CBSE Class 10 Maths HOTs Pair Of Linear Equations In Two Variables Set 06

Question. The pair of linear equations \( 2x + 3y = 5 \) and \( 4x + 6y = 10 \) is
(a) inconsistent
(b) consistent
(c) dependent and consistent
(d) None of the options
Answer: (c) dependent and consistent
Sol. (c) By Comparing \( \frac{a_1}{a_2}, \frac{b_1}{b_2} \) and \( \frac{c_1}{c_2} \), we can say that the given equations are dependent and Consistent.

Question. Match the Column:
(1) \( 2x + 5y = 10 \), \( 3x + 4y = 7 \) (A) Unique solution
(2) \( 2x + 5y = 10 \), \( 6x + 15y = 20 \) (B) Infinitely many solutions
(3) \( 5x + 2y = 10 \), \( 10x + 4y = 20 \) (C) No common solution
(a) 1 – A, 2 – B, 3 – C
(b) 1 – B, 2 – C, 3 – A
(c) 1 – C, 2 – B, 3 – A
(d) 1 – A, 2 – C, 3 – B
Answer: (d) 1 – A, 2 – C, 3 – B
Sol. (d) (1) \( \frac{2}{3} \neq \frac{5}{4} \) \( \therefore \) unique solution.
(2) \( \frac{2}{6} = \frac{5}{15} \neq \frac{10}{20} \) \( \therefore \) no solution.
(3) \( \frac{5}{10} = \frac{2}{4} = \frac{10}{20} \) \( \therefore \) dependent and consistent.

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
Sol. (d), For no solution \( \frac{1}{2} = \frac{1}{k} \neq \frac{-4}{-3} \)
\( \implies \) \( k = 2 \).

Question. The number of solutions of the following pair of linear equations: \( x + 2y - 8 = 0 \), \( 2x + 4y = 16 \) is
(a) no solution
(b) unique solution
(c) infinite many solutions
(d) None of the options
Answer: (c) infinite many solutions
Sol. (c) \( x + 2y - 8 = 0 \) ... (i)
\( 2x + 4y - 16 = 0 \) ... (ii)
Here, \( a_1 = 1, b_1 = 2, c_1 = -8 \)
and \( a_2 = 2, b_2 = 4, c_2 = -16 \)
Now, \( \frac{a_1}{a_2} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2} \), \( \frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2} \)

\( \implies \) \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
\( \therefore \) Given pair of linear equations has infinite many solutions.

Question. If \( am \neq bl \), then the pair of equations \( ax + by = c \), \( lx + my = n \)
(a) has a unique solution
(b) has no solution
(c) has infinitely many solution
(d) May or may not have a solution
Answer: (a) has a unique solution
Sol. (a) Give equations of lines are \( ax + by - c = 0 \) and \( lx + my - n = 0 \)
Since, \( am \neq bl \)
\( \implies \) \( \frac{a}{l} \neq \frac{b}{m} \)
\( \therefore \) The pair of equations has a unique solution.

Very Short Answer Type Questions

Question. Find the value of \( k \) for which the given pair of linear equations has no common solution. \( 4x + ky = 5 \), \( 8x + 12y = 10 \)
Answer: Given equations are \( 4x + ky = 5 \) ...(i) and \( 8x + 12y = 10 \) ...(ii)
\( a_1 = 4, b_1 = k, c_1 = 5 \)
\( a_2 = 8, b_2 = 12, c_2 = 10 \)
For no common solution \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)

\( \implies \) \( \frac{4}{8} = \frac{k}{12} \neq \frac{5}{10} \)

\( \implies \) \( \frac{1}{2} = \frac{k}{12} \) and \( \frac{k}{12} \neq \frac{1}{2} \)

\( \implies \) \( k = 6 \) and \( k \neq 6 \)
\( \therefore \) There is no value of \( k \) for which given pair of linear equations has no common solution.

Question. Find the value(s) of \( k \) so that the pair of equations \( x + 2y = 5 \) and \( 3x + ky + 15 = 0 \) has a unique solution. 
Answer: \( x + 2y - 5 = 0 \) ...(i)
\( 3x + ky + 15 = 0 \) ...(ii)
\( a_1 = 1, b_1 = 2, c_1 = -5 \)
\( a_2 = 3, b_2 = k, c_2 = 15 \)
For unique solution,

\( \implies \) \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)

\( \implies \) \( \frac{1}{3} \neq \frac{2}{k} \)

\( \implies \) \( k \neq 6 \)
So, given system of equations is consistent with unique solution for all values of \( k \) other than 6.

Question. Find \( c \) if the system of equations \( cx + 3y + (3 - c) = 0 \); \( 12x + cy - c = 0 \) has infinitely many solutions? 
Answer: The system of equations are \( cx + 3y + (3 - c) = 0 \) ...(i) and \( 12x + cy - c = 0 \) ...(ii)
For infinitely many solutions, we have \( \frac{c}{12} = \frac{3}{c} = \frac{3 - c}{-c} \)

\( \implies \) \( c^2 = 36 \) and \( -3c = 3c - c^2 \)

\( \implies \) \( c^2 = 6c \)
\( \implies \) \( c^2 - 6c = 0 \)

\( \implies \) \( c(c - 6) = 0 \)
\( \implies \) \( c = 0 \) or \( c = 6 \)
Hence, possible value is \( c = 6 \).

PRACTICE QUESTIONS

Question. The pair of equations \( x = 4 \) and \( y = 3 \) graphically represents lines which are
(a) parallel
(b) intersecting at (3, 4)
(c) coincident
(d) intersecting at (4, 3)
Answer: (d) intersecting at (4, 3)

Question. \( a_1x + b_1 y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), where \( a_1, b_1, c_1, a_2, b_2, c_2 \) are all real numbers and \( a_1^2 + b_1^2 \neq 0, a_2^2 + b_2^2 \neq 0 \), is called a
(a) family of two different straight lines
(b) family of two coincident lines
(c) pair of linear equations in two variables
(d) None of the options
Answer: (c) pair of linear equations in two variables

Question. If a pair of linear equations in two variables is consistent, then the lines represented by two equations are:
(a) intersecting
(b) parallel
(c) always coincident
(d) intersecting or coincident
Answer: (d) intersecting or coincident

Question. If \( ax + by = c \) and \( lx + my = n \) has unique solution then the relation between the coefficient will be:
(a) \( am \neq lb \)
(b) \( am = lb \)
(c) \( ab = lm \)
(d) \( ab \neq lm \)
Answer: (a) \( am \neq lb \)

Question. Match the Column:
(1) \( 2x + 3y = 40; 6x + 5y = 10 \) | (A) Coincident lines
(2) \( 2x + 3y = 40; 6x + 9y = 50 \) | (B) Intersecting lines
(3) \( 2x + 3y = 10; 4x + 6y = 20 \) | (C) Parallel lines
(a) 1 – A, 2 – B, 3 – C
(b) 1 – B, 2 – A, 3 – C
(c) 1 – B, 2 – C, 3 – A
(d) 1 – C, 2 – A, 3 – B
Answer: (c) 1 – B, 2 – C, 3 – A

Represent the situation as pair of linear equations in two variables.

Question. Half the perimeter of a rectangular garden, whose length is 12 m more than its width is 60 m. Find the dimensions of the garden.
Answer: Let the width of the garden be \( x \) m and the length be \( y \) m.
According to the question:
\( y = x + 12 \)
\( \implies y - x = 12 \)...(i)
Half of the perimeter = 60 m
\( \implies \frac{1}{2} [2(x + y)] = 60 \)
\( \implies x + y = 60 \)...(ii)
Adding (i) and (ii):
\( 2y = 72 \)
\( \implies y = 36 \)
Substituting \( y = 36 \) in (ii):
\( x + 36 = 60 \)
\( \implies x = 24 \)
Thus, the dimensions of the garden are: Length = 36 m, Width = 24 m.

Question. The larger of two supplementary angles exceeds the smaller by 20 degrees. Find the angles.
Answer: Let the two supplementary angles be \( x \) and \( y \), where \( x > y \).
Since they are supplementary:
\( x + y = 180 \)...(i)
According to the question:
\( x = y + 20 \)
\( \implies x - y = 20 \)...(ii)
Adding (i) and (ii):
\( 2x = 200 \)
\( \implies x = 100^\circ \)
Substituting \( x = 100^\circ \) in (i):
\( 100 + y = 180 \)
\( \implies y = 80^\circ \)
The angles are \( 100^\circ \) and \( 80^\circ \).

Question. Without drawing the graph, find out whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincident. \( 18x - 7y = 24; \frac{9}{5}x - \frac{7}{10}y = \frac{9}{10} \)
Answer: Given equations are:
\( 18x - 7y = 24 \)
\( \frac{9}{5}x - \frac{7}{10}y = \frac{9}{10} \)
Multiplying the second equation by 10:
\( 18x - 7y = 9 \)
Comparing ratios:
\( \frac{a_1}{a_2} = \frac{18}{18} = 1 \)
\( \frac{b_1}{b_2} = \frac{-7}{-7} = 1 \)
\( \frac{c_1}{c_2} = \frac{24}{9} = \frac{8}{3} \)
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the lines are parallel.

Question. Find whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincident: \( 2x - 3y + 6 = 0; 4x - 5y + 2 = 0 \)
Answer: Comparing ratios:
\( \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \)
\( \frac{b_1}{b_2} = \frac{-3}{-5} = \frac{3}{5} \)
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the lines intersect at a point.

Question. Determine by drawing the graph, whether the following system of linear equations has a unique solution or not \( 3x - 2y = 12; x - \frac{2}{3}y - 4 = 0 \)
Answer: The first equation is \( 3x - 2y = 12 \).
The second equation is \( x - \frac{2}{3}y - 4 = 0 \)
\( \implies 3x - 2y - 12 = 0 \)
\( \implies 3x - 2y = 12 \)
Both equations are identical. Graphically, they represent coincident lines. Thus, the system does not have a unique solution; it has infinitely many solutions.

Question. Given a linear equation \( 3x - 5y = 1 \) form another linear equation in these variables such that the geometric representation of pair so formed is: (i) intersecting lines (ii) coincident lines (iii) parallel lines
Answer:
(i) For intersecting lines (\( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)): \( 2x + 3y = 5 \)
(ii) For coincident lines (\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)): \( 6x - 10y = 2 \)
(iii) For parallel lines (\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)): \( 3x - 5y = 10 \)

Question. Solve the following system of equations graphically for \( x \) and \( y \): \( 3x + 2y = 12; 5x - 2y = 4 \). Find the co-ordinates of the points where the lines meet the y-axis.
Answer: By plotting the lines, we find the intersection point is \( (2, 3) \).
\( \implies x = 2, y = 3 \).
Line \( 3x + 2y = 12 \) meets y-axis at \( (0, 6) \).
Line \( 5x - 2y = 4 \) meets y-axis at \( (0, -2) \).

Question. Solve graphically: \( x + 4y = 10, y - 2 = 0 \)
Answer: From \( y - 2 = 0 \), we get \( y = 2 \).
Substituting \( y = 2 \) in \( x + 4y = 10 \):
\( x + 4(2) = 10 \)
\( \implies x + 8 = 10 \)
\( \implies x = 2 \)
Graphically, the lines intersect at \( (2, 2) \). Solution: \( x = 2, y = 2 \).

Question. Solve graphically: \( 2x - 3y = 6, x - 6 = 0 \)
Answer: From \( x - 6 = 0 \), we get \( x = 6 \).
Substituting \( x = 6 \) in \( 2x - 3y = 6 \):
\( 2(6) - 3y = 6 \)
\( \implies 12 - 3y = 6 \)
\( \implies -3y = -6 \)
\( \implies y = 2 \)
Graphically, the lines intersect at \( (6, 2) \). Solution: \( x = 6, y = 2 \).

Question. Solve the given system of equations graphically \( x - 5y = 6, 2x - 10y = 10 \).
Answer: For \( x - 5y = 6 \), we have \( \frac{a_1}{a_2} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{-5}{-10} = \frac{1}{2} \), and \( \frac{c_1}{c_2} = \frac{6}{10} = \frac{3}{5} \).
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the lines are parallel. Therefore, the system has no solution.

Question. The sum of the digits of a two digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. Find the number. (Graphically)
Answer: Let the digit at tens place be \( x \) and units place be \( y \).
Number = \( 10x + y \)
\( x + y = 15 \)...(i)
\( (10y + x) - (10x + y) = 9 \)
\( \implies 9y - 9x = 9 \)
\( \implies y - x = 1 \)...(ii)
From the graph of these lines, the intersection is \( (7, 8) \).
\( \implies x = 7, y = 8 \).
The number is 78.

Question. The sum of the digits of a two-digit number is 12. The number obtained by interchanging the digits exceeds the given number by 18. Find the number. (Graphically)
Answer: Let tens digit be \( x \) and units digit be \( y \).
\( x + y = 12 \)...(i)
\( (10y + x) - (10x + y) = 18 \)
\( \implies 9y - 9x = 18 \)
\( \implies y - x = 2 \)...(ii)
Graphically, the intersection point is \( (5, 7) \).
The number is 57.

Question. The sum of numerator and denominator of a fraction is 3 less than twice the denominator. If each of the numerator and denominator is decreased by 1, the fraction becomes \( \frac{1}{2} \). Find the fraction. (Graphically)
Answer: Let numerator be \( x \) and denominator be \( y \).
\( x + y = 2y - 3 \)
\( \implies x - y = -3 \)...(i)
\( \frac{x - 1}{y - 1} = \frac{1}{2} \)
\( \implies 2x - 2 = y - 1 \)
\( \implies 2x - y = 1 \)...(ii)
Graphically, the intersection of \( x - y = -3 \) and \( 2x - y = 1 \) is \( (4, 7) \).
The fraction is \( \frac{4}{7} \).

Question. Solve the following pair of linear equations graphically. \( x + 3y = 6; 2x - 3y = 12 \) Also find the area of the triangle formed by the lines representing the given equations with y-axis.
Answer: The solution (intersection point) is \( (6, 0) \).
The vertices of the triangle with the y-axis are \( (6, 0) \), \( (0, 2) \), and \( (0, -4) \).
Base of triangle (on y-axis) = \( |2 - (-4)| = 6 \) units.
Height (from y-axis to vertex \( (6,0) \)) = 6 units.
Area = \( \frac{1}{2} \times 6 \times 6 = 18 \) sq. units.

Question. Determine graphically the coordinates of the vertices of a triangle, the equations of whose sides are given by \( 2y - x = 8, 5y - x = 14 \) and \( y - 2x = 1 \).
Answer: By plotting the three lines:
1. Intersection of \( 2y - x = 8 \) and \( 5y - x = 14 \) is \( (4, 6) \). (Wait, checking intersection: \( 3y = 6 \implies y=2, x=-4 \)). Correct intersection: \( (-4, 2) \).
2. Intersection of \( 5y - x = 14 \) and \( y - 2x = 1 \): \( y = 2x + 1 \implies 5(2x + 1) - x = 14 \implies 9x = 9 \implies x = 1, y = 3 \). Point: \( (1, 3) \).
3. Intersection of \( 2y - x = 8 \) and \( y - 2x = 1 \): \( x = 2y - 8 \implies y - 2(2y - 8) = 1 \implies -3y = -15 \implies y = 5, x = 2 \). Point: \( (2, 5) \).
Vertices are \( (-4, 2), (1, 3), (2, 5) \).

Question. Solve the following system of linear equations graphically: \( 3x + y - 12 = 0; x - 3y + 6 = 0 \) Shade the region bounded by the lines and x-axis. Also, find the area of shaded region.
Answer: The intersection point is \( (3, 3) \).
Line \( 3x + y = 12 \) meets the x-axis at \( (4, 0) \).
Line \( x - 3y = -6 \) meets the x-axis at \( (-6, 0) \).
The vertices of the shaded triangle are \( (3, 3), (4, 0), \) and \( (-6, 0) \).
Base on x-axis = \( 4 - (-6) = 10 \) units.
Height = \( y \)-coordinate of intersection = 3 units.
Area = \( \frac{1}{2} \times 10 \times 3 = 15 \) sq. units.

Multiple Choice Questions 

Question. A pair of linear equations which has a unique solution \( x = 2, y = -3 \) is
(a) \( x + y = -1; 2x - 3y = -5 \)
(b) \( 2x + 5y = -11; 4x + 10y = -22 \)
(c) \( 2x - y = 1; 3x + 2y = 0 \)
(d) \( x - 4y - 14 = 0; 5x - y - 13 = 0 \)
Answer: (b) \( 2x + 5y = -11; 4x + 10y = -22 \)

Question. 5 chairs and 4 tables together cost Rs. 2800 while 4 chairs and 3 tables together cost Rs. 2170. Algebraic representation of the situation can be
(a) \( 5x - 4y = 2800, 4x - 3y = 2170 \)
(b) \( 5x + 4y = 2800, 4x + 3y = 2170 \)
(c) \( 5x + 3y = 2800, 4x + 3y = 2170 \)
(d) \( 5x - 3y = 2800, 4x - 3y = 2170 \)
Answer: (b) \( 5x + 4y = 2800, 4x + 3y = 2170 \)

Question. The sum of two numbers is 137 and their difference is 43. Algebraic representation of the situation can be
(a) \( x - y = 137, x + y = 180 \)
(b) \( 2(x + y) = 137, 2(x - y) = 43 \)
(c) \( x + y = 137, x - y = 43 \)
(d) \( x + y = 43, x - y = 137 \)
Answer: (c) \( x + y = 137, x - y = 43 \)

Very Short Answer Type Question 

Question. Solve for \( x \) and \( y \) using substitution method. \( x + 2y - 3 = 0; 3x - 2y + 7 = 0 \)
Answer: Given equations are
\( x + 2y - 3 = 0 \) ...(i)
\( 3x - 2y + 7 = 0 \) ...(ii)
From equation (i), we get
\( x + 2y - 3 = 0 \)

\( \implies x = 3 - 2y \) ...(iii)
Substituting \( x = 3 - 2y \) in equation (ii), we get
\( 3(3 - 2y) - 2y + 7 = 0 \)

\( \implies 9 - 6y - 2y + 7 = 0 \)

\( \implies -8y = -16 \)

\( \implies y = 2 \)
When \( y = 2 \), equation (iii) becomes
\( x = 3 - 2 \times 2 \)

\( \implies x = -1 \)
\( \therefore x = -1, y = 2 \)

Short Answer Type Questions 

Question. Two digit number is divisible by 9. Number when multiplied by sum of its digits is equal to 486. Find the number.
Answer: Let digit at unit’s place = \( x \) and digit at ten’s place = \( y \)
\( \therefore \) Number = \( 10y + x \)
Since the number is divisible by 9, the sum of digits should be divisible by 9

\( \implies x + y = 9q \), where \( q \) is any positive integer.
A.T.Q.
\( (10y + x)(x + y) = 486 \) ...(i)
Substitute the value of \( (x + y) \) in (i)

\( \implies (10y + x) \times 9q = 486 \)

\( \implies 10y + x = \frac{486}{9q} = \frac{54}{q} \)
Since \( 10y + x \) is a multiple of 9
\( \therefore q = 1, 2 \) or \( 3 \)
when \( q = 1, 10y + x = 54 \)
when \( q = 2, 10y + x = 27 \)
when \( q = 3, 10y + x = 18 \)
Only \( 10y + x = 54 \) satisfies the equation (i) where \( x+y = 5+4 = 9 \) and \( 54 \times 9 = 486 \).
\( \therefore \) Number = 54

Question. Solve for \( x \) and \( y \) using substitution method. \( \frac{ax}{b} - \frac{by}{a} = a + b; ax - by = 2ab \)
Answer: Given equations are
\( \frac{ax}{b} - \frac{by}{a} = a + b \) ...(i)
\( ax - by = 2ab \) ...(ii)
From equation (ii), we get
\( ax - by = 2ab \)

\( \implies ax = 2ab + by \)

\( \implies x = \frac{2ab + by}{a} \) ...(iii)
Substituting \( x = \frac{2ab + by}{a} \) in equation (i), we get
\( \frac{a}{b} \left( \frac{2ab + by}{a} \right) - \frac{by}{a} = a + b \)

\( \implies \frac{2ab + by}{b} - \frac{by}{a} = a + b \)

\( \implies \frac{2ab}{b} + \frac{by}{b} - \frac{by}{a} = a + b \)

\( \implies 2a + y - \frac{by}{a} = a + b \)

\( \implies \frac{ay - by}{a} = a + b - 2a \)

\( \implies \frac{(a - b)y}{a} = b - a \)

\( \implies y = (b - a) \times \frac{a}{a - b} \)

\( \implies y = -a \)
When \( y = -a \), equation (iii) becomes
\( x = \frac{2ab + b(-a)}{a} \)

\( \implies x = \frac{ab}{a} \)

\( \implies x = b \)
\( \therefore x = b, y = -a \) is the solution of given pair of linear equations.

Long Answer Type Question 

Question. Two vessels A and B contain mixtures of Boric acid and water. A mixture of 3 parts from A and 2 parts from B is found to contain 29% of boric acid and a mixture of 1 part from A and 9 parts from B is found to contain 34% of boric acid. Find the percentage of boric acid in A and B.
Answer: Let percentage of boric acid in A = \( x\% \) and percentage of boric acid in B = \( y\% \)
A.T.Q. \( \frac{x}{100} \times 3 + \frac{y}{100} \times 2 = \frac{29}{100} \times 5 \)

\( \implies 3x + 2y = 145 \) ...(i)
and \( \frac{x}{100} \times 1 + \frac{y}{100} \times 9 = \frac{34}{100} \times 10 \)

\( \implies x + 9y = 340 \)

\( \implies x = 340 - 9y \) ...(ii)
Substituting the value of \( x \) in eq. (i), we get
\( 3(340 - 9y) + 2y = 145 \)

\( \implies 1020 - 27y + 2y = 145 \)

\( \implies -25y = -875 \)

\( \implies y = 35 \)
When \( y = 35 \), eq. (ii) becomes
\( x = 340 - 9 \times 35 = 340 - 315 = 25 \)
\( \therefore \) A contains 25% of boric acid and B contains 35% of boric acid.