CBSE Class 10

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. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-3-ex-3-5/

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:

(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

⇒ 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

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

∴ 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

We hope the NCERT Solutions for Class 10 Mathematics 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 Mathematics 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 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. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-3-ex-3-4/

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


(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 :

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

(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

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 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. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-3-ex-3-2/

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

Class 10th Exercise 3.2 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.
(ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.
Solution:
(i) Let the number of girls be x and number of boys be y.
A.T.Q.
1st Condition :
x +y =10
Table :

2nd Condition :
x = y + 4 ⇒ x – y = 4
Table :

Solving
(i) and
(ii) graphically

Both the lines cut at (7, 3)
Hence, solutions is (7, 3), i.e.x = 7,y = 3
Number of girls – 7 and Number of boys = 3

(ii) Let cost of 1 pencil = ₹ x and cost of 1 pen = ₹ y.
A.T.Q.
1st Condition :
5x + 7y = 50
Table :

2nd Condition :
7x + 5y = 46
Table :

Class 10 Maths Ex 3.2 Question 2.
On comparing the ratios and find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x – 4y + 8 = 7x + 6y – 9 = 0
(ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0
(iii) 6 – 3y + 10 = 0, Zx – v + 9 = 0
Solution:
(i) Equations are 5x – 4y + 8 = 7x + 6y – 9 = 0

∴ Pair of lines represented by given equations intersect at one point. So, the system has exactly one solution.

(ii) 9x + 3y + 12 = 0, 18 + 6y + 24 = 0

∴ Pair of equations represents coincident lines and having infinitely many solutions.

(iii) 6x – 3y + 10 = 0, 2x -y + 9 = 0

∴ It represents parallel lines and having no solution.

Ex 3.2 NCERT Class 10 Question 3.
On comparing the ratios and 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; -10c + 6y = -22
(v) 4/3x + 2y = 8; 2x + 3y = 12

Solution:


∴ Pair of equations is consistent with infinitely many solutions.

Ex 3.2 Class 10 NCERT Solutions 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) x – y = 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

Solution:
(i) x + y – 5, 2x + 2y = 10

Here,

∴ System of equations is consistent and the graph gr represents coincident lines.
Table for equation (i),

Table for equation (ii),

(ii) x – y = 8, 3x – 3y = 16
Here \(\frac { a_{ 1 } }{ a_{ 2 } } =\frac { 1 }{ 3 } ,\)

Pair of equations is inconsistent. Hence, lines are parallel and

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0


Pair of equations is consistent.
Table for equation 2x + y – 6 = ()

Table for equation 4x – 2y – 4 = 0


(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

∴ Pair of equations is inconsistent. Hence, lines are parallel and system has no solution.

Class 10 Maths Chapter 3 Exercise 3.2 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 length of garden = x m and width of garden = y m
Perimeter of rectangular garden = 2(x + y)
A.T.Q.
1st Condition :
\(\frac { 2(x+y) }{ 2 }\) = 36 ⇒ x + y = 36
2nd Condition :
x = y + 4 ⇒ x -y = 4 … (ii)
Adding equation (i) and (ii), we get
2x = 40 ⇒ x = 20
Putting x = 20 in equation (i), we get
20 + y = 36
y = 16
Hence, dimensions of the garden are 20 m and 16 m.

Exercise 3.2 Class 10 Question 6.
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables, such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines
Solution:
Given equation is 2x + 3y – 8 – 0
We have 2x + 3y = 8
Let required equation be ax + by = c
Condition :
(i) For intersecting lines

\(\frac { 2 }{ a } \neq \frac { 3 }{ b } \neq \frac { 8 }{ c }\) where a, b, c can have any value which satisfy the above condition.
Let a = 3, b = 2, c = 4
so, \(\frac { 2 }{ 3 } \neq \frac { 3 }{ 2 } \neq \frac { 8 }{ 4 }\)
∴ Equations are 2v + 3y = 8 and 3A + 2y = 4 have unique solution and their geometrical representation shows intersecting lines.

(ii) Given equation is 2x + 3y = 8 Required equation be ax + by = c
Condition :
For parallel lines
\(\frac { 2 }{ a } \neq \frac { 3 }{ b } \neq \frac { 8 }{ c }\)
where a, b, c can have any value which satisfy the above condition.
Let, a = 2, b = 3, c = 4
Required equation will be 2x + 3y = 4

Equations 2x + 3y = 8 and 2x + 3y = 4 have no solution and their geometrical representation shows parallel lines.

(iii) For coincident lines:

Given equation is 2x + 3y = 8 Let required equation be ax + by = c For coincident lines.
\(\frac { 2 }{ 3 } =\frac { 3 }{ 2 } =\frac { 8 }{ 4 } \) where a, b, c can have any a b c possible value which satisfy the above condition.
Let a = 4, b = 6, c = 16 Required equation will be 4x + 6y = 16.
Equations 2x + 3y = 8 and 4x + 6y = 16 have infinitely many solutions and their geometrical representation shows coincident lines.

NCERT Ex 3.2 Class 10 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:
First equation is x – y + 1 = 0.
Table for 1st equation x = y – 1

Second equation is 3x + 2y = 12

⇒ 3x = 12 – 2y ⇒ x = \(x=\frac { 12-2y }{ 3 }\)

Required triangle is ABC. Coordinates of its vertices are A(2, 3), B(-1, 0), C(4, 0).

We hope the NCERT Solutions for Class 10 Mathematics 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 Mathematics 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.

NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 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 3.1. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-3/

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1

Ex 3.1 Class 10 Question 1.
Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting)? Represent this situation algebraically and graphically.
Solution:
Let present age of Aftab = x years and present age of Aftab’s daughter = y years.

1st Condition :
Seven years ago
x – 7 = 7(y – 7)
⇒ x – 7 = 7y – 49
⇒ x – 7y = – 42
Table :


2nd Condition :
Three years later,
x + 3 = 3(y + 3)
x + 3 = 3y + 9
x – 3y = 6
Table :

Thus, the algebraic equations are
x – 7y + 42 = 0 and x – 3y – 6 = 0

Students can also visit Linear Equations In Two Variables Class 10 Practice Set 1.1 here you can see more solved problems.

Math Class 10 Ex 3.1 Question 2.
The coach of a cricket team buys 3 bats and 6 balls for ₹ 3900. Later, she buys another bat and 3 more balls of the same kind for ₹ 1300. Represent this situation algebraically and geometrically.
Solution:
Let cost of one bat = ₹ x
and the cost of one ball = ₹y
A.T.Q.
1st Condition :
3x + 6y = 3900
Table :

2nd Condition :
x + 3y = 1300
Table :

Thus, the algebraic equations are 3x + 6y = 3900 and x + 3y – 1300

Class 10 Maths Chapter 3 Question 3.
The cost of 2 kg of apples and 1 kg of grapes on a day was found to be ₹ 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹ 300. Represent the situation algebraically and geometrically.
Solution:
Let cost of one kg of apples = ₹ x and the cost of one kg of grapes = ₹y

A.T.Q.
1st Condition :
2x + y = 160
Table :

2nd Condition :
4x + 2y = 300
Table :

Thus, algebraic situations are 2x + y = 160 and 4x + 2y = 300

We hope the NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 help you. If you have any query regarding NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1, drop a comment below and we will get back to you at the earliest.