CBSE Class 10 – Page 20

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2

March 13, 2023

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2.

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 6
Chapter Name	Triangles
Exercise	Ex 6.2
Number of Questions Solved	10
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2

Question 1.
In the given figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 1
Solution:

Question 2.
E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EF || QR:
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 3

Question 3.
In the given figure, if LM || CB and LN || CD.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 4

Solution:

Question 4.
In the given figure, DE || AC and DF || AE.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 8

Solution:

Question 5.
In the given figure, DE || OQ and DF || OR. Show that EF || QR.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 11
Solution:

Question 6.
In the given figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 14
Solution:

Question 7.
Using B.P.T., prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
Solution:
Given: A ∆ABC in which D is the mid-point of AB and DE || BC
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 17

Question 8.
Using converse of B.P.T., prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
Solution:
Given: A ΔABC in which D and E are mid-points of sides AB and AC respectively.
To Prove: DE || BC
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 18

Question 9.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 20

Question 10.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that \(\frac { AO }{ BO }\) = \(\frac { CO }{ DO }\) Show that ABCD is a trapezium.
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 22

We hope the NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

March 13, 2023

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3.

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 5
Chapter Name	Arithmetic Progressions
Exercise	Ex 5.3
Number of Questions Solved	20
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 1.
Find the sum of the following APs:
(i) 2, 7, 12,…… to 10 terms.
(ii) -37, -33, -29, …… to 12 terms.
(iii) 0.6, 1.7, 2.8, ……, to 100 terms.
(iv) \(\frac { 1 }{ 15 }\), \(\frac { 1 }{ 12 }\), \(\frac { 1 }{ 10 }\), …….., to 11 terms.
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 1

Question 2.
Find the sums given below:
(i) 7 + 10\(\frac { 1 }{ 2 }\) + 14 + … + 84
(ii) 34 + 32 + 30 + … + 10
(iii) -5 + (-8) + (-11) + ….. + (-230)
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 3

Question 3.
In an AP:
(i) given a = 5, d = 3, a_n = 50, find n and S_n.
(ii) given a = 7, a₁₃ = 35, find d and S₁₃.
(iii) given a₁₂ = 37, d = 3, find a and S₁₂.
(iv) given a₃ = -15, S₁₀ = 125, find d and a₁₀.
(v) given d = 5, S₉ = 75, find a and a₉.
(vi) given a = 2, d = 8, S_n = 90, find n and a_n.
(vii) given a = 8, a_n = 62, S_n = 210, find n and d.
(viii) given a_n = 4, d = 2, S_n = -14, find n and a.
(ix) given a = 3, n = 8, S = 192, find d.
(x) given l = 28, S = 144, and there are total 9 terms. Find a.
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 4

Question 4.
How many terms of AP: 9, 17, 25, … must be taken to give a sum of 636?
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 8

Question 5.
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 9

Question 6.
The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 10

Question 7.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 11

Question 8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 12

Question 9.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 13

Question 10.
Show that a₁, a₂, ……. a_n,…… form an AP where an is defined as below:
(i) a_n = 3 + 4n
(ii) a_n = 9 – 5n
Also find the sum of the first 15 terms in each case.
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 15

Question 11.
If the sum of the first n terms of an AP is 4n – n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 16

Question 12.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 17

Question 13.
Find the sum of the first 15 multiples of 8.
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 18

Question 14.
Find the sum of the odd numbers between 0 and 50.
Solution:
Let odd numbers between 0 and 50 be 1, 3, 5, 7,…….., 49.
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 19

Question 15.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows:
₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc. the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Solution:
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 20

Question 16.
A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.
Solution:
Let 1st prize be of ₹ a
2nd prize be ₹ (a – 20) and
3rd prize be ₹ (a – 20 – 20) = ₹ (a – 40)
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 21

Question 17.
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, eg. a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Solution:
Let the trees be planted 1, 2, 3, 4, 5 , …… 12
Here, a = 1, d = 1, n = 12
Total number of trees planted by each section
S₁₂ = \(\frac { 12 }{ 2 }\) [2a + (n – 1) d] = 6 [2 x 1 + (12 – 1) x 1]
= 6 [2 + 11] = 6 x 13 = 78
Total number of trees planted by 3 sections = 78 x 3 = 234

Question 18.
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm,… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles?
(Take π = \(\frac { 22 }{ 7 }\))
[Hint: Length of successive semicircles is l₁, l₂, l₃, l₄, … with centres at A, B, respectively.]
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 22
Solution:

Question 19.
200 logs are stacked in the following manner 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Figure). In how many rows are the 200 logs placed and how many logs are in the top row?
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 24
Solution:

Question 20.
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig.) A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 x 5 + 2 x (5 + 3)]
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 26

Solution:
Distance between the first potato and the bucket = 5 m
Distance between next 2 potatoes = 3 m each
So, series is 5 m, 8 m, 11 m,
Here, a = 5 m, d = (8 – 5) m = 3 m
Total distance travelled for 10 potatoes = 2 [5 + 8 + 11 + …….. + 10 terms]
= 2[\(\frac { 10 }{ 2 }\){2 x 5 + (10 – 1) 3}]
= 2[5{10 + 27}] = 2(37 x 5) = 37 x 10 = 370 m.

We hope the NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4

March 13, 2023

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4.

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 4
Chapter Name	Quadratic Equations
Exercise	Ex 4.4
Number of Questions Solved	5
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4

Question 1.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2x² -3x + 5 = 0
(ii) 3x² – 4√3x + 4 = 0
(iii) 2x²-6x + 3 = 0
Solution:
(i) 2x² – 3x + 5 = 0
This is of the form ax² + bx + c = 0,
where a = 2, b = -3 and c = 5
Discriminant, D = b²-4ac
= (-3) ²-4 x 2 x 5 = 9 – 40 = -31
Since, D < 0
Hence, no real roots exist.

(ii) 3x² – 4√3x + 4 = 0
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4 1
(iii) 2x²-6x + 3 = 0

Question 2.
Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(1) 2x² + kx + 3 = 0
(2) kx (x – 2) + 6 = 0
Solution:
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4 3

Question 3.
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.
Solution:
Let breadth of the rectangular mango grove be x m
Then, the length of rectangular mango grove be 2xm
According to question,
x x 2x = 800
⇒ 2x² = 800
⇒ x² = 400
⇒ x = ±20 [-20 is rejected]
Hence, breadth = 20 m and length = 2 x 20 = 40 m
So, it is possible to design a rectangular mango grove whose length is twice its breadth.

Question 4.
Is the following situation possible? If so, determine their present ages.
The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Solution:
Let the present age of one friend be x years
Then, the present age of other friend be (20 -x) years
4 years ago, one friend’s age was (x – 4) years
4 years ago, other friend’s age was (20 -x – 4) = (16 -x) years
According to question,
(x-4) (16-x) = 48
⇒ 16x – x² – 64 + 4x = 48
⇒ x² – 20x + 112 = 0
This is of the form ax² + bx + c = 0, where, a = 1, b = -20 and c = 112
Discriminant, D = b² – 4ac
= (-20)² – 4 x 1 x 112 = 400 – 448 = – 48 < 0
Since, no real roots exist.
So, the given situation is not possible.

Question 5.
Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.
Solution:
Let the length of rectangular park be x m and breadth be y m.
Given: area = 400 m²
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4 4

We hope the NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3

March 13, 2023

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3.

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 4
Chapter Name	Quadratic Equations
Exercise	Ex 4.3
Number of Questions Solved	11
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3

Question 1.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2x² – 7x + 3 = 0
(ii) 2x² + x – 4 = 0
(iii) 4x² + 4√3x + 3 = 0
(iv) 2x² + x + 4 = 0
Solution:
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3 1

Question 2.
Find the roots of the quadratic equations by applying the quadratic formula.
(i) 2x² – 7x + 3 = 0
(ii) 2x² – x + 4 = 0
(iii) 4x² – 4√3x + 3 = 0
(iv) 2x² – x + 4 = 0
Solution:
(i) 2x² – 7x + 3 = 0
This is of the form ax² + bx + c = 0,
where a = 2, b = -7 and c = 3
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3 5

(ii) 2x² – x + 4 = 0
This is of the form ax² + bx + c = 0,
where a = 2, b = 1 and c = – 4
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3 7

(iii) 4x² – 4√3x + 3 = 0
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3 8

(iv) 2x² – x + 4 = 0

Question 3.
Find the roots of the following equations:

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3 11
Solution:

Question 4.
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is \(\frac { 1 }{ 3 }\) Find his present age.
Solution:
Let the present age of Rehman be x years
3 years ago Rehman’s age was = (x – 3) years
5 years from now Rehman’s age will be = (x + 5) years
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3 13

Question 5.
In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
Solution:
Let the marks secured by Shefali in Mathematics = x Then,
According to question,
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3 14

Question 6.
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution:
Let the shorter side of rectangle = x m
Then, longer side = (x + 30) m and diagonal = (x + 60) m
In ΔABC, (x + 60)² = (x + 30)² + (x)² [Pythagoras Theorem]
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3 15

Question 7.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution:
Let the smaller number = x
and the larger number = y
According to question,
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3 17

Question 8.
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:
Total distance travelled = 360 km
Let uniform speed be x km/h
Then, increased speed = (x + 5) km/h
According to question,
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3 18

Question 9.
Two water taps together can fill a tank in 9\(\frac { 3 }{ 8 }\) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution:
Let larger pipe fills the tank in x hours and the smaller pipe fills the tank in y hours.
The tank filled by the larger pipe in 1 hour = \(\frac { 1 }{ x }\)
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3 19

Question 10.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bengaluru (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
Solution:
Let the speed of passenger train = x km/h
Then, the speed of express train be (x + 11) km/h
According to question,
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3 20

Question 11.
Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution:
Let the sides of two squares be x m and y m
According to question,
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3 21

We hope the NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3, drop a comment below and we will get back to you at the earliest.

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

March 13, 2023

NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7.

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 3
Chapter Name	Pair of Linear Equations in Two Variables
Exercise	Ex 3.7
Number of Questions Solved	8
Category	NCERT Solutions

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

Question 1.
The age 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.
Solution:
Let the ages of Ani and Biju be x years and y years respectively.
If Ani is older than Biju
x – y =3
If Biju is older than Ani
y – x = 3
-x + y =3 [Given]
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 1
Subtracting equation (i) from equation (ii), we get:
3x – 57
⇒ x = 19
Putting x = 19 in equation (i), we get
19-y = 3
⇒ y = 16
Again subtracting equation (iv) from equation (iii), we get
3x = 63
⇒ x = 21
Putting x = 21 in equation (iii) we get
21 -y= -3
⇒ y = 24
Hence, Ani’s age is either 19 years or 21 years and Biju’s age is either 16 years or 24 years.

Question 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?
Solution:
Let the two friends have ₹ x and ₹ y.
According to the first condition:
One friend has an amount = ₹(x + 100)
Other has an amount = ₹ (y – 100
∴ (x + 100) =2 (y – 100)
⇒ x + 100 = 2y – 200
⇒ x – 2y = -300       …(i)
According to the second condition:
One friend has an amount = ₹(x – 10)
Other friend has an amount =₹ (y + 10)
∴ 6(x – 10) = y + 10
⇒ 6x – 60 = y + 10
⇒ 6x-y = 70                                        …(ii)
Multiplying (ii) equation by 2 and subtracting the result from equation (i), we get:
x – 12x = – 300 – 140
⇒ -11x = -440
⇒ x = 40
Substituting x = 40 in equation (ii), we get
6 x 40 – y = 70
⇒ -y = 70- 24
⇒ y   = 170
Thus, the two friends have ₹ 40 and ₹ 170.

Question 3.
A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Solution:
Let the original speed of the train be x km/h
and the time taken to complete the journey be y hours. ‘
Then the distance covered = xy km

Case I: When speed = (x + 10) km/h and time taken = (y – 2) h
Distance = (x + 10) (y – 2) km
⇒ xy = (x + 10) (y – 2)
⇒ 10y – 2x = 20
⇒ 5y – x = 10
⇒ -x + 5y = 10 …(i)

Case II: When speed = (x – 10) km/h and time taken = (y + 3) h
Distance = (x – 10) (y + 3) km
⇒ xy = (x – 10) (y + 3)
⇒ 3x- 10y = 30 …(ii)
Multiplying equation (i) by 3 and adding the result to equation (ii), we get
15y – 10y = 30 f 30
⇒ 5y = 60
⇒ y   = 12
Putting y = 12 in equation (ii), we get
3x- 10 x 12= 30
⇒ 3x   = 150
⇒ x   = 50
∴ x = 50 and y = 12
Thus, original speed of train is 50 km/h and time taken by it is 12 h.
Distance covered by train = Speed x Time
= 50 x 12 = 600 km.

Question 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 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Solution:
Let the number of rows be x and the number of students in each row be y.
Then the total number of students = xy
Case I: When there are 3 more students in each row
Then the number of students in a row = (y + 3)
and the number of rows = (x – 1)
Total number of students = (x – 1) (y + 3)
∴ (x – 1) (y + 3) = xy
⇒ 3x -y =3 …(i)
Case II: When 3 students are removed from each row
Then the number of students in each row = (y-3)
and the number of rows = (x + 2)
Total number of students = (x + 2) (y – 3)
∴ (x + 2) (y – 3) = xy
⇒ -3x + 2y = 6 …(ii)
Adding the equations (i) and (ii), we get
-y + 2y = 3 + 6
⇒ y = 9
Putting y = 9 in the equation (ii), we get
-3x + 18 = 6
⇒ x = 4
∴ x = 4 and y = 9
Hence, the total number of students in the class is 9 x 4 = 36.

Question 5.
In a ∆ABC, ∠C = 3 ∠B = 2(∠A + ∠B). Find the three angles.
Solution:
Let ∠A = x° and ∠B = y°.
Then ∠C = 3∠B = (3y)°.
Now ∠A + ∠B + ∠C = 180°
⇒ x + y + 3y = 180°
⇒ x + 4y = 180° …(i)
Also, ∠C = 2(∠A + ∠B)
⇒ 3y – 2(x + y)
⇒ 2x – y = 0° …(ii)
Multiplying (ii) by 4 and adding the result to equation (i), we get:
9x = 180°
⇒ x = 20°
Putting x = 20 in equation (i), we get:
20 + 4y = 180°
⇒ 4y = 160°
⇒ y = \(\frac { 160 }{ 40 }\) = 40°
∴ ∠A = 20°, ∠B = 40° and ∠C = 3 x 40° = 120°.

Question 6.
Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the coordinates of the vertices of the triangle formed by these lines and the y-axis.
Solution:
5x – y = 5 …(i)
3x-y = 3 …(ii)
For graphical representation:
From equation (i), we get: y = 5x – 5
When x = 0, then y -5
When x = 2, then y = 10 – 5 = 5
When x = 1, then y = 5 – 5 = 10
Thus, we have the following table of solutions:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 2
From equation (ii), we get:
⇒ y = 3x – 3
When x = 0, then y = -3
When x = 2, then y = 6 – 3 = 3
When x = 1, then y = 3 – 3 = 0
Thus, we have the following table of solutions:

Plotting the points of each table of solutions, we obtain the graphs of two lines intersecting each other at a point C(1, 0).
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 4
The vertices of ΔABC formed by these lines and the y-axis are A(0, -5), B(0, -3) and C(1, 0).

Question 7.
Solve the following pairs of linear equations:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 5
Solution:
(i) The given equations are
px + qy = p – q …(1)
qx – py = p + q …(2)
Multiplying equation (1) byp and equation (2) by q and then adding the results, we get:
x(p² + q²) = p(p – q) + q(p + q)
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 6

(ii) The given equations are
ax + by = c …(1)
bx – ay = 1 + c …(2)
Multiplying equation (1) by b and equation (2) by a, we get:
abx + b²y = cb …(3)
abx + a²y = a(1+ c) …(4)
Subtracting (3) from (4), we get:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 8

(iii) The given equations may be written as: bx – ay = 0 …(1)
ax + by = a² + b² …(2)
Multiplying equation (1) by b and equation (2) by a, we get:
b²x + aby = 0 ….(3)
a²x + aby = a(a² + b²) …..(4)
Adding equation (3) and equation (4), we get:
(a² + b²)x = a (a² + b²) a(a² + b²)
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 10

(iv) The given equations may be written as:
(a – b)x + (a + b)y = a² – 2ab – b² …(1)
(a + b)x + (a + b)y = a² + b² …(2)
Subtracting equation (2) from equation (1), we get:
(a – b)x – (a + b)x
= (a² – 2ab – b²) – (a² + b²)
⇒ x(a – b- a-b) = a² – 2ab – b² – a² – b²⇒ -2bx = -2ab – 2b²⇒ 2bx = 2b² + 2ab
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 11

(v) The given equations may be written as:
76x – 189y = -37 …(1)
-189x + 76y = -302 …(2)
Multiplying equation (1) by 76 and equation (2) by 189, we get:
5776x – 14364y = -2812 …(3)
-35721x + 14364y = -57078 …(4)
Adding equations (3) and (4), we get:
5776x – 35721x = -2812 – 57078
⇒ – 29945x = -59890
⇒ x = 2
Putting x = 2 in equation (1), we get:
76 x 2 – 189y = -37
⇒ 152 – 189y = -37
⇒ -189y = -189
⇒ y = 1
Thus, x = 2 and y = 1 is the required solution.

Question 8.
ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 13
Solution:

We hope the NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 help you. If you have any query regarding NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5

March 13, 2023

NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5.

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 3
Chapter Name	Pair of Linear Equations in Two Variables
Exercise	Ex 3.5
Number of Questions Solved	4
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5

Question 1.
Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
(i) x – 3y – 3 = 0
3x – 9y – 2 = 0
(ii) 2x + y = 5
3x + 2y = 8
(iii) 3x – Sy = 20
6x – 10y = 40
(iv) x – 3y – 7 = 0
3x – 3y – 15 = 0
Solution:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5 1
(iii) Equations are 3x – 5y = 20 and 6x – 10y = 40
Here,

(iv) Equations are x – 3y = 7 and 3x – 3y = 15

Question 2.
(i) for which values of a and b does the following pai of linea equation have an infinite number of solutions₹
2x + 3y =7
(a – b)x + (a + b)y = 3a + b – 2
(ii) For which value of K will the following pair of linear equation have no solution₹
3x + y = 1
(2k – 1)x + (k – 1)y = 2k + 1
Solution:
(i) Equations are
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5 5
⇒ a – 2b = 3 Solving (iii) and (iv) for a and b
By cross multiplication method.

Question 3.
Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8x + 5y = 9
3x + 2y = 4
Solution:
Equations are
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5 7
By coss multiplication Method :
Equations are

Question 4.
Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay ₹ 1000 as hostel charges whereas a student B, who takes food for 26 days, pays ₹ 1180 as hostel charges. Find the fixed charges and the cost of food per day.
(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.
(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test₹
(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars ₹
(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Solution:
(i) Let fixed monthly hostel charges = and charges per day = ₹ y
A.T.Q.
As per condition of student A
x + 20y = 1000
As per condition of student B
x + 26y = 1180

By cross multiplication method
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5 9
∴ Fixed monthly hostel charges = ₹ 400 and charges per day = ₹ 30

By cross multiplication method

Solving (i) and (ii) for x and y
By cross multiplication method

By cross multiplication method
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5 13

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NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4

March 13, 2023

NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4.

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 3
Chapter Name	Pair of Linear Equations in Two Variables
Exercise	Ex 3.4
Number of Questions Solved	2
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4

Question 1.
Solve the following pair of linear equations by the elimination method and the substitution
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
(iv) x/2 + 2y/3 = -1 and x – y/3 = 3
Solution:
(i) By Elimination Method:
Fquations are x + y = 5
and 2x – 3y = 4
Multiply equation (i) by 2 and subtract equation (ii) from it, we have
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 1

(ii) By Elimination method:
Equations are 3x + 4y = 10
and 2x – 2y = 2
Multiplying equation (ii) by 2 and adding to equation (i), we

(iii) By Elimination Method:

(iv) By Elimination Method:
1st equation :
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 5

Question 2.
Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes – if we only add 1 to the denominator. What is the fraction₹
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu₹
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution:
(i) Let numerator be x and denominator be y.
Fraction = x/y
A.T.Q.

(ii) Let present age of Nuri be x years and Sonu’s present age bey years.
A.T.Q.
1st Condition :

2nd Condition :
Subtractomg equation (ii) from equetion (i), we get

Hence, present age of Nuri is 50 years and sonu’s present age is 20 years.

(iii) Let digit at unit place = x and digit at ten’s place = y.
Two digit number is lOy + x
A.T.Q.
1st Condition :
x + y = 9
2nd Condition :
9(10y + x) = 2(10k + y) ⇒ 90y + 9x = 20x + 2y
⇒ 88y – 11x = 0 ⇒ -11y + 88y = 0
⇒ -x + 8y = 0
Adding equestion (i) and (ii), we get
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 12

(iv) Let the number of notes of ₹ 50 = x and the number of notes of ₹ 100 = y
A.T.Q
1st Condition :
50x + 100y = 2000
⇒ x + 2y = 40
2nd Condition :

(v) Let, fixed charge for first 3 days be ₹ x and additional charge per day after 3 days be y.
A.T.Q.
1st Condition : as per Saritha
x + 4y = 27
2nd Condition : as per Susy
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 14
Putting y = 3 in equation (i),
x + 4(3) = 27 ⇒ x + 12 = 27 ⇒ x = 15
Hence, fixed charge is ₹ 15 and charge for each extra day is ₹ 3.

We hope the NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 help you. If you have any query regarding NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

March 13, 2023

NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2.

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 3
Chapter Name	Pair of Linear Equations in Two Variables
Exercise	Ex 3.3
Number of Questions Solved	3
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Question 1.
Solve the following pair of linear equations by the substitution method,
(i) x + y = 14, x – y = 4
(ii) s – t = 3, s/3 + t/2 = 6
(iii) 3x – y = 3, 9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 1
Solution:
From equation (i),
x + y – 14 ⇒ y = 14x
Putting the value ofy in equation (ii), we get
x – (14 – x) = 4 ⇒ x – 14 + x = 4 ⇒ 2x = 4 + 14
2x = 18 ⇒ x = 9
Now, puttingx = 9 in equation (i), we have
9 + y = 14 ⇒ y = 14 – 9 ⇒ y = 5
so, x = 9, y = 5
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 2
∴ y can have infinite real values
∴ x can have infinite real values because x = \(\frac { y+3 }{ 3 }\)

Question 2.
Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.
Solution:
Equations are 2x + 3y = 11
and 2x – 4y = -24
From equation (i)
2x = 11 – 3y
Putting this value in equation (ii), we get
11 – 3y – 4y = -24 ⇒ 11 – 7y = -24 ⇒ – 7y = – 35
y = \(\frac { 35 }{ 7 }\) ⇒ y = 5
Putting y = 5 in equation (i). we have
2x + 3 x 5 = 11 ⇒ 2x + 15 = 11 ⇒ 2x = 11 – 15 ⇒ 2x = -4 ⇒ x = -2
Now. putting the value of x andy in equation
y = mx + 3 ⇒ 5 = -2m + 3 ⇒ 2 = -2m ⇒ m = -1

Question 3.
Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball,
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charge per km₹ How much does a person have to pay for travelling a distance of 25 km₹
(v) A fraction becomes 9/2, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and denominator it becomes 5/6, Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution:
(i) Let 1st number be x and 2nd number be y.
Let x >y
1st condition :
x – y = 26
2nd condition :
x = 3y
Putting x = 3y in equation (i)
3y – y = 26 ⇒ 2y = 26 ⇒ y = 13
From (ii)
x = 3 x 13 = 39
∴ One number is 13 and the other number is 39.

(ii) Let one angle be x and its supplementary angle = y
Let x > y
1st Condition :
x + y = 180°
2nd Condition :
x – y = 18° ⇒ X = 18° + y
From equation (ii), putting the value ofx in equation (i),
18° + y + y = 180° ⇒ 18° + 2y = 180°
2y = 162° ⇒ y = 81°
From (ii) x = 18° + 81° = 99° ⇒ x = 99°
∴ One angle is 81° and another angle is 99°.

(iii) Let cost of 1 bat = ₹x and cost of 1 ball = ₹y
1st Condition:
7x + 6y = 3800
2nd Condition:
3x + 5y = 1750
From equation (ii), we get
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 6
putting x = 1750-5y/3 in equation (i), we get
Cost of one bat = ₹ 500 and cost of one ball = ₹ 50.

(iv) Let fixed charges be ₹.v and charge for per km be ₹y.
A.T.Q.
1st Condition :
x + lOy = 105
2nd Condition :
x + 15y = 155
From equation (i), we get
x= 105 – 10y
Putting this value in equation (ii), we have
105 – 10y + 15y = 155 ⇒ 105 + 5y = 155
⇒ 5y = 155 – 105 ⇒ 5y = 50 ⇒ y = 10
Now, puttingy = 10 in equation (i), we have
x + 10(10) = 105 ⇒ x + 100 = 105 ⇒ x = 5
Fixed charges is ₹ 5 and charges per km is ₹ 10.

3rd Condition :
For distance of 25 km
x + 25y = 5 + 25(10) = 5 + 250 = 255
Amount paid for travelling 25 km is ₹ 255.

(v) Let numerator be x and denominator be y.
∴ Fraction is x/y
A.T.Q.
1st condition :

2nd condition :

(vi) Let present age of Jacob be x years and that of his son bey years.
A.T.Q.
1st Condition :
x + 5 = 3(y + 5) ⇒ x + 5 = 3y + 15 ⇒ x – 3y = 15 – 5 ⇒ x – 3y = 10
2nd Condition:
x – 5 = 7(y – 5) ⇒ x – 5 = 7y – 35 ⇒ x = 7y – 35 + 5
⇒ x = 7y – 30
Putting the value of ‘x’ in equation (i), we get
7y – 30 – 3y = 10
4y – 30 = 10
4y = 40 y = 10 ⇒ y = 10
putting the value of y in equation(ii), we get
x = 7(10) – 30 = 70 – 30 ⇒ x = 40
Hence, the present age of Jacob is 40 years and that of his son is 10 years.

We hope the NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 help you. If you have any query regarding NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4

March 13, 2023

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4.

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 2
Chapter Name	Polynomials
Exercise	Ex 2.4
Number of Questions Solved	5
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4

Question 1.
Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and the coefficients in each case:
(i) 2x³ + x² – 5x + 2; \(\frac { 1 }{ 4 }\), 1, -2
(ii) x³ – 4x² + 5x – 2; 2, 1, 1
Solution:
(i) Comparing the given polynomial with ax³ + bx² + cx + d, we get:
a = 2, b – 1, c = -5 and d = 2.
∴ p(x) = 2x³ + x² – 5x + 2
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4 1

(ii) Compearing the given polynomial with ax³ + bx² + cx + d, we get:
a = 1, b = -4, c = 5 and d = – 2.
∴ p (x) = x³ – 4x² + 5x – 2
⇒ p(2) = (2)³ – 4(2)² + 5 x 2 – 2
= 8 – 16+ 10 – 2 = 0
p(1) = (1)³ – 4(1)² + 5 x 1-2
= 1 – 4 + 1 – 2
= 6-6 = 0
Hence, 2, 1 and 1 are the zeroes of x³ – 4x²+ 5x – 2.
Hence verified.
Now we take α = 2, β = 1 and γ = 1.
α + β + γ = 2 + 1 + 1 = \(\frac { 4 }{ 1 }\) = \(\frac { -b }{ a }\)
αβ + βγ + γα = 2 + 1 + 2 = \(\frac { 5 }{ 1 }\) = \(\frac { c }{ a }\)
αβγ = 2 x 1 x 1 = \(\frac { 2 }{ 1 }\) = \(\frac { -d }{ a }\).
Hence verified.

Question 2.
Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.
Solution:
Let α , β and γ be the zeroes of the required polynomial.
Then α + β + γ = 2, αβ + βγ + γα = -7 and αβγ = -14.
∴ Cubic polynomial
= x³ – (α + β + γ)x² + (αβ + βγ + γα)x – αβγ
= x³ – 2x² – 1x + 14
Hence, the required cubic polynomial is x³ – 2x² – 7x + 14.

Question 3.
If the zeroes of the polynomial x³ – 3x² + x + 1 are a-b, a, a + b, find a and b.
Solution:
Let α , β and γ be the zeroes of polynomial x³ – 3x² + x + 1.
Then α = a-b, β = a and γ = a + b.
∴ Sum of zeroes = α + β + γ
⇒   3 = (a – b) + a + (a + b)
⇒ (a – b) + a + (a + b) = 3
⇒ a-b + a + a + b = 3
⇒ 3a = 3
⇒ a = \(\frac { 3 }{ 3 }\) = 1 …(i)
Product of zeroes = αβγ
⇒ -1 = (a – b) a (a + b)
⇒ (a – b) a (a + b) = -1
⇒   (a² – b²)a = -1
⇒ a³ – ab² = -1 … (ii)
Putting the value of a from equation (i) in equation (ii), we get:
(1)³-(1)b² = -1
⇒ 1 – b² = -1
⇒ – b² = -1 – 1
⇒  b² = 2
⇒ b = ±√2
Hence, a = 1 and b = ±√2.

Question 4.
If two zeroes of the polynomial x⁴ – 6x³ – 26x²+ 138x – 35 are 2 ± √3, finnd other zeroes.
Solution:
Since two zeroes are 2 + √3 and 2 – √3,
∴ [x-(2 + √3)] [x- (2 – √3)]
= (x-2- √3)(x-2 + √3)
= (x-2)²– (√3)²x² – 4x + 1 is a factor of the given polynomial.
Now, we divide the given polynomial by x² – 4x + 1.
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4 3
So, x⁴ – 6x³ – 26x² + 138x – 35
= (x² – 4x + 1) (x² – 2x – 35)
= (x² – 4x + 1) (x² – 7x + 5x – 35)
= (x²-4x + 1) [x(x- 7) + 5 (x-7)]
= (x² – 4x + 1) (x – 7) (x + 5)
Hence, the other zeroes of the given polynomial are 7 and -5.

Question 5.
If the polynomial x⁴ – 6x³ + 16x² – 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a, find k and a.
Solution:
We have
p(x) = x⁴ – 6x³ + 16x² – 25x + 10
Remainder = x + a   … (i)
Now, we divide the given polynomial 6x³ + 16x² – 25x + 10 by x² – 2x + k.
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4 4
Using equation (i), we get:
(-9 + 2k)x + 10-8 k + k² = x + a
On comparing the like coefficients, we have:
-9 + 2k = 1
⇒ 2k = 10
⇒ k = \(\frac { 10 }{ 2 }\) = 5 ….(ii)
and 10 -8k + k²– a ….(iii)
Substituting the value of k = 5, we get:
10 – 8(5) + (5)² = a
⇒   10 – 40 + 25 = a
⇒ 35 – 40 = a
⇒ a =   -5
Hence, k = 5 and a = -5.

We hope the NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2

March 13, 2023

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2.

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 2
Chapter Name	Polynomials
Exercise	Ex 2.2
Number of Questions Solved	2
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2

Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x² – 2x – 8
(ii) 4s² – 4s + 1
(iii) 6x² – 3 – 7x
(iv) 4u² + 8u
(v) t² -15
(vi) 3x² – x – 4
Solution:
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2 1

Question 2.
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2 5
Solution:
(i) Zeroes of polynomial are not given, sum of zeroes = \(\frac { 1 }{ 4 }\) and product of zeroes = -1
If ax² + bx + c is a quadratic polynomial, then
α + β = sum of zeroes = \(\frac { -b }{ a }\) = \(\frac { 1 }{ 4 }\) and αβ = product of zeroes = \(\frac { c }{ a }\) = -1
Quadratic polynomial is ax² + bx + c
Let a = k, ∴ b = \(\frac { -k }{ 4 }\) and c = -k
Putting these values, we get

For different values of k, we can have quadratic polynomials all having sum of zeroes as \(\frac { 1 }{ 4 }\) and product of zeroes as -1.

(ii) Sum of zeroes = α + β = √2 = \(\frac { -b }{ a }\); product of zeroes = αβ = \(\frac { 1 }{ 3 }\) = \(\frac { c }{ a }\)
Quadratic polynomial is ax² + bx + c
Let a = k,b = -√2k and c = \(\frac { k }{ 3 }\)
Putting these values we get
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2 7
For all different real values of k, we can have different quadratic polynomials of the form 3×2 – 3√2x +1 having sum of zeroes = √2 and product of zeroes = \(\frac { 1 }{ 3 }\)

(iii) Sum of zeroes = α + β = 0 = \(\frac { -b }{ a }\); product of zeroes = αβ = √5 = \(\frac { c }{ a }\)
Let quadratic polynomial is ax² + bx + c
Let a = k,b = 0, c = √5 k
Putting these values, we get
k[x² – 0x + √5 ] = k(x² + √5).
For different real values of k, we can have different quadratic polynomials of the form
x² + √5, having sum of zeroes = 0 and product of zeroes = √5

(iv) Sum of zeroes = α + β = 1= \(\frac { -b }{ a }\); product of zeroes = αβ = 1 = \(\frac { c }{ a }\)
Let quadratic polynomial is ax² + bx + c.
Let a=k, c = k, b = -k
Putting these values, we get k[x² -x +1]
Quadratic polynomial is of the form x² -x + 1 for different values of k.

(v) Sum of zeroes = α + β = \(\frac { -1 }{ 4 }\)= \(\frac { -b }{ a }\); product of zeroes = αβ = \(\frac { 1 }{ 4 }\) = \(\frac { c }{ a }\)
Let quadratic polynomial is ax² + bx + c
Let a=k, b= \(\frac { k }{ 4 }\), c= \(\frac { k }{ 4 }\)
Putting these values, we get k

Quadratic polynomial is of the form 4x² +x + 1 for different values of k.

(vi) Sum of zeroes = α + β = 4 = \(\frac { -b }{ a }\); product of zeroes = αβ = 1 = \(\frac { c }{ a }\)
Let quadratic polynomial is ax² + bx + c
Let a = k,b = -4k and c = k
Putting these values, we get
k[x² – 4x + 1]
Quadratic polynomial is of the form x² – 4x + 1 for different values of k.

We hope the NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 10 Social Science Paper 3

March 13, 2023

CBSE Sample Papers for Class 10 Social Science Paper 3 are part of CBSE Sample Papers for Class 10 Social Science. Here we have given CBSE Sample Papers for Class 10 Social Science Paper 3.

CBSE Sample Papers for Class 10 Social Science Paper 3

Board	CBSE
Class	X
Subject	Social Science
Sample Paper Set	Paper 3
Category	CBSE Sample Papers

Students who are going to appear for CBSE Class 10 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 3 of Solved CBSE Sample Paper for Class 10 Social Science is given below with free PDF download solutions.

General Instructions:

The question paper has 27 questions in all. All questions are compulsory.
Marks are indicated against each question.
Questions from serial number 1 to 7 are very short answer questions. Each question carries 1 mark.
Questions from serial number 8 to 18 are 3 marks questions. Answer of these questions should not exceed 80 words each.
Questions from serial number 19 to 25 are 5 marks questions. Answer of these questions should not exceed 100 words each.
Question number 26 and 27 are map questions of 2 marks from History and 3 marks from Geography. After completion, attach the maps inside the answer book

Question 1.
What was ‘El Dorado’?
OR
State the contribution of James Watt towards industrialisation.
OR
Who wrote the novel ‘Debganer Martye Aagaman’ in Bengali?

Question 2.
What was the traditional book used by the Chinese from 594 AD?
OR
“Coketown, a fictitious industrial town, was a grim place full of machinery, smoking chimneys, rivers polluted purple, and buildings that all looked the same”. Who described about Coketown, an industrial city in his novel? Name the novel.

Question 3.
What was religion according to Gandhi?

Question 4.
What are community resources? Give an example.

Question 5.
Which of the following neighbouring countries has better performance in terms of human
development than India?

Question 6.
What are ‘Terms of Credit’?

Question 7.
Under which economic sector does the production of a commodity through the natural processes
come?

Question 8.
‘Satyagraha is not physical force. A satyagrahi does not inflict pain on the adversary; he does not
seek his destmction … In the use of satyagraha, there is no ill-will whatever. ‘Satyagraha is pure soul-force. Truth is the very substance of the soul. That is why this force is called satyagraha. The soul is informed with knowledge. In it bums the flame of love.Nonviolence is the supreme dharma…
Read the above passage and answer the following questions:
(1) Who spoke these words?
(2) Explain two differences between physical force and soul force, with examples.

Question 9.
Explain the three types of flows within the international economic exchanges during 1815-1914.
OR
In Victorian Britain, the upper classes preferred things produced by hand. Why?
OR
Name any two cities that first appeared along river valleys. State under what circumstances did these cities develop? Point out the main feature of these cities.

Question 10.
What are ‘residuary’ subjects? Give examples.

Question 11.
What are the origins of social difference?

Question 12.
Examine what is called‘a system of checks and balances’?

Question 13.
Explain three reasons to justify water scarcity in most cases.

Question 14.
What was the main focus of the First Five Year Plan? Explain any three institutional reforms
that were introduced after independence.

Question 15.
Describe any three public facilities needed for development.

Question 16.
Analyse the quick measure adopted by Central Government of India to provide employment
to the unemployed in the rural area.

Question 17.
“Consumers are exploited in the marketplace in various ways”. Discuss with example.

Question 18.
Describe how technology enabled globalisation? Explain with examples.

Question 19.
Who invented the printing press? How did he develop the printing technology?
OR
Explain the main theme of Charles Dickens ‘Oliver Twist’.

Question 20.
‘Napoleon was a great administrator.’ Explain the statement in the light to the changes he
brought about in France.
OR
Highlight the contribution of women in the anti-imperial struggle in Vietnam. Did they succeed?

Question 21.
“As Indian federation is ‘holding together’ federation, all States in the Indian Union do not
have identical powers”. Support your answer with two suitable examples.

Question 22.
What do you mean ‘Political Parties’? Explain the ideology of BJP.

Question 23.
“India is fortunate to have fairly rich and varied mineral resources. However, these are unevenly
distributed”. Explain the statement giving examples. Also state any two reasons for uneven distribution of minerals in India.

Question 24.
When and where was the first jute mill set up in India? Explain the reasons why most of the
jute mills concentrated in the Hugh river basin.

Question 25.
Which are the two groups of various sources of credit in India? Write any three features of
each group.

Question 26.
Two features A and B are marked on the given political outline map of India:
Identify these features with the help of the following information and write their correct names on the lines marked in the map:
A. The place where the Indian National Congress Session was held in 1927.
B. The place where Gandhiji led the indigo planter’s agitation in 1916.
OR
Locate and label on the same map given:
(1) The place where peasants organized a Satyagraha in 1918
(2) Amritsar

Question 27.
On the given same political outline map of India locate and label/identify the following with appropriate symbols:
(1) Identify the type of soil found in shaded area.
(2) Largest producer of Jute (Mark the state)
(3) Narora Atomic Power Station

Answer

Answer 1.
El Dorado was a fabled city of gold.
OR
James Watt improved the steam engine produced by Newcomen and patented the new engine in 1781. ‘
OR
In 1880, Durgacharan Ray wrote a novel, Debganer Martye Aagaman (The Gods Visit Earth).

Answer 2.
The traditional Chinese ‘accordion book’ was folded and stitched at the side. Superbly skilled craftsmen could duplicate, with remarkable accuracy, the beauty of calligraphy.
OR
Charles Dickens in his novel Hard Times (1854).

Answer 3.
Gandhiji used to say that religion can never be separated from politics. He meant religion was
not any particular religion like Hinduism or Islam but moral values that inform all religions. He believed that politics must be guided by ethics drawn from religion.

Answer 4.
Resources which are accessible to all the members of the community. Examples: Village grazing grounds, burial grounds, public parks, picnic spots etc.

Answer 5.
Sri Lanka.

Answer 6.
Interest rate, collateral and documentation requirement, and the mode of repayment together
comprise what is called the terms of credit.

Answer 7.
Primary

Answer 8.
(1) Mahatma Gandhi
(2) Physical force seeks vengeance and it is aggressive. The British worship the war-god and they carry arms.
(3) Soul Force: Satyagraha is soul force. Without seeking vengeance or being aggressive, a satyagrahi could win the battle through nonviolence. This could be done by appealing to the conscience of the oppressor. People – including the oppressors – had to be persuaded to see the truth, instead of being forced to accept truth through the use of violence. By this struggle, truth was bound to ultimately triumph.

Answer 9.
(1) Economists identify three types of movement or flows’ within international economic exchanges. The first is the flow of trade which in the nineteenth century referred largely to trade in goods (e.g., cloth or wheat).
(2) The second is the flow of labour – the migration of people in search of employment.
(3) The third is the movement of capital for short-term or long-term investments over long distances.
OR
(1) In Victorian Britain, the upper classes – the aristocrats and the bourgeoisie – preferred things produced by hand because handmade products came to symbolise refinement and class.
(2) They were better finished, individually produced, and carefully designed.
(3) Machine made goods were for export to the colonies.
OR
(1) Towns and cities that first appeared along river valleys, such as Ur, Nippur and Mohenjodaro.
(2) Ancient cities could develop only when an increase in food supplies made it possible to support a wide range of non-food producers.
(3) Cites were often the centres of political power, administrative network, trade and industry, religious institutions, and intellectual activity, and supported various social groups such as artisans, merchants and priests.

Answer 10.
(1) The subjects like computer software that came up after the constitution was made.
(2) According to our constitution, the Union Government has the power to legislate on these ‘residuary’ subjects.
(3) These subjects do not fall in any of the three subjects.

Answer 11.
(1) The social differences are mostly based on accident of birth. Normally we don’t choose
to belong to our community. We belong to it simply because we were born into it.
(2) We all experience social differences based on accident of birth in our everyday lives. People around us are male or female, they are tall and short, have different kinds of complexions, or have different physical abilities or disabilities.
(3) But all kinds of social differences are not based on accident of birth. Some of the differences are based on our choices. For example, some people are atheists. They don’t believe in God or any religion. Some people choose to follow a religion other than the one in which they were born.

Answer 12.
(1) Power is shared among different organs of government, such as the legislature, executive
and judiciary. This is called horizontal distribution of power because it allows different organs of government placed at the same level
(2) In a democracy, even though ministers and government officials exercise power, they are responsible to the Parliament or State Assemblies. Similarly, although judges are appointed by the executive, they can check the functioning of executive or laws made by the legislatures. This arrangement is also called a system of checks and balances.

Answer 13.
(1) Over exploitation: The people who have access to water resources exploit water resources
more than those who do not have access to water resources. For example; industrial use.
(2) Excessive use: there are people who exploit water resources more than their requirement.
(3) Unequal access to water among different social groups: there is a tendency that in the society rich people have more access to water than that of poor people.

Answer 14.
(1) Land reform
(2) Institutional reforms:

Consolidation of land holdings
Collectivisation
Cooperation
Abolition of zamindari system

Answer 15.
Public facilities refer to facilities, which a person cannot arrange at individual level, these are
provided by government. Following are the main public facilities: Pollution free environment

Good infrastructure like transport.
Collective security for the whole locality
Opening schools, colleges and hospitals
Taking preventive steps from infectious diseases
Provision for safe drinking water, sanitation facilities etc.
Provision for public distribution system.

Answer 16.
(1) For the short-term, as a quick measure, the central government in India made a law implementing the Right to Work in 200 districts of India and then extended to an additional 130 districts.
(2) The remaining districts in rural areas were brought under the act with effect from 1 April, 2008. It is called National Rural Employment Guarantee Act 2005 (NREGA 2005).
(3) Under NREGA 2005, all those who are able to, and are in need of, work are guaranteed 100 days of employment in a year by the government. If the government fails in its duty to provide employment, it will give unemployment allowances to the people.
(4) The types of work that would in future help to increase the production from land will be given preference under the Act.

Answer 17.
Sometimes traders indulge in unfair trade practices such as when shopkeepers weigh less
than what they should or when traders add charges that were not mentioned before, or when adulterated/defective goods are sold.

Answer 18.
Technology: Rapid improvement in technology has been one major factor that has stimulated
the globalisation process. For instance, the past fifty years have seen several improvements in transportation technology. This has made much faster delivery of goods across long distances possible at lower costs.
Example: Containers for transport of goods.

Answer 19.
Gutenberg:
(1) Gutenberg was the son of a merchant and grew up on a large agricultural estate.
(2) From his childhood he had seen wine and olive presses. He learnt the art of polishing stones, became a master goldsmith, and also acquired the expertise to create lead moulds used for making trinkets. Drawing on this knowledge, Gutenberg adapted existing technology to design his innovation. The olive press provided the model for the printing press, and moulds were used for casting the metal types for the letters of the alphabet.
(3) By 1448, Gutenberg perfected the system. The first book he printed was the Bible. About 180 copies were printed and it took three years to produce them. By the standards of the time this was fast production.
OR
Dickens focused on the terrible conditions of urban life under industrial capitalism. His Oliver Twist (1838) is the tale of a poor orphan who lived in a world of petty criminals and beggars. Brought up in a cruel workhouse. Oliver was finally adopted by a wealthy man and lived happily ever after.

Answer 20.
(1) In the administrative field he had incorporated revolutionary principles in order to make the whole system more rational and efficient. The Civil Code of 1804 – usually known as the Napoleonic Code- did away with all privileges based on birth, established equality before the law and secured the right to property.
(2) This Code was exported to the regions under French control. In the Dutch Republic, in Switzerland, in Italy and Germany, Napoleon simplified administrative divisions, abolished the feudal system and freed peasants from serfdom and manorial dues. In the towns too, guild restrictions were removed.
(3) Transport and communication systems were improved.
(4) Peasants, artisans, workers and new businessmen enj oyed a new-found freedom. Businessmen and small-scale producers of goods, in particular, began to realise that uniform laws, standardised weights and measures, and a common national currency would facilitate the movement and exchange of goods and capital from one region to another.
OR
(1) Women in Vietnam traditionally enjoyed greater equality in comparison to that in China. They had only limited freedom to determine their future. They enjoyed no public life. But with the growth of nationalist movement the status of women improved. Writers and political thinkers began idealising women who rebelled against social norms.
(2) This rebellion against social conventions marked the arrival of the new woman in Vietnamese society. A play was written by the nationalist Pharr Boi Chau in 1913 on the lives of the Trung Sisters who had fought against Chinese domination in 39-43 CE. In this play he depicted these sisters as patriots fighting to save the Vietnamese nation from the Chinese.
(3) They were portrayed as young, brave and dedicated. Nguyen Thi Xuan was reputed to have shot down a jet with just twenty bullets.
(4) They helped in nursing the wounded, constructing underground rooms and tunnels and fighting the enemy. Between 1965 and 1975, of the 17,000 youth worked on the trail, 70 to 80 percent were women.

Answer 21.
(1) Some States enjoy a special status. Jammu and Kashmir has its own Constitution. Many provisions of the Indian Constitution are not applicable to this State without the approval of the State Assembly. Indians who are not permanent residents of this State cannot buy land or house here. Similar special provisions exist for some other States of India as well.
(2) There are some units of the Indian Union which enjoy very little power. These are areas which are too small to become an independent State but which could not be merged with any of the existing States. These areas, like Chandigarh, or Lakshadweep or the capital city of Delhi, are called Union Territories. These territories do not have the powers of a State. The Central Government has special powers in running these areas.

Answer 22.
A political party is a group of people who come together to contest elections and hold power
in the government. They agree on some policies and programmes for the society with a view to promote the collective good:

Wants full territorial and political integration of Jammu and Kashmir with India.
A uniform civil code for all people living in the country irrespective of religion.
Cultural nationalism.
Party Symbol-Lotus

Answer 23.
(1) Peninsular rocks contain most of the reserves of coal, metallic minerals, mica and many other non-metallic minerals.
(2) Sedimentary rocks on the western and eastern flanks of the peninsula, in Gujarat and Assam have most of the petroleum deposits.
(3) Rajasthan with the rock systems of the peninsula, has reserves of many non-ferrous minerals.
(4) The vast alluvial plains of north India are almost devoid of economic minerals.
Reasons: Differences in the geological structure, processes and time involved in the formation of minerals.

Answer 24.
(1) Rishra, Near Kolkata in 1859.
(2) Factors responsible for location in the Hugh river basin:

(a) Availability of raw material: The Hugh river basin is rich in extremely good quality alluvial soil. Hence, jute is grown abundantly in this region and raw material is available in this region.

(b) Abundant water: For jute industry huge amount of water is required. Abundant water is supplied from the Hugh river to the jute industries.

(c) Network of transport: Inexpensive water transport is provided by the Hugh river itself. Moreover, the region is rich in railways, roadways and water transport. This helped in the growth of jute industry in this region.

Answer 25.
(1) Formal sector Loans:
Features:

Formal credit sector is known as organised sector.
This sector provides loan at low rate of interest.
This sector is supervised and controlled by Reserve Bank of India. It mainly includes: Commercial Banks and Cooperative Society.

(2) Informal Sector Loans:
Features:

Informal credit sectors in India include credit given by unorganised sectors.
These sectors meet the credit needs of poor households.
They charge high interest rate.
There is no organisation which supervises and controls the lending activities of informal sector.
The main informal credit sources are: Local Money lenders, Traders, Employers, Relatives and Friends etc.

Answer 26.
CBSE Sample Papers for Class 10 Social Science Paper 3

Answer 27.
CBSE Sample Papers for Class 10 Social Science Paper 3.1

We hope the CBSE Sample Papers for Class 10 Social Science Paper 3 help you. If you have any query regarding CBSE Sample Papers for Class 10 Social Science Paper 3, drop a comment below and we will get back to you at the earliest.

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