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

- Pair of Linear Equations in Two Variables Class 10 Ex 3.1
- Pair of Linear Equations in Two Variables Class 10 Ex 3.2
- Pair of Linear Equations in Two Variables Class 10 Ex 3.3
- Pair of Linear Equations in Two Variables Class 10 Ex 3.4
- Pair of Linear Equations in Two Variables Class 10 Ex 3.5
- Pair of Linear Equations in Two Variables Class 10 Ex 3.6
- Pair of Linear Equations in Two Variables Class 10 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 3.5 |

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

**Ex 3.5 Class 10 Maths Question 1.
**Which of the following pairs of linear equations has unique solution, no solution or infinitely many solutions. In case there is 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:**

**(i)**x – 3y – 3 = 0 ….(1)

3x – 9y – 2 = 0 …..(2)

Therefore, the pair of linear equations has no solution.

**2x + y – 5 = 0 …..(1)**

(ii)

(ii)

3x + 2y – 8 = 0 …..(2)

Therefore, the pair of linear equations has a unique solution.

Now by cross-multiplication method, we have:

Hence, the required solution is x and y = 1.

**3x – 5y = 20 …….(1)**

(iii)

(iii)

6x – 10y – 40 = 0 …..(2)

Therefore, the pair of linear equations has infinitely many solutions.

**(iv)** x – 3y – 7 = 0 …(1)

3x- 3y- 15 = 0 …(2)

Therefore, the pair of linear equations has a unique solution.

Now by cross-multiplication method, we have:

Hence, the required solution is x = 4 and y = -1.

**Ex 3.5 Class 10 Maths 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

(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)

**Solution:**

**(i)**We have:

2x + 3y – 7 = 0

(a – b)x + (a + b)y – (3a + b -2) = 0

Here, a

_{1}= 2, b

_{1}= 3, c

_{1}= -7

a

_{2}= (a – b), b

_{2}= (a + b), c

_{2}= -(3a + b – 2)

We know that for a pair of linear equations to have infinite numbers of solutions

**(ii)** We have: 3x + y – 1 = 0

(2k – 1)x + (k – 1)y – (2k + 1) = 0

Here, a_{1} = 3, b_{1} = 1, c_{1} = -1

a_{2} = 2k – 1, b_{2} = k – 1 , c_{1} = -(2k +1)

Since the pair of linear equations has no solution

Thus, the given pair of linear equations has no solution when k = 2.

**Ex 3.5 Class 10 Maths Question 3.
**Solve the following pair of linear equations by the substitution and cross-multiplication methods:

8x + 5y = 9, 3x + 2y = 4

**Solution:**

We have: 8x + 5y = 9 ……(i)

3x + 2y = 4 …..(ii)

Substitution method:

Substitution method:

Multiplying equation (i) by 2 and equation (ii) by 5 and then subtracting the results, we get:

16x – 15x = 18-20

⇒ x = -2

Substituting x = -2 in equation (i), we get:

8(-2) + 5y = 9

⇒ -16 + 5y = 9

⇒ 5y = 25

⇒ y = 5

Thus, the required solution is x = -2 and y = 5.

**8x + 5y – 9 = 0**

Cross multiplication method:

Cross multiplication method:

3x + 2y – 4 = 0

Thus, the required solution is x = -2 and y = 5.

**Ex 3.5 Class 10 Maths 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 \(\frac { 1 }{ 3 }\) when 1 is subtracted from the numerator and it becomes \(\frac { 1 }{ 4 }\) 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 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:**

[

**Note:**These problems can be solved by any algebraic method. Here, we have solved by the elimination method.]

**(i)**Let the fixed monthly charges of hostel be ₹ x and cost of food for one day be ₹ y.

Then according to the question, we have:

Total charges for student A:

x + 20y = 1000 …(1)

Total charges for student B:

x + 26y = 1180 …(2)

Subtracting equation (1) from equation (2), we get:

Substituting the value ofy in equation (1), we get:

x + 20 x 30 = 1000

⇒ x = 1000 – 20 x 30

= 1000 – 600 = 400

Thus, the fixed monthly charges are ? 400 and cost of food per day is ? 30.

**(ii)** Let the fraction be \(\frac { x }{ y }\).

Then according to the question, we have:

**(iii)** Let the number of right answers and wrong answers be x and y respectively

Then according to the question, we have:

3x – y = 40 …(1)

and 4x – 2y = 50 …(2)

Multiplying equation (1) by 2 and then subtracting the result from (2), we get:

4x – 6x = 50 – 80

⇒ – 2x = -30

⇒ x = 15

Substituting x = 15 in equation (1), we get:

3 x 15 -y = 40

⇒ 45 -y = 40

⇒ y = 5

Thus, the total number of questions in the test = x + y

= 15 + 5 = 20.

**(iv)** Let the speed of first car be x km/h and that of second car be y km/h.

Case I: When both the cars travel in the same direction

Distance travelled by first car = 5x km

Distance travelled by second car = 5y km

Distance between both the cars

5x – 5y = 100 [Given]

⇒ x- y = 20 …(1)

Case II: When both the cars travel in opposite directions

Distance between both the cars

= x + y = 100 …(2) [Given]

Adding equations (1) and (2), we get:

2x= 120

⇒ x = 60

Substituting x = 60 in equation (2), we get:

60 + y = 100

⇒ y = 100 – 60 = 40

Thus, the speeds of the cars are 60 km/h and 40 km/h.

**(v)** Let the length and breadth of a rectangle be x units and y units respectively.

Then the area of the rectangle = xy sq. units.

**Case 1:** When the length is reduced by 5 units and breadth is increased by 3 units

New length = (x – 5) units

New breadth = (y + 3) units

∴ New area = (x – 5) (y + 3) sq.

units According to the question, we have:

xy – (x – 5) (y + 3) =9

⇒ xy – (xy + 3x – 5y – 15) = 9

⇒ xy-xy-3x + 5y+15=9

⇒ -3x + 5y = -6 …(1)

**Case 2:** When the length is increased by 3 units and breadth is increased by 2 units

New length = (x + 3) units New breadth

New breadth (y + 2) units

∴ New area = (x + 3) (y + 2) sq units

According to the question, we have:

(x+3) (y+2)- xy =67

⇒ xy + 3y + 2x + 6 – xy = 67

⇒ 2x + 3y = 61 …(2)

Multiplying equation (1) by 2 and (2) by 3 and then adding the results, we get:

10y + 9y = -12+ 183

⇒ 19y =171

⇒ y=9

Substituting y = 9 in equation (1), we get:

-3x + 5 x 9 = -6

⇒ -3x = -6-45

⇒ x = 17

Thus, the length of the rectangle is 17 units and its breadth is 9 units.

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