Ans: (c)
The given pair of linear equations is 2x = 5y+ 6 and 15y = 6x  18
i.e ., 2x  5y  6 = 0 and 6x 15y 18 = 0
As,
i.e.. 1 /3 = 1 / 3 = 1/3
Therefore, the lines are coincident.
Q2: If the pair of linear equations x  y = 1, x + ky = 5 has a unique solution x = 2, y = 1. then the value of k [Year 2023, 1 Mark]
(a) 2
(b) 3
(c) 3
(d) 4
Ans: (c)
x + ky = 5
At x = 2, y = 1
2 + k.1 = 5
∴ k = 3
Q3: The pair of linear equations x + 2y + 5 = 0 and 3x  6y + 1 = 0 has [Year 2023, 1 Mark]
(a) A unique solution
(b) Exactly two solutions
(c) Infinitely many solutions
(d) No solution
Ans: (d)
Q4: Solve the pair of equations x = 5 and y = 7 graphically. [Year 2023, 2 Marks]
► View AnswerAns: Given equations are
x = 5 (i)
y = 7 (ii)
Draw the line x = 5 parallel to the yaxis and y= 7 parallel to the xaxis.
∴ The graph of equation (i) and (ii) is as follows
The lines x = 5 and y = 7 intersect each other at (5, 7).
Q5: Using the graphical method, find whether pair of equations x = 0 and y = 3 is consistent or not. [Year 2023, 2 Marks]
► View AnswerAns: Given pair of equations are
x = 0 (i)
and y = 3 (ii)
x = 0 means yaxis and draw a line y = 3 parallel to xaxis. The graph of given equations (i) and (ii) is
The lines intersect each other at (0, 3). Therefore, the given pair of equations Is consistent.
Q6: Half of the difference between two numbers is 2. The sum of the greater number and twice the smaller number is 13. Find the numbers. [Year 2023, 3 Marks]
► View AnswerAns: Let x and y be two numbers such that x> y
According to the question,
and x + 2y = 13  (ii)
Subtracting (i) from (ii), we get
3y = 9
⇒ y = 3
Substitute y = 3 in (i) we get
x  3 = 4
⇒ x = 7
Q7: (A) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1 It becomes 1/2 if we only add 1 to the denominator. What is the fraction? [Year 2023, 3 Marks]
OR
(B) For which value of 'k' will the following pair of linear equations have no solution? [Year 2023, 3 Marks]
3x + y = 1
(2k  1 )x + (k  1 )y = 2k + 1
Ans: (A) Let required fraction be x/y
According to question,
⇒ x + 1 = y  1
⇒ x = y2 ...(i)
Also,
⇒ 2x = y + 1 ...(ii)
From equations (i) and (ii), we get
2y — 4 = y + 1
y = 5
∴ x = 3
Required fraction x/y is 3/5
OR
(B) 3x + y = 1
(2k  1 )x + (k  1 )y = 2k + 1
For no solution;
2k  1 = 3k  3
⇒ k = 2
Also,
2k + 1 ≠ k  1
⇒ k ≠ 2
Q8: Two schools 'P' and 'Q' decided to award prizes to their students for two games of Hockey Rs. x per student and Cricket Rs. y per student. School 'P' decided to award a total of Rs. 9,500 for the two games to 5 and 4 students respectively, while school 'Q' decided to award Rs. 7,370 for the two games to 4 and 3 students respectively.
Based on the given information, answer the following questions.
(i) Represent the following information algebraically (in terms of x and y).
(ii) (a) What is the prize amount for hockey?
(b) Prize amount on which game is more and by how much?
(iii) What will be the total prize amount if there are 2 students each from two games? [Year 2023, 4,5,6 Marks]
Ans: (i) For Hockey, the amount given to per student = x
For cricket, the amount given to per student = y
From the question,
5x + 4y =9500 (i)
4x + 3y = 7370 (ii)
(ii) (a) Multiply (1) by 3 and (2) by 4 and then subtracting, we get
15x + 12y (16x + 12y) = 28500  29480
⇒  x =  980
⇒ x = 980
The prize amount given for hockey is Rs. 980 per student
(b) Multiply (1) by 4 and (2) by 5 and then subtracting, we get
20x + 16y 20x  15y = 38000  36850
⇒ y = 1150
The prize amount given for cricket is more than hockey by (1150  980) = 170.
(iii) Total prize amount = 2 x 980 + 2 x 1150
= Rs. (1960 + 2300) = Rs. 4260
Q1: The pair of lines represented by the linear equations 3x + 2y = 7 and 4x + 8y 11 = 0 are
(a) perpendicular
(b) parallel
(c) intersecting
(d) coincident [Year 2022, 1 Mark]
Ans: (c)
Clearly, from the graph, we can see that both lines intersect each other.
Q2: The pair of equations y = 2 and y =  3 has
(a) one solution
(b) two solutions
(c) infinitely many solutions
(d) no solution [Year 2022, 1 Mark]
Ans: (d)
Given equations are, y = 2 and y =  3.
Clearly, from the graph, we can see that both equations are parallel to each other.
So, there will be no solution.
Q3: A father is three times as old as his son. In 12 years time, he will be twice as old as his son. The sum of the present ages of the father and the son is
(a) 36 years
(b) 48 years
(c) 60 years
(d) 42 years [Year 2022, 1 Mark]
Ans: (b)
Let age of father be 'x' years and age of son be 'y' years.
According to the question, x = 3y ..(i)
and x + 12 = 2 (y + 12) ⇒ x  2y = 12 ..(ii)
From (i) and (ii), we get x = 36, y = 12
∴ x + y = 48 years
Q4: If 17x  19y = 53 and 19x  17y = 55, then the value of (x + y) is
(a) 1
(b) 1
(c) 3
(d) 3 [Year 2022, 1 Mark]
Ans: (a)
Given,17x  19y = 53 ...(i)
and 19x  17y = 55 _(ii)
Multiplying (i) by 19 and (ii) by 17, and by subtracting we get,
323x  361y (323x  289y) = 1007  935
⇒  72 y = 72
⇒ y =  1
Putting y =  1 in (i), we get,
17x  19 (1) = 53
⇒ 17x = 53  19
⇒ = 17x = 34
x = 2
∴ x + y = 2  1
= 1
Ans: (b)
For no solution.
Q2: Perimeter of a rectangle whose length (l) is 4 cm more than twice its breadth (b) is 14 cm. The pair of linear equations representing the above information is [Year 2021, 1 Mark]
(a)
(b)
(c)
(d)
Ans: (d)
Perimeter, 2(l + b) = 14 ...(i)
l = 2b + 4 ...(ii)
Q3: The solution of the pair of linear equations x = 5 and y = 6 is [Year 2021, 1 Mark]
(a) (5, 6)
(b) (5, 0)
(c) (0, 6)
(d) (0, 0)
Ans: (a)
(5, 6) is the solution of x = 5 and y = 6.
Q4: The value of k for which the pair of linear equations 3x + 5y = 8 and kx + 15y = 24 has infinitely many solutions, is [Year 2021, 1 Mark]
(a) 3
(b) 9
(c) 5
(d) 15
Ans: (b)
For. infinitely many solutions
Q5: The values of x and y satisfying the two equations 32x + 33y = 34, 33x + 32y = 31 respectively are: [Year 2021, 1 Mark]
(a) 1, 2
(b) 1, 4
(c) 1, 2
(d) 1, 4
Ans: (a)
32x + 33y = 34 ...(i)
33x + 32y = 31 ...(ii)
Adding equation (i) and (ii) and subtracting equation (ii) from (i),
we get 65x + 65y = 65 or x + y = 1 ...(iii)
and  x + y = 3 ...(iv)
Adding equation (iii) and (iv),
we get y = 2
Substituting the value of y in equation (iii),
x = 1
Q6: Two lines are given to be parallel. The equation of one of the lines is 3x  2y = 5. The equation of the second line can be [Year 2021, 1 Mark]
(a) 9x + 8 y = 7
(b)  12 x  8 y = 7
(c)  12 x + 8y = 7
(d) 12x + 8y = 7
Ans: (c)
If two lines a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 are parallel, then
It can only possible between 3x  2y = 5 and 12x + 8y = 7.
Q7: The sum of the numerator and the denominator of a fraction is 18. If the denominator is increased by 2, the fraction reduces to 1/3. Find the fraction. [Year 2021, 2 Marks]
► View AnswerAns: 5/13
Let the numerator be x and the denominator be y of the fractions. Then, the fraction = x /y.
Given , x + y = 13  (i)
and
⇒ 3x  y = 2 . . (ii)
Adding (i) and (ii), we get
4x = 20 ⇒ x = 5
Put the value of x in (i), we get
5+ y= 18
⇒ y = 13
∴ The required fraction is 5/13
Q8: Find the value of K for which the system of equations x + 2y = 5 and 3x + ky + 15 = 0 has no solution. [Year 2021, 2 Marks]
► View AnswerAns: Given, the system of equations
x + 2y = 5
3k + ky =  15 has no solution.
∴
For K = 6 the given system of equations has no solution.
Q9: Case studybased question is compulsory.
Attempt any 4 subpart from the question.
Each subpart carries 1 mark.
The residents of a housing society, on the occasion of Environment Day, decided to build two straight paths in the central park of the society and also plant trees along the boundary lines of each path.
Taking one corner of the park as the origin and the two mutually perpendicular lines as the xaxis and yaxis, the paths were represented by the two linear equations 2x  3y = 5 and 6x + 9y = 7.
Based on the above, answer the following questions:
(i) Two paths represented by the two equations here are
(a) intersecting
(b) overlapping
(c) parallel
(d) mutually perpendicular [Year 2021, 1 Mark]
Ans: (c)
Given equations are
2x  3y = 5 and 6x + 9y = 7
Hence,
Now,
As
∴ Equations are inconsistent and lines are parallel.
(ii) Which one of the following points lies on the line 2x  3y = 5?
(a) ( 4, 1)
(b) ( 4, 1)
(c) (4, 1 )
(d) ( 4, 1) [Year 2021, 1 Mark]
Ans: (c)
If any point lies on the line ax + by + c, then it must satisfy the equation of the line.
Here, (4, 1) satisfies the equation of line 2x  3y = 5.
As 2 x 4  3 x 1 = 5
⇒ 8  3 = 5
⇒5 = 5
(iii) If the line  6 x + 9y = 7 intersects the yaxis at a point, then its coordinates are
(a) (0, 7/9)
(b) (7/9, 0)
(c) (7/6, 0)
(d) (0, 7/6) [Year 2021, 1 Mark]
Ans: (a)
If a line ax + by = c intersects the yaxis at a particular point, then x = 0 at that point
In this case, if x =0. then we have
∴ The required coordinates are (0, 7/9)
(iv) If a pair of equations a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 has a unique solution, then
(a)
(b)
(c)
(d) [Year 2021, 1 Mark]
Ans: (b)
If a pair of equations a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 has a unique solution, then
(v) If , then the two lines a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 are
(a) parallel
(b) coincident
(c) intersecting
(d) perpendicular to each other [Year 2021, 1 Mark]
Ans: (b)
If , then the two lines a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 are coincident and the system of equations has infinite solutions.
Q10: Case studybased question is compulsory.
A book store shopkeeper gives books on rent for reading. He has variety of books in his store related to fiction, stories and quizzes etc. He takes a fixed charge for the first two days and an additional charge for subsequent day Amruta paid ₹22 for a book and kept for 6 days: while Radhika paid ₹16 for keeping the book for 4 days.
Assume that the fixed charge be ₹x and additional charge (per day) be ₹y.
Based on the above information, answer any four of the following questions.
(i) The situation of amount paid by Radhika. is algebraically represented by [Year 2021, 1 Mark]
(a) x  4 y = 16
(b) x + 4 y = 16
(c) x  2 y = 16
(d) x + 2 y = 16
Ans: (d)
For Amruta, x + (6  2)y = 22
i. e., x + 4y = 22 ...(i)
For Radhika, x + (4  2)y = 16 i.e.,x + 2y = 16 ...(ii)
Solving equation (i) and (ii). we get
x = 10 and y = 3
i.e., Fixed charges (x) = 710 ...(iii)
and additional charges per subsequent day
(y) = ₹ 3 ...(iv)
x + 2 y = 16 [From equation (ii)]
(ii) The situation of amount paid by Amruta. is algebraically represented by [Year 2021, 1 Mark]
(a) x  2y = 11
(b) x  2y = 22
(c) x + 4 y = 22
(d) x  4 y = 11
Ans: (c)
For Amruta, x + (6  2)y = 22
i. e., x + 4y = 22 ...(i)
For Radhika, x + (4  2)y = 16 i.e.,x + 2y = 16 ...(ii)
Solving equation (i) and (ii). we get
x = 10 and y = 3
i.e., Fixed charges (x) = 710 ...(iii)
and additional charges per subsequent day
(y) = ₹ 3 ...(iv)
x + 4 y = 22 [From equation (i)]
(iii) What are the fixed charges for a book? [Year 2021, 1 Mark]
(a) ₹ 9
(b) ₹ 10
(c) ₹ 13
(d) ₹ 15
Ans: (b)
For Amruta, x + (6  2)y = 22
i. e., x + 4y = 22 ...(i)
For Radhika, x + (4  2)y = 16 i.e.,x + 2y = 16 ...(ii)
Solving equation (i) and (ii). we get
x = 10 and y = 3
i.e., Fixed charges (x) = 710 ...(iii)
and additional charges per subsequent day
(y) = ₹ 3 ...(iv)
x = ₹ 10 [From equation (iii)]
(iv) What are the additional charges for each subsequent day for a book? [Year 2021, 1 Mark]
(a) ₹ 6
(b) ₹ 5
(c) ₹ 4
(d) ₹ 3
Ans: (d)
For Amruta, x + (6  2)y = 22
i. e., x + 4y = 22 ...(i)
For Radhika, x + (4  2)y = 16 i.e.,x + 2y = 16 ...(ii)
Solving equation (i) and (ii). we get
x = 10 and y = 3
i.e., Fixed charges (x) = 710 ...(iii)
and additional charges per subsequent day
(y) = ₹ 3 ...(iv)
y = ₹ 3 [From equation (iv)]
(v) What is the total amount paid by both, if both of them have kept the book for 2 more days? [Year 2021, 1 Mark]
(a) ₹ 35
(b) ₹ 52
(c) ₹ 50
(d) ₹ 58
Ans: (c)
For Amruta, x + (6  2)y = 22
i. e., x + 4y = 22 ...(i)
For Radhika, x + (4  2)y = 16 i.e.,x + 2y = 16 ...(ii)
Solving equation (i) and (ii). we get
x = 10 and y = 3
i.e., Fixed charges (x) = 710 ...(iii)
and additional charges per subsequent day
(y) = ₹ 3 ...(iv)
Total amount paid for 2 more days by both
= (x + 4 y) + 2 y + (x + 2y ) + 2 y
= 2 x + 10y
= 2 x 10 + 10 x 3
= ₹ 50
Ans: (b)
The given pair of linear equations is
So, the given pair of equations is inconsistent.
Q2: The pair of equations x = a and y = b graphically represent lines which are
(a) Intersecting at (a, b)
(b) Intersecting at (b, a)
(c) Coincident
(d) Parallel [Year 2020, 1 Mark]
Ans: (a)
The pair of equations x = a and y = b graphically represent lines which are parallel to the yaxis and xaxis respectively.
The lines will intersect each other at (a, b).
Q3: If the equations kx  2y = 3 and 3x + y = 5 represent two intersecting lines at unique point then the value of k is _________.
[Year 2020, 1 Mark]
Ans: For any real number except k = 6
kx  2y = 3 and 3x + y = 5 represent lines intersecting at a unique point.
For any real number except fc = 6
The given equation represent two intersecting lines at unique point.
Q4: The value of k for which the system of equations x + y  4 = 0 and 2x + ky = 3, has no solution. is [Year 2020, 1 Mark]
(a) 2
(b) ≠2
(c) 3
(d) 2
Ans: (d)
For no solution;
Hence, option (d) is correct
Q5: 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. [Year 2020, 3 Marks]
► View AnswerAns: Solutions of linear equations
2y  x = 8 ..(i)
5y  x = 14 ...(ii)
and y  2x = 1 ...(iii)
are given below:
From the graph of lines represented by given equations, we observe that
Lines (i) and (iii) intersect each other at C(2, 5),
Lines (ii) and {iii) intersect each other at B(1, 3) and Lines (i) and (ii) intersect each other at 4(4, 2).
Coordinates of the vertices of the triangle are A(4, 2), B(1, 3) and C(2, 5).
Q6: Solve the equations x + 2y = 6 and 2x  5y = 12 graphically.
► View AnswerAns: Solution of linear equations
x + 2y = 6 and 2x  5y = 12
are given below
From the graph, the two lines intersect each other at point (6, 0)
∴ x = 6 and y = 0
Q7: A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction. [Year 2020, 3 Marks]
► View AnswerAns: Let the required fraction be x/y.
According to question, we have
From (ii), 4x = y +8
so, 4x  y  8 = 0 ... (iv)
Subtracting (iii) from (iv),
we get x = 5
Substituting the value of x in (iii),
we get y = 12
Thus, the required fraction is 5/12
Q8: The present age of a father is three years more than three times the age of his son. Three years hence the father's age will be 10 years more than twice the age of the son. Determine their present ages. [Year 2020, 3 Marks]
► View AnswerAns: Let the present age of son be x years and that of father be y years.
According to question, we have
y = 3x+ 3 = ⇒ 3x  y+ 3 = 0 (i)
And y + 3 = 2(x + 3) + 10
⇒ y + 3 = 2x + 6 +10
⇒ 2x  y + 13 = 0 (ii)
Subtracting (ii) from (i), we get x = 10
Substituting the value of x in (ii). we get y = 33
So. the present age of the son is 10 years and that of the father is 33 years.
Q9: Solve graphically : 2x + 3y = 2, x  2y = 8
► View AnswerAns: Given lines are 2x + 3y = 2 and x  2y = 8 2x + 3y = 2
and x  2y = 8
∴ We will plot the points (1, 0), (2, 2) and (4,  2 ) and join them to get the graph o f 2x + 3y = 2 and we will plot the points (0, 4), (8, 0) and (2, 3) and join them to get the graph of x  2y = 8
Ans: Solutions of linear equation
x  y + 1 = 0 ...(i)
and 3x + 2y  12 = 0 ...(ii)
are given below:
From the graph, the two lines intersect each other at the point (2, 3)
∴ x = 2, y = 3.
Q2: The larger of two supplementary angles exceeds the smaller by 18°. Find the angles. [Year 2019, 2 Marks]
► View AnswerAns: Let the larger angle be x° and the smaller angle be y°. We know that the sum of two supplementary pairs of angles is always 180°.
We have x° + y° = 180° (i)
and x°  y° = 18° (ii) [Given]
By (1), we have x° = 180°  y° _(iii)
Put the value of x° in (ii), we get
180°  y°  y° = 18°
⇒ 162° = 2y°
⇒ y = 81
From (3), we have x° = 180°  81° = 99°
The angles are 99° and 81°
Q3: Solve the following pair of linear equations: 3x  5y =4, 2y+ 7 = 9x. [Year 2019, 2 Marks]
► View AnswerAns: Given, pair of linear equations:
3x  5y =4, (i)
2y+ 7 = 9x
9x  2y = 7 (ii)
Multiply (i) by 3 and subtract from (ii), as
Hence, x = 9/13 and y = 5/13
Q4: A father's age is three times the sum of the ages of his two children. After 5 years his age will be two times the sum of their ages. Find the present age of the father. [Year 2019, 3 Marks]
► View AnswerAns: Let the ages of two children be x and y respectively.
Father's present age = 3(x +y)
After 5 years, sum of ages of children = x + 5 + y + 5
= x + y + 10
and age of father = 3(x + y) + 5
According to the question,
3(x + y) + 5 = 2(x + y+ 10)
3x + 3y + 5 = 2x + 2y + 20
⇒ x + y = 15
Hence, present age of father = 3(x + y)
= 3 x 15 = 45 years
Q5: A fraction becomes 1/3 when 2 is subtracted from the numerator and ii becomes 1/2 when 1 is subtracted from the denominator. Find the fraction. [Year 2019, 3 Marks]
► View AnswerAns: Let the fraction be x/y
Then, according to question.
Subtracting (ii) from (i), we get x  7 = 0
So, x = 7
From (i) ,3[7]  y  6 = 0
⇒ 21  6 = y
⇒ y = 15
Therefore required fraction is 7/15
Q6: Find the value(s) of k so that the pair of equations x + 2y = 5 and 3x + ky + 15 = 0 has a unique solution. [Year 2019, 2 Marks]
► View AnswerAns: The given pair of linear equations is
x + 2y = 5
3x + ky= 15
Since the system of equations has a unique solution
∴ For all values of k except k = 6, the given pair of linear equations will have a unique solution.
Q7: Find the relation between p and q if x = 3 and y = 1 is the solution of the pair of equations x  4y + p = 0 and 2x + y  q 2 = 0. [Year 2019, 2 Marks]
► View AnswerAns: Given pair of equations are
x  4y + p = 0 (i)
and 2x + y  q  2 = 0 (ii)
It is given that x = 3 and y = 1 is the solution of (i) and (ii)
∴ 3  4 x 1+ p = 0
⇒ p = 1
and 2x 3 + 1  q  2 = 0
⇒ q = 5
∴ q = 5p
Q8: Find c if the system of equations cx + 3y + (3  c) = 0; 12x + cy  c = 0 has infinitely many solutions? [Year 2019, 2 Marks]
► View AnswerAns: Given system of equations
cx + 3y + (3  c) = 0
and 12x + cy  c = 0 has infinitely many solutions.
The value of c, that satisfies both the condition is c = 6.
Q9: Find the value of k for which the following pair of linear equations have infinitely many solutions. 2x + 3y = 7, (k +1 )x + (2k 1)y = 4k + 1 [Year 2019, 2 Marks]
► View AnswerAns: The given pair of ii near equations is
2x + 3y = 7,
(k +1 )x + (2k 1)y = 4k + 1
Since given equations have infinitely many solutions.
Now,
⇒ 4k  3k = 3 + 2
⇒ k = 5
Q10: For what value of k. does the system of linear equations 2x + 3y = 7, (k 1)x + (k + 2)y = 3k have an infinite number of solutions? [Year 2019, 2 Marks]
► View AnswerAns: The given pair of linear equations is
2x + 3y = 7
(k 1)x + (k + 2)y = 3k
Since, given equations have infinitely many solutions
Now,
⇒ 6k = 7k  7
⇒ k = 7
Q11: Find the value(s) of k for which the pair of equations has a unique solution. [Year 2019, 2 Marks]
► View AnswerAns: The given pair of linear equations
kx + 2y = 3
3x + 6y = 10
Since, the system of equations has unique solution
∴ For all values of k except k = 1. the given pair of linear equations will have unique solution.
126 videos477 docs105 tests

1. What is a pair of linear equations in two variables? 
2. How can we determine the solution to a pair of linear equations in two variables? 
3. Can a pair of linear equations in two variables have no solution? 
4. Is it possible for a pair of linear equations in two variables to have infinitely many solutions? 
5. How can we graphically represent a pair of linear equations in two variables? 

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