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

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# NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

Edited By Dinesh | Updated on Sep 04, 2023 08:55 PM IST | #CBSE Class 10th

## NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables are discussed here. Class 10 Maths chapter 3 is an important chapter of Algebra and this chapter are created by expert team at careers360 keeping in mind of latest CBSE syllabus 2023. In Class 10 Maths chapter 3 solutions, students will learn to solve the linear equation with two variables. Class 10 Maths chapter 3 NCERT solutions contain the answers of all exercise NCERT questions.

NCERT Class 10 Maths chapter 3 solutions are helpful to know the answers to the questions asked in NCERT class 10 maths book. Apart from this, by going through NCERT solutions for class 10 maths chapter 3, they will come to know about various methods of solving questions. NCERT solutions for class 10 are also available for other subjects.

## Pair of Linear Equations in Two Variables Class 10- Important Formulae

Types of Linear Equations:

 S. No. Types of Linear Equation General form Description Solutions 1. Linear Equation in one Variable ax + b = 0 Where a ≠ 0 and a & b are real numbers One Solution 2. Linear Equation in Two Variables ax + by + c = 0 Where a ≠ 0 & b ≠ 0 and a, b & c are real numbers Infinite Solutions possible 3. Linear Equation in Three Variables ax + by + cz + d = 0 Where a ≠ 0, b ≠ 0, c ≠ 0 and a, b, c, d are real numbers Infinite Solutions possible

Simultaneous System of Linear Equations: A pair of equations in the format:

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

This arrangement is visually depicted through the use of two straight lines on the Cartesian plane, as explained in the following context:

When there is a unique solution then

x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b2)

This formula can be remembered as the diagram given below

Free download NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables PDF for CBSE Exam.

## NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables (Intext Questions and Exercise)

Class 10 Maths Chapter 3 solutions Pair of Linear Equations in two variables Excercise: 3.1

Answer: Let x be the age of Aftab and y be the age of his daughter

Now, According to the question,

$(x-7)=7\times(y-7)$

$\Rightarrow x-7=7y-49$

$\Rightarrow x-7y=-42.....(1)$

Also,

$(x+3)=3(y+3)$

$\Rightarrow x+3=3y+9$

$\Rightarrow x-3y=6.....(2)$

Now, let's represent both equations graphically,

From (1), we get

$y=\frac{x+42}{7}$

So, Putting different values of x we get corresponding values of y

 X 0 7 -7 Y 6 7 5

And From (2) we get,

$y=\frac{x-6}{3}$

So, Putting different values of x we get corresponding values of y

 X 0 3 6 Y -2 -1 0

GRAPH:

Let the price of one Bat be x and the price of one ball be y,

Now, According to the question,

$3x+6y=3900.....(1)$

$x+3y=1300.....(2)$

From(1) we have

$y=\frac{3900-3x}{6}$

By putting different values of x, we get different corresponding values of y.So

 X 100 300 -100 Y 600 500 700

Now, From (2), we have,

$y=\frac{1300-x}{3}$

By putting different values of x, we get different corresponding values of y.So

 X 100 400 -200 Y 400 300 500

GRAPH:

Let, x be the cost of 1kg apple and y be the cost of 1kg grapes.

Now, According to the question,

On a day:

$2x+y=160.....(1)$

After One Month:

$4x+2y=300.....(2)$

Now, From (1) we have

$y=160-2x$

Putting different values of x we get corresponding values of y, so,'

 X 80 60 50 Y 0 40 60

And From (2) we have,

$y=\frac{300-4x}{2}=150-2x$

Putting different values of x we get corresponding values of y, so,

 X 50 60 70 Y 50 30 10

Graph:

Class 10 Maths Chapter 3 solutions Pair of Linear Equations in Two Variables Excercise: 3.2

Q1 Form the pair of linear equations in 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.

Let the number of boys is x and the number of girls is y.

Now, According to the question,

Total number of students in the class = 10, i.e.

$\Rightarrow x+y=10.....(1)$

And

the number of girls is 4 more than the number of boys,i.e.

$x=y+4$

$\Rightarrow x-y=4..........(2)$

Different points (x, y) for equation (1)

 X 5 6 4 Y 5 4 6

Different points (x,y) satisfying (2)

 X 5 6 7 y 1 2 3

Graph,

As we can see from the graph, both lines intersect at the point (7,3). that is x= 7 and y = 3 which means the number of boys in the class is 7 and the number of girls in the class is 3.

Q1 Form the pair of linear equations in the following problems and find their solutions graphically.

(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.

Let x be the price of 1 pencil and y be the price of 1 pen,

Now, According to the question

$5x+7y=50......(1)$

And

$7x+5y=46......(2)$

Now, the points (x,y), that satisfies the equation (1) are

 X 3 -4 10 Y 5 10 0

And, the points(x,y) that satisfies the equation (2) are

 X 3 8 -2 Y 5 -2 12

The Graph,

As we can see from the Graph, both line intersects at point (3,5) that is, x = 3 and y = 5 which means cost of 1 pencil is 3 and the cost of 1 pen is 5.

Give, Equations,

$\\5x - 4y + 8 = 0\\ 7x + 6y - 9 = 0$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{a_1}{a_2}=\frac{5}{7},\:\frac{b_1}{b_2}=\frac{-4}{6}\:and\:\frac{c_1}{c_2}=\frac{8}{-9}$

As we can see

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$

It means that both lines intersect at exactly one point.

Given, Equations,

$\\9x + 3y + 12 = 0\\ 18x + 6y + 24 = 0$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}\:and \\\:\frac{c_1}{c_2}=\frac{12}{24}=\frac{1}{2}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

It means that both lines are coincident.

Give, Equations,

$\\6x - 3y + 10 = 0\\ 2x - y+ 9 = 0$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{a_1}{a_2}=\frac{6}{2}=3,\:\frac{b_1}{b_2}=\frac{-3}{-1}=3\:and\:\frac{c_1}{c_2}=\frac{10}{9}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

It means that both lines are parallel to each other.

Give, Equations,

$\\3x + 2y = 5;\qquad\\ 2x - 3y = 7$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{a_1}{a_2}=\frac{3}{2},\:\frac{b_1}{b_2}=\frac{2}{-3}\:and\:\frac{c_1}{c_2}=\frac{5}{7}$

As we can see

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$

It means the given equations have exactly one solution and thus pair of linear equations is consistent.

Given, Equations,

$\\2x - 3y = 8;\qquad \\4x - 6y = 9$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{-3}{-6}=\frac{1}{2}\:and\\\:\frac{c_1}{c_2}=\frac{8}{9}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

It means the given equations have no solution and thus pair of linear equations is inconsistent.

Given, Equations,

$\\\frac{3}{2}x + \frac{5}{3}y = 7;\qquad\\ \\ 9x -10y = 14$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{3/2}{9}=\frac{3}{18}=\frac{1}{6},\\\:\frac{b_1}{b_2}=\frac{5/3}{-10}=\frac{5}{-30}=-\frac{1}{6}\:and\\\:\frac{c_1}{c_2}=\frac{7}{14}=\frac{1}{2}$

As we can see

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$

It means the given equations have exactly one solution and thus pair of linear equations is consistent.

Given, Equations,

$5x - 3y = 11;\qquad \\-10x + 6y =-22$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{5}{-10}=-\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{-3}{6}=-\frac{1}{2}\:and\\\:\frac{c_1}{c_2}=\frac{11}{-22}=-\frac{1}{2}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

It means the given equations have an infinite number of solutions and thus pair of linear equations is consistent.

Given, Equations,

$\\\frac{4}{3}x + 2y = 8; \qquad\\\\ 2x + 3y = 12$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{4/3}{2}=\frac{4}{6}=\frac{2}{3},\\\:\frac{b_1}{b_2}=\frac{2}{3}\:\:and\\\:\frac{c_1}{c_2}=\frac{8}{12}=\frac{2}{3}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

It means the given equations have an infinite number of solutions and thus pair of linear equations is consistent.

Given, Equations,

$\\x + y = 5 \qquad\\ 2x + 2 y = 10$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{1}{2}\:and\\\:\frac{c_1}{c_2}=\frac{5}{10}=\frac{1}{2}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

It means the given equations have an infinite number of solutions and thus pair of linear equations is consistent.

The points (x,y) which satisfies in both equations are

 X 1 3 5 Y 4 2 0

Given, Equations,

$\\x - y = 8,\qquad\\ 3x - 3y = 16$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{1}{3},\\\:\frac{b_1}{b_2}=\frac{-1}{-3}=\frac{1}{3}\:and\\\:\frac{c_1}{c_2}=\frac{8}{16}=\frac{1}{2}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

It means the given equations have no solution and thus pair of linear equations is inconsistent.

Given, Equations,

$\\2x + y - 6 =0, \qquad \\4x - 2 y - 4 = 0$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{1}{-2}=-\frac{1}{2}\:and\\\:\frac{c_1}{c_2}=\frac{-6}{-4}=\frac{3}{2}$

As we can see

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$

It means the given equations have exactly one solution and thus pair of linear equations is consistent.

Now The points(x, y) satisfying the equation are,

 X 0 2 3 Y 6 2 0

And The points(x,y) satisfying the equation $\\4x - 2 y - 4 = 0$ are,

 X 0 1 2 Y -2 0 2

GRAPH:

As we can see both lines intersects at point (2,2) and hence the solution of both equations is x = 2 and y = 2.

Given, Equations,

$\\2x - 2y - 2 =0, \qquad\\ 4x - 4y -5 = 0$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{-2}{-4}=\frac{1}{2}\:and\\\:\frac{c_1}{c_2}=\frac{-2}{-5}=\frac{2}{5}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

It means the given equations have no solution and thus pair of linear equations is inconsistent.

Let $l$be the length of the rectangular garden and $b$ be the width.

Now, According to the question, the length is 4 m more than its width.i.e.

$l=b+4$

$l-b=4....(1)$

Also Given Half Parameter of the rectangle = 36 i.e.

$l+b=36....(2)$

Now, as we have two equations, on adding both equations, we get,

$l+b+l-b=4+36$

$\Rightarrow 2l=40$

$\Rightarrow l=20$

Putting this in equation (1),

$\Rightarrow 20-b=4$

$\Rightarrow b=20-4$

$\Rightarrow b=16$

Hence Length and width of the rectangle are 20m and 16 respectively.

Given the equation,

$2x + 3y -8 =0$

As we know that the condition for the intersection of lines $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , is ,

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$

So Any line with this condition can be $4x+3y-16=0$

Here,

$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{3}{3}=1$

As

$\frac{1}{2}\neq1$ the line satisfies the given condition.

Given the equation,

$2x + 3y -8 =0$

As we know that the condition for the lines $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , for being parallel is,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

So Any line with this condition can be $4x+6y-8=0$

Here,

$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}$

$\frac{c_1}{c_2}=\frac{-8}{-8}=1$

As

$\frac{1}{2}=\frac{1}{2}\neq1$ the line satisfies the given condition.

Given the equation,

$2x + 3y -8 =0$

As we know that the condition for the coincidence of the lines $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , is,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

So any line with this condition can be $4x+6y-16=0$

Here,

$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}$

$\frac{c_1}{c_2}=\frac{-8}{-16}=\frac{1}{2}$

As

$\frac{1}{2}=\frac{1}{2}=\frac{1}{2}$ the line satisfies the given condition.

Given, two equations,

$x - y + 1=0.........(1)$

And

$3x +2 y - 12=0.........(2)$

The points (x,y) satisfying (1) are

 X 0 3 6 Y 1 4 7

And The points(x,y) satisfying (2) are,

 X 0 2 4 Y 6 3 0

GRAPH:

As we can see from the graph that both lines intersect at the point (2,3), And the vertices of the Triangle are ( -1,0), (2,3) and (4,0). The area of the triangle is shaded with a green color.

Pair of linear equations in two variables class 10 solutions Excercise: 3.3

Given, two equations,

$\\x + y = 14.......(1)\\ x - y = 4........(2)$

Now, from (1), we have

$y=14-x........(3)$

Substituting this in (2), we get

$x-(14-x)=4$

$\Rightarrow x-14+x=4$

$\Rightarrow 2x=4+14=18$

$\Rightarrow x=9$

Substituting this value of x in (3)

$\Rightarrow y=14-x=14-9=5$

Hence, Solution of the given equations is x = 9 and y = 5.

Given, two equations,

$\\s - t = 3.........(1)\\ \frac{s}{3} + \frac{t}{2} = 6.......(2)$

Now, from (1), we have

$s=t+3........(3)$

Substituting this in (2), we get

$\frac{t+3}{3}+\frac{t}{2}=6$

$\Rightarrow \frac{2t+6+3t}{6}=6$

$\Rightarrow 5t+6=36$

$\Rightarrow 5t=30$

$\Rightarrow t=6$

Substituting this value of t in (3)

$\Rightarrow s=t+3 = 6+3=9$

Hence, Solution of the given equations is s = 9 and t = 6.

Given, two equations,

$\\ 3 x - y = 3......(1)\\ 9x - 3y = 9.....(2)$

Now, from (1), we have

$y=3x-3........(3)$

Substituting this in (2), we get

$9x-3(3x-3)=9$

$\Rightarrow 9x-9x+9=9$

$\Rightarrow 9=9$

This is always true, and hence this pair of the equation has infinite solutions.

As we have

$y=3x-3$ ,

One of many possible solutions is $x=1,\:and\:y=0$ .

Given, two equations,

$\\0.2x + 0.3y = 1.3\\ 0.4x + 0.5y = 2.3$

Now, from (1), we have

$y=\frac{1.3-0.2x}{0.3}........(3)$

Substituting this in (2), we get

$0.4x+0.5\left(\frac{1.3-0.2x}{0.3}\right)=2.3$

$\Rightarrow 0.12x+0.65-0.1x=0.69$

$\Rightarrow 0.02x=0.69-0.65=0.04$

$\Rightarrow x=2$

Substituting this value of x in (3)

$\Rightarrow y=\left ( \frac{1.3-0.2x}{0.3} \right )=\left ( \frac{1.3-0.4}{0.3} \right )=\frac{0.9}{0.3}=3$

Hence, Solution of the given equations is,

$x=2\:and\:y=3$ .

Given, two equations,

$\\\sqrt2x + \sqrt3y = 0\\ \sqrt3x - \sqrt8y= 0$

Now, from (1), we have

$y=-\frac{\sqrt{2}x}{\sqrt{3}}........(3)$

Substituting this in (2), we get

$\sqrt{3}x-\sqrt{8}\left ( -\frac{\sqrt{2}x}{\sqrt{3}} \right )=0$

$\Rightarrow \sqrt{3}x=\sqrt{8}\left ( -\frac{\sqrt{2}x}{\sqrt{3}} \right )$

$\Rightarrow \3x=-4x$

$\Rightarrow7x=0$

$\Rightarrow x=0$

Substituting this value of x in (3)

$\Rightarrow y=-\frac{\sqrt{2}x}{\sqrt{3}}=0$

Hence, Solution of the given equations is,

$x=0,\:and \:y=0$ .

Given,

$\\\frac{3x}{2} - \frac{5y}{3}= - 2.........(1)\\ \frac{x}{3} + \frac{y}{2} = \frac{13}{6}.............(2)$

From (1) we have,

$x=\frac{2}{3}\left ( \frac{5y}{3}-2 \right )........(3)$

Putting this in (2) we get,

$\frac{1}{3}\times\frac{2}{3}\left ( \frac{5y}{3}-2 \right )+\frac{y}{2}=\frac{13}{6}$

$\frac{10y}{27}-\frac{4}{9}+\frac{y}{2}=\frac{13}{6}$

$\frac{20y}{54}-\frac{4}{9}+\frac{27y}{54}=\frac{13}{6}$

$\frac{47y}{54}=\frac{13}{6}+\frac{4}{9}$

$\frac{47y}{54}=\frac{117}{54}+\frac{24}{54}$

$47y=117+24$

$47y=141$

$y=\frac{141}{47}$

$y=3$

putting this value in (3) we get,

$x=\frac{2}{3}\left ( \frac{5y}{3}-2 \right )$

$x=\frac{2}{3}\left ( \frac{5\times3}{3}-2 \right )$

$x=\frac{2}{3}\left (5-2 \right )$

$x=\frac{2}{3}\times 3$

$x=2$

Hence $x=2\:\:and\:\:y=3.$

Given, two equations,

$2x + 3y = 11......(1)$

$2x - 4y = -24.......(2)$

Now, from (1), we have

$y=\frac{11-2x}{3}........(3)$

Substituting this in (2), we get

$2x-4\left ( \frac{11-2x}{3} \right )=-24$

$\Rightarrow 6x-44+8x=-72$

$\Rightarrow 14x=44-72$

$\Rightarrow 14x=-28$

$\Rightarrow x=-2$

Substituting this value of x in (3)

$\Rightarrow y=\left ( \frac{11-2x}{3} \right )=\frac{11-2\times(-2)}{3}=\frac{15}{3}=5$

Hence, Solution of the given equations is,

$x=-2,\:and\:y=5.$

Now,

As it satisfies $y=mx+3$ ,

$\Rightarrow 5=m(-2)+3$

$\Rightarrow 2m=3-5$

$\Rightarrow 2m=-2$

$\Rightarrow m=-1$

Hence Value of m is -1.

(i) The difference between the two numbers is 26 and one number is three times the other. Find them.

Let two numbers be x and y and let the bigger number is y.

Now, According to the question,

$y-x=26......(1)$

And

$y=3x......(2)$

Now, the substituting value of y from (2) in (1) we get,

$3x-x=26$

$\Rightarrow 2x=26$

$\Rightarrow x=13$

Substituting this in (2)

$\Rightarrow y=3x=3(13)=39$

Hence the two numbers are 13 and 39.

Let the larger angle be x and smaller angle be y

Now, As we know the sum of supplementary angles is 180. so,

$x+y=180.......(1)$

Also given in the question,

$x-y=18.......(2)$

Now, From (2) we have,

$y=x-18.......(3)$

Substituting this value in (1)

$x+x-18=180$

$\Rightarrow 2x=180+18$

$\Rightarrow 2x=198$

$\Rightarrow x=99$

Now, Substituting this value of x in (3), we get

$\Rightarrow y=x-18=99-18=81$

Hence the two supplementary angles are

$99^0\:and\:81^0.$

Let the cost of 1 bat is x and the cost of 1 ball is y.

Now, According to the question,

$7x+6y=3800......(1)$

$3x+5y=1750......(2)$

Now, From (1) we have

$y=\frac{3800-7x}{6}........(3)$

Substituting this value of y in (2)

$3x+5\left ( \frac{3800-7x}{6} \right )=1750$

$\Rightarrow 18x+19000-35x=1750\times6$

$\Rightarrow -17x=10500-19000$

$\Rightarrow -17x=-8500$

$\Rightarrow x=\frac{8500}{17}$

$\Rightarrow x=500$

Now, Substituting this value of x in (3)

$y=\frac{3800-7x}{6}=\frac{3800-7\times500}{6}=\frac{3800-3500}{6}=\frac{300}{6}=50$

Hence, The cost of one bat is 500 Rs and the cost of one ball 50 Rs.

Let the fixed charge is x and the per km charge is y.

Now According to the question

$x+10y=105.......(1)$

And

$x+15y=155.......(2)$

Now, From (1) we have,

$x=105-10y........(3)$

Substituting this value of x in (2), we have

$105-10y+15y=155$

$\Rightarrow 5y=155-105$

$\Rightarrow 5y=50$

$\Rightarrow y=10$

Now, Substituting this value in (3)

$x=105-10y=105-10(10)=105-100=5$

Hence, the fixed charge is 5 Rs and the per km charge is 10 Rs.

Now, Fair For 25 km :

$\Rightarrow x+25y=5+25(10)=5+250=255$

Hence fair for 25km is 255 Rs.

Let the numerator of the fraction be x and denominator of the fraction is y

Now According to the question,

$\frac{x+2}{y+2}=\frac{9}{11}$

$\Rightarrow 11(x+2)=9(y+2)$

$\Rightarrow 11x+22=9y+18$

$\Rightarrow 11x-9y=-4...........(1)$

Also,

$\frac{x+3}{y+3}=\frac{5}{6}$

$\Rightarrow 6(x+3)=5(y+3)$

$\Rightarrow 6x+18=5y+15$

$\Rightarrow 6x-5y=-3...........(2)$

Now, From (1) we have

$y=\frac{11x+4}{9}.............(3)$

Substituting this value of y in (2)

$6x-5\left ( \frac{11x+4}{9} \right )=-3$

$\Rightarrow 54x-55x-20=-27$

$\Rightarrow -x=20-27$

$\Rightarrow x=7$

Substituting this value of x in (3)

$y=\frac{11x+4}{9}=\frac{11(7)+4}{9}=\frac{81}{9}=9$

Hence the required fraction is

$\frac{x}{y}=\frac{7}{9}.$

Let x be the age of Jacob and y be the age of Jacob's son,

Now, According to the question

$x+5=3(y+5)$

$\Rightarrow x+5=3y+15$

$\Rightarrow x-3y=10..........(1)$

Also,

$x-5=7(y-5)$

$\Rightarrow x-5=7y-35$

$\Rightarrow x-7y=-30.........(2)$

Now,

From (1) we have,

$x=10+3y...........(3)$

Substituting this value of x in (2)

$10+3y-7y=-30$

$\Rightarrow -4y=-30-10$

$\Rightarrow 4y=40$

$\Rightarrow y=10$

Substituting this value of y in (3),

$x=10+3y=10+3(10)=10+30=40$

Hence, Present age of Jacob is 40 years and the present age of Jacob's son is 10 years.

Pair of linear equations in two variables class 10 solutions Excercise: 3.4

(i) $x + y =5 \ \textup{and} \ 2x - 3y = 4$

Elimination Method:

Given, equations

$\\x + y =5............(1) \ \textup{and} \\ \ 2x - 3y = 4........(2)$

Now, multiplying (1) by 3 we, get

$\\3x +3 y =15............(3)$

Now, Adding (2) and (3), we get

$\\2x-3y+3x +3 y =4+15$

$\Rightarrow 5x=19$

$\Rightarrow x=\frac{19}{5}$

Substituting this value in (1) we, get

$\frac{19}{5}+y=5$

$\Rightarrow y=5-\frac{19}{5}$

$\Rightarrow y=\frac{6}{5}$

Hence,

$x=\frac{19}{5}\:and\:y=\frac{6}{5}$

Substitution method :

Given, equations

$\\x + y =5............(1) \ \textup{and} \\ \ 2x - 3y = 4........(2)$

Now, from (1) we have,

$y=5-x.......(3)$

substituting this value in (2)

$2x-3(5-x)=4$

$\Rightarrow 2x-15+3x=4$

$\Rightarrow 5x=19$

$\Rightarrow x=\frac{19}{5}$

Substituting this value of x in (3)

$\Rightarrow y=5-x=5-\frac{19}{5}=\frac{6}{5}$

Hence,

$x=\frac{19}{5}\:and\:y=\frac{6}{5}$

(ii) $3x + 4 y = 10 \ \textup{and} \ 2x - 2y = 2$

Elimination Method:

Given, equations

$\\3x + 4 y = 10............(1) \ \textup{and}\\ \ 2x - 2y = 2..............(2)$

Now, multiplying (2) by 2 we, get

$\\4x -4 y =4............(3)$

Now, Adding (1) and (3), we get

$\\3x+4y+4x -4 y =10+4$

$\Rightarrow 7x=14$

$\Rightarrow x=2$

Putting this value in (2) we, get

$2(2)-2y=2$

$\Rightarrow 2y=2$

$\Rightarrow y=1$

Hence,

$x=2\:and\:y=1$

Substitution method :

Given, equations

$\\3x + 4 y = 10............(1) \ \textup{and}\\ \ 2x - 2y = 2..............(2)$

Now, from (2) we have,

$y=\frac{2x-2}{2}=x-1.......(3)$

substituting this value in (1)

$3x+4(x-1)=10$

$\Rightarrow 3x+4x-4=10$

$\Rightarrow 7x=14$

$\Rightarrow x=2$

Substituting this value of x in (3)

$\Rightarrow y=x-1=2-1=1$

Hence,

$x=2\:and\:y=1$

Elimination Method:

Given, equations

$\\3x - 5y -4 = 0..........(1)\ \textup{and}\ \\9x = 2y + 7$

$\\\Rightarrow 9x - 2y -7=0........(2)$

Now, multiplying (1) by 3 we, get

$\\9x -15 y -12=0............(3)$

Now, Subtracting (3) from (2), we get

$9x-2y-7-9x+15y+12=0$

$\Rightarrow 13y+5=0$

$\Rightarrow y=\frac{-5}{13}$

Putting this value in (1) we, get

$3x-5(\frac{-5}{13})-4=0$

$\Rightarrow 3x=4-\frac{25}{13}$

$\Rightarrow 3x=\frac{27}{13}$

$\Rightarrow x=\frac{9}{13}$

Hence,

$x=\frac{9}{13}\:and\:y=-\frac{5}{13}$

Substitution method :

Given, equations

$\\3x - 5y -4 = 0..........(1)\ \textup{and}\ \\9x = 2y + 7$

$\\\Rightarrow 9x - 2y -7=0........(2)$

Now, from (2) we have,

$y=\frac{9x-7}{2}.......(3)$

substituting this value in (1)

$3x-5\left(\frac{9x-7}{2} \right )-4=0$

$\Rightarrow 6x-45x+35-8=0$

$\Rightarrow -39x+27=0$

$\Rightarrow x=\frac{27}{39}=\frac{9}{13}$

Substituting this value of x in (3)

$\Rightarrow y=\frac{9(9/13)-7}{2}=\frac{81/13-7}{2}=\frac{-5}{13}$

Hence,

$x=\frac{9}{13}\:and\:y=-\frac{5}{13}$

Elimination Method:

Given, equations

$\\\frac{x}{2} + \frac{2y}{3} = -1........(1)\ \textup{and} \ \\ \\x - \frac{y}{3} = 3............(2)$

Now, multiplying (2) by 2 we, get

$\\2x - \frac{2y}{3} =6............(3)$

Now, Adding (1) and (3), we get

$\\\frac{x}{2}+\frac{2y}{3}+2x-\frac{2y}{3}=-1+6$

$\Rightarrow \frac{5x}{2}=5$

$\Rightarrow x=2$

Putting this value in (2) we, get

$2-\frac{y}{3}=3$

$\Rightarrow \frac{y}{3}=-1$

$\Rightarrow y=-3$

Hence,

$x=2\:and\:y=-3$

Substitution method :

Given, equations

$\\\frac{x}{2} + \frac{2y}{3} = -1........(1)\ \textup{and} \ \\ \\x - \frac{y}{3} = 3............(2)$

Now, from (2) we have,

$y=3(x-3)......(3)$

substituting this value in (1)

$\frac{x}{2}+\frac{2(3(x-3))}{3}=-1$

$\Rightarrow \frac{x}{2}+2x-6=-1$

$\Rightarrow \frac{5x}{2}=5$

$\Rightarrow x=2$

Substituting this value of x in (3)

$\Rightarrow y=3(x-3)=3(2-1)=-3$

Hence,

$x=2\:and\:y=-3$

Let the numerator of the fraction be x and denominator is y,

Now, According to the question,

$\frac{x+1}{y-1}=1$

$\Rightarrow x+1=y-1$

$\Rightarrow x-y=-2.........(1)$

Also,

$\frac{x}{y+1}=\frac{1}{2}$

$\Rightarrow 2x=y+1$

$\Rightarrow 2x-y=1..........(2)$

Now, Subtracting (1) from (2) we get

$x=3$

Putting this value in (1)

$3-y=-2$

$\Rightarrow y=5$

Hence

$x=3\:and\:y=5$

And the fraction is

$\frac{3}{5}$

Let the age of Nuri be x and age of Sonu be y.

Now, According to the question

$x-5=3(y-5)$

$\Rightarrow x-5=3y-15$

$\Rightarrow x-3y=-10.........(1)$

Also,

$x+10=2 (y+10)$

$\Rightarrow x+10=2y+20$

$\Rightarrow x-2y=10........(2)$

Now, Subtracting (1) from (2), we get

$y=20$

putting this value in (2)

$x-2(20)=10$

$\Rightarrow x=50$

Hence the age of Nuri is 50 and the age of Nuri is 20.

Let the unit digit of the number be x and 10's digit be y.

Now, According to the question,

$x+y=9.......(1)$ 24323

Also

$9(10y+x)=2(10x+y)$

$\Rightarrow 90y+9x=20x+2y$

$\Rightarrow 88y-11x=0$

$\Rightarrow 8y-x=0.........(2)$

Now adding (1) and (2) we get,

$\Rightarrow 9y=9$

$\Rightarrow y=1$

now putting this value in (1)

$x+1=9$

$\Rightarrow x=8$

Hence the number is 18.

Let the number of Rs 50 notes be x and the number of Rs 100 notes be y.

Now, According to the question,

$x+y=25..........(1)$

And

$50x+100y=2000$

$\Rightarrow x+2y=40.............(2)$

Now, Subtracting(1) from (2), we get

$y=15$

Putting this value in (1).

$x+15=25$

$\Rightarrow x=10$

Hence Meena received 10 50 Rs notes and 15 100 Rs notes.

Let fixed charge be x and per day charge is y.

Now, According to the question,

$x+4y=27...........(1)$

And

$x+2y=21...........(2)$

Now, Subtracting (2) from (1). we get,

$4y-2y=27-21$

$\Rightarrow 2y=6$

$\Rightarrow y=3$

Putting this in (1)

$x+4(3)=27$

$\Rightarrow x=27-12=15$

Hence the fixed charge is 15 Rs and per day charge is 3 Rs.

Pair of linear equations in two variables class 10 solutions Excercise: 3.5

(i) $\\x - 3y -3 = 0\\ 3x - 9y -2 = 0$

Given, two equations,

$\\x - 3y -3 = 0.........(1)\\ 3x - 9y -2 = 0........(2)$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{a_1}{a_2}=\frac{1}{3}$

$\frac{b_1}{b_2}=\frac{-3}{-9}=\frac{1}{3}$

$\frac{c_1}{c_2}=\frac{-3}{-2}=\frac{3}{2}$

As we can see,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

Hence, the pair of equations has no solution.

(ii) $\\2x + y = 5 \\ 3x + 2y = 8$

Given, two equations,

$\\2x + y = 5.........(1) \\ 3x + 2y = 8..........(2)$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{a_1}{a_2}=\frac{2}{3}$

$\frac{b_1}{b_2}=\frac{1}{2}$

$\frac{c_1}{c_2}=\frac{5}{8}$

As we can see,

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

Hence, the pair of equations has exactly one solution.

By Cross multiplication method,

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(1)(-8)-(2)(-5)}=\frac{y}{(-5)(3)-(-8)(2)}=\frac{1}{(2)(2)-(3)(1)}$

$\frac{x}{2}=\frac{y}{1}=\frac{1}{1}$

$x=2,\:and\:y=1$

(iii) $\\3x -5 y = 20\\ 6x - 10y = 40$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{a_1}{a_2}=\frac{3}{6}=\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{-5}{-10}=\frac{1}{2}$

$\frac{c_1}{c_2}=\frac{20}{40}=\frac{1}{2}$

As we can see,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Hence, the pair of equations has infinitely many solutions.

(iv) $\\x - 3y -7 = 0\\ 3x -3y -15 =0$

Given the equations,

$\\x - 3y -7 = 0.........(1)\\ 3x -3y -15 =0........(2)$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{a_1}{a_2}=\frac{1}{3}$

$\frac{b_1}{b_2}=\frac{-3}{-3}=1$

$\frac{c_1}{c_2}=\frac{-7}{-15}=\frac{7}{15}$

As we can see,

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

Hence, the pair of equations has exactly one solution.

By Cross multiplication method,

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(-3)(-15)-(-3)(-7)}=\frac{y}{(-7)(3)-(-15)(1)}=\frac{1}{(1)(-3)-(3)(-3)}$

$\frac{x}{45-21}=\frac{y}{-21+15}=\frac{1}{-3+9}$

$\frac{x}{24}=\frac{y}{-6}=\frac{1}{6}$

$x=\frac{24}{6}=4,\:and\:y=-1$

Given equations,

$\\2x + 3y = 7 \\(a - b) x + (a + b) y = 3a + b - 2$

As we know, the condition for equations $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ to have an infinite solution is

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

So, Comparing these equations with, $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{2}{a-b}=\frac{3}{a+b}=\frac{7}{3a+b-2}$

From here we get,

$\frac{2}{a-b}=\frac{3}{a+b}$

$\Rightarrow 2(a+b)=3(a-b)$

$\Rightarrow 2a+2b=3a-3b$

$\Rightarrow a-5b=0.........(1)$

Also,

$\frac{2}{a-b}=\frac{7}{3a+b-2}$

$\Rightarrow 2(3a+b-2)=7(a-b)$

$\Rightarrow 6a+2b-4=7a-7b$

$\Rightarrow a-9b+4=0...........(2)$

Now, Subtracting (2) from (1) we get

$\Rightarrow 4b-4=0$

$\Rightarrow b=1$

Substituting this value in (1)

$\Rightarrow a-5(1)=0$

$\Rightarrow a=5$

Hence, $a=5\:and\:b=1$ .

Given, the equations,

$\\3x + y = 1 \\(2k - 1) x + (k - 1) y = 2k + 1$

As we know, the condition for equations $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ to have no solution is

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

So, Comparing these equations with, $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{3}{2k-1}=\frac{1}{k-1}\neq\frac{1}{2k+1}$

From here we get,

$\frac{3}{2k-1}=\frac{1}{k-1}$

$\Rightarrow 3(k-1)=2k-1$

$\Rightarrow 3k-3=2k-1$

$\Rightarrow 3k-2k=3-1$

$\Rightarrow k=2$

Hence, the value of K is 2.

Given the equations

$\\8x + 5y = 9........(1) \\3x + 2y = 4........(2)$

By Substitution Method,

From (1) we have

$y=\frac{9-8x}{5}.........(3)$

Substituting this in (2),

$3x+2\left ( \frac{9-8x}{5} \right )=4$

$\Rightarrow 15x+18-16x=20$

$\Rightarrow -x=20-18$

$\Rightarrow x=-2$

Substituting this in (3)

$y=\frac{9-8x}{5}=\frac{9-8(-2)}{5}=\frac{25}{5}=5$

Hence $x=-2\:and\:y=5$ .

By Cross Multiplication Method

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(5)(-4)-(2)(-9)}=\frac{y}{(3)(-9)-(8)(-4)}=\frac{1}{(8)(2)-(3)(5)}$

$\frac{x}{-20+18}=\frac{y}{32-27}=\frac{1}{16-15}$

$\frac{x}{-2}=\frac{y}{5}=\frac{1}{1}$

$x=-2,\:and\:y=5$

Let the fixed charge be x and the cost of food per day is y,

Now, According to the question

$x+20y=1000.........(1)$

Also

$x+26y=1180.........(2)$

Now subtracting (1) from (2),

$x+26y-x-20y=1180-100$

$\Rightarrow 6y=180$

$\Rightarrow y=30$

Putting this value in (1)

$x+20(30)=1000$

$\Rightarrow x=1000-600$

$\Rightarrow x=400$

Hence, the Fixed charge is Rs 400 and the cost of food per day is Rs 30.

Let numerator of a fraction be x and the denominator is y.

Now, According to the question,

$\frac{x-1}{y}=\frac{1}{3}$

$\Rightarrow 3(x-1)=y$

$\Rightarrow 3x-3=y$

$\Rightarrow 3x-y=3........(1)$

Also,

$\frac{x}{y+8}=\frac{1}{4}$

$\Rightarrow 4x=y+8$

$\Rightarrow 4x-y=8.........(2)$

Now, Subtracting (1) from (2) we get,

$4x-3x=8-3$

$\Rightarrow x=5$

Putting this value in (2) we get,

$4(5)-y=8$

$\Rightarrow y=20-8$

$\Rightarrow y=12$

Hence, the fraction is

$\frac{x}{y}=\frac{5}{12}$ .

Let the number of right answer and wrong answer be x and y respectively

Now, According to the question,

$3x-y=40..........(1)$

And

$\\4x-2y=50\\\Rightarrow 2x-y=25..........(2)$

Now, subtracting (2) from (1) we get,

$x=40-25$

$x=15$

Putting this value in (1)

$3(15)-y=40$

$\Rightarrow y=45-40$

$\Rightarrow y=5$

Hence the total number of question is $x+y=15+5=20.$

Let the speed of the first car is x and the speed of the second car is y.

Let's solve this problem by using relative motion concept,

the relative speed when they are going in the same direction= x - y

the relative speed when they are going in the opposite direction= x + y

The given relative distance between them = 100 km.

Now, As we know,

Relative distance = Relative speed * time .

So, According to the question,

$5\times(x-y)=100$

$\Rightarrow 5x-5y=100$

$\Rightarrow x-y=20.........(1)$

Also,

$1(x+y)=100$

$\Rightarrow x+y=100........(2)$

Now Adding (1) and (2) we get

$2x=120$

$\Rightarrow x=60$

putting this in (1)

$60-y=20$

$\Rightarrow y=60-20$

$\Rightarrow y=40$

Hence the speeds of the cars are 40 km/hour and 60 km/hour.

Let $l$be the length of the rectangle and $b$ be the width,

Now, According to the question,

$(l-5)(b+3)=lb-9$

$\Rightarrow lb+3l-5b-15=lb-9$

$\Rightarrow 3l-5b-6=0..........(1)$

Also,

$(l+3)(b+2)=lb+67$

$\Rightarrow lb+2l+3b+6=lb+67$

$\Rightarrow 2l+3b-61=0..........(2)$

By Cross multiplication method,

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(-5)(-61)-(3)(-6)}=\frac{y}{(-6)(2)-(-61)(3)}=\frac{1}{(3)(3)-(2)(-5)}$

$\frac{x}{305+18}=\frac{y}{-12+183}=\frac{1}{9+10}$

$\frac{x}{323}=\frac{y}{171}=\frac{1}{19}$

$x=17,\:and\:y=9$

Hence the length and width of the rectangle are 17 units and 9 units respectively.

NCERT Solutions for class 10 math chapter 3 Pair of Linear Equations in Two Variables Excercise: 3.6

(i) $\\\frac{1}{2x} +\frac{1}{3y} = 2\\ \frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}$

Given Equations,

$\\\frac{1}{2x} +\frac{1}{3y} = 2\\ \frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}$

Let,

$\frac{1}{x}=p\:and\:\frac{1}{y}=q$

Now, our equation becomes

$\frac{p}{2}+\frac{q}{3}=2$

$\Rightarrow 3p+2q=12........(1)$

And

$\frac{p}{3}+\frac{q}{2}=\frac{13}{6}$

$\Rightarrow 2p+3q=13..........(2)$

By Cross Multiplication method,

$\frac{p}{b_1c_2-b_2c_1}=\frac{q}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{p}{(2)(-13)-(3)(-12)}=\frac{q}{(-12)(2)-(-13)(3)}=\frac{1}{(3)(3)-(2)(2)}$

$\frac{p}{-26+36}=\frac{q}{-24+39}=\frac{1}{9-4}$

$\frac{p}{10}=\frac{q}{15}=\frac{1}{5}$

$p=2,\:and\:q=3$

And Hence,

$x=\frac{1}{2}\:and\:y=\frac{1}{3}.$

(ii) $\\ \frac{2}{\sqrt x} + \frac{3}{\sqrt y} = 2\\ \frac{4}{\sqrt x} - \frac{9}{\sqrt y} = -1$

Given Equations,

$\\ \frac{2}{\sqrt x} + \frac{3}{\sqrt y} = 2\\ \frac{4}{\sqrt x} - \frac{9}{\sqrt y} = -1$

Let,

$\frac{1}{\sqrt{x}}=p\:and\:\frac{1}{\sqrt{y}}=q$

Now, our equation becomes

$2p+3q=2........(1)$

And

$4p-9q=-1..........(2)$

By Cross Multiplication method,

$\frac{p}{b_1c_2-b_2c_1}=\frac{q}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{p}{(3)(1)-(-9)(-2)}=\frac{q}{(-2)(4)-(1)(2)}=\frac{1}{(2)(-9)-(4)(3)}$

$\frac{p}{3-18}=\frac{q}{-8-2}=\frac{1}{-18-12}$

$\frac{p}{-15}=\frac{q}{-10}=\frac{1}{-30}$

$p=\frac{1}{2},\:and\:q=\frac{1}{3}$

So,

$p=\frac{1}{2}=\frac{1}{\sqrt{x}}\Rightarrow x=4$

$q=\frac{1}{3}=\frac{1}{\sqrt{y}}\Rightarrow y=9$ .

And hence

$x=4\:and\:y=9.$

(iii) $\\\frac{4}{x} + 3y = 14\\ \frac{3}{x} - 4y = 23$

Given Equations,

$\\\frac{4}{x} + 3y = 14\\ \frac{3}{x} - 4y = 23$

Let,

$\frac{1}{x}=p\:and\:y=q$

Now, our equation becomes

$\Rightarrow 4p+3q=14........(1)$

And

$\Rightarrow 3p-4q=23..........(2)$

By Cross Multiplication method,

$\frac{p}{b_1c_2-b_2c_1}=\frac{q}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{p}{(3)(-23)-(-4)(-14)}=\frac{q}{(-14)(3)-(-23)(4)}=\frac{1}{(4)(-4)-(3)(3)}$

$\frac{p}{-69-56}=\frac{q}{-42+92}=\frac{1}{-16-9}$

$\frac{p}{-125}=\frac{q}{50}=-\frac{1}{25}$

$p=5,\:and\:q=-2$

And Hence,

$x=\frac{1}{5}\:and\:y=-2.$

(iv) $\\\frac{5}{x - 1} + \frac{1}{y -2} = 2\\ \frac{6}{x-1} - \frac{3}{y -2} =1$

Given Equations,

$\\\frac{5}{x - 1} + \frac{1}{y -2} = 2\\ \frac{6}{x-1} - \frac{3}{y -2} =1$

Let,

$\frac{1}{x-1}=p\:and\:\frac{1}{y-2}=q$

Now, our equation becomes

$5p+q=2........(1)$

And

$6p-3q=1..........(2)$

Multiplying (1) by 3 we get

$15p+3q=6..........(3)$

Now, adding (2) and (3) we get

$21p=7$

$\Rightarrow p=\frac{1}{3}$

Putting this in (2)

$6\left ( \frac{1}{3} \right )-3q=1$

$\Rightarrow 3q=1$

$\Rightarrow q=\frac{1}{3}$

Now,

$p=\frac{1}{3}=\frac{1}{x-1}\Rightarrow x-1=3\Rightarrow x=4$

$q=\frac{1}{3}=\frac{1}{y-2}\Rightarrow y-2=3\Rightarrow x=5$

Hence,

$x=4,\:and\:y=5.$

$\\\frac{7x - 2y}{xy} = 5\\ \frac{8x + 7y}{xy} = 15$

Given Equations,

$\\\frac{7x - 2y}{xy} = 5\\\\\Rightarrow\frac{7}{y} -\frac{2}{x}=5\\ \frac{8x + 7y}{xy} = 15\\\Rightarrow \frac{8}{y}+\frac{7}{x}=15$

Let,

$\frac{1}{x}=p\:and\:\frac{1}{y}=q$

Now, our equation becomes

$7q-2p=5........(1)$

And

$8q+7p=15..........(2)$

By Cross Multiplication method,

$\frac{q}{b_1c_2-b_2c_1}=\frac{p}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{q}{(-2)(-15)-(7)(-5)}=\frac{p}{(-5)(8)-(-15)(7)}=\frac{1}{(7)(7)-(8)(-2)}$

$\frac{q}{30+35}=\frac{p}{-40+105}=\frac{1}{49 +16}$

$\frac{q}{65}=\frac{p}{65}=\frac{1}{65}$

$p=1,\:and\:q=1$

And Hence,

$x=1\:and\:y=1.$

(vi) $\\6x + 3y = 6xy\\ 2x + 4y = 5 xy$

Given Equations,

$\\6x + 3y = 6xy\\\Rightarrow \frac{6x}{xy}+\frac{3y}{xy}=6\\\\\Rightarrow \frac{6}{y}+\frac{3}{x}=6\\and\\\ 2x + 4y = 5 xy\\\Rightarrow \frac{2x}{xy}+\frac{4y}{xy}=5\\\Rightarrow \frac{2}{y}+\frac{4}{x}=5$

Let,

$\frac{1}{x}=p\:and\:\frac{1}{y}=q$

Now, our equation becomes

$6q+3p=6........(1)$

And

$2q+4p=5..........(2)$

By Cross Multiplication method,

$\frac{q}{b_1c_2-b_2c_1}=\frac{p}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{q}{(3)(-5)-(-6)(4)}=\frac{p}{(6)(2)-(6)(-5)}=\frac{1}{(6)(4)-(3)(2)}$

$\frac{q}{-15+24}=\frac{p}{-12+30}=\frac{1}{24 -6}$

$\frac{q}{9}=\frac{p}{18}=\frac{1}{18}$

$q=\frac{1}{2}\:and\:p=1$

And Hence,

$x=1\:and\:y=2.$

Given Equations,

$\\\frac{10}{x + y} + \frac{2}{x - y}= 4\\ \frac{15}{x+y} - \frac{5}{x - y} = -2$

Let,

$\frac{1}{x+y}=p\:and\:\frac{1}{x-y}=q$

Now, our equation becomes

$10p+2q=4........(1)$

And

$15p-5q=-2..........(2)$

By Cross Multiplication method,

$\frac{p}{b_1c_2-b_2c_1}=\frac{q}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{p}{(2)(2)-(-5)(-4)}=\frac{q}{(-4)(15)-(2)(10)}=\frac{1}{(10)(-5)-(15)(2)}$

$\frac{p}{4-20}=\frac{q}{-60-20}=\frac{1}{-50-30}$

$\frac{p}{-16}=\frac{q}{-80}=\frac{1}{-80}$

$p=\frac{1}{5},\:and\:q=1$

Now,

$p=\frac{1}{5}=\frac{1}{x+y}$

$\Rightarrow x+y=5........(3)$

And,

$q=1=\frac{1}{x-y}$

$\Rightarrow x-y=1...........(4)$

Adding (3) and (4) we get,

$\Rightarrow 2x=6$

$\Rightarrow x=3$

Putting this value in (3) we get,

$3+y=5$

$\Rightarrow y=2$

And Hence,

$x=3\:and\:y=2.$

(viii) $\\\frac{1}{3x + y} + \frac{1}{3x -y} = \frac{3}{4}\\ \frac{1}{2(3x+y)} - \frac{1}{2(3x -y)} = \frac{-1}{8}$

Given Equations,

$\\\frac{1}{3x + y} + \frac{1}{3x -y} = \frac{3}{4}\\ \frac{1}{2(3x+y)} - \frac{1}{2(3x -y)} = \frac{-1}{8}$

Let,

$\frac{1}{3x+y}=p\:and\:\frac{1}{3x-y}=q$

Now, our equation becomes

$p+q=\frac{3}{4}.........(1)$

And

$\\\frac{p}{2}-\frac{q}{2}=\frac{-1}{8}\\\\p-q=\frac{-1}{4}..........(2)$

Now, Adding (1) and (2), we get

$2p=\frac{3}{4}-\frac{1}{4}$

$\Rightarrow 2p=\frac{2}{4}$

$\Rightarrow p=\frac{1}{4}$

Putting this value in (1)

$\frac{1}{4}+q=\frac{3}{4}$

$\Rightarrow q=\frac{3}{4}-\frac{1}{4}$

$\Rightarrow q=\frac{2}{4}$

$\Rightarrow q=\frac{1}{2}$

Now,

$p=\frac{1}{4}=\frac{1}{3x+y}$

$\Rightarrow 3x+y=4...........(3)$

And

$q=\frac{1}{2}=\frac{1}{3x-y}$

$\Rightarrow 3x-y=2............(4)$

Now, Adding (3) and (4), we get

$6x=4+2$

$\Rightarrow 6x=6$

$\Rightarrow x=1$

Putting this value in (3),

$3(1)+y=4$

$\Rightarrow y=4-3$

$\Rightarrow y=1$

Hence,

$x=1,\:and\:y=1$

Let the speed of Ritu in still water be x and speed of current be y,

Let's solve this problem by using relative motion concept,

the relative speed when they are going in the same direction (downstream)= x +y

the relative speed when they are going in the opposite direction (upstream)= x - y

Now, As we know,

Relative distance = Relative speed * time .

So, According to the question,

$x+y=\frac{20}{2}$

$\Rightarrow x+y=10.........(1)$

And,

$x-y=\frac{4}{2}$

$\Rightarrow x-y=2...........(2)$

Now, Adding (1) and (2), we get

$2x=10+2$

$\Rightarrow 2x=12$

$\Rightarrow x=6$

Putting this in (2)

$6-y=2$

$\Rightarrow y=6-2$

$\Rightarrow y=4$

Hence,

$x=6\:and\:y=4.$

Hence Speed of Ritu in still water is 6 km/hour and speed of the current is 4 km/hour

Let the number of days taken by woman and man be x and y respectively,

The proportion of Work done by a woman in a single day

$=\frac{1}{x }$

The proportion of Work done by a man in a single day

$=\frac{1}{y }$

Now, According to the question,

$4\left ( \frac{2}{x}+\frac{5}{y} \right )=1$

$\Rightarrow \left ( \frac{2}{x}+\frac{5}{y} \right )=\frac{1}{4}$

Also,

$3\left ( \frac{3}{x}+\frac{6}{y} \right )=1$

$\Rightarrow \left ( \frac{3}{x}+\frac{6}{y} \right )=\frac{1}{3}$

Let,

$\frac{1}{x}=p\:and\:\frac{1}{y}=q$

Now, our equation becomes

$2p+5q=\frac{1}{4}$

$8p+20q=1........(1)$

And

$3p+6p=\frac{1}{3}$

$\Rightarrow 9p+18p=1.............(2)$

By Cross Multiplication method,

$\frac{p}{b_1c_2-b_2c_1}=\frac{q}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{p}{(20)(-1)-(18)(-1)}=\frac{q}{(-1)(9)-(8)(-1)}=\frac{1}{(8)(18)-(20)(9)}$

$\frac{p}{-20+18}=\frac{q}{-9+8}=\frac{1}{146 -60}$

$\frac{p}{-2}=\frac{q}{-1}=\frac{1}{-36}$

$p=\frac{1}{18},\:and\:q=\frac{1}{36}$

So,

$x=18\:and\:y=36.$

Let the speed of the train and bus be u and v respectively

Now According to the question,

$\frac{60}{u}+\frac{240}{v}=4$

And

$\frac{100}{u}+\frac{200}{v}=4+\frac{1}{6}$

$\Rightarrow \frac{100}{u}+\frac{200}{v}=\frac{25}{6}$

Let,

$\frac{1}{u}=p\:and\:\frac{1}{v}=q$

Now, our equation becomes

$60p+140q=4$

$\Rightarrow 15p+60q=1.........(1)$

And

$100p+200q=\frac{25}{6}$

$\Rightarrow 4p+8q=\frac{1}{6}$

$\Rightarrow 24p+48q=1..........(2)$

By Cross Multiplication method,

$\frac{p}{b_1c_2-b_2c_1}=\frac{q}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{q}{(60)(-1)-(48)(-1)}=\frac{p}{(-1)(24)-(-1)(15)}=\frac{1}{(15)(48)-(60)(24)}$

$\frac{p}{-60+48}=\frac{q}{-24+15}=\frac{1}{720-1440}$

$\frac{p}{-12}=\frac{q}{-9}=\frac{1}{-720}$

$p=\frac{12}{720}=\frac{1}{60},\:and\:q=\frac{9}{720}=\frac{1}{80}$

And Hence,

$x=60\:and\:y=80$

Hence the speed of the train and bus are 60 km/hour and 80 km/hour respectively.

Class 10 math chapter 3 solutions Pair of Linear Equations in Two Variables Excercise: 3.7

Let the age of Ani be $a$, age of Biju be $b$ ,

Case 1: when Ani is older than Biju

age of Ani's father Dharam:

$d=2a$ and

age of his sister Cathy :

$c=\frac{b}{2}$

Now According to the question,

$a-b=3...........(1)$

Also,

$\\d-c=30 \\\Rightarrow 2a-\frac{b}{2}=30$

$\Rightarrow 4a-b=60..............(2)$

Now subtracting (1) from (2), we get,

$3a=60-3$

$\Rightarrow a=19$

putting this in (1)

$19-b=3$

$\Rightarrow b=16$

Hence the age of Ani and Biju is 19 years and 16 years respectively.

Case 2:

$b-a=3..........(3)$

And

$\\d-c=30 \\\Rightarrow 2a-\frac{b}{2}=30$

$\Rightarrow 4a-b=60..............(4)$

Now Adding (3) and (4), we get,

$3a=63$

$\Rightarrow a=21$

putting it in (3)

$b-21=3$

$\Rightarrow b=24.$

Hence the age of Ani and Biju is 21 years and 24 years respectively.

[Hint : $x + 100 = 2(y - 100), y + 10 = 6(x - 10)$ ]

Let the amount of money the first person and the second person having is x and y respectively

Noe, According to the question.

$x + 100 = 2(y - 100)$

$\Rightarrow x - 2y =-300...........(1)$

Also

$y + 10 = 6(x - 10)$

$\Rightarrow y - 6x =-70..........(2)$

Multiplying (2) by 2 we get,

$2y - 12x =-140..........(3)$

Now adding (1) and (3), we get

$-11x=-140-300$

$\Rightarrow 11x=440$

$\Rightarrow x=40$

Putting this value in (1)

$40-2y=-300$

$\Rightarrow 2y=340$

$\Rightarrow y=170$

Thus two friends had 40 Rs and 170 Rs respectively.

Let the speed of the train be v km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km.

Now As we Know,

$speed=\frac{distance }{time}$

$\Rightarrow v=\frac{d}{t}$

$\Rightarrow d=vt..........(1)$
Now, According to the question,
$(v+10)=\frac{d}{t-2}$

$\Rightarrow (v+10){t-2}=d$

$\Rightarrow vt +10t-2v-20=d$

Now, Using equation (1), we have

$\Rightarrow -2v+10t=20............(2)$
Also,

$(v-10)=\frac{d}{t+3}$

$\Rightarrow (v-10)({t+3})=d$

$\Rightarrow vt+3v-10t-30=d$
$\Rightarrow3v-10t=30..........(3)$

Adding equations (2) and (3), we obtain:
$v=50.$
Substituting the value of x in equation (2), we obtain:
$(-2)(50)+10t=20$

$\Rightarrow -100+10t=20$

$\Rightarrow 10t=120$

$\Rightarrow t=12$
Putting this value in (1) we get,

$d=vt=(50)(12)=600$

Hence the distance covered by train is 600km.

Let the number of rows is x and the number of students in a row is y.
Total number of students in the class = Number of rows * Number of students in a row
$=xy$

Now, According to the question,

$\\xy = (x - 1) (y + 3) \\\Rightarrow xy= xy - y + 3x - 3 \\\Rightarrow 3x - y - 3 = 0 \\ \Rightarrow 3x - y = 3 ...... ... (1)$

Also,

$\\xy=(x+2)(y-3)\\\Rightarrow xy = xy + 2y - 3x - 6 \\ \Rightarrow 3x - 2y = -6 ... (2)$
Subtracting equation (2) from (1), we get:
$y=9$
Substituting the value of y in equation (1), we obtain:
$\\3x - 9 = 3 \\\Rightarrow 3x = 9 + 3 = 12 \\\Rightarrow x = 4$
Hence,
The number of rows is 4 and the Number of students in a row is 9.

Total number of students in a class

: $xy=(4)(9)=36$

Hence there are 36 students in the class.

Given,

Also, As we know that the sum of angles of a triangle is 180, so

$\angle A +\angle B+ \angle C=180$

$\angle A +\angle B+ 3\angle B=180^0$

$\angle A + 4\angle B=180^0..........(2)$

Now From (1) we have

$\angle B = 2 \angle A.......(3)$

Putting this value in (2) we have

$\angle A + 4(2\angle A)=180^0.$

$\Rightarrow 9\angle A=180^0.$

$\Rightarrow \angle A=20^0.$

Putting this in (3)

$\angle B = 2 (20)=40^0$

And

$\angle C = 3 \angle B =3(40)=120^0$

Hence three angles of triangles $20^0,40^0\:and\:120^0.$

Given two equations,

$5x - y =5.........(1)$

And

$3x - y = 3........(2)$

Points(x,y) which satisfies equation (1) are:

 X 0 1 5 Y -5 0 20

Points(x,y) which satisfies equation (1) are:

 X 0 1 2 Y -3 0 3

GRAPH:

As we can see from the graph, the three points of the triangle are, (0,-3),(0,-5) and (1,0).

Given Equations,

$\\px + qy = p - q.........(1)\\ qx - py = p + q.........(2)$

Now By Cross multiplication method,

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(q)(-p-q)-(q-p)(-p)}=\frac{y}{(q-p)(q)-(p)(-p-q)}=\frac{1}{(p)(-p)-(q)(q)}$

$\frac{x}{-p^2-q^2}=\frac{y}{p^2+q^2}=\frac{1}{-p^2-q^2}$

$x=1,\:and\:y=-1$

Given two equations,

$\\ax + by = c.........(1)\\ bx +ay = 1 + c.........(2)$

Now By Cross multiplication method,

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(b)(-1-c)-(-c)(a)}=\frac{y}{(-c)(b)-(a)(-1-c)}=\frac{1}{(a)(a)-(b)(b)}$

$\frac{x}{-b-bc+ac}=\frac{y}{-cb+a+ac}=\frac{1}{a^2-b^2}$

$x=\frac{-b-bc+ac}{a^2-b^2},\:and\:y=\frac{a-bc+ac}{a^2-b^2}$

Given equation,

$\\\frac{x}{a} - \frac{y}{b} = 0\\\Rightarrow bx-ay=0...........(1)\\ ax + by = a^2 +b^2...............(2)$

Now By Cross multiplication method,

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(-a)(-a^2-b^2)-(b)(0)}=\frac{y}{(0)(a)-(b)(-a^2-b^2)}=\frac{1}{(b)(b)-(-a)(a)}$

$\frac{x}{a(a^2+b^2)}=\frac{y}{b(a^2+b^2)}=\frac{1}{a^2+b^2}$

$x=a,\:and\:y=b$

Given,

$\\(a-b)x + (a+b)y = a^2 -2ab - b^2..........(1)$

And

$\\ (a+b)(x+y) = a^2 +b^2\\\Rightarrow (a+b)x+(a+b)y=a^2+b^2...........(2)$

Now, Subtracting (1) from (2), we get

$(a+b)x-(a-b)x=a^2+b^2-a^2+2ab+b^2$

$\Rightarrow(a+b-a+b)x=2b^2+2ab$

$\Rightarrow 2bx=2b(b+2a)$

$\Rightarrow x=(a+b)$

Substituting this in (1), we get,

$(a-b)(a+b)+(a+b)y=a^2-2ab-b^2$

$\Rightarrow a^2-b^2+(a+b)y=a^2-2ab-b^2$

$\Rightarrow (a+b)y=-2ab$

$\Rightarrow y=\frac{-2ab}{a+b}$ .

Hence,

$x=(a+b),\:and\:y=\frac{-2ab}{a+b}$

Given Equations,

$\\152x - 378y = -74............(1)\\ -378x + 152y = -604............(2)$

As we can see by adding and subtracting both equations we can make our equations simple to solve.

So,

Adding (1) and )2) we get,

$-226x-226y=-678$

$\Rightarrow x+y=3...........(3)$

Subtracting (2) from (1) we get,

$530x-530y=530$

$\Rightarrow x-y=1...........(4)$

Now, Adding (3) and (4) we get,

$2x=4$

$\Rightarrow x=2$

Putting this value in (3)

$2+y=3$

$\Rightarrow y=1$

Hence,

$x=2\:and\:y=1$

As we know that in a quadrilateral the sum of opposite angles is 180 degrees.

So, From Here,

$4y+20-4x=180$

$\Rightarrow 4y-4x=160$

$\Rightarrow y-x=40............(1)$

Also,

$3y-5-7x+5=180$

$\Rightarrow 3y-7x=180........(2)$

Multiplying (1) by 3 we get,

$\Rightarrow 3y-3x=120........(3)$

Now,

Subtracting, (2) from (3) we get,

$4x=-60$

$\Rightarrow x=-15$

Substituting this value in (1) we get,

$y-(-15)=40$

$\Rightarrow y=40-15$

$\Rightarrow y=25$

Hence four angles of a quadrilateral are :

$\angle A =4y+20=4(25)+20=100+20=120^0$

$\angle B =3y-5=3(25)-5=75-5=70^0$

$\angle C =-4x=-4(-15)=60^0$

$\angle D =-7x+5=-7(-15)+5=105+5=110^0$

### Class 10 Maths Chapter 3 Topics

• Solving a linear equation with two variables.

• Representation of linear equation in a graph.

• Solutions of linear equations using the graph.

• Algebraic interpretation of linear equations.

• Formation of linear equations using statements.

Also get the solutions exercise wise-

## Key Features of NCERT Class 10 Maths Solutions Chapter 3

• The questions and their answers given in Chapter 3 Class 10 Maths NCERT solutions are very interesting and important for board and competitive exams.
• NCERT solutions for Maths chapter 3 Linear Equations in Two Variables Class 10 will help to boost preparation for all of the examinations.
• Many real-life situations can be formulated using Mathematical equations given in this chapter 3 NCERT Class 10 Maths solutions.

• For example, consider the statement "cost of 1 Kg Apple and 2Kg orange is 120 and the cost of 3 Kg Apple and 1 Kg orange is 210". The statement can be formulated using the Mathematical equation, for this consider the cost of Apple as x and that of orange as y. Then we can write two equations as x+2y=120 and 3x+y=210.

## NCERT Books and NCERT Syllabus

### NCERT Solutions for Class 10 Maths - Chapter Wise

 Solution Link NCERT solutions for class 10 maths chapter 1 NCERT solutions for class 10 maths chapter 2 NCERT solutions for class 10 maths chapter 3 NCERT solutions for class 10 maths chapter 4 NCERT solutions for class 10 chapter 5 NCERT solutions for class 10 maths chapter 6 NCERT solutions for class 10 maths chapter 7 NCERT solutions for class 10 maths chapter 8 NCERT solutions for class 10 maths chapter 9 NCERT solutions class 10 maths chapter 10 NCERT solutions for class 10 maths chapter 11 NCERT solutions for class 10 chapter maths chapter 12 NCERT solutions class 10 maths chapter 13 NCERT solutions for class 10 maths chapter 14 NCERT solutions for class 10 maths chapter 15

## NCERT Exemplar solutions - Subject wise

### How to use NCERT Solutions for Class 10 Maths Chapter 3?

Follow the given tips to make the most of NCERT solutions Class 10 Maths Chapter 3 PDF download:

• NCERT solutions Class 10 Maths Chapter 3 are the most important tool when you are appearing for board examinations. 90% paper of CBSE board examinations, directly come from the NCERT.

• Now you have done the NCERT solutions for Class 10 Maths chapter 3 and learned the approach to solving questions in the step by step method.

• After covering Linear Equations in Two Variables Class 10 solutions you should target the past year papers of CBSE board examinations. The previous year papers will cover the rest 10% part of the Class 10 board exams.

1. What is a pair of linear equations in two variables Class 10 Maths?

In Class 10 Maths, a pair of linear equations in two variables is a set of two linear equations that contain two variables, typically represented by x and y. Linear equations are equations of the form ax + by = c, where a, b, and c are constants.

2. How can I obtain the solutions for Chapter 3 maths Class 10?

Students can find NCERT solutions for maths to class 10 linear equations in two variables above in this article. Our experts in simple and easy format develop these maths class 10 chapter 3 pair of linear equations in two variables class 10 solutions. These will be very helpful for students.

3. Is it necessary to work through all of the questions provided in the NCERT solutions for class 10 Chapter 3 maths?

It is very important to go through Class 10 maths linear equations in two variables questions as they are designed to help you understand the material and practice your problem-solving skills. Students can find ch 3 maths class 10 ncert solutions at careers360 official website. However, it is not necessarily required to work through class 10 chapter 3 maths solutions by every question in order to understand the concepts covered in the chapter.

4. What key concepts are covered in the NCERT solutions for Chapter 3 Class 10 Maths?

NCERT class chapter 3 pair of linear equations in two variables class 10 solutions

Contains the concepts of pairs of linear equations and solving them using methods like graphing method, substitution method, elimination method, Consistency and inconsistency of a pair of linear equations, and Dependent and independent linear equations. As  Class 10 linear equations in two variables questions are used to solve real-world problems.

5. What are the most crucial questions for ch3 maths class 10 linear equations in two variables?

In chapter 3 Pair of linear equations class 10 Exercise 3.1, Question 1 is important. All questions in Exercise 3.2 are important. In Exercise 3.3, Question 1 (parts iv, v, vi), Question 2, and Question 3 are important. Exercise 3.4's Question 2 is important, and Exercise 3.5's Questions 2 and 4 are important. All questions in Exercise 3.6 are important. All questions in the optional Exercise 3.7 are important. Students can download a pair of linear equations in two variables class 10 pdf.

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### Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.

According to CBSE norms, a student must be 14 years old by the end of the year in which the exam will be held in order to sit for the 10th board exam. If your age is greater than 14, however, there are no limits. Therefore, based on your DOB, you will be 12 years and 6 months old by the end of December 2022. After the actual 16 June 2024, you'll be qualified to take your 10th board.

Hello aspirant,

Central Board Of Secondary Education(CBSE) is likely to declare class 10 and 12 terms 2 board result 2022 by July 15. The evaluation process is underway. Students are demanding good results and don't want to lack behind. They are requesting the board to use their best scores in Term 1 and Term 2 exams to prepare for the results.

CBSE concluded board exams 2022 for 10, and 12 on June 15 and May 24. Exams for both classes began on April 26. A total of 35 lakh students including 21 lakh class 10 students and 14 lakh class 12 students appeared in exams and are awaiting their results.

You can look for your results on websites- cbse.gov (//cbse.gov) .in, cbseresults.nic.in

Thank you

Sir, did you get any problem in result
I have also done this same mistake in board 2023 exam

Hello SIR CBSE board 9th class admission 2022 Kara Raha hu entrance ki problem hai please show the Entrance for Baal vidya mandir school sambhal

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9

Database professionals use software to store and organise data such as financial information, and customer shipping records. Individuals who opt for a career as data administrators ensure that data is available for users and secured from unauthorised sales. DB administrators may work in various types of industries. It may involve computer systems design, service firms, insurance companies, banks and hospitals.

4 Jobs Available
##### Bio Medical Engineer

The field of biomedical engineering opens up a universe of expert chances. An Individual in the biomedical engineering career path work in the field of engineering as well as medicine, in order to find out solutions to common problems of the two fields. The biomedical engineering job opportunities are to collaborate with doctors and researchers to develop medical systems, equipment, or devices that can solve clinical problems. Here we will be discussing jobs after biomedical engineering, how to get a job in biomedical engineering, biomedical engineering scope, and salary.

4 Jobs Available
##### Ethical Hacker

A career as ethical hacker involves various challenges and provides lucrative opportunities in the digital era where every giant business and startup owns its cyberspace on the world wide web. Individuals in the ethical hacker career path try to find the vulnerabilities in the cyber system to get its authority. If he or she succeeds in it then he or she gets its illegal authority. Individuals in the ethical hacker career path then steal information or delete the file that could affect the business, functioning, or services of the organization.

3 Jobs Available
##### Database Architect

If you are intrigued by the programming world and are interested in developing communications networks then a career as database architect may be a good option for you. Data architect roles and responsibilities include building design models for data communication networks. Wide Area Networks (WANs), local area networks (LANs), and intranets are included in the database networks. It is expected that database architects will have in-depth knowledge of a company's business to develop a network to fulfil the requirements of the organisation. Stay tuned as we look at the larger picture and give you more information on what is db architecture, why you should pursue database architecture, what to expect from such a degree and what your job opportunities will be after graduation. Here, we will be discussing how to become a data architect. Students can visit NIT Trichy, IIT Kharagpur, JMI New Delhi

3 Jobs Available
##### Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

3 Jobs Available
##### Geothermal Engineer

Individuals who opt for a career as geothermal engineers are the professionals involved in the processing of geothermal energy. The responsibilities of geothermal engineers may vary depending on the workplace location. Those who work in fields design facilities to process and distribute geothermal energy. They oversee the functioning of machinery used in the field.

3 Jobs Available
##### Geotechnical engineer

The role of geotechnical engineer starts with reviewing the projects needed to define the required material properties. The work responsibilities are followed by a site investigation of rock, soil, fault distribution and bedrock properties on and below an area of interest. The investigation is aimed to improve the ground engineering design and determine their engineering properties that include how they will interact with, on or in a proposed construction.

The role of geotechnical engineer in mining includes designing and determining the type of foundations, earthworks, and or pavement subgrades required for the intended man-made structures to be made. Geotechnical engineering jobs are involved in earthen and concrete dam construction projects, working under a range of normal and extreme loading conditions.

3 Jobs Available
##### Cartographer

How fascinating it is to represent the whole world on just a piece of paper or a sphere. With the help of maps, we are able to represent the real world on a much smaller scale. Individuals who opt for a career as a cartographer are those who make maps. But, cartography is not just limited to maps, it is about a mixture of art, science, and technology. As a cartographer, not only you will create maps but use various geodetic surveys and remote sensing systems to measure, analyse, and create different maps for political, cultural or educational purposes.

3 Jobs Available
##### Bank Probationary Officer (PO)

A career as Bank Probationary Officer (PO) is seen as a promising career opportunity and a white-collar career. Each year aspirants take the Bank PO exam. This career provides plenty of career development and opportunities for a successful banking future. If you have more questions about a career as  Bank Probationary Officer (PO), what is probationary officer or how to become a Bank Probationary Officer (PO) then you can read the article and clear all your doubts.

3 Jobs Available
##### Operations Manager

Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

3 Jobs Available
##### Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

3 Jobs Available
##### Finance Executive

A career as a Finance Executive requires one to be responsible for monitoring an organisation's income, investments and expenses to create and evaluate financial reports. His or her role involves performing audits, invoices, and budget preparations. He or she manages accounting activities, bank reconciliations, and payable and receivable accounts.

3 Jobs Available
##### Investment Banker

An Investment Banking career involves the invention and generation of capital for other organizations, governments, and other entities. Individuals who opt for a career as Investment Bankers are the head of a team dedicated to raising capital by issuing bonds. Investment bankers are termed as the experts who have their fingers on the pulse of the current financial and investing climate. Students can pursue various Investment Banker courses, such as Banking and Insurance, and Economics to opt for an Investment Banking career path.

3 Jobs Available
##### Bank Branch Manager

Bank Branch Managers work in a specific section of banking related to the invention and generation of capital for other organisations, governments, and other entities. Bank Branch Managers work for the organisations and underwrite new debts and equity securities for all type of companies, aid in the sale of securities, as well as help to facilitate mergers and acquisitions, reorganisations, and broker trades for both institutions and private investors.

3 Jobs Available
##### Treasurer

Treasury analyst career path is often regarded as certified treasury specialist in some business situations, is a finance expert who specifically manages a company or organisation's long-term and short-term financial targets. Treasurer synonym could be a financial officer, which is one of the reputed positions in the corporate world. In a large company, the corporate treasury jobs hold power over the financial decision-making of the total investment and development strategy of the organisation.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available
##### Conservation Architect

A Conservation Architect is a professional responsible for conserving and restoring buildings or monuments having a historic value. He or she applies techniques to document and stabilise the object’s state without any further damage. A Conservation Architect restores the monuments and heritage buildings to bring them back to their original state.

2 Jobs Available
##### Safety Manager

A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

2 Jobs Available

A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

2 Jobs Available
##### Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software.

2 Jobs Available
##### Architect

Individuals in the architecture career are the building designers who plan the whole construction keeping the safety and requirements of the people. Individuals in architect career in India provides professional services for new constructions, alterations, renovations and several other activities. Individuals in architectural careers in India visit site locations to visualize their projects and prepare scaled drawings to submit to a client or employer as a design. Individuals in architecture careers also estimate build costs, materials needed, and the projected time frame to complete a build.

2 Jobs Available
##### Landscape Architect

Having a landscape architecture career, you are involved in site analysis, site inventory, land planning, planting design, grading, stormwater management, suitable design, and construction specification. Frederick Law Olmsted, the designer of Central Park in New York introduced the title “landscape architect”. The Australian Institute of Landscape Architects (AILA) proclaims that "Landscape Architects research, plan, design and advise on the stewardship, conservation and sustainability of development of the environment and spaces, both within and beyond the built environment". Therefore, individuals who opt for a career as a landscape architect are those who are educated and experienced in landscape architecture. Students need to pursue various landscape architecture degrees, such as M.Des, M.Plan to become landscape architects. If you have more questions regarding a career as a landscape architect or how to become a landscape architect then you can read the article to get your doubts cleared.

2 Jobs Available
##### Plumber

An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.

2 Jobs Available
##### Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
##### Veterinary Doctor

A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.

5 Jobs Available
##### Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
##### Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth.

4 Jobs Available
##### Surgical Technologist

When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications.

3 Jobs Available
##### Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
##### Chemical Pathologist

Are you searching for a chemical pathologist job description? A chemical pathologist is a skilled professional in healthcare who utilises biochemical laboratory tests to diagnose disease by analysing the levels of various components or constituents in the patient’s body fluid.

2 Jobs Available
##### Biochemical Engineer

A Biochemical Engineer is a professional involved in the study of proteins, viruses, cells and other biological substances. He or she utilises his or her scientific knowledge to develop products, medicines or ways to improve quality and refine processes. A Biochemical Engineer studies chemical functions occurring in a living organism’s body. He or she utilises the observed knowledge to alter the composition of products and develop new processes. A Biochemical Engineer may develop biofuels or environmentally friendly methods to dispose of waste generated by industries.

2 Jobs Available
##### Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.

4 Jobs Available
##### Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
##### Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
##### Talent Agent

The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

3 Jobs Available

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
##### Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
##### Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
##### Talent Director

Individuals who opt for a career as a talent director are professionals who work in the entertainment industry. He or she is responsible for finding out the right talent through auditions for films, theatre productions, or shows. A talented director possesses strong knowledge of computer software used in filmmaking, CGI and animation. A talent acquisition director keeps himself or herself updated on various technical aspects such as lighting, camera angles and shots.

2 Jobs Available
##### Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

5 Jobs Available
##### Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
##### Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
##### Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

3 Jobs Available
##### Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
##### Content Writer

Content writing is meant to speak directly with a particular audience, such as customers, potential customers, investors, employees, or other stakeholders. The main aim of professional content writers is to speak to their targeted audience and if it is not then it is not doing its job. There are numerous kinds of the content present on the website and each is different based on the service or the product it is used for.

2 Jobs Available
##### Reporter

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.

2 Jobs Available
##### Linguist

Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning).

Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian

2 Jobs Available
##### Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
##### Production Engineer

A career as Production Engineer is crucial in the manufacturing industry. He or she ensures the functionality of production equipment and machinery to improve productivity and minimize production costs in order to drive revenues and increase profitability.

2 Jobs Available
##### Automation Test Engineer

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process.

2 Jobs Available
##### Product Designer

Individuals who opt for a career as product designers are responsible for designing the components and overall product concerning its shape, size, and material used in manufacturing. They are responsible for the aesthetic appearance of the product. A product designer uses his or her creative skills to give a product its final outlook and ensures the functionality of the design.

Students can opt for various product design degrees such as B.Des and M.Des to become product designers. Industrial product designer prepares 3D models of designs for approval and discusses them with clients and other colleagues. Individuals who opt for a career as a product designer estimate the total cost involved in designing.

2 Jobs Available
##### R&D Personnel

A career as R&D Personnel requires researching, planning, and implementing new programs and protocols into their organization and overseeing new products’ development. He or she uses his or her creative abilities to improve the existing products as per the requirements of the target market.

2 Jobs Available
##### Commercial Manager

A Commercial Manager negotiates, advises and secures information about pricing for commercial contracts. He or she is responsible for developing financial plans in order to maximise the business's profitability.

2 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available
##### Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
##### Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
##### Computer System Analyst

Individuals in the computer systems analyst career path study the hardware and applications that are part of an organization's computer systems, as well as how they are used. They collaborate closely with managers and end-users to identify system specifications and business priorities, as well as to assess the efficiency of computer systems and create techniques to boost IT efficiency. Individuals who opt for a career as a computer system analyst support the implementation, modification, and debugging of new systems after they have been installed.

2 Jobs Available
##### Test Manager

A Test Manager is a professional responsible for planning, coordinating and controlling test activities. He or she develops test processes and strategies to analyse and determine test methods and tools for test activities. The test manager jobs involve documenting tests that have been carried out, analysing and evaluating software quality to determine further recommended procedures.

2 Jobs Available
##### Azure Developer

A career as Azure Developer comes with the responsibility of designing and developing cloud-based applications and maintaining software components. He or she possesses an in-depth knowledge of cloud computing and Azure app service.

2 Jobs Available
##### Deep Learning Engineer

A Deep Learning Engineer is an IT professional who is responsible for developing and managing data pipelines. He or she is knowledgeable about analyzing and storing data collected from various sources.  A Career as a Deep Learning Engineer needs to help the  data scientists and analysts to create effective data sets.

2 Jobs Available