CBSE Class 10 Mathematics Pairs of Linear Equations in Two Variables MCQs Set 13

Question. The pair of linear equations \( \frac{3x}{2} + \frac{5y}{3} = 7 \) and \( 9x + 10y = 14 \) is
(a) consistent
(b) inconsistent
(c) consistent with one solution
(d) consistent with many solutions
Answer: (b) inconsistent
Explanation:
For the given pair of equations, we have:
\( \frac{a_1}{a_2} = \frac{3/2}{9} = \frac{1}{6} \)
\( \frac{b_1}{b_2} = \frac{5/3}{10} = \frac{1}{6} \)
\( \frac{c_1}{c_2} = \frac{7}{14} = \frac{1}{2} \)
\( \therefore \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
Hence, the pair of equations is inconsistent.

Question. Graphically, the pair of equations \( 6x - 3y + 10 = 0 \) and \( 2x - y + 9 = 0 \) represents two lines which are:
(a) intersecting at exactly one point.
(b) intersecting at exactly two points.
(c) coincident
(d) parallel
Answer: (d) parallel

Question. If a pair of linear equations is consistent, then the lines will be:
(a) parallel
(b) always coincident
(c) intersecting or coincident
(d) always intersecting
Answer: (c) intersecting or coincident
Explanation: The conditions for a pair of linear equations to be consistent are:
• Intersecting lines having unique solution, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
OR
• Coincident or dependent lines having infinitely many solutions, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)

Question. The value of k for which the system of linear equation \( x + 2y = 3, 5x + ky + 7 = 0 \) is inconsistent is
(a) \( -\frac{14}{3} \)
(b) \( \frac{2}{5} \)
(c) 5
(d) 10
Answer: (d) 10
Explanation:
The system of equations will be inconsistent if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
Here, \( a_1 = 1, b_1 = 2, c_1 = 3 \) (after rewriting as \(x+2y-3=0\), \(c_1=-3\))
\( a_2 = 5, b_2 = k, c_2 = 7 \)
\( \frac{1}{5} = \frac{2}{k} \neq \frac{-3}{7} \)
i.e., when \( k = 10 \)

Question. The value of k for which the system of equations \( x + y - 4 = 0 \) and \( 2x + ky = 3 \), has no solution, is
(a) -2
(b) \( \neq 2 \)
(c) 3
(d) 2
Answer: (d) 2
Explanation: \( x + y - 4 = 0 \) and \( 2x + ky = 3 \) has no solution, when:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \therefore \frac{1}{2} = \frac{1}{k} \neq \frac{-4}{-3} \)
\( \Rightarrow k = 2 \)

Question. The pair of equations \( y = 0 \) and \( y = -7 \) has:
(a) one solution
(b) two solutions
(c) infinitely many solutions
(d) no solution
Answer: (d) no solution
Explanation: We know that equation of the form \( y = 'a' \) is a line parallel to the x-axis at a distance 'a' from it.
The given pair of equations are \( y = 0 \) and \( y = -7 \).
\( y = 0 \) is the equation of the x-axis and \( y = -7 \) is the equation of the line parallel to the x-axis. So, these two equations represent two parallel lines.
We know that parallel lines never intersect. So, there is no solution for these lines.

Question. The pair of equations \( x = a \) and \( y = b \) graphically represents lines which are:
(a) parallel
(b) intersecting at (b, a)
(c) coincident
(d) intersecting at (a, b)
Answer: (d) intersecting at (a, b)

Question. For which value(s) of p, will the lines represented by the following pair of linear equations be parallel: \( 3x - y - 5 = 0 \) \( 6x - 2y - p = 0 \)
(a) all real values except 10
(b) 10
(c) \( \frac{5}{2} \)
(d) \( \frac{1}{2} \)
Answer: (a) all real values except 10
Explanation: Conditions for parallel lines:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \frac{3}{6} = \frac{-1}{-2} \neq \frac{-5}{-p} \)
\( \frac{1}{2} = \frac{1}{2} \neq \frac{5}{p} \)
\( \Rightarrow p \neq 10 \)

Question. If the lines given by \( 3x + 2ky = 2 \) and \( 2x + 5y + 1 = 0 \) are parallel, then the value of k is:
(a) \( -\frac{5}{4} \)
(b) \( \frac{2}{5} \)
(c) \( \frac{15}{4} \)
(d) \( \frac{3}{2} \)
Answer: (c) \( \frac{15}{4} \)
Explanation: The given equation of lines are \( 3x + 2ky = 2 \) and \( 2x + 5y + 1 = 0 \)
Comparing with \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), we have
\( a_1 = 3; b_1 = 2k; c_1 = -2 \)
\( a_2 = 2; b_2 = 5; c_2 = 1 \)
We know that the condition for parallel lines is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \Rightarrow \frac{3}{2} = \frac{2k}{5} \)
\( \Rightarrow 15 = 4k \)
\( \Rightarrow k = \frac{15}{4} \)
For \( k = \frac{15}{4} \), \( \frac{2k}{5} = \frac{30}{20} = \frac{3}{2} \neq \frac{-2}{1} \)
Thus, \( k = \frac{15}{4} \)

Question. The pair of equations, \( x = 0 \) and \( x = -4 \) has
(a) a unique solution
(b) no solution
(c) infinitely many solution
(d) only solution (0, 0)
Answer: (b) no solution

Question. One equation of a pair of dependent linear equations is \( -5x + 7y = 2 \). The second equation can be:
(a) \( 10x + 14y + 4 = 0 \)
(b) \( -10x - 14y + 4 = 0 \)
(c) \( -10x + 14y + 4 = 0 \)
(d) \( 10x - 14y = -4 \)
Answer: (d) \( 10x - 14y = -4 \)
Explanation: In a pair of dependent linear equation, one equation is just a multiple of another equation. Thus, the second equation is \( k(-5x + 7y - 2) = 0 \)
Putting \( k = -2 \), we get
\( \Rightarrow 10x - 14y + 4 = 0 \)
On moving 4 to the other side, we get
\( \Rightarrow 10x - 14y = -4 \)
\( \therefore \) (D) option is correct.

Question. A pair of linear equations which has a unique solution \( x = 2, y = -3 \) is:
(a) \( x + y = -1 \) and \( 2x - 3y = -5 \)
(b) \( 2x + 5y = -11 \) and \( 4x + 10y = -22 \)
(c) \( 2x - y = 1 \) and \( 3x + 2y = 0 \)
(d) \( x - 4y - 14 = 0 \) and \( 5x - y - 13 = 0 \)
Answer: (d) \( x - 4y - 14 = 0 \) and \( 5x - y - 13 = 0 \)
Explanation: If \( x = 2 \) and \( y = -3 \) is a unique solution of any pair of equation, then these values must satisfy that pair of equations.
Putting the values in the equations for every option and checking it -
For case (A): \( x + y = -1 \Rightarrow 2 - 3 = -1 \) (True). \( 2x - 3y = -5 \Rightarrow 4 + 9 = 13 \neq -5 \) (False).
For case (B): Dependent linear equations, infinitely many solutions.
For case (C): \( 2x - y = 1 \Rightarrow 4 + 3 = 7 \neq 1 \) (False).
For case (D):
\( x - 4y - 14 = 0 \Rightarrow 2 - 4(-3) - 14 = 2 + 12 - 14 = 0 \) (True)
\( 5x - y - 13 = 0 \Rightarrow 5(2) - (-3) - 13 = 10 + 3 - 13 = 0 \) (True)
Since \( x = 2, y = -3 \) is satisfying both the equations, option (D) is true.

Question. If \( x = a, y = b \) is the solution of the equations \( x - y = 2 \) and \( x + y = 4 \), then the values of a and b respectively are:
(a) 3 and 5
(b) 5 and 3
(c) 3 and 1
(d) -1 and -3
Answer: (c) 3 and 1
Explanation: Since \( x = a, y = b \) is the solution of the equations \( x - y = 2 \) and \( x + y = 4 \), these values must satisfy the given pair of equations.
Putting the values in the equations, we have
\( a - b = 2 \) ...(i)
and \( a + b = 4 \) ...(ii)
Adding equations (i) and (ii), we get
\( 2a = 6 \) or \( a = 3 \)
Putting the value of a in equation (ii), we get
\( 3 + b = 4 \) or \( b = 1 \)
Hence, option (C) is correct.

Question. Aruna has only ₹ 1 and ₹ 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is ₹ 75, then the number of ₹ 1 and ₹ 2 coins respectively are:
(a) 35 and 15
(b) 35 and 20
(c) 15 and 35
(d) 25 and 25
Answer: (d) 25 and 25
Explanation: Let number of ₹ 1 coins = x and number of ₹ 2 coins = y.
It is given that,
Total number of coins = \( x + y = 50 \) ...(i)
Also, amount of money with her = (Number of ₹ 1 coins × 1) + (Number of ₹ 2 coins × 2)
\( \Rightarrow x(1) + y(2) = 75 \)
\( \Rightarrow x + 2y = 75 \) ...(ii)
On subtracting eq. (i) from eq. (ii), we get
\( \Rightarrow (x + 2y) - (x + y) = (75 - 50) \)
\( \Rightarrow y = 25 \)
Putting \( y = 25 \) in eq. (i), we get
\( x + 25 = 50 \Rightarrow x = 25 \)
Hence, Aruna has 25 ₹ 1 coins and 25 ₹ 2 coins.

Fill in the Blanks/True False

Fill in the blanks/tables with suitable information or answer whether True/False

Question. The value of k for which pair of linear equations \( 3x + 2y = 5 \) and \( x - ky = 2 \) has a unique solution is .................... .
Answer: \( k \neq - 2/3 \)
Explanation: For unique solution \( \frac{3}{1} \neq \frac{2}{-k} \Rightarrow k \neq - 2/3 \)

Question. The value of a so that the point (3, a), lies on the line represented by \( 2x - 3y = 5 \) is .................... .
Answer: \( \frac{1}{3} \)
Explanation: Given (3, a) lies on \( 2x - 3y = 5 \)
\( \Rightarrow 2 \times 3 - 3a = 5 \Rightarrow 3a = 6 - 5 = 1 \)
\( \Rightarrow a = \frac{1}{3} \)

Question. The co-ordinate where the line \( x - y = 8 \) will intersect y-axis is .................... .
Answer: (0, - 8)
Explanation: At y-axis, \( x = 0 \)
\( \therefore 0 - y = 8 \Rightarrow y = - 8 \)
Point = (0, - 8)

Question. The value of k for which the pair of linear equations \( kx + 3y = k - 2 \) and \( 12x + ky = k \) has no solution is .................... .
Answer: \( k = \pm 6 \)
Explanation: Since, pair of linear equations has no solution
Then \( \frac{k}{12} = \frac{3}{k} \Rightarrow \frac{k-2}{k} \) i.e., \( k^2 = 36 \Rightarrow k = \pm 6 \)

Question. The graphical representation of the pair of equations \( x + 2y - 4 = 0 \) and \( 2x + 4y - 12 = 0 \) represents .................... .
Answer: Parallel lines
Explanation: \( x + 2y - 4 = 0 \) and \( 2x + 4y - 12 = 0 \)
Here \( \frac{a_1}{a_2} = \frac{1}{2}, \frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2} \) and \( \frac{c_1}{c_2} = \frac{-4}{-12} = \frac{1}{3} \)
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
Therefore, lines are parallel or non-intersecting

Question. If \( x + y = 2 \) and \( \frac{1}{x+y} = \frac{2}{5} \) then x = ................... .
Answer: \( \frac{9}{4} \)
Explanation: \( x - y = 2 \) and \( \frac{1}{x+y} = \frac{2}{5} \)
\( x - y = 2 \) ...(i)
\( x + y = \frac{5}{2} \) ...(ii)
Adding equations (i) and (ii), we get
\( (x - y) + (x + y) = 2 + \frac{5}{2} \)
\( 2x = \frac{9}{2} \Rightarrow x = \frac{9}{4} \)

Question. The value of p for the following pair of linear equations \( (p - 3)x + 3y = p; px + py = 12 \) have infinitely many solutions is .................... .
Answer: p = 6
Explanation: Given equations are \( (p - 3) x + 3y = p \) and \( px + py = 12 \)
For infinitely many solution
\( \frac{p-3}{p} = \frac{3}{p} = \frac{p}{12} \)
\( \Rightarrow \frac{p-3}{p} = \frac{3}{p} \) and \( \frac{3}{p} = \frac{p}{12} \)
\( \Rightarrow p - 3 = 3 \) and \( p^2 = 36 \)
\( \Rightarrow p = 6 \) and \( p = \pm 6 \)
Common value is 6.
So, \( p = 6 \)

Question. If \( x = a, y = b \) is the solution of the pair of equation \( x - y = 2 \) and \( x + y = 4 \) then the value of 3a + 4b is .................... .
Answer: 13
Explanation: \( x - y = 2 \) ...(i)
\( x + y = 4 \) ...(ii)
On adding (i) and (ii), we get
\( 2x = 6 \Rightarrow x = 3 \)
and putting x = 3 in (i),
\( 3 - y = 2 \Rightarrow y = 1 \)
\( 3a + 4b = 3x + 4y = 3 \times 3 + 4 \times 1 = 13 \)

Question. For the pair of equations \( \lambda x - 3y = -7 \) and \( 2x + 6y = 14 \) to have infinitely many solutions, the value of \( \lambda \) should be 1. Is this statement true? Give reasons.
Answer: No
Explanation: No, for no value of \( \lambda \) will the given pair of linear equations has infinitely many solutions.
The given pair of linear equations is \( \lambda x + 3y + 7 = 0 \) and \( 2x + 6y - 14 = 0 \).
Comparing with \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), we have
\( a_1 = \lambda, b_1 = 3, c_1 = 7 \)
\( a_2 = 2, b_2 = 6, c_2 = -14 \)
\( \frac{a_1}{a_2} = \frac{\lambda}{2} ; \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2} ; \frac{c_1}{c_2} = \frac{7}{-14} = -\frac{1}{2} \)
For infinitely many solutions \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
\( \Rightarrow \frac{\lambda}{2} = \frac{1}{2} \) and \( \frac{\lambda}{2} = -\frac{1}{2} \)
\( \Rightarrow \lambda = 1 \) and \( \lambda = -1 \)
Since, \( \lambda \) does not have a unique value so for no value of \( \lambda \) will the given pair of linear equations have infinitely many solutions.

Question. For all real values of c, the pair of equations \( x - 2y = 8 \) and \( 5x - 10y = c \) have a unique solution. Justify whether it is true or false.
Answer: False
Explanation: The given pair of equations will not have a unique solution for any value of c.
The given pair of linear equations is \( x - 2y - 8 = 0 \) and \( 5x - 10y - c = 0 \)
Comparing with \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), we have
\( a_1 = 1, b_1 = -2, c_1 = -8 \);
\( a_2 = 5, b_2 = -10, c_2 = -c \);
\( \frac{a_1}{a_2} = \frac{1}{5} ; \frac{b_1}{b_2} = \frac{-2}{-10} = \frac{1}{5} ; \frac{c_1}{c_2} = \frac{-8}{-c} = \frac{8}{c} \)
But for c = 40 (any real value), the ratio will be \( \frac{c_1}{c_2} = \frac{8}{40} = \frac{1}{5} \).
When c = 40, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{5} \). Thus, the given pair of linear equations will have infinitely many solutions for c = 40.
Also, when \( c \neq 40 \), \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \). Thus, the given pair of linear equations will have no solution for \( c \neq 40 \).
Hence, for any value of c, the system of linear equations does not have a unique solution.