RD Sharma Solutions for CBSE Class 10 Mathematics chapter 3 - Pairs of Linear Equations in Two Variables

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Chapter 3 - Pairs of Linear Equations in Two Variables Excercise Ex. 3.1

Question 1
Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel and play Hoopla (a game in which you throw a rig on the items kept in the stall, and if the ring covers any object completely you get it). The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. Each ride costs Rs 3, and a game of Hoopla costs Rs 4. If she spent Rs 20 in the fair, represent this situation algebraically and graphically.
Solution 1
Question 2
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' this interesting?) Represent this situation algebraically and graphically.
Solution 2
Let the present age of Aftab and his daughter be x and y respectively.

Seven years ago,
Age of Aftab = x - 7
Age of his daughter = y - 7

According to the given condition,

 
Three years hence,
Age of Aftab = x + 3
Age of his daughter = y + 3

According to the given condition,

 

Thus, the given conditions can be algebraically represented as:
x - 7y = -42
x - 3y = 6

 
Three solutions of this equation can be written in a table as follows:
x -7 0 7
y 5 6 7


 

Three solutions of this equation can be written in a table as follows:
x 6 3 0
y 0 -1 -2


The graphical representation is as follows:



Concept insight: In order to represent a given situation mathematically, first see what we need to find out in the problem. Here, Aftab and his daughter's present age needs to be found so, so the ages will be represented by variables x and y. The problem talks about their ages seven years ago and three years from now. Here, the words 'seven years ago' means we have to subtract 7 from their present ages, and 'three years from now' or 'three years hence' means we have to add 3 to their present ages. Remember in order to represent the algebraic equations graphically the solution set of equations must be taken as whole numbers only for the accuracy. Graph of the two linear equations will be represented by a straight line.
Question 3
Solution 3

Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
Solution 10
Question 11
The cost of 2 kg of apples and 1 kg 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.
Solution 11
Let the cost of 1 kg of apples and 1 kg grapes be Rsx and Rsy.
The given conditions can be algebraically represented as:


Three solutions of this equation can be written in a table as follows:


x
50 60 70
y 60 40 20




Three solutions of this equation can be written in a table as follows:


x
70 80 75
y 10
-10
0


The graphical representation is as follows:



Concept insight: cost of apples and grapes needs to be found so the cost of 1 kg apples and 1 kg grapes will be taken as the variables. From the given conditions of collective cost of apples and grapes, a pair of linear equations in two variables will be obtained. Then, in order to represent the obtained equations graphically, take the values of variables as whole numbers only. Since these values are large so take the suitable scale.

Chapter 3 - Pairs of Linear Equations in Two Variables Excercise Ex. 3.2

Question 1
Solution 1

Question 2
Solution 2

Question 3
Solution 3




Question 4
Solution 4

Question 5
Solution 5

Question 6
Solution 6




Question 7
Solution 7

 

Question 8
Solution 8

Question 9
Solution 9

Question 10
Solution 10

Question 11
Solution 11

Question 12
Solution 12

Since, the graph of the two lines coincide, the given system of equations have infinitely many solutions.
Question 13
Solution 13


Question 14

Solution 14

Question 15
Show graphically that each one of the following systems of equations is in-coinsistent (i.e. has no solution):

3x - 5y = 20

6x - 10y = - 40
Solution 15

Question 16
Solution 16

Question 17
Solution 17

Question 18
Solution 18




Question 19
Solution 19




Question 20
Solution 20




Question 21

Solution 21




Question 22
Solution 22

Question 23
Solution 23

Question 24
Solution 24




Question 25
Solution 25




Question 26
Solve graphically each of the following systems of linear equations. Also find the coordinates of the points where the lines meet axis of y.

2x + y - 11 = 0

x - y - 1 = 0
Solution 26




Question 27
Solve graphically each of the following systems of linear equations. Also find the coordinates of the points where the lines meet axis of y.

x + 2y - 7 = 0

2x - y - 4 = 0
Solution 27

Question 28
Solution 28




Question 29
Solution 29




Question 30

Solution 30





Question 31

Solution 31





Question 32

Solution 32





Question 33

Solution 33

Question 34

Solution 34

 
Question 35

Solution 35




Question 36

Solution 36



Question 37

Solution 37

Question 38

Solution 38





Question 39

Solution 39





Question 40

Solution 40





Question 41

Solution 41



Question 42

Solution 42

Question 43

Solution 43






Question 44

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. Calculate the area of the triangle so formed.

Solution 44



Three solutions of this equation can be written in a table as follows:

x 0 1 2
y -5 0 5


x 0 1 2
y -3 0 3


The graphical representation of the two lines will be as follows:

 



It can be observed that the required triangle is ABC.
The coordinates of its vertices are A (1, 0), B (0, -3), C (0, -5).

Area space of space Triangle space increment ABC space equals space 1 half cross times BC cross times AO
equals 1 half cross times 2 cross times 1
equals 1 space sq. space unit

Concept insight: In order to find the coordinates of the vertices of the triangle so formed, find the points where the two lines intersects the y-axis and also where the two lines intersect each other. Here, note that the coordinates of the intersection of lines with y-axis is taken and not with x-axis, this is because the question says to find the triangle formed by the two lines and the y-axis.

Question 45

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

(iii) Champa went to a 'Sale' to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, "The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased". Help her friends to find how many pants and skirts Champa a bought.

Solution 45

(i) Let the number of girls and boys in the class be x and y respectively.
According to the given conditions, we have:
x + y = 10
x - y = 4
x + y = 10  x = 10 - y
Three solutions of this equation can be written in a table as follows:

x 4 5 6
y 6 5 4



x - y = 4 x = 4 + y
Three solutions of this equation can be written in a table as follows:

x 5 4 3
y 1 0 -1



The graphical representation is as follows:

From the graph, it can be observed that the two lines intersect each other at the point (7, 3).
So, x = 7 and y = 3.

Thus, the number of girls and boys in the class are 7 and 3 respectively.


(ii)    Let the cost of one pencil and one pen be Rs x and Rs y respectively.

According to the given conditions, we have:
5x + 7y = 50
7x + 5y = 46

Three solutions of this equation can be written in a table as follows:

x 3 10 -4
y 5 0 10




Three solutions of this equation can be written in a table as follows:

x 8 3 -2
y -2 5 12



The graphical representation is as follows:


From the graph, it can be observed that the two lines intersect each other at the point (3, 5).
So, x = 3 and y = 5.

Therefore, the cost of one pencil and one pen are Rs 3 and Rs 5 respectively.



(iii)

Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are:

y = 2x - 2 ...(1)

and y = 4x - 4 ...(2)

Let us draw the graphs of Equations (1) and (2) by finding two solutions for each of the equations.

They are given in Table

x 2 0
y = 2x - 2 2 -2



x 0 1
y = 4x - 4 -4 0



Plot the points and draw the lines passing through them to represent the equations, as shown in fig.,

The two lines intersect at the point (1,0). So, x = 1, y = 0 is the required solution of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did not buy any skirt.

Concept insight: Read the question carefully and examine what are the unknowns. Represent the given conditions with the help of equations by taking the unknowns quantities as variables. Also carefully state the variables as whole solution is based on it. On the graph paper, mark the points accurately and neatly using a sharp pencil. Also, take at least three points satisfying the two equations in order to obtain the correct straight line of the equation. Since joining any two points gives a straight line and if one of the points is computed incorrect will give a wrong line and taking third point will give a correct line. The point where the two straight lines will intersect will give the values of the two variables, i.e., the solution of the two linear equations. State the solution point.

Question 46

Solution 46





Question 47

Solution 47





Question 48

Solution 48

Question 49

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

Solution 49

(i) For the two lines a1x + b1x + c1 = 0 and a2x + b2x + c2 = 0, to be intersecting, we must have

So, the other linear equation can be 5x + 6y - 16 = 0
 
(ii)  For the two lines a1x + b1x + c1 = 0 and a2x + b2x + c2 = 0, to be parallel, we must have

So, the other linear equation can be 6x + 9y + 24 = 0,

(iii)  For the two lines a1x + b1x + c1 = 0 and a2x + b2x + c2 = 0 to be coincident, we must have

So, the other linear equation can be 8x + 12y - 32 = 0,


Concept insight: In order to answer such type of problems, just remember the conditions for two lines to be intersecting, parallel, and coincident. This problem will have multiple answers as their can be many equations satisfying the required conditions.

Question 50

Solution 50

Question 51

Solution 51

Question 52

Graphically, solve the following pair of equations:

 

2x + y = 6

2x - y + 2 = 0

Find the ratio of the areas of the two triangles formed by the lines representing these equations with the x-axis and the lines with the y-axis.

Solution 52

The lines AB and CD intersect at point R(1, 4). Hence, the solution of the given pair of linear equations is x = 1, y = 4.

 

From R, draw RM X-axis and RN Y-axis.

Then, from graph, we have

RM = 4 units, RN = 1 unit, AP = 4 units, BQ = 4 units

 

Question 53

Determine, graphically, the vertices of the triangle formed by the lines y = x, 3y = x, x + y = 8.

Solution 53

From the graph, the vertices of the triangle AOP formed by the given lines are A(4, 4), O(0, 0) and P(6, 2).

Question 54

Draw the graph of the equations x = 3, x = 5 and 2x - y - 4 = 0. Also, find the area of the quadrilateral formed by the lines and the x-axis.

Solution 54

The graph of x = 3 is a straight line parallel to Y-axis at a distance of 3 units to the right of Y-axis.

 

The graph of x = 5 is a straight line parallel to Y-axis at a distance of 5 units to the right of Y-axis.

 

Question 55

Draw the graphs of the lines x = -2, and y = 3. Write the vertices of the figure formed by these lines, the x-axis and the y-axis. Also, find the area of the figure.

Solution 55

The graph of x = -2 is a straight line parallel to Y-axis at a distance of 2 units to the left of Y-axis.

 

The graph of y = 3 is a straight line parallel to X-axis at a distance of 3 units above X-axis.

Question 56

Draw the graphs of the pair of linear equations x - y + 2 = 0 and 4x - y - 4 = 0. Calculate the area of the triangle formed by the lines so drawn and the x-axis.

Solution 56

  

Chapter 3 - Pairs of Linear Equations in Two Variables Excercise Ex. 3.3

Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3

Question 4

Solution 4

Question 5

Solution 5
Question 6
Solution 6
Question 7
Solution 7

Question 8

Solution 8
Question 9

Solution 9

Question 10
Solution 10

Question 11

Solution 11

Question 12
Solution 12

Question 13
Solution 13

Question 14
Solution 14
Question 15
Solution 15

Question 16
Solution 16
Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solve the following systems of equation:

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solve the pair of equations:

Solution 39

Question 40

Solution 40

fraction numerator 10 over denominator straight x plus straight y end fraction plus fraction numerator 2 over denominator straight x minus straight y end fraction equals 4
fraction numerator 15 over denominator straight x plus straight y end fraction minus fraction numerator 9 over denominator straight x minus straight y end fraction equals negative 2

Let space fraction numerator 1 over denominator straight x plus straight y end fraction equals straight p space and space fraction numerator 1 over denominator straight x minus straight y end fraction equals straight q
10 straight p plus 2 straight q minus 4 equals 0....... left parenthesis straight i right parenthesis
15 straight p minus 9 straight q plus 2 equals 0........ left parenthesis ii right parenthesis

Using space cross minus multiplication space method comma space we space obtain colon
fraction numerator straight p over denominator 4 minus 36 end fraction equals fraction numerator straight q over denominator negative 60 minus 20 end fraction equals fraction numerator 1 over denominator negative 90 minus 30 end fraction
fraction numerator straight p over denominator negative 32 end fraction equals fraction numerator straight q over denominator negative 80 end fraction equals fraction numerator 1 over denominator negative 120 end fraction
fraction numerator straight p over denominator negative 4 end fraction equals fraction numerator straight q over denominator negative 10 end fraction equals fraction numerator 1 over denominator negative 15 end fraction
straight p equals 4 over 15 comma straight q equals 10 over 15 equals 2 over 3

Substituting space the space values space of space straight p space and space straight q comma
straight x plus straight y equals 15 over 4....... left parenthesis straight i right parenthesis space
straight x minus straight y equals 3 over 2...... left parenthesis ii right parenthesis
left parenthesis straight i right parenthesis plus left parenthesis ii right parenthesis rightwards double arrow
2 straight x equals 21 over 4
straight x equals 21 over 8 comma straight y equals 9 over 8

Question 41

Solution 41

Question 42

Solution 42

Question 43

152 x space minus space 378 y space equals space minus 74 space space
minus 378 x space plus space 152 y space equals space minus 604

Solution 43

152 straight x minus 378 straight y equals negative 74....... left parenthesis straight i right parenthesis
minus 378 straight x plus 152 straight y equals negative 604....... left parenthesis ii right parenthesis

Adding space the space equations space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis comma space we space obtain colon
minus 226 straight x minus 226 straight y equals negative 678
rightwards double arrow straight x plus straight y equals 3........... left parenthesis 3 right parenthesis

Subtracting space the space equation space left parenthesis 2 right parenthesis space from space equation space left parenthesis 1 right parenthesis comma space we space obtain colon
530 straight x minus 530 straight y equals 530
rightwards double arrow straight x minus straight y equals 1........... left parenthesis 4 right parenthesis

Adding space equations space left parenthesis 3 right parenthesis space and space left parenthesis 4 right parenthesis comma space we space obtain colon
straight x equals 2
Substituting space straight x equals 2 space in space equation space left parenthesis 3 right parenthesis comma space we space get
straight y equals 1

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solve the following systems of equation:

 

21x + 47y = 110

47x + 21y = 162

Solution 48

Question 49

If x + 1 is a factor of 2x3 + ax2 + 2bx + 1, the find the values of a and b given that 2a - 3b = 4.

Solution 49

Question 50

Find the solution of the pair of equations   and  . Hence, find λ, if y = λx + 5.

Solution 50

Question 51

Find the values of x and y in the following rectangle.

 

Solution 51

Question 52

Write an equation of a line passing through the point representing solution of the pair of linear equations x + y = 2 and 2x - y = 1. How many such lines can we find?

Solution 52

Question 53

Write a pair of linear equations which has the unique solution x = -1, y = 3. How many such pairs can you write?

Solution 53

Chapter 3 - Pairs of Linear Equations in Two Variables Excercise Ex. 3.4

Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4

Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solve each of the following systems of equations by the method of cross-multiplication:

Solution 14

Question 15
Solution 15
Question 16
Solution 16
Question 17
Solution 17
Question 18
Solution 18
Question 19
Solution 19
Question 20
Solution 20

 

Question 21
Solution 21
Question 22

Solution 22

Question 23
Solution 23

Question 24
Solution 24

Question 25
Solution 25

Question 26

Solution 26

Question 27
Solution 27

Question 28

Solution 28

Chapter 3 - Pairs of Linear Equations in Two Variables Excercise Ex. 3.5

Question 1

Determine whether the following system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:

x - 3y = 3

3x - 9y = 2

Solution 1
Question 2

Determine whether the following system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:

2x + y = 5

4x + 2y = 10

Solution 2
Question 3

Determine whether the following system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:

3x - 5y = 20

6x - 10y = 40

Solution 3

Question 4

Determine whether the following system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:

x - 2y = 8

5x - 10y = 10

Solution 4
Question 5

Find the value of k for which the following system of equations has a unique solution:

kx + 2y = 5

3x + y = 1

Solution 5
Question 6
Find the value of k for which the following system of equations has a unique solution:

4x + ky + 8 = 0

2x + 2y + 2 = 0
Solution 6

Question 7

Find the value of k for which the following system of equations has a unique solution:

4x - 5y = k

2x - 3y = 12

Solution 7
Question 8

Find the value of k for which the following system of equations has a unique solution:

x + 2y = 3

5x + ky + 7 = 0

Solution 8
Question 9

Find the value of k for which the following systems of equations have infinitely many solutions:

2x + 3y - 5 = 0

6x + ky - 15 = 0

Solution 9
Question 10

Find the value of k for which the following systems of equations have infinitely many solutions:

4x + 5y = 3

kx + 15y = 9

Solution 10
Question 11

Find the value of k for which the following systems of equations have infinitely many solutions:

kx - 2y + 6 = 0

4x - 3y + 9 = 0

Solution 11
Question 12

Find the value of k for which the following systems of equations have infinitely many solutions:

8x + 5y = 9

kx + 10y = 18

Solution 12
Question 13

Find the value of k for which the following systems of equations have infinitely many solutions:

2x - 3y = 7

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

Solution 13
Question 14

Find the value of k for which the following systems of equations have infinitely many solutions:

2x + 3y = 2

(k + 2)x + (2k + 1)y = 2(k - 1)

Solution 14
Question 15

Find the value of k for which the following systems of equations have infinitely many solutions:

x + (k + 1)y = 4

(k + 1)x + 9y = (5k + 2)

Solution 15
Question 16

Find the value of k for which the following systems of equations have infinitely many solutions:

kx + 3y = 2k + 1

2(k + 1)x + 9y = 7k + 1

Solution 16
Question 17

Find the value of k for which the following systems of equations have infinitely many solutions:

2x + (k - 2)y = k

6x + (2k - 1)y = 2k + 5

Solution 17

Question 18

Find the value of k for which the following systems of equations have infinitely many solutions:

2x + 3y = 7

(k + 1)x + (2k - 1)y = 4k + 1

Solution 18

Question 19

Find the value of k for which the following systems of equations have infinitely many solutions:

2x + 3y = k

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

Solution 19

Question 20
Find the value of k for which the following system of equations has no solution:

kx - 5y = 2

6x + 2y = 7
Solution 20

Question 21
Solution 21
Question 22
Solution 22
Question 23
Solution 23
Question 24
Solution 24
Question 25

Find he value of k for which of the following system of equation has no solution:

 

kx + 3y = k - 3

12x + ky = 6

Solution 25

Question 26
Solution 26
Question 27
Solution 27
Question 28
Solution 28

Question 29
Solution 29
Question 30
Solution 30

Question 31
For what value of k, the following system of equations will represent the coincident lines?

x + 2y + 7 = 0

2x + ky + 14 = 0
Solution 31
Question 32
Solution 32
Question 33
Solution 33

Question 34
Solution 34

Question 35

Solution 35
Question 36
Solution 36
Question 37
Solution 37
Question 38
Solution 38
Question 39
Solution 39

Question 40

Find the values of a and b for which the following system of equations has infinitely many solutions:

2x + 3y = 7

(a - b) x + (a + b)y = 3a + b - 2

Solution 40

2 straight x plus 3 straight y minus 7 equals 0
left parenthesis straight a minus straight b right parenthesis straight x plus left parenthesis straight a minus straight b right parenthesis straight y minus left parenthesis 3 straight a plus straight b minus 2 right parenthesis equals 0
Here comma
straight a subscript 1 equals 2 comma space straight b subscript 1 equals 3 comma straight c subscript 1 equals negative 7
straight a subscript 2 equals left parenthesis straight a minus straight b right parenthesis comma straight b subscript 2 equals left parenthesis straight a plus straight b right parenthesis comma straight c subscript 2 equals negative left parenthesis 3 straight a plus straight b minus 2 right parenthesis

straight a subscript 1 over straight a subscript 2 equals fraction numerator 2 over denominator left parenthesis straight a minus straight b right parenthesis end fraction comma space straight b subscript 1 over straight b subscript 2 equals fraction numerator 3 over denominator left parenthesis straight a plus straight b right parenthesis end fraction comma space straight c subscript 1 over straight c subscript 2 equals fraction numerator negative 7 over denominator negative left parenthesis 3 straight a plus straight b minus 2 right parenthesis end fraction equals fraction numerator 7 over denominator left parenthesis 3 straight a plus straight b minus 2 right parenthesis end fraction
For space the space equations space to space have space infinitely space many space solutions comma space we space have
straight a subscript 1 over straight a subscript 2 equals space straight b subscript 1 over straight b subscript 2 equals straight c subscript 1 over straight c subscript 2
fraction numerator 2 over denominator left parenthesis straight a minus straight b right parenthesis end fraction equals fraction numerator 7 over denominator left parenthesis 3 straight a plus straight b minus 2 right parenthesis end fraction
6 straight a plus 2 straight b minus 4 equals 7 straight a minus 7 straight b
straight a minus 9 straight b equals negative 4........... left parenthesis straight i right parenthesis

fraction numerator 2 over denominator left parenthesis straight a minus straight b right parenthesis end fraction equals fraction numerator 3 over denominator left parenthesis straight a plus straight b right parenthesis end fraction
2 straight a plus 2 straight b equals 3 straight a minus 3 straight b
straight a minus 5 straight b equals 0............ left parenthesis ii right parenthesis
Subtracting space left parenthesis straight i right parenthesis space from space left parenthesis ii right parenthesis comma space we space get
4 straight b equals 4
straight b equals 1
Substituting space the space value space of space straight b space in space equation space left parenthesis 2 right parenthesis comma space we space obtain colon
straight a space minus space 5 cross times 1 equals 0
straight a equals 5
Thus space the space values space of space straight a space and space straight b space are space 5 space and space 1 space respectively.

Question 41
Solution 41

Question 42
Solution 42

Question 43

Find the values of a and b for which the following system of equations has infinitely many solutions:

 

x + 2y = 1

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

Solution 43

Question 44

Find the values of a and b for which the following system of equations has infinitely many solutions:

 

2x + 3y = 7

2ax + ay = 28 - by

Solution 44

Question 45

For which value(s) of λ, do the pair of linear equations λx + y = λ2 and x + λy = 1 have no solution?

Solution 45

Question 46

For which value(s) of λ, do the pair of linear equations λx + y = λ2 and x + λy = 1 have infinitely many solutions?

Solution 46

Question 47

For which value(s) of λ, do the pair of linear equations λx + y = λ2 and x + λy = 1 have a unique solution?

Solution 47

Chapter 3 - Pairs of Linear Equations in Two Variables Excercise Ex. 3.6

Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7

Jamila sold a table and a chair for Rs.1050, thereby making a profit of 10% on a table and 25% on the chair. If she had taken profit of 25% on the table and 10% on the chair she would have got Rs.1065. Find the cost price of each.

Solution 7

Question 8

Susan invested certain amount of money in two schemes A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively. She received Rs.1860 as annual interest. However, had she interchanged the amount of investment in the two schemes, she would have received Rs.20 more as annual interest. How much money did she invest in each scheme?

Solution 8

Question 9

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

Solution 9

Let space the space cost space of space straight a space bat space be space straight x space and space straight y space respectively.
According space to space the space given space information comma
7 straight x plus 6 straight y equals 3800....... left parenthesis 1 right parenthesis
3 straight x plus 5 straight y equals 1750....... left parenthesis 2 right parenthesis
From space left parenthesis 1 right parenthesis comma space we space obtain comma
straight y equals fraction numerator 3800 minus 7 straight x over denominator 6 end fraction........ left parenthesis 3 right parenthesis
Substituting space this space value space in space equation space left parenthesis 2 right parenthesis comma space we space obtain
3 straight x plus 5 open parentheses fraction numerator 3800 minus 7 straight x over denominator 6 end fraction close parentheses equals 1750
3 straight x plus fraction numerator 19000 minus 35 straight x over denominator 6 end fraction equals 1750
3 straight x minus fraction numerator 35 straight x over denominator 6 end fraction equals 1750 minus 19000 over 6
fraction numerator 18 straight x minus 35 straight x over denominator 6 end fraction equals fraction numerator 10500 minus 19000 over denominator 6 end fraction
fraction numerator 17 straight x over denominator 6 end fraction equals 8500 over 6
straight x equals 500........ left parenthesis 4 right parenthesis
Substituting space this space equation left parenthesis 3 right parenthesis comma space we space obtain comma
straight y equals fraction numerator 3800 minus 7 cross times 500 over denominator 6 end fraction
equals 300 over 6 equals 50
Hence comma space the space cost space of space straight a space bat space is space Rs space 500 space and space that space of space straight a space ball space is space Rs space 50.

bold Concept bold space bold Insight colon space Cost space of space bats space and space balls space need space to space be space found space so space the space cost space of space straight a space ball
and space bat space will space be space taken space as space the space variables. space Apply space the space conditions space of space total space cost space of space
bats space and space balls space algebraic space equations space will space be space obtained. space The space pair space of space equations space
can space then space be space solved space by space suitable space substitution.

Question 10

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.

Solution 10

Question 11

The cost of 4 pens and 4 pencils boxes is Rs.100. Three times the cost of a pen is Rs.15 more than the cost of a pencil box. Form the pair of linear equations for the above situation. Find the cost of a pen and a pencil box.

Solution 11

Question 12

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?

Solution 12

Question 13

A and B each have a certain number of mangoes. A says to B, "if you give 30 of your mangoes, I will have twice as many as left with you. "B replies, "if you give me 10, I will have thrice as many as left with you. "How many mangoes does each have?

Solution 13

Question 14

Vijay had some bananas, and he divided them into two lots A and B. He sold first lot at the rate of Rs.2 for 3 bananas and the second lot at the rate of Rs.1 per banana and got a total of Rs.400. If he had sold the first lot at the rate of Rs.1 per banana and the second lot at the rate of Rs.4 per five bananas, his total collection would have been Rs.460. Find the total number of bananas he had.

Solution 14

Question 15

Solution 15

Chapter 3 - Pairs of Linear Equations in Two Variables Excercise Ex. 3.7

Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
The sum of two-digit number and the number formed by reversing the order of digits is 66. If the two digits differ by 2, find the number. How many such numbers are there?
Solution 5

Question 6

The space sum space of space two space numbers space is space 1000 space and space the space difference space between space their space squares space is space 256000.
Find space the space numbers.

Solution 6

Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
Solution 10
Question 11
Solution 11
Question 12
Solution 12
Question 13
Solution 13

Question 14
The sum of 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.
Solution 14
Question 15
Solution 15
Question 16

Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5. Find the numbers.

Solution 16

Question 17

A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. Find the number.

Solution 17

Chapter 3 - Pairs of Linear Equations in Two Variables Excercise Ex. 3.8

Question 1
Solution 1
Question 2
A fraction becomes 9/11 if 2 is added to both numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.
Solution 2

Question 3
Solution 3
Question 4
If we add 1 to the numerator and subtract 1 from the denominator, a fraction becomes 1. It also becomes 1/2 if we only add 1 to the denominator. What is the fraction?
Solution 4
Question 5

The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes 1/2. Find the fraction.

Solution 5

Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10

Solution 10

Question 11

Solution 11

Chapter 3 - Pairs of Linear Equations in Two Variables Excercise Ex. 3.9

Question 1
Solution 1
Question 2
Solution 2
Question 3

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?

Solution 3

Question 4
Solution 4
Question 5
Solution 5
Question 6
The present age of a father is three more than three times the age of the son. Three years hence father's age will be 10 years more than twice the age of the son. Determine their present ages.
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9

Question 10

Solution 10

Question 11

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 differs by 30 years. Find the ages of Ani and Biju.

Solution 11

The difference between the ages of Ani and Biju is given as 3 years. So, either Biju is 3 years older than Ani or Ani is 3 years older than Biju.

Let the age of Ani and Biju be x years and y years respectively.
Age of Dharam = 2 × x = 2x years

Case I: Ani is older than Biju by 3 years
x - y = 3        ... (1)

begin mathsize 12px style 2 straight x minus straight y over 2 equals 30 end style
4x - y = 60     ....(2)

Subtracting (1) from (2), we obtain:
 
3x = 60 - 3 = 57

begin mathsize 12px style straight x equals 57 over 3 equals 19 end style
Age of Ani = 19 years
Age of Biju = 19 - 3 = 16 years

Case II: Biju is older than Ani by 3 years
y - x = 3        ... (3)

begin mathsize 12px style 2 straight x minus straight y over 2 equals 30 end style

4x - y = 60        ... (4)

Adding (3) and (4), we obtain:
3x = 63
x = 21

Age of Ani = 21 years
Age of Biju = 21 + 3 = 24 years

Concept Insight: In this problem, ages of Ani and Biju are the unknown quantities. So, we represent them by variables x and y. Now, note that here it is given that the ages of Ani and Biju differ by 3 years. So, it is not mentioned that which one is older. So, the most important point in this question is to consider both cases  Ani is older than Biju and  Biju is older than Ani. For second condition the relation  on the ages of Dharam and Cathy can be implemented . Pair of linear equations can be solved using a suitable algebraic method.

Question 12

Two years ago, Salim was thrice as old as his daughter and six years later, he will be four years older than twice her age. How old are they now?

Solution 12

Question 13

The age of the father is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father.

Solution 13

Chapter 3 - Pairs of Linear Equations in Two Variables Excercise Ex. 3.10

Question 1
Solution 1
Question 2
Solution 2
Question 3
The boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of stream and that of the boat in still water.
Solution 3

Question 4

Solution 4
Question 5

Solution 5

Question 6

A person rowing at the rate of 5 km/h in still water, takes thrice as much time in going 40 km upstream as in going 40 km downstream. Find the speed of stream.

Solution 6

Question 7

Solution 7

Question 8
Solution 8

Question 9
Solution 9

Question 10
Solution 10

 

 

Question 11
Solution 11

Question 12
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.
Solution 12
Question 13

A motor boat can travel 30 km upstream and 28 km downstream in 7 hours. It can travel 21 km upstream and return in 5 hours. Find the speed of the boat in still water and the speed of the stream.

Solution 13

Question 14
Solution 14

Question 15
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.
Solution 15
Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km.
Or, d = xt        ... (1)

According to the question,
 



By using equation (1), we obtain:
3x - 10t = 30        ... (3)

Adding equations (2) and (3), we obtain:
x = 50
Substituting the value of x in equation (2), we obtain:
(-2) x (50) + 10t = 20
-100 + 10t = 20
10t = 120
t = 12
From equation (1), we obtain:
d = xt = 50 x 12 = 600

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

Concept insight: To solve this problem, it is very important to remember the relation . Now, all these three quantities are unknown. So, we will represent these

by three different
variables. By using the given conditions, a pair of equations will be obtained. Mind one thing that the equations obtained will not be linear. But they can be reduced to linear form by using the fact that . Then two linear equations can be formed which can

be solved easily by elimination method.

Question 16
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 hours. What are the speeds of two cars?
Solution 16

Question 17

Solution 17

Question 18

Solution 18

Chapter 3 - Pairs of Linear Equations in Two Variables Excercise Ex. 3.11

Question 1
Solution 1

Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6

ABCD is a cyclic quadrilateral such that A = (4y + 20)o, B = (3y - 5)o, C = (-4x)o and D = (7x + 5)o. Find the four angles.

Solution 6

We know that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°.

A + C = 180
4y + 20 - 4x = 180
-4x + 4y = 160
x - y = -40            ... (1)
Also, B + D = 180
3y - 5 - 7x + 5 = 180
-7x + 3y = 180        ... (2)
Multiplying equation (1) by 3, we obtain:

3x - 3y = -120        ... (3)
Adding equations (2) and (3), we obtain:
-4x = 60
x = -15
Substituting the value of x in equation (1), we obtain:
-15 - y = -40
y = -15 + 40 = 25

A = 4y + 20 = 4(25) + 20 = 120o
B = 3y - 5 = 3(25) - 5 = 70o
C = -4x = -4(-15) = 60o
D = -7x + 5 = -7(-15) + 5 = 110o

Concept insight: The most important idea to solve this problem is by using the fact that the sum of the measures of opposite angles in a cyclic quadrilateral is 180o. By using this relation, two linear equations can be obtained which can be solved easily by eliminating a suitable variable.

Question 7
Solution 7

Question 8
Solution 8

Question 9

Solution 9
Question 10

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

Solution 10

Let space the space number space of space right space answers space and space wrong space answers space be space straight x space and space straight y space respectively.
According space to space the space question comma
3 straight x minus straight y equals 40......... left parenthesis 1 right parenthesis
4 straight x minus 2 straight y equals 50
rightwards double arrow 2 straight x minus straight y equals 25......... left parenthesis 2 right parenthesis
Subtracting space equation space left parenthesis 2 right parenthesis space from space equation space left parenthesis 1 right parenthesis comma space we space obtian colon
straight x equals 15
Substituting space the space value space of space straight x space in space equation space left parenthesis 2 right parenthesis comma space we space obtain colon
30 minus straight y equals 25
straight y equals 5

Thus comma space the space number space of space right space answers space and space the space number space of space wrong space answers space is space 15
and space 5 space respectively. space Therefore space the space total space number space of space questions space is space 20.

Concept space insight colon space In space this space problem comma space the space number space of space write space answers space and space the space
number space of space wrong space answers space answered space by space Yash space are space the space unknown space variable space straight y
has space the space same space coefficient space in space both space the space equations comma space so space it space will space be space easier space to space find space the
solution space by space eliminating space straight y space from space both space the space equations.

Question 11

Solution 11

Question 12
The car hire charges in a city comprise of fixed charges together with the charge for the distance covered. For a journey of 12 km, the charge paid is Rs 89 and for a journey of 20 km, the charge paid is Rs. 145. What will a person have to pay for travelling a distance of 30 km?
Solution 12

Question 13
A part of monthly hostel charges in a college are 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, he 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 charge and the cost of food per day.
Solution 13

Question 14
Solution 14
Question 15
The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Solution 15

Question 16
2 Women and 5 men can together finish a piece of embroidery 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 embroideery, and that taken by 1 man alone.
Solution 16

Question 17

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 Rs 50 and Rs 100 she received.

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

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 number of students in the class.

Solution 21

Question 22

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?

Solution 22

Let the money with the first person and second person be Rs x and Rs y respectively.

According to the question,
x + 100 = 2(y - 100)
x + 100 = 2y - 200
x - 2y = -300        ... (1)

6(x - 10) = (y + 10)
6x - 60 = y + 10
6x - y = 70            ... (2)

Multiplying equation (2) by 2, we obtain:
12x - 2y = 140        ... (3)
Subtracting equation (1) from equation (3), we obtain:
11x = 140 + 300
11x = 440
x = 40
Putting the value of x in equation (1), we obtain:
40 - 2y = -300
40 + 300 = 2y
2y = 340
y = 170

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

Concept insight: This problem talks about the amount of capital with two friends. So, we will represent them by variables x and y respectively. Now, using the given conditions, a pair of linear equations can be formed which can then be solved easily using elimination method.

Question 23

A shopkeeper sells a saree at 8% profit and a sweater at 10% discount, thereby getting a sum of Rs.1008. If she had sold the saree at 10% profit and sweater at 8% discount, she would have got Rs.1028. Find the cost price of the saree and the list price (price before discount) of the sweater.

Solution 23

Question 24

In a competitive examination, one mark is awarded for each correct answer while ½ mark is deducted for every wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly?

Solution 24

Question 25

A shopkeeper gives book on rent for reading. She takes a fixed charge for the first two days, and an additional charge for each day thereafter. Latika paid Rs.22 for a book kept for 6 days, while Rs.16 for the book kept for four days. Find the fixed charges and charge for each extraday.

Solution 25

Chapter 3 - Pairs of Linear Equations in Two Variables Excercise 3.114

Question 1

begin mathsize 11px style The space value space of space straight k space for space which space the space system space of space equations
table attributes columnalign left end attributes row cell table attributes columnalign left end attributes row cell kx minus straight y equals 2 end cell row cell 6 straight x minus 2 straight y equals 3 end cell row cell has text   end text straight a text   end text unique text   end text solution comma text   end text is end cell end table end cell row cell left parenthesis straight a right parenthesis equals 3 space space space space left parenthesis straight b right parenthesis not equal to 3 space space space space space space left parenthesis straight c right parenthesis not equal to 0 space space space left parenthesis straight d right parenthesis 0 space end cell end table end style

Solution 1

begin mathsize 11px style table attributes columnalign left end attributes row cell We text  know , if  a end text subscript text 1 end text end subscript straight x plus straight b subscript 2 straight y plus straight c subscript 1 equals 0 text    ---- end text box enclose 1 end cell row cell text                    a end text subscript 2 straight x plus straight b subscript 2 straight y plus straight c subscript 2 equals 0 text    ---- end text box enclose 2 end cell row cell text for unique solution end text end cell row cell straight a subscript 1 over straight a subscript 2 not equal to straight b subscript 1 over straight b subscript 2 text        ---- end text box enclose 3 end cell row cell text Given equations are end text end cell row cell text kx - y = 2  ---- end text box enclose 4 end cell row cell text 6x - 2y =3 ---- end text box enclose 5 end cell row cell text from  end text box enclose 1 text  &  end text box enclose 4 end cell row cell text a end text subscript text 1 end text end subscript text =k    b end text subscript 1 equals negative 1 end cell row cell from text   end text box enclose 2 text  &  end text box enclose 5 end cell row cell text a end text subscript text 2 end text end subscript text =6    b end text subscript text 2 end text end subscript equals negative 2 end cell row cell from text   end text box enclose 3 end cell row cell straight k over 6 not equal to fraction numerator negative 1 over denominator negative 2 end fraction end cell row cell rightwards double arrow box enclose straight k not equal to 3 end enclose end cell end table end style

So, the correct option is (b).

Question 2

begin mathsize 11px style The space value space of space straight k space for space which space the space system space of space equations
table attributes columnalign left end attributes row cell table attributes columnalign left end attributes row cell 2 straight x plus 3 straight y equals 5 end cell row cell 4 straight x plus ky equals 10 end cell row cell has text   end text infinite text   end text number text   end text of text   end text solutions comma text   end text is end cell end table end cell row cell left parenthesis straight a right parenthesis 1 space space text   end text left parenthesis straight b right parenthesis text   end text 3 space space left parenthesis straight c right parenthesis text   end text 6 space space space space left parenthesis straight d right parenthesis text   end text 0 end cell end table end style

Solution 2

begin mathsize 11px style table attributes columnalign left end attributes row cell we text  know, if  a end text subscript text 1 end text end subscript straight x plus straight b subscript 1 straight y plus straight c subscript 1 equals 0 text     --- end text box enclose 1 end cell row cell text                   a end text subscript text 2 end text end subscript straight x plus straight b subscript 2 straight y plus straight c subscript 2 equals 0 text    --- end text box enclose 2 text   end text end cell row cell text for infinite solution  end text end cell row cell text                   end text text a end text subscript text 1 end text end subscript over text a end text subscript text 2 end text end subscript equals straight b subscript 1 over straight b subscript 2 equals straight c subscript 1 over straight c subscript 2 text        --- end text box enclose 3 end cell row cell Given text  equation are end text end cell row cell text                   2x+3y=5     --- end text box enclose 4 end cell row cell text                   4x+ky=10   --- end text box enclose 5 text    end text end cell row cell text from equations   end text box enclose 1 text  &  end text box enclose 4 text  a end text subscript text 1 end text end subscript equals 2 text      end text straight b subscript 1 equals 3 text      end text straight c subscript 1 equals negative 5 end cell row cell text from equations  end text box enclose 2 text  &  end text box enclose 5 text  a end text subscript text 2 end text end subscript equals 4 text      end text straight b subscript 2 equals straight k text      end text straight c subscript 2 equals negative 10 end cell row cell text from equation  end text left parenthesis 3 right parenthesis end cell row cell 2 over 4 text  =  end text 3 over straight k text  =  end text fraction numerator negative 5 over denominator negative 10 end fraction end cell row cell rightwards double arrow 3 over straight k equals 1 half end cell row cell rightwards double arrow box enclose straight k equals 6 end enclose end cell end table
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 3

begin mathsize 11px style The space value space of space straight k space for space which space the space system space of space equations
table attributes columnalign left end attributes row cell straight x plus 2 straight y minus 3 equals 0 space text   end text and text    end text 5 straight x plus ky plus 7 equals 0 space text   end text has text   end text no text   end text solution comma text   end text is end cell row cell left parenthesis straight a right parenthesis 10 space space space space left parenthesis straight b right parenthesis text   end text 6 space space space space left parenthesis straight c right parenthesis text   end text 3 space space left parenthesis straight d right parenthesis text   end text 1 end cell end table end style

Solution 3

begin mathsize 11px style table attributes columnalign left end attributes row cell We text  know if   a end text subscript text 1 end text end subscript straight x plus straight b subscript 1 straight y plus straight c subscript 1 equals 0 text     ---- end text box enclose 1 end cell row cell text                   a end text subscript 2 straight x plus straight b subscript 2 straight y plus straight c subscript 2 equals 0 text     ---- end text box enclose 2 end cell row cell text for no solution  end text end cell row cell text                   end text text a end text subscript text 1 end text end subscript over text a end text subscript 2 equals text b end text subscript text 1 end text end subscript over text b end text subscript 2 not equal to straight c subscript 1 over straight c subscript 2 text         ---- end text box enclose 3 end cell row cell Given text  equations are end text end cell row cell text                  x+2y-3=0        ---- end text box enclose 4 end cell row cell text                  5x+ky+7=0     ---- end text box enclose 5 end cell row cell text from  end text box enclose 1 text  &  end text box enclose 4 text     a end text subscript text 1 end text end subscript text = 1      b end text subscript text 2 end text end subscript text =2    c end text subscript text 1 end text end subscript equals negative 3 end cell row cell text from  end text box enclose 2 text  &  end text box enclose 5 text     a end text subscript 2 text = 5      b end text subscript text 2 end text end subscript text =k    c end text subscript 2 equals 7 end cell row cell text from  end text box enclose 3 text   end text 1 fifth equals 2 over straight k not equal to fraction numerator negative 3 over denominator 7 end fraction end cell row cell rightwards double arrow box enclose straight k equals 10 end enclose end cell row cell text  So, the correct option is (a). end text end cell end table end style

Question 4

begin mathsize 11px style table attributes columnalign left end attributes row cell table attributes columnalign left end attributes row cell The text   end text value text   end text of text   end text straight k text   end text for text   end text which text   end text the text   end text system text   end text of text   end text equations end cell row cell table attributes columnalign left end attributes row cell 3 straight x plus 5 straight y equals 0 text   end text and space text    end text kx plus 10 straight y equals 0 space text   end text has text   end text straight a text   end text non minus zero text   end text solution comma text   end text end cell row is end table end cell end table end cell row cell left parenthesis straight a right parenthesis 0 space space space space left parenthesis straight b right parenthesis 2 space space space space left parenthesis straight c right parenthesis text   end text 6 space space text    end text left parenthesis straight d right parenthesis 8 end cell end table end style

Solution 4

begin mathsize 11px style table attributes columnalign left end attributes row cell text if the equations are   a end text subscript text 1 end text end subscript straight x plus straight b subscript 1 straight y plus straight c subscript 1 equals 0 text     ---- end text box enclose 1 end cell row cell text                a end text subscript 2 straight x plus straight b subscript 2 straight y plus straight c subscript 2 equals 0 text     ---- end text box enclose 2 end cell row blank row cell text for infinite solutions   end text end cell row cell text                   end text text a end text subscript text 1 end text end subscript over text a end text subscript 2 equals text b end text subscript text 1 end text end subscript over text b end text subscript 2 text         ---- end text box enclose 3 end cell row blank row cell Given text  equations are end text end cell row cell text                  3x + 5y = 0        ---- end text box enclose 4 end cell row cell text                  kx + 10y = 0     ---- end text box enclose 5 end cell row blank row cell on text   end text comparing text    end text box enclose 1 text ,  end text box enclose 2 text ,  end text box enclose 4 text ,  end text box enclose 5 end cell row cell text   end text rightwards double arrow text     a end text subscript text 1 end text end subscript text  = 3      b end text subscript text 1 end text end subscript text = 5     end text end cell row cell text           a end text subscript 2 equals straight k text       b end text subscript 2 text = 10 end text end cell row blank row cell text from  end text box enclose 3 text      end text 3 over straight k equals 5 over 10 end cell row cell rightwards double arrow box enclose straight k equals 6 end enclose text   end text end cell end table
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 5

begin mathsize 11px style table attributes columnalign left end attributes row cell If text   end text the space system text   end text of text   end text equations end cell row cell 2 straight x plus 3 straight y text   end text equals text   end text 7 end cell row cell left parenthesis straight a plus straight b right parenthesis straight x text   end text plus text   end text left parenthesis 2 straight a minus by right parenthesis equals 21 text   end text end cell row cell has text   end text infinitely text   end text many text   end text solutions comma text   end text then end cell row cell left parenthesis straight a right parenthesis space straight a equals 1 text   end text comma space text    end text straight b equals 5 space space space space space left parenthesis straight b right parenthesis text   end text straight a equals 5 comma text   end text straight b equals 1 space space space space end cell row cell left parenthesis straight c right parenthesis space straight a text  = - end text 1 comma text   end text straight b equals 5 space space space space space left parenthesis straight d right parenthesis text   end text straight a equals 5 comma space text   end text straight b equals negative text   end text 1 end cell end table end style

Solution 5

begin mathsize 11px style table attributes columnalign left end attributes row cell for text   end text infinite text  solution end text end cell row cell straight a subscript 1 over straight a subscript 2 equals straight b subscript 1 over straight b subscript 2 equals straight c subscript 1 over straight c subscript 2 end cell row cell rightwards double arrow text   end text fraction numerator 2 over denominator straight a plus straight b end fraction text   end text equals text   end text fraction numerator 3 over denominator 2 straight a minus straight b end fraction text   end text equals text   end text 7 over 21 end cell row cell rightwards double arrow text   end text fraction numerator 2 over denominator straight a plus straight b end fraction text   end text equals text   end text 7 over 21 text      &     end text fraction numerator 3 over denominator 2 straight a minus straight b end fraction text   end text equals text   end text 7 over 21 end cell row cell rightwards double arrow straight a text   end text plus text   end text straight b equals 6 text    -- end text box enclose 1 text            end text rightwards double arrow 9 text   end text equals text  2 end text straight a minus straight b text    -- end text box enclose 2 end cell row cell text on adding  end text box enclose 1 text   &   end text box enclose 2 end cell row cell rightwards double arrow 3 straight a equals 15 end cell row cell rightwards double arrow box enclose straight a equals 5 end enclose text     &    end text box enclose straight b equals 1 end enclose text      end text end cell row cell So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end cell end table end style

Chapter 3 - Pairs of Linear Equations in Two Variables Excercise 3.115

Question 1

begin mathsize 11px style If space the space system space of space equations space
table attributes columnalign left end attributes row cell table attributes columnalign left end attributes row cell 3 straight x text   end text plus text   end text straight y text   end text equals text   end text 1 end cell row cell left parenthesis 2 straight k minus 1 right parenthesis straight x text   end text plus left parenthesis straight k minus 1 right parenthesis text   end text equals text   end text 2 straight k plus 1 end cell row cell is text   end text inconsistent comma text   end text then text   end text straight k equals end cell end table end cell row cell left parenthesis straight a right parenthesis 1 space space space text   end text left parenthesis straight b right parenthesis 0 space space text    end text left parenthesis straight c right parenthesis minus 1 space space space text   end text left parenthesis straight d right parenthesis text   end text 2 end cell end table end style

Solution 1

begin mathsize 11px style table attributes columnalign left end attributes row cell if text        end text straight a subscript 1 straight x text  + b end text subscript text 1 end text end subscript text y + c end text subscript text 1 end text end subscript text =0   --- end text box enclose 1 end cell row cell text          end text straight a subscript 2 straight x text  + b end text subscript 2 text y + c end text subscript 2 text =0   --- end text box enclose 2 end cell row blank row cell text for inconsistent solution end text end cell row cell text         end text straight a subscript 1 over straight a subscript 2 text  =   end text straight b subscript 1 over straight b subscript 2 text     --- end text box enclose 3 end cell row cell text Given equations are end text end cell row cell text            3x + y =1   --- end text box enclose 4 text   end text end cell row cell text             end text left parenthesis 2 straight k minus 1 right parenthesis text x +  end text left parenthesis straight k minus 1 right parenthesis text y = 2k+1   --- end text box enclose 5 end cell row cell text from  end text left parenthesis 1 right parenthesis text   &  end text left parenthesis 4 right parenthesis end cell row cell rightwards double arrow text  a end text subscript text 1 end text end subscript text  = 3       b end text subscript text 1 end text end subscript equals 1 text      c end text subscript text 1 end text end subscript equals negative 1 end cell row cell from text    end text left parenthesis 2 right parenthesis text   &  end text left parenthesis 5 right parenthesis end cell row cell text        end text straight a subscript 2 equals 2 straight k minus 1 text      b end text subscript text 2 end text end subscript equals straight k minus 1 text     c end text subscript 2 text = 2k+1 end text end cell row cell text from  end text box enclose 3 end cell row cell text                 end text fraction numerator 3 over denominator 2 straight k minus 1 end fraction equals fraction numerator 1 over denominator straight k minus 1 end fraction end cell row cell text          end text rightwards double arrow text   3k - 3 = 2k-1 end text end cell row cell text          end text rightwards double arrow text     end text box enclose straight k equals 2 end enclose end cell row cell text  So, the correct option is (d). end text end cell end table end style

Question 2

If   am ≠ bl, then the system of equations

ax + by = c

lx + my = n

(a)  has a unique solution        

(b) has no solution

(c) has infinite many solution    

(d) may or may not have a solution

Solution 2

begin mathsize 11px style table attributes columnalign left end attributes row cell for text  unique solution end text end cell row cell text             end text straight a over straight l not equal to straight b over straight m end cell row cell text          end text rightwards double arrow text am  end text not equal to text  bl end text end cell row cell text which is the given condition  end text end cell row cell text Hence the given equation end text end cell row cell text            am end text not equal to text bl end text end cell row cell text is the condition to a unique solution. end text end cell end table end style

So, the correct option is (a).

Question 3

begin mathsize 11px style If space the space system space of space equations space
table attributes columnalign left end attributes row cell 2 straight x plus 3 straight y equals 7 end cell row cell 2 ax plus left parenthesis straight a plus straight b right parenthesis straight y equals 28 end cell row cell has text  infinite many solutions, then end text end cell row cell left parenthesis straight a right parenthesis text  a = 2b   end text left parenthesis straight b right parenthesis text  b = 2a   end text left parenthesis straight c right parenthesis text  a+2b = 0    end text left parenthesis straight d right parenthesis text  2a+b=0 end text end cell end table end style

Solution 3

begin mathsize 11px style table attributes columnalign left end attributes row cell if text   a end text subscript text 1 end text end subscript straight x plus text  b end text subscript text 1 end text end subscript straight y plus straight c subscript 1 equals 0 text     --- end text box enclose 1 end cell row cell text     a end text subscript text 2 end text end subscript straight x plus text  b end text subscript 2 straight y plus straight c subscript 2 equals 0 text     --- end text box enclose 2 end cell row cell text    for infinite many soluations end text end cell row cell text           end text straight a subscript 1 over text a end text subscript text 2 end text end subscript equals text b end text subscript text 1 end text end subscript over text b end text subscript 2 equals straight c subscript 1 over straight c subscript 2 text      --- end text box enclose 3 end cell row cell Given text  equations are end text end cell row cell text          2x + 3y = 7      --- end text box enclose 4 end cell row cell text          2ax + (a+b)y = 28     --- end text box enclose 5 end cell row cell from text    end text box enclose 1 text   &   end text box enclose 4 end cell row cell text          a end text subscript text 1 end text end subscript text  =  end text 2 text       b end text subscript 1 text  = 3     end text straight c subscript 1 equals text  -7  end text end cell row cell from text    end text box enclose 2 text   &   end text box enclose 5 end cell row cell text          a end text subscript 2 text  =  end text 2 straight a text       b end text subscript 2 text  = a+b    end text straight c subscript 2 equals text  -28 end text end cell row cell text from  end text box enclose 3 text    end text end cell row cell rightwards double arrow text      end text fraction numerator 2 over denominator 2 straight a end fraction text  =  end text fraction numerator 3 over denominator straight a plus straight b end fraction text  =  end text fraction numerator negative 7 over denominator negative 28 end fraction text    end text end cell row cell rightwards double arrow text      end text 1 over straight a text  =  end text fraction numerator 3 over denominator straight a plus straight b end fraction text    end text rightwards double arrow text    end text box enclose 2 straight a equals straight b end enclose end cell end table end style

So, the correct option is (b).

Question 4

begin mathsize 11px style The space value space of space straight k space for space which space the space system space of space equations
table attributes columnalign left end attributes row cell straight x plus 2 straight y equals 5 end cell row cell 3 straight x plus ky plus 15 equals 0 end cell row cell has text  no solution is end text end cell row cell left parenthesis straight a right parenthesis text  6      end text left parenthesis straight b right parenthesis text  -6      end text left parenthesis straight c right parenthesis text   end text 3 over 2 text      end text left parenthesis straight d right parenthesis text  none of these end text end cell end table end style

Solution 4

begin mathsize 11px style table attributes columnalign left end attributes row cell for text  equation a end text subscript text 1 end text end subscript straight x plus straight b subscript 1 straight y plus straight c subscript 1 equals 0 text      --- end text box enclose 1 end cell row cell text            a end text subscript 2 straight x plus straight b subscript 2 straight y plus straight c subscript 2 equals 0 text      --- end text box enclose 2 end cell row cell for space no text  solution end text end cell row cell text             end text text a end text subscript text 1 end text end subscript over straight b subscript 2 text  =  end text text a end text subscript text 1 end text end subscript over straight b subscript 2 not equal to text   end text text c end text subscript text 1 end text end subscript over straight c subscript 2 text    --- end text box enclose 3 end cell row cell given text  equations are end text end cell row cell text                    x+2y=5     --- end text box enclose 4 end cell row cell text            3x+ky+15=0     --- end text box enclose 5 end cell row cell from text   end text box enclose 1 text   &   end text box enclose 4 end cell row cell text        a end text subscript text 1 end text end subscript equals 1 text      b end text subscript text 1 end text end subscript text =2      c end text subscript text 1 end text end subscript text  =-5 end text end cell row cell text   end text from text   end text box enclose 2 text   &   end text box enclose 5 end cell row cell text        a end text subscript 2 equals 3 text      b end text subscript 2 text =k      c end text subscript 2 text  =15 end text end cell row cell from text   end text box enclose 3 text    end text end cell row cell text         end text 1 third equals 2 over straight k not equal to fraction numerator negative 5 over denominator 15 end fraction end cell row cell rightwards double arrow box enclose straight k equals 6 end enclose end cell end table end style

So, the correct option is (a).

Question 5

If 2x-3y=7 and (a+b)x – (a+b-3)y = 4a+b represent coincident lines than a and b satisfy the equation

(a) a+5b=0    (b) 51+b=0    (c) a-5b=0     (d) 5a-b=0

Solution 5

begin mathsize 11px style table attributes columnalign left end attributes row cell for text  conicident lines end text end cell row cell text        end text fraction numerator 2 over denominator straight a plus straight b end fraction text   end text equals text   end text fraction numerator negative 3 over denominator negative left parenthesis straight a plus straight b minus 3 right parenthesis end fraction text  =  end text fraction numerator 7 over denominator 4 straight a plus straight b end fraction end cell row cell rightwards double arrow fraction numerator 2 over denominator straight a plus straight b end fraction equals fraction numerator 7 over denominator 4 straight a plus straight b end fraction end cell row cell rightwards double arrow 8 straight a plus 2 straight b equals 7 straight a plus 7 straight b end cell row cell rightwards double arrow text   end text box enclose straight a equals 5 straight b end enclose end cell end table
rightwards double arrow straight a space minus space 5 straight b equals 0 end style

So, the correct option is (c).

Question 6

If a pair of linear equations in two variables is consistent, then the lines represented by two equations are

(a) Intersecting             (b) parallel

(c) always coincident       (d) intersecting or coincident

Solution 6

Consistent solution means either linear equations have unique solutions or infinite solutions.

⇒ In case of unique solution; lines are intersecting

⇒ If solutions are infinite, lines are coincident.

So, lines are either intersecting or coincident

So, the correct option is (d).

Question 7

begin mathsize 11px style The space area space of space the space triangle space formed space by space the space line space straight x over straight a plus straight y over straight b equals 1 space
with space the space coordinate space axes space is
left parenthesis straight a right parenthesis text ab      end text left parenthesis straight b right parenthesis text 2ab      end text left parenthesis straight c right parenthesis 1 half ab text        end text left parenthesis straight d right parenthesis 1 fourth ab
end style

Solution 7

begin mathsize 11px style Intercept space on space straight x space minus space axis space equals space straight a
Intercept space on space straight y minus space axis space equals space straight b

table attributes columnalign left end attributes row cell space space space area text   end text of text   end text triangle equals 1 half cross times straight a cross times straight b end cell row cell text                               end text equals 1 half ab end cell end table end style

So, the correct option is (c).

Question 8

The area of the triangle formed by the lines

y=x,  x=6, and  y=0  is

(a) 36 sq. units      

(b) 18 sq. units

(c) 9 sq. units         

(d) 72 sq. units

Solution 8

begin mathsize 11px style table attributes columnalign left end attributes row cell area text  of triangle =  end text 1 half cross times 6 cross times 6 end cell row cell text                             = 18 sq. units    end text end cell end table end style

So, the correct option is (b).

Question 9

If the system of equations 2x + 3y=5,  4x + ky =10 has infinitely many solutions, then k=

(a) 1     

(b) begin mathsize 12px style 1 half end style    

(c) 3    

(d) 6

Solution 9

begin mathsize 11px style table attributes columnalign left end attributes row cell If text  the equations,       a end text subscript text 1 end text end subscript straight x plus straight b subscript 1 straight y plus straight c subscript 1 equals 0 text       --- end text box enclose 1 end cell row cell text                                 a end text subscript 2 straight x plus straight b subscript 2 straight y plus straight c subscript 2 equals 0 text       --- end text box enclose 2 end cell row cell For space infinitely text   end text many text  solution    end text text a end text subscript blank to the power of text 1 end text end exponent end subscript over straight a subscript 2 equals text b end text subscript blank to the power of text 1 end text end exponent end subscript over straight b subscript 2 equals text c end text subscript blank to the power of text 1 end text end exponent end subscript over straight c subscript 2 text  --- end text box enclose 3 end cell row cell Comparing text  the given equations end text to the power of blank text  to  end text box enclose 1 text   &   end text box enclose 2 end cell row cell We text  get, end text end cell row cell text           a end text subscript text 1 end text end subscript equals 2 text          b end text subscript text 1 end text end subscript equals 3 text          c end text subscript text 1 end text end subscript equals negative 5 end cell row cell text           a end text subscript 2 equals 4 text          b end text subscript 2 equals straight k text          c end text subscript text 1 end text end subscript equals negative 10 end cell row cell rightwards double arrow text     end text from text   end text box enclose 3 end cell row cell text                    end text 2 over 4 equals 3 over straight k equals fraction numerator negative 5 over denominator negative 10 end fraction end cell row cell text               end text rightwards double arrow text   end text 3 over straight k equals 1 half end cell row cell text              end text rightwards double arrow text     end text box enclose straight k equals 6 end enclose end cell end table end style

So, the correct option is (d).

Question 10

If the system of equations  kx - 5y = 2,  6x +2y=7 has no solution, then k=

(a) -10     

(b) -5      

(c) -6     

(d)-15

Solution 10

begin mathsize 11px style table attributes columnalign left end attributes row cell If text  equations end text to the power of blank text           a end text subscript text 1 end text end subscript straight x text  + b end text subscript text 1 end text end subscript text y + c end text subscript text 1 end text end subscript text  = 0         --- end text box enclose 1 end cell row cell text                      and   a end text subscript 2 straight x text  + b end text subscript 2 text y + c end text subscript text 2 end text end subscript text  = 0         --- end text box enclose 2 end cell row cell have text  no solution, then end text end cell row cell text                  end text straight a subscript 1 over straight a subscript 2 equals straight b subscript 1 over straight b subscript 2 not equal to straight c subscript 1 over straight c subscript 2 text       --- end text box enclose 3 end cell row cell On text  comparing with given equation end text to the power of blank text   to   end text box enclose 1 text    &   end text box enclose 2 end cell row cell we text   end text get end cell row cell text          a end text subscript text 1 end text end subscript equals straight k text        b end text subscript text 1 end text end subscript equals text  -5      c end text subscript text 1  end text end subscript equals text  -2 end text end cell row cell text          a end text subscript 2 equals 6 text        b end text subscript 2 equals text  2      c end text subscript 2 equals text  -7 end text end cell row cell text From equation  end text box enclose 3 end cell row cell text             end text straight k over 6 equals text   end text fraction numerator negative 5 over denominator 2 end fraction text   end text not equal to text   end text 2 over 7 end cell row cell text       end text rightwards double arrow text    end text box enclose straight k equals negative 15 end enclose end cell end table end style

So, the correct option is (d).

Question 11

The area of the triangle formed by the lines

x = 3, y = 4 and x = y is

(a) begin mathsize 12px style 1 half end style sq. unit    

(b) 1 sq. unit

(c) 2 sq. unit     

(d) None of these

Solution 11

begin mathsize 11px style table attributes columnalign left end attributes row cell area text  of triangle  =  end text 1 half cross times 1 cross times 1 end cell row cell text                              =  end text 1 half text  sq. unit end text end cell end table end style

So, the correct option is (a).

Chapter 3 - Pairs of Linear Equations in Two Variables Excercise 3.116

Question 1

The area of the triangle formed by the lines

2x + 3y = 12,     x – y – 1= 0  and x = 0

(a) 7 sq. units          

(b) 7.5 sq. units

(c) 6.5 sq. units        

(d) 6 sq. units   

Solution 1

begin mathsize 11px style From space figure space shown comma space the space required space triangle space is space as space shown space above.
BD space equals space 5 space Units space space space and space space space space space PO space equals space 3 space Units
table attributes columnalign left end attributes row cell area text   end text of space triangle text   end text equals 1 half cross times 5 cross times 3 end cell row cell text                               =  end text 15 over 2 equals text  7.5 sq. Units end text end cell end table end style

So, the correct option is (b).

Question 2

The sum of the digits of a two digit number is 9. If 27 is added to it, the digits of the number get reversed. The number is

 

  1. 25
  2. 72
  3. 63
  4. 36
Solution 2

Question 3

If x = a, y = b is the solution of the system of equations x - y = 2 and x + y = 4, then the values of a and b are, respectively

 

  1. 3 and 1
  2. 3 and 5
  3. 5 and 3
  4. -1 and -3
Solution 3

Since x = a and y = b is the solution of given system of equations x - y = 2 and x + y = 4, we have

a - b = 2 ….(i)

a + b = 4 ….(ii)

Adding (i) and (ii), we have

2a = 6 a = 3

b = 4 - 3 = 1

Hence, correct option is (a).

Question 4

For what value k, do the equations 3x - y + 8 = 0 and 6x - ky + 16 = 0 represent coincident lines?

 

  1. 2
  2. -2
Solution 4

Question 5

Aruna has only Rs.1 and Rs.2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is Rs.75, then the number of Rs.1 and Rs.2 coins are, respectively

 

  1. 35 and 15
  2. 35 and 20
  3. 15 and 35
  4. 25 and 25
Solution 5