Mathematics Chapter 3 Pair Of Linear Equations In Two Variables
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    NCERT Solution For Class 10 Mathematics

    Pair Of Linear Equations In Two Variables Here is the CBSE Mathematics Chapter 3 for Class 10 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 10 Mathematics Pair Of Linear Equations In Two Variables Chapter 3 NCERT Solutions for Class 10 Mathematics Pair Of Linear Equations In Two Variables Chapter 3 The following is a summary in Hindi and English for the academic year 2021-2022. You can save these solutions to your computer or use the Class 10 Mathematics.

    Question 1
    CBSEENMA10006417

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

    Solution

    Let present age of Aftab be x years and present age of his daughter be y years.

    Case I. Seven years ago,

    Age of Aftab = (x - 7) years

    Age of his daughter = (y - 7) years

    According to question :

    (x - 7) = 7 (y - 7)

    ⇒ x - 7 = 7y - 49

    ⇒ x - 7y = -42

    Case II.

    Three years later,

    Age of Aftab = (x + 3) years

    Age of his daughter = (y + 3) years

    Accoring to questions,

    x + 3 = 3 (y + 3)

    ⇒ x + 3 = 3y + 9

    ⇒ x — 3y = 6

    So, algebraic expression be

    x - 7y = -42    ...(i)

    x - 3y = 6    ...(ii)

    Graphical representation

    For eq. (i), we have

    x - 7y = -42

    ⇒    x — 7y — 42

    Thus, we have following table :

    From eqn. (ii), we have

    x -3y = 6

    ⇒    x = 3y + 6

    Thus, we have following table

    When we plot the graph of equations. We find that both the lines intersect at the point (42, 12). Therefore, x = 42, y = 12 is the solution of the given system of equations.

    Fig. 3.1.

    Question 2
    CBSEENMA10006422

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

    Solution

    Let the cost of 1 bat be Rs. x and cost of I ball be Rs.y

    Case I. Cost of 3 bats = 3x

    Cost of 6 balls = 6y

    According to question,

    3x + 6y = 3900

    Case II. Cost of I bat = x

    Cost of 3 more balls = 3y

    According to question,

    x + 3y = 1300

    So, algebraically representation be

    3x + 6y = 3900

    x + 3y = 1300

    Graphical representation :

    We have,    3x + 6y = 3900

    ⇒    3(x + 2y) = 3900

    ⇒    x + 2y = 1300

    ⇒    a = 1300 - 2y

    Thus, we have following table :

    We have,    x + 3y = 1300

    ⇒    x = 1300 - 3y

    Thus, we have following table :

    When we plot the graph of equations, we find that both the lines intersect at the point (1300. 0). Therefore, a = 1300, y = 0 is the solution of the given system of equations.

    Question 3
    CBSEENMA10006425

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

    Solution

    Let the cost of 1 kg of apples be Rs. x and of 1 kg of grapes be Rs. y. So, algebraic representation

    <>2x + y = 160

     

    4x + 2y = 300

    ⇒ 2x + y = 150

    Graphical representation, we have

    2x + y = 160

    ⇒    y = 160 - 2x

    We have,

    2x + y = 150

    ⇒    y = 150 - 2x

    When we plot the graph of the equation we find that two lines do not intersect i.e. they are parallel.

    Fig. 3.3.

    Question 4
    CBSEENMA10006439

    (i) 10 students of Class X took part in Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

    Solution

    (i) Let the number of boys be x and number of girls be y.

    Case I.    x + y = 10    ...(i)

    Case II.    y = x + 4

    ⇒    x - y = -4    ...(ii)

    We have, x + y = 10

    ⇒    x = 10 - y

    Thus, we have following table :

    Fig. 3.4.

    We have, x - y = -4

    ⇒    x = y - 4

    Thus we have following table :

    When we plot the graph of the given equation, we find that both the lines intersect at me point (3, 7). So.r = 3,y = 7 is the required solution of the pair of linear equation.

    Hence, the number of boys be 3 and the number of girls be 7, who took part in quiz.

    Question 5
    CBSEENMA10006440

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

    Solution

    Let cost of 1 pencil be Rs. ,v and cost of 1 pen be Rs. y

    Case I. Cost of 5 pencils = 5x

    Cost of 7 pens = 7y

    <>According to question,

     

    5x + 7y = 50

    Case II. Cost of 7 pencils = 7x

    Cost of 5 pens = 5y

    According to question,

    7x + 5y = 46

    Thus, we have


    Thus, we have following table :

    Fig. 3.5.

    When we plot the graph of the given equation, we find that both the lines intersect at the point (3, 5). So, x = 3,y = 5 is the required solution of the pair of linear equation.

    Hence, the cost of 1 pencil be Rs. 3 cost of 1 pen be Rs. 5.

    Question 6
    CBSEENMA10006441

    On comparing the ratios  straight a subscript 1 over straight a subscript 2 comma straight b subscript 1 over straight b subscript 2 space and space straight c subscript 1 over straight c subscript 2 and without drawing them, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincide.

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

    Solution
    Comparing the given equations with standard forms of equations a1x + b1y + c1, = 0 and a2x + b2y + c2 = 0 we have,
    straight a subscript 1 equals 5 comma space space straight b subscript 1 equals negative 4 comma space space straight c subscript 1 equals negative 8
straight a subscript 2 equals 7 comma space space straight b subscript 2 equals 6. space space straight c subscript 2 equals negative 9
therefore space space space straight a subscript 1 over straight a subscript 2 space equals space 5 over 7 comma space space straight b subscript 1 over straight b subscript 2 equals fraction numerator negative 4 over denominator 6 end fraction rightwards double arrow straight a subscript 1 over straight a subscript 2 not equal to straight b subscript 1 over straight b subscript 2
    Thus, the lines representing the pair of linear equations are intersecting.
    Question 8
    CBSEENMA10006452

    On comparing the ratios  straight a subscript 1 over straight a subscript 2 comma space straight b subscript 1 over straight b subscript 2 space and space straight c subscript 1 over straight c subscript 2 and without drawing them, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincide.

    6x - 3y + 10 = 0, 2x - y + 9 = 0.

    Solution
    Comparing the given equations with standard forms of equations a1x + b1y + c1, = 0 and a2x + b2y + c2 = 0 we have,
    
straight a subscript 1 equals 6 comma space straight b subscript 1 equals 3 comma space straight c subscript 1 equals 10
straight a subscript 2 equals 2 comma space straight b subscript 2 equals negative 1 comma space straight c subscript 2 equals 9
therefore space space straight a subscript 1 over straight a subscript 2 equals 6 over 2 equals 3 comma space straight b subscript 1 over straight b subscript 2 equals fraction numerator negative 3 over denominator negative 1 end fraction equals 3 comma space straight c subscript 1 over straight c subscript 2 equals 10 over 9
rightwards double arrow 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
    Thus, the lines representing the pair of linear equations are parallel. 
    Question 9
    CBSEENMA10006454

    On comparing the ratios straight a subscript 1 over straight a subscript 2 comma straight b subscript 1 over straight b subscript 2 space and space straight c subscript 1 over straight c subscript 2 find out whether the following pair of linear equations are consistent, or inconsistent.

    3x + 2y = 5; 2x - 3y = 7

    Solution

    3x + 2y = 5 ; 2x - 3y = 7
    Here comma space space space space straight a subscript 1 equals 3 comma space straight b subscript 1 equals 2 comma space straight c subscript 1 equals 5
space space space space space space space space space space space space straight a subscript 2 equals 2 comma space straight b subscript 2 equals negative 3 comma space straight c subscript 2 equals 7
therefore space straight a subscript 1 over straight a subscript 2 equals 3 over 2 comma space straight b subscript 1 over straight b subscript 2 equals fraction numerator negative 2 over denominator 3 end fraction comma space straight c subscript 1 over straight c subscript 2 equals 5 over 7
Clearly comma space straight a subscript 1 over straight a subscript 2 not equal to straight b subscript 1 over straight b subscript 2
    Hence, the given lines are intersecting. So, the given pair of linear equations has exactly one solution and therefore it is consistent.

    Question 10
    CBSEENMA10006456

    On comparing the ratios straight a subscript 1 over straight a subscript 2 comma straight b subscript 1 over straight b subscript 2 space and space straight c subscript 1 over straight c subscript 2 find out whether the following pair of linear equations are consistent, or inconsistent.

    2x - 3y = 8; 4x - 6y = 9

    Solution

    2x - 3y = 8; 4x - 6y = 9
    Here comma space space straight a subscript 1 equals 2 comma space straight b subscript 1 equals 3 comma space straight c subscript 1 equals 8
space space space space space space space space space space straight a subscript 2 equals 4 comma space straight b subscript 2 equals negative 6 comma space straight c subscript 2 equals 9
therefore space space straight a subscript 1 over straight a subscript 2 equals 2 over 4 equals 1 half equals straight b subscript 1 over straight b subscript 2 equals fraction numerator negative 3 over denominator negative 6 end fraction equals 1 half comma space and space straight c subscript 1 over straight c subscript 2 equals 8 over 9
Clearly comma 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
    Hence, the given lines are parallel. So, the given pair of linear equation it has no solution and therefore it is inconsistent.

    Question 11
    CBSEENMA10006457

    On comparing the ratios straight a subscript 1 over straight a subscript 2 comma straight b subscript 1 over straight b subscript 2 space and space straight c subscript 1 over straight c subscript 2 find out whether the following pair of linear equations are consistent, or inconsistent.

    space space space space space space 3 over 2 straight x plus 5 over 3 straight y equals 7 semicolon space 9 straight x minus 10 straight y equals 14

    Solution
    space space space space space space 3 over 2 straight x plus 5 over 3 straight y equals 7 semicolon space 9 straight x minus 10 straight y equals 14
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#6 {main}</pre>
    Hence, the given lines are intersecting. So the given pair of linear equations has exactly one solution and therefore it is consistent.
    Question 12
    CBSEENMA10006459

    On comparing the ratios straight a subscript 1 over straight a subscript 2 comma straight b subscript 1 over straight b subscript 2 space and space straight c subscript 1 over straight c subscript 2 find out whether the following pair of linear equations are consistent, or inconsistent.

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

    Solution

         5x-3y=11 ; -10x + 6y = -22

    Hence, the given lines are consistent. So, the given pair of linear equations has infinitely many solutions and therefore it is consistent.

    Question 13
    CBSEENMA10006460

    On comparing the ratios straight a subscript 1 over straight a subscript 2 comma straight b subscript 1 over straight b subscript 2 space and space straight c subscript 1 over straight c subscript 2 find out whether the following pair of linear equations are consistent, or inconsistent.

    4 over 3 straight x space plus space 2 straight y equals 8 space semicolon space 2 straight x space plus space 3 straight y equals space 12

    Solution
    4 over 3 straight x space plus space 2 straight y equals 8 space semicolon space 2 straight x space plus space 3 straight y equals space 12


    Hence, the given lines are consistent. So the given pair of linear equations has infinitely many solutions and therefore it is consistent.
    Question 14
    CBSEENMA10006464

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

    x + y = 5, 2x + 2y = 10

    Solution

    (i) We have,

    x + y = 5    ...(i)

    ⇒    y = 5 - x

    Thus, we have following table

    2x + 2y = 10                      ...(ii)
    rightwards double arrow space space space space space space space space space straight y space equals space fraction numerator 10 minus 2 straight x over denominator 2 end fraction
    Thus, we have following table

    When we plot the the graph of the equations we find that both the lines are coincident.

    Hence, pair of linear equations has infinitely many solutions

     

    Question 15
    CBSEENMA10006466

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

    x – y = 8, 3x – 3y = 16

    Solution

    x - y = 8

    ⇒    y = x - 8

    Thus we have following table

    3 straight x minus 3 straight y equals 16
rightwards double arrow space space space straight y space space equals space fraction numerator 3 straight x minus 16 over denominator 3 end fraction
    Thus, we have following table :

    When we plot the graph of the equations. We find the both the lines never meet.

    Hence, lines are parallel and equation has no solution.

    Question 16
    CBSEENMA10006467

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

    2x + y – 6 = 0,      4x – 2y – 4 = 0

    Solution

    We have,

    2x + y - 6 = 0

    ⇒    y = 6 - 2x

    and 4x - 2y - 4 = 0

    ⇒    y = 2x - 2

    Thus we have following table

    When we plot the graph of the equations, we find that both the lines intersect at point (2, 2).

    Hence the solution of the given equation is x = 2, y = 2.

    Fig. 3.8.

    Therefore, the pair of equation is consistent at point (2, 2).



    Question 17
    CBSEENMA10006470

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

    2x – 2y – 2 = 0, 4x – 4y – 5 = 0

    Solution

    2x - 2y - 2 = 0

    ⇒    y = x - 1

    Thus we have following tables

    4x -4y - 5 = 0
    rightwards double arrow space space space space space straight y space equals space straight x space minus space fraction numerator negative 5 over denominator 4 end fraction
when space space space straight x space equals space 0 comma space space straight y space equals space minus 5 over 4
when space space space straight x space equals space 5 over 4 comma space space space straight y space equals space 0

    When we plot graph of the given equations, we find that both the lines never meet.

    Hence lines are parallel and equations has no solutions.

    Fig. 3.9.

    Sponsor Area

    Question 18
    CBSEENMA10006473

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

    Solution

    Let the length of the garden be x m and width bey m.

    Case I.    x = y + 4

    ⇒    x - y = 4

    Case II.

    Half perimeter = 36

    ⇒    x + y =36

    So algebraic representation be

    x - y = 4

    x + y = 36

    Graphical representation :

    We have, x - y = 4

    ⇒    x = 4 + y

    Thus, we have following table

    x + y = 36

    ⇒    x = 36 - y

    Thus, we have following table :

    Fig. 3.10.
    If we plot the graph of both the equations, we find that the two lines intersect at the point (20, 16). So, x = 20, y = 16 is the required solution of the given equation i.e., the length of the garden is 20 m and breadth be 16 m.

    Question 19
    CBSEENMA10006474

    Given the linear equations 2x + 3y - 8 = 0, write another linear equation in two variables such that the geometrical representing of the pair so formed is :

    (i) intersecting lines

    (ii) parallel lines

    (iii) coincident lines.

    Solution

    We have,

    2x + 3y - 8 = 0

    (i) Another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines is

    3x - 2y - 8 = 0

    (ii) Another parallel lines to above line is

    4x + 6y - 22 = 0

    (iii) Another coincident line to above line is

    6x + 9y-24 = 0. Ans.

    Question 20
    CBSEENMA10006476

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

    Solution

    We have,

    x - y + 1 = 0

    ⇒    y = x + 1

    Thus, we have following table :

    We have,  
           3x + 2y-12 = 0
    
rightwards double arrow space space space space space space space space space 2 straight y space equals space 12 space minus 3 straight x
rightwards double arrow space space space space space space space space space space straight y space equals space fraction numerator 12 minus 3 straight x over denominator 2 end fraction
    Thus, we have following table :

    Fig. 3.11.

    When we plot the graph of the given equations, we find that both the lines intersect at the point (2, 3), therefore x = 2, y = 3 is the solution of the given system of equations.

    Vertices of triangle (-1, 0) and (4, 0).

    Question 27
    CBSEENMA10006483

    Solve for x and y : 147x - 231y = 525; 77x - 49y = 203.

    Solution
    Solution not provided.
    Ans.  x=2,  y = -1
     
    Question 33
    CBSEENMA10006490
    Question 34
    CBSEENMA10006491
    Question 36
    CBSEENMA10006494

    Solve the following pairs of linear equations by the substitution method.

    x + y = 14
    x – y = 4

    Solution

    (i) x + y = 14

    x - y = 4

    The given pair of linear equations is

    x + y = 14    ...(i)

    x - y = 4    ...(ii)

    From equation (i), we have

    y = 14 - x    ...(iii)

    Substitute this value ot’y in equation (ii), we get
    x - (14 - x) = 4
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#6 {main}</pre>

    Substituting this value of jc in equation (iii), we get
    y = 14 - 9 = 5
    Therefore the solution is
    x = 9, y = 5
    Verification. Substituting x = 9 and y = 5. we find that both the equations (i) and (ii) are satisfied as shown below
    x + y = 9 + 5 = 14
    x - y = 9 - 5 = 4
    This verifies the solution.


    Question 37
    CBSEENMA10006500

    Solve the following pair of linear equations by the substitution method

    s - t = 3

    "<pre

    Solution

    We have
    s - t = 3
    "<pre
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#6 {main}</pre>

    The given pair of linear equation is
    s - t = 3    ...(i)
    2s + 3t = 36    ...(ii)
    From equation (i), we have
    s = 3 + t    ...(iii)

    Substitute this value of s in equation (ii), we get
    2s + 3t = 36
    ⇐ 2(3 + t) + 3t = 36
    ⇒ 6 + 2t + 3t = 36
    ⇒ 5t + 6 = 36
    ⇒ 5t = 30
    ⇒ t = 6
    Therefore, the solution is
    s = 9,    t =6
    
straight s over 3 plus straight t over 2 equals 9 over 3 plus 6 over 2 equals 3 plus 3 equals 6
    This verifies the solution.

     

    Question 38
    CBSEENMA10006502

    Solve the following pairs of linear equations by the substitution method.
    3x – y = 3
    9x – 3y = 9

    Solution

    3x – y = 3 
    9x – 3y = 9
    The given pair of linear equation is
    3x - y = 3    ...(i)
    9x - 3y = 9    ...(ii)
    From eqn. (i), we have

    y = 3x - 3    ...(iii)
    Substitute this value ofy in eq. (ii) we get
    9x - 3(3x - 3) = 9
    ⇒ 9a - 9x + 9 = 9
    ⇒ 9 = 9
    which is true. Therefore eqn. (i) and (ii) have infinitely many solutions.

     

    Question 39
    CBSEENMA10006505

    Solve the following pairs of linear equations by the substitution method.
    0.2x + 0.3y = 1.3    
    0.4x + 0.5y = 2.3    

     

    Solution

    0.2x + 0.3y = 1.3
    0.4x + 0.5y = 2.3
    The given system of linear equations
    0.2x + 0.3y = 1.3    ...(i)
    0.4x + 0.5y = 2.3    ...(ii)
    From equation (i), we have
    0.3y =1.3 - 0.2x
    rightwards double arrow space space space space straight y space equals fraction numerator 1.3 minus 0.2 straight x over denominator 0.3 end fraction space space space space space space space..... left parenthesis iii right parenthesis
    Substituting this value ofy in eqn. (ii) we get
    0.4x + 0.5  open parentheses fraction numerator 1.3 minus 0.2 straight x over denominator 0.3 end fraction close parentheses equals 2.3
    rightwards double arrow space 0.21 space straight x plus 0.65 minus 0.1 straight x space equals space 0.69
rightwards double arrow 0.12 straight x minus 0.1 straight x minus 0.69 minus 0.65
rightwards double arrow 0.02 straight x equals 0.04
rightwards double arrow space space space space space space straight x equals fraction numerator 0.04 over denominator 0.02 end fraction equals 2
    Substituting this value of x in eqn. (iii), we get
    y equals fraction numerator 1.3 minus 0.2 left parenthesis 2 right parenthesis over denominator 0.3 end fraction
space equals fraction numerator 1.3 minus 0.4 over denominator 0.3 end fraction equals fraction numerator 0.9 over denominator 0.3 end fraction equals 3

    Therefore, the solution is x = 2, y = 3,
    Verification. Substituting x = 2 and y = 3, we find that both the equations (i) and (ii) are satisfied as shown below :
    0.2a + 0.3y = (0.2) (2) + (0.3) (3)
    = 0.4 + 0.9 = 1.3
    0.4x + 0.5y = (0.4) (2) + (0.5) (3)
    = 0.8 + 1.5 = 2.23
    This verifies the solution.



    Sponsor Area

    Question 40
    CBSEENMA10006508

    Solve the following pairs of linear equations by the substitution method.

    square root of 2 straight x end root plus square root of 3 straight y end root equals 0
square root of 3 straight x end root minus square root of 8 straight y end root equals 0

    Solution
    square root of 2 straight x end root plus square root of 3 straight y end root equals 0
square root of 3 straight x end root minus square root of 8 straight y end root equals 0
    The given pair oflinear eqn. is
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#6 {main}</pre>

    From eq. (ii), we have
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#6 {main}</pre>
    Substituting this value of a in equation (i), we get
    
square root of 2. fraction numerator square root of 8 over denominator square root of 3 end fraction straight y plus square root of 3 straight y equals 0
rightwards double arrow space fraction numerator 4 over denominator square root of 3 end fraction straight y plus square root of 3 straight y end root equals 0
rightwards double arrow space open parentheses fraction numerator 4 over denominator square root of 3 end fraction plus square root of 3 close parentheses straight y equals 0
rightwards double arrow space space space space space space straight y equals 0
    Substituting this value of y in eqn. (iii), we get
    straight x equals fraction numerator square root of 8 over denominator square root of 3 end fraction left parenthesis 0 right parenthesis equals 0

    Therefore, the solution is

    x = 0, y = 0

    Verification. Substituting x = 0 and y = 0, we find that both the equation (i) and (ii) are satisfied as shown below :
    square root of 2 straight x end root plus square root of 3 straight y end root equals square root of 2 left parenthesis 0 right parenthesis plus square root of 3 left parenthesis 0 right parenthesis equals 0
square root of 3 straight x end root minus square root of 8 straight y end root equals square root of 3 left parenthesis 0 right parenthesis minus square root of 8 left parenthesis 0 right parenthesis equals 0
    This verifies the solution.

    Question 41
    CBSEENMA10006510

    Solve the following pairs of linear equations by the substitution method.

    
space space space space space space space space fraction numerator 3 straight x over denominator 2 end fraction minus fraction numerator 5 straight y over denominator 3 end fraction equals negative 2

    space space space space straight x over 3 plus straight y over 2 equals 13 over 6


        
         

     

    Solution

    We have
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#6 {main}</pre>

    The given pair of linear equation is
    9x - 10y = - 12    ...(i)
    2x - 3y = 13    ...(ii)
    From (ii), we have

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#6 {main}</pre>

    Substituting the value of x in (i), we get
    9x - 10y = - 12

    rightwards double arrow space 9 open parentheses fraction numerator 13 plus 3 y over denominator 2 end fraction close parentheses minus 10 y equals negative 12
rightwards double arrow space space space fraction numerator 9 left parenthesis 13 plus 3 y right parenthesis minus 20 y over denominator 2 end fraction equals negative 12
rightwards double arrow space space 11 plus 27 y minus 20 y equals negative 24
rightwards double arrow space space space minus 47 y equals negative 24 minus 117
rightwards double arrow space space space space space space space y equals 141 over 47 equals 3
    Substituting the value ofy in (iii), we get
    straight x equals fraction numerator 13 minus 3 straight y over denominator 2 end fraction
equals fraction numerator 13 minus 3 cross times 3 over denominator 2 end fraction
equals fraction numerator 13 minus 9 over denominator 2 end fraction equals 4 over 2 equals 2

    Therefore, the solution is
    x = 2,y = 3.
    Verification. Substituting x = 2 and y = 3, we find that both the equations (i) and (ii) are satisfied as shown below :

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#6 {main}</pre>
    This verifies the solution.
    Question 42
    CBSEENMA10006515

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

    Solution

    The given equations are

    2x + 3y = 11    ...(i)

    2x - 4y = - 24    ...(ii)

    From (i), we have
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#6 {main}</pre>
    Putting the value of ‘x’ in (ii), we get  
    open parentheses fraction numerator 11 minus 3 straight y over denominator 2 end fraction close parentheses minus 4 straight y equals negative 24
rightwards double arrow fraction numerator 22 minus 6 straight y over denominator 2 end fraction minus 4 straight y equals negative 24
rightwards double arrow space fraction numerator 22 minus 6 straight y minus 8 straight y over denominator 2 end fraction equals negative 24
rightwards double arrow space space 22 minus 14 straight y space equals space minus 48
rightwards double arrow space space space minus 14 straight y space equals space minus 48 space equals space minus 22
rightwards double arrow space space space space minus 14 straight y space equals space minus 70
rightwards double arrow space space space space space space space space straight y space equals space 5
    Putting the value ofy in (iii), we get
    
space space space space straight x equals fraction numerator 11 minus 3 straight y over denominator 2 end fraction equals fraction numerator 11 minus 15 over denominator 2 end fraction
space space space rightwards double arrow space space space space straight x equals fraction numerator negative 4 over denominator 2 end fraction equals negative 2

    Hence, the solution is x = -2,y = 5.
    It is given that: y = mx + 3 Putting the values of.v andy in given condition we gel
    5 = m(-2) + 3
    ⇒    5 = - 2m + 3
    ⇒    -2m = 2
    ⇒ m = -1.


    Question 43
    CBSEENMA10006516

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

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

    Solution

    Let the two numbers be x and y
    ∴ By the given conditions,
    x - y = 26    ...(i)
    and    x = 3y    ...(ii)
    Substituting x = 3y in (i), we get
    3y - y = 26
    ⇒    2y = 26
    ⇒    y = 13
    From (ii), x = 3y = 3 × 13 = 39
    Thus the numbers are 39 and 13.

    Question 44
    CBSEENMA10006519

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

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

    Solution

    Let the larger angle be x and smaller angle bey
    According to given conditions,
    x + y = 180    ...(i)
    and    x = y + 18    ...(ii)
    Substituting the value of (ii) in (i), we get
    x + y = 180
    ⇒ (y + 18) + y = 180
    ⇒ y + 18 + y = 180
    ⇒    2y = 180 - 18
    ⇒    2y = 162
    ⇒    y = 81°    ...(iii)
    Putting the value of (iii) in (ii)
    x = y + 18
    = 81 + 18 = 99°
    Thus, the angles arc 81° and 99°.

    Question 45
    CBSEENMA10006529

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

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

    Solution

    Let the cost of each bat and each ball be Rs. x and Rs. y respectively.
    According to given conditions,
    7x + 6y = 3800    ...(i)
    3x + 5y = 1750    ...(ii)
    From eqn. (ii), we get
    3x + 5y = = 1750
    rightwards double arrow space space space space space 5 straight y space equals space 1750 space minus space 3 straight x
rightwards double arrow space space space space space straight y space equals space 1750 space minus space 3 straight x
rightwards double arrow space space space space space straight y space equals space fraction numerator 1750 space minus space 3 straight x over denominator 5 end fraction space space space space space space space space space space.... left parenthesis iii right parenthesis
    Substitute the value of eqn. (iii) in (i), we get
    7x + 6y = 3800
    rightwards double arrow space 7 straight x space plus 6 space open parentheses fraction numerator 1750 minus 3 straight x over denominator 5 end fraction close parentheses equals 3800
rightwards double arrow fraction numerator 35 straight x plus 6 left parenthesis 1750 minus 3 straight x right parenthesis over denominator 5 end fraction equals 3800
rightwards double arrow space 35 straight x space plus space 10500 minus 18 straight x equals 19000
rightwards double arrow space 17 straight x plus 10500 equals 19000
rightwards double arrow 17 straight x space equals space 1900 minus 10500
rightwards double arrow space 17 straight x space equals space 8500
rightwards double arrow space space space space straight x space equals space 8500 over 17 equals 500
    Substituting the value of x in {iii) we get,
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#6 {main}</pre>
    Hence, the cost of one bat = Rs. 500 and cost of one ball = Rs. 50.

    Question 46
    CBSEENMA10006530

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

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

    Solution

     Let fixed charge be Rs. x and charge per km be Rs. y
    According to the given conditions,
    x + 10y = 105    ...(i)
    and x + 15y = 155    ...(ii)
    Subtracting (i) from (ii), we get
    (x + 15y)-(x + 10y) = 155 - 105
    ⇒ x + 15y - x — 10y = 50
    ⇒    5y = 50 ⇒ y = 10
    Putting the value ofy in (i), we get
    x + 10y = 105
    ⇒    x + 10(10) = 105
    ⇒    x + 100 = 105 ⇒ x = 5
    ∴ Fixed charge (x) = Rs. 5
    and charges per km (y) = Rs. 10
    Thus, charges for 25 km
    = x + 25y = 5 + 25(10)
    = 5 + 250 = Rs. 255.

    Question 47
    CBSEENMA10006537

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

    A fraction becomes 9 over 11 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes space 5 over 6 . Find the fraction.

    Solution
    Let the numerator of the fraction be x and denominator bey.

    Thus, we have following equations
    11x - 9y = -4    ...(i)
    6x - 5y = -3    ...(ii)
    From (ii), we have
    6x - 5y=-3
    rightwards double arrow space space space 6 straight x space equals space 5 straight y minus 3
rightwards double arrow space space space straight x equals fraction numerator 5 straight y minus 3 over denominator 6 end fraction space space space space space space... left parenthesis iii right parenthesis
    Substituting the value of (iii) in (i), we get
    rightwards double arrow space space 11 straight x minus 9 straight y equals negative 4
rightwards double arrow space space space 11 open parentheses fraction numerator 5 straight y minus 3 over denominator 6 end fraction close parentheses minus 9 straight y equals negative 4
rightwards double arrow space space space fraction numerator 11 left parenthesis 5 straight y minus 3 right parenthesis minus 54 straight y over denominator 6 end fraction minus 4
rightwards double arrow space space space 55 straight y minus 33 minus 54 straight y equals negative 24
rightwards double arrow space space space space space straight y minus 33 equals negative 24
rightwards double arrow space space space space space straight y equals negative 24 plus 33
rightwards double arrow space straight y space equals space 9
    Now, substituting the value ofy in (iii), we get
    straight x equals fraction numerator 5 straight y minus 3 over denominator 6 end fraction
rightwards double arrow space space straight x equals fraction numerator 5 cross times 9 minus 3 over denominator 6 end fraction
equals fraction numerator 45 minus 3 over denominator 6 end fraction equals 42 over 6 equals 7
    Hence, the required fraction.

    Question 48
    CBSEENMA10006538

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

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

    Solution

    Let the present age of Jacob = x years and present age of his son = y years.
    Case I.
    5 years hence.
    Age of Jacob = (x + 5) yearsand age of his son = (y + 5) years
    According to given conditions,
    x + 5 = 3 3(y + 5)
    ⇒ x + 5 = 3y + 15
    ⇒ x - 3y = 10
    Case II.
    5 years ago,
    Age of Jacob = (x - 5) years and age of his son = (y - 5) years
    According to given conditions,
    x - 5 = 7(y - 5)
    ⇒ x - 5 = 7y - 35
    ⇒ x - 7y = -35 + 5
    ⇒ x - 7y = -30
    Thus, we have following equations
    x - 3y = 10    ...(i)
    x - 7y = -30    ...(ii)
    From (i), we have
    x - 3y = 10
    ⇒    x = 3y + 10    ...{iii)
    Substituting the value ofx in (ii), we get
    x - 7y = -30
    ⇒ 3y + 10 - 7y = -30
    ⇒ -4y + 10 = -30
    ⇒    -4y = -40
    ⇒    y = 10
    Now, substituting the value of y in (iii), we get
    x = 3y + 10
    = 3(10) + 10
    = 30+10 = 40
    Hence,
    Age of Jacob = 40 years
    and Age of his son = 10 years.

    Question 49
    CBSEENMA10006547

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

    x + y = 5 and 2x - 3y = 4

    Solution

    x + y = 5    ...(i)

    2x - 3y = 4    ...(ii)

    For making the cocfficient of y in (i) and (ii) equal, we multiply (i) by 3 and adding, we get

    Now, putting the value of x in equation (i), we get
    x + y = 5
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>
    Hence,  straight x equals 19 over 5 comma space space straight y equals 6 over 5

    Substitution method :
    We have following equations :
    x + y = 5    ...(i)
    2x - 3y = 4    ...(ii)
    From (i), we have
    x + y = 5
    ⇒    x = 5 -y    ...(iii)
    Substituting the value ofx in (ii), we get
    2x - 3y = 4
    rightwards double arrow 2(5 -y)-3y = 4
    rightwards double arrow 10 - 2y -3y = 4
    rightwards double arrow 10 - 5y = 4
    rightwards double arrow   -5y = 4 - 10
    rightwards double arrow space   -5y = -6
    space space space rightwards double arrow space 6 over 5

    Substitution method :

    We have following equations :

    x + y = 5    ...(i)

    2x - 3y = 4    ...(ii)

    From (i), we have

    x + y = 5

    ⇒    x = 5 -y    ...(iii)

    Substituting the value ofx in (ii), we get
       x  = 5 - y
    rightwards double arrow space space space straight x space space equals space 5 minus 6 over 5
rightwards double arrow space space straight x space equals space fraction numerator 25 minus 6 over denominator 5 end fraction equals 19 over 5
Hence comma space space straight x equals 19 over 5 comma space straight y equals 6 over 5


    Question 53
    CBSEENMA10006553

    Solve for ‘x’ and ‘y’:
    
space space space space space bx plus ay equals straight a plus straight b space semicolon space space space ax open square brackets fraction numerator 1 over denominator straight a minus straight b end fraction minus fraction numerator 1 over denominator straight a plus straight b end fraction close square brackets plus by open square brackets fraction numerator 1 over denominator straight b minus straight a end fraction minus fraction numerator 1 over denominator straight b plus straight a end fraction close square brackets equals 2

    Solution

    Solution not provided
    Ans. space space straight x equals straight a over straight b comma comma space space straight y equals straight b over straight a

    Question 54
    CBSEENMA10006554

    Solve for ‘x’ and ‘y’:
    straight b left parenthesis straight a minus straight b right parenthesis straight x minus straight a left parenthesis straight b minus straight a right parenthesis straight y equals straight a squared minus straight b squared semicolon space space ax open square brackets fraction numerator 1 over denominator straight a minus straight b end fraction minus fraction numerator 1 over denominator straight a plus straight b end fraction close square brackets plus by open square brackets fraction numerator 1 over denominator straight b minus straight a end fraction minus fraction numerator 1 over denominator straight b plus straight a end fraction close square brackets equals 2

    Solution

    Solution is not provided.
    
space space Ans. space space straight x equals straight a over straight b comma space straight y equals straight a over straight b

    Question 55
    CBSEENMA10006556

    Solve the equation for ‘x’ and ‘y’ :

    148 over straight x plus 231 over straight y equals 257 over xy semicolon space space 231 over straight x plus 148 over straight y equals 610 over xy

    Solution

    Solution is not provided.
    Ans.   x = 1,  y =2

    Question 56
    CBSEENMA10006559

    Solve for ‘x’ and ‘y’ ;

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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>

    where 2x + 3y ≠ 0; 3x - 2y ≠ 0.

    Solution

    Solution not provided.
    Ans  x = 2, y = 1

    Question 59
    CBSEENMA10006565

    Solve for ‘x’ and ‘y’ : ax +  by  = 1 ; bx space plus space ay space equals space fraction numerator left parenthesis straight a plus straight b right parenthesis squared over denominator straight a squared plus straight b squared minus 1. end fraction

    Solution

    Solution not provided.
    
x equals fraction numerator a over denominator a squared plus b squared end fraction comma space space y space equals space fraction numerator a over denominator a squared plus b squared end fraction

    Question 63
    CBSEENMA10006573
    Question 67
    CBSEENMA10006579
    Question 78
    CBSEENMA10006594

    Sponsor Area

    Question 86
    CBSEENMA10006604
    Question 91
    CBSEENMA10006610

    Solve for x and y:

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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>

    Solution

    Solution not provided.
    Ans. space straight x space equals 1 fifth comma space straight Y space equals negative 2

    Question 93
    CBSEENMA10006613

    Solve for x and y:

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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>

    Solution

    Solution not provided.
    Ans.  x = ab ,    y = ab

    Question 95
    CBSEENMA10006619

    Solve for x and y :
    (a - b) x + (a + b) y = a2 - 2 ab - b2

    Solution

    Solution not provided.
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>

    Question 97
    CBSEENMA10006621

    Solve the following systems of equations for x and y.

    fraction numerator 5 over denominator straight x minus 1 end fraction plus fraction numerator 1 over denominator straight y minus 2 end fraction equals 2 semicolon space fraction numerator 1 over denominator straight x minus 1 end fraction minus fraction numerator 3 over denominator straight y minus 2 end fraction equals 1

    Solution

    Solution not provided.
    Ans.  x = 4, y =5

    Question 98
    CBSEENMA10006622

    Solve the following pairs of equations for x and y.

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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>

    Solution

    Solution not provided.
    Ans.  x = 8, y = 3

    Question 100
    CBSEENMA10006626

    Find the number of solutions of the following pair of linear equations :
    x + 2y - 8 = 0; 2x + 4y = 16

    Solution

    Solution not provided.
    Ans.  Infinite many solution.

    Question 103
    CBSEENMA10006631
    Question 106
    CBSEENMA10006642

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

    Solution

    Let the cost of 1 bat be Rs. x and cost of I ball be Rs.y
    Case I. Cost of 3 bats = 3x
    Cost of 6 balls = 6y
    According to question,
    3x + 6y = 3900
    Case II. Cost of I bat = x
    Cost of 3 more balls = 3y
    According to question,
    x + 3y = 1300
    So, algebraically representation be
    3x + 6y = 3900
    x + 3y = 1300
    Graphical representation :
    We have,    3x + 6y = 3900
    ⇒    3(x + 2y) = 3900
    ⇒    x + 2y = 1300
    ⇒    a = 1300 - 2y
    Thus, we have following table :

    We have,    x + 3y = 1300
    ⇒    x = 1300 - 3y
    Thus, we have following table :

    When we plot the graph of equations, we find that both the lines intersect at the point (1300. 0). Therefore, a = 1300, y = 0 is the solution of the given system of equations.

    Question 108
    CBSEENMA10006645
    Question 112
    CBSEENMA10006677

    Solve the following pair of linear equations by the elimination method and the substitution method
    x + y = 5 and 2x – 3y = 4

    Solution

    Elimination Method :
    (i)    x + y = 5    ...(i)
    2x - 3y = 4    ...(ii)
    For making the cocfficient of y in (i) and (ii) equal, we multiply (i) by 3 and adding, we get

    Now, putting the value ofx in (i), we get
    3x + 4y = 10
    ⇒ 3(2) + 4y = 10
    ⇒ 6 + 4y = 10
    ⇒ 4y = 4
    ⇒ y = 1
    Hence, x = 2, y = 1
    Substitution method :
    We have,
    3x + 4y = 10    ...(i)
    2x - 2y = 2    ...(ii)
    From (i), we have

    Now, putting the value of x in equation (i), we get
    x + y = 5

    rightwards double arrow space space space space space 19 over 5 plus straight y equals 5
rightwards double arrow space space space space space space space straight y space equals space 5 minus 19 over 5 equals fraction numerator 25 minus 19 over denominator 5 end fraction equals 6 over 5
Hence comma space straight x space equals space 19 over 5 comma space space space straight y equals 6 over 5

    Now, putting the value of x in equation (i), we get
    x + y = 5
    rightwards double arrow space space space space space straight x space equals space 5 space minus straight y space... left parenthesis iii right parenthesis

    Substituting the value ofx in (ii), we get
    2x - 3y = 4
    
rightwards double arrow space space straight x space equals space 5 minus 6 over 5
rightwards double arrow space straight x space equals space fraction numerator 25 minus 6 over denominator 5 end fraction equals 19 over 5
Hence comma space space space straight x equals 19 over 5 comma space space space straight y space equals 6 over 5

    Question 113
    CBSEENMA10006683

    Solve the following pair of linear equations by the elimination method and the substitution method
    3x + 4y = 10 and 2x – 2y = 2

    Solution

    3x + 4y = 10    ...(i)
    2x - 2y = 2    ...(ii)
    For making the coefficient of y in (i) and (ii) equal, we multiply (ii) by 2 and adding we get.
    
3 straight x plus 4 straight y equals 10
bottom enclose 4 straight x minus 4 straight y equals 4 space space end enclose
7 straight x space space space space space space space equals space 14

rightwards double arrow space space space space straight X space equals space 14 over 7 equals 2

    Now, putting the value ofx in (i), we get
    3x + 4y = 10
    ⇒ 3(2) + 4y = 10
    ⇒ 6 + 4y = 10
    ⇒ 4y = 4
    ⇒ y = 1
    Hence, x = 2, y = 1
    Substitution method :
    We have,
    3x + 4y = 10    ...(i)
    2x - 2y = 2    ...(ii)
    From (i), we have
    3x + 3y = 10
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>
    Substituting the value of (iii) in (ii), we get
    2x - 2y = 2
    rightwards double arrow space 2 open parentheses fraction numerator 10 minus 4 straight y over denominator 3 end fraction close parentheses minus 2 straight y equals 2
rightwards double arrow space fraction numerator 2 left parenthesis 10 minus 4 straight y right parenthesis minus 6 straight y over denominator 3 end fraction equals 2
rightwards double arrow space 20 minus 8 straight y minus 6 straight y equals 6
rightwards double arrow space minus 14 straight y equals 6 minus 20
rightwards double arrow negative 14 straight y equals negative 14
rightwards double arrow space straight y equals 1
    Now, substituting the value ofy in (iii), we get
    straight x equals fraction numerator 10 minus 4 straight y over denominator 3 end fraction equals fraction numerator 10 minus 4 left parenthesis 1 right parenthesis over denominator 3 end fraction equals fraction numerator 10 minus 4 over denominator 3 end fraction equals 6 over 3 equals 2
    Hence,   x=2, y=1

    Question 114
    CBSEENMA10006690

    Solve the following pair of linear equations by the elimination method and the substitution method
    3x – 5y – 4 = 0 and 9x = 2y + 7

    Solution

    3x - 5y = 4    ...(i)
    9x - 2y = 7    ...(ii)
    For making the coefficient of x in (i) and (ii) equal, we multiply eqn. (i) by 3 and subtracting, we get
       9x - 15y = 12
       9x - 2y = 7
      -    +        -
    -13y    = 5
    rightwards double arrow fraction numerator negative 5 over denominator 13 end fraction
    Now putting the value of y in (i), we get
    bottom enclose space space space space space space space space space space 3 straight x space minus space 5 straight y space equals end enclose space 4
rightwards double arrow space 3 straight x minus 5 open parentheses fraction numerator negative 5 over denominator 13 end fraction close parentheses space equals space 4
rightwards double arrow space 3 straight x plus 25 over 13 space equals space 4
rightwards double arrow space 3 straight x equals 4 over 1 minus 25 over 13
rightwards double arrow space space 3 straight x space equals space fraction numerator 52 minus 25 over denominator 13 end fraction
rightwards double arrow space 3 straight x space equals space 27 over 13
rightwards double arrow space space space space space straight x space equals space 9 over 13
Hence comma space space straight x equals space 9 over 13 comma space space straight y space equals space fraction numerator negative 5 over denominator 13 end fraction

    Substitution method :
    We have,
    3x - 5y = 4    ...(i)
    9x - 2y = 7    ...(ii)
    From (i), we have
    3x - 5y = 4
    
rightwards double arrow space space 3 straight x space equals space 4 space plus space 5 straight y
rightwards double arrow space space space straight x space equals space fraction numerator 4 plus 5 straight y over denominator 3 end fraction space space space space space space... left parenthesis iii right parenthesis
    Substituting the value of (iii) in (ii), we get

    Now, substituting the value ofy in (iii), we get


    Question 115
    CBSEENMA10006693

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

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#6 {main}</pre>

    Solution
    straight x over 2 plus fraction numerator 2 straight y over denominator 3 end fraction equals negative 1 space
space straight x minus straight y over 3 equals 3
    Considering equation
    straight x over 2 plus fraction numerator 2 straight y over denominator 3 end fraction equals negative 1
    rightwards double arrow fraction numerator 3 straight x plus 4 straight y over denominator 6 end fraction equals negative 1
rightwards double arrow space 3 straight x plus 4 straight y equals negative 6
and space space space space straight x minus straight y over 3 equals 3
rightwards double arrow fraction numerator 3 straight x minus straight y over denominator 3 end fraction equals 3
rightwards double arrow space 3 straight x minus straight y equals 9

    Now, we have following pairs of equations
    3x + 4y = - 6    ...(i)
    3x -y = 9    ...(ii)
    Since the coefficients of ‘x’ in (i) and (ii) are equal. So simply by subtracting we can eliminate the variable i.e., x.

    3 x space plus space 4 y space equals space minus 6
3 x space minus space space y space equals space 9
minus space space space space plus space space space space space minus
5 y space equals space minus 15
rightwards double arrow space space space bottom enclose space space space space space space space space y space equals space fraction numerator negative 15 over denominator 5 end fraction space space space end enclose space space
rightwards double arrow space space space space space space space space space space space space space y space equals space minus space 3

    Now putting the value ofy in eqn. (i), we get
    3x + 4y = - 6
    3x + 4(-3) = - 6
    ⇒    3x - 12 = -6
    ⇒    3x = 6
    ⇒    x = 2
    Hence, x = 2, y = -3.
    Substitution method :
    We have,
    straight x over 2 plus fraction numerator 2 straight y over denominator 3 end fraction equals negative 1
rightwards double arrow space 3 straight x plus 4 straight y space equals space minus 6
and space space space space straight x minus straight y over 3 equals 3
rightwards double arrow space 3 straight x space minus space straight y space equals space 9

    Now, we have following pairs of equations
    3x + 4y = -6    ...(i)
    3x - y = 9    ...(ii)
    From (ii), we have
    3x - y = 9
    ⇒    y = 3x - 9    ...(iii)
    Substituting the value of (iii) in (i), we get
    3x + 4y = -6
    ⇒ 3x + 4(3x - 9) = -6
    ⇒ 3x + 12x - 36 = -6
    ⇒    15x = -6 + 36
    ⇒    15x = 30
    ⇒    x = 2
    Now, substituting the value ofx in (iii), we get
    y = 3x - 9
    = 3(2) - 9 = 6 - 9= -3
    Hence, x = 2, y = -3.

    Question 116
    CBSEENMA10006702

    Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :
    If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to It becomes   1 half if we only add I to the denominator. What is the fraction?

    Solution
    Let the numerator of the fraction x and denominator be y. Then
    
space space space Fraction space equals space straight x over straight y
    Then, according to given conditions,
    Case space straight I. space space space space space space space fraction numerator straight x plus 1 over denominator straight y minus 1 end fraction equals 1
rightwards double arrow space space space space space space space straight x plus 1 equals straight y minus 1
rightwards double arrow space space space space space space space straight x minus straight y equals negative 2
space
    
Case space II. space fraction numerator straight x over denominator straight y plus 1 end fraction equals 1 half
rightwards double arrow space space 2 straight x space equals space straight y plus 1
rightwards double arrow space 2 straight x minus straight y equals 1

    Thus, we have following equation
    x - y = -2    ...(i)
    2x - y = 1    ...(ii)
    From (i), we have
    x - y = -2
    ⇒    x = y - 2    ...(iii)
    Substituting the value of x in (ii), we get
    2x - y = 1
    ⇒ 2(y - 2) - y = 1
    ⇒ 2y - 4 - y = 1
    ⇒    y - 4 = 1⇒    >> y = 5
    Now, substituting the value ofy in (iii), we get
    x = y - 2
    = 5 - 2 = 3
    Hence, the required fraction equals 3 over 5

    Question 117
    CBSEENMA10006703

    Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :
    Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

    Solution

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

    Case I.

    5 years ago,
    Age of Nuri = (x - 5) years
    and Age of Sonu = (y - 5) years
    According to the given conditions,
    x - 5 = 3(y - 5)
    ⇒ x - 5 = 3y - 15
    ⇒ x - 3y = -10
    Case II.
    Ten years later,
    Age of Nuri = (x + 10) years
    Age of Sonu = (y + 10) years
    According to the given conditions,
    x + 10 = 2(y + 10)
    ⇒ x + 10 = 2y + 20
    ⇒ x - 2y = 10

    Thus, we have following equations :
    x - 3 y = -10    ...(i)
    x - 2y = 10    ...(ii)
    From (i), we have
    x - 3y = -10
    ⇒    x = 3y - 10    ...(iii)
    Substituting the value of (iii) in (ii), we get
    x - 2y = 10
    ⇒ 3y - 10 - 2y = 10
    ⇒    y = 20
    Now, substituting the value ofy in (iii), we get
    x = 3y - 10
    = 3(20)- 10
    = 60 - 10 = 50
    Hence, Age of Nuri = 50 years
    and Age of Sonu = 20 years.

    Question 118
    CBSEENMA10006704

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

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

    Solution

    Let the digit at Unit’s place be x and digit at ten’s place be y
    Then,the number= 10y + x
    Also, the number obtained by reversing the order of the digits = 10x + y
    According to the given condition,
    x + y = 9    ...(i)
    And 9(10y + x) = 2(10x + y)
    ⇒ 90y + 9x = 20x + 2y
    ⇒ 11x - 88y = 0
    ⇒ x - 8y = 0    ...(ii)
    Subtracting equation (ii) from equation (i), we get
           9y = 9
    rightwards double arrow space space space space space space space space straight y space equals space 9 over 9 space equals space 1

    Substituting this value ofy in equation (i), we get

    x + 1 = 9

    ⇒    x = 9 - 1 = 8
    Hence, the required number
    = 10y + x
    = 10(1)+ 8
    = 10 + 8 = 18.

    Question 119
    CBSEENMA10006706

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

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

    Solution

    Suppose that Meena received x notes of Rs. 50 andy notes Rs. 100.
    Then, according to the given conditions
    x + y = 25    ...(i)
    And 50x + 100y = 2000
    ⇒ x + 2y = 40    ...(ii)
    Subtracting equation (i) from equation (ii), we get
    y = 15
    Substituting this value of y in equation (i), we get
    x + 15 = 25
    ⇒    x = 25 - 15 = 10
    Hence, Meena received 10 notes of Rs. 50 and 15 notes of Rs. 100

    Sponsor Area

    Question 120
    CBSEENMA10006708

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

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

    Solution

    Let the fixed charge be Rs. a and the charge for each extra day be Rs. b.
    Then, according to the given conditions,
    a + 4b = 27    ...(i)
    [Extra days = 7 - 3 = 4]
    a + 2b = 21    ...(ii)
    [Extra days = 5 - 3 = 2]
    Subtracting equation (ii) from equation (i), we get
    2b = 6
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#6 {main}</pre>

    Substituting this value of b in equation (ii), we get
    a + 2(3) = 21
    ⇒ a + 6 = 21
    ⇒ a = 21 - 6 = 15
    Hence, the fixed charges one Rs. 15 and the charge for each extra day is Rs. 3.

    Question 121
    CBSEENMA10006711

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

    x – 3y – 3 = 0
    3x – 9y – 2 = 0

    Solution

    The given pair of linear equation is
    x - 3y - 3 = 0
    3x - 9y - 2 = 0
    Here,    a1 = 1, b1 = -3
    a2 = 3, b2 = -9, c2= -2
    Wee see that
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#6 {main}</pre>
    Hence, the given pair of linear equations has no solutions.

    Question 122
    CBSEENMA10006715

    Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
    2x + y = 5
    3x + 2y = 8

    Solution

    2 x + y = 5
    3x + 2y = 8
    The given pair of linear equations is
    2x + y = 5
    3x + 2y = 8
    ⇒ 2x + y - 5 = 0
    3x + 2y - 8 = 0
    Here,    a1 = 2, b1 = 1, c, = -5
    a2 = 3, b2 = 2, c2 = -8
    We see that,
    straight a subscript 1 over straight a subscript 2 not equal to straight b subscript 1 over straight b subscript 2

    Hence, the given pair of linear equations has a unique solution.
    To solve the given equation by cross multiplication method, we draw the diagram below :

    Hence, the required solution of the given pair of linear equation is x = 2, y = 1.

    Question 123
    CBSEENMA10006717

    Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
    3x - 5y = 20
    6x - 10y = 40

    Solution

    3x - 5y = 20
    6x - 10y = 40
    The given pair of linear equations is
    3x - 5y = 20
    6x - 10y = 40
    ⇒ 3x - 5y - 20 = 0
    6x - 10y - 40 = 0
    Here,    a1 = 3, b1 = -5, c1 = -20
    a2 = 6, b2 10, c2 = —40
    We see that
    straight a subscript 1 over straight a subscript 2 equals straight b subscript 1 over straight b subscript 2 equals straight c subscript 1 over straight c subscript 2
    Hence, the given pair of linear equation has infinitely many solution.

    Question 124
    CBSEENMA10006718

    Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
    x - 3y - 7 = 0
    3x - 3y - 15 = 0

    Solution

    x - 3y - 7 = 0
    3x - 3y - 15 = 0
    The given pair of linear equations is
    x - 3y - 7 = 0
    3x - 3y - 15 = 0
    Here,    a1 = 1, b2 = -3, c1 = -7
    a2 = 3, b2 = -3, c2 = -15
    We see that
    straight a subscript 1 over straight a subscript 2 not equal to straight b subscript 1 over straight b subscript 2

    Hence, the given pair of linear equations has a unique solution.
    To solve the given equations by cross multiplication method, we draw the diagram below :

    Question 125
    CBSEENMA10006727

    For which values of a and b does the following pair of linear equations have an infinite number of solutions ?

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

    Solution

    We have following equations
    2x + 3y = 7
    (a - b)x + (a + b)y = 3a + b - 2>
    Here,    a1 = 2, b1 = 3, c1 = 7
    and    a2 = a - b, b2 = a + b,
    c2 = 3a + b -2
    For having an infinite number of solutions, we must have
    straight a subscript 1 over straight a subscript 2 equals straight b subscript 1 over straight b subscript 2 equals straight c subscript 1 over straight c subscript 2
rightwards double arrow space space fraction numerator 2 over denominator straight a minus straight b end fraction equals fraction numerator 3 over denominator straight a plus straight b end fraction equals fraction numerator 7 over denominator 3 straight a plus straight b minus 2 end fraction
    From First two.
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#6 {main}</pre>

    Thus, we have following equations
    a - 5b = 0    (i)
    a - 2b - 3 = 0    (ii)

    Question 126
    CBSEENMA10006728

    For which value of k will the following pair of linear equations have no solution?
    3x + y = 1 (2k – 1) x + (k – 1) y = 2k + 1

    Solution

    The given pair of linear equations is
    3x + y = 1
    (2k - 1) x + (k - 1) y= 2k + 1
    ⇒ 3x + y - 1 = 0
    Here, (2k - 1) x + (k - 1) y - (2k + 1) = 0
    a1 = 3, b1 = 1, c1 = -1)
    a2 = 2 k - 1, b2 = k - 1 c2 = - (2k + 1)
    For having no solution, we must have
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#6 {main}</pre>
    rightwards double arrow space fraction numerator 3 over denominator 2 straight k minus 1 end fraction equals fraction numerator 1 over denominator straight k minus 1 end fraction not equal to fraction numerator negative 1 over denominator negative left parenthesis 2 straight k plus 1 right parenthesis end fraction
rightwards double arrow fraction numerator 3 over denominator 2 straight k minus 1 end fraction equals fraction numerator 1 over denominator straight k minus 1 end fraction
rightwards double arrow 3 left parenthesis straight k minus 1 right parenthesis space equals space 2 straight k minus 1
rightwards double arrow 3 straight k minus 3 equals 2 straight k minus 1
rightwards double arrow 3 straight k minus 2 straight k equals 3 minus 1
rightwards double arrow 2 straight k equals 2
Hence comma space space straight k equals 2

    Question 127
    CBSEENMA10006730

    Solve the following pair of linear equations by the substitution and cross-multiplication methods :
    8x + 5y = 9
    3x + 2y = 4

    Solution

    The given pair of linear equations is
    8x + 5y = 9    ...(i)
    3x + 2y = 4    ...(ii)
    (I) By substitution method

    From equation (ii), we have
    2y = 4 - 3x
    rightwards double arrow space space space space space space space space space space straight y space equals space fraction numerator 4 minus 3 straight x over denominator 2 end fraction space space space space space space space space space space space space space space space space space... left parenthesis iii right parenthesis
    Substitute this value ofy in equation (i), we get
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#6 {main}</pre>
    Substituting this value of x in equation (iii), we get
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#6 {main}</pre>

    So, the solution of the given pair of linear equations is x = -2,y = 5.

    (II) By cross-multiplication method
    Let us write the given pair of linear equations is
    8x + 5y - 9 = 0    ...(i)
    3x + 2y - 4 = 0    ... (ii)
    Solving the equations, we get

    Hence, the required solution of the given pair of linear equations is x = -2, y - 5.

    Question 128
    CBSEENMA10006732

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

    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 ` 1000 as hostel charges whereas a student B, who takes food for 26 days, pays ` 1180 as hostel charges. Find the fixed charges and the cost of food per day

    Solution

    (i) Let the fixed hostel charges be Rs. x and the cost of food per day Rs. y (Running charges)
    Case I. Hostel charges of ‘A’
    Fixed charges be Rs. x and cost of food for 20 days = 20y According to Question,
    x + 20y = 1000
    Case II. Hostel charges of ‘B’
    Fixed charges be Rs. x and cost of food for 26 days = 26y According to question,
    x + 26y = 1180
    Thus, we have following equations
    x + 20y = 1000
    x + 2 6y = 1180
    or x + 20y - 1000 = 0
    x + 26y- 1180 = 0
    Now, by using cross multiplication method,

    Hence, fixed charges is Rs. 400 and cost of food per day is Rs. 30.

    Question 129
    CBSEENMA10006735

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

    A fraction becomes 1 half when 1 is subtracted from the numerator and it becomes 1 fourth when 8 is added to its denominator. Find the fraction.

    Solution
    Let the numerator of the fraction be x and denominator be y. Then

    Thus, we have following equations

    Hence, numerator (x) of the fraction be 5 and denominator (y) be 12.
    So Fraction = 5 over 12

    Question 130
    CBSEENMA10006737

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

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

    Solution

    Let the number of correct answers of Yash be x and number of wrong answers be y. Then according to question :

    Case I. He gets 40 marks if 3 marks are given for correct answer and 1 mark is deduced for incorrect answers.
    3x - y = 40
    Case II. He gets 50 marks if 4 marks are given for correct answer and 2 marks ar deducted for incorrect answers.
    4x - 2y = 50
    Thus, we have following equation
    3x - y - 40 = 0
    4x -2y - 50 = 0
    Now, using cross multiplication method,

    Hence, total number of questions
    = number of correct answers
    + number of incorrect answers
    = 15 + 5 = 20.

    Question 131
    CBSEENMA10006739

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

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

    Solution

    Let the speed of the car starting from point A be .v km/h and speed of the car starting from point B be y km/hr.

    Case I. When they travel in the same direction, let they meet at point P (Fig. 3.12).
    So, AP = distance covered by first car from point A to P
    = speed x time
    = x × 5 = 5
    and    BP = distance covered by second car from point B to P
    = speed × time
    = y × 5 = 5y
    Therefore, AB = AP - BP
    ⇒ 100 = 5x - 5y
    ⇒ 5a - 5y = 100
    ⇒ x - y = 20
    Case II. (Fig. 3.13)
    When they travel in opposite direction, let they meet at point Q.
    So, AQ = Distance covered by first car from point A to Q
    = Speed × Tinte
    = x × 1 = x
    and    BQ = Distance covered by second car from point B to Q
    = Speed × Time
    = y × 1 = y
    Therefore, AB = AQ + BQ
    ⇒    100 = x + y
    ⇒ x + y = 100
    Thus, we have following eqn.
    x - y = 20
    x + y = 100
    ⇒ x - y - 20 = 0
    x + y - 100 = 0

    Hence, speed of the car that starts from point A = 60 km/hr
    Speed of the car that starts from point B = 40 km/hr.

    Question 132
    CBSEENMA10006740

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

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

    Solution

    Let the dimensions (i.e., the length and the breadth) of the rectangle be x units and y units respectively.
    Then, area of the rectangle
    = length × breadth
    = xy square units
    According to the question,
    xy - 9 = (x - 5) (y + 3)
    ⇒ xy - 9 = xy + 3x - 5y - 15
    ⇒ 3x - 5y - 6 = 0
    and    xy + 67 = (x + 3) (y + 2)
    ⇒ xy + 67 = xy + 2x + 3y + 6
    ⇒ 2x + 3y - 61 =0
    Thus, we have following equatins
    3x - 5y - 6 = 0    (i)
    2x + 3y - 61 =0    (ii)


    Hence, the dimensions (i.e., the length and the breadth) of the rectangle are 17 units and 9 units respectively.

    Question 133
    CBSEENMA10006743

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

    fraction numerator 1 over denominator 2 straight x end fraction plus fraction numerator 1 over denominator 3 straight y end fraction equals 2
fraction numerator 1 over denominator 3 straight x end fraction plus fraction numerator 1 over denominator 2 straight y end fraction equals 13 over 6

    Solution
    Let space 1 over straight x equals straight u space and space 1 over straight y equals straight v
    Then, the given system of equation becomes
    1 half straight u plus 1 third straight v equals 2
rightwards double arrow space space space 3 straight u plus 2 straight v equals 12
and space 1 third straight u plus 1 half straight v equals 13 over 6
rightwards double arrow space fraction numerator 2 straight u plus 3 straight v over denominator 6 end fraction equals 13 over 6
rightwards double arrow space 2 straight u plus 3 straight v equals 13

    Thus, we have two equations
    3u + 2v = 12    ...(i)
    2u + 3v = 13    ...(ii)
    From (i), we have
    straight u space equals space fraction numerator 12 minus 2 straight v over denominator 3 end fraction space space space... left parenthesis iii right parenthesis
    Putting the value of ‘u’ in (ii), we get
    2 open parentheses fraction numerator 12 minus 2 v over denominator 3 end fraction close parentheses plus 3 v equals 13
rightwards double arrow space 24 minus 4 v plus 9 v equals 39
rightwards double arrow space 5 v space equals space 39 minus 24
rightwards double arrow space space 5 v equals 15
rightwards double arrow space space space v space equals space 3
    Putting the value of ‘v’ in (iii), we get
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#6 {main}</pre>
    H e n c e comma space straight x space equals space 1 half comma space space straight y space equals 1 third space space is the solution of the given equation.

    Question 134
    CBSEENMA10006746

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

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#6 {main}</pre>
    1
    
fraction numerator 4 over denominator square root of straight x end fraction minus fraction numerator 9 over denominator square root of 9 end fraction equals negative 1



    Solution
    Let space fraction numerator 1 over denominator square root of straight x end fraction equals straight u space and space fraction numerator 1 over denominator square root of straight y end fraction equals straight v

    Then the given system of equation becomes
    2u + 3v = 2    ....(i)
    4u - 9v = -1    ...(ii)
    For making the co-efficients of ‘u’ in (i) and (ii) equal, we multiply (i) by 2 and then subtracting


    Putting the value of ‘v’ in (i), we get

    Hence, x = 4,y = 9 is the solution of the given equation.
    Question 135
    CBSEENMA10006748

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

    4 over straight x plus 3 straight y equals 14
3 over straight x minus 4 straight y equals 23




    Solution
    Let space space space space space space space 1 over straight x equals straight a

    Thus, the given equations become
    4a + 3y = 14    ...(i)
    3a - 4y = 23    ...(ii)
    For making the coefficient of ‘y’ in (i) and (ii), we multiply (i) by 4 and (ii) by 3 and then adding, we get

    space space space 16 straight a plus 12 straight y equals 56
bottom enclose space space space 9 straight a minus 12 straight y space equals 69 space space space space space end enclose
25 straight a space space space space space space space space space space equals space 125
rightwards double arrow space space space space space space straight a space equals space 125 over 25
rightwards double arrow space space space space space space straight a equals 5
    Putting the value of ‘a’ in (i), we get
    4a + 3y = 14
    rightwards double arrow space space space space space space space 4 left parenthesis 5 right parenthesis space plus space 3 straight y space equals space 14
rightwards double arrow space space space space space space space space 20 straight y space plus space 3 straight y space equals space 14
rightwards double arrow space space space space space space space space 3 straight y space equals space minus 6
rightwards double arrow space space space space space space space space space straight y space equals space minus 2
Now space space space space space space space straight a space equals space 5
rightwards double arrow space space space space space space space space space space space 1 over straight x equals 5
rightwards double arrow space space space space space space space space space space space space straight x space equals 1 fifth

    Hence,  space space space space space space space space space space space space space space straight x equals 1 fifth space and space straight y space equals negative 2, is the solution of the given equation.
    Question 136
    CBSEENMA10006750

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

    fraction numerator 5 over denominator x minus 1 end fraction plus fraction numerator 1 over denominator y minus 2 end fraction equals 2

    fraction numerator 6 over denominator straight x minus 1 end fraction minus fraction numerator 3 over denominator straight y minus 2 end fraction equals 1

    Solution
    Let space fraction numerator 1 over denominator straight x minus 1 end fraction equals straight u space and space fraction numerator 1 over denominator straight y minus 2 end fraction equals straight v

    Thus, the given equations becomes
    5u + v = 2    ...(i)
    6u - 3v = 1    ...(ii)
    For making the coefficient of v in (i) and (ii) equal, we multiply (i) by 3 and adding, we get
    "<pre
    Putting the value of u in (i), we get
    5u + v = 2
    rightwards double arrow space space space 5 open parentheses 1 third close parentheses plus v space equals space 2

    Question 137
    CBSEENMA10006754

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

    
space space space space space space space space space fraction numerator 7 straight x minus 2 straight y over denominator xy end fraction equals 5

space space space space space space space fraction numerator 8 straight x plus 7 straight x over denominator x y end fraction equals 15

    Solution
    Considering equation
    space space space space space space space space space fraction numerator 7 straight x minus 2 straight y over denominator x y end fraction equals 5
    rightwards double arrow space space space space space space 7 straight x minus 2 straight y equals 5 xy
    Dividing both sides by xy, we get
    fraction numerator 7 straight x over denominator xy end fraction minus fraction numerator 2 straight y over denominator xy end fraction equals fraction numerator 5 xy over denominator xy end fraction
rightwards double arrow space space 7 over straight y minus 2 over straight x equals 5 space space space space space space space space space space space space space space space space... left parenthesis straight i right parenthesis
    Considering equation
    fraction numerator 8 straight x plus 7 straight y over denominator xy end fraction equals 15
rightwards double arrow space 8 straight x plus 7 straight y equals 15 straight y
    Dividing both sides by xy, we get
    
space space space space fraction numerator 8 straight x over denominator xy end fraction plus fraction numerator 7 straight y over denominator xy end fraction equals fraction numerator 15 xy over denominator xy end fraction
space rightwards double arrow space space 8 over straight y plus 7 over straight x equals 15 space space space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis ii right parenthesis
    Let space 1 over straight x equals straight u comma space 1 over straight y equals straight v Then the given system of equations become
    7v - 2u = 5        ...(iii)
    8v + 7u = 15     ...(iv)
    For making the coefficient of ‘u’ in (iii) and (iv) equal, we multiply (iii) by ‘7’ and (iv) by ‘2’ and then adding
    space space space space 49 v minus 14 u space equals space 35
bottom enclose space space space 16 v space plus 14 u space equals space 30 space space space end enclose
space space space space 65 v space plus space 0 space space equals space 65
rightwards double arrow space space space space space v space equals space 1
    Putting the value of V in (iii), we get

    Hence, x = 1 ,y = 1 is the solution of the given equation.
    Question 138
    CBSEENMA10006756

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

    6x + 3y = 6xy
    2x + 4y = 5x

    Solution

    Considering equation
    6x + 3y = 6xy
    Dividing both side by xy, we get
    
space space space space space space fraction numerator 6 straight x over denominator xy end fraction plus fraction numerator 3 straight y over denominator xy end fraction equals fraction numerator 6 xy over denominator xy end fraction
rightwards double arrow space space 6 over straight y plus 3 over straight x equals space 6 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis straight i right parenthesis

    Considering equation
    2x + 4y = 5xy
    Dividing both side by xy, we get
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#6 {main}</pre>
    Let space 1 over straight x equals straight u comma space 1 over straight y equals straight v. The given system of equation becomes
    6v + 3u  = 6    ....(iii)
    2v + 4u  = 5    ....(iv)
    For making the coefficient of u in (iii) and (iv) equal, we multiply (iii) by 4 and (iv) by 3 and by subtracting, we get
    
24 straight v space plus space 12 straight u space equals space 24
space space 6 straight v space plus space 12 straight u space equals space 15
bottom enclose negative space space space space space minus space space space space space space space space space space space minus space space end enclose
18 straight v space space space space space space space space space space space equals space space 9
rightwards double arrow space space space straight v space equals space 9 over 18 equals 1 half
    Putting the value of v in (iii), we get

    Question 139
    CBSEENMA10006758

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

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

    fraction numerator 15 over denominator straight x plus straight y end fraction minus fraction numerator 5 over denominator straight x minus straight y end fraction equals negative 2

    Solution
    Let space fraction numerator 1 over denominator straight x plus straight y end fraction equals straight u space and space fraction numerator 1 over denominator straight x minus straight y end fraction equals straight v

    Then the given system of equation becomes
    10u + 2v = 4    ...(i)
    15u - 5v = -2    ...(ii)
    For making the coefficient of v in (i) and (ii) equal, we multiply (i) by 5 and (ii) by 2 and then adding, we get

    
50 straight u space plus space 10 straight v space equals space 20
bottom enclose 30 straight u space minus space 10 straight v space equals space minus 4 space space end enclose
80 straight u space space space space space space space space space space space equals space 16
rightwards double arrow space space space space space space space space straight u space equals space 16 over 80 equals 1 fifth space space

    Putting the value of u in (i), we get
    10u + 2v = 4

    Substituting the value of x in (iv), we get
    x - y = 1
    ⇒ 5 - y - y = 1
    ⇒ 5 - 2y = 1
    ⇒ -2y = 1 - 5 ⇒ -2y = -4
    ⇒ y = 2
    Now, substituting the value of in (v), we get
    x = 5 - y
    = 5 - 2 = 3
    Hence, x = 3, y = 2.

    Question 140
    CBSEENMA10006761

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

    fraction numerator 1 over denominator 3 x plus y end fraction plus fraction numerator 1 over denominator 3 x minus y end fraction space equals space 3 over 4
fraction numerator 1 over denominator 2 left parenthesis 3 x plus y right parenthesis end fraction plus fraction numerator 1 over denominator 2 left parenthesis 3 x minus y right parenthesis end fraction space equals space fraction numerator negative 1 over denominator space space 8 end fraction

    Solution
    Let space space fraction numerator 1 over denominator 3 straight x plus straight y end fraction equals straight u space and space space fraction numerator 1 over denominator 3 straight x minus straight y end fraction space equals space straight v
    Then the given system of equations become
    space space space space space
space space space space space space space straight u plus straight v space equals space 3 over 4
and space space space space space space space straight u over 2 minus straight v over 2 equals fraction numerator negative 1 over denominator 8 end fraction
rightwards double arrow space space space space space space space space space straight u minus straight v equals fraction numerator negative 1 over denominator 4 end fraction space
    Thus, we have following equations
    straight u space plus space straight v equals 3 over 4 space space space space space space space space space space space space space space space space space space space space space space space space space.... left parenthesis straight i right parenthesis
straight u minus straight v equals fraction numerator negative 1 over denominator 4 end fraction space space space space space space space space space space space space space space space space space space space space space space space space.... left parenthesis ii right parenthesis
    Adding equation (i) and (ii), we get
    
2 straight u space equals space 3 over 4 plus open parentheses fraction numerator negative 1 over denominator 4 end fraction close parentheses
2 straight u space equals space fraction numerator 3 minus 1 over denominator 4 end fraction equals 2 over 4
rightwards double arrow space space space space space straight u space equals space 1 fourth
    Putting the value of ‘u’ in eqn. (i), we get


    Adding (iii) and (iv), we get
    6x = 6
    ⇒    x = 1
    Putting the value of ‘x’ in (iii), we get
    3 x + y = 4
    ⇒ 3(1) + y = 4
    ⇒    y = 1
    Hence, x = 1, y = 1 is the solution of the given equation.

    Question 141
    CBSEENMA10006762

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

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

    Solution

    Let the speed of the boat in still water = x km/hr
    and speed of the current = y km/hr
    Now, the speed of the boat upstream = (x - y) km/hr
    and the speed of the boat downstream = (x + y) km/hr
    Case I. Time taken to cover 20 km downstream
    equals fraction numerator 20 over denominator straight x plus straight y end fraction
rightwards double arrow space space 2 equals fraction numerator 20 over denominator straight x plus straight y end fraction
rightwards double arrow space space straight x plus straight y space equals space 10 space space space space space space space space space... left parenthesis straight i right parenthesis
    Case II. Time taken to cover 4 km upstream
    equals fraction numerator 4 over denominator straight x minus straight y end fraction
rightwards double arrow space space space space space space space space 2 equals fraction numerator 4 over denominator straight x minus straight y end fraction
rightwards double arrow space space space space space space straight x minus straight y equals 2 space space space space space space space space.... left parenthesis ii right parenthesis
From left parenthesis straight i right parenthesis comma space we space have
space space space space space space space space space space space space straight x space equals space 10 space minus space straight y space space space space space space space... left parenthesis iii right parenthesis

    Putting the value of ‘x’ in (ii), we get
    10 - y - y = 2
    ⇒ 10 - 2y = 2
    ⇒    -2y = -8
    ⇒    y = 4
    Now, substituting the value ot y in (i), we get
    x + y = 10
    ⇒    x + 4 = 10
    ⇒    x = 6
    Hence, speed of rowing her boat in still water = 6 km/hr
    and speed of the current = 4 km/hr.


    Question 142
    CBSEENMA10006764

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

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

    Solution

    Let 1 woman can finish embroidery work in x days

    and 1 man can finish embroidery work in y days
    ∴ 1 woman’s 1 day’s work   "<pre 
    and    1 man’s 1 day’s work space equals 1 over straight y
    Case I. 2 women’s 1 day’s work equals 2 over straight x
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#6 {main}</pre>
    According to question <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre>
    Case II. 3 women’s 1 day’s work <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre>
    and    6 men’s 1 day’s work equals 6 over straight y
    According to question  3 over straight x plus 6 over straight y equals 1 third
    Taking 1 over straight x straight u comma space 1 over straight y equals straight v  we  have
    2 straight u plus 5 straight v equals 1 fourth space space space space space space space space space space space space space space space space... left parenthesis straight i right parenthesis
3 straight u plus 6 straight v equals 1 third space space space space space space space space space space space space space space... left parenthesis ii right parenthesis
    For making the coefficient of ‘u’ we multiply equation (i) by 3 and equation (ii) by ‘2’ and then subtracting, we get


    Hence, 1 woman alone can finish the work in 18 days and 1 man alone can finish the work in 36 days.

    Question 143
    CBSEENMA10006766
    Question 144
    CBSEENMA10006775

    Formulate the following problems as a pair of equations, and hence find their solutions:
    Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

    Solution

    Let the speed of the train and the bus be x km/hour and y km/hour respectively.

    Case I. When she travels 60 km by train and the remaining (300 - 60) km, i.e., 240 km by bus, the time taken is 4 hours.
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#6 {main}</pre>
    Case II. When she travels 100 km by train and the remaining (300 - 100) km, i.e., 200 km by bus.
    the time taken is 4 hours 10 minutes, i.e., space space 25 over 6
 hours.
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#6 {main}</pre>

    [Dividing by 25]
    Multiplying equation (i) by (ii), we get
    2 over straight x plus 8 over straight y equals 2 over 15 space space space space space space... left parenthesis iii right parenthesis
    Subtracting equation (iii) from equation (ii), we get
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#6 {main}</pre>
    Substituting this value of x in equation (iii), we get
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#6 {main}</pre>

    So, the solution of the equations (i) and (ii) is x = 60 and y = 80.
    Hence, the speed of the train is 60 km/hour and the speed of the bus is 80 km/hour.

    Question 145
    CBSEENMA10006776

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

    Solution

    Let the ages of Ani and Biju be x years and y years respectively. Then, according to the question,
    x - y = ±3    ...(i)
    Age of Ani’s father Dharam = 2x years
    Age of Biju’s sister space space space space equals straight y over 2 space years.
    According to the question,

    2 straight x minus straight y over 2 equals 30
rightwards double arrow 4 straight x minus straight y equals 60 space space space space space space space space space space space space space... left parenthesis ii right parenthesis

    Case I. When x - y = 3
    Then, we have
    x - y = 3    ...(i)
    4x - y = 60    ...(ii)
    Subtracting equation (i) from eq. (ii)
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#6 {main}</pre>

    Substituting the vaiue of x in eqn. (i)
    19 - y = 3
    y = 19 - 3 = 16
    Ani’s age = 19 years
    Biju’s age = 16 years.

     

    Question 146
    CBSEENMA10006777

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

    Solution

    Let the amounts of their respective capitals be Rs. x and Rs. y respectively.
    Then, according to the question,
    x + 100 = 2(y - 100)
    ⇒ x - 2y = -300    ...(i)
    and 6(x - 10) = y + 10
    ⇒ 6x — y = 70    ...(ii)
    From equation (i), we have
    x = 2y - 300    ...(iii)
    Substitute the value of x in equation (ii), we get
    6 left parenthesis 2 straight y minus 300 right parenthesis minus straight y space equals space 70
rightwards double arrow space 12 straight y minus 1800 minus straight y equals 70
rightwards double arrow space space space space space space 11 straight y space equals space 1870
rightwards double arrow space space space space space space space space space straight y space equals space 1870 over 11 equals 170

    Substituting the value of>> in equation (iii), we get
    x = 2(170) - 300
    = 340 - 300 = 40
    So, the solution of the equations (i) and (ii) is x = 40 and y - 170. Hence, the amounts of their respective capitals are Rs. 40 and Rs. 170 respectively.


    Question 147
    CBSEENMA10006778

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

    Solution

    Let the actual speed of the train be x km/hr and the actual time taken by y hours. Then,
    Distance covered = (xy) km    ...(i)
    Case I. If the speed is increased by 10 km/hr then time of journey is reduced by 2 hrs i.e. when speed is (x + 10) km/hr, time of journey is (y - 2) hrs.
    ∴ Distance covered = (x + 10) (y - 2)
    ⇒    xy = xy - 2x + 10y - 20
    [Using (i)]
    ⇒    2x - 10y = -20
    Case II. When the speed is reduced by 10 km/hr then time of journey is by 3 hrs i.e. when speed is (x - 10) km/hr, time of journey is (y + 3) hrs.
    ∴ Distance covered = (x - 10) (y + 3)
    ⇒    xy = (x - 10) (y + 3)
    [Using (i)]
    ⇒    xy = xy + 3x - 10y - 30
    ⇒ -3x + 10y = -30
    Thus, we have following eqn.
    2x - 10y = -20    ...(i)
    -3x + 10y = -30    ...(ii)
    Since, the coefficient of y in both the equations are same, so we can eliminate it
    directly by adding i.e.
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#6 {main}</pre>

    Putting the value of x in (i), we get
    2x - 10y = -20
    ⇒ 2(50) - 10y = -20
    ⇒    100 - 10y = -20
    ⇒    -10y = -120
    ⇒    y = 12
    Hence, the distance covered by the train
    = (xy) km
    = (50 × 12) km
    = 600 km.


    Question 148
    CBSEENMA10006780

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


    Solution

    Let the total no. of students in each row be x and the total no. of rows by y
    ∴ Total no. of students in the class = xy ...(i)
    Case I. If 3 students are extra in a row, there would be 1 row less. i.e.
    total no. of students in each row = (a + 3)
    and    the total no. of rows = (y - 1)
    ∴ Total no. of students in the class
    = (x + 3)(y - 1) ...(ii)
    Comparing (i) and (ii), we get
    xy = (x + 3) (y - 1)
    ⇒    xy = xy - x + 3y - 3
    ⇒ x - 3y = -3
    Case II. If 3 students are less in a row, there would be 2 rows more. i.e.
    total no. of students in each row = (x - 3)and    the total no. of rows = (y + 2)
    ∴ Total no. of students in the class
    = (x - 3)(y + 2) ...(iii)
    Comparing (i) and (iii), we get
    xy = (x - 3) 0 + 2)
    ⇒    xy = xy + 2x - 3y - 6
    ⇒ -2x + 3y = -6
    Thus, we have following equations
    x - 3y = -3    ...(iv)
    -2x + 3y = -6    ...(v)
    From (iv), we have
    x - 3y = -3
    ⇒    x = 3y - 3    ...(vi)
    Substituting the value of (vi) in (v), we get
    -2x + 3y = -6
    ⇒ -2(3y - 3) + 3y = -6
    ⇒ -6y + 6 + 3y = -6
    ⇒    -3y = -12
    ⇒    y - 4
    Now, substituting the value of ‘y’ in (vi), we get
    x = 3y - 3
    = 3(4) - 3 = 12 - 3 = 9
    Hence, total no. of students in the class
    = xy
    = 9 × 4 = 36.

    Question 149
    CBSEENMA10006781

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

    Solution

    Let ∠A = x°, ∠B = y°. Then,
    ∠C = 3∠B = ⇒ ∠C = 3y°
    We have. ∠C = 3 ∠B = 2 (∠A + ∠B). ...(i)
    Now,0    3∠B = 2 (∠A + ∠B) From (i)
    ⇒    3y = 2(x + y)
    ⇒    y = 2x
    Since ∠A, ∠B and ∠C are angles of a triangle.
    ∴ ∠A + ∠B + ∠C = 180°
    ⇒ x + y + 3y = 180
    ⇒    x + 4y = 180    ...(ii)
    Putting y = 2x in equation (ii), we get
    x + 4(2x) = 180
    ⇒    x + 8x = 180
    ⇒    9x = 180
    ⇒    x = 20°
    Putting the value of x equation (i), we get y = 40°
    Hence,    ∠A = 20°, ∠B = 40°
    and    ∠C = 3y° = (3 × 40°)= 120°.

    Question 150
    CBSEENMA10006783

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

    Solution

    The given equations are
    5x - y = 5    ...(i)
    3x - y = 3    ...(ii)
    For equation (i), we have
    = 5x - 5
    Thus, we have following table :

    For equation (ii), we have
    y = 3x - 3
    Thus, we have following table :

    Fig. 3.14.

    Question 151
    CBSEENMA10006785

    Solve the following pair of linear equations:

    px + qy = p - q
    qx - py = p + q

    Solution

    we get

    
straight x equals fraction numerator straight p minus qqy over denominator straight p end fraction space space space.. left parenthesis iii right parenthesis
    Putting the value of ‘x’ in (ii), we get

    Putting the value of ‘y’ in (iii), we get
    
straight x equals fraction numerator straight p minus straight q minus straight q left parenthesis negative 1 right parenthesis over denominator straight p end fraction
rightwards double arrow space space space straight x equals fraction numerator straight p minus straight q plus straight q over denominator straight p end fraction equals straight p over straight p equals 1
rightwards double arrow space space space straight x space equals space 1

    Hence, the solution is

    x = 1, y = -1.

     

    Question 152
    CBSEENMA10006787

    Solve the following pair of linear equations:

    ax + by = c   
    bx + ay = 1 + c   

    Solution

    We have,
    ax + by = c    ...(i)
    bx + ay = 1 + c    ...(ii)
    From (i), we have
    straight x equals fraction numerator straight c minus by over denominator straight a end fraction space space space space space space... left parenthesis iii right parenthesis
    Putting the value of (iii) in (ii), we get

    Putting the value of ‘y’ in (iii), we get

    Hence, the solution is
    straight x space equals space fraction numerator ac minus straight b minus bc over denominator straight a squared minus straight b squared end fraction
and space space straight y space equals space fraction numerator straight a plus ac minus bc over denominator straight a squared minus straight b squared end fraction

    Question 153
    CBSEENMA10006789

    straight x over straight a minus straight y over straight b equals 0
ax plus by equals straight a squared plus straight b squared

    Solution
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#6 {main}</pre>
    The given pair of linear equations is
    straight x over straight a minus straight y over straight b equals 0 space space space space space space space space space space space space space space space space space space space space... left parenthesis straight i right parenthesis
ax plus bt equals straight a squared plus straight b squared space space space space space space space space space space space space space space space... left parenthesis ii right parenthesis
    From equation (i), we have
    straight y over straight b equals straight x over straight a
rightwards double arrow space space space straight y equals straight b over straight a straight x space space space space space space space space space space space space space space space space space space.... left parenthesis iii right parenthesis
    Substituting the value of y in equation (ii), we get
    
ax plus straight b open parentheses straight b over straight a straight x close parentheses equals straight a squared plus straight b squared
rightwards double arrow space space ax plus straight b squared over straight a straight x equals straight a squared plus straight b squared
rightwards double arrow space straight a squared straight x plus straight b squared straight x equals straight a left parenthesis straight a squared plus straight b squared right parenthesis
rightwards double arrow space space straight x equals fraction numerator straight a open parentheses straight a squared plus straight b squared close parentheses over denominator straight a squared plus straight b squared end fraction equals straight a
    Substituting the value of x in (iii), we get
    straight y equals straight b over straight a straight x
rightwards double arrow space space straight y equals straight b over straight a cross times straight a equals straight b
    Hence, X= a,  y = b
    Question 154
    CBSEENMA10006793

    Solve the following pair of linear equations:

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#6 {main}</pre>

    Solution
    The given system of equations may be written as

    Putting the value of x in (ii), we get
    (a + b)x + (a + b) y = a2 + b2
    ⇒(a + b) (a + b) + (a + b)y = a2 + b2
    ⇒    (a + b)2 (a + b)y = a2 + b2
    ⇒(a + b)y = (a2 + b2) - (a + b)2
    ⇒ (a + b)y = (a2 + b2) - (a2 + b2 + 2ab)
    ⇒    (a + b)y = a2 + b2 - a2 - b2 - 2ab
    ⇒ (a + b)y = - 2ab
    
rightwards double arrow space space space straight y equals fraction numerator negative 2 ab over denominator straight a plus straight b end fraction
    Hence,
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#6 {main}</pre>

    Question 155
    CBSEENMA10006800

    Solve the following pair of linear equations:

    152x – 378y = – 74
    –378x + 152y = – 604

    Solution

    152x - 378y = -74
    -378x + 152y = -604
    The given pair of linear equations is
    152x - 378y = -74    ...(i)
    -378x + 152y = -604    ...(ii)
    Adding equation (i) and equation (ii), we get
    -226x - 226y = -678
    ⇒    x + y = 3    ...(iii)
    [Dividing throughout by -226]
    Subtracting equation (ii) from equation (i), we get
    530x - 530y = 530
    ⇒    x - y = 1    ...(iv)
    [Dividing throughout by 530]
    Adding equation (iii) and equation (iv), we get
    2x = 4
    rightwards double arrow space space space space straight x space 4 over 2 equals 2
    Subtracting equation (iv) from equation (iii), we get
      2y = 2
    rightwards double arrow space space space y equals 2 over 2 equals 1
    Hence, the solution of the given pair of linear equations is x = 2, y = 1.

    Question 156
    CBSEENMA10006802

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

    Solution

    We know that the opposite angles of a cyclic quadrilateral are supplementary, therefore,
    ∠A + ∠C = 180°
    ⇒ 4y + 20 + 4x = 480°
    ⇒ 4x + 4y = 60°
    ⇒    x + y = 40°    ...(i)
    [Dividing throughout by 4]
    and ∠B + ∠D = 180°
    ⇒3y - 5 + 7x + 5= 180°
    ⇒ 7x + 3y = 180°    ...(ii)
    From equation (i), we have
    y = 40 - x    ...(iii)
    Substituting this value of y in equation (ii), we get
    7x + 3(40 - x) = 180°
    ⇒7x + 120 - 3x = 180°
    ⇒    4x = 60
    rightwards double arrow space space space straight x space equals 60 over 4 equals 15 degree

    Substituting x = 15 in equation (iii), we get
    y = 40 - x
    = 40 - 15 = 25°
    Hence, required angles be
    ∠A = 4y + 20 = 4 × 25 + 20 = 120
    ∠B = 3y - 5 = 3 × 25 - 5 = 75 - 5 = 70
    ∠C = 4x = 4 × 15 = 60°
    ∠D = 7x + 5 = 7 × 15 + 5
    = 105 + 5 = 110°

    Question 157
    CBSEENMA10006805

    Find the number of solutions of the follow ing pair of linear equations :
    x + 2y - 8 = 0
    2x + 4y = 16

    Solution

    The given pair of linear equations can be written as
    x + 2y - 8 = 0
    and 2x + 4y - 16 = 0
    Here comma space space straight a subscript 1 over straight a subscript 2 equals 1 half comma space straight b subscript 1 over straight b subscript 2 equals 2 over 4 equals 1 half space and space straight c subscript 1 over straight c subscript 2 equals fraction numerator negative 8 over denominator negative 16 end fraction equals 1 half
therefore space space straight a subscript 1 over straight a subscript 2 equals space straight b subscript 1 over straight b subscript 2 equals straight c subscript 1 over straight c subscript 2
    ∴ There are infinitely many solutions.

    Question 158
    CBSEENMA10006809

    Is the pair of linear equations consistent:
    2x - 3y + 2 = 0,3x - 5y + 4 = 0.

    Solution

     We have
    2x - 3y + 2 = 0 and 3x - 5y + 4 = 0
    Here, a1 = 2, b1 = -3 and c1 = 2
    and    a2 = 3, b2 = -5 and c2 = 4
    Conditions for consistency :
    straight a subscript 1 over straight a subscript 1 not equal to straight b subscript 1 over straight b subscript 2
rightwards double arrow 2 over 3 not equal to fraction numerator negative 3 over denominator negative 5 end fraction rightwards double arrow 2 over 3 not equal to 3 over 5
    So, the pair of linear equation are consistent.

    Question 159
    CBSEENMA10006810

    Find the value of k so that the followings systemk of a equations has no solution.
    3x - y - 5 = 0; 6x - 2y - k = 0 

    Solution

    We have,
    Here,    a1 = 3, b1 = -1 and c1 = -5
    and    a2 = 6, b2 = -2 and c2 = -k
    For no solution, we have

    Hence the given system, we have no solution if k = 10.

    Question 160
    CBSEENMA10006813

    For what value of m, the pair of linear equations mx = 2y, 2x - y + 5 = 0 has unique solution.

    Solution
    Given pair of linear equations are :
    mx space equals space 2 straight y
rightwards double arrow space mx space minus space 2 straight y space equals space 0
and space 2 straight x minus straight y plus 5 space equals space 0
    Here,  we  have
    straight a subscript 1 equals straight m subscript 1 comma space space straight b subscript 1 equals negative 2
and space straight a subscript 2 equals 2 comma space straight b subscript 2 equals negative 1
    For unique solution :
    straight a subscript 1 over straight a subscript 2 not equal to straight b subscript 1 over straight b subscript 2
rightwards double arrow space space straight m over 2 not equal to fraction numerator negative 2 over denominator negative 1 end fraction
rightwards double arrow space straight m not equal to 4
    ∴ The given system of equations has unique solution for all real values of k other than 4.
    Question 161
    CBSEENMA10006816

    For what value of k, the pair of linear equations 3x - ky + 7 = 0,x - 2y + 5 = 0 has unique solution.

    Solution

    Given pair of linear equations are :
    3x - ky + 7 = 0
    and x - 2y + 5 = 0
    Here, we have
    a1 = 3, b1 = -k
    a2 = 1, b22 = -2
    For unique solution
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#6 {main}</pre>
    ∴ The given system of linear equations has unique solution for all real values of k other than 6.

    Question 162
    CBSEENMA10006833

    Given below are three equations. Two of them have infinite solutions and two of them have no solution. State the two pairs :

    Solution

    2x - 3y = 4; 4x - 6y = 7; 6x - 9y - 12.
    Sol. Pair of infinite solutions :
    2x - 3y = 4; 6x - 9y = 12
    Here,    a1 = 2, b1 = -3 and c1 = 4
    and    a2 = 6, b2 = -9 and c2 = 12
    The given values satisfy the condition :
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#6 {main}</pre>
    So, given equations have infinite solutions pair of no solutions :
    2x - 3y = 4; 4x - 6y = 7
    Here,    a1 = 2, b1 = -3 and c1 = 4
    and    a2 = 4, b2 = -6 and c2 = 7
    The given values satisfy the condition:
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#6 {main}</pre>
    So, given equations have no solutions.

    Question 163
    CBSEENMA10006834

    Show that x = 1 and y = 1 is a solution of the pair of equations 2x + 3y = 5. 5x - 2y = 3.

    Solution

    The given pair of equation is
    2x + 3y = 5
    5x - 2y = 3
    Putting x = 1,y = 1 in eq. (i) and (ii), we get
    2x + 3y = 5
    ⇒ 2(1) + 3(1) = 5
    ⇒ 5 = 5 which is true.
    Also, 5a - 2y = 3
    ⇒ 5(1) - 2(1) = 3
    ⇒    3 = 3 which is true.
    Thus, x = 1, y - 1 is the solution of the given pair of equations.

    Question 164
    CBSEENMA10006837

    Write whether the following pair of linear equations is consistent or not.

    x + y = 14
    x - y = 4 

    Solution

    x + y = 14
    x - y = 4
    H e r e comma space space a subscript 1 over a subscript 2 equals 1 comma space space b subscript 1 over b subscript 2 equals negative 1
S o comma space space space space comma space space a subscript 1 over a subscript 2 space not equal to space b subscript 1 over b subscript 2

    the equation have unique solution.
    Pair of linear equations are consistent.

     
    Question 165
    CBSEENMA10006840

    Find the value of k so that the following system of equations has infinite solutions:
    3x - y - 5 = 0;     6x - 2y + k = 0

    Solution

    Here, a1 = 3, b1 = -1 and c1 = -5
    a2 = 6, b2 = -2 and c2 = k
    therefore space straight a subscript 1 over straight a subscript 1 equals 3 over 6 equals 1 half comma space space space straight b subscript 1 over straight b subscript 2 equals fraction numerator negative 1 over denominator negative 2 end fraction equals 1 half comma space straight c subscript 1 over straight c subscript 2 equals fraction numerator negative 5 over denominator straight k end fraction
Now comma space space space
space space space space space space space space space space space space space space fraction numerator negative 5 over denominator straight k end fraction equals 1 half
rightwards double arrow space space space space straight k space space equals space minus 10
    Hence the given system will have infinite no. of solution if k = -10.

    Question 166
    CBSEENMA10006844

    Find the value of k for which the pair of equations kx - 4y = 3; 6x - 12y = 9 has an infinite number of solution.

    Solution

    We have,
    kx - 4y = 3; 6x - 12y = 9
    Here,    a1 = k, b = -4,c1 = 3
    a2 = 6, b2 = - 12, c2 = 9
    For infinitely many solutions,
    
straight a subscript 1 over straight a subscript 2 equals straight b subscript 1 over straight b subscript 2 equals straight c subscript 1 over straight c subscript 2
rightwards double arrow space space space space space space space space space space space space space space straight k over 6 equals fraction numerator negative 4 over denominator negative 12 end fraction equals 3 over 9
rightwards double arrow space space space space space space space space space space space space space space space straight k over 6 equals 1 third equals 1 third
rightwards double arrow space space space space space space space space space space space space space space space space straight k over 6 equals 1 third rightwards double arrow 3 straight k space rightwards double arrow box enclose straight k equals 2 end enclose space space space
    Hence, the given system of equation has infinitely many solution if, k = 2.

    Question 167
    CBSEENMA10006846

    Check whether x = 3, y = -1 and x = -6, y = 5 are solutions of the system of linear equations 2x + 3y = 3 and 6x + 9y - 9 = 0. Does the given system has an infinite number of solutions ? Give reason.

    Solution

    For x = 3 and y = -1
    2x + 3y = 3
    ⇒ 2 × 3 + 3 × -1 = 3
    ⇒    6-3 = 3; which is true.
    And    6x + 9y - 9 = 0
    ⇒ 6 × 3 + 9 × -l -9 = 0
    ⇒    18 - 9 - 9 = 0; which is true.
    ∴ x = 3 and y = -1 is a solution of the given system.
    For x = -6 and y = 5
    2x + 3y = 3
    ⇒ 2 × -6 + 3 × 5 = 3
    ⇒    -12 + 15 = 3; which is true.
    ⇒    6x + 9y - 9 = 0
    ⇒ 6 × -6 + 9 x 5 - 9 = 0
    ⇒    -36 + 45 - 9 = 0; which is true.
    ∴ x = -6 and y = 5 is also a solution of the given system.
    Yes, the given system of simultaneous linear equations has infinite solutions.

    Question 168
    CBSEENMA10006848

    Obtain the condition for the following system of linear equations to have a unique solution ax + by = c and lx + my = n.

    Solution
    We know, for a unique solution straight a subscript 1 over straight a subscript 2 not equal to straight b subscript 1 over straight b subscript 2
    rightwards double arrow space space space space straight a over straight l not equal to straight b over straight m
rightwards double arrow space space space space am space not equal to bl apostrophe
    which is the required condition.
    Question 169
    CBSEENMA10006851

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

    (i) parallel lines

    (ii) intersecting lines

    (iii) coincident lines.

    Solution
    (i) Required line is 4x + 6y - 2 = 0; so that
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#6 {main}</pre>
    (ii) Required line is 3x + 2y - 4 = 0; so that
    straight a subscript 1 over straight a subscript 2 equals straight b subscript 1 over straight b subscript 2
    (iii) Required line is 6x + 9y - 24 = 0; so that
    straight a subscript 1 over straight a subscript 2 equals straight b subscript 1 over straight b subscript 2 equals straight c subscript 1 over straight c subscript 2


    Question 170
    CBSEENMA10006859

    For what value of k, the following pair of linear equations has infinitely many solutions?

    10x + 5y - (k - 5) = 0
    20x + 10y - k = 0.

    Solution

    Here, we have
    a1 = 10 b1 = 5 c1 = - (k - 5)
    a2 = 20 b2 = 10 c2 = - k
    For infinitely many solutions, we have

    Hence, the given system of equations has infinitely many solutions, when k = 10.

    Question 171
    CBSEENMA10006861

    Does (1, -1) lie on the linear equation 2x - 3y - 5 = 0.

    Solution

    Solution not provided.
    Ans.  Yes

    Question 172
    CBSEENMA10006865
    Question 173
    CBSEENMA10006867

    For what value of k the pair of linear equations 7x - 3y = 4, 14x + ky + 5 = 0 has unique solution.

    Solution

    Solution not provided.
    Ans.  All real numbers except-6 

    Question 174
    CBSEENMA10006868

    Is the pair of linear equations 2a + 3y - 6 = 0, 4x + 6y = 24 is consistent.

    Solution

    Solution not provided.
    Ans.  No

    Question 175
    CBSEENMA10006879

    For what value of k, the pair of linear equations 2a + 3y - 5 = 0; kx - by - 8 = 0 has unique solution.

    Solution

    Solution not provided.
    Ans.All real numbers except -4 

    Question 176
    CBSEENMA10006883
    Question 177
    CBSEENMA10006884

    Given below are three equations. Two of them have infinite solutions and two have a unique solution. State the two pairs : 4x - 5y - 3; 5a - 4y = 5; 8x - 10y = 6.

    Solution

    Solution not provided.
    Ans. Infinite solution, equation : 1 and 3; Unique solution, equation : 1 and 2.

    Question 178
    CBSEENMA10006885

    Given below are three equations two of them have infinite solutions and two have a unique solution state the two pairs : 3x - 2y = 4; 6x + 2y = 4; 9x - 6y = 12.

    Solution

    Solution not provided.
    Ans.Infinite solution, equation : 1 and 3; Unique solution, equation : 1 and 2.

    Question 179
    CBSEENMA10006886

    Is a = -3 and y = 4 a solution of 3x + 2y + 1 = 0 ?

    Solution

    Solution not provided.
    Ans.   Yes

    Question 180
    CBSEENMA10006887

    Is x = 5 and y = 2 a solution of linear equation 3x - 5y - 5 = 0.

    Solution

    Solution not provided.
    Ans.   Yes

    Question 184
    CBSEENMA10006892
    Question 186
    CBSEENMA10006901

    If x = 2, y = 4, find the value of p when 7x - 4y = p.

    Solution

    Solution not provided.
    Ans.  p = -2.

    Question 187
    CBSEENMA10006902

    Manish says to Rahim “I am three times as old as you are.” Later Manish says to Rahim “After 5 years I shall be two times as old as you will be.” Represent this situation algebraically and graphically.

    Solution

    Let present age of Manish be x years and present age of Rahim be y years.
    Case I.    x = 3y
    ⇒    x - 3y = 0
    Case II.
    After 5 years :
    Age of Manish = (x + 5) years
    Age of Rahim = {y + 5) year

    Fig. 3.18.
    According to question
    x + 5 = 2(y + 5)
    ⇒    x + 5 = 2y + 10
    ⇒    x - 2y = 10 - 5
    ⇒    x - 2y = 5
    So, algebraic representation be
    x - 3y = 0    ...(i)
    x - 2y = 5    ...(ii)
    Graphical representation
    For eqn. (i), we have
    x - 3y = 0
    ⇒    x = 3y
    Thus, we have following table :

    For eqn. (ii), we have
    x - 2y = 5
    ⇒    x = 5 + 2y
    Thus, we have following tables :

    When we plot the graph of the equation, we find that both the lines intersect at one point.
    Hence, given system of equations has a unique solution.
    So, we can say that system is consistent.

    Question 188
    CBSEENMA10006906

    Represent the following system of linear equations graphically from the graph find the points where the lines intersect y-axis.

    3x + y - 5 = 0, 2x - y - 5 = 0

    Solution

    We have,
    3x + y - 5 = 0
    ⇒ y = 5 - 3x
    Thus we have following table :

    We have, 2x - y - 5 = 0
    ⇒    y = 2x - 5
    Thus, we have following table :

    Fig. 3.19.
    When we plot the graph of the given equation, we find that both the lines intersect at the point (-1, 2), therefore x = -1, y = 2 is the solution of the given system of equations.
    From the graph we observe that lines intersect y-axis at (-5, 0) and (5,0)

    Question 189
    CBSEENMA10006913

    Determine vertically the co-ordinates of vertices of the triangle formed by the lines whose equations are y = x, y = 2x and x + y = 6.

    Solution

    We have x = y
    Thus, we have following table :

    We have,    y = 2x
    Thus, we have following table :

    We have x + y = 6
    ⇒    x = 6 - y

    Fig. 3.20.

    Thus, we have following table :

    Co-ordinates of the vertices are (0, 0), (2, 4), (3, 3).

    Question 190
    CBSEENMA10006917

    Draw the graphs of the equations :
    x - y = 1
    and    2x + y = 8
    Determine the vertices of the triangle formed by these lines and x-axis.

    Solution

    We have :
    x - y = 1
    ⇒    x = y + 1
    Thus, we have following table :

    Thus, we have following table :

    Fig. 3.21.

    When we plot the graph of the given equations, we find that both the lines intersect at the point (3, 2), therefore x = 3, y = 2 is the solution of the given system of equations.

    Vertices of triangle are A(3, 2), 13(1, 0), C(4, 0).



    Question 191
    CBSEENMA10006920

    Draw the graph of the following equation x + y = 5, 3x - y = 3. Shade the region bounded by these lines and x-axis.

    Solution

    We have
    x + y = 5
    ⇒    x = 5 - y
    Thus, we have the following table :

    We have 3x - y = 3
    ⇒    y = 3x - 3
    Thus, we have following table :

    Fig. 3.22.

    When we plot the graph of the given equations, we find that both the lines intersect at the point (2, 3), therefore, x = 3, y = 2 is the solution of the given system of equation.


    Question 192
    CBSEENMA10006928

    Nitish went to winter sale at Kamla Nagar to purchase some shirts and jackcts. When his friend asked him how many of each he had bought, he answered. “The number of jackets is three less than four times the number of shirts purchased. Also the number of jackets is four less than five times the numbers of shirts purchased.” Help his friends to find how many shirts and jackets Nitish bought. Express the solution graphically.

    Solution

    Let the number of shirts be x and the number of jackets be y. Then equations formed are
    y = 4x - 3    ...(i)
    y = 5x - 4    ...(ii)
    We have    y = 4x - 3
    Thus, we have following table :

    We have    y = 5x - 4
    Thus, we have following table :

    Fig. 3.23.

    When we plot the graph of both the equations, we find that the two lines intersect at the point (1, 1). So,x = 1,y - 1 is the required solution of the pair of linear equation i.e., the number of jackets he purchased is 1 and the number ofshirts he purchased is 1.

    Question 193
    CBSEENMA10006932

    Solve graphically the following system of linear equation, 2x + y = 6, x - 2y = -2. Also find the co-ordinates of the points where the lines meet the A-axis.

    Solution

    We  have,    2x + y = 6
    rightwards double arrow space space space space space space space space space space space space straight x space equals space fraction numerator 6 minus straight y over denominator 2 end fraction
    Thus, we have following table ;

    Fig. 3.24.

    II. We have,
    x - 2y = -2
    ⇒    x = 2y - 2
    Thus, we have following table :


    When we plot the graph of the given equation, we find that both the lines intersect at the point (2, 2).
    From the graph it is clear than the co-ordinates of the points where the lines meet the x-axis are (-2, 0), (3, 0).

    Tips: -

     
    Question 194
    CBSEENMA10006936

    Represent the following pair of equations graphically and write the coordinates of points where the lines intersect y-axis :
    x + 3y = 6
    2x - 3y = 12

    Solution

    The given equations are :
    x + 3y = 6    ...(i)
    and 2x - 3 y = 12    ...(ii)
    From equation (i), we get
    x - 6 - 3y
    Thus, we have following table

    From equation (ii), we get
       2x = 3y + 12
    rightwards double arrow space space space space space space space space straight x space equals space fraction numerator 3 straight y plus 12 over denominator 2 end fraction
    Thus, we have following table

    When we plot the graph of the given equation, we Find that both the lines intersect at the point (6, 0). So, a = 6, y = 0 is the solution of the given equation. From the graph it is clear that the vertices of triangle. So formed by lines representing given equations and y-axis are (6, 0).

    Problems Based on Substitution Method


    Question 195
    CBSEENMA10006940

    Solve the following system of equation by substitution method :

    fraction numerator 3 straight x over denominator 2 end fraction minus fraction numerator 5 straight y over denominator 3 end fraction equals negative 2
straight x over 3 plus straight y over 2 equals 13 over 6

    Solution

    The given equation are 
    fraction numerator 3 straight x over denominator 2 end fraction minus fraction numerator 5 straight y over denominator 3 end fraction equals negative 2 space comma space space space space space space space straight x over 3 plus straight y over 2 equals 13 over 6
    rightwards double arrow space space space space space space 9 x minus 10 y equals negative 2 cross times 6 rightwards double arrow fraction numerator 2 x plus 3 y over denominator 6 end fraction equals 13 over 6
rightwards double arrow space 9 x minus 10 y equals negative 12 space space space space rightwards double arrow 2 x plus 3 y equals 13

    Thus, we have,
    9x - 10y = -12    ...(i)
    2x + 3y = 13    ...(ii)
    From (ii), we get

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#6 {main}</pre>
    Substitution the value of eqn. (iii) in (i). we get
    9 open parentheses fraction numerator 13 minus 3 straight y over denominator 2 end fraction close parentheses minus 10 straight y equals negative 12
rightwards double arrow space space space space fraction numerator 9 left parenthesis 13 minus 3 straight y right parenthesis minus 20 straight y over denominator 2 end fraction equals negative 12
rightwards double arrow 9 left parenthesis 13 minus 3 straight y right parenthesis minus 20 straight y space equals negative 24
rightwards double arrow space 117 minus 27 straight y minus 20 straight y equals negative 24
rightwards double arrow space 117 minus 47 straight y space equals negative 24
rightwards double arrow space minus 47 straight y space equals negative 24 minus 117
rightwards double arrow space minus 47 straight y space equals negative 141
rightwards double arrow space straight y space equals space 141 over 47 equals 3
    Putting the value of ‘y’ in (iii), we get
    
straight x space equals space fraction numerator 13 minus 3 cross times 3 over denominator 2 end fraction
rightwards double arrow space space space straight x equals fraction numerator 13 minus 9 over denominator 2 end fraction equals 4 over 2 equals 2
    Hence, the solution is x = 2, y = 3.
    Question 196
    CBSEENMA10006944

    Solve the following system of equation by substitution method :

    square root of 2 straight x end root plus square root of 3 straight y end root equals 9 space space space space space space space space space space space
square root of 3 straight x end root minus square root of 8 straight y end root equals 0 space space space space space space space space space space space

    Solution

    
square root of 2 straight x end root plus square root of 3 straight y end root equals 9 space space space space space space space space space space space... left parenthesis straight i right parenthesis
square root of 3 straight x end root minus square root of 8 straight y end root equals 0 space space space space space space space space space space space... left parenthesis i i right parenthesis
    From eqn. (i), we have
    straight x equals fraction numerator 9 minus square root of 3 straight y end root over denominator square root of 2 end fraction space space space space space space.. left parenthesis iii right parenthesis
    Substituting the value of eqn. (iii) in (ii), we get

    Substituting the value of ‘y’ in (iii), we get

    Hence, the solution is
    straight x equals fraction numerator 36 over denominator 3 plus 4 square root of 2 end fraction comma space space space space straight y space equals fraction numerator 9 square root of 3 over denominator 3 plus 4 square root of 2 end fraction
    Question 197
    CBSEENMA10006946

    A part of monthly expenses of a family is constant and the remaining varies with the price of wheat. When the rate of wheat is Rs. 250 a quintal, the total monthly expenses of the family are Rs. 1000 and when it is 240 a quintal, the total monthly expenses are Rs. 980. Find the total monthly expenses of the family when the cost of wheat is Rs. 350 a quintal.

    Solution

    Let the constant expenditure be Rs. x
    and consumption of wheat = y quintals
    When rate per quintal = Rs. 250 Then,
    Total expenditure = constant expenditure +
    (consumption × rate per quintal)
    Case I.
    1000 = x + (y × 250)
    ⇒    1000 = x + 250y
    Case II. 980 = x + (y × 240)
    ⇒    980 = x + 240y
    Thus, we have following equations
    x + 250y = 1000    ...(i)
    x + 240y = 980    ...(ii)
    From (i), we have
    x = 1000 - 250y ...(iii)
    Substituting this value in (ii), we get
    (1000 - 250y) + 240y = 980
    ⇒ 1000 - 250y + 240y = 980
    ⇒    1000 - 10y = 980
    -10y = 980 - 1000
    ⇒    -10y = -20
    ⇒    y = 2
    Substituting this value in (iii), we get
    x = 1000 - 250 × 2
    ⇒    x = 1000 - 500
    ⇒    x = 500
    Hence, total monthly expenses = Rs. 500
    Now, total expenses when the price of wheat is Rs. 350 per quintal
    = x + 350y
    = 500 + 350 × 2
    = 500 + 700
    = Rs. 1200. Ans.

    Question 198
    CBSEENMA10006956

    A two digit number is obtained by either multiplying the sum of digits by 8 and adding 1 or by multiplying the difference of the digits by 13 and adding 2. Find the number.

    Solution

    Let the digit at 10’s place by x.
    And, digit of unit’s place be y.
    Then, Number = 10x + y
    Case I. 10x + y = 8(x + y) + 1
    ⇒10x + y = 8x + 8y + 1
    ⇒    10x - 8x + y - 8y = 1
    ⇒    2x - 7y = 1
    Case II.    10x + y = 13(x - y) + z
    ⇒    10x + y = 13x - 13y + 2
    ⇒ 10x - 13x + y + 13y = 2
    ⇒    -3x + 14y = 2
    Thus, we have, 2x - 7y = 1    ...(i)
    -3x + 14y = 2    ...(ii)
    From (i), we have 2x - 7y = 1
    rightwards double arrow space space space space space space space space space space space space space space space space space space 2 x space equals space 7 y plus 1
rightwards double arrow space space space space space space space space space space space space x space equals space fraction numerator 7 y plus 1 over denominator 2 end fraction space space left parenthesis i i i right parenthesis
    Subtituting the value of (ii) and (iii), we get
    rightwards double arrow space space space space space space space minus 3 open parentheses fraction numerator 7 y minus 1 over denominator 2 end fraction close parentheses plus 14 y equals 2
rightwards double arrow space space space space space space space space space fraction numerator negative 3 left parenthesis 7 y plus 1 right parenthesis plus 28 y over denominator 2 end fraction equals 2
rightwards double arrow space space space space space space minus 21 y minus 3 plus 28 y equals 4
rightwards double arrow space space space space space space space space space 7 y space equals space 7
rightwards double arrow space space space space space space space space space y space equals space 1
    Putting the value of y (iii), we get
    straight x equals fraction numerator 7 straight y plus 1 over denominator 2 end fraction equals fraction numerator 7 left parenthesis 1 right parenthesis plus 1 over denominator 2 end fraction equals fraction numerator 7 plus 1 over denominator 2 end fraction equals 8 over 2 equals 4

    Hence, Number = 10x + y
    = 10(4) + 1 =4 0 + 1 = 41

    Question 199
    CBSEENMA10006965

    A and B are friends and their ages differ by 2 years. A’s father D is twice as old as A and B is hvicc as old as his sister C. The age of D and C differ by 40 years. Find the ages of A and B.

    Solution
    Let the ages of A and B be x and y years respectively. Then,
    space space space space space space straight x minus straight y minus plus-or-minus 2 space space space space space space space left square bracket Given right square bracket
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#6 {main}</pre>
    Clearly, D is older than C 
    therefore space space space space space 2 straight x minus straight y over 2 equals 40 rightwards double arrow 4 straight x minus straight y equals 80

    Thus, we have the following systems of linear equations
    x - y = 2    ...(i)
    4x - y = 80    ...(ii)
    and    x - y = -2    ...(iii)
    Case I : From (i), we have
    [Considering (i) and (ii)]
    x - y = 2
    ⇒    x = 2 + y    ...(iv)
    Substituting the value of x in (ii), we get
    4x - y - 80
    ⇒ 4(2 + y) - y = 80
    ⇒ 8 + 4y - y = 80
    ⇒ 8 + 3y = 80 ⇒ 3y = 72
    ⇒    y = 24
    Substituting the value ofy in (iv), we get
    x = 2 + y = 2 + 24=    26
    Hence, Age of A = 26 years
    and Age of B = 24 years
    Case II : From (iii), we have
    [Considering (ii) and (iii)]
    x - y = -2 ⇒ x = y - 2 ...(v)
    Putting the value of (v) in (ii), we get
    4x - y = 80
    rightwards double arrow 4 left parenthesis straight y minus 2 right parenthesis minus straight y space equals space 80
rightwards double arrow 4 straight y minus 8 minus straight y equals 80
rightwards double arrow space 3 straight y space equals space 88
rightwards double arrow space space space space straight y space equals space 88 over 3 equals 29 1 third
    Substituting the value ofy in (v), we get
    

straight x equals 88 over 3 minus 2 equals fraction numerator 88 minus 6 over denominator 3 end fraction equals 82 over 3 equals 27 1 third
Hence comma space Age space of space straight A space equals space 29 1 third space space years
Age space of space straight B space equals space 27 1 third space years

    Question 200
    CBSEENMA10006967

    A man sold a chair and a table together for Rs. 1520 thereby making a profit of 25% on the 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

    Let the cost price of one chair be Rs. x and that of one table be Rs. y. Profit on chair = 25%.

    ∴ Selling price of one chair equals space straight x plus 25 over 100 straight x equals 125 over 100 straight x
    Profit on a table = 10%
    ∴ Selling price of one table = straight y plus fraction numerator 10 straight y over denominator 100 end fraction equals 110 over 100 straight y
    According to the given condition, we have
    125 over 100 straight x plus 110 over 100 straight y equals 1520
rightwards double arrow 125 straight x plus 110 straight y equals 152000
rightwards double arrow space 25 straight x space plus space 22 straight y space equals space 30400 space space space space space space space space space space space... left parenthesis straight i right parenthesis

    If profit on a chair is 10% and on a table is 25%, then total selling price is Rs. 1535.
    
therefore space space space open parentheses straight x plus 10 over 100 straight x close parentheses plus open parentheses straight y plus 25 over 100 straight y close parentheses equals 1535
rightwards double arrow space space space 110 over 100 straight x plus 125 over 100 straight y equals 1535
rightwards double arrow space space 110 straight x plus 125 straight y equals 153500
rightwards double arrow space 22 straight x space plus space 25 straight y space equals space 30700 space space space space space space space space space space space space... left parenthesis ii right parenthesis

    Subtracting equation (ii) from equation (i), we get
    3x - 3y = -300 ⇒ x - y = -100
    Adding equation (ii) and (i), we get
    47x + 47y = 61100 ⇒ x + y = 1300
    Thus, we have following equations
    x - y = -100    ...(iii)
    x + y = 1300    ...(iv)
    From (iii), we have
    x - y = - i 00
    ⇒    x = y - 100    ...(v)
    Substituting the value of x in (iv), we get
    x + y = 1300
    ⇒ y - 100 + y - 1300
    2y - 100 = 1300
    ⇒    2y - 1400 ⇒ y = 700
    Substituting the value of y in (v), we get
    x = y - 100
    ⇒ x = 700 - 100 = 600
    Hence, cost price of one chair = Rs. 600 and cost price of one table = Rs. 700

    Question 201
    CBSEENMA10006969

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

    Solution



    Fig. 3.26.
    Let the speeds of two cars be x km./hr. and y km/hr.
    Then, according to the given problem,
    5x - 5y = 100
    ⇒ x - y = 20 ...(i)
    and    x + y = 100 ...(ii)
    By adding, we get    2x = 120
    ⇒    x = 60
    By subtracting, we get 2y = 80
    ⇒    y = 40
    Hence, the required speeds of the two cars are: 60 km/hr. and 40 km/hr.

    Question 202
    CBSEENMA10006970

    The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each saves Rs. 200 per month find their monthly incomes.

    Solution

    Let the income of first person be Rs. 9x and the income of second person be Rs. 7x. Again, let the expenditures of first and second person be 4y and 3y respectively. Then,
    Saving of first person = 9x - 4y
    Saving of second person = 7x - 3y
    ∴ 9x - 4y = 200    ...(i)
    and    7x - 3y = 200    ...(ii)
    From (i), we have
    9x - 4y = 200
    rightwards double arrow space space space straight x space equals space fraction numerator 200 plus 4 straight y over denominator 9 end fraction space space space space space space space space space... left parenthesis i i straight i right parenthesis

     

    Substituting the value of x in (ii), we have
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#6 {main}</pre>
    Substituting the value of y in (iii), we get
    x equals fraction numerator 200 plus 4 y over denominator 9 end fraction
rightwards double arrow space x space equals space fraction numerator 200 plus 4 left parenthesis 400 right parenthesis over denominator 9 end fraction
equals fraction numerator 200 plus 1600 over denominator 9 end fraction equals 1800 over 9 equals 200

    Hence, monthly income of First person
    = Rs. 9x = Rs. (9 × 200) = Rs. 1800
    and monthly income of second person
    = Rs. 7x = Rs. (7 × 200) = Rs. 1400.
    Problems Based on Elimination Method


    Question 203
    CBSEENMA10006971

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

    Solution

    Let the speed of the train be x km/hr and that of the car be y km/hr.
    We have following cases :
    Case I. When he travels 250 km by train and the rest by car:
    In this case, we have
    Time taken by the man to travel 250 km by train
    equals 250 over straight x hrs
    Time taken by the man to travel (370 - 250)
    
space space space space space space space space space equals 120 space km space by space car space equals space 120 over straight y hrs
    According to the given condition
    250 over straight x plus 120 over straight y equals 4
rightwards double arrow space space 125 over straight x plus 60 over straight y equals 2

    Case II. When he travels 130 km by train and the rest by car:
    Time taken by the man to travel 130 km by train
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#6 {main}</pre>

    Time taken by the man to travel (370 - 130)
    equals space 240 space km space by space car space equals space 240 over straight y hrs

    According to the given condition

    
130 over straight x plus 240 over straight y equals 4 space space hrs space 18 space minutes
rightwards double arrow space 130 over straight x plus 240 over straight y equals 4 18 over 60
rightwards double arrow space space 130 over straight x plus 240 over straight y equals 43 over 10

    Thus, we have following system of equations :
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#6 {main}</pre>
    
Putting space 1 over straight x equals straight u space and space 1 over straight y equals straight v comma space t h e space a b o v e space s y s t e m space r e d u c e s space t o
    125u + 60 v = 2                         ....(i)
    130u + 240v = 43 over 10 space space space space space space space space space space space space space space space space space.... left parenthesis ii right parenthesis
    Multiplying equation (iii) by 4 the above system of equations becomes
    500u + 240v = 8                  ...(iii)
    130u  + 240v = space space space 43 over 10 space space space space space space space space space space space space... left parenthesis iv right parenthesis
    Subtracting equation (vi) from equation (iv), we get
    370 straight u space equals space 8 space minus 43 over 10
rightwards double arrow space space 370 straight u space equals space 37 over 10 rightwards double arrow space straight u equals 1 over 100
Putting space straight u space equals space 1 over 100 space in space equatiion space left parenthesis iii right parenthesis comma space we space get
    5 + 240v = 8
    
space space
space space space space rightwards double arrow space 240 straight v space equals space 3 space rightwards double arrow space straight v space equals space 1 over 80
space space space space Now comma space space space space straight u space equals space 1 over 100 space and space straight v space 1 over 80
space space space space rightwards double arrow space space space space space space space 1 over straight x equals 1 over 100 space and space 1 over straight y equals space 1 over 80
space space space space rightwards double arrow space space space space space space space straight x space equals space 100 space space and space straight y space equals space 80

    Hence, Speed of the train = 100 km/hr
    Speed of the car = 80 km/hr.

    Question 204
    CBSEENMA10006972

    The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?

    Solution

    Let the digit at 10’s place be x
    and digit at unit’s place be y.
    ∴ Number = 10x + y
    Number obtained after reversing the order of the digits = 10y + x
    Case I.
    10x + y + 10y + x = 66
    ⇒ 11x + 11y = 66
    ⇒ 11(x + y) = 66
    ⇒ x + y = 6    ...(i)
    Case II. x - y = ±2
    ⇒    x - y = + 2    ...(ii)
    and    x - y = - 2    ...(iii)
    Considering (i) and (ii), we get
    space space space space space space space space x space plus space y space equals space 6
space space space space space space space space x space minus space y space equals space 2
space bottom enclose space space space space space space space minus space space plus space space space space space space minus space space space space end enclose
space space space space space space space space space space 2 y space equals space 4
space rightwards double arrow space space space space space space y space equals space 2

    Putting the value of ‘y’ in (i), we get
    x + 2 = 6
    ⇒    x = 4
    So,    Number = 10x + y
    = 42
    Considering (i) and (iii), we get
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#6 {main}</pre>

    Putting the value of ‘y’ in (i), we get
    x + y = 6
    ⇒    x + 4 = 6
    ⇒    x = 2
    So,    Number = 10x + y = 24
    Thus, there are two such numbers 42 and 24.

    Question 205
    CBSEENMA10006973

    Solve the following system of equations by elimination method

    6(ax + by) = 3a + 2b
    6(bx - ay) =3b - 2a.

    Solution

    We have
    6 ax + 6by = 3a + 2b    ...(i)
    6bx - 6ay = 3b - 2a    ...(ii)
    For making the coefficient of ‘x’ in (i) and (ii) equal, we multiply eq. (i) by ‘b’ and (ii) by ‘a’ and then subtracting, we get

    Question 206
    CBSEENMA10006974

     A man has only 20 paisc coins and 25 paise coins in his purse. If he has 50 coins in all, totalling Rs. 11.25. How many coins of each type does he share ? (Use Elimination Method).

    Solution

    Let the number of 20 paise coins = x
    and    number of 25 paise coins = y
    Then,    value of 20 paise coins = 20x
    and    value of 25 paise coins = 25y
    Case I.    x + y = 50
    Case II. 20x + 25y = 1125
    [∵ 11.25 = 1125 Paise]
    Thus, we have
    x + y = 50    ...(i)
    20x + 25y = 1125    ...(ii)
    For making the coefficient of ‘x’ in (i) and (ii) equal, we multiply the (i) by 20 and then subtracting, we get
    space space space
space space space bottom enclose space space space space space space space 20 x space plus space 20 y space equals space 1000
space space space space space space space 20 x space plus space 25 y space equals space 1125
space space space space space space space minus space space space space minus space space space space space space space space space space minus space space space space space space space space end enclose
space space space space space space space space space space space space space minus 5 y space equals space minus 125
space rightwards double arrow space space space space space space space space space space space space space space space space y space equals space 25

    Putting the value of ‘y’ in (i), we get
    x + y = 50
    ⇒    x + 25 = 50
    ⇒    x = 25

    Hence, the number of 20 paise coins = 25 and the number of 25 paise coins = 25.

    Question 207
    CBSEENMA10006975

    A person starts his job with a monthly income and earns a fixed increment every year. If his salary was Rs. 4500 after 4 years of service and Rs. 5400 after 10 years of service, find the initial salary and the annual increment by using elimination method.

    Solution

    Let the initial salary be ‘x’ and increment per year be ‘y’
    Case I. Initial salary = x
    Increment after 4 years = 4y
    According to question
    x + 4y = 4500
    Case II. Initial salary = x
    Increment after 10 years = 10y
    According to question
    x + 10y = 5400
    Thus, we have x + 4y = 4500 ...(i)
    x + 10y = 5400 ...(ii)
    Since the coefficient of ‘x’ in (i) and (ii) are equal.
    So we can simply eliminate the variable ‘x’ by subtracting.
    bottom enclose x space plus space 4 y space equals space 4500
x space plus space 10 y space equals space 5400
minus space minus space space space space space space space space space space minus
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space end enclose
space space space space space space space minus 6 y space equals space minus 900
space rightwards double arrow space space space space space space space y space equals space 150 space

    Putting this value in (i), we get
    x + 4y = 4500
    ⇒ x + 4(150) = 4500
    ⇒ x + 600 = 4500
    ⇒ x = 3900
    Hence, initial salary = Rs. 3900 and annual Increment = Rs. 150.
    Problems Based on Cross-Multiplication Method

    Question 208
    CBSEENMA10009612

    A motor boat whose speed is 24 km/h in still water takes 1 hour more to go 32km upstream than to return downstream to the same spot. Find the speed of the stream.

    Solution

    Let the speed of the stream be s km/h.
    Speed of the motor boat 24 km / h
    Speed of the motor -boat upstream 24 s
    Speed of the motor boat downstream 24 s
    According to the given condition,
    fraction numerator 32 over denominator 24 minus straight s end fraction minus fraction numerator 32 over denominator 24 plus s end fraction space equals 1
therefore space 32 space open parentheses fraction numerator 1 over denominator 24 minus s end fraction minus fraction numerator 1 over denominator 24 plus s end fraction close parentheses equals 1
therefore space 32 space open parentheses fraction numerator 24 plus straight s minus 24 plus straight s over denominator 576 minus straight s squared end fraction close parentheses space equals 1
therefore space 32 space straight x space 2 straight s space equals 576 minus straight s squared
therefore space straight s squared plus 64 straight s minus 576 space equals 0
therefore space left parenthesis straight s plus 72 right parenthesis left parenthesis straight s minus 8 right parenthesis space equals 0
Therefore comma space
straight s equals negative 72 space ors equals 8
    Since, speed of the stream cannot be negative, the speed of the stream is 8 km/h.

    Question 209
    CBSEENMA10009636

    A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average speed of 6 km/h more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed?

    Solution

    Let the average speed of train for the first 54 km be x km/h
    ⇒ Average speed for the next 63 km = ( x + 6) km/h
    We know

    Time space equals fraction numerator space Distance over denominator Speed end fraction

therefore space Time space taken space by space the space train space to space cover space 54 space km space equals space 54 over straight x space straight h
Also comma space time space taken space by space the space train space to space cover space the space next space 63 space km space equals space fraction numerator 63 over denominator straight x space plus space 6 end fraction space straight h

According space to space the space question comma

54 over straight x space plus space fraction numerator 63 over denominator straight x space plus space 6 end fraction space equals space 63

rightwards double arrow space fraction numerator 54 space left parenthesis straight x space plus space 6 space right parenthesis space plus 63 straight x over denominator straight x space left parenthesis straight x space plus space 6 right parenthesis end fraction space space equals space 3
    ⇒ 117x + 324 = 3 ( x2 + 6x)
    ⇒ 117x + 324 = 3x2 + 18x
    ⇒ 3x2 - 99x -324 = 0 ..(i)
    Taking common from the above equation (i)
    we have
    x2 - 33x - 108 = 0
    ⇒ x2 - 36x + 3x -108 = 0
    ⇒ x (x - 36) + 3 (x - 36) = 0
    ⇒ (x - 36) (x +3 )=0
    ⇒ x -36 = 0 Or x + 3 =0
    x = 36 or x = -3
     The speed of the train cannot be negative. Thus, x = 36
    Hence, the speed of the train to cover 54 km or its first speed is 36 km/h.

    Question 210
    CBSEENMA10009798

    If the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear, then find the value of k.

    Solution

    Given points are A( k+1, 2k), B(3k, 2k+3), and C(5k-1, 5k)

    These points will be collinear, if area of the triangle formed by them is zero.

    We have,

                  

    i.e. 

    k+1 x 2k+3 + 3k x 2k + 5k - 1  x 2k - 3k x 2k  + 5k-1 x 2k+3 + k+1 x 5k = 02k2 + 5k  + 3  + 15k2 + 10k2 - 2k -[ 6k2 + 10k2 + 13k - 3 + 5k2 + 5k ] = 027k2 + 3k + 3 -[ 21k2 + 18k - 3 ]= 027 k2 + 3k + 3 - 21k2 - 18k + 3 = 06k2 - 15k + 6 = 02k2 - 5k + 2 = 02k2 - 4k -k + 2 = 0(k-2) (2k-1) = 0k-2 = 0  or  2k-1 = 0 k = 2   or   k = 12

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