NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 4 Quadratic Equations Class 10 NCERT Solutions Ex 4.2. 

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 4
Chapter NameQuadratic Equations
ExerciseEx 4.2
Number of Questions Solved6
CategoryNCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2

Page No: 76

Question 1. Find the roots of the following quadratic equations by factorisation:
(i) x2 – 3x – 10 = 0
(ii) 2x2 + x – 6 = 0
(iii) √2 x2 + 7x + 5√2 = 0
(iv) 2x2x + 1/8 = 0
(v) 100x2 – 20x + 1 = 0

Solution :
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations 5
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations 6

Concept Insight: Two solve a quadratic equation by factorisation method two key results are used (i) To split the middle term find the numbers a and b such that their sum is equal to the coeffident of x and product is equal to the product of coefficients of x2 and constant term. (ii) Product of num bers zero means either or both are zero.

Question 2.
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. Find out how many marbles they had to start with.

Solution :
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations 7

(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs 750. Find out the number of toys produced on that day.

Solution :
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations 8
Concept Insight: Read the question carefully and choose the variable to represent what needs to be found. Only one variable must be there in the final equation.

Question 3. Find two numbers whose sum is 27 and product is 182.

Solution :
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations 9

Question 4. Find two consecutive positive integers, sum of whose squares is 365.

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

Question 5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

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

Question 6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.

Solution :
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations 12

 

We hope the NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2, 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

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

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 3
Chapter NamePair of Linear Equations in Two Variables
ExerciseEx 3.5
Number of Questions Solved4
CategoryNCERT Solutions

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

Page No: 62

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 – 5y = 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 61
Thus, the given pair of equations has no solution.
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 62
Thus, the given pair of equations has unique solution.
By cross-multiplication method,
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 63
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 64
Thus, the given pair of equations has infinite solutions.
The solutions can be obtained by assuming the value of x to be k, where k is any
constant. So, the ordered pairs NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 65, where k is a constant, are the solutions of
the given pair of equations.
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 66
Thus, the given pair of equations has unique solution.
By cross-multiplication,
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 67

Concept Insight: In order to answer such questions, remember the condition for the pair of linear equations to have unique, infinitely many or no solution relating the coefficients. Also, remember the formula used to solve by cross multiplication method. While applying the formula, be careful about the signs of the coefficients. And above all don’t forget to first write the linear equations in standard form which is ax + by + c = 0.

Question 2.
(i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?
2x + 3y =7
(ab)x + (a + b)y = 3a +b –2

Solution:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 68

(ii) For which value of k will the following pair of linear equations have no solution?
3x + y = 1
(2k –1)x + (k –1)y = 2k + 1

Solution:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 69

Concept Insight: In order to solve such problems, firstly write the linear equations in standard form which is ax + by + c = 0.To find the value of the unknowns, the key idea is to remember the conditions for a given pair of equations to have infinite solutions and no solution.  In case of infinite solutions rule out the values which does not satisfy all the ratios. Also, be careful about the signs of the coefficients.

Question 3. Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8x +5y = 9
3x +2y = 4

Solution:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 70
Substituting this value of y in equation (1), we obtain:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 71
∴ x = -2, y = 5
Cross-multiplication method:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 72

Concept Insight: In order to solve the given pair of equations by cross multiplication method, remember the formula to be used and convert the system of equations to standard form. While applying the formula, be careful about the signs of the coefficients. In order to solve the given pair of equations by substitution method, substitute the value of any one of the variable from any one of the equation. Make sure you substitute the value of that variable which simplifies your calculations.
Solution will be same in both cases.

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 Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.

Solution:
Let the fixed charge of the food and the charge for food per day be x and y respectively.
According to the question,
x + 20y = 1000     …(1)
x + 26y = 1180   …(2)
Substituting this value of y in equation (1) from equation (2), we obtain
6y = 180
y = 30
Substituting this value of y in equation (1), we obtain:
x+ 20 x 30 = 1000
x = 1000 – 600
x = 400
Thus, the fixed charge of the food and the charge per day are Rs 400 and Rs 30 respectively.Concept Insight: Here, the fixed charge of the food and charge for food per day are the unknown quantities. So they are taken as variables x and y. The two equations can then be obtained by using the given conditions. You will observe that the variable x has the same coefficient in both the equations, so it will be easier to find the solution by eliminating x from both the equations. Also, one can solve the system by other methods.
(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.
Solution:
Let the fraction be x/y According to the question, 
According to the question, 
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 73
Subtracting equation (1) from equation (2), we obtain:
x = 5
Putting the value of x in equation (1), we obtain:
15 – y = 3
y = 12
Thus the fraction is 5/12.

Concept Insight: Since the problem asks for a fraction. The numerator and denominator of the fraction need to be represented by variables. A pair of linear equations can be obtained from the given conditions. Observe that the variable y has the same coefficient in both the equations, so it will be easier to find the solution by eliminating y from both the equations.

We hope the NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5, 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 3.4

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

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 3
Chapter NamePair of Linear Equations in Two Variables
ExerciseEx 3.4
Number of Questions Solved2
CategoryNCERT Solutions

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

Page No: 56

Question 1. Solve the following pair of linear equations by the elimination method and the substitution method:
(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 xy/3 = 3

Solution:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 55
(ii)      Elimination method:
3x + 4y = 10      …(1)
2x – 2y = 2         …(2)
Multiplying equation (2) by 2, we obtain:
4x – 4y = 4         …(3)
Adding equation (1) and (3), we obtain
7x = 14
x = 2
Substituting the value of x in equation (1), we obtain:
6 + 4y = 10
4y = 4
y = 1
Hence, x = 2, y = 1
Substitution method:
From equation (2), we obtain:
x = 1 + y            …(4)
Putting this value in equation (1), we obtain:
3(1 + y) + 4y = 10.
7y = 7
y = 1
Substituting the value of y in equation (4), we obtain:
x = 1+1 = 2
∴ x = 2, y = 1
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 56
Substitution method:
From equation (1), we obtain:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 57
Substitution method:
From equation (2), we obtain:
y = 3x – 9        … (3)
Putting this value in equation (1), we obtain:
3x + 4(3x – 9) = -6
15x = 30
x = 2
Substituting the value of x in equation (3), we obtain:
y = 6 – 9 = -3
∴ x = 2 , y = -3

Concept Insight:In order to solve the given pairs of equations by elimination method, it is required to eliminate one variable to get a linear equation in one variable. For elimination, we multiply one equation (or sometimes even both equations) by suitable constant to make the coefficient of one variable same in both the equations. The trick is to eliminate that variable which involves lesser computations.
In order to solve the given pair of equations by substitution method, we need to substitute the value of any one of the variable from any one of the equation. The trick here is to make sure you substitute the value of that variable which simplifies your calculations. Note that the solution must be same in both the cases.

Page No: 57

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 1/2 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 bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 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 Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution:
(i)    Let the fraction be x/y
According to the question,
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 58
Subtracting equation (1) from equation (2), we obtain:
x = 3
Substituting this value of x in equation (1), we obtain:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 59

Concept Insight: This problem talks about a fraction. The numerator and denominator of the fraction are not known so we represent these as variables x and y respectively where variable y must be non zero. Then, a pair of linear equations can be formed from the given conditions. The pair of equations can then be solved by eliminating a suitable variable.

(ii)  Let present age of Nuri and Sonu be x and y respectively.
According to the question,
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 60
Subtracting equation (1) from equation (2), we obtain:
y = 20
Substituting the value of y in equation (1), we obtain:
x – 60 = -10
x = 50
Thus, the age of Nuri and Sonu are 50 years and 20 years respectively.

Concept Insight: Here, Nuri’s and Sonu’s present age are not known. So, we will write both these in terms of variables. Then, using the given conditions, a pair of linear equations can be formed. The pair of equations can then be solved by eliminating a suitable variable.

(iii)  Let the units digit and tens digit of the number be x and y respectively.
Number = 10y + x
Number after reversing the digits = 10x + y
According to the question,
x + y = 9          … (1)
9(10y + x) = 2(10x + y)
88y – 11x = 0
– x + 8y =0          … (2)
Adding equations (1) and (2), we obtain:
9y = 9
y = 1
Substituting the value of y in equation (1), we obtain:
x = 8
Thus, the number is 10y + x = 10 x 1 + 8 = 18

Concept Insight: This problem talks about a two digit number. Here, remember that a two digit number xy can be expanded as 10x + y. Then, using the two given conditions, a pair of linear equations can be formed which can be solved by eliminating one of the variables.

(iv)  Let the number of Rs 50 notes and Rs 100 notes be x and y respectively.
According to the question,
x + y = 25                 —- (1)
50x + 100y = 2000 —-(2)
Multiplying equation (1) by 50, we obtain:
50x + 50 y = 1250
Subtracting equation (3) from equation (2), we obtain:
50y = 750
y = 15
Substituting the value of y in equation (1), we obtain:
x = 10
Hence, Meena received 10 notes of Rs 50 and 15 notes of Rs 100.

Concept Insight:This problem talks about two types of notes, Rs 50 notes and Rs 100 notes. And the number of both these notes with Meena is not known. So, we denote the number of Rs 50 notes and Rs 100 notes by variables x and y respectively. Now two linear equations can be formed by the given conditions which can be solved by eliminating one of the variables.

(v)    Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y respectively.
According to the question,
x + 4y = 27 —–(1)
x + 2y = 21 —–(2)
Subtracting equation (2) from equation (1), we obtain:
2y = 6
y = 3
Substituting the value of y in equation (1), we obtain:
x + 12 = 27
x = 15
Hence, the fixed charge is Rs 15 and the charge per day is Rs 3.

Concept Insight: Here, the fixed charges for the first three days and per day charges are not known so, they will be represented  using two different variables. The two equations can then be obtained by using the given conditions which can be solved by eliminating one of the variables.

We hope the NCERT Solutions for Class 10 Maths 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 Maths 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

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 3 Pair of Linear Equations in Two Variables Class 10 NCERT Solutions Ex 3.3. 

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 3
Chapter NamePair of Linear Equations in Two Variables
ExerciseEx 3.3
Number of Questions Solved3
CategoryNCERT Solutions

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

Page No: 53

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) 3xy = 3 ; 9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3 ; 0.4x + 0.5y = 2.3
(v) √2x+ √3y = 0 ; √3x – √8y = 0
(vi) 3/2x – 5/3y = -2 ; x/3 + y/2 = 13/6

Solution:
(i) x + y = 14      … (i)
x – y = 4        … (ii)
From (i), we obtain:
x = 14 – y      … (iii)
Substituting this value in equation (ii), we obtain:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 42
Substituting the value of y in equation (iii), we obtain:
x = 9
∴ x = 9 , y = 5
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 43
From (i), we obtain:
s = t + 3                   …(iii)
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 44
Substituting this value in equation (ii), we obtain:
Substituting the value of t in equation (iii), we obtain:
s = 9
∴ s = 9 , t = 6
(iii)  3x – y = 3        … (i)
9x – 3y = 9        … (ii)
From (i), we obtain
y = 3x – 3        … (iii)
Substituting this value in equation (ii), we obtain:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 45
9 = 9
This is always true.
Thus, the given pair of equations has infinitely many solutions and the relation between these variables can be given by
y  = 3x – 3
So, one of the possible solutions can x = 3, y = 6.

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 46

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 47
Substituting the value of y in equation (iii), we obtain:
x = 0
∴ x = 0, y = 0
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 48

Concept Insight: In order to solve the given pairs of equations, we need to substitute the value of any one of the variable from any one of the equation. But make sure to substitute the value of that variable which simplifies calculations. For example, in part (iv) it is most convenient to substitute the value of x from the first equation to the second equation, as the division by 0.2 is more easier than the division by 0.3, 0.4 and 0.5.

Question 2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m‘ for which y =mx + 3.

Solution:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 49
Thus, the value of m is -1.Concept Insight: Firstly the solution of the given pair of linear equations can be found out by substituting the value of one variable, say x, from one equation into the other equation. Then after finding out the values of x and y, substitute them in the equation y = m x + 3 in order to find the value of m.

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.

Solution:
Let one number be x and the other number be y such that y > x.
According to the question:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 50
Substituting this in equation (1), we obtain
y = 39
Hence, the numbers are 13 and 39.

Concept Insight: In this problem, two relations between two numbers are given. So, the two numbers have to be found out here. So the two numbers will be represented by variables x and y explicitly state the greater variable.
A pair of equations can be obtained from the given conditions. The pair of equations can then be solved by suitable substitution.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Solution:
Let the larger angle be x and smaller angle be y.
We know that the sum of the measures of angles of a supplementary pair is always 180º.
According to the given information,
x + y = 180º — (1)
x – y = 180º — (2)
From (1), we obtain
x = 180º – y — (3)
Substituting this in equation (2), we obtain
180º – y – y = 18º
162º = 2y
81º = y —- (4)
Putting this in equation (3), we obtain
x = 180º – 81º
= 99º
Hence, the angles are 99º and 81º.

Concept Insight: This problem talks about the measure of two supplementary angles. So, the two angles will be written as variables. The pair of equations can be formed using the fact that the sum of two supplementary angles is 180° and using the condition given in the problem. The pair of equations can then be solved by suitable substitution. 

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

Solution:
Let the cost of a bat and a ball be x and y respectively.
According to the given information,
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 51
Hence, the cost of a bat is Rs 500 and that of a ball is Rs 50.

Concept Insight: Cost of bats and balls needs to be found so the cost of a ball and bat will be taken as the variables. Applying the conditions of total cost of bats and balls algebraic equations will be obtained. The pair of equations can then be solved by suitable substitution.

(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 Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for traveling a distance of 25 km?

Solution:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 52

Concept Insight: In this problem, we are required to find out the fixed charge and the charge per km. So, we will represent these two by using different variables. Now, two linear equations can be written by using the conditions given in the problem. The pair of equations can then be solved by suitable substitution.

(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6 . Find the fraction.

Solution:
Let the fraction be x/y.
According to the given information,
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 53

Concept Insight: This problem talks about a fraction which is not known to us. So numerator and denominators will be taken to be variables x and y respectively and y will be strictly non zero. Then, a pair of linear equations can be formed from the given conditions. The pair of equations can then be solved by suitable substitution.

(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:
Let the age of Jacob be x and the age of his son be y.
According to the given information,
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 54
Hence, the present age of Jacob is 40 years whereas the present age of his son is 10 years.

Concept insight: Here, Jacob’s and his son’s present age are not known. So, we will write both these in terms of variables. The problem talks about their ages five years ago and five years hence. Here, five years ago means we have to subtract 5 from their present ages, and five years hence means we have to add 3 to their present ages. So, using the given conditions, a pair of linear equations can be formed. The pair of equations can then be solved by suitable substitution. 

 

We hope the NCERT Solutions for Class 10 Maths 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 Maths 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 3 Pair of Linear Equations in Two Variables Ex 3.2

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

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 3
Chapter NamePair of Linear Equations in Two Variables
ExerciseEx 3.2
Number of Questions Solved7
CategoryNCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

Page No: 49

Question 1. Form the pair of linear equations in the following problems, and find their solutions graphically.

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

Solution:
Let the number of girls and boys in the class be x and y respectively.
According to the given conditions, we have:
x + y = 10
x – y = 4
x + y = 10 ⇒ x = 10 – y
Three solutions of this equation can be written in a table as follows:

x546
y564

x – y = 4 ⇒ x = 4 + y
Three solutions of this equation can be written in a table as follows:

x543
y10-1

The graphical representation is as follows:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 9
From the graph, it can be observed that the two lines intersect each other at the point (7, 3).
So, x = 7 and y = 3.
Thus, the number of girls and boys in the class are 7 and 3 respectively.

(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.

Solution:

Let the cost of one pencil and one pen be Rs x and Rs y respectively.
According to the given conditions, we have:
5x + 7y = 50
7x + 5y = 46
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 10
Three solutions of this equation can be written in a table as follows:

x310-4
y5010

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 11
Three solutions of this equation can be written in a table as follows:

x83-2
y-2512

The graphical representation is as follows:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 12
From the graph, it can be observed that the two lines intersect each other at the point (3, 5).
So, x = 3 and y = 5.
Therefore, the cost of one pencil and one pen are Rs 3 and Rs 5 respectively.

Concept Insight: Read the question carefully and examine what are the unknowns. Represent the given conditions with the help of equations by taking the unknowns quantities as variables. Also carefully state the variables as whole solution is based on it. On the graph paper, mark the points accurately and neatly using a sharp pencil. Also, take at least three points satisfying the two equations in order to obtain the correct straight line of the equation. Since joining any two points gives a straight line and if one of the points is computed incorrect will give a wrong line and taking third point will give a correct line. The point where the two straight lines will intersect will give the values of the two variables, i.e., the solution of the two linear equations. State the solution point.

Question 2. On comparing the ratios a1/a2 , b1/b2 and c1/c2, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident.
(i)  5x – 4y + 8 = 0 , 7x + 6y – 9 = 0
(ii)  9x + 3y + 12 = 0 , 18x + 6y + 24 = 0
(iii) 6x – 3y + 10 = 0 , 2x – y + 9 = 0

Solution:
(i)  5x – 4y + 8 = 0
7x + 6y – 9 = 0
Comparing these equations with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0,
we get:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 13
Since NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 14, the given pair of equations intersect at exactly one point.

(ii)  9x + 3y + 12 = 0
18x + 6y + 24 = 0
Comparing these equations with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, we get:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 15
Since NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 16, the given pair of equation are coincident.

(iii)  6x – 3y + 10 = 0
2x – y + 9 = 0
Comparing these equations with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, we get:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 17
Since NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 18, the given pair of equation are parallel to each other.

Concept Insight:In order to answer such questions, remember the condition for the pair of linear equations to be intersecting, parallel or coincident. Also, while writing the coefficients, don’t forget to take the signs.

Question 3. On comparing the ratios \(\frac { { a }_{ 1 } }{ { b }_{ 1 } } =\frac { { a }_{ 2 } }{ { b }_{ 2 } } =\frac { { a }_{ 3 } }{ { b }_{ 3 } } \) find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5 ; 2x – 3y = 7
(ii) 2x – 3y = 8 ; 4x – 6y = 9
(iii) 3/2x + 5/3y = 7 ; 9x – 10y = 14
(iv) 5x – 3y = 11 ; – 10x + 6y = –22
(v) 4/3x + 2y =8 ; 2x + 3y = 12

Solution:

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 19
Since NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 20, the given pair of equation has only one solution.
Thus, the pair of linear equations is consistent.
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 21
Thus, the pair of linear equations is inconsistent.
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 22
Thus, the pair of linear equations is consistent.
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 23
Thus, the pair of linear equations is consistent.
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 24
Thus, the pair of linear equations is consistent.

Concept Insight: If a pair of linear equations has one or more than one solution then they are said to be consistent and if they have no solution then they are said to be inconsistent. So, to identify the consistency of a given pair of equations, apply the conditions involving the coefficients of the given pair of equations.  In case, two consistent linear equations are plotted, they will either intersect or overlap each other.

Question 4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) x + y = 5, 2x + 2y = 10
(ii) xy = 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

Solution:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 25
Thus, the pair of linear equations is consistent.
Now, x + y = 5 ⇒ x = 5 – y
Three solutions of this equation can be written in a table as follows:

x432
y123

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 26
Three solutions of this equation can be written in a table as follows:

x432
y123

Thus, the graphical representation is as follows:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 27
From the graph, it can be observed that the two lines coincide. Thus, the given pair of equations has infinite solutions.
Let x = k, then y = 5 – k. So, the ordered pair (k, 5 – k) , where k is a constant, will be the solution of the given pair of linear equations.
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 28
Thus, the pair of linear equations is inconsistent.
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 29
Thus, the pair of linear equations is consistent.
Now, 2x + y – 6 = 0 ⇒ y = 6 – 2x
Three solutions of this equation can be written in a table as follows:

x012
y642

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 30
Three solutions of this equation can be written in a table as follows:

x123
y024

Thus, the graphical representation is as follows:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 31
From the graph, it can be observed that the two lines intersect each other at the point (2, 2). Thus, the solution of the given pair of equations is (2, 2).
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 32
Thus, the pair of linear equations is inconsistent.

Concept Insight: If a pair of linear equations has one or more than one solutions then they are said to be consistent and if they have no solution then they are said to be inconsistent. The graph of each equation can be plotted by taking at least three ordered pairs which are the solutions of the equations. The point where both the lines intersect will be the solution of the given pair of equations. Remember two overlapping lines intersect each other at infinitely many points. State the solution explicitly.

Page No: 50

Question 5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Solution:

Let the width and length of the rectangular garden be x and y respectively.
According to the given conditions,
y – x = 4
y + x = 36
y – x = 4 ⇒ y = x + 4
Three solutions of this equation can be written in a table as follows:

x0812
y41216

y + x = 36
Three solutions of this equation can be written in a table as follows:

x03616
y36020

Thus, the graphical representation is as follows:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 33
From the graph, it can be observed that the two lines intersect each other at the point (16, 20). So, x = 16 and y = 20.
Thus, the length and width of the rectangular garden is 20 m and 16 m respectively.

Concept Insight: Here dimensions of the rectangular garden needs to be found. Since opposite sides of the rectangle are equal so length and breadth can be taken as variables. Applying conditions given in the problem two linear equations in the 2 variables can be obtained. Now, in order to represent the obtained equations graphically, take the values of variables as whole numbers only, because then it will be easier to represent the values on the graph. The point where the two equations intersect will give the required dimensions. State the dimensions length and breadth from the values of the variables.

Question 6. Given the linear equation 2x + 3y – 8 = 0, write another linear equations in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines

Solution:
(i) For the two lines a1x + b1x + c1 = 0 and a2x + b2x + c2 = 0, to be intersecting, we must have
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 34
So, the other linear equation can be 5x + 6y – 16 = 0
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 35
(ii) For the two lines a1x + b1x + c1 = 0 and a2x + b2x + c2 = 0, to be parallel, we must have
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 36
So, the other linear equation can be 6x + 9y + 24 = 0,
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 37
(iii) For the two lines a1x + b1x + c1 = 0 and a2x + b2x + c2 = 0 to be coincident, we must have
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 38
So, the other linear equation can be 8x + 12y – 32 = 0,
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 39
Concept Insight: In order to answer such type of problems, just remember the conditions for two lines to be intersecting, parallel, and coincident. This problem will have multiple answers as their can be many equations satisfying the required conditions.

Question 7. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Solution:
x – y + 1 = 0 ⇒ x = y – 1
Three solutions of this equation can be written in a table as follows:

x012
y123

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 40
Three solutions of this equation can be written in a table as follows:

x420
y036

Now, these equations can be drawn on a graph. The triangle formed by the two lines and the x-axis can be shown by the shaded part as:
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 41
From the graph, it can be observed that the coordinates of the vertices of the triangle so formed are (2, 3), (-1, 0), and (4, 0).

Concept Insight: In order to find the coordinates of the vertices of the triangle so formed, find the points where the two lines intersects the x-axis and also where the two lines intersect each other. Note here that the coordinates of the intersection of lines with x-axis is taken and not with y-axis, this is because the question says to find the triangle formed by the two lines and the x-axis.

 

We hope the NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2, drop a comment below and we will get back to you at the earliest.