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MATHEMATICS–X PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 25

CHAPTER 3PAIR OF LINEAR EQUATIONS

IN TWO VARIABLES

Points to Remember :1. A pair of linear equations in two variables x and y can be represented as follows :

a1x + b1y + c1 = 0; a2x + b2y + c2 = 0,

where a1, a2, b1, b2, c1, c2 are real numbers such that 2 2 2 21 1 2 20, 0.a b a b

2. Graphically, a pair of linear equations a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 in two variables represents a pairof straight lines which are :

(i) Intersecting, if 1 1

2 2

a ba b

here, the equations have a unique solution, and pair of equations is said to be consistent.

(ii) parallel, if 1 1 1

2 2 2

a b ca b c

here, the equations have No solution, and pair of equations is said to be inconsistent.

(iii) Coincident, if 1 1 1

2 2 2

a b ca b c

here, the equations have infinitely many solutions, and pair of equations is said to be consistent.3. A pair of linear equations in two variables can be solved by the :

(i) Graphical method(ii) Algebraic Methods; which are of three types :

(a) Substitution method (b) Elimination method (c) Cross-multiplication method

ILLUSTRATIVE EXAMPLES

Example 1. Draw the graph of linear equation 2x + 3y = 7.

Solution. 2x + 3y = 7 3y = 7 – 2x 7 23

xy

Give atleast two suitable values to x to find the corresponding value of y.

If 7 2(2) 32, 13 3

x y

If 7 2(5) 35, 1

3 3x y

AMIT BAJA

J

26 PAIR OF LINEAR EQUATIONS IN TWO VARIABLES MATHEMATICS–X

Representing it in a tabular form, we get

2 51 1

xy

We now plot the points (2, 1) and (5, –1) on graph paper to obtain a straight line.

Y-axis

X-axis

4

3

2

1

–1

–3

–4

–2

–1–2–3 0 1 2 3 4 5 6

(5, –1)

(2, 1)

2 +3 = 7

xy

Example 2. Aftab tells his daughter, ‘‘Seven years ago, I was seven times as old as you were then. Also, threeyears from now, I shall be three times as old as you will be. Represent this situation algebraicallyand graphically. [NCERT]

Solution. Let the present age of the daughter = x years.7 years ago daughter’s age = (x – 7) years3 years from now, daughter’s age = (x + 3) yearsLet the present age of father = y years7 years ago father’s age = ( y – 7) years3 years from now, fathers age = (y + 3) years.According to given question, two algebraic equations are :

y – 7 = 7 (x – 7) y = 7x – 42 ...(1)and y + 3 = 3 (x + 3) y = 3x + 6 ...(2)Required table for y = 7x – 42

12 1842 84

xy

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and, required table for y = 3x + 6 is

6 1224 42

xy

X6 12 18 24 30

12

24

36

48

60

72

84

96

YA

ge o

f Fat

her

Age of daughter

(12, 42)

(6, 24)

(18, 84)

yx

= 3

+ 6

yx

= 7

– 4

2

From graph, we observe that :Present age of father is 42 years and, present age of his daughter is 12 years.

Example 3. Use a single graph paper and draw the graph of the following equations :2y – x = 8; 5y – x = 14; y – 2x = 1.

Obtain the vertices of the triangle so obtained.Solution. For equation 2y – x = 8

We have, 2y – x = 8 x = 2y – 8When y = 2, we have x = 2 (2) – 8 = 4 – 8 = – 4when y = 3, we have x = 2(3) – 8 = 6 – 8 = – 2Thus, we have the following table :

–4 –22 3

xy

For equation 5y – x = 14We have, 5y – x = 14 x = 5y – 14when y = 3, we have x = 5(3) – 14 = 15 – 14 = 1

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when y = 4, we have x = 5 (4) – 14 = 20 – 14 = 6Thus, we have the following table :

1 63 4

xy

For equation y – 2x = 1We have, y – 2x = 1 y = 2x + 1when x = – 1, we have y = 2 (–1) + 1 = –2 + 1 = – 1when x = 0, we have y = 2 (0) + 1 = 0 + 1 = 1Thus, we have the following table :

1 01 1

xy

The graph for the given equations is shown below :

X

4

3

2

1

–1

–3

–4

–2

–1–2–3 0 1 2 3 4 5 6 7 8

5

6

–5 –4–6–7

Y

(2,5)

(6, 4)

(1,3)(–2,3)

(–4, 2)

2 – = 8y xy

x–2

=15 – = 14y x

From graph, we observe that the vertices of the triangle are (–4, 2), (1, 3) and (2, 5). Example 4. Draw the graphs of the equations :

x – y + 1 = 0; 3x + 2y – 12 = 0.Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis andshade the triangular region. Also, find area of this triangular region.

Solution. For equation x – y + 1 = 0We have, x – y + 1 = 0 y = x + 1when, x = 0, we have y = 0 + 1 = 1when, x = –1, we have y = –1 + 1 = 0Thus, we have the following table :

0 11 0

xy

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For equation 3x + 2y – 12 = 0

we have, 3x + 2y – 12 = 0 12 3

2xy

when x = 0, we have 12 3(0) 12 6

2 2y

when x = 4, we have 12 3(4) 12 12 0

2 2y

Thus, we have the following table :

0 46 0

xy

The graph for the given equations is shown below.

X1 2 3 4 5 6 7

1

2

3

4

5

6

7

Y

A (2, 3)

(0, 1)

(0, 6)

–2 –1

B (4, 0)C (–1, 0)0

3+2

–12=0

xy

x y– +1=0

From graph, we observe that the vertices of ABC are A(2, 3), B(4, 0) and C(–1, 0).

Also, Area of 1ABC 5 32

sq. units = 7.5 sq. units.

Example 5. Find the values of a and b so that the following system of linear equations has an infinite numberof solutions : 2x – 3y = 7

(a + b)x – (a + b – 3) y = 4a + b [NCERT, CBSE 2002]

Solution. Here, 1 1 1

2 2 2

2 3 7, and( 3) 4

a b ca a b b a b c a b

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For infinite solutions, we have 1 1 1

2 2 2

a b ca b c

2 3 7

3 4a b a b a b

Taking Ist two terms : Taking last two terms :

2 3

3a b a b

3 7

3 4a b a b

3a + 3b = 2a + 2b – 6 12a + 3b = 7a + 7b – 21 a + b = – 6 ...(1) 5a – 4b = – 21 ...(2)From equation (1) we have a = – 6 – bUsing the value of a in equation (2), we get

5(–6 ) 4 21b b

–30 – 5b – 4b = – 21 –9b = 9 b = – 1and a = – 6 – (–1) = – 6 + 1= – 5Hence, a = – 5 and b = – 1 Ans.

Example 6. Use the method of substitution to solve the following system of equations :

1 8 1 3 54x+ y = ; x+ y = –3 3 2 4 2

Solution. Consider, 1 843 3

x y

12x + y = 8 ...(1)

and,1 3 52 4 2

x y

2 3 10x y ...(2)

Now, we will solve equation (1) and (2) by substitution method.

From (1), we get y = 8 – 12 x.

Now, substituting the value of y in eqn. (2), we get

2x + 3 (8 – 12x) = –10

2x + 24 – 36 x = –10

– 34x = – 34 x = 1

Now, substitute x = 1 in y = 8 – 12 x, we get

y = 8 – 12 (1) = 8 – 12 = – 4

Thus x = 1, y = – 4 is the required solution.AMIT B

AJAJ


Example 7. Solve the following system of linear equations by elimination method (equating the coeffi-cients).6 (ax + by) = 3a + 2b6 (bx – ay) = 3b – 2a [CBSE 2004, 2006 outside]

Solution. Given equations :6 (ax + by) = 3a + 2b ...(1)6 (bx – ay) = 3b – 2a ...(2)

multiplying eqn. (1) by a and equation (2) by b, and adding, we get6a2x + 6aby = 3a2 + 2ab6b2x – 6aby = 3b2 – 2ab

6 (a2 + b2) x = 3 (a2 + b2)

2 2

2 2

3 ( ) 126 ( )

a bxa b

Putting 12

x in eqn. (1), we get

16 6 3 22

a by a b

3a + 6by = 3a + 2b

2 16 3byb

1 1= , =2 3

x y is the required solution.

Example 8. Solve the following system of linear equations by using the method of cross-multiplication :ax + by = a – bbx – ay = a + b [CBSE 2000, 2005]

Solution. The given system of equations may be written asax + by – (a – b) = 0bx – ay – (a + b) = 0

Using cross-multiplication method, we getx

b a b – ( – )

1

b a –

–y

a a b – ( – )= =

–a a b – ( + ) b a b – ( + )

a b

2 2

1[ ( )] ( ) [ ( )] [ ( )] [ ( )]

x yb a b a a b a a b b a b a b

2 2

1( ) ( ) ( ) ( ) ( )

x yb a b a a b a a b b a b a b

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2 2 2 2 2 2

1( )

x yb a a b a b

2 2 2 2

2 2 2 2

( ) ( )1 1( ) ( )a b a bx and ya b a b

Thus, x =1 , y = – 1 is the required solution.Example 9. Solve for x and y :

2 1 –5+ =x + 2y 2x - y 9

9 6+ = –4x+ 2y 2x - y

Solution. Let 1 1and .2 2

u vx y x y

Putting in the given equations, we get

529

u v ...(1)

9 6 4u v ...(2)multiplying eqn. (1) by 6, and eqn. (2) by 1, we get

1012 63

u v ...(3)

9 6 4u v ...(4)Subtracting eqn. (4) from eqn. (3), we get

12u – 9u = 10 43

2 233 9

u u

Putting 29

u in eqn. (2), we get

29 6 4 19

v v

Now, 2 , 19

u v

x + 2y = 92

...(5)

and 2x – y = – 1 ...(6)multiplying eqn. (5) by 1, and eqn. (6) by – 2, we get

922

x y ...(7)

–4x + 2y = 2 ...(8)AMIT B

AJAJ


Subtracting eqn. (8) from eqn. (7), we get

9 5 15 22 2 2

x x

Putting 12

x in eqn. (6), we get

1 – y = – 1 y = 2

Thus, 1= , = 22

x y is the required solution.

Example 10. 2 tables and 3 chairs together cost Rs. 2000 whereas 3 tables and 2 chairs together cost Rs.2500. Find the total cost of 1 table and 5 chairs.

Solution. Let the cost of a table be Rs. x and that of a chair be Rs. y. According to given question,2x + 3y = 2000 ...(1)

and 3x + 2y = 2500 ...(2)Adding eqn. (1) and (2), we get

5x + 5y = 4500 x + y = 900 ...(3)Subtracting eqn. (1) from eqn. (2), we get

x – y = 500 ...(4)Adding eqn. (3) and eqn. (4), we get

2x = 1400 x = 700Using x = 700 in eqn. (3), we get

700 + y = 900 y = 200 Cost of 1 table = Rs. 700 and cost of 1 chair = Rs. 200.Hence, cost of 1 table and 5 chairs

= Rs. (x + 5y) = Rs. 700 + 5 (200) = Rs. 1700 Ans.Example 11. The sum of a two-digit number and the number obtained by reversing the order of digits is 165.

If the digits differ by 3, find the number. [CBSE 2002]Solution. Let the unit digit be x and tens digit be y.

Then, number = 10 y + x.Number obtained by reversing the order of the digits = 10x + y.According to the given question, we have

(10y + x) + (10 x + y) = 165 ...(1)and, x – y = 3 ...(2)

ORy – x = 3 ...(3)

From eqn. (1), we get11x + 11y = 165 x + y = 15 ...(4)

Solving eqn. (2) and (4) together, we getx = 9, y = 6

Solving eqn. (3) and (4) together, we getx = 6, y = 9

Substituting the values of x and y, for the number 10 y + x, we get number = 69 or 96. Ans.

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Example 12. A fraction is such that if the numerator is multiplied by 3 and the denominator is reduced by 3,

we get 18 ,11

but if the numerator is increased by 8 and the denominator is doubled, we get 25 .

Find the fraction.

Solution. Let the fraction be .xy

Then, according to given equestion, we have :

3 18 8 2and3 11 2 5

x xy y

11x = 6y – 18 and 5x + 40 = 4y 11x – 6y = – 18 ...(1)and 5x – 4y = – 40 ...(2)multiplying eqn. (1) by 2 and eqn. (2) by 3, we get :

22x – 12 y = – 36 ...(3)15x – 12y = – 120 ...(4)

Subtracting eqn. (4) from eqn. (3), we get7x = 84 x = 12

Substituting x = 12 in eqn. (1), we get11(12) – 6y = – 18

–6y = – 150 y = 25.

Hence, the required fraction is 12 . Ans.35

Example 13. A man travels 370 km partly by train and partly by car. If he covers 250 km by train and the restby car, it takes him 4 hours. But, if he travels 130 km by train and the rest by car, he takes 18minutes longer. Find the speed of the train and that of the car. [CBSE 2001]

Solution. Let the speed of the train be x km/h and that of the car be y km/hr.Case I : When he travels 250 km by train and rest by car :In this case, we have

Time taken by the man to travel 250 km by train 250x

hrs .

Time taken by the man to travel (370 – 250) km = 120 km by car 120y

hrs.

Total time taken by man to cover 370 km 250 120 hrs.x y

It is given that the total time taken is 4 hours.

250 120 125 604 2

x y x y ...(1)

Case II. When he travels 130 km by train and rest by car :

Time taken by the man to travel 130 km by train 130 hrs.x

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Time taken by the man to travel (370 – 130) km = 240 km by car = 240 hrs.y

In this case, total time of the journey is 4 hrs. 18 minutes.

130 240x y

4 hr 18 minutes

130 240 184

60x y

130 240 43

10x y ...(2)

Now, we will solve eqn. (1) and (2).

Putting 1 1, ,u vx y the above equations reduces to

125 u + 60 v = 2 ...(3)

130 u + 240 v = 4310 ...(4)

multiplying eqn. (3) by 4, and eqn. (4) by 1, we get500 u + 240 v = 8 ...(5)

130 u + 240 v = 4310 ...(6)

subtracting eqn. (6) from eqn. (5), we get

43 37 1370 8 37010 10 100

u u u

Putting 1100

u in eqn. (5), we get

5 + 240 v = 8 240 v = 3 180

v

Now, 1 1and100 80

u v

1 1 1 1and 100, 80

100 80x y

x y

Hence, the speed of train = 100 km/hr and the speed of car = 80 km/hr.Example 14. A part of monthly hostel charges is fixed and the remaining depends on the number of days one

has taken food in the mess. When a student A takes food for 20 days she has to pay Rs. 1000 ashostel charges whereas a student B, who takes food for 26 days, pays Rs. 1180 as hostel charges.Find the fixed charges and the cost of food per day. [CBSE 2000, NCERT]

Solution. Let the fixed charges be Rs. x and charges per day be Rs. y.According to given question,

x + 20 y = 1000 ...(1)x + 26 y = 1180 ...(2)

Subtracting eq. (1) from eqn. (2), we get6y = 180 y = 30

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from eqn. (1), x + 20(30) = 1000 x = 1000 – 600 = 400 Fixed charges = Rs. 400 and cost of food per day = Rs. 30.

Example 15. The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units andbreadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units,the area increases by 67 square units. Find the dimensions of the rectangle. [NCERT]

Solution. Let the length and breadth of the rectangle be x and y units respectively.Then, Area = xy sq. unitsIf the length is reduced by 5 units and breadth is increased by 3 units, then area is reduced by 9 sq.units. xy – 9 = (x – 5) (y + 3) xy – 9 = xy + 3x – 5y – 15 3x – 5y = 6 ...(1)When the length is increased by 3 units and breadth by 2 units, the area is increased by 67 sq.units. xy + 67 = (x + 3) (y + 2) xy + 67 = xy + 2x + 3y + 6 2x + 3y = 61 ...(2)Solving Eq. (1) and eq. (2), we get

1305 18 183 12 9 10

x y

323 17117, 919 19

x y

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

PRACTICE EXERCISE

Section - A

1. Draw the graph of the equation 2y – x = 7 and determine from the graph whether x = 3, y = 2 is a solutionor not?

Draw the graph of the following equations (2-10)

2. 3x + 2y = 5 3. x – 3y = 2 4. x = 4 5. y = – 56. 2x – 3y = 12 7. 3x + 5y = – 18 8. 3y – 2x = 30 9. x + 3y = – 1

10. y – 4x = 12

Solve the following system of equations graphically (11-20)

11. 3x + 2y = 6 12. 5x – y = 7 13. x – 2y = 6 14. x – 4y + 14 = 03x – 4y = – 12 x – y = – 1 3x – 6y = 9 3x + 2y – 14 = 0

15. 2x + 3y = – 5 16. 3x + 2y = 5 17. 5x + 4y + 6 = 0 18. 4x + 2y = 1 x – 2y = 8 5x – 3y = 2 2x + 7y – 3 = 0 2x + y = – 4AMIT B

AJAJ


19. 5x + y – 7 = 0 20. 2x + 3y = 10

2x + 5y = 12 3 52

x y

In each of the following system of linear equations, determine graphically whether the given system has aunique solution, no solution or infinitely many solutions (21-30)

21. 6x + 4y = 7 22. 3x – 5y + 1 = 0 23. x – 4y = 10 24. –2x + y = 63x + 2y = 10 4x – y + 8 = 0 3x – 12y = 30 4x + 2y = 12

25. 2y – 3x = 21 26. 3x – y = 7 27. 7x – 3y = 9 28. x – y = 8

–x + 23

y = 7 7x – 2y = 15 x + y = 4 4x – 4y = 12

29. 3x + y = 5 30. –y + 3x = 76x + 2y = 10 7x – 2y – 15 = 0

Show graphically that each one of the following systems of equations has infinitely many solutions (31-33)

31. 2 4 62 3

y xx y

32. 3 2 89 6 24

x yx y

33. 2 6 02 4 12 0x yx y

Show graphically that each one of the following systems of equations is in-consistent (i.e. has no solution)(34-36)

34. 2 96 3 21

x yy x

35. 2 6 04 8 5x yx y

36. 2 3 66 9 10

x yx y

Solve graphically each of the following systems of linear equations. Also find the co-ordinates of the pointswhere the lines meet y-axis. (37-39)

37. 2 5 4 02 8 0

x yx y [CBSE 2005]

38. 3 2 125 2 4

x yx y [CBSE 2006(C)]

39. 3 5 02 5 0x yx y

[CBSE 2002(C)]

Solve graphically each of the following systems of linear equations. Also find the co-ordinates of the pointswhere the lines meet x-axis. (40-42)

40. 2 3 42 5

x yx y

41. 2 62 2 0

x yx y

42. 2 03 2 9 0x yx y

Solve graphically the following system of equations. Shade the region bounded by these lines and x-axis.Find the area of the shaded region. (43-45)

43. 2 62 2 0

x yx y

[CBSE 2002]

44. 52 2 0x yx y

45. 1 03 2 12 0x yx y

[CBSE 2002]

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Solve graphically the following system of equations. Shade the region bounded by these lines and y-axis.Find the area of the shaded region. (46-48).

46. 33 2 4x yx y

47. 4 5 20 03 5 15 0

x yx y

48. 12 8x yx y

[CBSE 2001]

Solve the following system of linear equations graphically. Also, find the vertices of the triangle formed bythese lines and x-axis. (49-51)

49. 5 6 30 05 4 20 0

x yx y

[CBSE 2004]

50. 3 9 03 4 6 0

x yx y

[CBSE 2006]

51. 4x – y – 8 = 02x – 3y + 6 = 0

Solve the following system of linear equations graphically. Also, find the vertices of the triangle formed bythese lines and y-axis (52-54) :

52. 3x + y – 11 = 0 53. 2x + 3y = 12

x – y – 1 = 0 [CBSE 2000 (C)] x – y = 1 [CBSE 2001]

54. 3x + y – 5 = 0

2x – y – 5 = 0 [CBSE 2002 (C)]

Determine graphically the co-ordinates of the vertices of a triangle, the equations of whose sidesare (55-60):

55.3

8

y xy x

x y

[CBSE 2000]

56. 2 3 6 02 3 18 0

2

x yx y

y

57. 2 85 14

2 1

y xy x

y x

58. 55

0

x yx yx

59. 2 52 73

x yx yx

60. 32 5 12

0

x yx y

y

Solve the following system of equations using the method of substitution (61-69)

61. 2x + 3y = 4 62. x + 2y = – 1 63. 2x – 7y = 15x + 8y = 9 2x – 3y = 12 4x + 3y = – 15

64. 4 3 13 4 18

x yx y

65. 0.4 0.3 1.70.7 0.2 0.8

x yx y

66. 113 45 76 3

x y

x y

67.322322

x y

x y

68. 6 5 119 10 21

x yx y

69. 2 5 37 2 4

x yx y AMIT BAJA

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Solve the following system of equations using the method of elimination (by equating the coefficients)(70-78) :

70. 712 5 7x y

x y

71. 7 8 118 7 7

x yx y

72. 3 5 711 13 9

x yx y

73. 6 24 52 3 2

x yx y

74. 3 2 115 8

x yx y

75. 2 7 35 3 13

x yx y

76. x + y = a + b 77. 3 (bx + ay) = a – 6b 78. ax + by = a – bax – by = a2 – b2 3 (ax – by) = – (6a + b) [CBSE 2004] bx – ay = a + b [CBSE 2000]

Solve the following system of equations using the method of cross-multiplication (79-87):

79. 2x + 3y = 17 80. 3x + 2y = – 25 81. 2x + y = 353x – 2y = 6 2x + y = – 10 3x + 4y = 65

82. 5 2 53 2

x yx y

83. 24 1 03

6 2 0

x y

x y

84. 2 3 10 03 4 2 0

x yx y

85.

2 2 2

x y a ba bx y

a b

86. ax by a bbx ay a b

[CBSE 2000, 2005]

87.

2 2

2x ya bax by a b

[CBSE 2005]

Solve the following system of equations (88-110) :

88. 1.5 0.1 6.23 0.4 11.2

x yx y

89. 34 6

22

x y

x y

90. 101 99 49999 101 501

x yx y

91. 1 1 121 1 8, 0, 0

2

x y

x yx y

92. 2 2

4

x ya bx ya b

93. 4 3 14

3 4 23

yx

yx

[CBSE 2004(C)]

94. 6 6

83 5

xy

xy

[CBSE 2007]

95. 1 1 82 3

1 1 93 2

x y

x y

[CBSE 2007]

96. 47 31 6331 47 15

x yx y

[CBSE 2006]AMIT BAJA

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97.3 3

x y a ba bax by a b

[CBSE 2005]

98. 3(2 ) 73( 3 ) 11

x y xyx y xy

99. 2

6

x yxy

x yxy

100. 1 1 125 61 3 83 7

x y

x y

101. 22 15 5

55 45 14

x y x y

x y x y

102.6 7 3

1 12( ) 3( )

x y x y

x y x y

103. 2

2

ax by aay bx b

104. 0.9

11 2

x y

x y

105. 1 5 32( 2 ) 3(3 2 ) 2

5 3 614( 2 ) 5(3 2 ) 60

x y x y

x y x y

106. 1 1 3

3 3 41 1 1

2(3 ) 2(3 ) 8

x y x y

x y x y

107.

2 22 2

0

; , 0

a bx yab a b a b x y

x y

108. 2 2

2 2

0

; , 0

a bx y

a b b a a b x yx y

[CBSE 2006 (C)]

109. 2 2

2mx ny m nx y m

[CBSE 2006 (C)]

110. 2 2

0

( ) ( )

x ya ba b x a b y a b

In each of the following systems of equations determine whether the system has a unique solution, nosolution or infinitely many solutions. In case there is a unique solution, find it (111-113) :

111. 3 5 206 10 40

x yx y

112. 2 85 10 10x yx y

113. 3 1 02 3 8 0x yx y

Find the value of k for which the following system of equations has a unique solution (114-116) :

114. 23 2 5x kyx y

115. 4 5 02 3 7 0

x kyx y

116. 10 3 1 05 3 2 0

x kyx y

Find the value of k for which the following system of equations have infinitely many solutions (117-122) :

117. 4 5 315 9

x ykx y

118. 2 3 4( 2) 6 3 2

x yk x y k

119. 4 3 3(2 3) (2 1) 4( 1)

x yk x k y k

120. 2 3 2( 2) (2 1) 2( 1)

x yk x k k

121. ( 1) 5( 1) 9 8 1x k yk x y k [CBSE 2003]

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122. 2 ( 2)6 (2 1) 2 5

x k y kx k y k

Find the value of k for which the following system of equations has no solution (123-127) :

123. 5 26 2 7kx y

x y

124. 2 22 5x yx ky

125. 6 9 03 2 1 0

x kyx y

126. 3 1(2 1) ( 1) 2 1

x yk x k y k [CBSE 2000]

127. 2

(2 1) 2 2 0( 1) ( 2) 5 0

k x yk x k y

Find the values of a and b for which the following system of equations has infinitely many solutions(128-130):

128. (2 1) 3 5 03 ( 1) 2

a x yx b y

129. 2 3 7( ) (2 ) 3( 1)

x ya b x a b y a b

[CBSE 2002]

130. 3 ( 1) 2 15 (1 2 ) 3

x a y bx a y b

Section - B (Word Problems)

Problems related to Ages (131-136) :131. Two years ago, a father 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 age of the father and son. [CBSE 2004]132. Father’s age is three times the sum of ages of his two children. After 5 years his age will be twice the sum

of ages of two children. Find the age of father.133. A father is three times as old as his son. In 12 years, he will be twice as old as his son. Find the present

age of the father and the son.134. Four years ago a mother was four times as old as her daughter. Six years later, the mother will be two-and-

a-half times as old as her daughter. Determine the present ages of mother and her daughter.135. Five years ago, I was thrice as old as my son. Five years hence I shall be twice as old as my son. How old

are we now?136. The present age of a father is three years 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.

Problems related to Articles and their costs (137-140) :

137. 3 bags and 4 pens together cost Rs. 79 whereas 4 bags and 3 pens together cost Rs. 324. Find the totalcost of 1 bag and 10 pens.

138. 4 chairs and 3 tables cost Rs. 2100 and 5 chairs and 2 tables cost Rs. 1750. Find the cost of a chair and atable seperately.

139. Two audio cassetes and three video cassettes cost Rs. 340. But three audio cassettes and two videocassettes cost Rs. 260. Find the price of an audio cassette and that of a video cassette.

140. On selling a T.V. at 5% gain and a fridge at 10% gain, a shopkeeper gains Rs. 2000. But if he sells the T.V.at 10% gain and the fridge at 5% loss, he gains Rs. 1500 on the transaction. Find the actual prices of T.V.and fridge.

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Problems related to Numbers (141-148) :

141. The sum of two numbers is 26 and three times one of them exceeds five times the other by 6. Find thenumbers.

142. Find two numbers such that one-third of the first number added to one-fifth of the second number gives20 and one-sixth of the first added to one-twelfth of the second gives 9.

143. The sum of the digits of a two digit number is 12. The number obtained by reversing the order of thedigits of the given number exceeds the given number by 18. Find the two digit number. [CBSE 2006]

144. In a two digit number, the ten’s digit is three times the unit’s digit. When the number is decreased by 54,the digits are reversed. Find the number.

145. A two-digit number is such that the product of the digits is 20. If 9 is added to the number, the digitsinterchange their places. Find the number. [CBSE 2005]

146. A two-digit number is four times the sum of its digits and twice the product of the digits. Find the number.[CBSE 2005]

147. The sum of a two digit number and the number obtained by reversing the order of its digits is 99. If thedigits differ by 3, find the number. [CBSE 2002]

148. A two-digit number is obtained by multiplying the sum of the digits by 8. Also it is obtained by multiply-ing the difference of the digits by 14 and adding 2. Find the number.

Problems related to Fractions (149-152) :149. The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the

fraction becomes 12

. Find the fraction. [CBSE 2006(C)]

150. A fraction becomes 13 if 1 is subtracted from both its numerator and denominator. If 1 is added to both

the numerator and denominator, it becomes 12

. Find the fraction.

151. The denominator of a fraction is 4 more than twice the numerator. When both the numerator anddenominator are decreased by 6, then the denominator becomes 12 times the numerator. Determine thefraction. [CBSE 2001 (C)]

152. If the numerator of a certain fraction is increased by 2 and denominator by 1, the fraction becomes 5 ,8

and if the numerator and denominator are each diminished by 1, the fraction becomes 1 .2

Find thefraction.

Problems related to Speed, Distance and Time (153-158) :153. Points A and B are 70 km apart on a highway. A car starts from A and another car starts from B at the same

time. If they travel in the same direction, they meet in 7 hours, but if they travel towards each other theymeet in one hour. What are their speeds? [CBSE 2002]

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

155. A sailor goes 8 km downstream in 40 minutes and returns in 1 hour. Determine the speed of the sailor instill water and the speed of the current.

156. A man travelled 300 km by train and 200 km by taxi, it took him 5 hours 30 minutes. But if he travels 260km by train and 240 km by taxi he takes 6 minutes longer. Find the speed of the train and that of the taxi.

157. Sumit travels 600 km to his home partly by train and partly by car. He takes 8 hours if he travels 120 kmby train and the rest by car. He takes 20 minutes longer if he travels 200 km by train and the rest by car.Find the speed of the train and the car.

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158. A boat can go 20 km upstream and 30 km downstream in 3 hours. It can go 10 km upstream and 20 km

downstream in 213

hours. Find the speed of the boat in still water and also the speed of the stream.

Problems related to Area and Perimeter (159-160):159. If in a rectangle, the length is increased and breadth reduced each by 2 units, the area is reduced by 28

sq. units. If, however the length is reduced by 1 unit and the breadth increased by 2 units, the areaincreases by 33 sq. units. Find the area of the rectangle.

160. The area of a rectangle gets reduced by 80 sq. units if its length is reduced by 5 units and the breadth isincreased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units, the areais increased by 50 sq. units. Find the length and breadth of the rectangle.

Problems related to Fixed Salary, Fare etc. (161-163) :161. The taxi charges in a city comprise of a fixed charge together with the charge for the distance covered.

For a journey of 10 km the charge paid is Rs. 75 and for a journey of 15 km the charge paid is Rs. 110. Whatwill a person have to pay for travelling a distance of 25 km? [CBSE 2000]

162. A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on fullticket. One reserved first class ticket from Mumbai to Ahmedabad costs Rs. 216 and one full and one halfreserved first class tickets cost Rs. 327. What is the basic first class full fare and what is the reservationcharge?

163. A part of the monthly expenditure of a family is constant and the remaining varies with the price of wheat.When the rate of wheat is Rs. 250 per quintal, the total monthly expenditure is Rs. 1,000 and when it is Rs.240 per quintal, the total monthly expenditure of the family is Rs. 980. Find the total monthly expenditureof the family when the cost of wheat is Rs. 350 per quintal.

Problems related to Geometry (164-165) :164. In a ABC, C = 3B and 3B = 2 (A + B). Find the angles of the triangle.165. In a cyclic quadrilateral ABCD, A = (2x + 4)°, B = (y + 3)°, C= (2y + 10)° and D = (4x – 5)°. Find the

angles of the cyclic quadrilateral.

Miscellaneous Problems (166-180) :166. A man sold a chair and a table together for Rs. 1520 thereby making a profit of 25% on the chair and 10%

on the table. By selling them together for Rs. 1535 he would have made a profit of 10% on the chair and25% on the table. Find the cost price of each.

167. The incomes of A and B are in the ratio of 8 : 7 and their expenditures are in the ratio 19 : 16. If each savesRs. 1250, find their incomes.

168. Students of a class are made to stand in a rows. If one student is extra in a row, there would be 2 rows less.If one student is less in a row, there would be 3 rows more. Find the number of students in the class.

169. Riya has only 50 p and 25 p coins in her purse. If in all she has 210 coins of the total value of Rs. 82.50,find the number of coins of each type.

170. A person invested some amount at the rate of 12% simple interest and some other amount at 10% simpleinterest. He received yearly interest of Rs. 130. If he had interchanged the amounts invested, he wouldhave received Rs. 4 as more interest. Find the amount invested in each case.

171. 8 women and 12 girls can finish a piece of work in 10 days, while 6 women and 8 girls can finish it in 14days. Find the time taken by one woman alone and that by one girl alone to finish the same work.

172. In a school, there are only two sections of X students A and B. If 10 students are sent from A to B, thenumber of students in each room is same. If 20 students are sent from B to A the number of students inA is double the number of students in B. Find the number of students in each room.

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173. A number consists of two digits. When it is divided by the sum of the digits, the quotient is 8. The sumof the reciprocals of digits is nine times the product of the reciprocals of the digits. Find number.

174. Bhavya scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wronganswer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrectanswer, then she would have scored 50 marks. How many questions were there in the test?

175. A and B each have certain number of oranges. A says to B, ‘‘If you give me 10 of your oranges, I will havetwice the number of oranges left with you. B replies, ‘‘If you give 10 of your oranges, I will have the samenumber of oranges as left with you.’’ Find the number of oranges with A and B seperately.

176. The age of father 8 years back was five times that of his son. After 8 years, his age will be 8 years morethan twice the age of his son. Find their present ages.

177. It takes 12 hours to fill a swimming pool using two pipes. If the larger pipe is used for 4 hours and thesmaller pipe for 9 hours, only half the pool is filled. How long would it take for each pipe alone to fill thepool?

178. A takes 3 hours more than B to walk 30 km. But if A doubles his pace, he is ahead of B by 112

hours. Find

their speeds of walking.179. A number consists of two digits. When it is divided by the sum of the digits, the quotient is 6 with no

remainder. When the number is divided by 9, the digits are reversed. Find the number.180. A man lent a part of money at 10% p.a. and the rest at 15% p.a. His annual income is Rs. 1900. If he had

interchanged the rate of interest on two sums, he would have earned Rs. 200 more. Find the amount lentin each case.

HINTS TO SELECTED QUESTIONS

90. Add and subtract given equations to get the value of (x + y) and (x – y).

95. Take LCM and them simplify the equations.

99. 1 12 2 2.x y x yxy xy xy y x

Now, let 1 1,a bx y etc.

104.11 1 112

11 2 2x y x y

x y

129. For infinitely many solutions, 2 3 72 3 ( 1)a b a b a b

Now, equate first two and last two expressions, and get two linear equations in a and b. Solve themnow.

134. Let present age of mother be x years and present age of daughter be y years. Then,

x – 4 = 4 (y – 4) and 16 2 ( 6)2

x y .

140. Let actual prices of TV and fridge be Rs. x and Rs. y respectively. Then,

5 10 10 52000 and 1500

100 100 100 100x y x y AMIT B

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146. Let unit digit be x and tens digit be y.Then, number = 10 y + x.here, 10 y + x = 4 (x + y) and 10y + x = 2xy.

147. Let unit digit be x and ten’s digit be y.Then, number = 10 y + x.Number obtained by reversing order of digits = 10 x + y.here, (10 y + x) + (10 x + y) = 99 and x – y = ± 3.

151. Let the fraction be .xy

Then, y = 2x + 4 and y – 6 = 12 (x – 6).158. Let speed of boat in still water be x km/hr and the speed of the stream be y km/hr.

Then, 20 30 10 20 23 and 13x y x y x y x y

162. Let basic first class full fare be Rs. x and the reservation charge be Rs. y.

Then, 216 and ( ) 3272xx y x y y

168. Let number of students be x and the number of rows be y. Then, number of students in each row .xy

here, Total number of students = No. of rows × No. of students in each row

1 ( 2) and 1 ( 3)x xx y x yy y

.

171. Let one woman along can finish the work in x days and one girl along can finish it in y days. Then,

One woman’s one day’s work 1x

and, one girl’s one day’s work 1y

here, 8 12 1 6 8 1 .and

10 14x y x y

175. Let no. of oranges with A and B be x and y respectively. Then, x + 10 = 2 (y – 10) and y + 10 = x – 10.177. Let larger and smaller pipes can fill the pool along in x and y hours respectively. Then,

4 9 1 1 1 1and2 12x y x y

178. Let A’s speed = x km/hr and B’s speed = y km/hr.

Then, 30 30 1 1 1310x y y x

...(1)

Again, A’s speed = 2x km/hr and B’s speed = y km/hr.AMIT B

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Then, 30 30 3 2 1 12 2 10y x y x

...(2)

Put 1 1,a bx y and solve now..

180. Let the amount lent bet Rs. x and Rs. y.

Then, 10 15 15 101900 and 1900 200 2100.100 100 100 100

x y x y

MULTIPLE CHOICE QUESTIONS

Mark the correct alternative in each of the following :

1. The system of equations 3x – 2y = 7 and 7x + 8y = 12 has :(a) unique solution (b) no solution(c) infinitely many solutions (d) can’t determine

2. The value of x and y satisfying the system of equations 3x + y = – 1 and 2x – 3y = – 8 is:(a) x =1, y = – 2 (b) x = –1, y = 2 (c) x = 2, y = – 3 (d) none of these

3. The value of x and y satisfying the system of equations 2x – 3y + 13 = 0 and 3x – 2y + 12 = 0 is :(a) x = 2, y = 3 (b) x = – 2, y = – 3 (c) x = – 2, y = 3 (d) none of these

4. The value of x and y satisfying the system of equations 2x ya b and ax – by = a2 – b2 is :

(a) x = a, y = b (b) x = – a, y = – b (c) x = a2, y = b2 (d) none of these

5. The value of x and y satisfying the systems of equations 57 6 5x y x y

and 38 21 9

x y x y

is

(a) x = 11, y = – 8 (b) x = – 11, y = 8 (c) x = 11, y = 8 (d) none of these6. The value of k for whcih the system of equations 2x + ky = 1, 3x – 5y = 7 has a unique solution, is :

(a) 103

k (b) 103

k (c) k = 0 (d) k 0

7. The value of k for which the system of equations 3x – 2y = 8, 6x – ky = 16 has infinitely many solutions,is(a) k = 1 (b) k = 2 (c) k = 3 (d) k = 4

8. The value of k for which the system of equations x + 2y = 3 and 5x + ky + 7 = 0 has no solution, is :(a) 1 (b) 10 (c) 3 (d) 6

9. If the system of equations 3x + y = 1, (2k – 1) x + (k – 1) y = 2k + 1 is inconsistent, then the value of k is:(a) 1 (b) 0 (c) –1 (d) 2

10. If 2x – 3y = 7 and (a + b) x – (a + b – 3) y = 4a + b represent coincident lines, then a and b satisfy theequation:(a) 5a + b = 0 (b) a – 5b = 0 (c) a + 5b = 0 (d) 5a – b = 0

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VERY SHORT ANSWER TYPE QUESTIONS (1 MARK QUESTIONS)

1. Form a pair of linear equations for the following problem :

Fourteen students of class X took part in a quiz and the number of boys is 2 more than the number ofgirls.

2. Is x = 2 and y = – 1 a solution of the pair of linear equations x + 2y = 0 and 3x + 4y = 20?

3. How many solutions do two linear equations in two variables have, if their graph are parallel?

4. What is the minimum and maximum number of solutions that a system of simultaneous linear equationscan have, if it is consistent system?

5. How many solutions will the following pair of linear equations have?

7x – 4y + 11 = 0

2x – 9y + 15 = 0

6. The graph of y = – 3 is a straight line parallel to which axis?

7. Given a linear equation 3x – 5y = 15. Write another linear equation in two variables, such that thegeometric representation of the pair so formed is coincident lines.

8. For what value of k, will the equations 2x – y + 8 = 0 and 4x – ky + 16 = 0 represent coincident lines?

9. For what value of k, will the equations 4x + my = 8 and 3x – 5y + 7 = 0 represent parallel lines?

10. For what value of k, will the equations 2x + ky = 7 and 3x + 9y = 13 may have a unique solution?

11. Find a point on y-axis satisfying 3x – 4y = 12.

12. At what point does the line 4x + 5y = 20 intersects x-axis?

13. Give an equation of a line which passes through the origin.

14. If a system of equation is inconsistent, then what type of graph the equations will have?

15. What are the coordinates of the point of intersection of two lines 3x + 2y = 0 and 2x – y = 0?

16. What are the points of intersection of the line 3 0x ya b with x-axis and with y-axis?

17. If x = – y, x = – 3, and x-axis form a triangle as shown, find the co-ordinates of the three vertices of thetriangle.

Y

XXO

Y

y x=–

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18. What is the area of the triangle formed by the lines 2x + y = 6, 2x – y + 2 = 0 and x-axis in the figure givenbelow ?

X

4

3

2

1

–1

–3

–4

–2

–1–2–3 0 1 2 3 4 5 6

(1, 4)

–4X

Y

Y

2–

+2=0

xy

2+

=6x

y

19. For what value of k, the equations 2x – ky + 3 = 0 and 3x + 2y – 1 = 0 has no solution?20. For what value of k, the equations kx + 3y = k – 3 and 12 x + ky = k has infinitely many solutions?

PRACTICE TESTM.M : 30 Time : 1 hour

General Instructions :

Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each.

1. Solve for x and y :64 15xy

43 7 0x yy

2. Solve for x and y using cross-multiplication method :x y a b ax – by = a2 – b2

3. For what value of k will the following system of linear equations has no solution?3x + y = 1(2k – 1) x + (k – 1) y = 2k + 1

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4. Show that the following system of equations has an infinite number of solutions :3x + 5y – 8 = 09x + 15y = 24

5. Solve for x and y :37 x + 53 y = 32053 x + 37 y = 400

6. In a two digit number, the units digit is twice the ten’s digit. If 27 is added to the number, the digitsinterchange their places. Find the number.

7. Solve graphically the system of equations :x + y = 3 ; 3x – 2y = 4

8. The larger of two supplementary angles exceeds the smaller by 18°. Find them.9. On the same axes, draw the graph of the following equations :

x – 5y + 14 = 0x – 2y + 8 = 02x – y + 1 = 0

Hence obtain the vertices of the triangle so formed.10. A man travels 600 km partly by train and partly by car. If he covers 400 km by train and the rest by the car,

it takes him 6 hours and 30 minutes. But, if he travels 200 km by train and the rest by the car, he takes halfan hour longer. Find the speed of the train and that of the car.

ANSWERS OF PRACTICE EXERCISE

1. No 11. x = 0, y = 3 12. x = 2, y = 3 13. No solution 14. x = 2, y = 415. x = 2, y = – 3 16. x = 1, y = 1 17. x = – 2, y = 1 18. No solutoin19. x = 1, y = 2 20. Infinitely many solutions 21. No solution 22. Unique solutoin23. Infinitely many solutions 24. Unique solution 25. Infinitely many soluions26. Unique solution 27. Unique solution 28. No solution

29. Infinitely many soluions 30. Unique solution 37. 43, 2; 0, , (0,8)5

x y

38. x = 2, y = 3; (0, 6), (0, –2) 39. x = 2, y = –1; (0, 5), (0, –5) 40. x = 1, y = 2; (–2, 0), (5, 0)41. x = 2, y = 2; (–2, 0), (3, 0) 42. x = 1, y = 3; (–2, 0), (3, 0) 43. x = 1, y = 4; Area = 8 sq. units44. x = 1, y = 4; Area = 12 sq. units 45. x = 2, y = 3; Area = 7.5 sq. units46. x = 2, y = 1; Area = 5 sq. units 47. x = 5, y = 0; Area = 17.5 sq. units48. x = 3, y = 2; Area = 13.5 sq. units 49. x = 0, y = 5; (–6, 0), (0, 5), (4, 0)50. x = – 2, y = 3; (–2, 3), (–3, 0), (2, 0) 51. x = 3, y = 4; (3, 4), (2, 0), (–3, 0)52. x = 3, y = 2; (3, 2), (0, –1), (0, 11) 53. x = 3, y = 2; (0, 4), (0, –1), (3, 2)54. x = 2, y = – 1; (2, –1), (0, 5), (0, –5) 55. (0, 0), (4, 4), (6, 2) 56. (3, 4), (0, 2), (6, 2)57. (–4, 2), (1, 3), (2, 5) 58. (0, 5), (0, –5), (5, 0) 59. (3, –5), (3, –11), (–1, –3)60. (6, 0), (3, 0), (1, 2) 61. x = 5, y = – 2 62. x = 3, y = – 263. x = – 3, y = – 1 64. x = – 2, y = – 3 65. x = 2, y = 3

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66. x = 6, y = 36 67. 1 1,2 2

x y 68. 1 9,3 5

x y

69.2 1,3 3

x y 70. x = 4, y = 3 71. 7 13,5 5

x y

72.68 25,47 47

x y 73. 1 1,2 3

x y 74. x = 3, y = – 1

75. x = 2, y = 1 76. x = a, y = b 77. x = – 2, 13

y

78. x = 1, y = – 1 79. x = 4, y = 3 80. x = 5, y = – 20

81. x = 15, y = 5 82. 1 25,11 11

x y 83. 1 7,

24 4x y

84. x = – 2, y = 2 85. x = a2, y = b2 86. x = 1, y = – 187. x = a, y = b 88. x = 4, y = 2 89. x = 20, y = 12

90. x = 2, y = 3 91. 1 1,6 4

x y 92. x = 2a, y = – 2b

93.1 , 25

x y 94. 3, 2x y 95. 23 83,5 5

x y

96. x = 2, y = – 1 97. 2 2,x a y b 98. x = 0, y = 0 or 31,2

x y

99.1 1,2 4

x y 100. 89 89,4080 1512

x y 101. x = 8, y = 3

102.5 1,

4 4x y 103.

2 2

,a b ab abx ya b a b

104. 3.2, 2.3x y

105.1 5,2 4

x y 106. 1, 1x y 107. x = a, y = b

108. x = a2, y = b2 109. ,x m n y m n 110. x = a, y = – b111. infinitely many solutions 112. No solution 113. Unique solution, x = – 1, y = 2

114.23

k 115. k 6 116. k 2

117. k = 12 118. k = 2 119. 52

k

120. k = 4 121. k = 2 122. k = 5123. k = –5 124. k = 4 125. k = – 4

126. k = 2 127. 43

k 128. 17 11,4 5

a b

129. a = 5, b = 1 130. a = 8, b = 5131. Father’s age = 42 years, son’s age = 10 years 132. 45 years133. Father’s age = 36 years, son’s age = 12 years134. Mother’s age = 44 years, daughter’s age = 14 years135. 35 years, 15 years 136. Father’s age = 33 years, son’s age = 10 years.137. Rs. 155 138. Rs. 150, Rs. 500 139. Rs. 20 and Rs. 100140. Rs. 20,000 and Rs. 10,000 141. 17 and 9 142. 24 and 60

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143. 57 144. 93 145. 45146. 36 147. 63 or 36 148. 72

149.57 150.

37 151.

718

152.8

15 153. 40 km/hr and 30 km/hr 154. 6 km/hr and 2 km/hr

155. 10 km/hr and 2 km/hr 156. 100 km/hr, 80 km/hr 157. 60 km/hr, 80 km/hr158. 20 km/hr, 10 km/hr 159. 253 sq. units 160. 40 units and 30 units161. Rs. 180 162. Fare = Rs. 210, Reservation charge = Rs. 6

163. Rs. 1200 164. 120°, 40° and 20° 165. 127°, 110°, 53° and 70°

166. CP of chair = Rs. 600, CP = table = Rs. 700

167. A’s income = Rs. 6000, B’s income = Rs. 5250

168. 60 169. 25 p coins = 90, 50 p coins = 120

170. Rs. 500, Rs. 700 171. 140 days, 280 days 172. 100, 80

173. 72 174. 20 175. 70, 50

176. 16 years and 48 years 177. 20 hours and 30 hours

178. A’s speed = 10 km/hr3 , B’s speed = 5 km/hr

179. 54 180. Rs. 10000 at 10% and Rs. 6000 at 15%

ANSWERS OF MULTIPLE CHOICE QUESTIONS1. (a) 2. (b) 3. (c) 4. (a) 5. (c)6. (b) 7. (d) 8. (b) 9. (d) 10. (b)

ANSWERS OF VERY SHORT ANSWER TYPE QUESTIONS1. Let x and y denote the number of boys and girls respectively. Then, x = y + 2 and x + y = 14.2. No 3. No solution 4. One and infinite 5. unique 6. x-axis

7. 6x – 10y = 30 8. k = 2 9. 203

k 10. All real numbers except 6.

11. (0, –3) 12. (5, 0) 13. y = mx 14. parallel lines15. x = 0, y = 0 16. (–3a, 0), (0, –3b) 17. (0, 0), (–3, 0), (–3, 3)

18. 8 sq. units 19. 4

3

20. 6

ANSWERS OF PRACTICE TEST1. x = 3, y = 2 2. x = a, y = b 3. k = 2 5. x = 6.5, y = 1.5 6. 36

7. x = 2, y = 1 8. 99°, 81° 9. (2, 5), (–4, 2), (1, 3) 10. 100 km/hr, 80 km/hr

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pair of linear equations in two variables · mathematics–x pair of linear equations in...

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