NCERT Solutions for Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables - Exercise 3.7

Exercise 3.7

1. The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

Ans:

The age gap between Biju and Ani is a whopping 3. Biju is either 3 older than Ani, or Biju is 3 older than Ani.

However, it is self-evident that Ani's father's age will be 30 more than Cathy's age in both situations.

Let us assume that the age of Ani and Biju be x and y years.

Therefore, age of Ani’s father i.e. Dharam will be 2 × x=2x years 

And age of Biju’s sister Cathy = \[\dfrac{\text{y}}{\text{2}}\]

 years

Using the data provided in the question,

Case (I) When Ani is older than Biju by 3 years,

\[\text{x-y}\ \text{=}\ \text{3}\]   (i)  

\[\text{4x-y}\ \text{=}\ 60\]    (ii)

On subtracting (i) from (ii), we get

\[\text{3x}\ \text{=}\ \text{60-3}\ \text{=}\ \text{57}\]

\[\text{x}\ \text{=}\ \dfrac{57}{3}\ =\ 19\]

Therefore, age of Ani = 19 years 

And age of Biju =19-3 = 16 years

Case (II) if Biju is older than Ani, \[\text{y-x}\,\text{=}\ \text{3}\]

\[\text{2x-}\dfrac{\text{y}}{\text{2}}\ \text{=}\ \text{30}\]    (i)    

\[\text{4x-y}\ \text{=}\ \text{60}\]         (ii)

Adding (i) and (ii), we get 

\[\text{3x}\ \text{=}\ \text{63}\]

\[\text{x}\ \text{=}\ \text{21}\]

Hence, age of Ani =

\[\text{21}\]

years 

And also the age of Biju will be

\[text{21+3}\ \text{=}\ \text{24}\]

years
 

2. One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? (From the Bijaganita of Bhaskara II)

(Hint: \[\text{x+100}\ \text{=}\ \text{2(y-100),}\ \text{y+10}\ \text{=}\ \text{6(x-10)}\])

Ans:

Let those friends have Rs. x and y with them. 

Using the data provided in the question,

\[\text{x+100}\ \text{=}\ \text{2(y-100)}\]

\[\text{x+100}\ \text{=}\ \text{2y-200}\]

\[\text{x-2y}\ \text{=}\ -3\text{00}\]

(i) 

And

\[\text{6(x-10)}\ \text{=}\ \text{y+10}\]

\[\text{6x-60}\ \text{=}\ \text{y+10}\]

\[\text{6x-y}\ \text{=}\ 7\text{0}\]

(ii)

Multiplying equation (ii) by

\[2\]

, we get 

\[\text{12x-2y}\ \text{=}\ \text{140}\]

On subtracting equation (i) from equation (iii), we get 

\[\text{11x}\ \text{=}\ \text{140+300}\ \text{=}\ \text{440}\]

\[\text{x}\ \text{=}\ \text{40}\]

Using this in equation (i), we get 

\[\text{40-2y}\ \text{=}\ \text{-300}\]

\[\text{40+300}\ \text{=}\ \text{2y}\]

\[\text{2y}\ \text{=}\ \text{340}\]

\[\text{y}\ \text{=}\ 170\]

Therefore, those friends had Rs. 40 and Rs. 170

with them respectively.

3. A train covered a certain distance at a uniform speed. If the train would have been \[\text{10}\ \text{km/hr}\] faster, it would have taken 2 hours less than the scheduled time. And if the train were slower by \[\text{10}\ \text{km/hr}\]. it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Ans:

Let the speed of the train be \[\text{x}\ \text{km/hr}\] and the time taken by train to travel the given distance be t hours and the distance to travel was \[\text{d}\ \text{km}\]. We know that,

\[\text{speed}\ \text{=}\ \dfrac{\text{distance}\ \text{travelled}}{\text{time}\ \text{taken}\ \text{to}\ \text{travel}\ \text{that}\ \text{distance}}\]

\[\text{x}\ \text{=}\ \dfrac{\text{d}}{\text{t}}\]

Or,

\[\text{d}\ \text{=}\ \text{xt}\]

(i)

Using the data provided in the question,

\[\text{(x+10)}\ \text{=}\ \dfrac{\text{d}}{\text{(t-2)}}\]

\[\text{(x+10)(t-2)}\ \text{=}\ \text{d}\]

\[\text{xt+10t+2x+20 = d}\]

By using equation (i), we get

\[\text{-2x+10t = 20}\]

(ii)

\[\text{(x-10)}\ \text{=}\ \dfrac{\text{d}}{\text{(t+3)}}\]

\[\text{(x-10)(t+3)}\ \text{=}\ \text{d}\]

\[\text{xt-10t+3x-30 = d}\]

By using equation (i), we get 

\[\text{3x-10t}\ \text{=}\ \text{30}\]

(iii)

On adding equation (ii) and (iii), we get 

\[\text{x = 50}\]

Using equation (ii), we get 

\[\text{(-2)(50)+10t = 20}\]

\[\text{-100+10t = 20}\]

\[\text{10t = 120}\]

\[\text{t = 12}\]

hours

From equation (i), we get 

Distance to travel =

\[\text{d}\ \text{=}\ \text{xt}\]

=

\[\text{50}\times \text{12}\]

=

\[\text{600}\ \text{km}\]

Hence, the distance covered by the train is

\[\text{600}\ \text{km}\].

4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3students are less in a row, there would be 2 rows more. Find the number of students in the class.

Ans:

Let the number of rows be x and number of students in a row be y 

Total students of the class = Number of rows × Number of students in a row = \[\text{xy}\]

Using the data provided in the question,

Condition 1 

Total number of students =

\[\text{(x-1)(y+3)}\]

\[\text{xy = xy-y+3x-3}\]

\[\text{3x-y = 3}\]

(i)

Condition 2 

Total number of students =

\[\text{(x+2)(y-3)}\]

\[\text{xy = xy+2y-3x-6}\]

\[\text{3x-2y = -6}\]

(ii)

On subtracting equation (ii) from (i), 

\[\text{(3x-y)-(3x-2y) = 3-(-6)}\]

\[\text{-y+2y}\ \text{=}\ \text{3+6}\]

\[\text{y = 9}\]

By using equation (i), we get 

\[\text{3x-9}\ \text{=}\ \text{3}\]

\[\text{3x}\ \text{=}\ 9+3\ =\ 12\ \Rightarrow \ \text{x}\ \text{=}\ \text{4}\]

Number of rows =

\[4\]

Number of students in a row =

\[9\] 

Number of total students in a class =

\[4\times 9\ =\ 36\]
 

5. In a triangle ABC, \[\angle \text{C}\ =\ 3\angle \text{B}\ =\ 2(\angle \text{A}+\angle \text{B})\]. Find the three angles.

Ans:

\[\angle \text{C}\ =\ 3\angle \text{B}\ =\ 2(\angle \text{A}+\angle \text{B})\] 

\[3\angle \text{B}\ =\ 2(\angle \text{A}+\angle \text{B})\]

\[3\angle \text{B}\ =\ 2\angle \text{A}+2\angle \text{B}\] 

\[\angle \text{B}\ =\ 2\angle \text{A}\]

\[2\angle \text{A}-\angle \text{B}\ =\ 0\]

(i)

As we already know, the sum of the measures of all angles of a triangle is

\[{{180}^{\bullet }}\].

Therefore,

\[\angle \text{A+}\angle \text{B+}\angle \text{C}\ \text{=}\ \text{18}{{\text{0}}^{\bullet }}\]

\[\angle \text{A+}\angle \text{B+3}\angle \text{B}\ \text{=}\ \text{18}{{\text{0}}^{\bullet }}\]

\[\angle \text{A+4}\angle \text{B}\ \text{=}\ \text{18}{{\text{0}}^{\bullet }}\]

(ii) 

Multiplying equation (i) by\[4\], we get

\[8\angle \text{A-4}\angle \text{B}\ \text{=}\ 0\]

(iii)

On adding equation (ii) and (iii), we get

\[9\angle \text{A}\ \text{=}\ \text{18}{{\text{0}}^{\bullet }}\]

\[\angle \text{A}\ \text{=}\ 2{{\text{0}}^{\bullet }}\]

From equation (ii), we get 

\[2{{\text{0}}^{\bullet }}+4\angle \text{B}\ \text{=}\ \text{18}{{\text{0}}^{\bullet }}\]

\[4\angle \text{B}\ \text{=}\ \text{16}{{\text{0}}^{\bullet }}\]

\[\angle \text{B}\ \text{=}\ 4{{\text{0}}^{\bullet }}\]

\[\angle \text{C}\ \text{=}\ 3\angle \text{B}\ \text{=}\ \text{3}\times \text{4}{{\text{0}}^{\bullet }}\ =\ {{120}^{\bullet }}\]

Therefore,

\[\angle \text{A}\ \text{=}\ 2{{\text{0}}^{\bullet }}\],

\[\angle \text{B}\ \text{=}\ 4{{\text{0}}^{\bullet }}\],

\[\angle \text{C}\ \text{=}\ \text{12}{{\text{0}}^{\bullet }}\]
 

6. Draw the graphs of the equation \[\text{5x-y}\ \text{=}\ \text{5}\] and\[\text{3x-y}\ \text{=}\ 3\]. Determine the coordinates of the vertices of the triangle formed by these lines and the \[\text{y}\]-axis.

Ans:

\[\text{5x-y}\ \text{=}\ \text{5}\] Or, \[\text{y}\ \text{=}\ \text{5x-5}\]

The solution table will be as follows: 

\[\text{3x-y}\ \text{=}\ 3\].

Or,

\[\text{y}\ \text{=}\ 3\text{x-3}\]

The solution table will be as follows: 

Graphical representation of these lines is given by

Graphical representation of Given Lines

It can be observed that the required triangle is SABC formed by these lines and \[\text{y}\] -axis 

The coordinates of vertices are \[\text{A(1,0),}\ \text{B(0,-3),}\ \text{C(0,-5)}\]

7. Solve the following pair of linear equation.

(i) \[\text{px+qy}\ \text{=}\ \text{p-q}\] and \[\text{qx-py}\ \text{=}\ \text{p+q}\]

Ans:

\[\text{px+qy}\ \text{=}\ \text{p-q}\]

(1)

\[\text{qx-py}\ \text{=}\ \text{p+q}\]

(2)

Multiplying equation (1) by p and equation (2) by q , we get

\[{{\text{p}}^{\text{2}}}\text{x+pqy}\ \text{=}\ {{\text{p}}^{\text{2}}}\text{-pq}\]

(3)

\[{{\text{q}}^{\text{2}}}\text{x-pqy}\ \text{=}\ {{\text{q}}^{\text{2}}}\text{+pq}\]

(4)

On adding equation (3) and (4), we get

\[{{\text{p}}^{\text{2}}}\text{x+}{{\text{q}}^{\text{2}}}\text{x}\ \text{=}\ {{\text{p}}^{\text{2}}}\text{+}{{\text{q}}^{\text{2}}}\]

\[\text{(}{{\text{p}}^{\text{2}}}\text{+}{{\text{q}}^{\text{2}}}\text{)x}\ \text{=}\ {{\text{p}}^{\text{2}}}\text{+}{{\text{q}}^{\text{2}}}\]

\[\text{x}\ =\ 1\]

From equation (1), we get

\[\text{p(1)+qy}\ \text{=}\ \text{p-q}\]

\[\text{qy}\ \text{=}\ \text{-q}\]

\[\text{y}\ \text{=}\ \text{-1}\]
 

(ii)

\[\text{ax+by}\ \text{=}\ \text{c}\] and \[\text{bx+ay}\ \text{=}\ 1+\text{c}\]

Ans:

\[\text{ax+by}\ \text{=}\ \text{c}\]

(1)

\[\text{bx+ay}\ \text{=}\ 1+\text{c}\]

(2)

Multiplying equation (1) by a and equation (2) by b, we get

\[{{\text{a}}^{\text{2}}}\text{x+aby}\ \text{=}\ \text{ac}\]

(3)

\[{{\text{b}}^{\text{2}}}\text{x+aby}\ \text{=}\ \text{b+bc}\]

(4)

On subtracting equation (4) from equation (3),

\[\text{(}{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}}\text{)x}\ \text{=}\ \text{ac-bc-b}\]

\[\text{x}\ \text{=}\ \dfrac{\text{c(a-b)-b}}{{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}}}\]

From equation (1), we get

\[\text{ax+by}\ \text{=}\ \text{c}\]

\[\text{a }\!\!\{\!\!\text{ }\dfrac{\text{c(a-b)-b}}{{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}}}\text{ }\!\!\}\!\!\text{ +by}\ \text{=}\ \text{c}\]

\[\dfrac{\text{ac(a-b)-ab}}{{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}}}\text{+by}\ \text{=}\ \text{c}\]

\[\text{by}\ \text{=}\,\text{c-}\dfrac{\text{ac(a-b)-ab}}{{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}}}\]

\[\text{by}\ \text{=}\,\dfrac{{{\text{a}}^{\text{2}}}\text{c-}{{\text{b}}^{\text{2}}}\text{c-}{{\text{a}}^{\text{2}}}\text{c+abc+ab}}{{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}}}\]

\[\text{by}\ \text{=}\,\dfrac{\text{-}{{\text{b}}^{\text{2}}}\text{c+abc+ab}}{{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}}}\]

\[\text{by}\ \text{=}\,\dfrac{\text{bc(a-b)+ab}}{{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}}}\]

\[\text{y}\ \text{=}\,\dfrac{\text{c(a-b)+a}}{{{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}}}\]
 

(iii) \[\dfrac{\text{x}}{\text{a}}\text{-}\dfrac{\text{y}}{\text{b}}\ \text{=}\ \text{0}\] and \[\text{ax+by}\ \text{=}\ {{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\]

Ans:

\[\dfrac{\text{x}}{\text{a}}\text{-}\dfrac{\text{y}}{\text{b}}\ \text{=}\ \text{0}\] or \[\text{bx-ay}\ \text{=}\ \text{0}\]

(1) \[\text{ax+by}\ \text{=}\ {{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\]

(2) On multiplying the equation (1) and (2) by b and a respectively, we get \[{{\text{b}}^{2}}\text{x-aby}\ \text{=}\ \text{0}\]

(3) \[{{\text{a}}^{\text{2}}}\text{x+aby}\ \text{=}\ {{\text{a}}^{\text{3}}}\text{+a}{{\text{b}}^{\text{2}}}\]

(4) On adding equation (3) and (4), we get \[{{\text{b}}^{2}}\text{x+}{{\text{a}}^{\text{2}}}\text{x}\ \text{=}\ {{\text{a}}^{\text{3}}}\text{+a}{{\text{b}}^{\text{2}}}\]

\[\text{x(}{{\text{b}}^{\text{2}}}\text{+}{{\text{a}}^{\text{2}}}\text{)}\ \text{=}\ \text{a(}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{)}\]

\[\text{x}\ \text{=}\ \text{a}\]

By using (1), we get

\[\text{b(a)-ay}\ \text{=}\ \text{0}\]

\[\text{ab-ay}\ \text{=}\ \text{0}\]

\[\text{ab}\ \text{=}\ \text{ay}\]

\[\text{y}\ \text{=}\ \text{b}\] 

(iv)

\[\text{(a-b)x+(a+b)y}\ \text{=}\,{{\text{a}}^{\text{2}}}\text{-2ab-}{{\text{b}}^{\text{2}}}\]

and

\[\text{(a+b)(x+y)}\ \text{=}\,{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\]

Ans:

\[\text{(a-b)x+(a+b)y}\ \text{=}\,{{\text{a}}^{\text{2}}}\text{-2ab-}{{\text{b}}^{\text{2}}}\]

(1) \[\text{(a+b)(x+y)}\ \text{=}\,{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\]

Or,

\[\text{(a+b)x+(a+b)y}\ \text{=}\ {{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\]

(2) On subtracting equation (2) from (1), we get \[\text{(a-b)x-(a+b)x}\ \text{=}\ \text{(}{{\text{a}}^{\text{2}}}\text{-2ab-}{{\text{b}}^{\text{2}}}\text{)-(}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{)}\]

\[\text{(a-b-a-b)x}\ \text{=}\ \text{-2ab-2}{{\text{b}}^{2}}\]

\[\text{-2bx}\ \text{=}\ \text{-2b(a+b)}\]

\[\text{x}\ \text{=}\ \text{a+b}\]

Using equation (1), we get 

\[\text{(a-b)(a+b)+(a+b)y}\ \text{=}\ {{\text{a}}^{\text{2}}}\text{-2ab-}{{\text{b}}^{\text{2}}}\]

\[{{\text{a}}^{2}}\text{-}{{\text{b}}^{2}}\text{+(a+b)y}\ \text{=}\ {{\text{a}}^{\text{2}}}\text{-2ab-}{{\text{b}}^{\text{2}}}\]

\[\text{(a+b)y}\ \text{=}\ \text{-2ab}\]

\[\text{y}\ \text{=}\ \dfrac{\text{-2ab}}{a+b}\]
 

(v) \[\text{152x-378y}\ \text{=}\ \text{-74}\] and \[\text{-378x+152y}\ \text{=}\ \text{-604}\]

Ans:

\[\text{152x-378y}\ \text{=}\ \text{-74}\]

\[\text{x}\ \text{=}\ \dfrac{\text{189y-37}}{\text{76}}\]

(1)

\[\text{-378x+152y}\ \text{=}\ \text{-604}\]

\[\text{-189x+76y}\ \text{=}\ \text{-302}\]

(2)

Substituting the value of \[\text{x}\] in equation (2), we get \[\text{-(189}{{\text{)}}^{\text{2}}}\text{y+189 }\!\!\times\!\!\text{ 37+(76}{{\text{)}}^{\text{2}}}\text{y}\ \text{=}\ \text{-302 }\!\!\times\!\!\text{ 76}\]

\[\text{189 }\!\!\times\!\!\text{ 37+302 }\!\!\times\!\!\text{ 76}\ \text{=}\ {{\text{(189)}}^{\text{2}}}\text{y-(76}{{\text{)}}^{\text{2}}}\text{y}\]

\[\text{6993+22952}\ =\ \text{(189-76)(189+76)y}\]

\[\text{29945}\ \text{=}\ \text{(113)(265)y}\]

\[\text{y}\ \text{=}\ \text{1}\]

From equation (1), we get

\[\text{x = }\dfrac{\text{189(1)-37}}{\text{76}}\ \text{=}\ \dfrac{\text{189-37}}{\text{76}}\ \text{=}\ \dfrac{\text{152}}{\text{76}}\]

\[\text{x}\ \text{=}\ \text{2}\]
 

8. ABCD is a cyclic quadrilateral that finds the angles of the cyclic quadrilateral.

Cyclic Quadrilateral

Ans:

We know that the sum of the measures of opposite angles in a cyclic quadrilateral is \[\text{18}{{\text{0}}^{\bullet }}\]

Therefore,

\[\angle \text{A+}\angle \text{C}\ \text{=}\ \text{18}{{\text{0}}^{\bullet }}\]

\[\text{4y+20-4x}\ \text{=}\ \text{18}{{\text{0}}^{\text{}}}\]

\[\text{4y-4x}\ \text{=}\ \text{16}{{\text{0}}^{\text{}}}\]

\[\text{x-y}\ \text{=}\ \text{-4}{{\text{0}}^{\text{}}}\]

(i) Also,

\[\angle \text{B+}\angle \text{D}\ \text{=}\ \text{18}{{\text{0}}^{\bullet }}\]

\[\text{3y-5-7x+5}\ \text{=}\ \text{18}{{\text{0}}^{\text{}}}\]

\[\text{3y-7x}\ \text{=}\ \text{18}{{\text{0}}^{\text{}}}\]

(ii) Multiplying equation (i) by \[3\] , we get \[\text{3x-3y}\ \text{=}\ -12{{\text{0}}^{\text{}}}\]

(iii) On adding equation (ii) and (iii), we get \[\text{3x-7x}\ \text{=}\ {{180}^{\bullet }}-12{{\text{0}}^{\text{}}}\]

\[\text{-4x}\ \text{=}\ \text{6}{{\text{0}}^{\text{}}}\]

\[\text{x}\ \text{=}\ -{{15}^{\text{}}}\]

By using equation (i), we get

\[\text{x-y}\ \text{=}\ -{{40}^{\text{}}}\]

\[\text{-15-y}\ \text{=}\ -{{40}^{\text{}}}\]

\[\text{y}\ \text{=}\ \text{-1}{{\text{5}}^{\text{}}}\text{+4}{{\text{0}}^{\text{}}}\ \text{=}\ \text{2}{{\text{5}}^{\text{}}}\]

\[\angle \text{A}\ \text{=}\ \text{4y+2}{{\text{0}}^{\bullet }}\ =\ 4({{25}^{\bullet }})+{{20}^{\bullet }}\ =\ {{120}^{\bullet }}\]

\[\angle \text{B}\ \text{=}\ 3\text{y-}{{\text{5}}^{\bullet }}\ =\ 3({{25}^{\bullet }})-{{5}^{\bullet }}\ =\ {{70}^{\bullet }}\]

\[\angle \text{C}\ \text{=}\ -4x\ =\ -4(-{{15}^{\bullet }})\ =\ {{60}^{\bullet }}\]

\[\angle \text{D}\ \text{=}\ \text{-7x+}{{\text{5}}^{\text{}}}\ \text{=}\ \text{-7(-1}{{\text{5}}^{\text{}}}\text{)+}{{\text{5}}^{\text{}}}\ \text{=}\ \text{11}{{\text{0}}^{\text{}}}\]

NCERT Solution Maths Class 10 Chapter 3 – An Overview

NCERT Solutions Class 10 Maths Chapter 3 Exercise 3.7 - Free PDF Download

Students can download the Class 10 Maths NCERT Ex 3.7 Solutions PDF to focus on understanding each type of sum covered in this chapter. The objective of Class 10 Maths Chapter 3 Exercise 3.7 NCERT Solutions PDF is to promote conceptual learning experience among the students of Class 10. The PDF contains a detailed explanation of every sum given in this exercise. You are marked for every step in the board exams, so there’s a good chance for you to score well if you practice these solutions. Hence, with Class 10 Maths Chapter 3 Exercise 3.7 NCERT Solutions students can prepare well for their upcoming examinations.

CBSE Class 10 Maths Exercise 3.7 Solutions are available for download for free of cost on Vedantu because we believe in quality education for all. These solutions will clear all your doubts. These solutions are prepared to give you an in-depth understanding of the concepts of this chapter. Linear equations in two variables are governed by several important concepts that have applications in higher mathematics as well.

Students of Class 10th preparing for their examinations should be careful while planning their study schedule. It requires an understanding of the syllabus and the weightage of each chapter to correctly plan your studies. Students should try to strengthen their strong points first and then bolster their weak points.

The mathematics syllabus is divided into 6 units. Each unit is important and connected to the other units. The weightage of each unit in the examination is different. Given below is the weightage of each unit.

Class 10th Maths Weightage Marks

The question paper of Class 10 is divided into four sections- Section A, B, C, and D. There are four different types of questions you will find in the question paper. These are- very short answer type, short answer type I, short answer type II, and long answer type questions. The weightage of these in the paper is given below.

• 20 questions x 1 marks each: Very Short Answer type questions.

• 6 questions x 2 marks each: Short Answer  type I questions.

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• 6 questions x 4 marks each: Long Answer type questions.
  
  

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