CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Worksheet Set 11

Question. On comparing the ratios \( \frac{a_1}{a_2}, \frac{b_1}{b_2} \) and \( \frac{c_1}{c_2} \), find out whether the lines representing the pair of linear equations \( 6x - 3y + 10 = 0 \) and \( 2x - y + 9 = 0 \) intersects at a point are parallel or coincident.
Answer: We have, \( 6x - 3y + 10 = 0 \) ...(i)
\( 2x - y + 9 = 0 \) ...(ii)
Here, \( a_1 = 6, b_1 = -3, c_1 = 10 \)
\( a_2 = 2, b_2 = -1, c_2 = 9 \)
and \( \frac{a_1}{a_2} = \frac{6}{2} = 3, \frac{b_1}{b_2} = \frac{-3}{-1} = 3, \frac{c_1}{c_2} = \frac{10}{9} \)
Since, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
So, equations (i) and (ii) represent parallel lines.

Question. On comparing the ratios \( \frac{a_1}{a_2}, \frac{b_1}{b_2} \) and \( \frac{c_1}{c_2} \), find out whether the pair of linear equations \( \frac{4}{3}x + 2y = 8 \) and \( 2x + 3y = 12 \), is consistent or inconsistent.
Answer: We have, \( \frac{4}{3}x + 2y - 8 = 0 \) ...(i)
\( 2x + 3y - 12 = 0 \) ...(ii)
Here, \( a_1 = \frac{4}{3}, b_1 = 2, c_1 = -8 \)
and \( a_2 = 2, b_2 = 3, c_2 = -12 \)
Thus, \( \frac{a_1}{a_2} = \frac{4}{3 \times 2} = \frac{2}{3}, \frac{b_1}{b_2} = \frac{2}{3}, \frac{c_1}{c_2} = \frac{-8}{-12} = \frac{2}{3} \)
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), so equations (i) and (ii) represent coincident lines.
Hence, the pair of linear equations is consistent with infinitely many solutions.

Question. 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.
Answer: Let length and width of rectangular garden be \( x \) and \( y \) respectively.
\(\therefore\) From question \( x = y + 4 \)
\( \implies \) \( x - y = 4 \) ...(i)
and \( \frac{1}{2}\{2x + 2y\} = 36 \)
\( \implies \) \( x + y = 36 \) ...(ii)
Adding (i) and (ii), we get
\( x - y = 4 \)
\( x + y = 36 \)
\( 2x = 40 \)
\( \implies \) \( x = 20 \)
\(\therefore\) \( y = 20 - 4 = 16 \)
Hence, dimensions are 20 m, 16 m.

Question. A fraction becomes \( \frac{9}{11} \), if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes \( \frac{5}{6} \). Find the fraction.
Answer: Let required fraction be \( \frac{x}{y} \).
According to question
\( \frac{x+2}{y+2} = \frac{9}{11} \)
\( \implies \) \( 11x + 22 = 9y + 18 \)
\( \implies \) \( 11x - 9y = -4 \)
and \( \frac{x+3}{y+3} = \frac{5}{6} \)
\( \implies \) \( 6x + 18 = 5y + 15 \)
\( \implies \) \( 6x - 5y = -3 \)
Now, we have a pair of linear equations
\( 11x - 9y = -4 \) ...(i)
\( 6x - 5y = -3 \) ...(ii)
Multiplying (i) by 5 and (ii) by 9, we get
\( 55x - 45y = -20 \) ...(iii)
\( 54x - 45y = -27 \) ...(iv)
Subtracting (iv) from (iii), we get
\( 55x - 45y - (54x - 45y) = -20 - (-27) \)
\( \implies \) \( 55x - 45y - 54x + 45y = -20 + 27 \)
\( \implies \) \( x = 7 \)
Putting \( x = 7 \) in (i), we have
\( 11 \times 7 - 9y = -4 \)
\( \implies \) \( 9y = 77 + 4 \)
\( \implies \) \( 9y = 81 \)
\( \implies \) \( y = 9 \)
Hence, required fraction is \( \frac{7}{9} \).

Question. 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?
Answer: Let the present age of Nuri and Sonu be \( x \) and \( y \) years respectively.
According to question \( x - 5 = 3(y - 5) \)
\( \implies \) \( x - 5 = 3y - 15 \)
\( \implies \) \( x - 3y = -10 \) ...(i)
and \( x + 10 = 2(y + 10) \)
\( \implies \) \( x + 10 = 2y + 20 \)
\( \implies \) \( x - 2y = 10 \) ...(ii)
From equation (i), we get
\( x = 3y - 10 \) ...(iii)
Putting it in (ii), we have,
\( 3y - 10 - 2y = 10 \)
\( \implies \) \( y = 10 + 10 \)
\( \implies \) \( y = 20 \)
Now, from (iii)
\( x = 3 \times 20 - 10 \)
\( \implies \) \( x = 60 - 10 \)
\( \implies \) \( x = 50 \)
Therefore, present age of Nuri is 50 years and of Sonu is 20 years.

Question. 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.
Answer: Let the digit at tens and unit place of the two digit number be \( x \) and \( y \) respectively.
Therefore required two digit number \( = 10x + y \)
According to question
\( x + y = 9 \) ...(i)
and \( 9(10x + y) = 2(10y + x) \)
\( \implies \) \( 90x + 9y = 20y + 2x \)
\( \implies \) \( 90x + 9y - 20y - 2x = 0 \)
\( \implies \) \( 88x - 11y = 0 \)
\( \implies \) \( 8x - y = 0 \) ...(ii)
Adding equation (i) and (ii), we get
\( x + y + 8x - y = 9 + 0 \)
\( \implies \) \( 9x = 9 \)
\( \implies \) \( x = 1 \)
Putting \( x = 1 \) in (i), we get
\( 1 + y = 9 \)
\( \implies \) \( y = 8 \)
Hence, required number \( = 10 \times 1 + 8 = 18 \)

Question. 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 Rs. 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs. 1180 as hostel charges. Find the fixed charges and the cost of food per day.
Answer: Let the fixed charge be Rs. \( x \) and the cost of food per day be Rs. \( y \).
Therefore, according to question,
\( x + 20y = 1000 \) ...(i)
\( x + 26y = 1180 \) ...(ii)
Now, subtracting equation (ii) from (i), we have
\( x + 20y = 1000 \)
\( x + 26y = 1180 \)
\( - 6y = -180 \)
\( y = \frac{-180}{-6} = 30 \)
Putting the value of \( y \) in equation (i), we have
\( x + 20 \times 30 = 1000 \)
\( \implies \) \( x + 600 = 1000 \)
\( \implies \) \( x = 1000 - 600 = 400 \)
Hence, fixed charge is Rs. 400 and cost of food per day is Rs. 30.

Question. A fraction becomes \( \frac{1}{3} \) when 1 is subtracted from the numerator and it becomes \( \frac{1}{4} \) when 8 is added to its denominator. Find the fraction.
Answer: Let \( x \) be the numerator and \( y \) be the denominator of the fraction.
\(\therefore\) Fraction \( = \frac{x}{y} \)
Now, according to question
\( \frac{x - 1}{y} = \frac{1}{3} \)
\( \implies \) \( 3x - 3 = y \)
\( \implies \) \( 3x - y = 3 \) ...(i)
Also, \( \frac{x}{y + 8} = \frac{1}{4} \)
\( \implies \) \( 4x = y + 8 \)
\( \implies \) \( 4x - y = 8 \) ...(ii)
Subtracting (ii) from (i) we get
\( 3x - y = 3 \)
\( 4x - y = 8 \)
\( -x = -5 \)
\( \implies \) \( x = 5 \)
Putting \( x = 5 \) in equation (i), we get
\( 3 \times 5 - y = 3 \)
\( \implies \) \( 15 - 3 = y \)
\( \implies \) \( y = 12 \)
\(\therefore\) Fraction \( = \frac{x}{y} = \frac{5}{12} \)

Question. 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 deduced for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
Answer: Let \( x \) be the number of questions of right answer and \( y \) be the number of questions of wrong answer.
\(\therefore\) According to question,
\( 3x - y = 40 \) ...(i)
and \( 4x - 2y = 50 \)
or \( 2x - y = 25 \) ...(ii)
Subtracting (ii) from (i), we have
\( 3x - y = 40 \)
\( 2x - y = 25 \)
\( x = 15 \)
Putting the value of \( x \) in equation (i), we have
\( 3 \times 15 - y = 40 \)
\( \implies \) \( 45 - y = 40 \)
\(\therefore\) \( y = 45 - 40 = 5 \)
Hence, total number of questions is \( x + y \) i.e., \( 5 + 15 = 20 \).

Question. 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.
Answer: Let the length and breadth of a rectangle be \( x \) and \( y \) respectively.
Then area of the rectangle \( = xy \)
According to question, we have
\( (x - 5)(y + 3) = xy - 9 \)
\( \implies \) \( xy + 3x - 5y - 15 = xy - 9 \)
\( \implies \) \( 3x - 5y = 15 - 9 = 6 \)
\( \implies \) \( 3x - 5y = 6 \) ...(i)
Again, we have
\( (x + 3)(y + 2) = xy + 67 \)
\( \implies \) \( xy + 2x + 3y + 6 = xy + 67 \)
\( \implies \) \( 2x + 3y = 67 - 6 = 61 \)
\( \implies \) \( 2x + 3y = 61 \) ...(ii)
Now, from equation (i), we express the value of \( x \) in terms of \( y \) as
\( x = \frac{6 + 5y}{3} \)
Substituting the value of \( x \) in equation (ii), we have
\( 2 \times \left( \frac{6 + 5y}{3} \right) + 3y = 61 \)
\( \implies \) \( \frac{12 + 10y + 9y}{3} = 61 \)
\( \implies \) \( 19y = 183 - 12 = 171 \)
\( \implies \) \( y = \frac{171}{19} = 9 \)
Putting the value of \( y \) in equation (i), we have
\( 3x - 5 \times 9 = 6 \)
\( \implies \) \( 3x = 6 + 45 = 51 \)
\( \implies \) \( x = \frac{51}{3} = 17 \)
Hence, the length of rectangle \( = 17 \) units and breadth of rectangle \( = 9 \) units.

Question. 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 distance covered by the train.
Answer: Let actual speed of the train be \( x \) km/h and actual time taken be \( y \) hours.
Then, distance covered \( = \text{speed} \times \text{time} = xy \) km ...(i)
Case I: When speed is \( (x + 10) \) km/h, then time taken is \( (y - 2) \) hours
\(\therefore\) Distance covered \( = (x + 10)(y - 2) \)
\( \implies \) \( xy = (x + 10)(y - 2) \) [From (i)]
\( \implies \) \( xy = xy - 2x + 10y - 20 \)
\( \implies \) \( 2x - 10y = -20 \)
\( \implies \) \( x - 5y = -10 \) ...(ii)
Case II: When speed is \( (x - 10) \) km/h, then time taken is \( (y + 3) \) hours.
\(\therefore\) Distance covered \( = (x - 10)(y + 3) \) [From (i)]
\( \implies \) \( xy = (x - 10)(y + 3) \)
\( \implies \) \( xy = xy + 3x - 10y - 30 \)
\( \implies \) \( 3x - 10y = 30 \) ...(iii)
Multiplying equation (ii) by 2 and subtracting it from (iii), we get
\( 3x - 10y = 30 \)
\( 2x - 10y = -20 \)
\( x = 50 \)
Putting \( x = 50 \) in equation (ii), we get
\( 50 - 5y = -10 \)
\( \implies \) \( 50 + 10 = 5y \)
\( \implies \) \( y = 12 \)
\(\therefore\) Distance covered by the train \( = xy \) km \( = 50 \times 12 \) km \( = 600 \) km

Question. Solve the following linear equations: \( 152x - 378y = -74 \) and \( -378x + 152y = -604 \) 
Answer: We have, \( 152x - 378y = -74 \) ...(i)
\( -378x + 152y = -604 \) ...(ii)
Adding equation (i) and (ii), we get
\( 152x - 378y = -74 \)
\( -378x + 152y = -604 \)
\( -226x - 226y = -678 \)
\( \implies \) \( -226(x + y) = -678 \)
\( \implies \) \( x + y = \frac{-678}{-226} \)
\( \implies \) \( x + y = 3 \) ...(iii)
Subtracting equation (ii) from (i), we get
\( 152x - 378y = -74 \)
\( -378x + 152y = -604 \)
\( 530x - 530y = 530 \)
\( \implies \) \( x - y = 1 \) ...(iv)
Adding equations (iii) and (iv), we get
\( x + y = 3 \)
\( x - y = 1 \)
\( 2x = 4 \)
\( \implies \) \( x = 2 \)
Putting the value of \( x \) in (iii), we get
\( 2 + y = 3 \)
\( \implies \) \( y = 1 \)
Hence, the solution of given system of equations is \( x = 2, y = 1 \).

Question. Solve the following pair of linear equations by the elimination method and the substitution method: \( \frac{x}{2} + \frac{2y}{3} = -1 \) and \( x - \frac{y}{3} = 3 \)
Answer: We have, \( \frac{x}{2} + \frac{2y}{3} = -1 \)
\( \implies \) \( \frac{3x + 4y}{6} = -1 \)
\(\therefore\) \( 3x + 4y = -6 \) ...(i)
and \( x - \frac{y}{3} = 3 \)
\( \implies \) \( \frac{3x - y}{3} = 3 \)
\(\therefore\) \( 3x - y = 9 \) ...(ii)
By elimination method:
Subtracting (ii) from (i), we have
\( 5y = -15 \) or \( y = -\frac{15}{5} = -3 \)
Putting the value of \( y \) in equation (i), we have
\( 3x + 4 \times (-3) = -6 \)
\( \implies \) \( 3x - 12 = -6 \)
\(\therefore\) \( 3x = -6 + 12 \)
\( \implies \) \( 3x = 6 \)
\(\therefore\) \( x = \frac{6}{3} = 2 \)
Hence, solution is \( x = 2, y = -3 \).
By substitution method:
Expressing \( x \) in terms of \( y \) from equation (i), we have
\( x = \frac{-6 - 4y}{3} \)
Substituting the value of \( x \) in equation (ii), we have
\( 3 \times \left( \frac{-6 - 4y}{3} \right) - y = 9 \)
\( \implies \) \( -6 - 4y - y = 9 \)
\( \implies \) \( -6 - 5y = 9 \)
\(\therefore\) \( -5y = 9 + 6 = 15 \)
\( y = \frac{15}{-5} = -3 \)
Putting the value of \( y \) in equation (i), we have
\( 3x + 4 \times (-3) = -6 \)
\( \implies \) \( 3x - 12 = -6 \)
\(\therefore\) \( 3x = 12 - 6 = 6 \)
\( x = \frac{6}{3} = 2 \)
Hence, the required solution is \( x = 2, y = -3 \).

Multiple Choice Questions

Question. Choose and write the correct option in the following questions.
Graphically, the pair of equations \( 6x - 3y + 10 = 0; 2x - y + 9 = 0 \) 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. The value of \( k \) for which the system of equations \( 2x + ky = 12, x + 3y - 4 = 0 \) are inconsistent is
(a) \( \frac{21}{4} \)
(b) \( \frac{1}{6} \)
(c) 6
(d) \( \frac{4}{21} \)
Answer: (c) 6

Question. If the lines given by \( 3x + 2ky = 2 \) and \( 2x + 5y + 1 = 0 \) are parallel, then value of \( k \) is 
(a) \( \frac{-5}{4} \)
(b) \( \frac{2}{5} \)
(c) \( \frac{15}{4} \)
(d) \( \frac{3}{2} \)
Answer: (c) \( \frac{15}{4} \)

Question. If \( 2x - 3y = 7 \) and \( (a + b)x - (a + b - 3)y = 4a + b \) represent coincident lines, then \( a \) and \( b \) satisfy the equation
(a) \( a + 5b = 0 \)
(b) \( 5a + b = 0 \)
(c) \( a - 5b = 0 \)
(d) \( 5a - b = 0 \)
Answer: (c) \( a - 5b = 0 \)

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

Question. The pair of equations \( x + 2y + 5 = 0 \) and \( -3x - 6y + 1 = 0 \) have
(a) unique solution
(b) exactly two solutions
(c) infinitely many solutions
(d) no solution
Answer: (d) no solution

Question. Consider the equations shown: \( ax + by = ab \) and \( 2ax + 3by = 3b \). Which of these is the value of \( y \) in terms of \( a \)?
(a) \( y = 5 - 3a \)
(b) \( y = 3 - 2a \)
(c) \( y = 9a - 35 \)
(d) \( y = 2ab - 3b \)
Answer: (c) \( y = 9a - 35 \)

Question. The pair of linear equations \( y = 0 \) and \( y = -6 \) has 
(a) unique solution
(b) no solution
(c) infinitely many solutions
(d) only solution (0, 0)
Answer: (b) no solution

Question. The value of \( k \), for which the pair of linear equations \( kx + y = k^2 \) and \( x + ky = 1 \) have infinitely many solution is 
(a) \( \pm 1 \)
(b) 1
(c) -1
(d) 2
Answer: (b) 1

Question. Consider the equations shown \( 4x + 3y = 41; x + 3y = 26 \). Which of these is the correct way of solving the given pair of equations? 
(a) \( 4x + x + 3y - 3y = 41 - 26 \)
(b) \( 4x + 3y + 3y = 41 - 26 \)
(c) \( 4x - x + 3y - 3y = 41 - 26 \)
(d) \( 4(x + 3y) + 3y = 41 \)
Answer: (c) \( 4x - x + 3y - 3y = 41 - 26 \)

Question. Gunjan 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 are respectively 
(a) 25 and 25
(b) 15 and 35
(c) 35 and 15
(d) 35 and 20
Answer: (a) 25 and 25

Question. The sum of the digits of a two-digits number is 9. If 27 is added to it, the digits of number get reversed. The number is 
(a) 27
(b) 72
(c) 63
(d) 36
Answer: (d) 36

Question. Shipra gave a note of Rs. 2,000 for a pair of jeans worth Rs. 500. She was returned 11 notes in denominations of Rs. 200 and Rs. 100. Which pair of equations can be used to find the number of Rs. 200 notes as \( x \), and the number of Rs. 100 notes as \( y \)? How many notes of Rs. 200 did she get? 
(a) \( x + y = 11 \) and \( 200x + 100y = 1500; 4 \)
(b) \( x = y + 11 \) and \( 200x + 100y = 2000; 4 \)
(c) \( x + y = 15 \) and \( 200x + 100y = 1800; 10 \)
(d) \( x + y = 15 \) and \( 100x + 200y = 1800; 12 \)
Answer: (a) \( x + y = 11 \) and \( 200x + 100y = 1500; 4 \)

Question. The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son. The present ages (in years) of the son and the father are, respectively 
(a) 4 and 24
(b) 5 and 30
(c) 6 and 36
(d) 3 and 24
Answer: (c) 6 and 36

Question. The sum of the digits of a two-digit number is 9. If 27 is subtracted from the number, its digits are interchanged. Which of these is the product of the digits of the number? 
(a) 8
(b) 14
(c) 18
(d) 20
Answer: (c) 18

Question. What is the value of \( q \) if \( \frac{p}{2} + 3q = 6 \) and \( 2p - 2q = 10 \)? 
(a) 1
(b) 4
(c) 6
(d) 16
Answer: (a) 1

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

Question. Consider the equations as shown: \( (x - a)(y - b) = (x - 2a)(y - \frac{b}{2}) \) and \( x(x + \frac{1}{2b}) + y(y + \frac{a}{2b}) - 2xy = 5 + (x - y)^2 \). On comparing the coefficients, a student says these pairs of equations is consistent. Is he/she correct? Which of these explains why?
(a) Yes; because they are parallel lines.
(b) Yes; because they are intersecting lines.
(c) No; because they are parallel lines.
(d) No; because they are intersecting lines.
Answer: (b) Yes; because they are intersecting lines.

Question. The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes \( \frac{1}{2} \), then the fraction is
(a) \( \frac{4}{7} \)
(b) \( \frac{5}{7} \)
(c) \( \frac{6}{7} \)
(d) \( \frac{8}{7} \)
Answer: (b) \( \frac{5}{7} \)

Very Short Answer Questions

Question. Find the value of \( c \) for which the pair of equations \( cx - y = 2 \) and \( 6x - 2y = 3 \) will have infinitely many solutions. 
Answer: If the system of linear equations has infinitely many solutions then
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
\( \implies \) \( \frac{c}{6} = \frac{-1}{-2} = \frac{-2}{-3} \)
\( \implies \) \( \frac{c}{6} = \frac{1}{2} \neq \frac{2}{3} \)
Here \( \frac{1}{2} \neq \frac{2}{3} \) i.e; \( \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\(\therefore\) There is no value of \( c \) for which lines have infinitely many solutions.