Monthly Archives: September 2022

Exercise 3.2 | Class 8th Mathematics

Exercise 3.2 | Class 8th Mathematics

Q1. Find x in the following figures.

Solution:
(a) We know that the sum of all the exterior angles of a polygon = 360°
125° + 125° + x = 360°
⇒ 250° + x = 360°

x = 360° – 250° = 110°
Hence x = 110°
(b) Here ∠y = 180° – 90° = 90°

and ∠z = 90° (given)
x + y + 60° + z + 70° = 360° [∵ Sum of all the exterior angles of a polygon = 360°]
⇒ x + 90° + 60° + 90° + 70° = 360°
⇒ x + 310° = 360°
⇒ x = 360° – 310° = 50°
Hence x = 50°

Q2. Find the measure of each exterior angle of a regular polygon of
(i) 9 sides
(ii) 15 sides
Solution:
(i) We know the sum of all the exterior angles of polygon = 360°
Measure of each angle of 9 sided regular polygon = 3609 = 40°
(ii) Sum of all the exterior angles of a polygon = 360°
Measure of each angle of 15 sided regular polygon = 36015 = 24°

Q3. How many sides does a regular polygon have if the measure of an exterior angle is 24°?
Solution:
Sum of all exterior angles of a regular polygon = 360°
Number of sides

Hence, the number of sides = 15

Q4. How many sides does a regular polygon have if each of its interior angles is 165°?
Solution:
Let re be the number of sides of a regular polygon.
Sum of all interior angles = (n – 2) × 180°
and, measure of its each angle

Hence, the number of sides = 24

Q5.  What is the minimum interior angle possible for a regular polygon? Why?
(b) What is the maximum exterior angle possible for a regular polygon?
Solution:
(a) Sum of all interior angles of a regular polygon of side n = (n – 2) × 180°
The measure of each interior angle

The minimum measure the angle of an equilateral triangle (n = 3) = 60°.
(b) From part (a) we can conclude that the maximum exterior angle of a regular polygon = 180° – 60° = 120°.

Exercise 3.3 |Class 8th Mathematics

Q1. Given a parallelogram ABCD. Complete each statement along with the definition or property used.
(i) AD = …………
(ii) ∠DCB = ………
(iii) OC = ………
(iv) m∠DAB + m∠CDA = ……..

Solution:
(i) AD = BC [Opposite sides of a parallelogram are equal]
(ii) ∠DCB = ∠DAB [Opposite angles of a parallelogram are equal]
(iii) OC = OA [Diagonals of a parallelogram bisect each other]
(iv) m∠DAB + m∠CDA = 180° [Adjacent angles of a parallelogram are supplementary]

Q2. Consider the following parallelograms. Find the values of the unknowns x, y, z.

Solution:
(i) ABCD is a parallelogram.

∠B = ∠D [Opposite angles of a parallelogram are equal]
∠D = 100°
⇒ y = 100°
∠A + ∠B = 180° [Adjacent angles of a parallelogram are supplementary]
⇒ z + 100° = 180°
⇒ z = 180° – 100° = 80°
∠A = ∠C [Opposite angles of a ||gm]
x = 80°
Hence x = 80°, y = 100° and z = 80°
(ii) PQRS is a parallelogram.

∠P + ∠S = 180° [Adjacent angles of parallelogram]
⇒ x + 50° = 180°
x = 180° – 50° = 130°
Now, ∠P = ∠R [Opposite angles are equal]
⇒ x = y
⇒ y = 130°
Also, y = z [Alternate angles]
z = 130°
Hence, x = 130°, y = 130° and z = 130°
(iii) ABCD is a rhombus.
[∵ Diagonals intersect at 90°]

x = 90°
Now in ∆OCB,
x + y + 30° = 180° (Angle sum property)
⇒ 90° + y + 30° = 180°
⇒ y + 120° = 180°
⇒ y = 180° – 120° = 60°
y = z (Alternate angles)
⇒ z = 60°
Hence, x = 90°, y = 60° and z = 60°.
(iv) ABCD is a parallelogram

∠A + ∠B = 180° (Adjacent angles of a parallelogram are supplementary)
⇒ x + 80° = 180°
⇒ x = 180° – 80° = 100°
Now, ∠D = ∠B [Opposite angles of a |jgm]
⇒ y = 80°
Also, z = ∠B = 80° (Alternate angles)
Hence x = 100°, y = 80° and z = 80°
(v) ABCD is a parallelogram.

∠D = ∠B [Opposite angles of a ||gm]
y = 112°
x + y + 40° = 180° [Angle sum property]
⇒ x + 112° + 40° = 180°
⇒ x + 152° = 180°
⇒ x = 180° – 152 = 28°
z = x = 28° (Alternate angles)
Hence x = 28°, y = 112°, z = 28°.

Q3. Can a quadrilateral ABCD be a parallelogram if
(i) ∠D + ∠B = 180°?
(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
(iii) ∠A = 70° and ∠C = 65°?
Solution:
(i) For ∠D + ∠B = 180, quadrilateral ABCD may be a parallelogram if following conditions are also fulfilled.
(a) The sum of measures of adjacent angles should be 180°.

Q4. Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.
Solution:
ABCD is a rough figure of a quadrilateral in which m∠A = m∠C but it is not a parallelogram. It is a kite.

Q5. The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram.
Solution:
Let ABCD is parallelogram such that
m∠B : m∠C = 3 : 2

Let m∠B = 3x° and m∠C = 2x°
m∠B + m∠C = 180° (Sum of adjacent angles = 180°)
3x + 2x = 180°
⇒ 5x = 180°
⇒ x = 36°
Thus, ∠B = 3 × 36 = 108°
∠C = 2 × 36° = 72°
∠B = ∠D = 108°
and ∠A = ∠C = 72°
Hence, the measures of the angles of the parallelogram are 108°, 72°, 108° and 72°.

Q6. Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Solution:
Let ABCD be a parallelogram in which
∠A = ∠B

We know ∠A + ∠B = 180° [Sum of adjacent angles = 180°]
∠A + ∠A = 180°
⇒ 2∠A = 180°
⇒ ∠A = 90°
Thus, ∠A = ∠C = 90° and ∠B = ∠D = 90°
[Opposite angles of a parallelogram are equal]

Q7. The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.

Solution:
∠y = 40° (Alternate angles)
∠z + 40° = 70° (Exterior angle property)
⇒ ∠z = 70° – 40° = 30°
z = ∠EPH (Alternate angle)
In ∆EPH
∠x + 40° + ∠z = 180° (Adjacent angles)
⇒ ∠x + 40° + 30° = 180°
⇒ ∠x + 70° = 180°
⇒ ∠x = 180° – 70° = 110°
Hence x = 110°, y = 40° and z = 30°.

Q8. The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)

Solution:
(i) GU = SN (Opposite sides of a parallelogram)

⇒ y + 7 = 20
⇒ y = 20 – 7 = 13
Also, ON = OR
⇒ x + y = 16
⇒ x + 13 = 16
x = 16 – 13 = 3
Hence, x = 3 cm and y = 13 cm.

Q9.

In the above figure both RISK and CLUE are parallelograms. Find the value of x.
Solution:
Here RISK and CLUE are two parallelograms.

∠1 = ∠L = 70° (Opposite angles of a parallelogram)
∠K + ∠2 = 180°
Sum of adjacent angles is 180°
120° + ∠2 = 180°
∠2 = 180° – 120° = 60°
In ∆OES,
∠x + ∠1 + ∠2 = 180° (Angle sum property)
⇒ ∠x + 70° + 60° = 180°
⇒ ∠x + 130° = 180°
⇒ ∠x = 180° – 130° = 50°
Hence x = 50°

Q10. Explain how this figure is a trapezium. Which of its two sides are parallel?

Solution:
∠M + ∠L = 100° + 80° = 180°
∠M and ∠L are the adjacent angles, and sum of adjacent interior angles is 180°
KL is parallel to NM
Hence KLMN is a trapezium.

Ex 3.3 Class 8 Maths Question 11.
Find m∠C in below figure if AB¯ || DC¯
Solution:
Given that AB¯ || DC¯
m∠B + m∠C = 180° (Sum of adjacent angles of a parallelogram is 180°)

120° + m∠C = 180°
m∠C = 180° – 120° = 60°
Hence m∠C = 60°

Exercise 3.4 | Class 8th Mathematics

Q1. State whether True or False.
(a) All rectangles are squares.
(b) All rhombuses are parallelograms.
(c) All squares are rhombuses and also rectangles.
(d) All squares are not parallelograms.
(e) All kites are rhombuses.
(f) All rhombuses are kites.
(g) All parallelograms are trapeziums.
(h) All squares are trapeziums.
Solution:
(a) False
(b) True
(c) True
(d) False
(e) False
(f) True
(g) True
(h) True

Q2. Identify all the quadrilaterals that have
(a) four sides of equal length
(b) four right angles
Solution:
(a) Squares and rhombuses.
(b) Rectangles and squares.

Q3. Explain how a square is
(i) a quadrilateral
(ii) a parallelogram
(iii) a rhombus
(iv) a rectangle
Solution:
(i) Square is a quadrilateral because it is closed with four line segments.
(ii) Square is a parallelogram due to the following properties:
(a) Opposite sides are equal and parallel.
(b) Opposite angles are equal.
(iii) Square is a rhombus because its all sides are equal and opposite sides are parallel.
(iv) Square is a rectangle because its opposite sides are equal and has equal diagonal.

Q4. Name the quadrilaterals whose diagonals
(i) bisect each other
(ii) are perpendicular bisectors of each other
(iii) are equal
Solution:
(i) Parallelogram, rectangle, square and rhombus
(ii) Square and rhombus
(iii) Rectangle and square

Q5. Explain why a rectangle is a convex quadrilateral.
Solution:
In a rectangle, both of its diagonal lie in its interior. Hence, it is a convex quadrilateral.

Q6. ABC is a right-angled triangle and O is the midpoint of the side opposite to the right angle. Explain why O is equidistant from A, B and C. (The dotted lines are drawn additionally to help you).

Solution:
Since the right-angled triangle ABC makes a rectangle ABCD by the dotted lines.
Therefore OA = OB = OC = OD [Diagonals of a rectangle are equal and bisect each other]
Hence, O is equidistant from A, B and C.

Extra Question | Class 8th Mathematics

Q1. In the given figure, ABCD is a parallelogram. Find x.

Solution:
AB = DC [Opposite sides of a parallelogram]
3x + 5 = 5x – 1
⇒ 3x – 5x = -1 – 5
⇒ -2x = -6
⇒ x = 3

Q2. In the given figure find x + y + z.

Solution:
We know that the sum of all the exterior angles of a polygon = 360°
x + y + z = 360°

Q3. In the given figure, find x.

Solution:
∠A + ∠B + ∠C = 180° [Angle sum property]
(x + 10)° + (3x + 5)° + (2x + 15)° = 180°
⇒ x + 10 + 3x + 5 + 2x + 15 = 180
⇒ 6x + 30 = 180
⇒ 6x = 180 – 30
⇒ 6x = 150
⇒ x = 25

Q4. The angles of a quadrilateral are in the ratio of 2 : 3 : 5 : 8. Find the measure of each angle.
Solution:
Sum of all interior angles of a quadrilateral = 360°
Let the angles of the quadrilateral be 2x°, 3x°, 5x° and 8x°.
2x + 3x + 5x + 8x = 360°
⇒ 18x = 360°
⇒ x = 20°
Hence the angles are
2 × 20 = 40°,
3 × 20 = 60°,
5 × 20 = 100°
and 8 × 20 = 160°.

Q5. Find the measure of an interior angle of a regular polygon of 9 sides.
Solution:
Measure of an interior angle of a regular polygon

Q6. Length and breadth of a rectangular wire are 9 cm and 7 cm respectively. If the wire is bent into a square, find the length of its side.
Solution:
Perimeter of the rectangle = 2 [length + breadth]
= 2[9 + 7] = 2 × 16 = 32 cm.

Now perimeter of the square = Perimeter of rectangle = 32 cm.
Side of the square = 324 = 8 cm.
Hence, the length of the side of square = 8 cm.

Q7. In the given figure ABCD, find the value of x.

Solution:
Sum of all the exterior angles of a polygon = 360°
x + 70° + 80° + 70° = 360°
⇒ x + 220° = 360°
⇒ x = 360° – 220° = 140°

Q8. In the parallelogram given alongside if m∠Q = 110°, find all the other angles.

Solution:
Given m∠Q = 110°
Then m∠S = 110° (Opposite angles are equal)
Since ∠P and ∠Q are supplementary.
Then m∠P + m∠Q = 180°
⇒ m∠P + 110° = 180°
⇒ m∠P = 180° – 110° = 70°
⇒ m∠P = m∠R = 70° (Opposite angles)
Hence m∠P = 70, m∠R = 70°
and m∠S = 110°

Q9. In the given figure, ABCD is a rhombus. Find the values of x, y and z.

Solution:
AB = BC (Sides of a rhombus)
x = 13 cm.
Since the diagonals of a rhombus bisect each other
z = 5 and y = 12
Hence, x = 13 cm, y = 12 cm and z = 5 cm.

Q10In the given figure, ABCD is a parallelogram. Find x, y and z.

Solution:
∠A + ∠D = 180° (Adjacent angles)
⇒ 125° + ∠D = 180°
⇒ ∠D = 180° – 125°
x = 55°
∠A = ∠C [Opposite angles of a parallelogram]
⇒ 125° = y + 56°
⇒ y = 125° – 56°
⇒ y = 69°
∠z + ∠y = 180° (Adjacent angles)
⇒ ∠z + 69° = 180°
⇒ ∠z = 180° – 69° = 111°
Hence the angles x = 55°, y = 69° and z = 111°

Exercise 2.1 | Class 8th Mathematics

1.Solve the equation: x – 2 = 7.
Solution:
Given: x – 2 = 7
⇒ x – 2 + 2 = 7 + 2 (adding 2 on both sides)
⇒ x = 9 (Required solution)

2.Solve the equation: y + 3 = 10.
Given: y + 3 = 10
⇒ y + 3 – 3 = 10 – 3 (subtracting 3 from each side)
⇒ y = 7 (Required solution)

3.Solve the equation: 6 = z + 2
Solution:
We have 6 = z + 2
⇒ 6 – 2 = z + 2 – 2 (subtracting 2 from each side)
⇒ 4 = z
Thus, z = 4 is the required solution.

4.Solve the equation 6x = 12.
Solution:
We have 6x = 12
⇒ 6x ÷ 6 = 12 ÷ 6 (dividing each side by 6)
⇒ x = 2
Thus, x = 2 is the required solution.

5.Solve the equation t5 = 10.
Solution:
Given t5 = 10
⇒ t5 × 5 = 10 × 5 (multiplying both sides by 5)
⇒ t = 50
Thus, t = 50 is the required solution.

6.Solve the equation 2×3 = 18.
Solution:
We have 2×3 = 18
⇒ 2×3 × 3 = 18 × 3 (multiplying both sides by 3)
⇒ 2x = 54
⇒ 2x ÷ 2 = 54 ÷ 2 (dividing both sides by 2)
⇒ x = 27
Thus, x = 27 is the required solution.

7.Solve the equation 1.6 = y1.5
Solution:
Given: 1.6 = y1.5
⇒ 1.6 × 1.5 = y1.5 × 1.5 (multiplying both sides by 1.5)
⇒ 2.40 = y
Thus, y = 2.40 is the required solution.

8.Solve the equation 7x – 9 = 16.
Solution:
We have 7x – 9 = 16
⇒ 7x – 9 + 9 = 16 + 9 (adding 9 to both sides)
⇒ 7x = 25
⇒ 7x ÷ 7 = 25 ÷ 7 (dividing both sides by 7)
⇒ x = 257
Thus, x = 257 is the required solution.

9.Solve the equation 14y – 8 = 13.
Solution:
We have 14y – 8 = 13
⇒ 14y – 8 + 8 = 13 + 8 (adding 8 to both sides)
⇒ 14y = 21
⇒ 14y ÷ 14 = 21 ÷ 14 (dividing both sides by 14)
⇒ y = 2114
⇒ y = 32
Thus, y = 32 is the required solution.

10.Solve the equation 17 + 6p = 9.
Solution:
We have, 17 + 6p = 9
⇒ 17 – 17 + 6p = 9 – 17 (subtracting 17 from both sides)
⇒ 6p = -8
⇒ 6p ÷ 6 = -8 ÷ 6 (dividing both sides by 6)
⇒ p = −86
⇒ p = −43
Thus, p = −43 is the required solution.

 Exercise 2.2 | Class 8th Mathematics

1.The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?
Solution:
Let the breadth of the pool be x m.
Condition I: Length = (2x + 2) m.
Condition II: Perimeter = 154 m.
We know that Perimeter of rectangle = 2 × [length + breadth]
2 × [2x + 2 + x] = 154
⇒ 2 × [3x + 2] = 154
⇒ 6x + 4 = 154 (solving the bracket)
⇒ 6x = 154 – 4 [Transposing 4 from (+) to (-)]

⇒ 6x = 150
⇒ x = 150 ÷ 6 [Transposing 6 from (×) to (÷)]
⇒ x = 25
Thus, the required breadth = 25 m

and the length = 2 × 25 + 2 = 50 + 2 = 52 m.

2.Sum of two numbers be 95. If one exceeds the other by 15, find the numbers.
Solution:
Let one number be x
Other number = x + 15
As per the condition of the question, we get
x + (x + 15) = 95
⇒ x + x + 15 = 95
⇒ 2x + 15 = 95
⇒ 2x = 95 – 15 [transposing 15 from (+) to (-)]
⇒ 2x = 80
⇒ x = 802 [transposing 2 from (×) to (÷)]
⇒ x = 40
Other number = 95 – 40 = 55
Thus, the required numbers are 40 and 55.

3.Two numbers are in the ratio 5 : 3. If they differ by 18, what are the numbers?
Solution:
Let the two numbers be 5x and 3x.
As per the conditions, we get
5x – 3x = 18
⇒ 2x = 18
⇒ x = 18 ÷ 2 [Transposing 2 from (×) to (÷)]
⇒ x = 9.
Thus, the required numbers are 5 × 9 = 45 and 3 × 9 = 27

4.Three consecutive integers add up to 51. What are these integers?
Solution:
Let the three consecutive integers be x, x + 1 and x + 2.
As per the condition, we get
x + (x + 1) + (x + 2) = 51
⇒ x + x + 1 + x + 2 = 51
⇒ 3x + 3 = 51
⇒ 3x = 51 – 3 [transposing 3 to RHS]
⇒ 3x = 48
⇒ x = 48 ÷ 3 [transposing 3 to RHS]
⇒ x = 16
Thus, the required integers are 16, 16 + 1 = 17 and 16 + 2 = 18, i.e., 16, 17 and 18.

5.The sum of three consecutive multiples of 8 is 888. Find the multiples.
Solution:
Let the three consecutive multiples of 8 be 8x, 8x + 8 and 8x + 16.
As per the conditions, we get
8x + (8x + 8) + (8x + 16) = 888
⇒ 8x + 8x + 8 + 8x + 16 = 888
⇒ 24x + 24 = 888
⇒ 24x = 888 – 24 (transposing 24 to RHS)
⇒ 24x = 864
⇒ x = 864 ÷ 24 (transposing 24 to RHS)

⇒ x = 36
Thus, the required multiples are
36 × 8 = 288, 36 × 8 + 8 = 296 and 36 × 8 + 16 = 304,
i.e., 288, 296 and 304.

6.Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3, and 4 respectively, they add up to 74. Find these numbers.
Solution:
Let the three consecutive integers be x, x + 1 and x + 2.
As per the condition, we have
2x + 3(x + 1) + 4(x + 2) = 74
⇒ 2x + 3x + 3 + 4x + 8 = 74
⇒ 9x + 11 = 74
⇒ 9x = 74 – 11 (transposing 11 to RHS)
⇒ 9x = 63
⇒ x = 63 ÷ 9
⇒ x = 7 (transposing 7 to RHS)
Thus, the required numbers are 7, 7 + 1 = 8 and 7 + 2 = 9, i.e., 7, 8 and 9.

7.The ages of Rahul and Haroon are in the ratio 5 : 7. Four years later the sum of their ages will be 56 years. What are their present ages?
Solution:
Let the present ages of Rahul and Haroon he 5x years and 7x years respectively.
4 years later, the age of Rahul will be (5x + 4) years.
4 years later, the age of Haroon will be (7x + 4) years.
As per the conditions, we get
(5x + 4) + (7x + 4) = 56
⇒ 5x + 4 + 7x + 4 = 56
⇒ 12x + 8 = 56
⇒ 12x = 56 – 8 (transposing 8 to RHS)
⇒ 12x = 48
⇒ x = 48 ÷ 12 = 4 (transposing 12 to RHS)
Hence, the required age of Rahul = 5 × 4 = 20 years.
and the required age of Haroon = 7 × 4 = 28 years.

8.The number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the numbers of girls. What is the total class strength?
Solution:
Let the number of boys be 7x
and the number of girls be 5x
As per the conditions, we get
7x – 5x = 8
⇒ 2x = 8
⇒ x = 8 ÷ 2 = 4 (transposing 2 to RHS)
the required number of boys = 7 × 4 = 28
and the number of girls = 5 × 4 = 20
Hence, total class strength = 28 + 20 = 48

9.Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?
Solution:
Let the age of Baichung be x years.
The age of his father = x + 29 years,
and the age of his grandfather = x + 29 + 26 = (x + 55) years.
As per the conditions, we get
x + x + 29 + x + 55 = 135
⇒ 3x + 84 = 135
⇒ 3x = 135 – 84 (transposing 84 to RHS)
⇒ 3x = 51
⇒ x = 51 ÷ 3 (transposing 3 to RHS)
⇒ x = 17
Hence Baichung’s age = 17 years
Baichung’s father’s age = 17 + 29 = 46 years,
and grand father’s age = 46 + 26 = 72 years.

10.Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?
Solution:
Let the present age of Ravi be x years.
After 15 years, his age will be = (x + 15) years
As per the conditions, we get
⇒ x + 15 = 4x
⇒ 15 = 4x – x (transposing x to RHS)
⇒ 15 = 3x
⇒ 15 ÷ 3 = x (transposing 3 to LHS)
⇒ x = 5
Hence, the present age of Ravi = 5 years.

Thus, the number of winners = 19

 Exercise 2.3 | Class 8th Mathematics

1.3x = 2x + 18
Solution:
We have 3x = 2x + 18
⇒ 3x – 2x = 18 (Transposing 2x to LHS)
⇒ x = 18
Hence, x = 18 is the required solution.
Check: 3x = 2x + 18
Putting x = 18, we have
LHS = 3 × 18 = 54
RHS = 2 × 18 + 18 = 36 + 18 = 54
LHS = RHS
Hence verified.

2.5t – 3 = 3t – 5
Solution:
We have 5t – 3 = 3t – 5
⇒ 5t – 3t – 3 = -5 (Transposing 3t to LHS)
⇒ 2t = -5 + 3 (Transposing -3 to RHS)
⇒ 2t = -2
⇒ t = -2 ÷ 2
⇒ t = -1
Hence t = -1 is the required solution.
Check: 5t – 3 = 3t – 5
Putting t = -1, we have
LHS = 5t – 3 = 5 × (-1)-3 = -5 – 3 = -8
RHS = 3t – 5 = 3 × (-1) – 5 = -3 – 5 = -8
LHS = RHS
Hence verified.

3. 5x + 9 = 5 + 3x
Solution:
We have 5x + 9 = 5 + 3x
⇒ 5x – 3x + 9 = 5 (Transposing 3x to LHS) => 2x + 9 = 5
⇒ 2x = 5 – 9 (Transposing 9 to RHS)
⇒ 2x = -4
⇒ x = -4 ÷ 2 = -2
Hence x = -2 is the required solution.
Check: 5x + 9 = 5 + 3x
Putting x = -2, we have
LHS = 5 × (-2) + 9 = -10 + 9 = -1
RHS = 5 + 3 × (-2) = 5 – 6 = -1
LHS = RHS
Hence verified.

4. 4z + 3 = 6 + 2z
Solution:
We have 4z + 3 = 6 + 2z
⇒ 4z – 2z + 3 = 6 (Transposing 2z to LHS)
⇒ 2z + 3 = 6
⇒ 2z = 6 – 3 (Transposing 3 to RHS)
⇒ 2z = 3
⇒ z = 32
Hence z = 32 is the required solution.
Check: 4z + 3 = 6 + 2z
Putting z = 32, we have
LHS = 4z + 3 = 4 × 32 + 3 = 6 + 3 = 9
RHS = 6 + 2z = 6 + 2 × 32 = 6 + 3 = 9
LHS = RHS
Hence verified.

5. 2x – 1 = 14 – x
Solution:
We have 2x – 1 = 14 – x
⇒ 2x + x = 14 + 1 (Transposing x to LHS and 1 to RHS)
⇒ 3x = 15
⇒ x = 15 ÷ 3 = 5
Hence x = 5 is the required solution.
Check: 2x – 1 = 14 – x
Putting x = 5
LHS we have 2x – 1 = 2 × 5 – 1 = 10 – 1 = 9
RHS = 14 – x = 14 – 5 = 9
LHS = RHS
Hence verified.

6. 8x + 4 = 3(x – 1) + 7
Solution:
We have 8x + 4 = 3(x – 1) + 7
⇒ 8x + 4 = 3x – 3 + 7 (Solving the bracket)
⇒ 8x + 4 = 3x + 4
⇒ 8x – 3x = 4 – 4 [Transposing 3x to LHS and 4 to RHS]
⇒ 5x = 0
⇒ x = 0 ÷ 5 [Transposing 5 to RHS]
or x = 0
Thus x = 0 is the required solution.
Check: 8x + 4 = 3(x – 1) + 7
Putting x = 0, we have
8 × 0 + 4 = 3(0 – 1) + 7
⇒ 0 + 4 = -3 + 7
⇒ 4 = 4
LHS = RHS
Hence verified.

7. x = 45 (x + 10)
Solution:
We have x = 45 (x + 10)
⇒ 5 × x = 4(x + 10) (Transposing 5 to LHS)
⇒ 5x = 4x + 40 (Solving the bracket)
⇒ 5x – 4x = 40 (Transposing 4x to LHS)
⇒ x = 40
Thus x = 40 is the required solution.
Check: x = 45 (x + 10)
Putting x = 40, we have
40 = 45 (40 + 10)
⇒ 40 = 45 × 50
⇒ 40 = 4 × 10
⇒ 40 = 40
LHS = RHS
Hence verified.

8. 2×3 + 1 = 7×15 + 3
Solution:
We have 2×3 + 1 = 7×15 + 3
15(2×3 + 1) = 15(7×15 + 3)
LCM of 3 and 15 is 15
2×3 × 15 + 1 × 15 = 7×15 × 15 + 3 × 15 [Multiplying both sides by 15]
⇒ 2x × 5 + 15 = 7x + 45
⇒ 10x + 15 = 7x + 45
⇒ 10x – 7x = 45 – 15 (Transposing 7x to LHS and 15 to RHS)
⇒ 3x = 30
⇒ x = 30 ÷ 3 = 10 (Transposing 3 to RHS)
Thus the required solution is x = 10

9. 2y + 53 = 263 – y
Solution:


10. 3m = 5m – 85
Solution:
We have

 Extra Questions Very Short Answer Type | Class 8th Mathematics

Question 1.
Identify the algebraic linear equations from the given expressions.
(a) x² + x = 2
(b) 3x + 5 = 11
(c) 5 + 7 = 12
(d) x + y² = 3
Solution:
(a) x² + x = 2 is not a linear equation.
(b) 3x + 5 = 11 is a linear equation.
(c) 5 + 7 = 12 is not a linear equation as it does not contain variable.
(d) x + y² = 3 is not a linear equation.

Question 2.
Check whether the linear equation 3x + 5 = 11 is true for x = 2.
Solution:
Given that 3x + 5 = 11
For x = 2, we get
LHS = 3 × 2 + 5 = 6 + 5 = 11
LHS = RHS = 11
Hence, the given equation is true for x = 2

Question 3.
Form a linear equation from the given statement: ‘When 5 is added to twice a number, it gives 11.’
Solution:
As per the given statement we have
2x + 5 = 11 which is the required linear equation.

Question 4.
If x = a, then which of the following is not always true for an integer k. (NCERT Exemplar)
(a) kx = ak
(b) xk = ak
(c) x – k = a – k
(d) x + k = a + k
Solution:
Correct answer is (b).

Question 5.
Solve the following linear equations:
(a) 4x + 5 = 9
(b) x + 32 = 2x
Solution:
(a) We have 4x + 5 = 9
⇒ 4x = 9 – 5 (Transposing 5 to RHS)
⇒ 4x = 4
⇒ x = 1 (Transposing 4 to RHS)
(b) We have x + 32 = 2x
⇒ 32 = 2x – x
⇒ x = 32

Question 6.
Solve the given equation 31x × 514 = 1712
Solution:
We have 31x × 514 = 1712

Question 7.
Verify that x = 2 is the solution of the equation 4.4x – 3.8 = 5.
Solution:
We have 4.4x – 3.8 = 5
Putting x = 2, we have
4.4 × 2 – 3.8 = 5
⇒ 8.8 – 3.8 = 5
⇒ 5 = 5
L.H.S. = R.H.S.
Hence verified.

Question 8.

Solution:

⇒ 3x × 3 – (2x + 5) × 4 = 5 × 6
⇒ 9x – 8x – 20 = 30 (Solving the bracket)
⇒ x – 20 = 30
⇒ x = 30 + 20 (Transposing 20 to RHS)
⇒ x = 50
Hence x = 50 is the required solution.

Question 9.
The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles of the triangle.
Solution:
Let the angles of a given triangle be 2x°, 3x° and 4x°.
2x + 3x + 4x = 180 (∵ Sum of the angles of a triangle is 180°)
⇒ 9x = 180
⇒ x = 20 (Transposing 9 to RHS)
Angles of the given triangles are
2 × 20 = 40°
3 × 20 = 60°
4 × 20 = 80°

Question 10.
The sum of two numbers is 11 and their difference is 5. Find the numbers.
Solution:
Let one of the two numbers be x.
Other number = 11 – x.
As per the conditions, we have
x – (11 – x) = 5
⇒ x – 11 + x = 5 (Solving the bracket)
⇒ 2x – 11 = 5
⇒ 2x = 5 + 11 (Transposing 11 to RHS)
⇒ 2x = 16
⇒ x = 8
Hence the required numbers are 8 and 11 – 8 = 3.