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Mathematics Class 10 Assignments Pdf Download

We have provided below the biggest collection of free CBSE NCERT KVS Assignments for Class 10 Mathematics . Students and teachers can download and save all free Mathematics assignments in Pdf for grade 10th. Our expert faculty have covered Class 10 important questions and answers for Mathematics as per the latest syllabus for the current academic year. All test papers and question banks for Class 10 Mathematics and CBSE Assignments for Mathematics Class 10 will be really helpful for standard 10th students to prepare for the class tests and school examinations. Class 10th students can easily free download in Pdf all printable practice worksheets given below.

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NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables  will help the students in understanding how the problems under this concept are solved. Maths is one subject that requires a lot of practice. The students appearing for the 10 th grade board examinations can turn to the NCERT Solutions Class 10 for reference. These solutions of the Chapter Pair of Linear Equations in Two Variables give step-wise answers to all the Maths problems in the NCERT textbook . An equation that can be of the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero, is called a linear equation in two variables x and y.

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Download most important questions for class 10 maths chapter – 3 pair of linear equations in two variables.

The NCERT Solutions for Class 10 Maths Chapter 3 also lets the students understand the fact that a solution of such an equation is a pair of values, one for x and the other for y, which makes the two sides of the equation equal. Students also learn that every solution of the equation is a point on the line representing it.

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• Chapter 1 Real Numbers
• Chapter 2 Polynomials
• Chapter 3 Pair of Linear Equations in Two Variables
• Chapter 4 Quadratic Equations
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• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Areas Related to Circles
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics
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• Exercise 3.1 Chapter 3 Pair of Linear Equations in Two Variables
• Exercise 3.2 Chapter 3 Pair of Linear Equations in Two Variables
• Exercise 3.3 Chapter 3 Pair of Linear Equations in Two Variables
• Exercise 3.4 Chapter 3 Pair of Linear Equations in Two Variables
• Exercise 3.5 Chapter 3 Pair of Linear Equations in Two Variables
• Exercise 3.6 Chapter 3 Pair of Linear Equations in Two Variables
• Exercise 3.7 Chapter 3 Pair of Linear Equations in Two Variables

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Exercise 3.1 page: 44.

1. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

Solutions: Let the present age of Aftab be ‘x’.

And, the present age of his daughter be ‘y’.

Now, we can write, seven years ago,

Age of Aftab = x-7

Age of his daughter = y-7

According to the question,

x−7 = 7(y−7)

⇒x−7 = 7y−49

⇒x−7y = −42         ………………………(i)

Also, three years from now or after three years,

Age of Aftab will become = x+3.

Age of his daughter will become = y+3

According to the situation given,

x+3 = 3(y+3)

⇒x+3 = 3y+9

⇒x−3y = 6       …………..…………………(ii)

Subtracting equation (i) from equation (ii) we have

(x−3y)−(x−7y) = 6−(−42)

⇒−3y+7y = 6+42

The algebraic equation is represented by

For, x−7y = −42 or x = −42+7y

The solution table is

For,  x−3y = 6   or     x = 6+3y

The graphical representation is:

2. The coach of a cricket team buys 3 bats and 6 balls for Rs.3900. Later, she buys another bat and 3 more balls of the same kind for Rs.1300. Represent this situation algebraically and geometrically.

Solutions: Let us assume that the cost of a bat be ‘Rs x’

And,the cost of a ball be ‘Rs y’

According to the question, the algebraic representation is

3x+6y = 3900

And x+3y = 1300

For, 3x+6y = 3900

Or x = (3900-6y)/3

For, x+3y = 1300

Or x = 1300-3y

The graphical representation is as follows.

3. The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs.160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs.300. Represent the situation algebraically and geometrically.

Solutions: Let the cost of 1 kg of apples be ‘Rs. x’

And, cost of 1 kg of grapes be ‘Rs. y’

And 4x+2y = 300

For, 2x+y = 160 or y = 160−2x, the solution table is;

For 4x+2y = 300 or y = (300-4x)/2, the solution table is;

The graphical representation is as follows;

Exercise 3.2 Page: 49

1. 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.

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

(i)Let there are x number of girls and y number of boys. As per the given question, the algebraic expression can be represented as follows.

Now, for x+y = 10 or x = 10−y, the solutions are;

For x – y = 4 or x = 4 + y, the solutions are;

From the graph, it can be seen that the given lines cross each other at point (7, 3). Therefore, there are 7 girls and 3 boys in the class.

(ii) Let 1 pencil costs Rs.x and 1 pen costs Rs.y.

According to the question, the algebraic expression cab be represented as;

5x + 7y = 50

7x + 5y = 46

For, 5x + 7y = 50 or  x = (50-7y)/5, the solutions are;

For 7x + 5y = 46 or x = (46-5y)/7, the solutions are;

Hence, the graphical representation is as follows;

From the graph, it is can be seen that the given lines cross each other at point (3, 5).

So, the cost of a pencil is 3/- and cost of a pen is 5/-.

2. On comparing the ratios a1/a2 , b1/b2 , c1/c2 find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

(i) 5x – 4y + 8 = 0

7x + 6y – 9 = 0

(ii) 9x + 3y + 12 = 0

18x + 6y + 24 = 0

(iii) 6x – 3y + 10 = 0

2x – y + 9 = 0

(i) Given expressions;

5x−4y+8 = 0

7x+6y−9 = 0

Comparing these equations with a1x+b1y+c1 = 0

And a2x+b2y+c2 = 0

a1 = 5, b1 = -4, c1 = 8

a2 = 7, b2 = 6, c2 = -9

(a1/a2) = 5/7

(b1/b2) = -4/6 = -2/3

(c1/c2) = 8/-9

Since, (a1/a2) ≠ (b1/b2)

So, the pairs of equations given in the question have a unique solution and the lines cross each other at exactly one point.

(ii) Given expressions;

9x + 3y + 12 = 0

a1 = 9, b1 = 3, c1 = 12

a2 = 18, b2 = 6, c2 = 24

(a1/a2) = 9/18 = 1/2

(b1/b2) = 3/6 = 1/2

(c1/c2) = 12/24 = 1/2

Since (a1/a2) = (b1/b2) = (c1/c2)

So, the pairs of equations given in the question have infinite possible solutions and the lines are coincident.

(iii) Given Expressions;

6x – 3y + 10 = 0

a1 = 6, b1 = -3, c1 = 10

a2 = 2, b2 = -1, c2 = 9

(a1/a2) = 6/2 = 3/1

(b1/b2) = -3/-1 = 3/1

(c1/c2) = 10/9

Since (a1/a2) = (b1/b2) ≠ (c1/c2)

So, the pairs of equations given in the question are parallel to each other and the lines never intersect each other at any point and there is no possible solution for the given pair of equations.

3. On comparing the ratio, (a1/a2) , (b1/b2) , (c1/c2) find out whether the following pair of linear equations are consistent, or inconsistent.

(i) 3x + 2y = 5 ; 2x – 3y = 7

(ii) 2x – 3y = 8 ; 4x – 6y = 9

(iii)(3/2)x+(5/3)y = 7; 9x – 10y = 14

(iv) 5x – 3y = 11 ; – 10x + 6y = –22

(v)(4/3)x+2y = 8 ; 2x + 3y = 12

(i) Given : 3x + 2y = 5 or 3x + 2y -5 = 0

and 2x – 3y = 7 or 2x – 3y -7 = 0

Comparing these equations with a1x+b1y+c1 = 0

a1 = 3, b1 = 2, c1 = -5

a2 = 2, b2 = -3, c2 = -7

(a1/a2) = 3/2

(b1/b2) = 2/-3

(c1/c2) = -5/-7 = 5/7

So, the given equations intersect each other at one point and they have only one possible solution. The equations are consistent.

(ii) Given 2x – 3y = 8 and 4x – 6y = 9

a1 = 2, b1 = -3, c1 = -8

a2 = 4, b2 = -6, c2 = -9

(a1/a2) = 2/4 = 1/2

(b1/b2) = -3/-6 = 1/2

(c1/c2) = -8/-9 = 8/9

Since , (a1/a2) = (b1/b2) ≠ (c1/c2)

So, the equations are parallel to each other and they have no possible solution. Hence, the equations are inconsistent.

(iii)Given (3/2)x + (5/3)y = 7 and 9x – 10y = 14

a1 = 3/2, b1 = 5/3, c1 = -7

a2 = 9, b2 = -10, c2 = -14

(a1/a2) = 3/(2×9) = 1/6

(b1/b2) = 5/(3× -10)= -1/6

(c1/c2) = -7/-14 = 1/2

So, the equations are intersecting  each other at one point and they have only one possible solution. Hence, the equations are consistent.

(iv) Given, 5x – 3y = 11 and – 10x + 6y = –22

a1 = 5, b1 = -3, c1 = -11

a2 = -10, b2 = 6, c2 = 22

(a1/a2) = 5/(-10) = -5/10 = -1/2

(b1/b2) = -3/6 = -1/2

(c1/c2) = -11/22 = -1/2

These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.

(v)Given, (4/3)x +2y = 8 and 2x + 3y = 12

a1 = 4/3 , b1= 2 , c1 = -8

a2 = 2, b2 = 3 , c2 = -12

(a1/a2) = 4/(3×2)= 4/6 = 2/3

(b1/b2) = 2/3

(c1/c2) = -8/-12 = 2/3

4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

(i) x + y = 5, 2x + 2y = 10

(ii) x – y = 8, 3x – 3y = 16

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

(i)Given, x + y = 5 and 2x + 2y = 10

(a1/a2) = 1/2

(b1/b2) = 1/2

(c1/c2) = 1/2

∴The equations are coincident and they have infinite number of possible solutions.

So, the equations are consistent.

For, x + y = 5 or x = 5 – y

For 2x + 2y = 10 or x = (10-2y)/2

So, the equations are represented in graphs as follows:

From the figure, we can see, that the lines are overlapping each other.

Therefore, the equations have infinite possible solutions.

(ii) Given, x – y = 8 and 3x – 3y = 16

(a1/a2) = 1/3

(b1/b2) = -1/-3 = 1/3

(c1/c2) = 8/16 = 1/2

Since, (a1/a2) = (b1/b2) ≠ (c1/c2)

The equations are parallel to each other and have no solutions. Hence, the pair of linear equations is inconsistent.

(iii) Given, 2x + y – 6 = 0 and 4x – 2y – 4 = 0

(a1/a2) = 2/4 = ½

(b1/b2) = 1/-2

(c1/c2) = -6/-4 = 3/2

The given linear equations are intersecting each other at one point and have only one solution. Hence, the pair of linear equations is consistent.

Now, for 2x + y – 6 = 0 or y = 6 – 2x

And for 4x – 2y – 4 = 0 or y = (4x-4)/2

From the graph, it can be seen that these lines are intersecting each other at only one point,(2,2).

(iv) Given, 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0

(b1/b2) = -2/-4 = 1/2

(c1/c2) = 2/5

Since, a1/a2 = b1/b2 ≠ c1/c2

Thus, these linear equations have parallel and have no possible solutions. Hence, the pair of linear equations are inconsistent.

5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Solutions: Let us consider.

The width of the garden is x and length is y.

Now, according to the question, we can express the given condition as;

Now, taking y – x = 4 or y = x + 4

For y + x = 36, y = 36 – x

The graphical representation of both the equation is as follows;

From the graph you can see, the lines intersects each other at a point(16, 20). Hence, the width of the garden is 16 and length is 20.

6. Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

(i) Intersecting lines

(ii) Parallel lines

(iii) Coincident lines

(i) Given the linear equation 2x + 3y – 8 = 0.

To find another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines, it should satisfy below condition;

(a1/a2) ≠ (b1/b2)

Thus, another equation could be 2x – 7y + 9 = 0, such that;

(a1/a2) = 2/2 = 1 and (b1/b2) = 3/-7

Clearly, you can see another equation satisfies the condition.

(ii) Given the linear equation 2x + 3y – 8 = 0.

To find another linear equation in two variables such that the geometrical representation of the pair so formed is parallel lines, it should satisfy below condition;

(a1/a2) = (b1/b2) ≠ (c1/c2)

Thus, another equation could be 6x + 9y + 9 = 0, such that;

(a1/a2) = 2/6 = 1/3

(b1/b2) = 3/9= 1/3

(c1/c2) = -8/9

(iii) Given the linear equation 2x + 3y – 8 = 0.

To find another linear equation in two variables such that the geometrical representation of the pair so formed is coincident lines, it should satisfy below condition;

(a1/a2) = (b1/b2) = (c1/c2)

Thus, another equation could be 4x + 6y – 16 = 0, such that;

(a1/a2) = 2/4 = 1/2 ,(b1/b2) = 3/6 = 1/2, (c1/c2) = -8/-16 = 1/2

7. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Solution: Given, the equations for graphs are x – y + 1 = 0 and 3x + 2y – 12 = 0.

For, x – y + 1 = 0 or x = -1+y

For, 3x + 2y – 12 = 0 or x = (12-2y)/3

Hence, the graphical representation of these equations is as follows;

From the figure, it can be seen that these lines are intersecting each other at point (2, 3) and x-axis at (−1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).

Exercise 3.3 Page: 53

1. Solve the following pair of linear equations by the substitution method

(i) x + y = 14

(ii) s – t = 3

(s/3) + (t/2) = 6

(iii) 3x – y = 3

9x – 3y = 9

(iv) 0.2x + 0.3y = 1.3

0.4x + 0.5y = 2.3

(v) √2 x+√3 y = 0

√3 x-√8 y = 0

(vi) (3x/2) – (5y/3) = -2

(x/3) + (y/2) = (13/6)

x + y = 14 and x – y = 4 are the two equations.

From 1 st equation, we get,

Now, substitute the value of x in second equation to get,

(14 – y) – y = 4

14 – 2y = 4

By the value of y, we can now find the exact value of x;

∵ x = 14 – y

∴ x = 14 – 5

Hence, x = 9 and y = 5.

(ii) Given,

s – t = 3 and (s/3) + (t/2) = 6 are the two equations.

s = 3 + t ________________(1)

Now, substitute the value of s in second equation to get,

(3+t)/3 + (t/2) = 6

⇒ (2(3+t) + 3t )/6 = 6

⇒ (6+2t+3t)/6 = 6

⇒ (6+5t) = 36

Now, substitute the value of t in equation (1)

s = 3 + 6 = 9

Therefore, s = 9 and t = 6.

(iii) Given,

3x – y = 3 and 9x – 3y = 9 are the two equations.

x = (3+y)/3

Now, substitute the value of x in the given second equation to get,

9(3+y)/3 – 3y = 9

⇒9 +3y -3y = 9

Therefore, y has infinite values and since, x = (3+y) /3, so x also has infinite values.

(iv) Given,

0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3are the two equations.

x = (1.3- 0.3y)/0.2 _________________(1)

0.4(1.3-0.3y)/0.2 + 0.5y = 2.3

⇒2(1.3 – 0.3y) + 0.5y = 2.3

⇒ 2.6 – 0.6y + 0.5y = 2.3

⇒ 2.6 – 0.1 y = 2.3

⇒ 0.1 y = 0.3

Now, substitute the value of y in equation (1), we get,

x = (1.3-0.3(3))/0.2 = (1.3-0.9)/0.2 = 0.4/0.2 = 2

Therefore, x = 2 and y = 3.

√2 x + √3 y = 0 and √3 x – √8 y = 0

are the two equations.

x = – (√3/√2)y __________________(1)

Putting the value of x in the given second equation to get,

√3(-√3/√2)y – √8y = 0 ⇒ (-3/√2)y- √8 y = 0

Therefore, x = 0 and y = 0.

(vi) Given,

(3x/2)-(5y/3) = -2 and (x/3) + (y/2) = 13/6 are the two equations.

(3/2)x = -2 + (5y/3)

⇒ x = 2(-6+5y)/9 = (-12+10y)/9 ………………………(1)

((-12+10y)/9)/3 + y/2 = 13/6

⇒y/2 = 13/6 –( (-12+10y)/27 ) + y/2 = 13/6

(3x/2) – 5(3)/3 = -2

⇒ (3x/2) – 5 = -2

2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

2x + 3y = 11…………………………..(I)

2x – 4y = -24………………………… (II)

From equation (II), we get

x = (11-3y)/2 ………………….(III)

Substituting the value of x in equation (II), we get

2(11-3y)/2 – 4y = 24

11 – 3y – 4y = -24

y = 5……………………………………..(IV)

Putting the value of y in equation (III), we get

x = (11-3×5)/2 = -4/2 = -2

Hence, x = -2, y = 5

Therefore the value of m is -1.

3. Form the pair of linear equations for the following problems and find their solution by substitution method.

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

Let the two numbers be x and y respectively, such that y > x.

y = 3x ……………… (1)

y – x = 26 …………..(2)

Substituting the value of (1) into (2), we get

3x – x = 26

x = 13 ……………. (3)

Substituting (3) in (1), we get y = 39

Hence, the numbers are 13 and 39.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Let the larger angle by x o  and smaller angle be y o .

We know that the sum of two supplementary pair of angles is always 180 o .

x + y = 180 o ……………. (1)

x – y = 18 o  ……………..(2)

From (1), we get x = 180 o  – y …………. (3)

Substituting (3) in (2), we get

180 o  – y – y =18 o

162 o  = 2y

y = 81 o  ………….. (4)

Using the value of y in (3), we get

x = 180 o  – 81 o

Hence, the angles are 99 o  and 81 o .

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later, she buys 3 bats and 5 balls for Rs.1750. Find the cost of each bat and each ball.

Let the cost a bat be x and cost of a ball be y.

7x + 6y = 3800 ………………. (I)

3x + 5y = 1750 ………………. (II)

From (I), we get

y = (3800-7x)/6………………..(III)

Substituting (III) in (II). we get,

3x+5(3800-7x)/6 =1750

⇒3x+ 9500/3 – 35x/6 = 1750

⇒3x- 35x/6 = 1750 – 9500/3

⇒(18x-35x)/6 = (5250 – 9500)/3

⇒-17x/6 = -4250/3

⇒-17x = -8500

x = 500 ……………………….. (IV)

Substituting the value of x in (III), we get

y = (3800-7 ×500)/6 = 300/6 = 50

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

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

Let the fixed charge be Rs x and per km charge be Rs y.

x + 10y = 105 …………….. (1)

x + 15y = 155 …………….. (2)

From (1), we get x = 105 – 10y ………………. (3)

Substituting the value of x in (2), we get

105 – 10y + 15y = 155

y = 10 …………….. (4)

Putting the value of y in (3), we get

x = 105 – 10 × 10 = 5

Hence, fixed charge is Rs 5 and per km charge = Rs 10

Charge for 25 km = x + 25y = 5 + 250 = Rs 255

(v) A fraction becomes 9/11 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.

Let the fraction be x/y.

(x+2) /(y+2) = 9/11

11x + 22 = 9y + 18

11x – 9y = -4 …………….. (1)

(x+3) /(y+3) = 5/6

6x + 18 = 5y +15

6x – 5y = -3 ………………. (2)

From (1), we get x = (-4+9y)/11 …………….. (3)

6(-4+9y)/11 -5y = -3

-24 + 54y – 55y = -33

y = 9 ………………… (4)

Substituting the value of y in (3), we get

x = (-4+9×9 )/11 = 7

Hence the fraction is 7/9.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Let the age of Jacob and his son be x and y respectively.

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

x – 3y = 10 …………………………………….. (1)

(x – 5) = 7(y – 5)

x – 7y = -30 ………………………………………. (2)

From (1), we get x = 3y + 10 ……………………. (3)

3y + 10 – 7y = -30

y = 10 ………………… (4)

x = 3 x 10 + 10 = 40

Hence, the present age of Jacob’s and his son is 40 years and 10 years respectively.

Exercise 3.4 Page: 56

1. Solve the following pair of linear equations by the elimination method and the substitution method:

(i) x + y = 5 and 2x – 3y = 4

(ii) 3x + 4y = 10 and 2x – 2y = 2

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

(iv) x/2+ 2y/3 = -1 and x-y/3 = 3

By the method of elimination.

x + y = 5 ……………………………….. (i)

2x – 3y = 4 ……………………………..(ii)

When the equation (i) is multiplied by 2, we get

2x + 2y = 10 ……………………………(iii)

When the equation (ii) is subtracted from (iii) we get,

y = 6/5 ………………………………………(iv)

Substituting the value of y in eq. (i) we get,

x=5−6/5 = 19/5

∴x = 19/5 , y = 6/5

By the method of substitution.

From the equation (i), we get:

x = 5 – y………………………………….. (v)

When the value is put in equation (ii) we get,

2(5 – y) – 3y = 4

When the values are substituted in equation (v), we get:

x =5− 6/5 = 19/5

∴x = 19/5 ,y = 6/5

3x + 4y = 10……………………….(i)

2x – 2y = 2 ………………………. (ii)

When the equation (i) and (ii) is multiplied by 2, we get:

4x – 4y = 4 ………………………..(iii)

When the Equation (i) and (iii) are added, we get:

x = 2 ……………………………….(iv)

Substituting equation (iv) in (i) we get,

6 + 4y = 10

Hence, x = 2 and y = 1

By the method of Substitution

From equation (ii) we get,

x = 1 + y……………………………… (v)

Substituting equation (v) in equation (i) we get,

3(1 + y) + 4y = 10

When y = 1 is substituted in equation (v) we get,

A = 1 + 1 = 2

Therefore, A = 2 and B = 1

By the method of elimination:

3x – 5y – 4 = 0 ………………………………… (i)

9x = 2y + 7

9x – 2y – 7 = 0 …………………………………(ii)

When the equation (i) and (iii) is multiplied we get,

9x – 15y – 12 = 0 ………………………………(iii)

When the equation (iii) is subtracted from equation (ii) we get,

y = -5/13 ………………………………………….(iv)

When equation (iv) is substituted in equation (i) we get,

3x +25/13 −4=0

∴x = 9/13 and y = -5/13 

By the method of Substitution:

From the equation (i) we get,

x = (5y+4)/3 …………………………………………… (v)

Putting the value (v) in equation (ii) we get,

9(5y+4)/3 −2y −7=0

Substituting this value in equation (v) we get,

x = (5(-5/13)+4)/3

∴x = 9/13 , y = -5/13

(iv) x/2 + 2y/3 = -1 and x-y/3 = 3

By the method of Elimination.

3x + 4y = -6 …………………………. (i)

3x – y = 9 ……………………………. (ii)

When the equation (ii) is subtracted from equation (i) we get,

y = -3 ………………………………….(iii)

When the equation (iii) is substituted in (i) we get,

3x – 12 = -6

Hence, x = 2 , y = -3

From the equation (ii) we get,

x = (y+9)/3…………………………………(v)

Putting the value obtained from equation (v) in equation (i) we get,

3(y+9)/3 +4y =−6

When y = -3 is substituted in equation (v) we get,

x = (-3+9)/3 = 2

Therefore, x = 2 and y = -3

2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes if we only add 1 to the denominator. What is the fraction?

Let the fraction be a/b

According to the given information,

(a+1)/(b-1) = 1

=> a – b = -2 ………………………………..(i)

a/(b+1) = 1/2

=> 2a-b = 1…………………………………(ii)

When equation (i) is subtracted from equation (ii) we get,

a = 3 …………………………………………………..(iii)

When a = 3 is substituted in equation (i) we get,

Hence, the fraction is 3/5.

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Let us assume, present age of Nuri is x

And present age of Sonu is y.

According to the given condition, we can write as;

x – 5 = 3(y – 5)

x – 3y = -10…………………………………..(1)

x + 10 = 2(y +10)

x – 2y = 10…………………………………….(2)

Subtract eq. 1 from 2, to get,

y = 20 ………………………………………….(3)

Substituting the value of y in eq.1, we get,

x – 3.20 = -10

x – 60 = -10

Age of Nuri is 50 years

Age of Sonu is 20 years.

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Let the unit digit and tens digit of a number be x and y respectively.

Then, Number (n) = 10B + A

N after reversing order of the digits = 10A + B

According to the given information, A + B = 9…………………….(i)

9(10B + A) = 2(10A + B)

88 B – 11 A = 0

-A + 8B = 0 ………………………………………………………….. (ii)

Adding the equations (i) and (ii) we get,

B = 1……………………………………………………………………….(3)

Substituting this value of B, in the equation (i) we get A= 8

Hence the number (N) is 10B + A = 10 x 1 +8 = 18

(iv) Meena went to a bank to withdraw Rs.2000. She asked the cashier to give her Rs.50 and Rs.100 notes only. Meena got 25 notes in all. Find how many notes of Rs.50 and Rs.100 she received.

Let the number of Rs.50 notes be A and the number of Rs.100 notes be B

A + B = 25 ……………………………………………………………………….. (i)

50A + 100B = 2000 ………………………………………………………………(ii)

When equation (i) is multiplied with (ii) we get,

50A + 50B = 1250 …………………………………………………………………..(iii)

Subtracting the equation (iii) from the equation (ii) we get,

Substituting in the equation (i) we get,

Hence, Meena has 10 notes of Rs.50 and 15 notes of Rs.100.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs.27 for a book kept for seven days, while Susy paid Rs.21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Let the fixed charge for the first three days be Rs.A and the charge for each day extra be Rs.B.

According to the information given,

A + 4B = 27 …………………………………….…………………………. (i)

A + 2B = 21 ……………………………………………………………….. (ii)

When equation (ii) is subtracted from equation (i) we get,

B = 3 …………………………………………………………………………(iii)

Substituting B = 3 in equation (i) we get,

A + 12 = 27

Hence, the fixed charge is Rs.15

And the Charge per day is Rs.3

Exercise 3.5 Page: 62

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

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

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

(i) Given, x – 3y – 3 =0 and 3x – 9y -2 =0

a 1 /a 2 =1/3 ,         b 1 /b 2 = -3/-9 =1/3,     c 1 /c 2 =-3/-2 = 3/2

(a 1 /a 2 ) = (b 1 /b 2 ) ≠ (c 1 /c 2 )

Since, the given set of lines are parallel to each other they will not intersect each other and therefore there is no solution for these equations.

(ii) Given, 2x + y = 5 and 3x +2y = 8

a 1 /a 2 = 2/3 , b 1 /b 2 = 1/2 , c 1 /c 2 = -5/-8

(a 1 /a 2 ) ≠ (b 1 /b 2 )

Since they intersect at a unique point these equations will have a unique solution by cross multiplication method:

x/(b 1 c 2 -c 1 b 2 ) = y/(c 1 a 2 – c 2 a1) = 1/(a 1 b 2 -a 2 b 1 )

x/(-8-(-10)) = y/(-15-(-16)) = 1/(4-3)

x/2 = y/1 = 1

∴ x = 2 and y =1

(iii) Given, 3x – 5y = 20 and 6x – 10y = 40

(a 1 /a 2 ) = 3/6 = 1/2

(b 1 /b 2 ) = -5/-10 = 1/2

(c 1 /c 2 ) = 20/40 = 1/2

a 1 /a 2 = b 1 /b 2 = c 1 /c 2

Since the given sets of lines are overlapping each other there will be infinite number of solutions for this pair of equation.

(iv) Given, x – 3y – 7 = 0 and 3x – 3y – 15 = 0

(a 1 /a 2 ) = 1/3

(b 1 /b 2 ) = -3/-3 = 1

(c 1 /c 2 ) = -7/-15

a 1 /a 2 ≠ b 1 /b 2

Since this pair of lines are intersecting each other at a unique point, there will be a unique solution.

By cross multiplication,

x/(45-21) = y/(-21+15) = 1/(-3+9)

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

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

∴ x = 4 and y = 1.

2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?

2x + 3y = 7

(a – b) x + (a + b) y = 3a + b – 2

(ii) For which value of k will the following pair of linear equations have no solution?

(2k – 1) x + (k – 1) y = 2k + 1

(i) 3y + 2x -7 =0

(a + b)y + (a-b)y – (3a + b -2) = 0

a 1 /a 2 = 2/(a-b) ,               b 1 /b 2 = 3/(a+b) ,               c 1 /c 2 = -7/-(3a + b -2)

For infinitely many solutions,

Thus 2/(a-b) = 7/(3a+b– 2)

6a + 2b – 4 = 7a – 7b

a – 9b = -4  ……………………………….(i)

2/(a-b) = 3/(a+b)

2a + 2b = 3a – 3b

a – 5b = 0 ……………………………….….(ii)

Subtracting (i) from (ii), we get

Substituting this eq. in (ii), we get

a -5 x 1= 0

Thus at a = 5 and b = 1 the given equations will have infinite solutions.

(ii) 3x + y -1 = 0

(2k -1)x  +  (k-1)y – 2k -1 = 0

a 1 /a 2 = 3/(2k -1) ,           b 1 /b 2 = 1/(k-1), c 1 /c 2  = -1/(-2k -1) = 1/( 2k +1)

For no solutions

a 1 /a 2 = b 1 /b 2 ≠ c 1 /c 2

3/(2k-1) = 1/(k -1)     ≠ 1/(2k +1)

3/(2k –1) = 1/(k -1)

3k -3 = 2k -1

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

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

8x + 5y = 9

3x + 2y = 4

8x + 5y = 9 …………………..(1)

3x + 2y = 4 ……………….….(2)

From equation (2) we get

x = (4 – 2y )/ 3  ……………………. (3)

Using this value in equation 1, we get

8(4-2y)/3 + 5y = 9

32 – 16y +15y = 27

y = 5 ……………………………….(4)

Using this value in equation (2), we get

3x + 10 = 4

Thus, x = -2 and y = 5.

Now, Using Cross Multiplication method:

8x +5y – 9 = 0

3x + 2y – 4 = 0

x/(-20+18) = y/(-27 + 32 ) = 1/(16-15)

-x/2 = y/5 =1/1

∴ x = -2 and y =5.

4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs.1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs.1180 as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

(i) Let x be the fixed charge and y be the charge of food per day.

x + 20y = 1000……………….. (i)

x + 26y = 1180………………..(ii)

Subtracting (i) from  (ii) we get

Using this value in equation (ii) we get

x = 1180 -26 x 30

Therefore, fixed charges is Rs.400 and charge per day is Rs.30.

(ii)  L et the fraction be x/y.

So, as per the question given,

(x -1)/y = 1/3 => 3x – y = 3…………………(1)

x/(y + 8) = 1/4  => 4x –y =8 ………………..(2)

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

x = 5 ………………………………………….(3)

Using this value in equation (2), we get,

(4×5)– y = 8

Therefore, the fraction is 5/12.

(iii) Let the number of right answers is x and number of wrong answers be y

According to the given question;

3x−y=40……..(1)

⇒2x−y=25…….(2)

Subtracting equation (2) from equation (1), we get;

x = 15 ….….(3)

Putting this in equation (2), we obtain;

30 – y = 25

Therefore, number of right answers = 15 and number of wrong answers = 5

Hence, total number of questions = 20

(iv)  Let x km/h be the speed of car from point A and y km/h be the speed of car from point B.

If the car travels in the same direction,

5x – 5y = 100

x – y = 20 …………………………………(i)

If the car travels in the opposite direction,

x + y = 100………………………………(ii)

Solving equation (i) and (ii), we get

x = 60 km/h………………………………………(iii)

Using this in equation (i), we get,

60 – y = 20

y = 40 km/h

Therefore, the speed of car from point A = 60 km/h

Speed of car from point B = 40 km/h.

The length of rectangle = x unit

And breadth of the rectangle = y unit

Now, as per the question given,

(x – 5) (y + 3) = xy -9

3x – 5y – 6 = 0……………………………(1)

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

2x + 3y – 61 = 0…………………………..(2)

Using cross multiplication method, we get,

x/(305 +18) = y/(-12+183) = 1/(9+10)

x/323 = y/171 = 1/19

Therefore, x = 17 and y = 9.

Hence, the length of rectangle = 17 units

And breadth of the rectangle = 9 units

Exercise 3.6 Page: 67

1. Solve the following pairs of equations by reducing them to a pair of linear equations:

(i) 1/2x + 1/3y = 2

1/3x + 1/2y = 13/6

Let us assume 1/x = m and 1/y = n  , then the equation will change as follows.

m/2 + n/3 = 2

⇒ 3m+2n-12 = 0…………………….(1)

m/3 + n/2 = 13/6

⇒ 2m+3n-13 = 0……………………….(2)

Now, using cross-multiplication method, we get,

m/(-26-(-36) ) = n/(-24-(-39)) = 1/(9-4)

m/10 = n/15 = 1/5

m/10 = 1/5 and n/15 = 1/5

So, m = 2 and n = 3

1/x = 2 and 1/y = 3

x = 1/2 and y = 1/3

(ii) 2/√x + 3/√y = 2

4/√x + 9/√y = -1

Substituting 1/√x = m and 1/√y = n in the given equations, we get

2m + 3n = 2 ………………………..(i)

4m – 9n = -1 ………………………(ii)

Multiplying equation (i) by 3, we get

6m + 9n = 6 ………………….…..(iii)

Adding equation (ii) and (iii), we get

m = 1/2…………………………….…(iv)

Now by putting the value of ‘m’ in equation (i), we get

2×1/2 + 3n = 2

Hence, x = 4 and y = 9

(iii) 4/x + 3y = 14

3/x -4y = 23

Putting in the given equation we get,

So, 4m + 3y = 14     => 4m + 3y – 14 = 0  ……………..…..(1)

3m – 4y = 23     => 3m – 4y – 23 = 0  ……………………….(2)

By cross-multiplication, we get,

m/(-69-56) = y/(-42-(-92)) = 1/(-16-9)

-m/125 = y/50 = -1/ 25

-m/125 = -1/25 and y/50 = -1/25

m = 5 and b = -2

m = 1/x = 5

So , x = 1/5

(iv) 5/(x-1) + 1/(y-2) = 2

6/(x-1) – 3/(y-2) = 1

Substituting 1/(x-1) = m and 1/(y-2) = n  in the given equations, we get,

5m + n = 2 …………………………(i)

6m – 3n = 1 ……………………….(ii)

15m + 3n = 6 …………………….(iii)

Adding (ii) and (iii) we get

Putting this value in equation (i), we get

5×1/3 + n = 2

n = 2- 5/3 = 1/3

m = 1/ (x-1)

⇒ 1/3 = 1/(x-1)

n = 1/(y-2)

⇒ 1/3 = 1/(y-2)

Hence, x = 4 and y = 5

(v) (7x-2y)/ xy = 5

(8x + 7y)/xy = 15

(7x-2y)/ xy = 5

7/y – 2/x = 5…………………………..(i)

8/y + 7/x = 15…………………………(ii)

Substituting 1/x =m in the given equation we get,

– 2m + 7n = 5     => -2 + 7n – 5 = 0  ……..(iii)

7m + 8n = 15     => 7m + 8n – 15 = 0 ……(iv)

By cross-multiplication method, we get,

m/(-105-(-40)) = n/(-35-30) = 1/(-16-49)

m/(-65) = n/(-65) = 1/(-65)

m/-65 = 1/-65

n/(-65) = 1/(-65)

m = 1 and n = 1

m = 1/x = 1        n = 1/x = 1

Therefore, x = 1 and y = 1

(vi) 6x + 3y = 6xy

2x + 4y = 5xy

6x + 3y = 6xy

6/y + 3/x = 6

Let 1/x = m and 1/y = n

=> 6n +3m = 6

=>3m + 6n-6 = 0…………………….(i)

=> 2/y + 4/x = 5

=> 2n +4m = 5

=> 4m+2n-5 = 0……………………..(ii)

3m + 6n – 6 = 0

4m + 2n – 5 = 0

By cross-multiplication method, we get

m/(-30 –(-12)) = n/(-24-(-15)) = 1/(6-24)

m/-18 = n/-9 = 1/-18

m/-18 = 1/-18

n/-9 = 1/-18

m = 1 and n = 1/2

m = 1/x = 1 and n = 1/y = 1/2

x = 1 and y = 2

Hence, x = 1 and y = 2

(vii) 10/(x+y) + 2/(x-y) = 4

15/(x+y) – 5/(x-y) = -2

Substituting 1/x+y = m and 1/x-y = n in the given equations, we get,

10m + 2n = 4      =>  10m + 2n – 4 = 0      ………………..…..(i)

15m – 5n = -2     =>   15m – 5n + 2 = 0    ……………………..(ii)

Using cross-multiplication method, we get,

m/(4-20) = n/(-60-(20)) = 1/(-50 -30)

m/-16 = n/-80 = 1/-80

m/-16 = 1/-80 and n/-80 = 1/-80

m = 1/5 and n = 1

m = 1/(x+y) = 1/5

x+y = 5 …………………………………………(iii)

n = 1/(x-y) = 1

x-y = 1……………………………………………(iv)

Adding equation (iii) and (iv), we get

2x = 6   => x = 3 …….(v)

Putting the value of x = 3 in equation (3), we get

Hence, x = 3 and y = 2

(viii) 1/(3x+y) + 1/(3x-y) = 3/4

1/2(3x+y) – 1/2(3x-y) = -1/8

Substituting 1/(3x+y) = m and 1/(3x-y) = n in the given equations, we get,

m + n = 3/4 …………………………….…… (1)

m/2 – n/2 = -1/8

m – n = -1/4  …………………………..…(2)

Adding (1) and (2), we get

2m = 3/4 – 1/4

Putting in (2), we get

1/4 – n = -1/4

n = 1/4 + 1/4 = 1/2

m = 1/(3x+y) = 1/4

3x + y = 4  …………………………………(3)

n = 1/( 3x-y) = 1/2

3x – y = 2 ………………………………(4)

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

x = 1 ……………………………….(5)

Putting in (3), we get

3(1) + y = 4

Hence, x = 1 and y = 1

2. Formulate the following problems as a pair of equations, and hence find their solutions:

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

(i) Let us consider,

Speed of Ritu in still water = x km/hr

Speed of Stream = y km/hr

Now, speed of Ritu during,

Downstream = x + y km/h

Upstream = x – y km/h

As per the question given,

2(x+y) = 20

Or x + y = 10……………………….(1)

And, 2(x-y) = 4

Or x – y = 2………………………(2)

Adding both the eq.1 and 2, we get,

Putting the value of x in eq.1, we get,

Speed of Ritu rowing in still water = 6 km/hr

Speed of Stream = 4 km/hr

(ii) Let us consider,

Number of days taken by women to finish the work = x

Number of days taken by men to finish the work = y

Work done by women in one day = 1/x

Work done by women in one day = 1/y

4(2/x + 5/y) = 1

(2/x + 5/y) = 1/4

And, 3(3/x + 6/y) = 1

(3/x + 6/y) = 1/3

Now, put 1/x=m and 1/y=n, we get,

2m + 5n = 1/4 => 8m + 20n = 1…………………(1)

3m + 6n =1/3 => 9m + 18n = 1………………….(2)

Now, by cross multiplication method, we get here,

m/(20-18) = n/(9-8) = 1/ (180-144)

m/2 = n/1 = 1/36

m = 1/x = 1/18

n = 1/y = 1/36

Number of days taken by women to finish the work = 18

Number of days taken by men to finish the work = 36.

(iii) Let us consider,

Speed of the train = x km/h

Speed of the bus = y km/h

According to the given question,

60/x + 240/y = 4 …………………(1)

100/x + 200/y = 25/6 …………….(2)

Put 1/x=m and 1/y=n, in the above two equations;

60m + 240n = 4……………………..(3)

100m + 200n = 25/6

600m + 1200n = 25 ………………….(4)

Multiply eq.3 by 10, to get,

600m + 2400n = 40 ……………………(5)

Now, subtract eq.4 from 5, to get,

n = 15/1200 = 1/80

Substitute the value of n in eq. 3, to get,

60m + 3 = 4

m = 1/x = 1/60

And y = 1/n

Speed of the train = 60 km/h

Speed of the bus = 80 km/h

Exercise 3.7 Page: 68

1. The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

The age difference between Ani and Biju is 3 yrs.

Either Biju is 3 years older than Ani or Ani is 3 years older than Biju. 

From both cases, we find out that Ani’s father’s age is 30 yrs more than that of Cathy’s age.

Let the ages of Ani and Biju be A and B, respectively.

Therefore, the age of Dharam = 2 x A = 2A yrs.

And the age of Biju’s sister Cathy is B/2 yrs.

By using the information that is given,

When Ani is older than Biju by 3 yrs, then A – B = 3 …..(1)

2A − B/2 = 30

4A – B = 60 ….(2)

By subtracting the equation (1) from (2), we get;

3A = 60 – 3 = 57

A = 57/3 = 19

Therefore, the age of Ani = 19 yrs

And the age of Biju is 19 – 3 = 16 yrs.

When Biju is older than Ani,

B – A = 3 ….(1)

Adding the equations (1) and (2), we get;

Therefore, the age of Ani is 21 yrs

And the age of Biju is 21 + 3 = 24 yrs.

2. One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II] [Hint : x + 100 = 2(y – 100), y + 10 = 6(x – 10)].

Let the capital amount with two friends be Rs. x and Rs. y, respectively. 

As per the given,

x + 100 = 2(y − 100)…..(i)

6(x − 10) = (y + 10)….(ii)

Consider the equation (i),

x + 100 = 2(y − 100)

x + 100 = 2y − 200

x − 2y = −300…..(iii)

From equation (ii),

6x − 60 = y + 10

6x − y = 70…..(iv)

(iv) × 2 – (iii)

12x – 2y – (x – 2y) = 140 – (-300)

Substituting x = 40 in equation (iii), we get;

40 – 2y = -300

Therefore, the two friends had Rs. 40 and Rs. 170 with them.

3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Let the speed of the train be x km/hr and the time taken by the train to travel a distance be t hours, and the d km be the distance.

Speed of the train = Distance travelled by train / Time taken to travel that distance

d = xt …..(i)

Case 1: When the speed of the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time.

(x + 10) = d/(t – 2)

(x + 10)(t – 2) = d

xt + 10t – 2x – 20 = d

d + 10t – 2x = 20 + d [From (i)]

10t – 2x = 20…..(ii)

Case 2: When the train was slower by 10 km/h, it would have taken 3 hours more than the scheduled time.

So, (x – 10) = d/(t + 3)

(x – 10)(t + 3) = d

xt – 10t + 3x – 30 = d

d – 10t + 3x = 30 + d [From (i)]

-10t + 3x = 30…..(iii)

Adding (ii) and (iii), we get;

Thus, the speed of the train is 50 km/h.

Substituting x = 50 in equation (ii), we get;

10t – 100 = 20

t = 12 hours

Distance travelled by train, d = xt

Hence, the distance covered by the train is 600 km.

4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Let x be the number of rows and y be the number of students in a row.

Total students in the class = Number of rows × Number of students in a row

Total number of students = (x − 1) (y + 3)

xy = (x − 1) (y + 3) 

xy = xy − y + 3x − 3

3x − y − 3 = 0

3x − y = 3…..(i)

Total number of students = (x + 2) (y − 3)

xy = xy + 2y − 3x − 6

3x − 2y = −6…..(ii)

Subtracting equation (ii) from (i), we get;

(3x − y) − (3x − 2y) = 3 − (−6)

− y + 2y = 9

Substituting y = 9 in equation (i), we get;

Therefore, the total number of students in a class = xy = 4 × 9 = 36

5. In a ∆ABC, ∠ C = 3 ∠ B = 2 (∠ A + ∠ B). Find the three angles.

∠C = 3 ∠B = 2(∠B + ∠A)

3∠B = 2 ∠A + 2 ∠B

2∠A – ∠B= 0- – – – – – – – – – – – (i)

We know that the sum of a triangle’s interior angles is 180°.

Thus, ∠ A +∠B + ∠C = 180°

∠A + ∠B +3 ∠B = 180°

∠A + 4 ∠B = 180°– – – – – – – – – – – – – – -(ii)

Multiplying equation (i) by 4, we get;

8 ∠A – 4 ∠B = 0- – – – – – – – – – – – (iii)

Adding equations (iii) and (ii), we get;

9 ∠A = 180°

Using this in equation (ii), we get;

20° + 4∠B = 180°

∠C = 3∠B = 3 x 40 = 120°

Therefore, ∠A = 20°, ∠B = 40°, and ∠C = 120°.

6. Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis.

⇒ y = 5x – 5

Its solution table will be.

Also given,3x – y = 3

The graphical representation of these lines will be as follows:

From the above graph, we can see that the coordinates of the vertices of the triangle formed by the lines and the y-axis are (1, 0), (0, -5) and (0, -3).

7. Solve the following pair of linear equations:

(i) px + qy = p – q

qx – py = p + q

(ii) ax + by = c

bx + ay = 1 + c

(iii) x/a – y/b = 0

ax + by = a 2 + b 2

(iv) (a – b)x + (a + b) y = a 2 – 2ab – b 2

(a + b)(x + y) = a 2 + b 2

(v) 152x – 378y = – 74

–378x + 152y = – 604

(i) px + qy = p – q……………(i)

qx – py = p + q……………….(ii)

Multiplying equation (i) by p and equation (ii) by q, we get;

p 2 x + pqy = p 2  − pq ………… (iii)

q 2 x − pqy = pq + q 2  ………… (iv)

Adding equations (iii) and (iv), we get;

p 2 x + q 2 x = p 2 + q 2

(p 2 + q 2 ) x = p 2  + q 2

x = (p 2 + q 2 )/ (p 2 + q 2 ) = 1

Substituting x = 1 in equation (i), we have;

p(1) + qy = p – q

qy = p – q – p

(ii) ax + by= c…………………(i)

bx + ay = 1+ c………… ..(ii)

Multiplying equation (i) by a and equation (ii) by b, we get;

a 2 x + aby = ac ………………… (iii)

b 2 x + aby = b + bc…………… (iv)

Subtracting equation (iv) from equation (iii),

(a 2 – b 2 ) x = ac − bc– b

x = (ac − bc – b)/ (a 2 – b 2 )

x = c(a – b) – b / (a 2  – b 2 )

From equation (i), we obtain

ax + by = c

a{c(a − b) − b)/ (a 2 – b 2 )} + by = c

{[ac(a−b)−ab]/ (a 2 – b 2 )} + by = c

by = c – {[ac(a − b) − ab]/(a 2  – b 2 )}

by = (a 2 c – b 2 c – a 2 c + abc + ab)/ (a 2 – b 2 )

by = [abc – b 2 c + ab]/ (a 2 – b 2 )

by = b(ac – bc + a)/(a 2 – b 2 )

y = [c(a – b) + a]/(a 2 – b 2 )

x/a – y/b = 0

⇒ bx − ay = 0 ……. (i)

ax + by = a 2 + b 2 …….. (ii)

Multiplying equations (i) and (ii) by b and a, respectively, we get;

b 2 x − aby = 0 …………… (iii)

a 2 x + aby = a 3 + ab 2  …… (iv)

b 2 x + a 2 x = a 3 + ab 2

x(b 2 + a 2 ) = a(a 2  + b 2 ) 

Substituting x = 1 in equation (i), we get;

b(a) − ay = 0

ab − ay = 0

(a + b) y + (a – b) x = a 2 − 2ab − b 2  …………… (i)

(x + y)(a + b) = a 2 + b 2

(a + b) y + (a + b) x = a 2 + b 2  ………………… (ii)

Subtracting equation (ii) from equation (i), we get;

(a − b) x − (a + b) x = (a 2 − 2ab − b 2 ) − (a 2 + b 2 )

x(a − b − a − b) = − 2ab − 2b 2

− 2bx = − 2b (a + b)

Substituting x = a + b in equation (i), we get;

y (a + b) + (a + b)(a − b) = a 2 − 2ab – b 2

a 2 − b 2  + y(a + b) = a 2 − 2ab – b 2

(a + b)y = −2ab

y = -2ab/(a + b)

(v) 152x – 378y = – 74 

152x – 378y = – 74 ….(i)

–378x + 152y = – 604….(ii)

From equation (i),

152x + 74 = 378y

y = (152x + 74)/378

y = (76x + 37)/189…..(iii)

Substituting the value of y in equation (ii), we get;

-378x + 152[(76x + 37)/189] = -604

(-378x)189  + [152(76x) + 152(37)] = (-604)(189)

-71442x + 11552x + 5624 = -114156

-59890x = -114156 – 5624 = -119780

x = -119780/-59890

Substituting x = 2 in equation (iii), we get;

y = [76(2) + 37]/189

= (152 + 37)/189

Therefore, x = 2 and y = 1

8. ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the angles of the cyclic quadrilateral.

Given that ABCD is a cyclic quadrilateral.

As we know, the opposite angles of a cyclic quadrilateral are supplementary.

∠A + ∠C = 180

4y + 20 + (-4x) = 180

-4x + 4y = 160

⇒ -x + y = 40….(i)

∠B + ∠D = 180

3y – 5 + (-7x + 5) = 180

⇒ -7x + 3y = 180…..(ii)

Equation (ii) – 3 × (i),

-7x + 3y – (-3x + 3y) = 180 – 120

Substituting x = -15 in equation (i), we get;

-(-15) + y = 40

y = 40 – 15 = 25

Therefore, x = -15 and y = 25.

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

Chapter 3 – Pair of Linear Equations in Two Variables holds a weightage of 11 marks in the examinations. This chapter gives an account of the various topics related to the linear equations in two variables . The topics discussed in the chapter are mentioned below:

3.1 Introduction In earlier classes, you have studied Linear Equations in Two Variables. You have also studied that a Linear Equation in Two Variables has infinitely many solutions. In this chapter, the knowledge of Linear Equations in Two Variables shall be recalled and extended to that of Pair of Linear Equations In Two Variables. 

3.2 Pair of Linear Equations in Two Variables An equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not zeros, is called a linear equation in two variables x and y. The solution of such a problem is a pair of values, one for x and the other for y, which makes the two sides of the equation equal. The topic also discusses the geometrical representation of Pair of Linear Equations in Two Variables along with suitable examples.

3.3 Graphical Method of Solution of a Pair of Linear Equations In the previous section, you have seen how we can graphically represent a pair of linear equations as two lines. You have also seen that the lines may intersect, or may be parallel, or may coincide. In this section, you will know how to solve it in each case from the geometrical point of view.

3.4 Algebraic Methods of Solving a Pair of Linear Equations In the previous section, we discussed how to solve a pair of linear equations graphically. In some of the cases, the graphical method is not convenient. In this topic, we shall discuss various algebraic methods such as the Substitution Method, Elimination Method, and Cross – Multiplication Method. Each subtopic is explained elaborately with suitable examples, for better understanding.

3.5 Equations Reducible to a Pair of Linear Equation in Two Variables In this section, we shall discuss the solution of such pairs of equations which are not linear but can be reduced to linear form by making some suitable substitutions. The process is explained through some examples related to the subtopic.

3.6 Summary The Summary section consists of the overall important points that need to be memorized while solving the exercise questions of the chapter Pair of Linear Equations in Two Variables. The points mentioned in this section will help you to revise all the concepts mentioned in the chapter.

Two linear equations in the same two variables are called pair of linear equations in two variables. The pair of linear equations in two variables can be represented graphically and algebraically. The graph can be represented by two lines:

• If the lines intersect at a point, the pair of equations is said to be consistent.
• If the lines coincide, the pair of equations is dependent.
• If the lines are parallel, the pair of equations is inconsistent.

Algebraically, the following methods can be used to solve the pair of linear equations in two variables:

• Substitution method
• Elimination method
• Cross-multiplication method

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

• NCERT Solutions are created by subject experts.
• The answers are provided after a lot of brainstorming and are accurate.
• These contain questions related to all the important topics.
• NCERT Solutions Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables includes the solutions to the exercises given in the textbook as well.

The faculty have curated the NCERT Class 10 Maths Solutions in a lucid manner to improve the problem-solving abilities of the students. For a more clear idea about Pair of Linear Equations in Two Variables, students can refer to the study materials available at BYJU’S.

• RD Sharma Solutions for Class 10 Maths Pair of Linear Equations in Two Variables

Disclaimer – 

Dropped Topics – 

3.2 Pair of linear equations in two variables 3.3 Graphical method of solution of a pair of linear equations

3.4.3 Cross-multiplication method 3.5 Equation reducible to a pair of linear equations in two variables

Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 3

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ISC Class 11th Maths Syllabus 2024-25: Download Revised PDF for ISC Class 11 Maths

Isc class 11 maths syllabus 2024–25: download the revised isc class 11 maths syllabus for the academic session 2024–25 here. check out the complete revised syllabus to know the list of topics and exam pattern prescribed by cicse for isc class 11 maths..

ISC Class 11 Maths Syllabus 2024-25: The ISC Board has released the 2024–25 ISC Class 11 Maths syllabus. This article provides students with direct access to the revised Maths syllabus for class 11. The curriculum and syllabus for ISC Class 11 Maths are available for students to see and download here. 860, is the subject code for the same. The syllabus is divided into three sections A, B and C. Section A carries the highest weightage followed by section B and C. Students can make choice between section B OR C, for attempting questions. The syllabus is thoughtfully created, with appropriate topic elaboration. There are two components to the ISC class 11 Maths paper for 2024–2025: theory and project work. To know the entire syllabus, read the entire article: 

ISC Class 11 Maths Syllabus Key Highlights

Isc class 11 maths syllabus 2024-25, paper i: theory - 80 marks .

The syllabus is divided into three sections A, B and C. Section A is compulsory for all candidates. Candidates will have a choice of attempting questions from EITHER Section B OR Section C.

PAPER II: PROJECT WORK - 20 Marks  

List of suggested assignments for project work: .

• Using a Venn diagram, find the number of subsets of a given set and verify that if a set has ‘n’ number of elements, the total number of subsets is 2n
• Verify that for two sets A and B, n(A × B) = pq, where n(A) = p and n(B)= q, the total number of relations from A to B is 2pq.
• Using Venn diagram, verify the distributive law for three given non-empty sets A, B and C. 
• Identify distinction between a relation and a function with suitable examples and illustrate graphically.
• Establish the relationship between the measure of an angle in degrees and in radians with suitable examples by drawing a rough sketch. 
• Illustrate with the help of a model, the values of sine and cosine functions for different angles which are multiples of π/2 and π.
• Draw the graphs of sin x, sin 2x, 2 sin x, and sin x/2 on the same graph using same coordinate axes and interpret the same.
• Draw the graph of cos x, cos 2x, 2 cos x, and cos x/2 on the same graph using same coordinate axes and interpret the same.
• Using argand plane, interpret geometrically, the meaning of and its integral powers.
• Draw the graph of quadratic function. From the graph find maximum/minimum value of the function. Also determine the sign of the expression.
• Construct a Pascal’s triangle to write a binomial expansion for a given positive integral exponent. 
• Obtain a formula for the sum of the squares/sum of cubes of ’n’ natural numbers.
• Obtain the equation of the straight line in the normal form, for (the angle between the perpendicular to the line from the origin and the x-axis) for each of the following, on the same graph: α < 90°, 90° < α < 180°, 180° < α < 270°,  270° < α < 360°
• Identify the variability and consistency of two sets of statistical data using the concept of coefficient     of variation.
• Construct the tree structure of the outcomes of a random experiment, when elementary events are  not equally likely. Also construct a sample space by taking a suitable example.
• Construct different types of conics by PowerPoint Presentation, or by making a model, using the concept of double cone and a plane.
• Use focal property of ellipse to construct ellipse.
• Use focal property of hyperbola to construct hyperbola.
• Write geometrical significance of X coordinate, Y coordinate, and Z coordinate in space. Using the above, find the distance of the point in space from x-axis/y-axis/z-axis. Explain the above using a three-dimensional model/ power point presentation.
• Obtain truth values of compound statements of the type by using switch connection in series.
• Obtain truth values of compound statements of the type by using switch connection in parallel.
• Explain the statistical significance of percentile and draw inferences of percentile for a given data.
• Find median from the point of intersection of cumulative frequency curves (less than and more than cumulative frequency curves).
• Describe the limitations of Spearman’s rank correlation coefficient and illustrate with suitable examples.
• Identify the purchasing power using the concept of cost of living index number.
• Identify the purchasing power using the concept of weighted aggregate price index number. 
• Calculate moving averages with the given even Periodicity. Plot them and as well as the original data on the same graph.

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• HPBOSE Class 12th Result 2024
• HPBOSE 12th Result 2024
• HP Board 12th Result 2024 Roll Number
• HPBOSE 12th Toppers List 2024
• एमपी बोर्ड 10 वीं टॉपर लिस्ट 2024
• AP SSC Results 2024 Manabadi by Jagran Josh
• AP SSC Topper List 2024

Latest Education News

JAC 12th Result 2024 Live: झारखंड बोर्ड 12वीं का रिजल्ट कल सुबह 11 बजे आएगा, रोल नंबर से jac.jharhand.gov.in पर चेक कर सकेंगे छात्र

TS SSC Results 2024 LIVE Updates: Manabadi BSE Telangana 10th Result Tomorrow at 11 AM, Check Website Link

[डायरेक्ट लिंक] hpbose.org, hpbose.org Result.aspx 12th Result 2024 OUT: जारी हुए हिमाचल बोर्ड 12वीं के नतीजे, ये रहा डायरेक्ट डाउनलोड लिंक?

Jharkhand Board 12th Result 2024 Date and Time: जारी हुई जैक 12वीं रिजल्ट की तारीखें, यहाँ देखें परिणाम से जुड़ी सभी जरुरी बातें

JAC 12th Result 2024 LIVE: Jharkhand Board Class 12 Results to be Announced Tomorrow at 11 AM, Check Online by Roll Number

Viral Optical Illusion: Help Tom Find His Gift Box Hidden In This Birthday Party Picture. 7 Seconds Left!

[Official] JAC Board 12th Result 2024 Date and Time Notification Released: Check Notice for Jharkhand Class 12 Results Here

JAC Result 2024 Class 12: Jharkhand Board Class 12th Result Date And Time at jacresults.com

UK Board Result 2024: Uttrakhand Board UBSE Class 10, 12 Result Date And Time at uaresults.nic.in

UK Board Result 2024 Class 12: Uttrakhand Board Class 12th Result Date And Time at ubse.uk.gov.in

UK Board Result 2024 Class 10: Uttrakhand Board Class 10th Result Date And Time at ubse.uk.gov.in

Picture Puzzle IQ Test: You Have 20/20 Vision If You Can Spot The Headphones In 8 Seconds!

HPBOSE 12th Result 2024 Out: एचपीबोस 12वीं परिणाम टर्म 2 hpbose.org पर जारी, 73.76 फीसदी पास, इस Direct Link से करें चेक

Ayushman Bharat Diwas 2024: Date, Theme and What are the Benefits?

[Official] PSEB Class 8th, 12th Result 2024 Date and Time Announced, Check Official Notice and Link will active on 1st May

JAC Board Result 2024: Check झारखंड बोर्ड रिजल्ट at Jagran Josh, jacresults.com and jac.jharkhand.gov.in

JAC 12th Science Result 2024: Jharkhand Board Class 12 Science Result Date And Time at jac.nic.in

JAC 12 Commerce Result 2024:Jharkhand Board Class 12 Commerce Result Date And Time at jacresults.com

JAC 12 Arts Result 2024:Jharkhand Board Class 12 Arts Result Date And Time at jacresults.com

TS ICET 2024 Application Without Late Fee Ends Tomorrow: April 30, Check Steps To Apply, Exam Schedule