CBSE Extra Questions For Class 8 Maths
CBSE Maths Extra Questions Class 8
With the help of Extra Questions for Class 8 Maths and some references from NCERT books, students can practise for the final exam. It is important for the students to also refer to CBSE revision notes and CBSE sample papers for a better analysis of how the overall pattern will be.
Importance Of CBSE Extra Questions in Class 8 Maths
Maths Class 8 Extra Questions from the CBSE syllabus should be practised in order to understand the pattern of the final exam. The CBSE past year question papers can be a great option to understand different long and short extra questions.
Chapter 1 – Rational Numbers
- What are the multiplicative and additive identities of rational numbers?
Solution: 0 and 1 are the additive and multiplicative identities of rational numbers respectively.
- Mention the commutativity, associative and distributive properties of rational numbers. Also, check a × b = b × a and a + b = b + a for a = ½ and b = ¾
Solution:
Commutative property:
For any two rational numbers a and b, a + b = b + a.
For any two rational numbers a and b, a × b = b × a.
Associative Property:
For any three rational numbers a, b and c,
(a + b) + c = a + (b + c)
Distributive property states that for any three numbers x, y and z,
x × ( y + z ) = (x × y) + ( x × z)
a*b = b*a
a*b = ½ * ¾ = 3/8
b*a = ¾ * ½ = 3/8
a + b = ¾ + ½ = 5/4
b + a = ½ + ¾ = 5/4
Chapter 2 – Linear Equations in One Variable
- Solve: 23x/2 = 46
Solution:
23x/2 = 46
x = (46 × 2)/23
x = 2 × 2
x = 4
Chapter 3 – Understanding Quadrilaterals
- A quadrilateral has three acute angles, each measuring 80°. What is the measure of the fourth angle?
Solution:
Let x be the measure of the fourth angle of a quadrilateral.
Sum of the four angles of a quadrilateral = 360°
80° + 80° + 80° + x = 360°
x = 360° – (80° + 80° + 80°)
x = 360° – 240°
x = 120°
Hence, the fourth angle is 120°.
Chapter 4 – Practical Geometry
- Construct a Rhombus BEND with BN = 5.6 cm DE = 6.5 cm.
Solution:
As we know, the diagonals of a rhombus bisect each other at a right angle.
Let us consider that these diagonals intersect each other at a point O in rhombus.
Hence, EO = OD = 3.25 cm
Chapter 5 – Data Handling
- Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box and mixed well. One slip is chosen from the box without looking into it. What is the probability of:
(i) getting a number 6?
(ii) getting a number less than 6?
(iii) getting a number greater than 6?
(iv) getting a 1-digit number?
Solution:
(i) The outcome of getting a number 6 from ten separate slips is one.
Therefore, the probability of getting a number 6 = 1/10
(ii) Numbers which are less than 6 are 1, 2, 3, 4 and 5. So in total, there are five numbers. Thus, there are 5 outcomes.
Therefore, the probability of getting a number less than 6 = 5/10 = ½
(iii) The number is greater than 6 out of ten that are 7, 8, 9, 10. So there are 4 possible outcomes.
Therefore, the probability of getting a number greater than 6 = 4/10 = ⅖
(iv) One digit numbers out of 1 to 10 are 1, 2, 3, 4, 5, 6, 7, 8, 9.
Therefore, the probability of getting a 1-digit number = 9/10
Chapter 6 – Squares and Square Roots
- (n+1)²-n² = ?
Solution:
(n+1)²-n²
= (n² + 2n + 1) – n²
= 2n + 1
Chapter 7 – Cubes and Cube Roots
- Find the cube of 3.5.
Solution:
3.5³ = 3.5 x 3.5 x 3.5
= 12.25 x 3.5
= 42.875
Chapter 8 – Comparing Quantities
- Express 25% and 12% as decimals.
Solution:
25% = 25/100 = 0.25
12% = 12/100 = 0.12
Chapter 9 – Algebraic Expressions and Identities
- Using suitable algebraic identity, solve 1092²
Solution:
Use the algebraic identity: (a + b)² = a² + 2ab + b²
Now, 1092 = 1000 + 92
So, 1092² = (1000 + 92)²
(1000 + 92)² = ( 1000 )² + 2 × 1000 × 92 + ( 92 )²
= 1000000 + 184000 + 8464
Thus, 1092² = 1192464.
Chapter 10 – Visualizing Solid Shapes
- What is Euler’s formula for a polyhedron?
Solution:
Euler’s formula for polyhedra states that for any solid shape, the number of vertices (V) minus the number of edges (E) plus the number of faces (F) always equals to 2.
So, V – E + F = 2
Chapter 11 – Mensuration
- The parallel sides of a trapezium measure 12 cm and 20 cm. Calculate its area if the distance between the parallel lines is 15 cm.
Solution:
Area of trapezium = ½ × perpendicular distance between parallel sides × sum of parallel sides
= ½ × 15 × (12 + 20)
= 1/2 × 15 × 32
= 15 × 16
= 240 cm²
Chapter 12 – Exponents and Powers
- Find the value of (40 + 4¯¹) × 22
Solution:
(4⁰ + 4¯¹) × 2² = (1 + ¼) × 4
= 5/4 x 4
= 5
- Solve 3¯⁴ and (½)¯²
Solution:
We know, b¯ⁿ = 1/bⁿ
So, 3¯ = 1/3⁴ = 1/81
And, (½)¯² = 1¯²/2¯² = 2²/1² = 4
Chapter 13 – Direct and Inverse Proportions
- In a certain situation, the increase in time causes a corresponding decrease in the price of a product. Identify the proportionality.
Solution:
As per the given question, the increase in time reduces the price of a product. Thus,
Time ∝ 1/Product Price
Hence, the time and price of the product are inversely proportional.
Chapter 14 – Factorisation
- Express the following as in the form of (a+b)(a-b)
(i) a² – 64
(ii) 20a² – 45b²
(iii) 32x²y² – 8
(iv) x² – 2xy + y² – z²
(v) 49x² – 1
Solution:
For representing the expressions in (a+b)(a-b) form, use the following formula
a² – b² = (a+b)(a-b)
(i) a² – 64 = a² – 8² = (a + 8)(a – 8)
(ii) 20a² – 45b² = 5(4a² – 9b²) = 5(2a + 3b)(2a – 3b)
(iii) 32x²y² – 8 = 8( 4x²y² – 1) = 8(2xy + 1)(2xy – 1)
(iv) x² – 2xy + y² – z² = (x – y)² – z² = (x – y – z)(x – y + z)
(v) 49x² – 1 = (7x)² – (1)² = (7x + 1)(7x – 1)
Chapter 15 – Introduction to Graphs
- Plot the following points and verify if they lie on a line. If they lie on a line, name it.
(i) (0, 2), (0, 5), (0, 6), (0, 3.5)
(ii) A (1, 1), B (1, 2), C (1, 3), D (1, 4)
(iii) K (1, 3), L (2, 3), M (3, 3), N (4, 3)
(iv) W (2, 6), X (3, 5), Y (5, 3), Z (6, 2)
Chapter 16 – Playing with Numbers
- Express 3458 in a generalised form.
Ans: 3458 will be written in a generalised way in the following manner,
3458 = 3 x 10³ + 4 x 10² + 5 x 10¹ + 8 x 10⁰
