RS Agarwal Solution | Class 11th | Chapter-7 | Linear Equations in Two Variable

Table of Contents

Exercise Ex. 7A

Question 1

Solve each of the following systems of equations graphically:

2x + 3y = 2

x – 2y = 8Solution 1

Question 3

Solve each of the following systems of equations graphically:

2x + 3y = 8

x – 2y + 3 = 0Solution 3

Question 4

Solve each of the following systems of equations graphically:

2x – 5y + 4 = 0

2x + y – 8 = 0Solution 4

Question 5

Solve each of the following systems of equations graphically:

3x + 2y = 12, 5x – 2y = 4.Solution 5

Since the two graphs intersect at (2, 3),

x = 2 and y = 3.Question 6

Solve each of the following systems of equations graphically:

3x + y + 1 = 0

2x – 3y + 8 = 0Solution 6

Question 7

Solve each of the following systems of equations graphically:

2x + 3y + 5 = 0, 3x – 2y – 12 = 0.Solution 7

Since the two graphs intersect at (2, -3),

x = 2 and y = -3.Question 8

Solve each of the following systems of equations graphically:

2x – 3y + 13 = 0, 3x – 2y + 12 = 0.Solution 8

Since the two graphs intersect at (-2, 3),

x = -2 and y = 3.Question 9

Solve each of the following systems of equations graphically:

2x + 3y – 4 = 0, 3x – y + 5 = 0.Solution 9

Since the two graphs intersect at (-1, 2),

x = -1 and y = 2.Question 10

Solve each of the following systems of equations graphically:

x + 2y + 2 = 0

3x + 2y – 2 = 0Solution 10

Question 11

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:

x – y + 3 = 0, 2x + 3y – 4 = 0.Solution 11

Question 12

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:

2x – 3y + 4 = 0, x + 2y – 5 = 0.Solution 12

Question 13

Solve the following system of linear equations graphically:

4x – 3y + 4 = 0, 4x + 3y – 20 = 0

Find the area of the region bounded by these lines and the x-axis.Solution 13

Question 14

Solve the following system of linear equation graphically:

x – y + 1 = 0, 3x + 2y – 12 = 0

Calculate the area bounded by these lines and x-axis.Solution 14

Question 15

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:

x – 2y + 2 = 0, 2x + y – 6 = 0.Solution 15

Question 16

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

2x – 3y + 6 = 0, 2x + 3y – 18 = 0.Solution 16

Question 17

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

4x – y – 4 = 0, 3x + 2y – 14 = 0.Solution 17

Question 18

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

x – y – 5 = 0, 3x + 5y – 15 = 0.Solution 18

Question 19

Solve the following system of linear equations graphically:

2x – 5y + 4 = 0, 2x + y – 8 = 0

Find the point, where these lines meet the y-axisSolution 19

Question 20

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

5x – y – 7 = 0, x – y + 1 = 0.Solution 20

Question 21

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

2x – 3y = 12, x + 3y = 6.Solution 21

Question 22

Show graphically that each of the following given systems of equations has infinitely many solutions:

2x + 3y = 6, 4x + 6y = 12.Solution 22

Since the graph of the system of equations is coincident lines, the system has infinitely many solutions.Question 23

Show graphically that the system of equations 3x – y = 5, 6x – 2y = 10 has infinitely many solutions.Solution 23

Question 24

Show graphically that the system of equations 2x + y = 6, 6x + 3y = 18 has infinitely many solutions.Solution 24

Question 25

Show graphically that each of the following given systems of equations has infinitely many solutions:

x – 2y = 5, 3x – 6y = 15.Solution 25

Since the graph of the system of equations is coincident lines, the system has infinitely many solutions. Question 26

Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solution:

x – 2y = 6, 3x – 6y = 0Solution 26

Since the graph of the system of equations is parallel lines, the system has no solution and hence is inconsistent.Question 27

Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solution:

2x + 3y = 4, 4x + 6y = 12.Solution 27

Since the graph of the system of equations is parallel lines, the system has no solution and hence is inconsistent.Question 28

Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solution:

2x + y = 6, 6x + 3y = 20.Solution 28

Since the graph of the system of equations is parallel lines, the system has no solution and hence is inconsistent.Question 29

Draw the graphs of the following equations on the same graph paper:

2x + y = 2, 2x + y = 6.

Find the coordinates of the vertices of the trapezium formed by these lines. Also, find the area of the trapezium so formed.Solution 29

Exercise Ex. 7B

Question 1

Solve for x and y

x + y = 3

4x – 3y = 26Solution 1

Question 2

Solve for x and y:

Solution 2

Question 3

Solve for x and y

2x + 3y = 0

3x + 4y = 5Solution 3

Question 4

Solve for x and y

2x – 3y = 13

7x – 2y = 20Solution 4

Question 5

Solve for x and y

3x – 5y – 19 = 0

-7x + 3y + 1 = 0Solution 5

Question 6

Solve for x and y:

2x – y + 3 = 0, 3x – 7y + 10 = 0.Solution 6

Question 7

Solve for x and y:

Solution 7

Question 8

Solve for x and y

Solution 8

Question 9

Solve for x and y

Solution 9

Question 10

Solve for x and y

Solution 10

Question 11

Solve for x and y

Solution 11

Question 12

Solve for x and y

Solution 12

Question 13

Solve for x and y:

0.4x + 0.3y = 1.7, 0.7x – 0.2y = 0.8.Solution 13

Question 14

Solve for x and y:

0.3x + 0.5y = 0.5, 0.5x + 0.7y = 0.74.Solution 14

Question 15

Solve for x and y

7(y + 3) – 2(x + 2) = 14

4(y – 2) + 3(x – 3) = 2Solution 15

Question 16

Solve for x and y

6x + 5y = 7x + 2y + 1 = 2(x + 6y – 1)Solution 16

Question 17

Solve for x and y

Solution 17

Question 18

×Solve for x and y

Solution 18

Putting the given equations become

5u + 6y = 13—(1)

3u + 4y = 7 —-(2)

Multiplying (1) by 4 and (2) by 6, we get

20u + 24y = 52—(3)

18u + 24y = 42—(4)

Subtracting (4) from (3), we get

2u = 10 u = 5

Putting u = 5 in (1), we get

5 × 5 + 6y = 13

6y = 13 – 25

6y = -12

y = -2

Question 19

Solve for x and y

Solution 19

The given equations are and

Putting

x + 6v = 6 —-(1)

3x – 8v = 5—(2)

Multiplying (1) by 4 and (2) by 3

4x + 24v = 24—(3)

9x – 24v = 15 —(4)

Adding (3) and (4)

13x = 24 + 15 = 39

Puttingx = 3 in (1)

3 + 6v = 6

6v = 6 – 3 = 3

solution is x = 3, y = 2Question 20

Solve for x and y

Solution 20

Putting in the given equation

2x – 3v = 9 —(1)

3x + 7v = 2 —(2)

Multiplying (1) by7 and (2) by 3

14x – 21v = 63 —(3)

9x + 21v = 6 —(4)

Adding (3) and (4), we get

Putting x= 3 in (1), we get

2 × 3 – 3v = 9

-3v = 9 – 6

-3v= 3

v = -1

the solution is x = 3, y = -1Question 21

Solve for x and y

Solution 21

Question 22

Solve for x and y

Solution 22

Putting in the equation

9u – 4v = 8 —(1)

13u + 7v = 101—(2)

Multiplying (1) by 7 and (2) by 4, we get

63u – 28v = 56—(3)

52u + 28v = 404—(4)

Adding (3) and (4), we get

Putting u = 4 in (1), we get

9 × 4 – 4v = 8

36 – 4v = 8

-4v = 8 – 36

-4v = -28

Question 23

Solve for x and y

Solution 23

Putting in the given equation, we get

5u – 3v = 1 —(1)

Multiplying (1) by 4 and (2) by 3, we get

20u – 12v = 4—-(3)

27u + 12v = 90—(4)

Adding (3) and (4), we get

Putting u = 2 in (1), we get

(5 × 2) – 3v = 1

10 – 3v = 1

-3v = 1 – 10 -3v = -9

v = 3

Question 24

Solve for x and y:

Solution 24

Question 25

Solve for x and y

4x + 6y = 3xy

8x + 9y = 5xy;Solution 25

4x + 6y = 3xy

Putting in (1) and (2), we get

4v + 6u = 3—(3)

8v + 9u = 5—(4)

Multiplying (3) by 9 and (4) by 6, we get

36v + 54u = 27 —(5)

48v + 54u = 30 —(6)

Subtracting (3) from (4), we get

12v = 3

Putting in (3), we get

the solution is x = 3, y = 4Question 26

Solve for x and y:

Solution 26

Question 27

Solve for x and y:

Solution 27

Question 28

Solve for x and y

Solution 28

Putting

3u + 2v = 2—-(1)

9u – 4v = 1—-(2)

Multiplying (1) by 2 and (2) by 1. We get

6u + 4v = 4—-(3)

9u – 4v = 1—-(4)

Adding (3) and (4), we get

Adding (5) and (6), we get

Putting in (5). We get

the solution is Question 29

Solve for x and y

Solution 29

The given equations are

Putting

Adding (1) and (2)

Putting value of u in (1)

Hence the required solution isx = 4, y = 5Question 30

Solve for x and y

Solution 30

Putting in the equation, we get

44u + 30v = 10—-(1)

55u + 40v = 13—-(2)

Multiplying (1) by 4 and (2) by 3, we get

176u + 120v = 40—(3)

165u + 120v = 39—(4)

Subtracting (4) from (3), we get

Putting in (1) we get

Adding (5) and (6), we get

Putting x = 8 in (5), we get

8 + y = 11 y = 11 – 8 = 3

the solution is x = 8, y = 3Question 31

Solve for x and y:

Solution 31

Question 32

Solve for x and y

71x + 37y = 253

37x + 71y = 287Solution 32

The given equations are

71x + 37y = 253—(1)

37x + 71y = 287—(2)

Adding (1) and (2)

108x + 108y = 540

108(x + y) = 540

—-(3)

Subtracting (2) from (1)

34x – 34y = 253 – 287 = -34

34(x – y) = -34

—(4)

Adding (3) and (4)

2x = 5 – 1= 4

Subtracting (4) from (3)

2y = 5 + 1 = 6

solution is x = 2, y = 3Question 33

Solve for x and y

217x + 131y = 913

131x + 217y = 827Solution 33

217x + 131y = 913—(1)

131x + 217y = 827—(2)

Adding (1) and (2), we get

348x + 348y = 1740

348(x + y) = 1740

x + y = 5—-(3)

Subtracting (2) from (1), we get

86x – 86y = 86

86(x – y) = 86

x – y = 1—(4)

Adding (3) and (4), we get

2x = 6

x = 3

putting x = 3 in (3), we get

3 + y = 5

y = 5 – 3 = 2

solution is x = 3, y = 2Question 34

Solve for x and y:

23x – 29y = 98, 29x – 23y = 110.Solution 34

Question 35

Solve for x and y:

Solution 35

Question 36

Solve for x and y:

Solution 36

Question 37

Solve for x and y

where Solution 37

The given equations are

Multiplying (1) by 6 and (2) by 20, we get

Multiplying (3) by 6 and (4) by 5, we get

18u + 60v = -54—(5)

125u – 60v = —(6)

Adding (5) and (6), we get

Question 38

Solve for x and y:

Solution 38

Question 39

Solve for x and y:

Solution 39

Question 40

Solve for x and y:

x + y = a + b, ax – by = a² – b².Solution 40

Question 41

Solve for x and y

Solution 41

Question 42

Solve for x and y:

px + py = p – q, qx – py = p + q.Solution 42

Question 43

Solve for x and y:

Solution 43

Question 44

Solve for x and y

6(ax + by) = 3a + 2b

6(bx – ay) = 3b – 2aSolution 44

6(ax + by) = 3a + 2b

6ax + 6by = 3a + 2b —(1)

6(bx – ay) = 3b – 2a

6bx – 6ay = 3b- 2a —(2)

6ax + 6by = 3a + 2b —(1)

6bx – 6ay = 3b – 2a —(2)

Multiplying (1) by a and (2) by b

Adding (3) and (4), we get

Substituting in(1), we get

Hence, the solution is Question 45

Solve for x and y:

ax – by = a² + b², x + y = 2a.Solution 45

Question 46

Solve for x and y

bx – ay + 2ab = 0Solution 46

Question 47

Solve for x and y

x + y = 2abSolution 47

Taking L.C.M, we get

Multiplying (1) by 1 and (2) by

Subtracting (4) from (3), we get

Substituting x = ab in (3), we get

solution is x = ab, y = abQuestion 48

Solve for x and y:

x + y = a + b, ax – by = a² – b².Solution 48

Question 49

Solve for x and y:

a²x + b²y = c², b²x + a²y = d².Solution 49

Question 50

Solve for x and y:

Solution 50

Exercise Ex. 7C

Question 1

Solve for x and y by method of cross multiplication:

x + 2y + 1 = 0

2x – 3y – 12 = 0Solution 1

x + 2y + 1 = 0 —(1)

2x – 3y – 12 = 0 —(2)

By cross multiplication, we have

Hence, x = 3 and y = -2 is the solutionQuestion 2

Solve for x and y by method of cross multiplication:

3x – 2y + 3 = 0

4x + 3y – 47 = 0Solution 2

3x – 2y + 3 = 0

4x + 3y – 47 = 0

By cross multiplication we have

the solution is x = 5, y = 9Question 3

Solve for x and y by method of cross multiplication:

6x – 5y – 16 = 0

7x – 13y + 10 = 0Solution 3

6x – 5y – 16 = 0

7x – 13y + 10 = 0

By cross multiplication we have

the solution is x = 6, y = 4Question 4

Solve for x and y by method of cross multiplication:

3x + 2y + 25 = 0

2x + y + 10 = 0Solution 4

3x + 2y + 25 = 0

2x + y + 10 = 0

By cross multiplication, we have

the solution is x = 5,y = -20Question 5

Solve for x and y by method of cross multiplication:

2x +5y = 1

2x + 3y = 3Solution 5

2x + 5y – 1 = 0 —(1)

2x + 3y – 3 = 0—(2)

By cross multiplication we have

the solution is x = 3, y = -1Question 6

Solve for x and y by method of cross multiplication:

2x + y – 35 = 0

3x + 4y – 65 = 0Solution 6

2x + y – 35 = 0

3x + 4y – 65 = 0

By cross multiplication, we have

Question 7

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

7x – 2y = 3, 22x – 3y = 16.Solution 7

Question 8

Solve for x and y by method of cross multiplication:

Solution 8

Question 9

Solve for x and y by method of cross multiplication:

Solution 9

Taking

u + v – 7 = 0

2u + 3v – 17 = 0

By cross multiplication, we have

the solution is Question 10

Solve for x and y by method of cross multiplication:

Solution 10

Let in the equation

5u – 2v + 1 = 0

15u + 7v – 10 = 0

Question 11

Solve for x and y by method of cross multiplication:

Solution 11

Question 12

Solve for x and y by method of cross multiplication:

2ax + 3by – (a + 2b) = 0

3ax+ 2by – (2a + b) = 0Solution 12

2ax + 3by – (a + 2b) = 0

3ax+ 2by – (2a + b) = 0

By cross multiplication, we have

Question 13

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

Solution 13

Exercise Ex. 7D

Question 1

Show that the following system of equations has a unique solution:

3x + 5y = 12, 5x + 3y = 4

Also, find the solution of the given system of equations.Solution 1

Question 2

Show that each of the following systems of equations has a unique solution and solve it:

2x – 3y = 17, 4x + y = 13.Solution 2

Question 3

Show that the following system of equations has a unique solution:

Also, find the solution of the given system of equationsSolution 3

Question 4

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

2x + 3y – 5 = 0, kx – 6y – 8 = 0.Solution 4

Question 5

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

x – ky = 2, 3x + 2y + 5 = 0.Solution 5

Question 6

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

5x – 7y – 5 = 0, 2x + ky – 1 = 0.Solution 6

Question 7

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

4x + ky + 8 = 0, x + y + 1 = 0.Solution 7

Question 8

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

4x – 5y = k , 2x – 3y = 12Solution 8

4x – 5y – k = 0, 2x – 3y – 12 = 0

These equations are of the form

Thus, for all real value of k the given system of equations will have a unique solutionQuestion 9

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

kx + 3y = (k – 3),12x + ky = kSolution 9

kx + 3y – (k – 3) = 0

12x + ky – k = 0

These equations are of the form

Thus, for all real value of k other than , the given system of equations will have a unique solutionQuestion 10

Show that the system of equations

2x – 3y = 5, 6x – 9y = 15

has an infinite number of solutions.Solution 10

2x – 3y – 5 = 0, 6x – 9y – 15 = 0

These equations are of the form

Hence the given system of equations has infinitely many solutionsQuestion 11

Show that the system of equations Solution 11

Question 12

For what value of k, the system of equations

kx + 2y = 5, 3x – 4y = 10

has (i) a unique solution (ii) no solution?Solution 12

kx + 2y – 5 = 0

3x – 4y – 10 = 0

These equations are of the form

This happens when

Thus, for all real value of k other that , the given system equations will have a unique solution

(ii) For no solution we must have

Hence, the given system of equations has no solution if Question 13

For what value of k, the system of equations

x + 2y = 5, 3x + ky + 15 = 0

has (i) a unique solution (ii) no solution?Solution 13

x + 2y – 5= 0

3x + ky + 15 = 0

These equations are of the form of

Thus for all real value of k other than 6, the given system ofequation will have unique solution

(ii) For no solution we must have

k = 6

Hence the given system will have no solution when k = 6.Question 14

For what value of k, the system of equations

x + 2y = 3, 5x + ky + 7 = 0

has (i) a unique solution (ii) no solution?

Is there any value of k for which the given system has an infinite number of solutions?Solution 14

x + 2y – 3 = 0, 5x + ky + 7 = 0

These equations are of the form

(i)For a unique solution we must have

Thus, for all real value of k other than 10

The given system of equation will have a unique solution.

(ii)For no solution we must have

Hence the given system of equations has no solution if

For infinite number of solutions we must have

This is never possible since

There is no value of k for which system of equations has infinitely many solutionsQuestion 15

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

2x + 3y = 7

(k – 1)x + (k + 2)y = 3kSolution 15

2x + 3y – 7 = 0

(k – 1)x + (k + 2)y – 3k = 0

These are of the form

This hold only when

Now the following cases arises

Case : I

Case: II

Case III

For k = 7, there are infinitely many solutions of the given system of equationsQuestion 16

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

2x + (k – 2)y =k

6x + (2k – 1)y = (2k + 5)Solution 16

2x + (k – 2)y – k = 0

6x + (2k – 1)y – (2k + 5) = 0

These are of the form

For infinite number of solutions, we have

This hold only when

Case (1)

Case (2)

Case (3)

Thus, for k = 5 there are infinitely many solutionsQuestion 17

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

kx + 3y = (2k +1)

2(k + 1)x + 9y = (7k + 1)Solution 17

kx + 3y – (2k +1) = 0

2(k + 1)x + 9y – (7k + 1) = 0

These are of the form

For infinitely many solutions, we must have

This hold only when

Now, the following cases arise

Case – (1)

Case (2)

Case (3)

Thus, k = 2, is the common value for which there are infinitely many solutionsQuestion 18

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

5x + 2y = 2k

2(k + 1)x + ky = (3k + 4)Solution 18

5x + 2y – 2k = 0

2(k +1)x + ky – (3k + 4) = 0

These are of the form

For infinitely many solutions, we must have

These hold only when

Case I

Thus, k = 4 is a common value for which there are infinitely by many solutions.Question 19

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

(k – 1)x – y = 5

(k + 1)x + (1 – k)y = (3k + 1)Solution 19

(k – 1)x – y – 5 = 0

(k + 1)x + (1 – k)y – (3k + 1) = 0

These are of the form

For infinitely many solution, we must now

k = 3 is common value for which the number of solutions is infinitely manyQuestion 20

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

(k – 3)x + 3y = k, kx + ky = 12.Solution 20

Question 21

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

(a – 1)x + 3y = 2

6x + (1 – 2b)y = 6Solution 21

(a – 1)x + 3y – 2 = 0

6x + (1 – 2b)y – 6 = 0

These equations are of the form

For infinite many solutions, we must have

Hence a = 3 and b = -4Question 22

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

(2a – 1)x + 3y = 5

3x + (b – 1)y = 2Solution 22

(2a – 1)x + 3y – 5 = 0

3x + (b – 1)y – 2 = 0

These equations are of the form

These holds only when

Question 23

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

2x – 3y = 7

(a + b)x + (a + b – 3)y = (4a + b)Solution 23

2x – 3y – 7 = 0

(a + b)x + (a + b – 3)y – (4a + b) = 0

These equation are of the form

For infinite number of solution

Putting a = 5b in (2), we get

Putting b = -1 in (1), we get

Thus, a = -5, b = -1Question 24

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

2x + 3y=7, (a + b + 1)x +(a + 2b + 2)y = 4(a + b)+ 1.Solution 24

Question 25

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

2x + 3y = 7, (a + b)x + (2a – b)y = 21.Solution 25

Question 26

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

2x + 3y = 7, 2ax + (a + b)y = 28.Solution 26

Question 27

Find the value of k for which each of the following systems of equations no solution:

8x + 5y = 9, kx + 10y = 15.Solution 27

Question 28

Find the value of k for which each of the following systems of equations no solution:

kx + 3y = 3, 12x + ky = 6.Solution 28

Question 29

Find the value of k for which each of the following systems of equations no solution:

Solution 29

Question 30

Find the value of k for which each of the following systems of equations no solution:

kx + 3y = k – 3, 12x + ky = k.Solution 30

Question 31

Find the value of k for which the system of equations

5x – 3y = 0;2x + ky = 0

has a nonzero solution.Solution 31

We have 5x – 3y = 0 —(1)

2x + ky = 0—(2)

Comparing the equation with

These equations have a non – zero solution if

Exercise Ex. 7E

Question 1

5 chairs and 4 tables together cost Rs.5600, while 4 chairs and 3 tables together cost Rs.4340. Find the cost of a chair and that of a table.Solution 1

Question 2

23 spoons and 17 forks together cost Rs.1770, while 17 spoons and 23 forks together cost Rs.1830. Find the cost of a spoon and that of a fork.Solution 2

Question 3

A lady has only 25-paisa and 50-paisa coins in her purse. If she has 50 coins in all Rs.19.50, how many coins of each kind does she have?Solution 3

Question 4

The sum of two numbers is 137 and their difference is 43. Find the numbers.Solution 4

Let the two numbers be x and y respectively.

Given:

x + y = 137 —(1)

x – y = 43 —(2)

Adding (1) and (2), we get

2x = 180

Putting x = 90 in (1), we get

90 + y = 137

y = 137 – 90

= 47

Hence, the two numbers are 90 and 47.Question 5

Find two numbers such that the sum of twice the first and thrice the second is 92, and four times the first exceeds seven times the second by 2.Solution 5

Let the first and second number be x and y respectively.

According to the question:

2x + 3y = 92 —(1)

4x – 7y = 2 —(2)

Multiplying (1) by 7 and (2) by 3, we get

14 x+ 21y = 644 —(3)

12x – 21y = 6 —(4)

Adding (3) and (4), we get

Putting x = 25 in (1), we get

2 × 25 + 3y = 92

50 + 3y = 92

3y = 92 – 50

y = 14

Hence, the first number is 25 and second is 14Question 6

Find two numbers such that the sum of thrice the first and the second is 142, and four times the first exceeds the second by 138.Solution 6

Let the first and second numbers be x and y respectively.

According to the question:

3x + y = 142 —(1)

4x – y = 138 —(2)

Adding (1) and (2), we get

Putting x = 40 in (1), we get

3 × 40 + y = 142

y = 142 – 120

y = 22

Hence, the first and second numbers are 40 and 22.Question 7

If 45 is subtracted from twice the greater of two numbers, it results in the other number. If 21 is subtracted from twice the smaller number, it results in the greater number. Find the number.Solution 7

Let the greater number be x and smaller be y respectively.

According to the question:

2x – 45 = y

2x – y = 45—(1)

and

2y – x = 21

-x + 2y = 21—(2)

Multiplying (1) by 2 and (2) by 1

4x – 2y = 90—(3)

-x + 2y = 21 —(4)

Adding (3) and (4), we get

3x = 111

Putting x = 37 in (1), we get

2 × 37 – y = 45

74 – y = 45

y = 29

Hence, the greater and the smaller numbers are 37 and 29.Question 8

If three times the larger of two numbers is divided by the smaller, we get 4 as the quotient and 8 as the remainder. If five times the smaller is divided by the larger, we get 3 as the quotient and 5 as the remainder. Find the numbers.Solution 8

Let the larger number be x and smaller be y respectively.

We know,

Dividend = Divisor × Quotient + Remainder

3x = y × 4 + 8

3x – 4y = 8 —(1)

And

5y = x × 3 + 5

-3x + 5y = 5 —(2)

Adding (1) and (2), we get

y = 13

putting y = 13 in (1)

Hence, the larger and smaller numbers are 20 and 13 respectively.Question 9

If 2 is added to each of two given numbers, their ratio becomes 1: 2. However, if 4 is subtracted from each of the given numbers, the ratio becomes 5: 11. Find the numbers.Solution 9

Let the required numbers be x and y respectively.

Then,

Therefore,

2x – y =-2—(1)

11x – 5y = 24 —(2)

Multiplying (1) by 5 and (2) by 1

10x – 5y = -10—(3)

11x – 5y = 24—(4)

Subtracting (3) and (4) we get

x = 34

putting x = 34 in (1), we get

2 × 34 – y = -2

68 – y = -2

-y = -2 – 68

y = 70

Hence, the required numbers are 34 and 70.Question 10

The difference between two numbers is 14 and the difference between their squares is 448. Find the numbers.Solution 10

Let the numbers be x and y respectively.

According to the question:

x – y = 14 —(1)

From (1), we get

x = 14 + y —(3)

putting x = 14 + y in (2), we get

Putting y = 9 in (1), we get

x – 9 = 14

x = 14 + 9 = 23

Hence the required numbers are 23 and 9Question 11

The sum of the digits of a two digit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number.Solution 11

Let the ten’s digit be x and units digit be y respectively.

Then,

x + y = 12—(1)

Required number = 10x + y

Number obtained on reversing digits = 10y + x

According to the question:

10y + x – (10x + y) = 18

10y + x – 10x – y = 18

9y – 9x = 18

y – x = 2 —-(2)

Adding (1) and (2), we get

Putting y = 7 in (1), we get

x + 7 = 12

x = 5

Number= 10x + y

= 10 × 5 + 7

= 50 + 7

= 57

Hence, the number is 57.
Question 12

A number consisting of two digits is seven times the sum of its digits. When 27 is subtracted from the number, the digits are reversed. Find the number.Solution 12

Let the ten’s digit of required number be x and its unit’s digit be y respectively

Required number = 10x + y

10x + y = 7(x + y)

10x + y = 7x + 7y

3x – 6y = 0—(1)

Number found on reversing the digits = 10y + x

(10x + y) – 27 = 10y + x

10x – x + y – 10y = 27

9x – 9y = 27

(x – y) = 27

x – y = 3—(2)

Multiplying (1) by 1 and (2) by 6

3x – 6y = 0—(3)

6x – 6y = 18 —(4)

Subtracting (3) from (4), we get

Putting x = 6 in(1), we get

3 × 6 – 6y = 0

18 – 6y = 0

Number = 10x + y

= 10 × 6 + 3

= 60 + 3

= 63

Hence the number is 63.
Question 13

The sum of the digits of a two-digit number is 15. The number of obtained by interchanging the digits exceeds the given number by 9. Find the number.Solution 13

Let the ten’s digit and unit’s digits of required number be x and y respectively.

Then,

x + y = 15—(1)

Required number = 10x + y

Number obtained by interchanging the digits = 10y + x

10y + x – (10x + y) = 9

10y + x – 10x – y = 9

9y – 9x = 9

Add (1) and (2), we get

Putting y = 8 in (1), we get

x + 8 = 15

x = 15 – 8 = 7

Required number = 10x + y

= 10 × 7 + 8

= 70 + 8

= 78

Hence the required number is 78.
Question 14

A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.Solution 14

Let the ten’s and unit’s of required number be x and y respectively.

Then,required number =10x + y

According to the given question:

10x + y = 4(x + y) + 3

10x + y = 4x + 4y + 3

6x – 3y = 3

2x – y = 1 —(1)

And

10x + y + 18 = 10y + x

9x – 9y = -18

x – y = -2—(2)

Subtracting (2) from (1), we get

x = 3

Putting x = 3 in (1), we get

2 × 3 – y = 1

y = 6 – 1 = 5

x = 3, y = 5

Required number = 10x + y

= 10 × 3 + 5

= 30 + 5

= 35

Hence, required number is 35.
Question 15

A number consists of two digits. When it is divided by the sum of its digits, the quotient is 6 with no remainder. When the number is diminished by 9, the digits are reversed. Find the number.Solution 15

Let the ten’s digit and unit’s digit of required number be x and y respectively.

We know,

Dividend = (divisor × quotient) + remainder

According to the given questiion:

10x + y = 6 × (x + y) + 0

10x – 6x + y – 6y = 0

4x – 5y = 0 —(1)

Number obtained by reversing the digits is 10y + x

10x + y – 9 = 10y + x

9x – 9y = 9

9(x – y) =9

(x – y) = 1—(2)

Multiplying (1) by 1 and (2) by 5, we get

4x – 5y = 0 —(3)

5x – 5y = 5 —(4)

Subtracting (3) from (4), we get

x = 5

Putting x = 5 in (1), we get

x =5 and y= 4

Hence, required number is 54.Question 16

A two – digit number is such that the product of its digits is 35. If 18 is added to the number, the digits interchanged their places. Find the number.Solution 16

Let the ten’s and unit’s digits of the required number be x and y respectively.

Then, xy = 35

Required number = 10x + y

Also,

(10x + y) + 18 = 10y + x

9x – 9y = -18

9(y – x) = 18—(1)

y – x = 2

Now,

Adding (1) and (2),

2y = 12 + 2 = 14

y = 7

Putting y = 7 in (1),

7 – x = 2

x = 5

Hence, the required number = 5 × 10 + 7

= 57Question 17

A two-digit number is such that the product of its digit is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.Solution 17

Let the ten’s and unit’s digits of the required number be x and y respectively.

Then, xy = 18

Required number = 10x + y

Number obtained on reversing its digits = 10y + x

(10x + y) – 63 = (10y + x)

9x – 9y = 63

x – y = 7—(1)

Now,

Adding (1) and (2), we get

Putting x = 9 in (1), we get

9 – y = 7

y = 9 – 7

y ₌2

x = 9, y = 2

Hence, the required number = 9 × 10 + 2

= 92.Question 18

The sum of a two-digit number and the number obtained by reversing the order of its digits is 121, and the two digits differ by 3. Find the number.Solution 18

Question 19

The sum of the numerator and denominator of a fraction is 8. If 3 is added to both of the numerator and the denominator, the fraction becomes Find the fraction.Solution 19

Let the numerator and denominator of fraction be x and y respectively.

According to the question:

x + y = 8—(1)

And

Multiplying (1) be 3 and (2) by 1

3x + 3y = 24—(3)

4x – 3y = -3 —(4)

Add (3) and (4), we get

Putting x = 3 in (1), we get

3 + y= 8

y = 8 – 3

y = 5

x = 3, y = 5

Hence, the fraction is Question 20

If 2 is added to the numerator of a fraction, it reduces to and if 1 is subtracted from the denominator, it reduces to . Find the fractionSolution 20

Let the numerator and denominator be x and y respectively.

Then the fraction is .

Subtracting (1) from (2), we get

x = 3

Putting x = 3 in (1), we get

2 × 3 – 4

-y = -4 -6

y = 10

x = 3 and y = 10

Hence the fraction isQuestion 21

The denominator of a fraction is greater than its numerator by 11. If 8 is added to both its numerator and denominator, it becomes Find the fraction.Solution 21

Let the numerator and denominator be x and y respectively.

Then the fraction is _.

According to the given question:

y = x + 11

y- x = 11—(1)

and

-3y + 4x = -8 —(2)

Multiplying (1) by 4 and (2) by 1

4y – 4x = 44—(3)

-3y + 4x = -8—(4)

Adding (3) and (4), we get

y = 36

Putting y = 36 in (1), we get

y – x = 11

36 – x = 11

x = 25

x = 25, y = 36

Hence the fraction is Question 22

Find a fraction which becomes when 1 is subtracted from the numerator and 2 is added to the denominator, and the fraction becomes when 7 is subtracted from the numerator and 2 is subtracted from the denominator.Solution 22

Let the numerator and denominator be x and y respectively.

Then the fraction is

Subtracting (1) from (2), we get

x = 15

Putting x = 15 in (1), we get

2 × 15 – y = 4

30 – y = 4

y = 26

x = 15 and y = 26

Hence the given fraction is Question 23

The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2 : 3. Determine the fraction.Solution 23

Question 24

The sum of two numbers is 16 and the sum of their reciprocals is Find the numbers.Solution 24

Let the two numbers be x and y respectively.

According to the given question:

x + y = 16—(1)

And

—(2)

From (2),

xy = 48

We know,

Adding (1) and (3), we get

2x = 24

x = 12

Putting x = 12 in (1),

y = 16 – x

= 16 – 12

= 4

The required numbers are 12 and 4Question 25

There are two classrooms A and B. If 10 students are sent from A to B, the number of students in each room becomes the same. If 20 students are sent from B to A, the number of students in A becomes double the number of students in B. Find the number of students in each room.Solution 25

Let the number of student in class room A and B be x and y respectively.

When 10 students are transferred from A to B:

x – 10 = y + 10

x – y = 20—(1)

When 20 students are transferred from B to A:

2(y – 20) = x + 20

2y – 40 = x + 20

-x + 2y = 60—(2)

Adding (1) and (2), we get

y = 80

Putting y = 80 in (1), we get

x – 80 = 20

x = 100

Hence, number of students of A and B are 100 and 80 respectively.Question 26

Taxi charges in a city consist of fixed charges and the remaining depending upon the distance travelled in kilometres. If a man travels 80 km, he pays Rs.1330, and travelling 90 km, he pays Rs.1490. Find the fixed charges and rate per km.Solution 26

Question 27

A part of monthly hostel charges in a college hostel 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 25 days, he has to pay Rs.4500, whereas a student B who takes food for 30 days, has to pay Rs.5200. Find the fixed charges per month and the cost of the food per day.Solution 27

Question 28

A man invested an amount at 10% per annum and another amount at 8% per annum simple interest. Thus, he received Rs. 1350 as an annual interest. Had he interchanged the amounts invested, he would have received Rs.45 less as interest. What amounts did he invest at different rates?Solution 28

Question 29

The monthly incomes of A and B are in the ratio 5 : 4 and their monthly expenditures are in the ratio 7 : 5. If each saves Rs.9000 per month, find the monthly income of eachSolution 29

Question 30

A man sold a chair and a table together for Rs.1520, thereby making a profit of 25% on chair and 10% on table. By selling them together for Rs.1535, he would have made a profit of 10% on the chair and 25% on the table. Find the cost price of each.Solution 30

Question 31

Points A and B are 70km apart on a highway. A car starts from A and another starts from B simultaneously. If they travel in the same direction, they meet in 7 hours. But, if they travel towards each other, they meet in 1 hour. Find the speed of each car.Solution 31

Let P and Q be the cars starting from A and B respectively and let their speeds be x km/hr and y km/hr respectively.

Case- I

When the cars P and Q move in the same direction.

Distance covered by the car P in 7 hours = 7x km

Distance covered by the car Q in 7 hours = 7y km

Let the cars meet at point M.

AM = 7x km and BM = 7y km

AM – BM = AB

7x – 7y = 70

7(x – y) = 70

x – y = 10 —-(1)

Case II

When the cars P and Q move in opposite directions.

Distance covered by P in 1 hour = x km

Distance covered by Q in 1 hour = y km

In this case let the cars meet at a point N.

AN = x km and BN = y km

AN + BN = AB

x + y = 70—(2)

Adding (1) and (2), we get

2x = 80

x = 40

Putting x = 40 in (1), we get

40 – y = 10

y = (40 – 10) = 30

x = 40, y = 30

Hence, the speeds of these cars are 40 km/ hr and 30 km/ hr respectively.Question 32

A train covered a certain distance at a uniform speed. If the train had been 5kmph faster, it would have taken 3 hours less than the scheduled time. And, if the train were slower by 4 kmph, it would have taken 3hours less than the scheduled time. Find the length of the journey.Solution 32

Let the original speed be x km/h and time taken be y hours

Then, length of journey = xy km

Case I:

Speed = (x + 5)km/h and time taken = (y – 3)hour

Distance covered = (x + 5)(y – 3)km

(x + 5) (y – 3) = xy

xy + 5y -3x -15 = xy

5y – 3x = 15 —(1)

Case II:

Speed (x – 4)km/hr and time taken = (y + 3)hours

Distance covered = (x – 4)(y + 3) km

(x – 4)(y + 3) = xy

xy -4y + 3x -12 = xy

3x – 4y = 12 —(2)

Multiplying (1) by 4 and (2) by 5, we get

20y – 12x = 60 —(3)

-20y + 15x = 60 —(4)

Adding (3) and (4), we get

3x = 120

or x = 40

Putting x = 40 in (1), we get

5y – 3 × 40 = 15

5y = 135

y = 27

Hence, length of the journey is (40 × 27) km = 1080 kmQuestion 33

Abdul travelled 300 km by train and 200 km by taxi taking 5 hours 30 minutes. But, if he travels 260 km by train and 240 km by taxi, he takes 6 minutes longer. Find the speed of the train and that of the taxi.Solution 33

Question 34

Places A and B are 160 km apart on a highway. One car starts from A and another from B at the same time. If they travel in the same direction, they meet in 8 hours. But, if they travel towards each other, they meet in 2 hours. Find the speed of each car.Solution 34

Question 35

A sailor goes 8 km downstream in 40 minutes and returns in 1 hour. Find the speed of the sailor in still water and the speed of the current.Solution 35

Question 36

A boat goes 12km upstream and 40km downstream in 8hours. It can go 16km upstream and 32 downstream in the same times. Find the speed of the boat in still water and the speed of the stream.Solution 36

Let the speed of the boat in still water be x km/hr and speed of the stream be y km/hr.

Then,

Speed upstream = (x – y)km/hr

Speed downstream = (x + y) km/hr

Time taken to cover 12 km upstream =

Time taken to cover 40 km downstream =

Total time taken = 8hrs

Again, time taken to cover 16 km upstream =

Time taken to taken to cover 32 km downstream =

Total time taken = 8hrs

Putting

12u + 40v = 8

3u + 10v = 2 —(1)

and

16u + 32v = 8

2u + 4v = 1—(2)

Multiplying (1) by 4 and (2) by 10, we get

12u + 40v = 8—(3)

20u + 40v = 10 —(4)

Subtracting (3) from (4), we get

Putting in (3), we get

On adding (5) and (6), we get

2x = 12

x = 6

Putting x = 6 in (6) we get

6 + y = 8

y = 8 – 6 = 2

x = 6, y = 2

Hence, the speed of the boat in still water = 6 km/hr and speedof the stream = 2km/hrQuestion 37

2 men and 5 boys can finish a piece of work in 4 days, while 3 men and 6 boys can finish it in 3days. Find the time taken by one man alone to finish the work and that taken by one boy alone to finish the work.Solution 37

Let man’s 1 day’s work be and 1 boy’s day’s work be

Also let and

Multiplying (1) by 6 and (2) by 5 we get

Subtracting (3) from (4), we get

Putting in (1), we get

x = 18, y = 36

The man will finish the work in 18 days and the boy will finish the work in 36 days when they work alone.Question 38

The length of a room exceeds its breadth by 3 meters. If the length is increased by 3 meters and the breadth is decreased by 2 meters, the area remains the same. Find the length and breadth of the room.Solution 38

Let the length = x meters and breadth = y meters

Then,

x = y + 3

x – y = 3 —-(1)

Also,

(x + 3)(y – 2) = xy

3y – 2x = 6 —-(2)

Multiplying (1) by 2 and (2) by 1

-2y + 2x = 6 —(3)

3y – 2x = 6 —(4)

Adding (3) and (4), we get

y = 12

Putting y = 12 in (1), we get

x – 12 = 3

x= 15

x = 15, y = 12

Hence length = 15 metres and breadth = 12 metresQuestion 39

The area of a rectangle gets reduced by 8 , when its length is reduced by 5m and its breadth is increased by 3m. If we increase the length by 3m and breadth by 2m, the area is increased by 74 . Find the length and breadth of the rectangle.Solution 39

Let the length of a rectangle be x meters and breadth be y meters.

Then, area = xy sq.m

Now,

xy – (x – 5)(y + 3) = 8

xy – [xy – 5y + 3x -15] = 8

xy – xy + 5y – 3x + 15 = 8

3x – 5y = 7 —(1)

And

(x + 3)(y + 2) – xy = 74

xy + 3y +2x + 6 – xy = 74

2x + 3y = 68—(2)

Multiplying (1) by 3 and (2) by 5, we get

9x – 15y = 21—(3)

10x + 15y = 340—(4)

Adding (3) and (4), we get

Putting x = 19 in (3) we get

x = 19 meters, y = 10 meters

Hence, length = 19m and breadth = 10mQuestion 40

The area of a rectangle gets reduced by 67 square metres, when its length is increased by 3 m and breadth is decreased by 4 m. If the length. is reduced by 1 m and breadth is increased by 4 m, the area is increased by 89 square metres. Find the dimensions of the rectangle.Solution 40

Question 41

A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket. One reserved first class ticket from Mumbai to Delhi costs Rs.4150 while one full and one half reserved first class tickets cost Rs.6255. What is the basic first class full fare and what is the reservation charge?Solution 41

Question 42

Five years hence, a man’s age will be three times the age of his son. Five years ago, the man was seven times as old as his son. Find their present ages.Solution 42

Question 43

Two years ago, a man was five times as old as his son. Two years later, his age will be 8 more than three times the age of the son. Find the present ages of man and his son.Solution 43

Let the present ages of the man and his son be x years and y years respectively.

Then,

Two years ago:

(x – 2) = 5(y – 2)

x – 2 = 5y – 10

x – 5y = -8 —(1)

Two years later:

(x + 2) = 3(y + 2) + 8

x + 2 = 3y + 6 + 8

x – 3y = 12 —(2)

Subtracting (2) from (1), we get

-2y = -20

y = 10

Putting y = 10 in (1), we get

x – 5 10 = -8

x – 50 = -8

x = 42

Hence the present ages of the man and the son are 42 years and 10 respectively.Question 44

If twice the son’s age in years is added to the mother’s age, the sum is 70 years. But, if twice the mother’s age is added to the son’s age, the sum is 95years. Find the age of the mother and that of the son.Solution 44

Let the present ages of the mother and her son be x and y respectively.

According to the given question:

x + 2y = 70—(1)

and

2x + y = 95—(2)

Multiplying (1) by 1 and (2) by 2, we get

x + 2y = 70 —(3)

4x + 2y = 190—(4)

Subtracting (3) from (4), we get

Putting x = 40 in (1), we get

40 + 2y = 70

2y = 30

y = 15

x = 40, y = 15

Hence, the ages of the mother and the son are 40 years and 15 years respectively.Question 45

The present age of a woman is 3 years more than three times the age of her daughter; three years hence, the woman’s age will be 10 years more than twice the age of her daughter. Find their present ages.Solution 45

Let the present ages of woman and daughter be x and y respectively.

Then,

Their present ages:

x = 3y + 3

x – 3y = 3—(1)

Three years later:

(x + 3) = 2(y + 3) + 10

x + 3 = 2y + 6 + 10

x – 2y = 13—(2)

Subtracting (2) from (1), we get

y = 10

Putting y = 10 in (1), we get

x – 3 × 10 = 3

x = 33

x = 33, y = 10

Hence, present ages of woman and daughter are 33 and 10 years.Question 46

On selling a tea set at 5% loss and a lemon set at 15% gain, a crockery seller gains Rs.7. If he sells the tea set at 5% gain and the lemon set at 10% gain, he gains Rs.13. Find the actual price of each of the tea set and the lemon set.Solution 46

Question 47

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Mona paid Rs.27 for a book kept for 7 days, while Tanvy paid Rs. 21 for the book she kept for 5 days. Find the fixed charge and the charge for each extra day.Solution 47

Question 48

A chemist has one solution containing 50% acid and a second one containing 25% arid. How much of each should be used to make 10 litres of a 40% arid solution?Solution 48

Question 49

A jeweller has bars of 18-carat gold and 12-carat gold. How much of each must be melted together to obtain a bar of 16-carat gold, weighing 120 g? (Given: Pure gold is 24-carat)Solution 49

Question 50

90% and 97% pure acid solutions are mixed to obtain 21 litres of 95% pure acid solution. Find the quantity of each type of adds to be mixed to form the mixture.Solution 50

Question 51

The larger of the two supplementary angles exceeds the smaller by 18^o. Find them.Solution 51

Question 52

Solution 52

Question 58

Solution 58

Exercise Ex. 7F

Question 1

Write the number of solutions of the following pair of linear equations:

x + 2y – 8 = 0, 2x + 4y = 16.Solution 1

Question 2

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

2x + 3y = 7, (k – 1) x + (k + 2) y = 3k.Solution 2

Question 3

For what value of k does the following pair of linear equations have infinitely many solutions?

10x + 5y – (k – 5) = 0 and 20x + 10y – k = 0.Solution 3

Question 4

For what value of k will the following pair of linear equations have no ‘solution?

2x + 3y = 9, 6x + (k – 2)y = (3k – 2).Solution 4

Question 5

Write the number of solutions of the following pair of linear equations:

x + 3y – 4 = 0 and 2x + 6y – 7 = 0.Solution 5

Question 6

Write the value of k for which the system of equations 3x + ky = 0, 2x – y = 0 has a unique solution.Solution 6

Question 7

The difference between two numbers is 5 and the difference between their squares is 65. Find the numbers.Solution 7

Question 8

The cost of 5 pens and 8 pencils is Rs.120, while the cost of 8 pens and 5 pencils is Rs.153. Find the cost of 1 pen and that of I pencil.Solution 8

Question 9

The sum of two numbers is 80. The larger number exceeds four times the smaller one by 5. Find the numbers.Solution 9

Question 10

A number consists of two digits whose sum is 10. If 18 is subtracted from the number, its digits are reversed. Find the number.Solution 10

Question 11

A man purchased 47 stamps of 20 p and 25 p for Rs.10. Find the number of each type of stamps.Solution 11

Question 12

A man has some hens and cows. If the number of heads be 48 and number of feet be 140, how many cows are there?Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

If 12x + 17y = 53 and 17x + 12y = 63 then find the value of (x +y).Solution 15

Question 16

Find the value of k for which the system 3x + 5y = 0, kx + 10y = 0 has a nonzero solution.Solution 16

Question 17

Find k for which the system kx – y = 2 and 6x – 2y = 3 has a unique solution.Solution 17

Question 18

Find k for which the system 2x + 3y – 5 = 0, 4x + ky – 10 = 0 has an infinite number of solutions.Solution 18

Question 19

Show that the system 2x + 3y – 1 = 0, 4x + 6y – 4 = 0 has no solution.Solution 19

Question 20

Find k for which the system x + 2y = 3 and 5x + ky + 7 = 0 is inconsistent.Solution 20

Question 21

Solution 21

Exercise MCQ

Question 1

If 2x + 3y = 12 and 3x – 2y = 5 then

(a) x = 2, y = 3

(b) x = 2, y = -3

(d) x = 3, y = -2Solution 1

Question 2

(a) x = 4, y = 2

(b) x = 5, y = 3

(d) x = 7, y = 5Solution 2

Question 3

(a) x = 2, y = 3

(b) x = -2, y = 3

(d) x = -2, y = -3Solution 3

Question 4

Solution 4

Question 5

(a) x = 1, y = 1

(b) x = -1, y = -1

(d) x = 2, y = 1Solution 5

Question 6

Solution 6

Question 7

If 4x+6y=3xy and 8x+9y=5xy then

(a) x=2, y=3

(b) x=1, y=2

(d) x=1, y=-1Solution 7

Question 8

If 29x+37y=103 and 37x+29y=95 then

(a) x=1, y=2

(b) x=2, y=1

(d) x=2, y=3Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

The system kx – y = 2 and 6x – 2y = 3 has a unique solution only when

(a) k = 0

(b) k ≠ 0

(d) k ≠ 3Solution 11

Question 12

The system x – 2y = 3 and 3x + ky = 1 has a unique solution only when

(a) k = -6

(b) k ≠ -6

(d) k ≠ 0Solution 12

Question 13

The system x+2y=3 and 5x+ky+7=0 has no solution, when

(a) k=10

(b) k≠10

(c)

(d) K=-21Solution 13

Question 14

If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel then the value of k is

Solution 14

Question 15

For what value of k do the equations kx – 2y = 3 and

3x + y = 5 represent two lines intersecting at a unique point?

(a) k=3

(b) k=-3

(d) all real values except -6Solution 15

Question 16

The pair of equations x + 2y + 5 = 0 and -3x – 6y + 1 = 0 has

(a) a unique solution

(b) exactly two solutions

(d) no solutionSolution 16

We space know space that comma
the space system space of space linear space equations space ax subscript 1 plus by subscript 1 plus straight c subscript 1 equals 0 space and space ax subscript 2 plus by subscript 2 plus straight c subscript blank subscript 2 end subscript equals 0
has space no space solution space if space 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.
So comma space the space pair space of space equations space has space no space solution.

Question 17

The pair of equations 2x + 3y = 5 and 4x + 6y = 15 has

(a) a unique solution

(b) exactly two solutions

(d) no solutionSolution 17

Question 18

If a pair of linear equations is consistent then their graph lines will be

(a) parallel

(b) always coincident

(d) intersecting or coincidentSolution 18

Question 19

If a pair of linear equations is inconsistent then their graph lines will be

(a) parallel

(b) always coincident

(d) intersecting or coincidentSolution 19

Question 20

In a ΔABC, ∠C = 3 ∠B = 2 (∠A + ∠B), then ∠B = ?

(a) 20°

(b) 40°

(d) 80° Solution 20

Question 21

In a cyclic quadrilateral ABCD, it is being given that

∠A = (x + y + 10) °, ∠B = (y + 20) °,

∠C = (x + y – 30)° and ∠D = (x + y)°. Then, ∠B = ?

(a) 70°

(b) 80°

(d) 110° Solution 21

Question 22

The sum of the digits of a two-digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. The number is

(a) 96

(b) 69

(d) 78Solution 22

Question 23

Solution 23

Question 24

5 years hence, the age of a man shall be 3 times the age of his son while 5 years earlier the age of the man was 7 times the age of his son. The present age of the man is

(a) 45 years

(b) 50 years

(d) 40 yearsSolution 24

Question 25

The graphs of the equations 6x – 2y + 9 = 0 and

3x – y + 12 = 0 are two lines which are

(a) coincident

(b) parallel

(d) perpendicular to each otherSolution 25

Question 26

The graphs of the equations 2x+3y-2=0 and x-2y-8=0 are two lines which are

(a) coincident

(b) parallel

(d) perpendicular to each otherSolution 26

Question 27

(a) coincident

(b) parallel

(d) perpendicular to each otherSolution 27

Exercise FA

Question 1

The graphic representation of the equations x+2y=3 and 2x+4y+7=0 gives a pair of

(a) parallel lines

(b) intersecting lines

(d) none of theseSolution 1

Question 2

If 2x – 3y = 7 and (a + b) x – (a + b – 3) y = 4a+b have an infinite number of solutions then

(a) a= 5, b = 1

(b) a = -5, b = 1

(d) a = -5, b = -1Solution 2

Question 3

The pair of equations 2x+y=5, 3x+2y=8 has

(a) a unique solution

(b) two solutions

(d) infinitely many solutionsSolution 3

Question 4

If x = -y and y > 0, which of the following is wrong?

(a) x²y > 0

(b) x + y = 0

(d) Solution 4

Question 5

Solution 5

Question 6

For what values of k is the system of equations kx + 3y = k – 2, 12x + ky = k inconsistent?Solution 6

Question 7

Solution 7

Question 8

Solve the system of equations x – 2y = 0, 3x + 4y = 20.Solution 8

Question 9

Show that the paths represented by the equations x – 3y = 2 and -2x + 6y = 5 are parallel.Solution 9

Question 10

The difference between two numbers is 26 and one number is three times the other. Find the numbers.Solution 10

Question 11

Solve : 23x+29y=98, 29x+23y=110.Solution 11

Question 12

Solve : 6x+3y=7xy and 3x+9y = 11xy.Solution 12

Question 13

Find the value of k for which the system of equations 3x+y=1 and kx+2y=5 has (i) a unique solution, (ii) no solution.Solution 13

Question 14

In a ΔABC, ∠C =3∠B =2(∠A+∠B). Find the measure of each one of the ∠A, ∠B and ∠C. Solution 14

Question 15

5 pencils and 7 pens together cost Rs. 195 while 7 pencils and 5 pens together cost Rs. 153. Find the cost of each one of the pencil and the pen.Solution 15