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NCERT Solutions for Class 10 Maths Exercise 3.1 of Chapter 3 Linear Equations in Two Variables
NCERT Solutions for Class 10 Maths Exercise 3.1 of Chapter 3 Linear Equations in Two Variables will give you an idea of solving two linear equations in two variables. In the questions of the exercise of chapter 3 -Linear Equation In two variables two conditions of two variables would be given and you are required to build two relations between the two variables in the form of two equations and then solve them for two or more than 2 values and draw the graphs wherever both graphs intersect each other, the corresponding values of x and y of the intersecting point is the solution of the linear equation.
NCERT Solutions for Class 10 Maths Exercise 3.1 of Chapter 3 Linear Equations in Two Variables
Q1.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 be interesting?)Represent this situation algebraically and graphically.
Ans. Let the age of Aftab is x and of his daughter is y
According to first condition, seven years ago the age of Aftab = x – 7 and of his daughter =y-7,further it is given that
Seven years back the age of Afteb = 7× Seven years back his daughter’s age
x – 7 = 7(y-7)
x -7 = 7y -49
x – 7y = -49 + 7
x – 7y = -42……(i)
According to second condition,after 3 years the age of Aftab is = x + 3 and of his daughter’s age = y +3,furher it is given that
After 3 years the age of Aftab = 3× His daughter’s age after 3 years
x + 3 = 3(y + 3)
x + 3 = 3y + 9
x – 3y = 6……(ii)
Solutions of the equation x – 7y = -42
| x | 0 | 7 | -7 |
| y | 6 | 7 | 5 |
Solutions of the equation x – 3y = 6
| x | 0 | 6 | -3 |
| y | -2 | 0 | -3 |
Graphical representation of both equations
Q2.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.
Ans. Let the cost of 1 bat is x and of 1 ball is y
According to fist condition
3 × cost of 1 bat + 6× cost of 1 ball = Rs 3900
3x + 6y = 3900…..(i)
According to second condition
Cost of 1 bat+ 3× cost of 1 ball = Rs 1300
x + 3y = 1300….(ii)
Solutions of equation (i) 3x + 6y = 3900
| x | 100 | -100 | 300 |
| y | 600 | 700 | 500 |
Solutions of equation (ii) x + 3y = 1300
| x | 100 | 400 | -200 |
| y | 400 | 300 | 500 |
Graphical representation of both equations
Q3.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.
Ans. Let the cost of 1 kg apple is = x and the cost of 1 kg grapes is= y
According to first condition of the question
2×cost of 1 kg apple +cost of 1 kg grapes = Rs 160
2x + y = 160…..(i)
According to second condition of the question
4×cost of 1 kg apple +2×cost of 1 kg grapes = Rs 300
4x + 2y = 300…..(ii)
Solutions of the first equation 2x + y = 160
| x | 80 | 60 | 40 |
| y | 0 | 40 | 80 |
Solutions of the second equation 4x + 2y = 300
| x | 60 | 40 | 20 |
| y | 30 | 70 | 110 |
Graphical representation of both equations
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| Chapter 1-Sets | Chapter 9-Sequences and Series |
| Chapter 2- Relations and functions | Chapter 10- Straight Lines |
| Chapter 3- Trigonometry | Chapter 11-Conic Sections |
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| 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 |
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| Chapter 8-Application of Integrals |
