**NCERT Solutions for Class 10 Maths Exercise 3.5 of Chapter 3 Pair of Linear Equations**

NCERT Solutions for Class 10 Maths Exercise 3.5 of Chapter 3 Pair of Linear Equations are helpful for the preparation of the class 10 maths exam and maths worksheet. NCERT Solutions for Class 10 Maths Exercise 3.5 of Chapter 3 Pair of Linear Equations will clear your all doubts whatever you face in doing class 10 maths NCERT chapter 3 Pair of Linear Equations. Chapter 3 of class 10 NCERT Maths is a basic algebraic chapter that is useful in evaluating variables in almost every subject like economics, chemistry, science, physics, business, and accounts. There are three ways of solving a pair of linear equations but here in this exercise, you will study the cross multiplication method of solving a pair of linear equations.

**NCERT Solutions for Class 10 Maths Chapter 3 Linear Equations in Two Variables**

**Class 10 Maths Sample Paper (Basic) with Solutions for Term 1 CBSE Board Exam 2021-22**

**Exercise 3.1 -Linear Equations in Two Variables**

**Exercise 3.2 -Linear Equations in Two Variables**

**Exercise 3.3-Linear Equation in Two Variables**

**Exercise 3.4-Pair of Linear Equations**

**Class 10 maths NCERT solutions of important questions of chapter 3 Pair of Linear Equations**

**Exercise 3.7 – Linear Equations in Two Variables**

**NCERT Solutions for Class 10 Maths Exercise 3.5 of Chapter 3 Pair of Linear Equations**

**Q1.Which of the following pairs of linear equations has unique solutions, no solutions, or infinitely many solutions? In case there is a unique solution, find it by using the cross multiplication method.**

**(i) x-3y-3=0 and 3x-9y-2=0 (ii) 2x +y =5 and 3x +2y =8**

**(iii) 3x -5y = 20 and 6x -10y =40 (iv) x -3y -7 =0 and 3x -3y -15 =0**

Ans.(i) Given pair of linear equations are x-3y-3=0 and 3x-9y-2=0

Evaluating the ratios of the coefficients and the constants of the equations as follows.

a_{1}/a_{2}= 1/3

b_{1}/b_{2}=-3/-9 =1/3

c_{1}/c_{2}=-3/-2 =3/2

We observe

a_{1}/a_{2 }=b_{1}/b_{2 }≠ c_{1}/c_{2}

Therefore the given pair of the linear equations has no solution

(ii) The given pair of linear equations is 2x +y =5 and 3x +2y =8

Rearranging the equations into the standard form

2x + y -5=0 and 3x +2y -8 =0

Evaluating the ratios of the coefficients and the constants of the equations as follows.

a_{1}/a_{2}= 2/3

b_{1}/b_{2} =1/2

c_{1}/c_{2}=-5/-8 =5/8

a_{1}/a_{2}≠b_{1}/b_{2}

The given pair of the linear equations has a unique solution

Arranging the coefficients and constants of both equations in a matrix as follows

x = 2, y = 1

(iii) The given equations are 3x -5y = 20 and 6x -10y =40

Arranging these equations into the standard form

3x -5y – 20=0 and 6x -10y -40 =0

Arranging the coefficients and constants of both equations in a matrix as follows

Evaluating the ratios of the coefficients and the constants of the equations as follows.

a_{1}/a_{2}= 3/6=1/2

b_{1}/b_{2} =-5/-10 = 1/2

c_{1}/c_{2}=-20/-40 =1/2

It is observed that

a_{1}/a_{2}=b_{1}/b_{2}= c_{1}/c_{2}

Therefore the pair of the equations have infinite solutions

(iv) The given equations are x -3y -7 =0 and 3x -3y -15 =0

Evaluating the ratios of the coefficients and the constants of the equations as follows.

a_{1}/a_{2}= 1/3

b_{1}/b_{2} =-3/-3 = 1

c_{1}/c_{2}=-7/-15

It is observed that

a_{1}/a_{2}≠b_{1}/b_{2}

Therefore the pair of the equations have unique solution

Arranging the coefficients and constants into a matrix as follows

x = 24/6 = 4 and y =-6/6 = -1

Hence the value of x is 4 and the value of y is -1

**NCERT Solutions for Class 10 Maths Exercise 3.5 of Chapter 3 Pair of Linear Equations**

**Q2. (i) For which value 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)**

We know two linear equations a_{1}x +b_{1}y+ c_{1}=0 and a_{2}x +b_{2}y+ c_{2}=0 have infinite solutions if the ratios between coefficients and constants are as follows.

Arranging the given equations into the standard form

2x +3y -7=0

(a -b)x + (a +b)y -3a -b +2=0

Where a_{1}=2, b_{1}=3, c_{1}=-7 and a_{2}=a -b, b_{2}=a +b, c_{2}=-2a -b +2

We get two equations

2a +2b = 3a -3b

-a +5b =0…..(i)

-9a -3b +6 = -7a -7b

-2a +4b +6 =0…..(ii)

Multiplying equation (i) by 2,we get equation (iii)

-2a + 10b =0…..(iii)

Subtracting the equation (iii) from the equation (ii)

-6b + 6=0

-6b = -6

b =1

Putting the value of b in equation (i)

-a +5×1 =0

-a +5 = 0

-a = -5

a = 5

Hence the required values of a,b are 5 and 1 respectively

(ii) The given pair of the equations is

3x + y = 1

(2k – 1)x + (k – 1)y = (2k + 1)

We know two linear equations a_{1}x +b_{1}y+ c_{1}=0 and a_{2}x +b_{2}y+ c_{2}=0 have no solutions if the ratios between coefficients and constants are as follows.

Arranging the given equations into the standard form

3x + y – 1=0

(2k – 1)x + (k – 1)y -(2k + 1)=0

Where a_{1}=3, b_{1}=1, c_{1}=-1 and a_{2}=2k-1, b_{2}=k-1, c_{2}=2k+1

3k -3 = 2k -1

k = -1 +3 =2

Therefore for k=2 , the given pair of linear equations have no solution

**NCERT Solutions for Class 10 Maths Exercise 3.5 of Chapter 3 Pair of Linear Equations**

**Q3.Solve the following pair of linear equations by the substitution and cross-multiplication methods:**

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

Ans. The given pair of linear equations is

8x + 5y = 9….(i)

3x + 2y = 4….(ii)

Taking equation (i) and solving it for the value of x

8x = 9-5y

x = (9 -5y)/8

Putting the value of x in equation (ii)

3×(9-5y)/8 + 2y =4

27/8 -15y/8 +2y = 4

27/8 – (15y -16y)/8 = 4

(27 -15y +16y)/8 = 4

(27 +y)/8 =4

27 +y = 32

y = 5

Putting the value of y in equation (i)

8x + 5×5 = 9

8x +25 =9

8x = -16

x = -2

Hence the value of x is -2 and the value of y is 5

(ii) Arranging the given pair of linear equations into standard form

8x + 5y – 9=0…..(i)

3x + 2y -4=0….(ii)

Arranging the coefficients and constants into a matrix as follows

x =-2 and y =5

Hence the value of x is -2 and the value of y is 5

**NCERT Solutions for Class 10 Maths Exercise 3.5 of Chapter 3 Pair of Linear Equations**

**Q4. 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.**

Ans.Let the monthly fixed charges of the hostel is x and the cost of food per day is y

According to first condition

x + 20y = 1000…(i)

According to second condition

x + 26y = 1180…(ii)

Substracting equation (ii) from equation (i)

-6y = -180

y = 30

Putting the value of y in equayion (i)

x + 20×30 = 1000

x + 600 = 1000

x = 1000 -600 =400

Hence fixed charges is Rs 400 and cost of food per day is Rs 30

**(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.**

Ans.Let the denominator of the fraction is x and the numerator is y

According to the first condition of the question

(y-1)/x = 1/3

3y-3 = x

3y -x = 3….(i)

According to the second condition of the question

y/(x +8) =1/4

4y = x +8

4y -x =8…..(ii)

Substracting equation (ii) from equation (i)

-y =-5

y =5

Putting the value of y in equation (ii)

4×5 -x = 8

x = 20 -8 =12

**(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?**

Ans.Let the correct answers Yash solved in the exam = x and wrong answers are =y

According to first condition of the question

Total marks of Yash in a test =Marks in one correct answer × Number of correct answers + Marks in wrong answer × Number of wrong answers

40 = 3x + (-1)y = 3x -y

3x -y = 40……(i)

According to second condition

4x -2y = 50…..(ii)

Multiplying equation (i) by 2,we get equation (iii)

6x – 2y = 80….(iii)

Substracting equation (iii) from equation (ii)

-2x = -30

x = 15

Putting the value of x in equation (i)

3×15 -y = 40

45 -y = 40

y = 45-40 = 5

Hence total questions in the test are = Number of correct answers + Number of wrong answers = x + y = 15 +5 = 20

**(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?**

Ans. Let the speed of the car which starts from the place A is = x km/h and the speed of the car which starts from the place B is =y km/h

According to first condition

The car from A travels the distance in 5 h = Speed ×Time = x×5 = 5x km

The car from B travels the distance in 5 h = Speed ×Time = y×5 = 5y km

As per the fig.

5x -5y = 100…..(i)

According to second condition,if both of the cars travels in opposite direction

The car from A travels the distance in 1 h = Speed ×Time = x×1 = x km

The car from B travels the distance in 1 h = Speed ×Time = y×1 = y km

As per the fig.

x +y =100…..(ii)

Multiplying the equation (ii) by 5,we get equation (iii)

5x + 5y = 500…..(iii)

10 x = 600

x = 60

Putting the value of x in equation (ii)

60 + y = 100

y =100 -60 =40

Hence the speeds of the cars which start from the place A is 60 km/h and from place B is 40 km/h

**(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.**

Ans.Let the length of the rectangle is x and the breadth of the rectangle is y

∴ Area of the rectangle is =Length ×Breadth = xy

According to first condition of the question

(x -5)(y+3) = xy -9

xy + 3x -5y -15 = xy -9

3x -5y = 6…..(i)

According to second condition of the question

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

xy + 2x +3y +6 = xy +67

2x +3y = 61…..(ii)

Multiplying equation (i) by 2 and equation (ii) by 3,we get equation (iii) and equation (iv)

6x -10y = 12…..(iii) and 6x + 9y =183….(iv)

Substracting equation (iv) from the equation (iii)

-19y = -171

y =171/19 = 9

Putting the value of y in equation (i)

3x -5×9 =6

3x -45 =6

3x = 6 + 45 = 51

x = 51/3 = 17

Hence the length of the rectangle is 17 units and the breadth is 9 units

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**NCERT Solutions of Science and Maths for Class 9,10,11 and 12**

**NCERT Solutions of class 9 maths**

Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |

Chapter 2-Polynomial | Chapter 10-Circles |

Chapter 3- Coordinate Geometry | Chapter 11-Construction |

Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |

Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |

Chapter 6-Lines and Angles | Chapter 14-Statistics |

Chapter 7-Triangles | Chapter 15-Probability |

Chapter 8- Quadrilateral |

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**CBSE Class 9-Question paper of science 2020 with solutions**

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**NCERT solutions of class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

**CBSE Class 11-Question paper of maths 2015**

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

**NCERT solutions of class 12 maths**

Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |

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