NCERT Solutions for Class 10 Maths Exercise 3.2 of Chapter 3 Linear Equations in Two Variables
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NCERT Solutions for Class 10 Maths Exercise 3.2 of Chapter 3 Linear Equations in Two Variables
Q1. Form the pair of linear equations of 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.
Ans. Let the number of girls is x and the number of boys is y who participated in the quiz.
It is given that total number of the students who took part in the quiz =10
x + y = 10……(i)
According to the second condition of question
Numer of girls = 4 more than the number of boys =4 + Number of the boys
x = y + 4
Rearranging the equation
x – y =4……(ii)
Solutions of the first equation x + y = 10
x | 4 | 5 | 6 |
y | 6 | 5 | 4 |
Solutions of the second equation x – y =4
x | 0 | 4 | 1 |
y | -4 | 0 | -3 |
It is clear from the graph that both lines intersect at (7,3) there fore x =7 and y = 3
Therefore number of girls 7 and number of boys 3 participated 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.
Ans. Let the cost of one pen is x and of one pencil is y
According to first condition
The cost of one pencil × 5 + The cost of one pen ×7 = Rs 50
5x + 7y =50…..(i)
According to the second condition
The cost of one pencil × 7 + The cost of one pen ×5 = Rs 50
7x + 5y =46…..(ii)
Solutions of the first equation 5x + 7y =50
x | 3 | -4 | 10 |
y | 5 | 10 | 0 |
Solutions of the second equation 7x + 5y =46
x | 3 | -2 | 8 |
y | 5 | 12 | -2 |
Both of the lines intersects at (3,5), x =3 and y =5,therefore cost of one pencil is Rs 3 and cost of one pen is Rs 5.
Q2. On comparing the ratios a_{1}/a_{2}, b_{1}/b_{2 }and c_{1}/c_{2}, find out whether the lines representing the following pairs of linear equations intersects at a point,are parallel or coincident:
(i) 5x – 4y + 8 = 0,7x + 6y -9 = 0
(ii)9x + 3y +12 = 0,18x + 6y +24 = 0
(iii)6x -3y +10 = 0, 2x -y +9 = 0
Ans. The given pair of linear equation is 5x – 4y + 8 = 0,7x + 6y -9 = 0
Comparing the given pair of linear equations with standard pair of linear equations a_{1}x+b_{1}y_{ }+ c_{1}=0 and a_{2}x+b_{2}y_{ }+ c_{2}=0
a_{1} =5, b_{1}=-4, c_{1}= 8, a_{2}=7, b_{2}=6, c_{2}= -9
a_{1}/a_{2},=5/7, b_{1}/b_{2 }= -4/6 =-2/3,c_{1}/c_{2 }= 8/-9 =-8/9
Here a_{1}/a_{2}≠b_{1}/b_{2}
Therefore both lines intersects at a point.
(ii) The given pair of linear equation is 9x + 3y +12 = 0,18x + 6y +24 = 0
Comparing the given pair of linear equations with standard pair of linear equations a_{1}x+b_{1}y_{ }+ c_{1}=0 and a_{2}x+b_{2}y_{ }+ c_{2}=0
a_{1} =9, b_{1}=3, c_{1}= 12, a_{2}=18, b_{2}=6, c_{2}= 24
a_{1}/a_{2},=9/18= 1/2, b_{1}/b_{2 }=3/6 = 1/2,c_{1}/c_{2 }= 12/24 =1/2
Therefore the given pair of linear equations are coincident
(iii) The given pair of linear equation is 6x -3y +10 = 0, 2x -y +9 = 0
Comparing the given pair of linear equations with standard pair of linear equations
Therefore the given linear equations are parallel to each other
Q3. On comparing the ratios a_{1}/a_{2}, b_{1}/b_{2 }and c_{1}/c_{2}, find out whether the following pairs of linear equations consistent or inconsistent .
(i) 3x + 2y =5, 2x -3y = 7
(ii)2x -3y = 8, 4x – 6y = 9
(iii)3x/2 +5y/3 = 7, 9x -10 y =14
Ans.
(i) Rearranging the given pair of linear equation is 3x + 2y -5=0, 2x -3y – 7=0
Comparing the given pair of linear equations with standard pair of linear equations a_{1}x+b_{1}y_{ }+ c_{1}=0 and a_{2}x+b_{2}y_{ }+ c_{2}=0
a_{1} =3, b_{1}=2, c_{1}= -5, a_{2}=2, b_{2}=-3, c_{2}= -7
a_{1}/a_{2}=3/2, b_{1}/b_{2 }= 2/-3=-2/3,c_{1}/c_{2 }= -5/-7 =5/7
a_{1}/a_{2}≠b_{1}/b_{2 }
Therefore the given pair of linear equations has unique solutions,thus it is consistent.
(ii) Rearranging the given pair of linear equation is 2x -3y – 8=0, 4x – 6y – 9=0
Comparing the given pair of linear equations with standard pair of linear equations a_{1}x+b_{1}y_{ }+ c_{1}=0 and a_{2}x+b_{2}y_{ }+ c_{2}=0
a_{1} =2, b_{1}=-3, c_{1}= -8, a_{2}=4, b_{2}=-6, c_{2}= -9
a_{1}/a_{2}=2/4=1/2, b_{1}/b_{2 }= -3/-6=1/2,c_{1}/c_{2 }= -8/-9 =8/9
a_{1}/a_{2}=b_{1}/b_{2}≠c_{1}/c_{2}
Therefore the given pair of linear equations are parallel ,it has no solutions thus it is Inconsistent.
(iii) Rearranging the given pair of linear equation 3x/2 +5y/3-7=0, 9x -10 y -14=0
Comparing the given pair of linear equations with standard pair of linear equations a_{1}x+b_{1}y_{ }+ c_{1}=0 and a_{2}x+b_{2}y_{ }+ c_{2}=0
a_{1} =3/2, b_{1}=5/3, c_{1}= -7, a_{2}=9, b_{2}=-10, c_{2}= -14
a_{1}/a_{2}=3/(2×9) = 1/6, b_{1}/b_{2 }= 5/(3×-10)=5/-30 =-1/6,c_{1}/c_{2 }= -7/-14=1/2
a_{1}/a_{2}≠b_{1}/b_{2}
Therefore the given pair of linear equations has unique solutions, it is consistent.
Q4. 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, 2x + 2y = 10
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
Ans.(i) Arranging the given pair of the linear equation x + y – 5=0, 2x + 2y – 10 = 0
Comparing the given pair of linear equations with standard pair of linear equations a_{1}x+b_{1}y_{ }+ c_{1}=0 and a_{2}x+b_{2}y_{ }+ c_{2}=0
a_{1}/a_{2 }= 1/2, b_{1}/b_{2 }= 1/2, c_{1}/c_{2 }= -5/-10 = 1/2,
∴ a_{1}/a_{2 }= b_{1}/b_{2 }= c_{1}/c_{2}
Therefore the given linear equations are coincident,so for every value of x there are infinite values of y
Solutions of the equation (i) x + y – 5=0
x | 1 | 4 | 2 | 3 |
y | 4 | 1 | 3 | 2 |
Solutions of the equation (ii) 2x + 2y – 10 = 0
x | 1 | 4 | 2 | 3 |
y | 4 | 1 | 3 | 2 |
Similarly (ii),(iii),(iv) and v part of the questions can be solved
Q5.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 graphically.
Ans.According to first condition of the question,the half of the perimeter of rectangle = 36
Let the length of the rectangle = x and the length of the breadh =y
Half of the perimeter = 36
2(x +y)/2 = 36
x + y = 36……..(i)
According to second condition,the length is 4 m more than the breadth
Length = Width + 4 m
x = y + 4
arranging the equation
x – y = 4…….(ii)
Solutions of the first equation x + y = 36
x | 20 | 16 | 12 |
y | 16 | 20 | 24 |
Solutions of the second equation x – y = 4
x | 4 | 0 | 8 |
y | 0 | -4 | 4 |
Graphical representation of both equations
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