CBSE Class 10 Mathematics Pairs of Linear Equations in Two Variables MCQs Set 14

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:
\( \frac{a_1}{a_2} = \frac{3/2}{9} = \frac{1}{6} \)
\( \frac{b_1}{b_2} = \frac{5/3}{10} = \frac{1}{6} \)
\( \frac{c_1}{c_2} = \frac{7}{14} = \frac{1}{2} \)
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), 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
Explanation: The given equations are:
\( 6x - 3y + 10 = 0 \) ...(i)
Also, \( 2x - y + 9 = 0 \) ...(ii)
The coefficients are \( \frac{a_1}{a_2} = \frac{6}{2} = 3 \), \( \frac{b_1}{b_2} = \frac{-3}{-1} = 3 \), \( \frac{c_1}{c_2} = \frac{10}{9} \).
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the lines are 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, \( a_1 = 1, b_1 = 2, c_1 = -3 \) and \( a_2 = 5, b_2 = k, c_2 = 7 \).
\( \frac{1}{5} = \frac{2}{k} \neq \frac{-3}{7} \)
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 \( 2x + ky - 3 = 0 \) has no solution, when:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \therefore \frac{1}{2} = \frac{1}{k} \neq \frac{-4}{-3} \)
\( \implies \) \( k = 2 \)

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. Since parallel lines never intersect, 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 \( (a, b) \)
Explanation: We know that \( x = a \) is the equation of a straight line parallel to the y-axis at a distance of \( a \) from it. Again, \( y = b \) is the equation of a straight line parallel to the x-axis at a distance of \( b \) from it. So, the pair of equations \( x = a \) and \( y = b \) graphically represents lines which are 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

Question. If the lines given by \( 3x + 2ky = 2 \) and \( 2x + 5y + 1 = 0 \) are parallel, then the value of \( k \) is:
(a) \( -\frac{4}{5} \)
(b) \( \frac{5}{2} \)
(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 \( a_1 = 3; b_1 = 2k; c_1 = -2 \) and \( a_2 = 2; b_2 = 5; c_2 = 1 \).
The condition for parallel lines is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \implies \frac{3}{2} = \frac{2k}{5} \)

\( \implies 15 = 4k \)

\( \implies k = \frac{15}{4} \)

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
Explanation: Since the lines represented by the given equations are parallel to each other, the pair of equations has 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 equations, one equation is just a multiple of another equation. Thus, the second equation is \( k(-5x + 7y - 2) = 0 \).
Putting \( k = -2 \), we get

\( \implies 10x - 14y + 4 = 0 \)
On moving it to the other side, we get

\( \implies 10x - 14y = -4 \)

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. For option (D):
Putting \( x = 2, y = -3 \) in \( x - 4y - 14 = 0 \), we get \( 2 - 4(-3) - 14 = 2 + 12 - 14 = 0 = RHS \).
Putting \( x = 2, y = -3 \) in \( 5x - y - 13 = 0 \), we get \( 5 \times 2 - (-3) - 13 = 10 + 3 - 13 = 0 = RHS \).

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: Putting the values in the equations, we have:
\( a - b = 2 \) ...(i)
\( a + b = 4 \) ...(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 \( b = 1 \).

Question. Aruna has only Rs. 1 and Rs. 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is Rs. 75, then the number of Rs. 1 and Rs. 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 Rs. 1 coins = \( x \) and number of Rs. 2 coins = \( y \).
\( x + y = 50 \) ...(i)
\( x + 2y = 75 \) ...(ii)
On subtracting eq. (i) from eq. (ii), we get

\( \implies y = 25 \)
Putting \( y = 25 \) in eq. (i), we get \( x + 25 = 50 \)

\( \implies x = 25 \)

Fill in the Blanks/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 \( \frac{3}{1} \neq \frac{2}{-k} \)

\( \implies k \neq -2/3 \)

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: \( 2(3) - 3a = 5 \)

\( \implies 6 - 3a = 5 \)

\( \implies 3a = 1 \)

\( \implies a = \frac{1}{3} \)

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

\( \implies y = -8 \). 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 \( \frac{k}{12} = \frac{3}{k} \neq \frac{k-2}{k} \).
\( \frac{k}{12} = \frac{3}{k} \)

\( \implies k^2 = 36 \)

\( \implies k = \pm 6 \)

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

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

\( \implies x = \frac{9}{4} \)

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: For infinitely many solution: \( \frac{p-3}{p} = \frac{3}{p} = \frac{p}{12} \).
From \( \frac{p-3}{p} = \frac{3}{p} \), we get \( p - 3 = 3 \)

\( \implies p = 6 \).
From \( \frac{3}{p} = \frac{p}{12} \), we get \( p^2 = 36 \)

\( \implies p = \pm 6 \).
Common value is \( p = 6 \).

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: \( a - b = 2 \) and \( a + b = 4 \). Adding them gives \( 2a = 6 \)

\( \implies a = 3 \).
Putting \( a = 3 \) in \( a + b = 4 \), we get \( b = 1 \).
\( 3a + 4b = 3(3) + 4(1) = 9 + 4 = 13 \).

Question. For the pair of equations \( \lambda x + 3y = -7, 2x + 6y = 14 \) to have infinitely many solutions, the value of \( \lambda \) should be 1. Is this statement true? Give reasons.
Answer: No, the statement is false.
Explanation: For infinitely many solutions, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
Here, \( \frac{\lambda}{2} = \frac{3}{6} = \frac{7}{-14} \).
This gives \( \frac{\lambda}{2} = \frac{1}{2} \)

\( \implies \lambda = 1 \).
However, \( \frac{b_1}{b_2} = \frac{1}{2} \) and \( \frac{c_1}{c_2} = -\frac{1}{2} \).
Since \( \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), there is no value of \( \lambda \) for which the system has infinitely many solutions.

Question. For all real values of \( c \), the pair of equations \( x - 2y = 8, 5x - 10y = c \) have a unique solution. Justify whether it is true or false.
Answer: False
Explanation: For unique solution, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).
Here \( \frac{a_1}{a_2} = \frac{1}{5} \) and \( \frac{b_1}{b_2} = \frac{-2}{-10} = \frac{1}{5} \).
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \), the system will either have no solution or infinitely many solutions, but never a unique solution.