NCERT Solutions for Class 10 Maths Exercise 3.5 Chapter 3 Pair of Linear Equations in Two Variables – FREE PDF Download
NCERT Class 10 Maths Ch 3 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths Chapter
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NCERT Solutions for Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables
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
3x − 9y – 2 = 0
Comparing equation x − 3y – 3 = 0 with a1x +b1y + c1 = 0 and 3x − 9y – 2 = 0 with 
,
We get 
Here 
this means that the two lines are parallel.
Therefore, there is no solution for the given equations i.e. it is inconsistent.
(ii) 2x + y = 5
3x + 2y = 8
Comparing equation 2x + y = 5 with 
and 3x + 2y = 8 with ,
We get 
Here
this means that there is unique solution for the given equations.
⇒ 
⇒ x = 2 and y = 1
(iii) 3x − 5y = 20
6x − 10y = 40
Comparing equation 3x − 5y = 20 with and 6x − 10y = 40 with ,
We get 
Here 
It means lines coincide with each other.
Hence, there are infinitely many solutions.
(iv) x − 3y – 7 = 0
3x − 3y – 15 = 0
Comparing equation x − 3y – 7 = 0 with and 3x − 3y – 15 = 0 with ,
We get 
Here this means that we have unique solution for these equations.
⇒ 
⇒ 
⇒ x = 4 and y = –1
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
(a − b) x + (a + b) y = 3a + b – 2
(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
and (a − b) x + (a + b) y − 3a – b + 2 = 0 with
We get 
and 
and 
Linear equations have infinite many solutions if
⇒ 
⇒ 
⇒ 2a + 2b = 3a − 3b and 6 − 3b − 9a = −7a − 7b
⇒ a = 5b… (1) and −2a = −4b – 6… (2)
Putting (1) in (2), we get
−2 (5b) = −4b – 6
⇒ −10b + 4b = −6
⇒ −6b = –6 ⇒ b = 1
Putting value of b in (1), we get
a = 5b = 5 (1) = 5
Therefore, a = 5 and b = 1
(ii) Comparing (3x + y – 1 = 0) with and (2k − 1)x + (k − 1)y −2k – 1 = 0) with ,
We get 
and 
and c2 = −2k − 1
Linear equations have no solution if
⇒ 
⇒ 
⇒ 3 (k − 1) = 2k – 1
⇒ 3k – 3 = 2k − 1
⇒ k = 2
3. Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8x + 5y = 9
3x + 2y = 4
8x + 5y = 9 … (1)
3x + 2y = 4 … (2)
From equation (1),
5y = 9 − 8x⇒ y = 
Putting this in equation (2), we get
3x + 2 
= 4
⇒ 3x + 
= 4
⇒ 3x − 
⇒ 15x − 16x = 20 – 18
⇒ x = −2
Putting value of x in (1), we get
8 (−2) + 5y = 9
⇒ 5y = 9 + 16 = 25⇒ y = 5
Therefore, x = −2 and y = 5
Cross multiplication method
8x + 5y = 9 … (1)
3x + 2y = 4 … (2)
⇒ 
⇒ 
⇒ x = −2 and y = 5
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.
(ii) A fraction becomes 
when 1 is subtracted from the numerator and it becomes 
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.
According to given conditions,
x + 20y = 1000 … (1),
and x + 26y = 1180 … (2)
Subtracting equation (1) from equation (2), we get
6y = 180
⇒ y = 30
Putting value of y in (1), we get
x + 20 (30) = 1000
⇒ x = 1000 – 600 = 400
Therefore, fixed monthly charges = Rs 400 and, charges of food for one day = Rs 30
(ii) Let numerator = x and let denominator = y
According to given conditions,
⇒ 3x – 3 = y … (1) 4x = y + 8 … (1)
⇒ 3x – y = 3 … (1) 4x – y = 8 … (2)
Subtracting equation (1) from (2), we get
4x – y − (3x − y) = 8 – 3
⇒ x = 5
Putting value of x in (1), we get
3 (5) – y = 3
⇒ 15 – y = 3
⇒ y = 12
Therefore, numerator = 5 and, denominator = 12
It means fraction = 
(iii) Let number of correct answers = x and let number of wrong answers = y
According to given conditions,
3x – y = 40 … (1)
And, 4x − 2y = 50 … (2)
From equation (1), y = 3x − 40
Putting this in (2), we get
4x – 2 (3x − 40) = 50
⇒ 4x − 6x + 80 = 50
⇒ −2x = −30
⇒ x = 15
Putting value of x in (1), we get
3 (15) – y = 40
⇒ 45 – y = 40
⇒ y = 45 – 40 = 5
Therefore, number of correct answers = x = 15and number of wrong answers = y = 5
Total questions = x + y = 15 + 5 = 20
(iv)Let speed of car which starts from part A = x km/hr
Let speed of car which starts from part B = y km/hr
According to given conditions,
(Assuming x > y)
⇒ 5x − 5y = 100
⇒ x – y = 20 … (1)
And, 
⇒ x + y = 100 … (2)
Adding (1) and (2), we get
2x = 120
⇒ x = 60 km/hr
Putting value of x in (1), we get
60 – y = 20
⇒ y = 60 – 20 = 40 km/hr
Therefore, speed of car starting from point A = 60 km/hr
And, Speed of car starting from point B = 40 km/hr
(v) Let length of rectangle = x units and Let breadth of rectangle = y units
Area =xy square units. According to given conditions,
xy – 9 = (x − 5) (y + 3)
⇒ xy – 9 = xy + 3x − 5y – 15
⇒ 3x − 5y = 6 … (1)
And, xy + 67 = (x + 3) (y + 2)
⇒ xy + 67 = xy + 2x + 3y + 6
⇒ 2x + 3y = 61 … (2)
From equation (1),
3x = 6 + 5y
⇒ x = 
Putting this in (2), we get
2 
+ 3y = 61
⇒ 12 + 10y + 9y = 183
⇒ 19y = 171
⇒ y = 9 units
Putting value of y in (2), we get
2x + 3 (9) = 61
⇒ 2x = 61 – 27 = 34
⇒ x = 17 units
Therefore, length = 17 units and, breadth = 9 units
