cm³?
Solution:
Let the height of cuboid be h cm.
Now, = Area of base × Height
∴ 900 = 180 × h
∴ h = \(\frac {900}{180}\) = 5cm
Hence, height of cuboid is 5 cm.

4. A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?
Solution:
Volume of a cuboid = 60 × 54 × 30 cm³
Volume of a cube = 6³ = 6 × 6 × 6 cm³
∴ Number of small cubes
= \(\frac{\text { Volume of cuboid }}{\text { Volume of one cube}}\)
= \(\frac{60 \times 54 \times 30}{6 \times 6 \times 6}\)
= 10 × 9 × 5 = 450
Hence, 450 cubes can be placed in the given cuboid.

5. Find the height of the cylinder whose volume is 1.54 m³ and diameter of the base is 140 cm?
Solution:
For given cylinder:
Volume = 1.54 m³
Radius(r) = \(\frac {diameter}{2}\) = \(\frac {140}{2}\) = 70cm = 0.7 m
Volume of cylinder = πr²h
∴ 1.54 = \(\frac {22}{7}\) × 0.7 × 0.7 × h
∴ h = \(\frac{1.54 \times 7}{22 \times 0.7 \times 0.7}\) = 1m
Hence, height of the cylinder is 1 m.

6. A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in litres that can be stored in the tank?
Solution:
For given cylindrical milk tank:
Radius (r) = 1.5 m = \(\frac {15}{10}\) m
Height (h) = 7m
Volume of cylindrical milk tank
= πr²h
= \(\frac {22}{7}\) × (\(\frac {15}{10}\))² × 7
= \(\frac {22}{7}\) × \(\frac {15}{10}\) × \(\frac {15}{10}\) × 7
= \(\frac{11 \times 3 \times 3}{2}=\frac{99}{2}\) = 49.5m³
= 49.5 m³
1 m³ = 1000 litres
∴ 49.5 m³ = 49.5 × 1000 = 49500 litres
Hence, 49,500 litres of milk can be stored in the tank.

7. If each edge of a cube is doubled:

Question (i)
How many times will its surface area increase?
Solution:
Let the edge of the original cube be x.
Then, its new edge (by doubling) = 2x
Original surface area of the cube
= 6 (side)²
= 6 (x)²
= 6x²
New surface area of the cube
= 6 (side)²
= 6 (2x)²
= 6 (2x × 2x)
= 6 (4x²) = 24X²
= \(\frac{\text { New surface area of the cube }}{\text { Original surface area of the cube }}\) = \(\frac{24 x^{2}}{6 x^{2}}\) = 4
Hence, surface area of a cube will increase 4 times.

Question (ii)
How many times will its volume increase ?
Solution:
Original volume of the cube
= (side)³
= (x)³
= x³
New volume of the cube = (side)³
= (2x)³
= (2x × 2x × 2x)
= 8x³
Now,
\(\frac{\text { New volume of the cube }}{\text { Original volume of the cube }}\) = \(\frac{8 x^{3}}{x^{3}}\) = 8
Hence, volume of the cube will increase 8 times.

8. Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m³, find the number of hours it will take to fill the reservoir.
Solution:
Volume of the cuboidal reservoir =108 m³ 1 m³ = 1000 litres
∴ 108 m³ = 108 × 1000 litres
= 1,08,000 litres
Water poured in a minute = 60 litres
∴ Water poured in an hour
(∵ 1 hour = 60 minutes) = 60 × 60
= 3600 litres
Time taken to fill reservoir = \(\frac{\text { volume of reservoir }}{\text { water poured in an hour }}\) = \(\frac {108000}{3600}\)
= 30 hours
Hence, 30 hours it will take to fill the reservoir.

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.