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NCERT Exemplar Solution for CBSE Class 10 Mathematics: Pair of Linear Equations (Part-IIIA)

Jun 22, 2017 15:46 IST

    Class 10 Maths NCERT Exemplar, Real Numbers NCERT Exemplar Problems, NCERT Exemplar Problems, Class 10 NCERT ExemplarHere you get the CBSE Class 10 Mathematics chapter 3, Pair of Linear Equations in Two Variables: NCERT Exemplar Problems and Solutions (Part-IIIA). This part of the chapter includes solutions of Question Number 1 to 11 from Exercise 3.1 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Pair of Linear Equations in Two Variables. This exercise comprises of only the Short Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.

    NCERT Exemplar Solution for CBSE Class 10 Mathematics: Pair of Linear Equations (Part-I)

    NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.

    NCERT Exemplar Solution for CBSE Class 10 Mathematics: Pair of Linear Equations (Part-II)

    Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Pair of Linear Equations in Two Variables:

    Exercise 3.3

    Short Answer Type Questions (Q. No. 1 to 11)

    Quesntion1. For which value(s) of l, do the pair of linear equations lx + y = l2 and

    x+ly = 1 have

    (i) no solution?           

    (ii) infinitely many solutions?

    (iii) a unique solution?

    Solution:

    Given, lx + y = l2 and x + ly -1                  ….(i)

    Here, a1 = l, b1, = 1, c, = -l2 and a2 = 1, b2 = l, c2 = -1

    (iii) Condition for a unique solution is:

    Therefore, all real values of l except  ±1 equation will have unique solution.

    Quesntion2. For which value (s) of k will the pair of equations

    kx + 3y = k -3,

    12x+ ky = k

    have no solution?

    Solution:

    Given,

    kx + 3y – (k -3) = 0 and 12x + ky – k = 0

    On comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 we have         

    Here, a1 = k, b1=3 and c1 = -(k -3)

    And a2 = 12, b2 = k and c2 = -k         …..(i)

    Condition for a pair of linear equations to have no solution is:

    Thus, from (iii) and (iv), it’s clear that at k = - 6 the given pair of linear equations will have no solution.

    Quesntion3. For which values of a and b will the following pair of linear equations has infinitely many solutions?

    x + 2y = 1

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

    Solution:

    Given equation are:

    x + 2y −1 = 0 and (a - b) x + (a + b) y – (a + b - 2) = 0 ….(i)

    On comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 we have:

    Quesntion4. Find the values of p in (i) to (iv) and p and q in (v) for the following pair of equations

    (i) 3x -y - 5 = 0 and 6x - 2y - p = 0, if the Lines represented by these

    equations are parallel.

    (ii) -x + py = 1 and px- y = 1, if the pair of equations has no solution.

    (iii) -3x + 5y = 7 and 2px - 3y =1

    if the lines represented by these equations are intersecting at a unique point.

    (iv) 2x + 3y -5 = 0 and px - 6y - 8 = 0,

    if the pair of equations has a unique solution.

    (v) 2x + 3y = 7 and 2px + py = 28 -qy,

    if the pair of equations has infinitely many solutions.

    Solution:

    (i) Given, 3x - y -5 = 0 and 6x-2y - p = 0

    On comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 we have:

    c1 = 5 and a2 = 6, b2 = -2, c2 = -p

    As, the lines represented by these equations are parallel, therefore

    Therefore, the given pair of linear equations are parallel for all real values of p except 10.

    (ii) Given, -x + py -1 = 0 and px-y-1 = 0      …..(i)

    We have, a1 = -1, b1 = p, c1= -1and a2 = p, b2 = -1 and c2 = -1

    As, the pair of linear equations has no solution i.e., both lines are parallel to each other.

    Therefore, the given pair of linear equations has no solution for p = 1.

    (iii) Given, -3x + 5y - 7 = 0 and 2px-3y-1 = 0

    Here, a1 = -3, b1 = 5, c1 = -7 and a2 = 2p, b2 = -3, c2 = -1

    As, the lines are intersecting at a unique point i.e., it has a unique solution.

    Therefore, the lines represented by these equations are intersecting at a unique point for all great values of p except 9/10.

    (iv) Given, 2x + 3y-5 = 0 and px - 6y -8 = 0

    Here,c1 = -5 and a2 = p, b2 = -6, c2 = −8

    As, the pair of linear equations has a unique solution.

    Given, the pair of linear equations has a unique solution for all values of p except -4.

    (v)   Given, 2x + 3y = 7 and 2px + py = 28 -qy

    Here, a1 = 2, b2, = 3, c1 = -7 and a2 = 2p, b2 = (p + q), c2 = -28

    Since, the pair of equations has infinitely many solutions i.e., both lines are coincident.

    ⟹ 4 + q =12

    Therefore, q = 8

    Therefore, the pair of equations has infinitely many solutions for the values of p = 4and q = 8

    Quesntion5. Two straight paths are represented by the equations x - 3y = 2 and -2x + 6y = 5. Check whether the paths cross each other or not.

    Solution:

    Hence, two straight paths represented by the given equations never cross each other because they are parallel to each other.

    Quesntion6. Write a pair of linear equations which has the unique solution

    x = - 1 and y 3. How many such pairs can you write?

    Solution:


    Hence, infinitely many pairs of linear equations are possible.

    Quesntion7. If 2x + y = 23 and 4x - y = 19, then find the values of 5y -2x and

    Solution:

    Quesntion8. Find the values of x and y in the following rectangle

     

    Quesntion9. Solve the following pairs of equations

                

    Solution:


    Hence, the required values of x and y are 1.2 and 2.1, respectively

    Hence, the required values of x and y are 6 and 8, respectively

    Hence, the required values of x and y are 3 and 2, respectively.

    (v) Given pair of linear equations is

              43x + 67y = -24     ...(i)

    And   67x + 43y = 24      ...(ii)

    Now taking 43 × (i) + 67 ×(ii), we get

    Hence, the required values of x and y are 1 and -1, respectively.

    (vi) Given pair of linear equations is

    Hence, the required values of x and y are a2 and b2, respectively.

    Now, multiplying both sides of equation (i) by LCM (10, 5) = 10, we get

    x + 2y – 10 = 0

    Þ x + 2y = 10 ... (iii)

    Again, multiplying both sides of (iv) by LCM (8, 6) = 24, we get

    3x + 4y = 360 ... (iv)

    Now, (iv) – 2 × (iii) gives:

    Hence, the solution of the pair of equations is x = 340, y = - 165 and the required value of l is

    Quesntion11. By the graphical method, find whether the following pair of equations are consistent or not. If consistent, solve them.

    (i) 3x + y + 4 = 0, 6x - 2y + 4 = 0

    (ii) x - 2y = 6, 3x - 6y = 0

    (iii) x + y = 3, 3x + 3y = 9

    Solution:

    (i) Given pair of equations is:

     3x+ y + 4= 0 and 6x - 2y + 4 = 0

    On comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 we have

    and c1 = 4  and a2 = 6, b2 = - 2 and c2 = 4

                                         

    So, the given pair of linear equations has a unique solution and thereby it is consistent.    

    We have,  3x + y + 4 = 0

    Þ y = - 4 - 3x                           ….(i)

    If x = 0, y = - 4

    x = - 1,  y = - 1

    x = - 2, y = 2                    

    x

    -1

    -2

    y

    -4

    -1

    2

    Points

    B

    C

    A

    And 6 - 2 + 4 = 0                     …..(ii)

    Þ 2y = 6x + 4

    Þ y = 3x + 2                                                                      

    If x = 0, y = 2

    x = -1, y = - 1

    x = 1, y = 5

     

    x

    -1

    1

    y

    -1

    2

    5

    Points

    C

    Q

    P

     

    Plotting (i) and (ii) as per the respective values of x and y, we get two lines AB and PQ respectively that intersect at C (-1.-1).

    Thus, the given pair of linear equations has no solution, i.e., inconsistent.

    (iii) Given pair of equations is:

    x + y = 3 and 3x + 3y = 9

    On comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 we have

    a1 = 1, b1 = 1 and c1, = - 3 and a2 = 3, b2 = 3 andc2 = - 9                         

    So, the given pair of lines is coincident. Therefore, these lines have. Hence, the given pair of linear equations has infinitely many solutions, i.e., consistent.

    Now, x + y = 3             ….(i)

    Or      y = 3 - x

    If x = 0, y = 3

    x = 3, y = 0

    x

     

    3

    y

    3

    Points

    A

    B

    Also 3x + 3y = 9           ….(ii)

    Or    3y = 9 - 3x 

    Plotting (i) and (ii) for respective set of values for x and y, we get two lines AB and CD respectively, that are coincident.

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