 NCERT Solutions for Class 10 Maths Exercise 3.2 of Chapter 3 Linear Equations in Two Variables - Future Study Point # NCERT Solutions for Class 10 Maths Exercise 3.2 of Chapter 3 Linear Equations in Two Variables # NCERT Solutions for Class 10 Maths Exercise 3.2 of Chapter 3 Linear Equations in Two Variables

NCERT Solutions for Class 10 Maths Exercise 3.2 of Chapter 3 Linear Equations in Two Variables are created here by an expert of maths that will boost your preparation of maths subject for your CBSE board exam and help you in doing your homework, preparing your worksheet etc. The NCERT Solutions of Class 10 Maths are the best study material in clearing Maths concept, here these NCERT Solutions for Class 10 Maths Exercise 3.2 of Chapter 3 Linear Equations in Two Variables are solved by the expert by a step by step method so every student can understand each solution clearly. You can subscribe us for getting the updated information regarding any kind of study material published by us. ## 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 a1/a2, b1/b2 and c1/c2, 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  a1x+b1y + c1=0 and   a2x+b2y + c2=0

a1 =5, b1=-4, c1= 8, a2=7, b2=6, c2= -9

a1/a2,=5/7, b1/b2 = -4/6 =-2/3,c1/c2 = 8/-9 =-8/9

Here a1/a2≠b1/b2

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  a1x+b1y + c1=0 and   a2x+b2y + c2=0

a1 =9, b1=3, c1= 12, a2=18, b2=6, c2= 24

a1/a2,=9/18= 1/2, b1/b2 =3/6 = 1/2,c1/c2 = 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 a1/a2, b1/b2 and c1/c2, 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  a1x+b1y + c1=0 and   a2x+b2y + c2=0

a1 =3, b1=2, c1= -5, a2=2, b2=-3, c2= -7

a1/a2=3/2, b1/b2 = 2/-3=-2/3,c1/c2 = -5/-7 =5/7

a1/a2≠b1/b

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  a1x+b1y + c1=0 and   a2x+b2y + c2=0

a1 =2, b1=-3, c1= -8, a2=4, b2=-6, c2= -9

a1/a2=2/4=1/2, b1/b2 = -3/-6=1/2,c1/c2 = -8/-9 =8/9

a1/a2=b1/b2≠c1/c2

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  a1x+b1y + c1=0 and   a2x+b2y + c2=0

a1 =3/2, b1=5/3, c1= -7, a2=9, b2=-10, c2= -14

a1/a2=3/(2×9) = 1/6, b1/b2 = 5/(3×-10)=5/-30 =-1/6,c1/c2 = -7/-14=1/2

a1/a2≠b1/b2

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  a1x+b1y + c1=0 and   a2x+b2y + c2=0

a1/a2 = 1/2, b1/b2 = 1/2, c1/c2 = -5/-10 = 1/2,

∴ a1/a2 = b1/b2 = c1/c2

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 You can compensate us by donating any amount of money for our survival

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