# Important Questions for CBSE Class 8 Maths Chapter 14 Factorisation

## Important Questions Class 8 Maths Chapter 14 – Factorisation

Maths is the study of numerical, theorems, assumptions, interpretations, axioms, postulates and laws. The vastness of Maths can only be tackled with proper study and the right amount of practice. As a result, Maths experts suggest students practice Maths consistently to excel in the subject. Having a firm grip on the concepts and gaining excellence in Maths is difficult without proper practice and the right guidance.

Chapter 14 of Class 8 Maths is about Factorisation. The essential topics covered in this chapter are:

• Introduction
• What is Factorisation?
• Division of Algebraic Expressions
• Division of Algebraic Expressions
• Can you Find the Error?

Extramarks is one of the popular websites that provide comprehensive and reliable NCERT-related study materials. Our platform gives students the best of NCERT solutions, revision notes, CBSE mock tests, sample papers, CBSE past years’ papers, etc. which helps students to build proficiency in Maths.

Our faculty experts acknowledge the significance of solving problems to attain mastery in Maths. After careful research and study of CBSE past years’ question papers and NCERT textbook and exemplar, our team has compiled the most critical questions in our question bank of class 8 maths chapter 14 extra questions. This set of important questions and solutions will help students to furnish themselves with the proper concepts of all the topics relevant to Factorisation. Our step-by-step solutions with formulas and explanations make it easy for students to understand the answers without much difficulty. Extramarks believes in incorporating joyful learning experiences through its own repository of resources. Besides providing Important Questions Class 8 Maths Chapter 14, Extramarks also gives many other exam-oriented study materials. All the solutions provided by us are strictly in accordance with the NCERT books following the latest CBSE syllabus and guidance, making them completely reliable for the students to enhance their learning experience and achieve excellent results.

## Get Access to CBSE Class 8 Maths Important Questions 2022-23 with Chapter-Wise Solutions

You can also find CBSE Class 8 Maths Chapter-by-Chapter Important Questions here:

 CBSE Class 8 Maths Important Questions Sr No. Chapters Chapters Name 1 Chapter 1 Rational Numbers 2 Chapter 2 Linear Equations in One Variable 3 Chapter 3 Understanding Quadrilaterals 4 Chapter 4 Practical Geometry 5 Chapter 5 Data Handling 6 Chapter 6 Squares and Square Roots 7 Chapter 7 Cubes and Cube Roots 8 Chapter 8 Comparing Quantities 9 Chapter 9 Algebraic Expressions and Identities 10 Chapter 10 Visualising Solid Shapes 11 Chapter 11 Mensuration 12 Chapter 12 Exponents and Powers 13 Chapter 13 Direct and Inverse Proportions 14 Chapter 14 Factorisation 15 Chapter 15 Introduction to Graphs 16 Chapter 16 Playing with Numbers

## Factorisation Class 8 Extra Questions with Solutions

Given below are a few questions and their solutions from factorization class 8 important questions. Students can get the question bank by registering on our Extramarks website.

Question 1:  Find the common factors of the given term:

• 6abc, 24ab², 12a²b
• 10pq, 20qr, 30rp
• 3x²y³, 10x³y², 6x²y²z

Answer 1: (a)  On factorising 6abc, 24ab² and 12a²b, we get

6abc = 2 × 3 × a × b × c

24ab² = 2 × 2 × 2 × 3 × a × b × b

12a²b = 2 × 2 × 3 × a × a × b

Hence, the common factors of 6abc, 24ab² and 12a²b are 2, 3, a and b

Therefore, multiplying the common factors we get

2 × 3 × a × b = 6ab

(b) On factorising 10pq, 20qr and 30rp, we get

10pq = 2 × 5 × p × q

20qr = 2 × 2 × 5 × q × r

30rp = 2 × 3 × 5 × r × p

Hence, the common factors are 2 and 5

Therefore, multiplying the common factors we get

2 × 5 = 10

(c ) On factorising 3x²y³, 10x³y², 6x²y²z, we get

3x²y³ = 3 × x × x × y × y × y

10x³y² = 2 × 5 × x × x × x × y × y

6x²y²z = 2 × 3 × x × x × y × y × z

Hence, the common factors are x, x, y and y

Therefore, multiplying the common factors we get

x × x × y × y = x²y²

Question 2: Factorise the following expressions

• ax²y + bxy² + cxyz
• z – 7 + 7xy – xyz

Answer 2: (a) On factorising ax²y,  bxy² and cxyz, we get

ax²y = a + x + x + y

bxy² = b × x × y × y

cxyz = c × x × y × z

Hence, the common factors are x and y

Therefore, ax²y + bxy² + cxyz = xy ( ax + by + cz )

(b) z – 7 + 7xy – xyz

=⟩ z -7 – z (xy) + 7 (xy)

=⟩ ( z – 7) – xy ( z – 7 )

=⟩ ( 1 – xy ) ( z – 7 )

Question 3: Factorise the following expressions.

• (l + m)² – 4lm (Hindi: Expand (l + m)² first)
• 25m² + 30m + 9
• 16x5 – 144x³
• (l + m)² – (l – m)²

Answer 3: (a) (l + m)² – 4lm

=⟩ l² + m² + 2lm – 4lm

[Using (x + y)² = x² + 2xy + y²]

=⟩ l² + m² – 2lm

=⟩ (l – m)²

[Using (x – y)² = x² – 2xy + y²]

(b) 25m² + 30m + 9

=⟩ (5m)² + 2 × 5m × 3 + 3²

=⟩ (5m + 3)²

[Using (x + y)² = x² + 2xy + y² ]

(c ) 16x5 – 144x³

=⟩ 16x³ (x² – 9)

=⟩ 16x³ (x – 3) (x + 3).  [Using (x² – y²) = (x + y)(x – y)

(d) (l + m)² – (l – m)²

=⟩ {(l + m) – (l – m)} {(l + m) + (l – m)}.

[Using x² – y² = (x + y) (x – y)]

=⟩ (l + m – l + m) (l + m + l – m)

=⟩ (2m) (2l)

= 4ml

Question 3: Factorise the following expressions:

• 10ab + 4a + 5b + 2
• a⁴ – 2a²b² + b⁴
• q² – 10q + 21

Answer 3: (a) 10ab + 4a + 5b + 2

=⟩ 5b (2a + 1) + 2(2a + 1)

=⟩ (5b +2) (2a + 1)

(b) a⁴ – 2a²b² + b⁴

=⟩ (a²)² – 2a²b² + (b²)²

=⟩ (a² – b²)²

=⟩ {(a – b) (a + b)}²

=⟩ (a – b) ² (a + b)²

(c ) q² – 10q + 21

Here we observe that,

21 = -7 × -3 and -7 + (-3) = -10

=⟩ q² – 10q + 21 = q² – 3q – 7q + 21

=⟩ q(q – 3) – 7(q – 3)

=⟩ (q – 7) (q- 3)

Question 4: Carry out the following divisions.

• 34x³y³z³ ÷ 51xy²z³
• (x³ + 2x² + 3x) ÷ 2x
• 9x²y²(3z – 24) ÷ 27xy(z – 8)

Answer 4: (a) 34x³y³z³ / 51xy²z³

= 2 × 17 × x × x × x × y × y × y × z × z × z / 3 × 17 × x × y × y × z × z × z

= 2 x²y /3

(b) (x³ + 2x² + 3x) = x(x² + 2x + 3)

Therefore, x(x² + 2x + 3) / 2x

= (x² + 2x + 3) / 2

(c ) 9x²y²(3z – 24) / 27xy(z – 8)

= 9x²y² × 3(z – 8) / 27xy (z – 8)

= xy

Question 5: Divide the following as directed

• 20(y + 4) (y² + 5y + 3) ÷ 5(y + 4)
• 39y³ (50y² – 98) ÷ 26y²(5y + 7)

Answer 5: (a) 20(y + 4) (y² + 5y + 3) / 5(y + 4)

= 4(y² + 5y + 3)

(b) In this case, first we have to factorise 50y² – 98

50y² – 98 = 2(25y² – 49) = 2(5y + 7) (5y – 7)

Therefore, 39y³(50y² – 98) / 26y² (5y + 7)

= 2 × 3 × 13 × y³ (5y + 7) (5y – 7) / 2 × 13 × y²(5y + 7)

= 3y (5y – 7)

Question 6:  Find and correct the errors in the statement

(3x + 2)² = 3x² + 6x + 4

Answer 6: L. H. S. = (3x + 2)²

= (3x)² + 2² + 2 × 2 × 3x

= 9x² + 4 + 12x

1. H. S. = 3x² + 6x + 4

Therefore, L. H. S. ≠ R. H. S.

Hence, correct statement is (3x + 2)² = 9x² + 4 + 12x

Question 7:Find and correct the errors in the statement

(2a + 3b) (a – b) = 2a² – 3b²

Answer 7: L. H. S. = (2a + 3b) (a – b)

= 2a(a – b) + 3b(a – b)

= 2a² – 2ab + 3ab – 3b²

= 2a² + ab – 3b²

1. H. S. = 2a² – 3b²

Therefore, L. H. S. ≠ R. H. S.

Hence, the correct statement is (2a + 3b) (a – b) = 2a² + ab – 3b²

Question 8:Find and correct the errors in the statement

(z + 5)² = z² + 25

Answer 8: L. H. S. = (z + 5)²

(z + 5)² = z² + 10z + 25

[Using identity (a + b)² =  a² + 2ab + b²]

1. H. S. = z² + 25

Hence, L. H. S. ≠  R. H. S.

Therefore, the correct statement is (z + 5)² = z² + 10z + 25

Question 9: Find and correct the errors in the statement

3x / (3x + 2) = 1 / 2

Answer 9: L. H. S. = 3x / (3x + 2)

1. H. S. = 1 / 2

Therefore, L. H. S. ≠  R. H. S.

Hence, 3x / (3x + 2) = 3x / (3x + 2)

Question 10:Find and correct the errors in the statement

(7x + 5) / 5 = 7x

Answer 10: L. H. S. = (7x + 5) / 5

= 7x/5 + 5/5

= 7x/5 + 1

1. H. S. = 7x

Therefore, L. H. S. ≠ R. H. S.

Hence, the correct statement is (7x + 5) / 5 = (7x/5) + 1

Question 11: Factorise 4x² – 20x + 25.

Answer 11: 4x² – 20x + 25

= (2x)² – 2 × 2x × 5 + (5)²

= (2x – 5)²

[Using the identity a² – 2ab + b² = (a – b)²]

Question 12: Verify that

(3x + 5y)² – 30xy = 9x² + 25y²

Answer 12: L. H. S. = (3x + 5y)² – 30xy

9x² + 30xy + 25y² – 30xy = 9x² + 25y²

1. H. S. = 9x² + 25y²

Therefore, L. H. S. = R. H. S. (verified)

Question 13: Verify that

(11pq + 4q)² – (11pq – 4q)² = 176pq²

Answer 13: L. H. S. = (11pq + 4q)² – (11pq – 4q)²

= 121p²q² + 88pq² + 16q² – (121p²q² – 88pq² + 16q²)

[ Using identities (a + b)² = (a² + 2ab + b²)

And (a – b)² = (a² – 2ab + b²) ]

= 121p²q² + 88pq² + 16q² – 121p²q² + 88pq² – 16q²

= 88pq² + 88pq²

= 176pq²

1. H. S. = 176pq²

Therefore, L. H. S. = R. H. S. (verified)

Question 14: The area of a rectangle is x² + 12xy + 27y² and its  length is (x + 9y). Find the breadth of the rectangle.

= (x² + 12xy + 27y²) / (x + 9y)

= x(x + 9y) + 3y(x + 9y) / (x + 9y)

= (x + 3y) (x + 9y) / (x + 9y)

= (x + 3y)

Hence, the breadth of the rectangle is (x + 3y)

Question 15: Divide 15(y + 3)(y² – 16) by 5(y² – y – 12).

Answer 15: On factorising 15(y + 3)(y² – 16), we get 5 × 3 × (y + 3)(y – 4)(y + 4).

On factorising 5(y² – 4y + 3y – 12)

= 5(y – 4)(y + 3)

Therefore, on dividing the first expression by second expression, we get

15(y + 3) (y² – 16) / 5 (y +3) (y – 4)

= 3(y + 4)

Question 16: Factorise 2ax² + 4axy + 3bx² + 2ay² + 6bxy + 3by².

Answer 16: 2ax² + 4axy + 3bx² + 2ay² + 6bxy + 3by²

= 2ax² + 4axy + 3bx² + 6bxy + 2ay² + 3by²

= 2ax(x + 2y) + 3bx(x + 2y) + 2y²(2a + 3b)

= x(2a + 3b)(x + 2y) + 2y²(2a + 3b)

= (2a + 3b) [x(x + 2y) + 2y²]

= (2a + 3b) ( x² + 2y² + 2xy]

Question 17: Factorise 4a² – 4ab + b²

Answer 17: 4a² – 4ab + b²

= (2a)² – 2(2a)(b) + b²

= (2a – b)²

[Using the identity a² – 2ab + b² = (a – b)² ]

Question 18: Factorise 3a²b³ – 27a⁴b

= 3a²b(b² – 9a²)

= 3a²b(b² – (3a)²)

= 3a²b(b + 3a)(b – 3a)

[Using the identity (a² – b²) = (a + b)(a – b)

Question 19: Factorise (4x² / 9) – (9y² / 16)

Answer 19: (4x² / 9) – (9y² / 16)

= (2x / 3)² – (3y / 4)²

= [(2x / 3) + (3y / 4)][(2x / 3) – (3y / 4)]

[Using the identity (a² – b²) = (a + b)(a – b)]

Question 20: Factorise 1331x³y – 11y³x

= 11xy (121x² – y²)

= 11xy [(11x)² – y²]

= 11xy (11x – y)(11x + y)

[Using the identity (a² – b²) = (a + b)(a – b)]

Question 21: The area of a rectangle is x² + 19x – 20. Find the possible length and the breadth of the rectangle.

= x² + 19x – 20

= x² + 20x – x – 20

= x(x + 20) – 1(x + 20)

= (x – 1) (x + 20)

Thus, the length and the breadth are (x – 1) and (x + 20)

Question 22: Perform the following division:

(3pqr – 6p²q²r²) ÷ 3pq

Answer (3pqr – 6p²q²r²) ÷ 3pq

= (3pqr – 6p²q²r²) / 3pq

= 3pqr (1 – 2pqr) / 3pq

= r(1 – 2pqr)

Question 23: Perform the following division:

(x³y)/9 – (xy³)/16

= xy(x²/9 – y²/16)

= xy [(x/3)² – (y/4)²]

= xy (x/3 – y/4)(x/3 + y/4)

[Using the identity (a² – b²) = (a + b)(a – b) ]

Question 24: The area of a rectangle is x² + 7x + 12. If the breadth is (x + 3), find its length.

=⟩ x² + 7x + 12 = Length × (x + 3)

=⟩ Length = (x² + 7x + 12) / (x +3)

=⟩ Length = (x² + 3x + 4x + 12) / (x + 3)

=⟩ Length = x(x + 3) + 4(x + 3) / (x + 3)

=⟩ Length = (x + 3)(x + 4) / (x + 3)

=⟩ Length = (x + 4)

Question 25: The area of a circle is given by the expression πx² + 6πx + 9π. Find the radius of the circle.

Answer 25: Area of a circle = πr²

Then, πx² + 6πx + 9π = πr²

=⟩ π(x² + 6x + 9) = πr²

=⟩ (x² + 6x + 9) = r²

=⟩ r² = (x² + 2.x.3 + 3²)

=⟩ r² = (x + 3)²

Therefore, r = x + 3

Question 26: The sum of the first n natural numbers is given by the expression n²/2 + n/2. Factorise this expression.

Answer 26: Given that the sum of the first n natural number = n²/2 + n/2 = n/2 (n + 1)

Question 27: The sum of (x + 5) observations is x⁴ – 625. Find the mean of the observations.

Answer 27: Mean = (x⁴ – 625) / (x + 5)

Mean = [( x²)² – (25)²] / (x + 5)

Mean = [(x² + 25)(x² – 5²)] / (x + 5)

Mean = [(x² + 25)(x – 5)(x + 5)] / (x + 5)

Mean = (x² + 25)(x – 5)

Question 28: The height of a triangle is x⁴ + y⁴ and its base is 14xy. Find the area of the triangle.

Answer 28: Area of the triangle = 1 /2 × height × base

=⟩ Area = 1 /2 × (x⁴ + y⁴) × (14xy)

=⟩ Area = 7xy (x⁴ + y⁴)

Question 29: The cost of a chocolate is Rs (x + y) and Rohit bought (x + y) chocolates. Find the total amount paid by him in terms of x. If x = 10, find the amount paid by him.

Answer 29: The cost of chocolate = Rs (x + y)

No. of chocolates Rohit bought = (x + y)

Therefore, total amount he paid = Rs (x + y)(x + y)

= Rs (x + y)²

If x = 10, then Rs (10 + y)²

Question 30: The base of a parallelogram is (2x + 3 units) and the corresponding height is (2x – 3 units). Find the area of the parallelogram in terms of x. What will be the area of the parallelogram of x = 30 units?

Answer: Area of Parallelogram = Base × Height

Therefore, Area = (2x + 3)(2x – 3)

Area = (2x)² – (3)² = 4x² – 9

Putting x = 30 units, we get

Area = 4 × (30)² – 9 = 4 × 900 – 9 = 3600 – 9 = 3591sq. units.

Question 31: The radius of a circle is 7ab – 7bc – 14ac. Find the circumference of the circle. (π = 22/7)

Answer 31: The circumference of the circle = 2πr

Therefore, Circumference = 2π (7ab – 7bc – 14ac)

Circumference = 2 × 22/7 (7ab – 7bc – 14ac)

= 2 × 22 (ab – bc – 2ac)

= 44(ab – bc – 2ac)

Question 32: Factorise p⁴ + q⁴ + p²q²

Answer 32: p⁴ + q⁴ + p²q²

= (p²)² + (q²)² + 2p²q² – p²q²

= (p²+ q²)² – (pq)²

[Using the identity a² + b² + 2ab = (a + b)²]

= (p² + q² + pq)(p² + q² – pq)

[Using the identity a² – b² = (a + b)(a – b)]

Question 33: Factorise the expression and divide them as directed:

(2x³ – 12x² + 16x) ÷ (x – 2)(x – 4)

Answer 33: (2x³ – 12x² + 16x) / [(x – 2)(x – 4)]

= [2x (x² – 6x + 8)] / [(x – 2)(x – 4)]

= [2x (x² – 2x – 4x + 8)] / [(x – 2)(x – 4)]

= [2x {x (x – 2) – 4 (x – 2)}] / [(x – 2) (x – 4)]

= [2x (x – 4)(x – 2)] / [(x – 2) (x – 4)]

= 2x

Question 34: Factorise x² + 1/x² + 2 – 3x – 3/x

Answer 34: x² + 1/x² + 2 – 3x – 3/x

=⟩ x² + 1/x² + 2 – 3 (x + 1/x²)

=⟩ (x + 1/x)² – 3 (x + 1/x²)

[Using the identity a² + b² + 2ab = (a + b)² ]

=⟩ (x + 1/x) (x + 1/x – 3)

## Benefits of solving Class 8 Factorisation Extra Questions

At Extramarks, we understand the importance of solving important questions and we take our role seriously to provide the best resource to the students and help them get excellent grades. Practice is mandatory to be an expert in Maths. Students gain a good grasp on the concepts and feel confident in Maths only through firm practice. By practising the Important Questions Class 8 Maths Chapter 14 provided by Extramarks, students shall be prepared and proficient in solving any perplexing problems during examinations.

Below are some benefits of solving Class 8 Maths Chapter 14 Important Questions:

• Important Questions Class 8 Maths Chapter 14 is a collection of some specially chosen questions by experienced  Maths faculty members. Our experts have prepared these questions after taking into account and analysing all the past years’ question papers and  NCERT  textbooks
• Every question and answer provided by Maths Class 8 Chapter 14 Important Questions strictly adheres to the latest CBSE syllabus and follows the CBSE guidelines so that students can completely rely on our solutions.
• All the Chapter 14 Class 8 Maths Important Questions have been solved and supported with the required formulas and explanations for the students to understand the concept well.

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Q.1 Divide 39x3 (50x2 98) by 26x2(5x + 7).

Marks:4
Ans

$\begin{array}{l}39{\mathrm{x}}^{3}\left(50{\mathrm{x}}^{2}\text{}98\right)\\ \\ 39\text{}×{\text{x}}^{\text{3}}\text{}×\text{2}\left({\text{25x}}^{\text{2}}\text{49}\right)\\ \\ \text{=39}×{\text{x}}^{\text{3}}\text{}×\text{2[}{\left(\text{5x}\right)}^{\text{2}}\text{}{\left(\text{7}\right)}^{\text{2}}\text{]}\\ \\ \text{=39}×{\text{x}}^{\text{3}}\text{}×\text{2}\left(\text{5x+7}\right)\left(\text{5×7}\right)\text{}\left\{\mathrm{Since},{a}^{2}{\text{b}}^{2}\text{=(a+b)(ab)}}\\ \\ \text{=13}×\text{3}×{\text{x}}^{\text{2}}\text{}×\text{x}×\text{2}×\text{}\left(\text{5x+7}\right)\left(\text{5×7}\right)\text{}\\ \\ {\text{=26x}}^{\text{2}}\text{}×\text{3}×\text{x}×\text{}\left(\text{5x+7}\right)\left(\text{5×7}\right)\\ \\ \mathrm{We}\mathrm{have}\mathrm{to}\mathrm{divide}39{\text{x}}^{\text{3}}\left({\text{50x}}^{\text{2}}\text{98}\right){\text{by26x}}^{\text{2}}\left(\text{5x+7}\right)\text{}\\ \\ \text{=}\frac{26{\text{x}}^{\text{2}}\text{}×\text{3}×\text{x}×\text{}\left(\text{5x+7}\right)\left(\text{5×7}\right)}{26{\mathrm{x}}^{2}\left(\text{5x+7}\right)\text{}}\\ \\ =3\mathrm{x}\left(\text{5×7}\right)\text{}\\ \\ {\text{=15x}}^{\text{2}}\text{21x}\end{array}$

Q.2 Factorise the polynomial given below.
7 + 10(ab) 8(ab)2

Marks:4
Ans

We have,

7 + 10(ab) 8(ab)2

Let x = (ab), then we get

7 + 10(ab) 8(ab)2 = 7 + 10x 8x2

We now find two numbers such that their sum is 10 and product is -56.

The numbers are 14 and -4.

Therefore, 7 + 10x 8x2 = 7 + 14x 4x 8x2

= 7(1 + 2x) 4x(1 + 2x)

= (1 + 2x)(7 4x)

= (1 + 2(a b))(7 4(a b))

[Put x = (a b)]

= (1 + 2a 2b)(7 4a + 4b)

Q.3 Factorise ax2 + by2 + bx2 + ay2.

Marks:2
Ans

ax2 + by2 + bx2 + ay2

= ax2 + bx2 + by2 + ay2

= x2(a + b) +y2(a + b)      [Taking x2 and y2 as common]

= (x2 + y2)(a + b)

Q.4 Factorise the polynomial given below

(2r 3s)2 7(2r 3s) 30

Marks:3
Ans

We have,

(2r 3s)2 7(2r 3s) 30

Let x = (2r 3s), then we get

(2r 3s)2 7(2r 3s) 30= x2 7x 30

We now find two numbers such that their sum is 7 and product is 30.

The numbers are 10 and 3.

Therefore, x2 7x 30 = x2 10x + 3x 30

= x(x 10) + 3(x 10)

= (x + 3)(x 10)

= (2r 3s + 3)(2r 3s 10) [Putting x = (2r 3s)]

Q.5 Find the factors of 25x2 4y2 + 28yz 49z2.

Marks:5
Ans

25x2 4y2 + 28yz 49z2
= 25x2 [4y2 28yz + 49z2]
= 25x2 [(2y)2 2 x 2y x 7z + (7z)2]
= (5x)2 (2y 7z)2
= (5x + 2y 7z)(5x 2y + 7z) [Using a2 b2 = (a+b)(a b)]

## 1. What are the topics covered in Important Questions Class 8 Maths Chapter 12?

Our experienced  Maths subject experts have carefully picked some best questions for our question bank Important Questions Class 8 Maths Chapter 14. They have chosen the questions after carefully analysing and discussing the CBSE past year question papers, NCERT books and exemplars. The questions provided in our question bank are in accordance with the following topics:

1. Introduction
2. What is Factorisation?
3. Division of Algebraic Expression
4. Can you Find the Error?

Hence, our experts have tried to cover the whole chapter to make it a reliable source for the students to understand, practice and master the chapter.

## 2. How can a student score above 90% marks in Maths?

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2. Next, they should try figuring out their weak and strong parts in maths. They should then focus and work on their weak part and strengthen it.  The topics which you are good at can be given less time as the concept is clear, but the daily practice is required.
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