NCERT Solutions Class 10 Maths Chapter 3

NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables

NCERT Class 10 Mathematics Chapter 3 defines and explains the line and how the students can plot on a graph. It also talks about geometric representations of the pair of linear equations in two variables, parallel, intersection, and coinciding lines, along with other relevant concepts.

Extramarks provides NCERT Solutions for Class 10 Mathematics Chapter 3 that have answers to all the exercise questions given at the end of NCERT Chapter 3. Students can refer to these solutions to cross-check if they have derived the right answers or understand how to solve a question accurately.

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables – Free PDF

Access NCERT Solutions for Class 10 Mathematics Chapter 3 – Pair of Linear Equations in Two Variables

Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables NCERT Solutions Exercises PDF Download

The chapter discusses how a linear equation in two variables can form a straight line. NCERT Solutions Class 10 Mathematics Chapter 3 covers all the exercise questions given in respective chapters. The subject-matter experts have prepared the solutions based on the latest CBSE syllabus. 

NCERT Solutions for Class 10 Mathematics Chapter 3 – Free Download

Students can refer to NCERT Solutions for Class 10 Mathematics Chapter 3 on Extramarks to get help in preparing for board exams. The answers in the solutions are explained in an elaborate manner while ensuring they are highly accurate and meet the guidelines of CBSE. 

Topics Covered Under Mathematics Chapter 10 NCERT Solutions Chapter 3

Let us look at the topics included in the chapter.

Section 3.1: The students learn that a linear equation in two variables has plenty of solutions. The section helps the students remember the concepts in a linear equation in two variables.

Section 3.2: The section discusses the topic in detail. The exercise includes the Geometric representations of the pair of linear equations in two variables.

Section 3.3: In this section, the Class 10 students learn how to draw the pair of linear equations in two variables in the form of lines. The lines can be parallel, intersecting, or coinciding.

Section 3.4: The section teaches the students: how to find solutions to the two equations with graphical representations. The topic has been explained in a step-by-step manner. Examples have been provided for elimination, cross-multiplication, and substitution.

Section 3.5: The students learn: how to find solutions to the linear equations in two variables. Students must practise this section thoroughly.

Section 3.6: The section presents the gist of all the points associated with the chapter. 

Importance of NCERT Solutions Mathematics Class 10 Chapter 3

NCERT Solutions aid the Class 10 students’ learning process comprehensively. The benefits of NCERT Solutions of Class 10 Mathematics Chapter 3 include:

  • Subject-matter experts have prepared solutions to benefit the students in their learning process.
  • The solutions have answers to all the questions given at the end of Chapter 3. 
  • NCERT Solutions follow the CBSE syllabus and guidelines.
  • The solutions are written in simple language making them easy to comprehend.

Related Question

Q1. How do you solve the system 5x−2y=3 and y=2x?

Ans. We are given the two equations 5x−2y=3 and y=2x

We know the steps required to solve a system of equations in two variables. Let’s take the second equation, we get

⇒y=2x

Substituting this in the equation 5x−2y=3, we get

⇒5x−2(2x)=3

Simplifying the above equation, we get

⇒x=3

Substituting this value in the second equation to find the value of y, we get

⇒y=2×3

 

Multiplying 2 and 3 we get 6, substituting this above

⇒y=6

Hence, the solution values for the system of equations are 

x=3 and y=6

Q.1 Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

Ans. Let the present age of Aftab and his daughter be x years and y years respectively.
So, seven years ago,
Aftab’s age = (x–7) years and his daughter’s age = (y–7) years
According to the question,
x–7 = 7(y–7)
or x – 7y + 42 = 0
Three years hence,
Age of Aftab = (x + 3) years
Age of daughter = (y + 3) years
According to the question,
x + 3 = 3(y + 3)
or x – 3y – 6 = 0
Hence, the given information is represented algebraically by the two equations below.
x – 7y + 42 = 0
x – 3y – 6 = 0
To represent these equations graphically, we need at least two solutions for each equation. We write these solutions in table.
x – 7y + 42 = 0
or x = 7y – 42

x 0 –7 7
y 6 5 7

Also,
x – 3y – 6 = 0
or x = 3y + 6

x 0 –3 3
y –2 –3 –1

The graphical representation is given below.

Q.2 The coach of a cricket team buys 3 bats and 6 balls for ₹ 3900. Later, she buys another bat and 2 more balls of the same kind for ₹ 1300. Represent this situation algebraically and geometrically.

Ans.

Let the price of a bat and a ball be x and y respectively.
According to the question,
3x + 6y = 3900 …(1)
or 3 (x + 2y) = 3900
or x + 2y = 3900/3 =1300 …(2)
Equation (1) represents the total price of 3 bats and 6 balls whereas equation (2) represents the total price of one bat and 2 balls.
To represent these equations graphically, we need at least two solutions for each equation. We write these solutions in table.
3x + 6y = 3900
or y = (3900 – 3x)/6

x 100 300
y 600 500

(i)
Also,
x + 2y =1300
or y = (1300 – x)/2

x 500 900
y 400 200

(ii)
The below given graphical representation shows that
graphs of both the equations coincide.

Q.3 The cost of 2 kg of apples and 1 kg of grapes on a day was found to be ₹ 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹ 300. Represent the situation algebraically and geometrically.

Ans. Let the price of 1 kg of apple and 1 kg of grapes be ₹ x and ₹ y respectively.
According to the question,
2x + y = 160 …(1)
And
4x + 2y = 300
or 2x + y = 150 …(2)
Equations (1) and (2) represent the given situation algebraically.
To represent the given situation graphically, we need at least two solutions for each equation. We write these solutions in table.
2x + y = 160 …(1)
or y = 160 – 2x

x 60 40
y = 160 – 2x 40 80

(i)
Also,
2x + y = 150 …(2)
or y = 150 – 2x

x 60 40
y = 150 – 2x 30 70

(ii)
The graphical representation of the situation is given below. The two lines never intersect each other, i.e., the two lines are parallel.

Q.4 Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.

Ans.

(i) Let the number of girls and the number of boys be x and y respectively.
According to question,
x + y = 10
x – y = 4
For x + y = 10,
y = 10 – x

x 4 6 8
y = 10 – x 6 4 2

For x – y = 4,
y = x – 4

x 4 6 8
y = x – 4 0 2 4

The graphs of equations are drawn below which shows that the two lines intersect at (7, 3).
Hence the number of boys and the number of girls are 7 and 3 respectively.

(ii)
Let the cost of 1 pencil and the cost of 1 pen be ₹ x and ₹ y respectively.
According to the question,
5x + 7y = 50
and
7x + 5y = 46
For 5x + 7y = 50,
y = (50 – 5x)/7

x 3 -4 10
y = (50 – 5x)/7 5 10 0

For 7x + 5y = 46,
y = (46 – 7x)/5

x 8 3 –2
y = (46 – 7x)/5 –2 5 12

The graphs of equations are drawn below which shows that the two lines intersect at (3, 5).
Hence the cost of 1 pencil and the cost of 1 pen are ₹ 3 and ₹ 5 respectively.

Q.5 On comparing the ratios a1a2,b1b2, c1c2, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: (i)  5x – 4y + 8 = 0    7x + 6y – 9 = 0(ii) 9x + 3y + 12 = 0    18x + 6y + 24 = 0(iii) 6x – 3y + 10 = 0    2x – y + 9 = 0

Ans.

(i) 5x-4y+8 = 0      7x+6y-9 = 0Comparing these equations with a1x+b1y+c1= 0and a2x+b2y+c2= 0, we geta1=5,      b1=-4,    c1=8a2=7,      b2=6,  c2=-9Also,a1a2=57,     b1 b2=-46=-23and soa1a2b1 b2Therefore, the given pairs of linear equations intersect at a point. (ii) 9x+3y+12=0       18x+6y+24=0Comparing these equations with a1x+b1y+c1= 0and a2x+b2y+c2= 0, we geta1=9,      b1=3,    c1=12a2=18,    b2=6, c2=24Also,a1a2=918=12,     b1 b2=36=12,    c1 c2=1224=12and soa1a2=b1 b2=c1 c2Therefore, the given pairs of linear equations are coincident. (iii) 6x-3y+10 = 0         2x-y+9 = 0Comparing these equations with a1x+b1y+c1=0and a2x+b2y+c2= 0, we geta1 = 6,      b1 = -3,    c1 = 10a2= 2,      b2 = -1, c2 = 9Also,a1a2=62= 3,     b1 b2=-3-1= 3,    c1 c2=109and soa1a2=b1 b2c1 c2Therefore, the given pairs of linear equations are parallel.

Q.6

On comparing the ratios a1a2, b1b2,  c1c2, find out whetherthe following pair of linear equations are consistent,or inconsistent.(i) 3x+2y=5; 2x3y=7     (ii) 2x3y=8; 4x6y=9(iii) 32x+53y=7;  9x10y=14    (iv)  5x3y=11;  10x+6y=22   (v) 43x+2y=8;   2x+3y=12

Ans.

(i) 3x+2y=5; 2x3y=7Comparing these equations with a1x+b1y+c1=0and a2x+b2y+c2=0, we geta1=3,      b1=2,    c1=5a2=2,      b2=3, c2=7Also,a1a2=32,     b1b2=23,    c1c2=57and soa1a2b1b2Therefore, the given pair of linear equations has a uniquesolution and hence the two linear equations are consistent. (ii) 2x3y=8; 4x6y=9 Comparing these equations with a 1 x+ b 1 y+ c 1 =0 and a 2 x+ b 2 y+ c 2 =0, we get a 1 =2, b 1 =3, c 1 =8 a 2 =4, b 2 =6, c 2 =9 Also, a 1 a 2 = 2 4 = 1 2 , b 1 b 2 = 3 6 = 1 2 , c 1 c 2 = 8 9 and so a 1 a 2 = b 1 b 2 c 1 c 2 Therefore, the given linear equations are parallel and hence the two linear equations are inconsistent. MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafeaakq aabeqaaiaabIcacaqGPbGaaeyAaiaabMcacaqGGaGaaGOmaiaadIha cqGHsislcaaIZaGaamyEaiabg2da9iaaiIdacaGG7aGaaeiiaiaabc cacaaI0aGaamiEaiabgkHiTiaaiAdacaWG5bGaeyypa0JaaGyoaaqa aiaaboeacaqGVbGaaeyBaiaabchacaqGHbGaaeOCaiaabMgacaqGUb Gaae4zaiaabccacaqG0bGaaeiAaiaabwgacaqGZbGaaeyzaiaabcca caqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBai aabohacaqGGaGaae4DaiaabMgacaqG0bGaaeiAaiaabccacaWGHbWa aSbaaSqaaiaaigdaaeqaaOGaamiEaiabgUcaRiaadkgadaWgaaWcba GaaGymaaqabaGccaWG5bGaey4kaSIaam4yamaaBaaaleaacaaIXaaa beaakiabg2da9iaaicdaaeaacaqGHbGaaeOBaiaabsgacaqGGaGaam yyamaaBaaaleaacaaIYaaabeaakiaadIhacqGHRaWkcaWGIbWaaSba aSqaaiaaikdaaeqaaOGaamyEaiabgUcaRiaadogadaWgaaWcbaGaaG OmaaqabaGccqGH9aqpcaaIWaGaaiilaiaabccacaqG3bGaaeyzaiaa bccacaqGNbGaaeyzaiaabshaaeaacaWGHbWaaSbaaSqaaiaaigdaae qaaOGaeyypa0JaaGOmaiaacYcacaaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caWGIbWaaSbaaSqaaiaaigdaaeqaaOGaeyypa0Jaey OeI0IaaG4maiaacYcacaaMc8UaaGPaVlaaykW7caaMc8Uaam4yamaa BaaaleaacaaIXaaabeaakiabg2da9iabgkHiTiaaiIdaaeaacaWGHb WaaSbaaSqaaiaaikdaaeqaaOGaeyypa0JaaGinaiaacYcacaaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWGIbWaaSbaaSqaaiaaik daaeqaaOGaeyypa0JaeyOeI0IaaGOnaiaacYcacaqGGaGaaeiiaiaa dogadaWgaaWcbaGaaGOmaaqabaGccqGH9aqpcqGHsislcaaI5aaaba GaaeyqaiaabYgacaqGZbGaae4BaiaabYcaaeaadaWcaaqaaiaadgga daWgaaWcbaGaaGymaaqabaaakeaacaWGHbWaaSbaaSqaaiaaikdaae qaaaaakiabg2da9maalaaabaGaaGOmaaqaaiaaisdaaaGaeyypa0Za aSaaaeaacaaIXaaabaGaaGOmaaaacaGGSaGaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7daWcaaqaaiaadkgadaWgaaWcbaGaaGymaaqabaaa keaacaaMc8UaamOyamaaBaaaleaacaaIYaaabeaaaaGccqGH9aqpda WcaaqaaiabgkHiTiaaiodaaeaacqGHsislcaaI2aaaaiabg2da9maa laaabaGaaGymaaqaaiaaikdaaaGaaiilaiaaykW7caaMc8UaaGPaVl aaykW7daWcaaqaaiaadogadaWgaaWcbaGaaGymaaqabaaakeaacaaM c8Uaam4yamaaBaaaleaacaaIYaaabeaaaaGccqGH9aqpdaWcaaqaai aaiIdaaeaacaaI5aaaaaqaaiaabggacaqGUbGaaeizaiaabccacaqG ZbGaae4BaaqaamaalaaabaGaamyyamaaBaaaleaacaaIXaaabeaaaO qaaiaadggadaWgaaWcbaGaaGOmaaqabaaaaOGaeyypa0ZaaSaaaeaa caWGIbWaaSbaaSqaaiaaigdaaeqaaaGcbaGaaGPaVlaadkgadaWgaa WcbaGaaGOmaaqabaaaaOGaeyiyIK7aaSaaaeaacaWGJbWaaSbaaSqa aiaaigdaaeqaaaGcbaGaaGPaVlaadogadaWgaaWcbaGaaGOmaaqaba aaaaGcbaGaaeivaiaabIgacaqGLbGaaeOCaiaabwgacaqGMbGaae4B aiaabkhacaqGLbGaaeilaiaabccacaqG0bGaaeiAaiaabwgacaqGGa Gaae4zaiaabMgacaqG2bGaaeyzaiaab6gacaqGGaGaaeiBaiaabMga caqGUbGaaeyzaiaabggacaqGYbGaaeiiaiaabwgacaqGXbGaaeyDai aabggacaqG0bGaaeyAaiaab+gacaqGUbGaae4CaiaabccacaqGHbGa aeOCaiaabwgacaqGGaGaaeiCaiaabggacaqGYbGaaeyyaiaabYgaca qGSbGaaeyzaiaabYgacaqGGaGaaeyyaiaab6gacaqGKbaabaGaaeiA aiaabwgacaqGUbGaae4yaiaabwgacaqGGaGaaeiDaiaabIgacaqGLb GaaeiiaiaabshacaqG3bGaae4BaiaabccacaqGSbGaaeyAaiaab6ga caqGLbGaaeyyaiaabkhacaqGGaGaaeyzaiaabghacaqG1bGaaeyyai aabshacaqGPbGaae4Baiaab6gacaqGZbGaaeiiaiaabggacaqGYbGa aeyzaiaabccacaqGPbGaaeOBaiaabogacaqGVbGaaeOBaiaabohaca qGPbGaae4CaiaabshacaqGLbGaaeOBaiaabshacaqGUaaaaaa@5EAD@ (iii) 32x+53y=7;  9x10y=14    Comparing these equations with a1x+b1y+c1=0and a2x+b2y+c2=0, we geta1=32,      b1=53,    c1=7a2=9,      b2=10, c2=14Also,a1a2=329=16,     b1b2=5310=16,    c1c2=714=12and soa1a2b1b2Therefore, the given pair of linear equations has a uniquesolution and hence the two linear equations are consistent. (iv)  5x3y=11;         10x+6y=22  Comparing these equations with a1x+b1y+c1=0and a2x+b2y+c2=0, we geta1=5,      b1=3,    c1=11a2=10,    b2=6, c2=22Also,a1a2=510=12,     b1b2=36=12,    c1c2=1122=12and soa1a2=b1b2=c1c2Therefore, the given linear equations have infinitely manysolutions and hence they consistent.  (v) 43x+2y=8;         2x+3y=12Comparing these equations with a1x+b1y+c1=0and a2x+b2y+c2=0, we geta1=43,      b1=2,    c1=8a2=2,    b2=3, c2=12Also,a1a2=432=23,     b1b2=23,    c1c2=812=23and soa1a2=b1b2=c1c2Therefore, the given linear equations have infinitely manysolutions and hence they are consistent.

Q.7

Which of the following pairs of linear equations are consistent/inconsistent?If consistent, obtain the solution graphically:(i)   x + y = 5,             2x + 2y = 10     (ii)  x – y = 8,              3x – 3y = 16(iii) 2x + y 6 = 0,       4x – 2y  4 = 0(iv) 2x – 2y – 2 = 0,      4x – 4y  5 = 0  

Ans.

 (i) x+y=5,                    2x+2y=10 Comparing these equations with a1x+b1y+c1=0and a2x+b2y+c2=0, we geta1=1,      b1=1,    c1=5a2=2,    b2=2, c2=10Now,a1a2=12,     b1b2=12,    c1c2=510=12and soa1a2=b1b2=c1c2Therefore, the given linear equations have infinitely manysolutions and hence they are consistent.Graphs of the two equations coincide and hence each andevery point on this graph is a solution of these equations.

(ii) xy=8,                    3x3y=16Comparing these equations with a1x+b1y+c1=0and a2x+b2y+c2=0, we geta1=1,      b1=1,    c1=8a2=3,    b2=3, c2=16Now,a1a2=13,     b1b2=13,    c1c2=816=12and soa1a2=b1b2c1c2Therefore, the given linear equations are paralleland hence they are inconsistent. (iii) 2x+y6=0,          4x2y4=0Comparing these equations with a1x+b1y+c1=0and a2x+b2y+c2=0, we geta1=2,      b1=1,       c1=6a2=4,     b2=2, c2=4Now,a1a2=24=12,     b1b2=12,    c1c2=64=32and soa1a2b1b2Therefore, the given linear equations are consistent.Now,2x+y6=0y=62xSome points which satisfy this equation are writtenin the following table.x023y=62x620Again,4x2y4=0y=2x2Some points which satisfy this equation are writtenin the following table.x023y=2x2224We get the following graphs of the given equations andfind that they intersect at (2, 2). Hence, x = 2 and y = 2.

(iv)  2x2y2=0,       4x4y5=0  Comparing these equations with a1x+b1y+c1=0and a2x+b2y+c2=0, we geta1=2,      b1=2,    c1=2a2=4,    b2=4, c2=5Now,a1a2=24=12,     b1b2=13=12,    c1c2=25=25and soa1a2=b1b2c1c2Therefore, the given linear equations are paralleland hence they are inconsistent.

Q.8 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.

Ans. Let the width of the garden be x and length be y.
According to the question, length is 4 m more than its width. Therefore,
y − x = 4 …(1)
Also, half the perimeter of a rectangular garden is 36 m.
Therefore,
x + y = 36 …(2)
Now,
y − x = 4 …(1)
or y = x + 4

x –4 0
y = x + 4 0 4

Similarly,
x + y = 36 …(2)
or y = 36 – x

x 28 20
y = 36 – x 8 16

Graphs of the equations (1) and (2) are drawn below and from there we observe that they intersect at (16, 20).
Therefore, x = 16 and y = 20. Hence, width of the garden is 16 m and length of the garden is 20 m.

Q.9 Given the linear equation 2x+ 3y– 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines

Ans.

(i) Intersecting lines: A line which will intersect the given line 2x+3y8=0       is 3x+2y8=0 as a1a2=23 and b1b2=32 impllies that a1a2b1b2.(ii) P lines: A line which is paralel to the given line 2x+3y8=0       is 4x+6y7=0 as a1a2=24=12 and b1b2=36=12 impllies that a1a2=b1b2.iii : A line which is   to the given line 2x+3y8=0       is 4x+6y16=0 because a1a2=b1b2=c1c2=12.

Q.10 Draw the graphs of the equations
x – y + 1 = 0 and 3x + 2y – 12 = 0.
Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Ans. We have,
x – y + 1 = 0
or y = x + 1
We have the following table.

x 0 –1
y = x + 1 1 0

Again,
3x + 2y – 12 = 0
or y = (12–3x)/2
and so we have the following table.

x 0 4
y = (12–3x)/2 6 0

Graphs of the given equations are drawn below and from there we find that coordinates of the vertices of the triangle formed by these lines and the x-axis are (-1, 0); (4, 0) and (2, 3).

Q.11

Solve the following pair of linear equations by the substitution method. (i) x+y=14                                          (ii) st=3              xy=4                                                     s3+t2=6      (iii) 3xy=3                                        (iv) 0.2x+0.3y=1.3 9x3y=9                                               0.4x+0.5y=2.3 (v) 2x+3y=0                                 (vi) 3x25y3=2               3x8y=0                                           x3+y2=136

Ans.

(i) The given pair of linear equations are:      x+y=14                               ...(1)                 xy=4                                 ...(2)We express x in terms of y from equation (1) to get x=14yWe substitute this value of x in equation (2) to get 14yy=4i.e., 2y=414=10i.e., y=5Putting this value of y in equation (2), we get              x5=4    i.e.,      x=9(ii) The given pair of linear equations are:      st=3                               ...(1)                 s3+t2=6                             ...(2)We express s in terms of t from equation (1) to get s=t+3We substitute this value of s in equation (2) to get t+33+t2=6i.e., 5t+66=6i.e., 5t+6=36i.e., t=6Putting this value of t in equation (1), we get              s6=3i.e.,      s=9   (iii) The given pair of linear equations are:      3xy=3                               ...(1)                 9x3y=9                            ...(2)We express y in terms of x from equation (1) to get y=3x3We substitute this value of y in equation (2) to get 9x9x+9=9i.e., 9=9This statement is true for all values of x and so we can notobtain a specific value of x. We observe that both the givenequations are the same as one is derived from another.Therefore, given equations have infinitely many solutions. (iv) The given pair of linear equations are:      0.2x+0.3y=1.3                            ...(1)      0.4x+0.5y=2.3                            ...(2)We express x in terms of y from equation (1) to get x=(1.30.3y)/(0.2)=(133y)/2We substitute this value of x in equation (2) to get 0.2(133y)+0.5y=2.3i.e., 2.60.6y+0.5y=2.3i.e., 0.1y=2.32.6=0.3i.e.,                    y=3Putting this value of y in equation (2), we get              0.4x+0.5×3=2.3i.e.,      x=(2.31.5)/0.4=8/4=2 (v) The given pair of linear equations are:     2x+3y=0                             ...(1)               3x8y=0                             ...(2)We express x in terms of y from equation (1) to get x=32yWe substitute this value of x in equation (2) to get 332y8y=0i.e., 342y=0i.e., y=0Putting this value of y in equation (2), we get             3x=0i.e.,           x=0(vi) The given pair of linear equations are:                                                              3x25y3=2                             or   9x10y=12                           ...(1)and       x3+y2=136                                 or 2x+3y=13                                ...(2)We express x in terms of y from equation (1) to get x=(12+10y)/9We substitute this value of x in equation (2) to get 2(12+10y9)+3y=13i.e., 24+20y+27y=117i.e., 47y=117+24=141i.e.,                  y=14147=3Putting this value of y in equation (2), we get            2x+9=13i.e.,           x=42=2

Q.12 Solve 2x + 3y = 11 and 2x − 4y = −24 and hence find the value of ‘m’ for which y = mx + 3.

Ans.

The given linear equations are:2x+3y=11 ...(1)2x4y=24...(2)We express x in terms of y from equation (1) to get x=113y2We substitute this value of x in equation (2) to get 113y4y=24i.e., 117y=24i.e., 7y=24+11=35i.e.,               y=5Putting this value of y in equation (1), we get              2x+3×5=11i.e.,       2x=1115=4i.e.,          x=2Now,                y=mx+3or 5=2m+3or m= 1

Q.13

Form the pair of linear equations for the followingproblems and find their solution by substitution method. (i) The difference betrween two numbers is 26 and one number is three times the other. Find them. (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. (iii) The coach of a cricket team buys a 7 bats and 6 balls for 3800. Later she buys the 3 bats and 5 balls for 1750. Find the cost of each bat and each ball. (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For  a distance of 10 km, the charge paid is 105 and for a journey of 15 km, the charge paid is 155. What are the  fixed charges and the charge per km? How much does  a person have to pay for travelling a           distance of 25 km? (v) A fraction becomes 911, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 56. Find the fraction. (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Ans.

(i)
Let the larger number is y and the smaller number is x.
According to question,
y = 3x …(1)
and
y – x = 26 …(2)
We substitute the value of y from equation (1) in equation (2) and get
3x – x = 26
or 2x = 26
or x = 13
Putting this value of x in equation (1), we get
y = 39
Hence, the numbers are 13 and 39.

(ii)
Let the larger angle is y and the smaller angle is x.
According to question,
y – x = 18° …(1)
and
y + x = 180° …(2)
We write y in terms of x from (1) to get
y = 18° + x
Putting this value of y in equation (2), we get
18° + x + x = 180°
or 2x = 180° – 18° = 162°
or x = 81°
Putting this value of x in equation (1), we get
y = 18° + 81° = 99°
Hence, the two supplementary angles are 81° and 99°.

(iii)
Let the cost of one bat and one ball be ₹ x and ₹ y respectively.
According to question,
7x + 6y = ₹ 3800 …(1)
and
3x + 5y = ₹ 1750 …(2)
We write y in terms of x from (1) to get
y= (3800 – 7x)/6
Putting this value of y in equation (2), we get
3x + 5(3800 – 7x)/6 = 1750
or 18x – 35x = 1750 × 6 – 19000 = –8500
or x = 8500/17 = 500
Putting this value of x in equation (1), we get
7×500 + 6y = 3800
or y = (3800 – 3500)/6 = 50
Hence, the cost of each bat is ₹ 500 and the cost of each ball is ₹ 50.

(iv)
Let the fixed charge be ₹ x and charge per km be ₹ y.
According to question,
x + 10y = ₹ 105 …(1)
and
x + 15y = ₹ 155 …(2)
We write x in terms of y from (1) to get
x = 105 – 10y
Putting this value of x in equation (2), we get
105 – 10y + 15y = 155
or 5y = 155 – 105 = 50
or y = 10
Putting this value of y in equation (1), we get
x + 10×10 = 105
or x = 105 – 100 = 5
Hence, the fixed charge is ₹ 5 and charge per km is ₹ 10.
Charge for 25 km = 5 + 25 ×10 = ₹ 255

vLet the fraction be xy.According to question,x+2y+2=911           11x9y=4      ...(1)andx+3y+3=56              6x5y=3      ...(2)From equation (2), we get   x=5y36 We put this value of x in equation (1) and get       115y369y=4  55y3354y=24y=24+33=9Putting this value of y in equation (1), we get           11x9×9=4       11x=814=77       x=7Hence the fraction is 79.

(vi)
Let the age of Jacob be x and the age of his son be y.
According to question,
x + 5 = 3(y + 5)
or x – 3y = 10 …(1)
Also,
x – 5 = 7(y – 5)
or x – 7y = –30 …(2)
From equation (1), we find
x = 10 + 3y
We substitute this value of x in equation (2) and get
10 + 3y – 7y = –30
or –4y = –40
or y = 10
Putting this value of y in (1), we get
x – 3×10 = 10
or x = 40
Hence the present age of Jacob is 40 years and that of his son is 10 years.

Q.14 

Solve the following pair of linear equations by theelimination method and the substitution method:(i)   x + y = 5 and 2x – 3y = 4(ii)  3x + 4y = 10 and 2x – 2y = 2(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7(iv) x2 + 2y3 = –1 and x –y3 = 3

Ans.

(i) By elimination method:Given that        x+y=5 ...(1) 2x3y=4 ...(2)We multiply equation (1) by 3 and then add it toequation (2) to eliminate the variable y and getthe value of x as follows.                  3x+3y=15                  2x3y=4 ¯                   5x          =19or                  x=195Using this value of x in equation (1), we get                      195+y=5or y=5195=65By substitution method:From equation (1), we get                   y=5xSubstituting this in equation (2), we get                  2x3  (5x)=4or 5x=4+15=19or x=195Substituting this value of x in equation (1), we get                    195+y=5or y=5195=65 (ii) By elimination method:Given that      3x+4y=10 ...(1) 2x2y=2          ...(2)We multiply equation (2) by 2 and then add it toequation (1) to eliminate the variable y and getthe value of x as follows.                  3x+4y=10                  4x4y=4 ¯                   7x          =14or               x=2Using this value of x in equation (1), we get                      6+4y=10or y=1064=1By substitution method:From equation (2), we get                   x=1+ySubstituting this in equation (1), we get                  3(1+y)+4y=10or    7y=103=7or y=1Substituting this value of y in equation (2), we get                    2x2=2or x=2+22=2 (iii) By elimination method:Given that      3x5y4=0             ...(1) 9x=2y+7                           ...(2)We multiply equation (1) by 3 and then subtract equation (2) from it to eliminate the variable x and getthe value of y as follows.                 9x15y12=0                  9x                      =2y+7                                   ¯                          15y12=2y7or               15y+2y=7+12=5or y=513Using this value of y in equation (2), we get                      9x=513×2+7or x=10117+79=10+91117=81117=913By substitution method:From equation (2), we get                   x=2y+79Substituting this in equation (1), we get                  3  ×2y+795y4=0or    2y+735y=4or 2y+715y=12or 13y=127or y=513Substituting this value of y in equation (2), we get                    9x=2×513+7=10+9113or x=8113×9=913 (iv) By elimination method:Given that      x2+2y3=1             ...(1) xy3=3                           ...(2)We multiply equation (2) by 2 and then add it toequation (1) to eliminate the variable y and getthe value of x as follows.                 x2+2y3=1                   2x2y3=6¯                  x2+2x=5or               5x2=5or x=2Using this value of x in equation (2), we get                      2y3=3or y=3  (32)=3or y=3By substitution method:From equation (2), we get                   x=3+y3Substituting this in equation (1), we get                  12  (3+y3)+2y3=1or   9+y6+2y3=1or 9+y+4y6=1 or 9+y+4y=6or y=155=3Substituting this value of y in equation (2), we get                     x=3+33=2

Q.15 Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes ½ if we only add 1 to the denominator. What is the fraction?

(ii) 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?

(iii) 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.

(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹50 and ₹100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹100 she received.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Ans.

(i)Let the fraction be xy.According to question,      x+1y1=1    x+1=y1xy=2 ...(1)Also,        xy+1=12 x=y+12 2x=y+1 2xy=1           ...(2)Subtracting equation (2) from equation (1), we get             x=3Substituting this value in equation (1), we gety=5Hence, the fraction is 35. (ii)Let the present age of Nuri and Sonu be x yearsand y years respectively.So, five years ago,Nuris age=(x5) yearsandSonu’s age=(y5) yearsAccording to question,       x5=3  (y5)  x5=3y15  x3y=10                ...(1)Again, Ten years hence,Nuris age=(x+10) yearsandSonu’s age=(y+10) yearsAccording to the question,   x+10=2(y+10)or x2y=10             ...(2)Subtracting equation (2) from equation (1), we gety=20Putting this value in (2), we getx=50Hence, Nuri and Sonu are 50 yearsand 20 years old respectively. (iii)Let the two digit number be 10x+y.According to question,            x+y=9               ...(1)and        9(10x+y)=2(10y+x) 90x+9y=20y+2x 90x2x+9y20y=0 88x11y=0 8xy=0 ...(2)Adding equations (1) and (2), we getx = 1Putting this value of x in equation (2), we gety=8Hence, Number=10x+y=10+8=18 (iv)Let number of notes of 50 and 100bexandyrespectively.According to question,            x+y=25               ...(1)and       50x+100y=2000 50 (x+2y)=2000 x+2y=40 ...(2)Subtracting equation (1) from equation(2), we getx+2y=40x+y=25              ¯        y=15Putting this value in equation (1), we getx=10Hence, Meena received 10 notes of 50 and 15 notes of 100. (v)Let the fixed charge for first three day be x and additionalcharge for each day thereafter be y.According to question,            x+4y=27               ...(1)and       x+2y=21      ...(2)Subtracting equation (2) from equation (1), we getx+4y  =27x+2y=21              ¯        2y=6or y=3Putting this value in equation (1), we getx=2712=15Hence, fixed charge is 15 and additional charge foreach extra day is 3.

Q.16 

Which of the following pairs of linear equations has unique solution,no solution, or infinitely many solutions. In case there is a uniquesolution, find it by using cross multiplication method.(i) x  3y 3 = 0                         (ii) 2x + y = 5    3x  9y  2 = 0                           3x + 2y = 8(iii) 3x  5y = 20                         (iv)  x  3y  7 = 0          6x  10y = 40                             3x  3y  15 = 0

Ans.

(i) x3y3=0      3x9y2=0Here, a1a2=13, b1b2=39=13, c1c2=32=32and soa1a2=b1b2c1c2Thus, the given pair of linear equations have no solution.(ii) 2x+y=5      i.e.,      2x+y5=0        3x+2y=8      i.e.,      3x+2y8=0Here, a1a2=23, b1b2=12, c1c2=58and soa1a2b1b2Thus, the given pair of linear equations have unique solution.By cross-multiplication method, we have        xb1c2b2c1=yc1a2a1c2=1a1b2b1a2    x8(10)=y15(16)=143    x2=y1=11    x=2 and y=1 (iii) 3x5y=20         i.e.,      3x5y20=0         6x10y=40        i.e.,       6x10y40=0Here, a1a2=36=12, b1b2=510=12, c1c2=2040=12and soa1a2=b1b2=c1c2Thus, the given pair of linear equations have infinitelymany solutions.(iv) x3y7=0                  3x3y15=0Here, a1a2=13, b1b2=33=1, c1c2=715=715and soa1a2b1b2Thus, the given pair of linear equations have unique solution.By cross-multiplication method, we have        xb1c2b2c1=yc1a2a1c2=1a1b2b1a2    x4521=y21(15)=13(9)    x24=y6=16    x=4 and y=1

Q.17

(i) For which values of a and b does the following pair of linear equations have an     infinite number of solutions?      2x + 3y =7     (a – b)x + (a + b)y = 3a + b  2(ii) For which value of k will the following pair of linear equations have no solution?     3x + y = 1    (2k – 1)x + (k – 1) y = 2k + 1

Ans.

(i) Given pair of linear equations are: 2x+3y=7(ab)x+ (a+b)y=3a+b2Here,  a1a2=2ab,   b1b2=3a+b,    c1c2=73a+b2For a pair of linear equations to have infinitely many solutions:              a1a2=b1b2=c1c2So, we need         2ab=3a+b=73a+b2or 2ab=73a+b2  and    2ab=3a+bor 2(3a+b2)=7(ab) and   2(a+b)=3(ab)or    6a+2b4=7a7b and   2a3a=3b2bor    a+9b4=0 and a=5bNow, we put a=5b in  a+9b4=0.Thus,         5b+9b=4 or           4b=4or             b=1Also,a=5b=5×1=5 (ii) Given pair of linear equations are: 3x+y=1(2k1)x+(k1)y=2k+1Here,  a1a2=32k1,   b1b2=1k1,    c1c2=12k+1For a pair of linear equations to have no solution:              a1a2=b1b2c1c2So, the given pair of linear equations have no solution when         32k1=1k1or 3k3=2k1or 3k2k=1+3 or    k=2

Q.18

Solve the following pair of linear equations by thesubstitution and cross-multiplication methods:                    8x+5y=9 3x+2y=4

Ans.

                    8x+5y=9                   ...(1)3x+2y=4                  ...(2)By substitution mwthod:From equation (2), we get                               y=43x2Substituting this value of y in equation (1), we get                     8x+5y=9or                 8x+5×43x2=9or                16x+2015x=18  or                x=1820=2Putting this value of x in equation (2), we get                       3×(2)+2y=4or 6+2y=4or 2y=4+6=10or y=5 By cross-multiplication method:8x+5y=93x+2y=4Or8x+5y9=0                   ...(1)3x+2y4=0                  ...(2)Here, a1= 8, b1= 5, c1=9a2=3, b2= 2, c2=4By cross-multiplication method, we have        xb1c2b2c1=yc1a2a1c2=1a1b2b1a2    x20+18=y27+32=11615    x2=y5=11    x=2 and y=5

Q.19

Form the pair of linear equations in the followingproblems and find their solutions (if they exist)by any algebraic method: (i) 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 1000 as hostel charges whereas a student B, who takes food for 26 days, pays 1180 as hostel charges. Find the fixed charges and the cost of food per day.(ii) A fraction becomes 13 when 1 is subtracted from the numerator and it becomes 14 when 8 is added to its denominator. Find the fraction.(iii) 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 deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?(v) 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.

Ans.

(i) Let the fixed charge be x and the cost of food per day be y.According to question,            x+20y=1000               ...(1)and            x+26y=1180...(2)Subtracting equation (1) from equation (2), we getx+26y=1180x+20y  =1000              ¯          6y=180or y=30Putting this value of y in equation (1), we get          x+20×30=1000or x=1000600=400Hence, fixed charge is 400 and the cost of food perday is 30. (ii)Let the fraction be xy.According to question,      x1y=13    3x3=y3xy=3 ...(1)Also,        xy+8=14 x=y+84 4x=y+8 4xy=8           ...(2)Subtracting equation (2) from equation (1), we get             x=5Substituting this value in equation (1), we get          3×5y=3or y=315 or y=12Therefore,xy=512Hence, the fraction is 512. (iii) Let the number of right and the number of wronganswers be x and y respectively.According to questions,              3xy=40                       ...(1)and             4x2y=50         2(2xy)=2×25         2xy=25                        ...(2)Subtracting equation (2) from equation (1), we get                x=15Putting this value of x in equation (2), we get             2×15y=25         y=2530=5            y=5Thus,Number of right answers=x=15Number of wrong answers=y=5andNumber of questions=x+y=15+5=20 (iv) Let the speeds of Ist and 2nd cars be x km/h and y km/hrespectively.According to question,              x+y=100                       ...(1)and             5x100=5y         5(x20)=5y         x20=y         xy=20                          ...(2)Adding equations (1) and (2), we get                2x=120x=60Putting this value of x in equation (1), we get             60+y=100         y=10060=40Thus,Speed of Ist car=x km/h=60  km/hSpeed of 2nd car=y km/h=40 km/h (v)Let length and breadth of the rectangle be x units and y unitsrespectively.According to question,              (x5)(y+3)=xy9          xy5y+3x15=xy9              5y+3x=159=6             3x5y6=0                 ...(1)Also,             (x+3)(y+2)=xy+67         xy+3y+2x+6=xy+67         3y+2x=676=61         2x+3y61=0                     ...(2)By cross-multiplication method, we have               x305(18)=y12(183)=19(10)          x=32319=17 and y=17119=9Hence, length and breadth of the rectangle are 17 units and9 units respectively.

Q.20

Solve the following pair of equations by reducing them to a pair of linear equations:(i) 12x + 13y=2                            (ii) 2x + 3y = 2   13x + 12y = 136                               4x9y =–1(iii) 4x + 3y = 14                         (iv) 5x – 1 + 1y – 2 = 2     3x 4y = 23                                6x – 13y – 2 = 1(v) 7x – 2yxy = 5                            (vi) 6x + 3y = 6xy    8x + 7yxy = 15                                 2x + 4y = 5xy(vii) 10x + y + 2x – y = 4                 (viii)  13x + y + 13x  y = 34      15x + y5x – y= –2                         12(3x + y)12(3x – y) = –18

Ans.

(i)Let 1x=p and 1y=qthen given equations can be written as:p2+q3=2 3p+2q12=0 ...(1)p3+q2=136 2p+3q13=0 ...(2)Using cross-multiplication method, we getp26(36)=q24(39)=194 p10=q15=15 p10=15 and q15=15 p=2 and q=3 1x=2 and 1y=3 x=12 and y=13(ii) Given  that2x+3y=2and4x9y=1Let 1x=p and 1y=q , then we get2p+3q=2 ...(1)4p9q=1 ...(2)Multiplying equation (1) by 3, we get6p + 9q = 6Adding this to equation (2), we get 10p = 5 p= 12Putting value of p in equation (1), we get 2×12+3q=2 3q=1 q=13 Now,p=1x=12x=2x=4q=1y=13y=3y=9Hence, x=4 and y=9.(iii) Given that4x+3y=143x4y=23Substituting 1x=p in above equations, we get4p+3y14=0 ...(1)3p4y23=0 ...(2)By cross-multiplication method, we get       p6956=y42(92)=1169   p125=y50=125  p=5 and y=2p=1x=5x=15    and  y=2

(iv)Giventhat5x1+1y2=2and6x13y2=1Putting 1x1=p and 1y2=q , we get5p+q=2 ...(1)6p3q=1 ...(2)multiplying equation (1) by 3, we get15p+3q=6 ...(3)Adding (2) and (3), we get 21p=7p=13Putting value of p in equation (1), we get 5×13+q=2q=253=13Now,           p=1x1=13x1=3x=4and q=1y2=13y2=3y=5   x=4 and y=5.

(v)We have      7x2yxy=5 7y2x=5 ...(1)    8x+7yxy=15 8y+7x=15 ...(2)Putting 1x=p and 1y=q in equation (1) and (2), we get 2p+7q5=0 ...(3) 7p+8q15=0 ...(4)By cross-multiplication method, we getp105(40)=q3530=11649p65=q65=165p65=165 and q65=165p=1 and q=1p=1x=1 and q=1y=1x=1 and y=1

(vi)We have6x+3y=6xy         6y+3x=6 ...(1)and2x+4y=5xy         2y+4x=5 ...(2)Putting 1x=p and 1y=q in above equations, we get3p+6q6=0 ...(3)4p+2q5=0 ...(4)By cross-multiplication method, we have      p30(12)=q24(15)=1624p18=q9=118p18=118 and q9=118p=1 and q=12p=1x=1 and q=1y=12x=1 and y=2 MathType@MTEF@5@5@+= feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xH8yiVC0xbbG8F4rqqrFfpe ea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0Firpe peKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabaqaamaaeaqbaaGcea qabeaacaGGOaGaamODaiaadMgacaGGPaaabaGaae4vaiaabwgacaqG GaGaaeiAaiaabggacaqG2bGaaeyzaaqaaiaaiAdacaWG4bGaey4kaS IaaG4maiaadMhacqGH9aqpcaaI2aGaamiEaiaadMhacaaMc8UaaGPa VlaaykW7caaMc8UaeyO0H4TaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7daWcaaqaaiaaiAdaaeaacaWG5baaaiabgUcaRmaalaaabaGaaG4m aaqaaiaadIhaaaGaeyypa0JaaGOnaiaabccacaqGGaGaaeiiaiaabc cacaqGGaGaaeOlaiaab6cacaqGUaGaaeikaiaabgdacaqGPaaabaGa aeyyaiaab6gacaqGKbaabaGaaGOmaiaadIhacqGHRaWkcaaI0aGaam yEaiabg2da9iaaiwdacaWG4bGaamyEaiaaykW7caaMc8UaaGPaVlaa ykW7cqGHshI3caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVpaalaaaba GaaGOmaaqaaiaadMhaaaGaey4kaSYaaSaaaeaacaaI0aaabaGaamiE aaaacqGH9aqpcaaI1aGaaeiiaiaabccacaqGGaGaaeiiaiaabccaca qGUaGaaeOlaiaab6cacaqGOaGaaeOmaiaabMcaaeaacaqGqbGaaeyD aiaabshacaqG0bGaaeyAaiaab6gacaqGNbGaaeiiamaalaaabaGaaG ymaaqaaiaadIhaaaGaeyypa0JaamiCaiaabccacaqGHbGaaeOBaiaa bsgacaqGGaWaaSaaaeaacaaIXaaabaGaamyEaaaacqGH9aqpcaWGXb GaaeiiaiaabMgacaqGUbGaaeiiaiaabggacaqGIbGaae4BaiaabAha caqGLbGaaeiiaiaabwgacaqGXbGaaeyDaiaabggacaqG0bGaaeyAai aab+gacaqGUbGaae4CaiaabYcacaqGGaGaae4DaiaabwgacaqGGaGa ae4zaiaabwgacaqG0baabaGaae4maiaabchacqGHRaWkcaaI2aGaam yCaiabgkHiTiaaiAdacqGH9aqpcaaIWaGaaeiiaiaabccacaqGGaGa aeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccaca qGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaa bccacaqGGaGaaeOlaiaab6cacaqGUaGaaeikaiaabodacaqGPaaaba GaaeinaiaabchacqGHRaWkcaaIYaGaamyCaiabgkHiTiaaiwdacqGH 9aqpcaaIWaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaae iiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqG GaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeOlaiaab6 cacaqGUaGaaeikaiaabsdacaqGPaaabaGaaeOqaiaabMhacaqGGaGa ae4yaiaabkhacaqGVbGaae4CaiaabohacaqGTaGaaeyBaiaabwhaca qGSbGaaeiDaiaabMgacaqGWbGaaeiBaiaabMgacaqGJbGaaeyyaiaa bshacaqGPbGaae4Baiaab6gacaqGGaGaaeyBaiaabwgacaqG0bGaae iAaiaab+gacaqGKbGaaeilaiaabccacaqG3bGaaeyzaiaabccacaqG ObGaaeyyaiaabAhacaqGLbaabaGaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8+aaSaaaeaacaWGWbaabaGaeyOeI0IaaG4maiaaicda cqGHsislcaGGOaGaeyOeI0IaaGymaiaaikdacaGGPaaaaiabg2da9m aalaaabaGaamyCaaqaaiabgkHiTiaaikdacaaI0aGaeyOeI0Iaaiik aiabgkHiTiaaigdacaaI1aGaaiykaaaacqGH9aqpdaWcaaqaaiaaig daaeaacaaI2aGaeyOeI0IaaGOmaiaaisdaaaaabaGaeyO0H49aaSaa aeaacaWGWbaabaGaeyOeI0IaaGymaiaaiIdaaaGaeyypa0ZaaSaaae aacaWGXbaabaGaeyOeI0IaaGyoaaaacqGH9aqpdaWcaaqaaiaaigda aeaacqGHsislcaaIXaGaaGioaaaaaeaacqGHshI3daWcaaqaaiaadc haaeaacqGHsislcaaIXaGaaGioaaaacqGH9aqpdaWcaaqaaiaaigda aeaacqGHsislcaaIXaGaaGioaaaacaqGGaGaaeyyaiaab6gacaqGKb GaaeiiamaalaaabaGaamyCaaqaaiabgkHiTiaaiMdaaaGaeyypa0Za aSaaaeaacaaIXaaabaGaeyOeI0IaaGymaiaaiIdaaaaabaGaeyO0H4 TaamiCaiabg2da9iaaigdacaqGGaGaaeiiaiaabggacaqGUbGaaeiz aiaabccacaqGGaGaaeyCaiabg2da9maalaaabaGaaGymaaqaaiaaik daaaaabaGaeyO0H4TaamiCaiabg2da9maalaaabaGaaGymaaqaaiaa dIhaaaGaeyypa0JaaGymaiaabccacaqGGaGaaeyyaiaab6gacaqGKb GaaeiiaiaabccacaqGGaGaaeyCaiabg2da9maalaaabaGaaGymaaqa aiaadMhaaaGaeyypa0ZaaSaaaeaacaaIXaaabaGaaGOmaaaaaeaacq GHshI3caWG4bGaeyypa0JaaGymaiaabccacaqGGaGaaeyyaiaab6ga caqGKbGaaeiiaiaadMhacqGH9aqpcaaIYaaaaaa@7FC7@

(vii) We have    10x+y+2xy=4and    15x+y5xy=2Putting 1x+y=p and 1xy=q in above equations, we get      10p+2q4=0 ...(3)and      15p5q+2=0 ...(4)By cross-multiplication method, we have      p420=q6020=15030p16=q80=180p16=180 and q80=180p=15 and q=1p=1x+y=15 and q=1xy=1Thus we have             x+y=5     ...(5) and xy=1     ...(6)Adding equations (5) and (6), we get 2x=6x=3 Subtracting equation (6) from equation (5), we get y=2Hence, x=3 and y=2.(viii) We have 13x+y+13xy=34and 12(3x+y)12(3x+y)=18Putting 13x+y=p and 13xy=q in above equations, we get                  p+q=34...(3)and p2q2=18               pq=14 ...(4)Adding equations (3) and (4), we get 2p=3414p=14 Now, p=13x+y=14                3x+y=4 ...(5)Also,                  q=13xy=12                3xy=2             ...(6)Adding equations (5) and (6), we get                    6x=6x=1Substitute x=1 in equation (5), we get                    3+y=4y=1Hence, x=1 and y=1.

Q.21 Formulate the following problem as a pair of linear equations and hence find their solution:
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Ans.

(i)
Let speed of Ritu in still water be x km/h and the speed of stream be y km/h.
Speed of Ritu while rowing upstream = (x– y) km/h
Speed of Ritu while rowing downstream = (x + y) km/h
According to question,
2(x + y) = 20
or x + y = 10 …(1)
Also,
2(x – y) = 4
or x – y = 2 …(2)
Adding equations (1) and (2), we get
2x = 12
or x = 6
Putting this in equation (1), we get
y = 4
Hence, Ritu’s speed in still water is 6 km/h and the speed of current is 4 km/h.

(ii)

Let the number of days taken by a woman to finish the workis x and the number of days taken by a man to finish thework is y.Therefore, Part of the work finished by a woman in 1 day = 1xandpart of the work finished by a man in 1 day = 1yGiven that the .Therefore, 4 (2x+5y) = 1 2x+5y=14 It is also given that the work .Therefore, 3  (3x+6y) = 1 3x+6y=13Putting 1x=p and 1y=q in the above equations, we get 2p + 5q = 148p+20q=1 3p + 6q = 139p+18q=1By cross-multiplication method, we get      p20(18)=q9(8)=1144180p2=q1=136p2=136 and q1=136p=118 and q=136p=1x=118 and q=1y=136x=18 and y=36Hence, number of days taken by a woman = 18 daysand         number of days taken by a man = 36 days. (iii)Let the speeds of train and bus be x km/h and y km/h respectively.According to question, 60x+240y=4 ...(1)and 100x+200y=256 ...(2)Putting 1x=p and 1y=q in these equations, we get 60p+240q=4 ...(3)and 100p+200q=256 600p+1200q=25 ...(4)Multiplying equation (3) by 10, we get 600p+2400q=40 ...(5)Subtracting equation (4) ​from (5), we get 1200q = 15q=180Substituting in equation (3), we get 60p = 1p=160Now, p=1x=160 and q = 1y=180

x = 60 km/h and y = 80 km/hHence, speed of train = 60 km/hand  speed of bus = 80 km/h

Q.22

The ages of two friends Ani and Biju differ by 3 years.Anis father Dharam is twice as old as Ani and Biju istwice as old as his sister Cathy. The ages of Cathy andDharam differ by 30 years. Find the ages of Ani and Biju.

Ans.

Let the ages of Ani and Biju be x and y years respectively.If Ani is older than Biju thenxy=3                          ...(1)Anis father Dharam is twice as old as Ani. So,Dharams age = 2xThe ages of Cathy and Dharam differ by 30 years.Therefore,  Cathys age = 2x30Biju is twice as old as Cathy. Therefore,y=2(2x30)            ...(2)Putting the value of y from equation (2) in equation (1), we get        x2(2x30)=3    x4x+60=3  3x=360=57      x=19From equation (1), we gety=193=16Hence, Ani is 19 years old and Biju is 16 years old.Again, if Biju is older than Ani thenyx=3                          ...(3)Again,  putting the value of y from (2) in (3), we get      2(2x30)x=3    4x60x=3    3x=3+60=63      x=21Putting this value of x in equation (3), we gety=21+3=24Hence, in this case Ani is of 21 years and Biju is of 24 years.

Q.23

One says, “Give me a hundred, friend! I shall thenbecome twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich asyou”. Tell me what is the amount of their(respective) capital? [From the Bijaganita of Bhaskara II][Hint: x+100=2(y100), y+10=6(x10)]

Ans.

Let their respective capitals are x and y.According to question,    x+100=2(y100)    x2y=300                 ...(1)Also,        y+10=6(x10)    y6x=70    y=6x70                      ...(2)Putting this value of y in equation (1), we get        x2(6x70)=300     x12x=300140     x=44011=40Putting this value of x in equation (2), we get          y=6×4070=24070=170     Hence, their respective capitals are 40 and 170.

Q.24 

A train covered a certain distance at a uniformspeed. If the train would have been 10 km/h faster,it would have taken 2 hours less than the scheduledtime. And, if the train were slower by 10 km/h; itwould have taken 3 hours more than the scheduledtime. Find the distance covered by the train.

Ans.

Let the distance covered by the train is x km and theuniform speed of the train is y km/h. Then time takento travel this distance is xy hour.According to question,    xy2=xy+10   (x2y)(y+10)=xy   xy+10x2y220y=xy   10x2y220y=0                                  ...(1)Also,       xy+3=xy10   (x+3y)(y10)=xy   xy10x+3y230y=xy   10x+3y230y=0                                ...(2)Adding equations (1) and (2), we get   10x2y220y=010x+3y230y=0¯                 y250y=0            y(y50)=0            y=0 or    y=50 Here, speed can not be taken as zero unit as then distancetravelled would be zero unit.So, y=50 Putting this value of y in equation (1), we get           10x2(50)220×50=0       10x50001000=0            x=600Hence, distance covered by the train is 600 km.

Q.25 The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Ans. Let the number of rows be x and the number of students in a row be y.
Total students in the class
= (Number of rows) × (Number of students in a row)
= xy
Given that if 3 students are extra in a row, then there would be 1 row less.
Therefore,
Total number of students = (x – 1)(y + 3)
or xy =(x-1)(y+3)=xy – y + 3x – 3
or 3x – y – 3 = 0 …(1)
It is also given that if 3 students are less in a row, then there would be 2 rows more.
Therefore,
Total number of students = (x+2)(y– 3)
or xy = (x+2)(y– 3) = xy + 2y – 3x – 6
or 3x – 2y + 6 = 0 …(2)
Subtracting equation (2) from equation (1),
– y + 2y = 3 + 6
or y= 9
By using equation (1)
3x – 9 = 3
or 3x = 12
or x = 4
Number of rows = x = 4
Number of students in each row = y = 9
Hence, number of students in the class = 9 × 4 = 36.

Q.26

In a ΔABC, C=3 B=2A+B. Find thethree angles.

Ans.

Given that,        C=3B=2(A+B) 3B=2A+2 B=2A       2AB = 0 ...(1)We know that the sum of measures of all angles ofa triangle is 180°. Therefore,     A+B+C=180°A+B+3B=180°A+4B=180° ...(2)Multiplying equation (1) by 4, we get     8A4B = 0 ...(3)Adding equations (2) and (3), we get 9A = 180°A=20°From equation (2), we get 20°+4B=180° 4B=160°              B=40°and C=3BC=120°.Therefore, A, B and C are respectively20°, 40° and 120°.

Q.27 Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y-axis

Ans. The given equations are
5x – y = 5 or y = 5(x – 1) …(1)
3x – y = 3 or y = 3(x – 1) …(2)
We find the value of y when x is zero and the value of x when y is zero for both equations and write the corresponding values in tables as below.

x 0 1
y = 5(x – 1) –5 0

(1)

x 0 1
y = 3(x – 1) –3 0

(2)
Now, we draw graphs of the given equations as given below.

From the graph above, we find that the co-ordinates of the vertices of the triangle formed by lines and the y-axis are (1, 0), (0, –3) and (0, –5).

Q.28

Solve the following pair of linear equations:(i)    px+qy = p-q                 (ii) ax+by = c        qx-py = p+q                        bx+ay = 1+c(iii)  xayb= 0                           (iv) (a-b)x+(a+b)y = a2-2ab-b2       ax+by = a2+b2                     (a+b)(x+y) = a2+b2(v)  152x-378y = -74       -378x+152y = -604

Ans.

(i)   The given equations can be written as:px+qy(pq)=0                 ...(1)qxpy(p+q)=0                 ...(2)By cross-multiplication method, we have       xpqq2(p2pq)=ypq+q2+p2+pq=1p2q2xq2p2=yq2+p2=1p2q2x=1 and y=1 (ii)   The given equations can be written as:ax+byc=0                          ...(1)bx+ay(1+c)=0                 ...(2)By cross-multiplication method, we have       xbbc+ac=ybc+a+ac=1a2b2xc(ab)b=yc(ab)+a=1a2b2x=c(ab)ba2b2 and   y=c(ab)+aa2b2(iii)The given equations can be written as:bxay=0                                ...(1)ax+by(a2+b2)=0            ...(2)By cross-multiplication method, we have       xa(a2+b2)=yb(a2+b2)=1a2+b2   x=a and   y=b (iv)The given equations can be written as:(ab)x+(a+b)y(a22abb2)=0                      ...(1)(a+b)x+(a+b)y(a2+b2)=0                                   ...(2)By cross-multiplication method, we have        x(a+b)(a2+b2)+(a+b)(a22abb2)       =y(a+b)(a22abb2)+(ab)(a2+b2)=1a2b2(a+b)2   x(a+b)(a2b2+a22abb2)              =ya3+2a2b+ab2a2b+2ab2+b3+a3+ab2a2bb3             =1a2b2a2b22ab   x2b(a+b)2=y4ab2=12b(a+b)   x=a+b and     y=2aba+b(v)   152x378y=74378x+152y=604The given equations can be written as:76x189y+37=0                      ...(1)189x+76y+302=0                  ...(2)By cross-multiplication method, we have        x37×76189×302=y189×3776×302=17621892          x=37×76+189×3027621892=2and       y=189×3776×3027621892=1

Q.29 ABCD is a cyclic quadrilateral. Find its all angles.

Ans.

We know that sum of the measures of the oppositeangles of a cyclic quadrilateral is 180°.Therefore,             A+C=180 4y+204x=180     xy=40 ...(i)Also,           B+D=180 3y57x+5=180 7x + 3y = 180 ...(ii)Multiplying equation (i) by 3, we get 3x3y=120 ...(iii)Adding equations (ii) and (iii), we get 7x+3x=60x=15Putting value of x in equation (i), we get 15 y=40 y=25Thus,A=4y+20=4×25+20=120°B=3y5=3×255=70°C=4x=4×15=60°D=7x+5=7×15+5=110°.

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