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NCERT Exemplar for Class 10 Maths Chapter 3 - Pair of Linear Equations in Two Variables (Book Solutions)

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

Sample Questions 

1. The pair of equations 5x15y=8 and 3x9y=245 has

(A) one solution

(B) two solutions

(C) infinitely many solutions

(D) no solution

Ans: D

Given equations are 5x15y=8and  3x9y=245

The equation 3x9y=245 can be written as 

15x45y=24

The equations 5x15y=8 and 15x45y=24

Comparing with a1x+b1y=c1anda2x+b2y=c2, we get:

a1a2=515=13

b1b2=1545=13

c1c2=824=13

a1a2=b1b2=c1c2

Therefore, the pair of equations 5x15y=8 and 3x9y=245 has infinitely many solutions.


2. The sum of the digits of a two-digit number is9. If 27 is added to it, the digits of the number get reversed. The number is

(A) 25 

(B) 72  

(C) 63

(D) 36 

Ans: D

Consider one digit number be x and tens digit number be y.

And, to find the number be x+10y 

We have:

The sum of the digits of a two-digit number is 9

i.e., x+y=9 …………..(i)

And if 27 is added to it, the digits of the number get reversed then,

(x+10y)=(10x+y)+27

x+10y=10x+y+27

9y9x=27

9(yx)=27

yx=279

y=3+x…………(ii)

Substitute the value of y in equation (i), we get:

3+x+x=9

2x=93 

x=62

x=3

Substitute the value of x in equation (ii) , we get:

y=3+3

y=6

Then ,

Number =(10x+y)

Number =10×3+6

Number =30+6

Number  =36 


EXERCISE 3.1

Choose the correct answer from the given four options:

1. Graphically, the pair of equations

6x3y+10=0

2xy+9=0

represents two lines which are

(A) intersecting at exactly one point.

(B) intersecting at exactly two points.

(C) coincident.

(D) parallel.

Ans: D

Given, the linear equations are 6x3y+10=0and 2xy+9=0

On comparing with a1x+b1y=c1anda2x+b2y=c2, we get:

So, a1=6,b1=3,c1=10

a2=2,b2=1,c2=9

a1a2=62=3

And 

b1b2=31=3

And 

c1c2=109

a1a2=b1b2c1c2

Therefore, the lines representing the given equations are parallel.


2. The pair of equations x+2y+5=0 and 3x6y+1=0 have

(A) a unique solution

(B) exactly two solutions

(C) infinitely many solutions

(D) no solution

Ans: D

From the equations x+2y+5=0 and 3x6y+1=0.

The above equations are of the form a1x+b1y+c1=0and a2x+b2y+c2=0.

a1a2=13

b1b2=26=13

c1c2=51

a1a2=b1b2c1c2

Therefore, the pair of equations x+2y+5=0 and 3x6y+1=0 have no solution.


3. If a pair of linear equations is consistent, then the lines will be

(A) parallel

(B) always coincident

(C) intersecting or coincident

(D) always intersecting

Ans: C

If a pair of linear equations is consistent, then it has unique solutions and infinite solutions.

  • If a1a2b1b2, then the graph will be a pair of lines intersecting at a unique point, which is the solution of the pair of equations.

  • If a1a2=b1b2=c1c2, then the graph will be a pair of coincident lines. Each point on the lines will be a solution, and so the pair of equations will have infinitely many solutions


4. The pair of equations y=0 and y=7 has

(A) one solution

(B) two solutions

(C) infinitely many solutions

(D) no solution

Ans: D

Given the pair of equations are y=0 and y=7 

Graph the pair of equations y=0and y=7 

seo images

From the above graph y=0 and y=7 are parallel, so the pair of equations  y=0 and y=7 have no solutions.


5. The pair of equations x=a and y=b graphically represents lines which are

(A) parallel

(B) intersecting at (b,a)

(C) coincident

(D) intersecting at (a,b)

Ans: D

Given pair of equations x=a and y=b 

seo images

From the graph, the lines are x=a and y=b intersecting lines and they intersect at the point (a,b) 


6. For what value of k, do the equations 3xy+8=0 and 6xky=16 represent coincident lines?

(A) 12         

(B) 12     

(C) 2      

(D) 2 

Ans:  C

Given equations 3xy+8=0 and 6xky=16

The above equations are of the form a1x+b1y+c1=0and a2x+b2y+c2=0

So, a1=3,b1=1,c1=8 and a2=6,b2=k,c2=16   

since, lines are coincident 

therefore, a1a2=b1b2=c1c2

36=1k=816 

1k=12 

k=2 


7. If the lines given by 3x+2ky=2 and 2x+5y+1=0 are parallel, then the value of k is

(A) 54

(B) 25

(C) 154

(D) 32

Ans: C

Given lines 3x+2ky2=0 and 2x+5y1=0are parallel.

On comparing the linear equations with a1x+b1y+c1=0and a2x+b2y+c2=0, we get:

a1=3,b1=2k,c1=2 and a2=2,b2=5,c2=1

Since, the lines are parallel.

a1a2=b1b2c1c2

32=2k5

k=154


8. The value of c for which the pair of equations cxy=2 and 6x2y=3 will have infinitely many solutions is

(A) 3 

(B) 3

(C) 12

(D) no value

Ans: D

Given a pair of equations cxy=2 and 6x2y=3 will have infinitely many solutions. 

On comparing the linear equations with a1x+b1y+c1=0and a2x+b2y+c2=0, we get:

a1=c,b1=1,c1=2 and a2=6,b2=2,c2=3.

since, 

a1a2=b1b2=c1c2

Then, 

c6=12=23

c6=12 and  c6=23

c=3  and  c=4

Here, the value of chas no value because cvalue has different values.


9. One equation of a pair of dependent linear equations is 5x+7y=2. The second equation can be

(A) 10x+14y+4=0

(B) 10x14y+4=0

(C) 10x+14y+4=0

(D) 10x14y=4

Ans: D

Given pair of dependent linear equation is 5x+7y2 

From the equation 5x+7y2

a1=5,b1=7,c1=2

Then, from the dependent linear condition a1a2=b1b2=c1c2=1k

5a2=7b2=2c2=1k

a2=5k,b2=7k,c2=2k

where, k is an arbitrary constant.

Substitutingk=2, then 

a2=10,b2=14and c2=4

The required second line becomes a2x+b2y+c2=0

10x+14y4=0

10x14y+4=0


10. A pair of linear equations which has a unique solution x=2,y=3 is

(A) x+y=1

2x3y=5

(B) 2x+5y=11

4x+10y=22

(C) 2xy=1

3x+2y=0

(D) x4y14=0

5xy13=0

Ans: (B, D)

The pair of linear equations which have a unique solution x=2,y=3 .

If x=2 and y=3 substitute in all equations it must satisfy the L.H.S = R.H.S 

Substitute x=2,y=3 in option (A) 

x+y=1;2x3y=5

23=1;2(2)3(3)=5

1=1;13=5

True;False

Substitute x=2,y=3 in option (B) 

2x+5y=11;4x+10y=22

2(2)+5(3)=11;4(2)+10(3)=22

415=11;830=22

11=11 True  ;22=22 True

Substitute x=2,y=3 in option (C) 

2xy=1;3x+2y=0

2(2)(3)=1;3(2)+3(3)=0

4+3=1;66=0

7=1(False) ;0=0(True) 

Substitute x=2,y=3 in option (D) 

x4y14=0;5xy13=0

24(3)14=0;5(2)(3)13=0

2+1214=0;10+313=0

0=0 (True)  ;1313=0

0=0 (True);0=0 (True)  

By substituting the values in x=2,y=3 in options(A),(B),(C) and  (D) , two options (B),(D) satisfy the L.H.S = R.H.S.


11. If x=a,y=b is the solution of the equations xy=2 and x+y=4, then the values of a and b are, respectively

(A) 3 and 5

(B) 5 and 3

(C) 3 and 1

(D) 1 and 3

Ans: C

Given  x=a and y=b 

Substitute the values of x=a,y=b in given equations xy=2 and x+y=4, then 

ab=2..(1)

a+b=4(2)

Add equations (1) and (2)

a+b=4

ab=2

2a=6

a=62

a=3

Substitute the value of a  in equation (1) 

3b=2

b=23

b=1

Therefore, the value of a=3 and b=1


12. Aruna has only Rs 1 and Rs 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is Rs75, then the number of Re1 and Rs 2 coins are, respectively

(A)35  and 15 

(B) 35 and 20 

(C) 15 and 35

(D) 25and 25 

Ans: D

Consider, 

x: Aruna has only Rs 1 coins with her.

y: Aruna has only Rs 1 coins with her. 

From the question, 

The total number of coins that she has is 50 

Total number of coins =x+y=50 …………. (i)

Also, Amount of money with her 

=(Number of Rs1×coin1)+ ( Number of Rs2×coin2)

=x+2y=75

Subtract Eq. (i) from Eq. (ii), then 

(x+2y)(x+y)=(7550)

So, y=25

Substituting  y=25 in eq(i) 

x+25=50

x=5025

x=25

Therefore, Aruna has 25 Re 1 coins and 25 Rs 2 coins.


13. 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's age. 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 

Ans: C

Consider the present age of father be ' x ' year and the present age of son be 'y' year .

From the given question, Four years hence, the age of the father will be four times his son's age then

(x+4)=4(y+4)

x+4=4y+16

x4y12=0 ……..(i)

And the father's age is six times his son's age, 

x=6y ………(ii)

Substituting the value of Eq. (ii) in Eq. (i) then

6y4y12=0

2y=12

Therefore, y=6

Substitutingy=6 in eq (ii), 

x=6(6)

x=36

Therefore, the present age of the father is 36 years and the age of the son is 6 years.


Sample Question

Short Answer Questions with Reasoning

1. Is it true to say that the pair of equationsx+2y+2=0and12x14y1=0has a unique solution? Justify your answer. 

Ans: The given equations are in the form a1x+b1y+c1=0and a2x+b2y+c2=0

So, a1=1,b1=2,c1=2 and a2=12,b2=14,c2=1

Consider,

a1a2=112=2

b1b2=214=8

Since,a1a2b1b2

Therefore, the pair of equations x+2y+2=0 and 12x14y1=0has a unique solution.

Hence, the given statement “Is it true to say that the pair of equationsx+2y+2=0 and 12x14y1=0has a unique solution?” is true.


2. Do the equations 4x+3y1=5 and 12x+9y=15 represent a pair of coincident lines? Justify your answer.

Ans: The equations 4x+3y1=5 and 12x+9y=15 represent a pair of coincident lines.

The above equations are of the form a1x+b1y+c1=0and a2x+b2y+c2=0

So, a1=4,b1=3,c1=6 and a2=12,b2=9,c2=15 

So that

a1a2=412=13

b1b2=39=13

c1c2=615=25

Here, a1a2=b1b2c1c2

Thus, The given equations 4x+3y1=5 and 12x+9y=15does not represent a pair of coincident lines.

Hence, the given statement “The statement is “the equations 4x+3y1=5 and 12x+9y=15 represent a pair of coincident lines?” is true.


3. Is the pair of equations x+2y3=0 and 6y+3x9=0consistent? Justify your answer.

Ans: Equations x+2y3=0 and 3x+6y9=0are consistent.

The above equations are of the form a1x+b1y+c1=0and a2x+b2y+c2=0

So, a1=1,b1=2,c1=3 and a2=3,b2=6,c2=9 

Then, 

a1a2=13,

b1b2=26=13,

c1c2=39=13

Therefore, a1a2=b1b2=c1c2then, the pair of equations is consistent.


EXERCISE 3.2

1. Do the following pair of linear equations have no solution? Justify your answer.

(i) 2x+4y=3

12y+6x=6

(ii) x=2y

y=2x

(iii) 3x+y3=0

2x+23y=2

Ans:

(i) Given  Linear equations 2x+4y=3and 12y+6x=6

The above equations are of the form a1x+b1y=c1and a2x+b2y=c2

So, a1=2,b1=4,c1=3 and a2=12,b2=6,c2=6 

a1a2= 26= 13 

b1b2= 46= 13 

And 

c1c2=36=12 

From the above a1a2=b1b2c1c2 i.e., parallel lines

Therefore, the given pair of linear equations 2x+4y=3and 12y+6x=6has no solutions.

(ii) Given, pair of Linear equations x=2yand y=2x.

Rewrite the equations as x2y=0and y2x=0.

The above equations are of the form a1x+b1y=c1and a2x+b2y=c2

So, a1=1,b1=2,c1=0 and a2=2,b2=1,c2=0 

a1a2=12 

And 

 b1b2=21=2 

Here, a1a2b1b2 

Therefore, the given pair of linear equationsx=2y and y=2xhas not unique solutions.

(iii) Given, the pair of Linear equations 3x+y3=0and 2x+23y=2

The above equations are of the form a1x+b1y+c1=0and a2x+b2y=c2

So, a1=3,b1=1,c1=3 and a2=2,b2=23,c2=2 

a1a2=32  

b1b2=123=32

And 

c1c2=32=32

From the above a1a2=b1b2=c1c2i.e., coincident lines 

Therefore, the given pair of linear equations 3x+y3=0 and 2x+23y=2has many solutions.


2. Do the following equations represent a pair of coincident lines? Justify your answer.

(i)3x+17y=3

7x+3y=7

(ii)2x3y=1

6y+4x=2

(iii) x2+y+25=0

4x+8y+516=0

Ans:

(i)Given, the pair of linear equations are3x+17y=3and 7x+3y=7.

The above equations are of the form a1x+b1y+c1=0and a2x+b2y+c2=0

So,a1=3,b1=17,c1=3;

And a2=7,b2=3,c2=7

a1a2=37     

b1b2=173=121

c1c2=37=37

From the above a1a2b1b2

Therefore, the given pair of linear equations has a unique solution.

A pair of equations 3x+17y=3and 7x+3y=7does not represent a pair of coincident lines.

(ii) Given, the pair of linear equations are 2x3y1=0 and 4x+6y+2=0.

The above equations are of the form a1x+b1y+c1=0and a2x+b2y+c2=0.

So, a1=2,b1=3,c1=1, and a2=4,b2=6,c2=2

Then, 

a1a2=24=12

b1b2=36=12

c1c2=12

Hence,a1a2=b1b2=c1c2, i.e. system has infinitely many solutions.

Therefore, the given pair of linear equations 2x3y1=0 and 4x+6y+2=0 represent a pair of coincident lines.

(iii) Given, the pair of linear equations are x2+y+25=0and 4x+8y+516=0.

The above equations are of the form a1x+b1y+c1=0and a2x+b2y+c2=0

So, a1=12,b1=1,c1=25; and a2=4,b2=8,c2=516;

Then 

a1a2=18

b1b2=18

c1c2=3225

Hence, a1a2=b1b2c1c2, i.e. system has no solution.

Therefore, the given pair of linear equations x2+y+25=0and 4x+8y+516=0 does not represent a pair of coincident lines 


3. Are the following pair of linear equations consistent? Justify your answer.

(i)3x4y=12

4y+3x=12

(ii)35xy=12

15x3y=16

(iii) 2ax+by=a

(iv) x+3y=11

2(2x+6y)=22

Ans:

(i) Given, Pair of linear equations are 3x4y12=0 and 4y+3x12=0.

The above equations are of the form a1x+b1y+c1=0and a2x+b2y+c2=0.

Then 

a1=3,b1=4,c1=12, a2=3,b2=4,c2=12

So, 

a1a2=33=1

b1b2=44=1

c1c2=1212=1

Hence, a1a2=b1b2c1c2, system has no solution.

Therefore, the pair of linear equations 3x4y12=0 and 4y+3x12=0 is not consistent. 

(ii) Given, the pair of linear equations are 35xy=12 and15x3y=16.

The above equations are of the form a1x+b1y+c1=0anda2x+b2y+c2=0.

So, a1=35,b1=1,c1=12 and a2=15,b2=3,c2=16

Then,

a1a2=3

b1b2=13=13

And c1c2=3

Hence, a1a2b1b2, system has unique solution.

Therefore, the pair of linear equations are 35xy=12 and15x3y=16is consistent.

(iii) Given , Pair of linear equations are 2ax+bya=0 and 4ax+2by2a=0.

The above equations are of the form a1x+b1y+c1=0and a2x+b2y+c2=0

So, a1=2a,b1=b,c1=a;

a2=4a,b2=2b,c2=2a;

Then, 

a1a2=2a4a=12

b1b2=b2b=12

c1c2=a2a=12

Hence, a1a2=b1b2=c1c2, system has infinitely many solution.

Therefore, the pairs of linear equations 2ax+bya=0 and 4ax+2by2a=0 are consistent.

(iv) Given , pairs of linear equations are x+3y=11 and 2x+6y=11.

The above equations are of the form a1x+b1y+c1=0and a2x+b2y+c2=0

So,  a1=1,b1=3,c1=11and  a2=2,b2=6,c2=11

Then, 

a1a2=12

b1b2=12

c1c2=1  $$ 

Hence, a1a2=b1b2c1c2, system has no solution.

Therefore, Pair of linear equations x+3y=11 and 2x+6y=11 is not consistent.


4. For the pair of equations

λx+3y=7

2x+6y=14to have infinitely many solutions, the value of λ should be 1 . Is the statement true? Give reasons.

Ans: Pair of linear equations are λx+3y+7=0 and 2x+6y14=0.

The above equations are of the form a1x+b1y+c1=0and a2x+b2y=c2.

So, a1=λ,b1=3,c1=7;a2=2,b2=6,c2=14.

If a1a2=b1b2=c1c2, then system has infinitely many solutions.

λ2=36=714

Since, 

λ2=36

λ=1

And

λ2=714

λ=1

Therefore, λ have different values. 

So, the value of λ should not be 1 .

Therefore, the given statement “For the pair of equations

λx+3y=7

2x+6y=14to have infinitely many solutions, the value of λ should be 1” is not true.


5. For all real values ofc, the pair of equations x2y=8and 5x10y=chave a unique solution. Justify whether it is true or false.

Ans: The statement for all real values of c, the pair of equations x2y=8and 5x10y=chave a unique solution  is true .

Given the pair of equations x2y=8and 5x10y=c

The above equations are of the form a1x+b1y=c1=0and a2x+b2y=c2

So, a1=1,b1=2,c1=8 and a2=5,b2=10,c2=c

a1a2=15

b1b2=210

b1b2=15

c1c2=8c

From the above a1a2b1b2, system has unique solution.

The statement that for all real values ofc, the pair of equations x2y=8and 5x10y=c have a unique solution  is true .


6. The line represented by x=7 is parallel to the x -axis. Justify whether the statement is true or not.

Ans: The line represented by x=7 is parallel to the x -axis.

In the given statement the line represented by x=7 is parallel to the x -axis is not true. 


SAMPLE QUESTIONS

1. For which values of p and q, will the following pair of linear equations have infinitely many solutions?

4x+5y=2

(2p+7q)x+(p+8q)y=2qp+1

Ans:

Given: A pair of linear equation

4x+5y=2

(2p+7q)x+(p+8q)y=2qp+1

If a1a2=b1b2=c1c2 , the system has finitely many solutions.

From the given Equations we can derive the below as, 

a1a2=42p+7q

b1b2=5p+8q

c1c2=22qp+1

If a1a2=b1b2=c1c2 , system has finitely many solutions.

Therefore,

42p+7q=5p+8q=22qp+1

So we can write that,

42p+7q=5p+8q and 42p+7q=22qp+1

4p + 32q = 10p + 35q and 8q  4p + 4 = 4p + 14q

6p + 3q = 0 and 8p + 6q =4

q = 2p .....eq(1) and 

4p + 3q = 2 .....eq(2)

By substituting the value of q in eq (2) we get:

4p  6p = 2 or p = 1

Substituting the value of p in eq(1), we get

q = 2

Hence, Forp=1 and q=2 the pair of linear equations have infinitely many solutions.


2. Solve the following pair of linear equations:

21x+47y=110.....(1)

47x+21y=162.....(2)

Ans: A pair of linear equations

21x + 47y = 110.....(1)

47x + 21y = 162.....(2)

Multiplying Equation (1) by 21 and Equation (2) by 47 then,

441x+987y=2310 ..... (3)

2209x+987y=7849 ..... (4)

Subtracting Equation (3) from Equation (4), we get

1768x=5548

Therefore, x=3

Substituting value of xin Equation (1) we get,

63+47y=110

47y=11063

47y=47

y=1


3. Draw the graphs of the pair of linear equations xy+2=0 and 4xy4=0. Calculate the area of the triangle formed by the lines so drawn and the x axis.

Ans:

(1) Considering the equationxy+2=0.

The line will cut the x axis aty=0, so at y=0 the value ofx=2. The coordinate is (2,0).

It will cut the y axis atx=0, so at x=0 the value of y=2. The coordinate is (0,2).

(2) Considering the equation4xy4=0.

The line will cut the x axis aty=0, so at y=0 the value ofx=1. The coordinate is (1,0).

It will cut the y axis atx=0, so at x=0 the value of y=4. The coordinate is (0,4).

Therefore, the graph of the two lines is shown below by taking 1 cm=1 unit on the x axis and 1 cm=1 unit on the y axis: -

seo images

The two lines intersect at the point (2,4), which can be obtained algebraically by solving the two equations. The shaded region is the area subtended by the two lines and the x axis.

The distance of the point (1,0) from the point (2,0) is 3 units, which is the base of the triangle.

Now, the distance of the point (2,4) from the x axis is 4 units, which is the height of the triangle.

Hence, the area of the triangle

=12×3×4 square units

=6 square units


EXERCISE 3.3

1. For which value(s) of λ, do the pair of linear equations

λx+y=λ2andx+λy=1have

(i)  No Solution?

(ii) Infinitely many solutions?

(iii) a unique solution

Ans: The given equations are as follows:

λx + yλ2=0

x + λy1=0

Comparing the above equations with ax+by+c=0 we get:

a1=λ, b1= 1, c1=  λ2 

a2=1, b2=λ, c2=1

(i)  For no solution

a1a2=b1b2c1c2

λ=1λ  λ2

 so,λ2= 1;

 andλ2λ 

Here, we take onlyλ= 1, 

Hence, the system of linear equations has no solution atλ= 1.

(ii) For infinitely many solutions

a1a2=b1b2=c1c2

λ=1λ = λ2

so,λ=1λ gives λ=±1;

andλ=λ2 gives λ=1,0

Hence, the system of linear equations has infinitely many solutions at λ=1.

(iii) For a unique solution

a1a2b1b2

soλ1λ

hence,λ21; 

λ ±1; 

Hence, the system of linear equation has infinitely many solution at all real values of λexcept ±1.


2. For which value(s) of kwill the pair of equations

kx+3y=k3

12x+ky=k

Have no solution?

Ans: Comparing the above equations with ax+by+c=0 we get:

a1=k, b1= 3, c1= (k3) 

a2=12, b2=k, c2=k

For no solution of the pair of linear equations, 

a1a2=b1b2c1c2 

k12=3k (k3)k 

Taking first two parts, we get 

k12=3k 

Then, a1/a2=k/12 

b1/b2= 3/k 

c1/c2= (k3)/k 

For no solution of the pair of linear equations, 

a1/a2= b1/b2c1/c2 

k/12= 3/k  (k3)/k 

Taking first two parts, we get 

k/12 = 3/k 

k2= 36 

k = +6 

Taking last two parts, we get 

3/k  (k3)/k 

3kk(k  3) 

k2 6k 0 

so, k0,6 

k=6

k2= 36 

k = ±6 

Taking last two parts, we get 

3k (k3)k

3kk(k  3) 

k2 6k 0 

so, k0,6 

Hence for k=6 the given pair of linear equations has no solution.


3. For which values of a and b will the following pair of linear equations have infinitely many solutions?x+2y=1
(ab)x+(a+b)y=a+b2

Ans: Given pair of linear equations are 

x+2y=1 

(ab)x+(a+b)y=a+b2

Rewrite the above equation, we get:

x+2y1=0 .....(i)

(ab)x+(a+b)y(a+b2)=0(ii)

On comparing with a1x+b1y+c1=0, and a2x+b2y+c2=0, we get

a1=1,b1=2 and c1=1 [from (i) ]

a2=(ab),b2=(a+b) and c2=(a+b2) [from (ii) ]

The system has infinitely many solutions if a1a2=b1b2=$c1c2$

1ab=2a+b=1(a+b2)

Choose the first two parts, we get:

1ab=2a+b

a+b=2a2b

2aa=2b+b

a=3b….(iii)

Taking last two parts,

2a+b=1(a+b2)

 2a+2b4=a+b

a+b=4 (iv)

Now, solve the equation (iii) and (iv), we get

3b+b=4

4b=4

b=1

Put the value of b in eq. (iii), we get 

a=3×1 

a=3

So, the value (a,b)=(3,1) satisfies all the parts.

Hence, required values of a and b are 3 and 1 respectively for which the given pair of linear equations has infinitely many solutions.


4. Find the value(s) of p in (i) to (iv) and p and q in (v) for the following pair of equations:

(i) 3xy5=0and 6x2yp=0, if the lines represented by these equations are parallel.

(ii) x+py=1andpxy=1 , if the pair of equations has no solution.

(iii)3x+5y=7and2px3y=1 , if the lines represented by these equations are intersecting at a unique point.

(iv) 2x+3y5=0andpx6y8=0 , if the pair of equations has a unique solution.

(v)2x+3y=7and2px+py=28qy , if the pair of equations have infinitely many solutions.

Ans: (i) Given pair of linear equations are 

3xy5=0.... (i)

and 6x2yp=0.... (ii)

On comparing with a1x+b1y+c1=0, and a2x+b2y+c2=0, we get

a1=3,b1=1 and c1=5 [from (i)] 

a2=6,b2=2,c2=p [from (ii) ]

Since, the lines represented by these equations are parallel, then

a1a2=b1b2c1c2

36=125p

Taking last two parts, we get 

125p

125p

p10.

Hence, the given pair of linear equations are parallel for all real values of p except 10 

(ii) Given pair of linear equation are

x + py  1 = 0 (i) 

px  y  1 = 0 (ii)

On comparing with a1x+b1y+c1=0, and a2x+b2y+c2=0, we get

a1=1,b1=p and c1=1 [from (i) ]

a2=p,b2=1 and c2=1 [from (ii) ]

Since, the pair of linear equations has no solution.

i.e., both lines are parallel to each other

a1a2=b1b2c1c2

1p=p111

Taking first two parts, we get 

1p=p1

p2=1

p=±1

Taking last two parts, we get

p111

p1

Hence, the given pair of linear equations has no solution for p=1.

(iii) Given, pair of linear equations are 

3x+5y7=0 ....(i)

2px3y1=0 ....(ii)

On comparing with a1x+b1y+c1=0, and a2x+b2y+c2=0, we get

a1=3,b1=5 and c1=7 [from (i) ]

a2=2p,b2=3 and c2=1 [from (ii) ]

Since, the lines are intersecting at a unique point,

i.e. it has a unique solution. 

a1a2b1b2

32p53

910p

p910

Hence, the lines represented by these equations are intersecting at a unique point for all real values of p except 910 

(iv)  Given pair of linear equations are

2x+3y5=0 (i)

pxby8=0 (ii)

On comparing with a1x+b1y+c1=0, and a2x+b2y+c2=0, we get

a1=2,b1=3 and c1=5 from (i)]

a2=p,b2=6 and c2=8 [from (ii)]

Since, the pair of linear equations has a unique solution, 

a1a2b1b2

2p36

p4

Hence, the pair of linear equations has a unique solution for all values of p except 4.

(v) Given pair of linear equation are 

2x+3y=7

2px+py=28qy

Rewrite the equations, we get:

2x+3y7=0            ....(i)

2px+(p+q)y28=0 ...(ii)

On comparing with a1x+b1y+c1=0, and a2x+b2y+c2=0, we get

a1=2,b1=3 and c1=7               [from (i)]

a2 =  2p,b2 =  (p  +  q) and c2=28 [from (ii)] 

Since, the pair of equations has infinitely many solutions

i.e., both lines are coincident.

a1a2=b1b2=c1c2

22p=3(p+q)=728

Taking first and third parts, we get 

22p=728

1p=14

p=4

Again, taking last two parts, we get 

3p+q=728

3p+q=14

p+q=12

4+q=12[p=4]

q=8

Hence, the pair of equations has infinitely many solutions for the values of p=4 and q=8.


5. Two straight paths are represented by the equations x  3y = 2 and 2x + 6y = 5 . Check whether the paths cross each other or not.

Ans: 

Given linear equations are 

x3y2=0 (i) 

and 2x+6y5=0 (ii)

On comparing with a1x+b1y+c1=0, and a2x+b2y+c2=0, we get

a1=1,b1=3 and c1=2 [from (i) ] 

a2=2,b2=6 and c2=5 [from (ii) ]

Here, a1a2=12

b1b2=36=12

and c1c2=25=25 

i.e., a1a2=b1b2c1c2 system has no solution.

Thus, lines are parallel. 

Hence, two straight paths represented by the given equations never cross each other, because they are parallel to each other.


6. Write a pair of linear equations which has the unique solutionx=1, y=3 . How many such pairs can you write?

Ans: Let the equations are a1x+b1y+c1=0 and a2x+b2y+c2=0 

Ifa1a2b1b2, then pair of system have unique solution.

Given, x=1 and y=3 is the unique solution of these two equations.

Then it must satisfy the given equation.

a1(1)+b1(3)+c1=0

a1+3b1+c1=0 (i)

and a2(1)+b2(3)+c2=0

a2+3b2+c2=0 (ii)

So, many values of a1,a2,b1,b2,c1 and c2 satisfy the eq. (i) and eq. (ii). 

Hence, infinitely many pairs of linear equations are possible.


7. If 2x+y=23and4xy=19 , find the values of 5y2xandyx2 .

Ans: Given equation are 

2x+y=23 (i)

and 4xy=19 (ii)

On adding both equations, we get

6x=42

x=7

Put the value of x in eq. (i), we get

2(7)+y=23

14+y=23

y=2314

y=9

We have, 5y2x=5×92×7=4514=31

and yx2=972=9147=57

Hence, the values of (5y2x) and (yx2) are 31 and 57 respectively.


8. Find the values of x and y in the following rectangle (see figure).

seo images

Ans:

seo images

By property of rectangle,

Lengths are equal i.e., CD=AB

x+3y=13 (i)

Breadth are equal i.e., AD=BC

3x+y=7 (ii)

On multiplying eq. (ii) by 3 and then subtracting eq. (i), we get 

8x=8

x=1

On putting x=1 in eq.(i), we get 

3y=12

y=4

Hence, the required values of x and y are 1 and 4 , respectively.


9. Solve the following pairs of equations: 

(i) x+y=3.3,0.63x2y=1,3x2y0

(ii) x3+y4=4,5x6y8=4

(iii) 4x+6y=15,6x8y=14,y10

(iv) 12x1y=1,1x+12y=8,x,y10

(v) 43x+67y=24,67x+43y=24

(vi) xa+yb=a+b,xa2+yb2=2,a,b10

(vii) 2xyx+y=32,xy2xy=310,x+y10,2xy10

Ans: Given pair of linear equations are 

x+y=3.3 (i) 

0.63x2y=1

0.6=3x+2y

3x2y=0.6 (ii)

Now, multiplying eq. (i) by 2 and then adding with eq. (ii), we get 

5x=6 

x=65=1.2

Now, put the value of x in eq. (i), we get 

1.2+y=3.3 

y=3.31.2

y=2.1

Hence, the required values of x and y are 1.2 and 2.1 respectively.

(ii) Given, pair of linear equations are

x3+y4=4 

4x+3y=48 ...(i)

and 5x6y8=4

20x3y=96 ...(ii)

Now, adding eq. (i) and eq. (ii), we get 

24x=144

x=6

Now, put the value of x in eq. (i), we get 

4×6+3y=48

3y=4824

3y=24

y=8

Hence, the required values of x and y are 6 and 8 , respectively.

(iii) Given pair of linear equations are

4x+6y=15....(i)

and 6x8y=14....(ii)

Let u=1y, then above equations become 

4x+6u=15 ...(iii)

and 6x8u=14...(iv) 

On multiplying eq. (iii) by 8 and eq. (iv) by 6 and then adding both of them, we get

68x=204

x=3

Now, put the value of x in eq. (iii), we get

4×3+6u=15

6u=1512

6u=3

u=12

1y=12 [u=1y]

y=2

Hence, the required values of x and y are 3 and 2 , respectively.

(iv) Given pair of linear equations are 

12x1y=1(i)

and 1x+12y=8...(ii)

Put u=1x and v=1y, then the above equations become 

u2v=1

u2v=2 (iii)

and u+v2=8

2u+v=16 (iv)

On multiplying eq. (iv) by 2 and then adding with eq. (iii), we get 

5u=30

u=6

Now, put the value of u in eq. (iv), we get 

2×6+v=16

v=1612=4

v=4

x=1u=16 and y=1v=14

Hence, the required values of x and y are 16  and 14 respectively.

(v) Given pair of linear equations are 

43x+67y=24...(i)

and 67x+43y=24 ...(ii)

On multiplying eq. (i) by 43 and eq. (ii) by 67 and then subtracting eq (i) from eq (ii), we get

{(67)2(43)2}x=24(67+43)

(67+43)(6743)x=24×110[(a2b2)=(ab)(a+b)]

110×24x=24×110

x=1

Now, put the value of x in eq. (i), we get 

43×1+67y=24

67y=2443

67y=67

y=1

Hence, the required values of x and y are 1 and 1, respectively.

(vi) Given pair of linear equations are

xa+yb=a+b(i)

and xa2+yb2=2...(ii)

On multiplying eq. (i) by 1a and then subtracting from eq. (ii), we get 

y(1b21ab)=21ba

y(abab2)=1ba=(aba)

y=ab2ay=b2

Now, put the value of y in eq. (ii), we get

xa2+b2b2=2

xa2=21=1

x=a2

Hence, the required values of x and zy are a2 and b2, respectively.

(vii) Given pair of equations are 

2xyx+y=32

x+y2xy=23

xxy+yxy=43

1y+1x=43

and xy2xy=310,

2xyxy=103

2xxyyxy=103

2y1x=103

Now, put 1x=u and 1y=v

then the above equations become 

v+u=43....(iii)

and 2vu=103...(iv)

On adding both equations, we get 

3v=43103=63

3v=2

v=23

Now, put the value of v in eq. (iii). we get

 23+u=43

u=43+23=63=2

x=1u=12

and 

y=1v

1(23)=32

Hence, the required values of x and y are 12 and 32 respectively.


10. Find the solution of the pair of equations
x10+y51=0andx8+y6=15. Hence, findλ,ify=λx+5 .

Ans: Given pair of equations are x10+y51=0

x+2y10=0(i)

x+2y=10

and x8+y6=15

3x+4y=360...(ii)

On multiplying eq. (i) by 2 and then subtracting from eq. (ii), we get

(3x+4y)(2x+4y)=36020

x=340

Put the value of x in eq. (i), we get 

340+2y=10 

2y=10340=330

y=165

Now, y=λx+5

Put the values of x and y in above relation, we get

165=λ(340)+5

340λ=170

λ=12

Hence, the solution of the pair of equations are x=340,y=165 and the required value of λ is 12.


11. By the graphical method, find whether the following pair of equations are consistent or not. If consistent, solve them.

(i) 3x+y+4=0,6x2y+4=0 

(ii) x2y=6,3x6y=0 

(iii) x+y=3,3x+3y=9 

Ans: (i) Given pair of equations are 

3x+y+4=0... (i)

6x2y+4=0 (ii)

On comparing with a1x+b1y+c1=0, and a2x+b2y+c2=0, we get

a1=3,b1=1 and c1=4 [from, (i)]

a2=6,b2=2 and c2=4 [from (ii)]

Here, a1a2=36=12

b1b2=12

and, c1c2=44=11

a1a2b1b2

So, the given pair of linear equations are intersecting at one point, therefore these lines have a unique solution. 

Hence, given pairs of linear equations are consistent. 

We have, 3x+y+4=0 

y=43xWe have ,3x+y+4=0

y=43x

x

0

-1 

-2

Y

-4

-1

2


And 6x+y+4=0

2y=6x+4y=3x+2

x

-1

0

1

Y

-1

2

5


Plot the points B(0,4) and A(2,2), and join it. We get the straight lineAB.

Plot the point Q(0,2) and P(1,5) and join it. We get the straight line PQ. The lines AB and PQ intersect at C(1,1).


seo images


(ii) Given pair of equation is 

x2y=6(i) 

and 3x6y=0(ii)

On comparing with a1x+b1y+c1=0, and a2x+b2y+c2=0, we get

a1=1,b1=2 and c1=6 [from (i)]

a2=3,b2=6 and c2=0 [from (ii)]

Here, a1a2=13 

b1b226=13

and c1c2=60

a1a2=b1b2c1c2

Hence, the lines represented by the given equations are parallel. 

Therefore, it has no solution. So, the given pair of lines is inconsistent.

(iii) Given pair of equation are

x+y=3

x+y3=0(i)

and 3x+3y=9

3x+3y9=0... (ii)

On comparing witha1x+b1y+c1=0, and a2x+b2y+c2=0, we get

a1=1,b1=1 and c1=3 [from (i) ]

a2=3,b2=3 and c2=9 [from (ii)]

Here, a1a2=13 

b1b2=13

and c1c2=39=13

a1a2=b1b2=c1c2

So, the given pair of lines is coincident. 

Therefore, these lines have infinitely many solutions. 

Hence, the given pair of linear equations are consistent. 

Now, x+y=3

y=3x

x

0

3

3

Y

3

0

1


and

3x+3y=9

3y=93x

y=93x3

x

0

1

3

Y

3

2

0


Plotting the points we get the graph of lines.

seo images

We observe that the lines represented by (i) and (ii) are coincident.


12. Draw the graph of the pair of equations 2x+y=4and 2x  y = 4. Write the vertices of the triangle formed by these lines and the y-axis. Also, find the area of this triangle?

Ans:

Consider the line 2x+y=4 

y=42x 

x

0

2

y

4

0


Now, consider the line

2x  y = 4 

y = 2x  4

x

0

2

y

-4

0


Graphical representation of both lines are as follows:

seo images

Here, both lines and y -axis form ΔABC

Hence, the vertices of a ΔABC are A(0,4),B(2,0) and C(0,4)

Required area of ΔABC=2× Area of ΔAOB=2×12×4×2=8 sq units

Hence, the required area of the triangle is 8 sq units.


13. Write an equation of a line passing through the point representing the solution of the pair of linear equations x+y=2and 2xy=1. How many such lines can we find?

Ans: Given pair of linear equation are 

x+y2=0 (i)

and 2xy1=0 (ii)

On comparing witha1x+b1y+c1=0, and a2x+b2y+c2=0, we get

a1=1,b1=1 and c1=2 [from (i) ]

a2=2,b2=1 and c2=1 [from (ii)]

a1a2b1b2, System is unique solution i.e. consistent. 

Hence, both lines intersect at a point. 

Consider the line x+y=2

y=2x

x

0

2

1

Y

2

0

1


The line 2xy1=0 

y=2x1 

x

0

1/2

1

Y

-1

0

1


seo images

The given lines intersect at E(1,1) .

Hence, infinite lines can pass through the intersection point of linear equations x+y=2 and 2xy=1 .


14. If x+1 is a factor of 2x3+ax2+2bx+1, then find the values of aandb given that 2a3b=4.

Ans: Let f(x)=2x3+ax2+2bx+1, and

x+1=0

x=1 

Putting x=1inf(x)

f(x)=2x3+ax2+bx+1

f(1)=2x3+ax2+2bx+1

2(1)3+a(1)2+2b(1)+1=0

2+a2b+1=0

a2b1=0 ...(i)

Now, 2a3b=4 (given)

2a3b=4

b=(2a43)

Putting the value ofb, in equation (i)

a2(2a43)1=0

32a(2a4)3=0

3a4+83=0

a+5=0

a=5

Putting the value of a, in equation (i)

52b1=0

2b=4

b=2

Hence, the values of a and b\]are\[5 and 2


15. The angles of a triangle are x,yand40. The difference between the two angles xandyis30 . Find xandy.

Ans: According to the given question

xy=30

x=y+30...(i)

Since, sum of all sides of a triangle is 180

x+y+40=180

y+30+y+40=180(fromequation(i))

2y+70=180

2y=110

y=55

Putting the value of y  in equation (i)

x=y+30

x=55+30     

 =85

The value of x and yis85 and 55, respectively.


16. Two years ago, Salim was thrice as old as his daughter and six years later, he will be four years older than twice her age. How old are they now?

Ans: Let Salim and his daughter’s age be x and y year respectively.
Now, according to question
Two years ago, Salim was thrice as old as his daughter

x2=3(y2)

x2=3y6

x3y=4

x=3y4

Again according to the given question

six years later, Salim will be four years older than her daughter’s age.

x+6=2(y+6)+4

x+6=2y+12+4

x+6=2y+16

x=2y+10...(i)

Now, putting the value of \[x\] in equation (i)

x2y=10

(3y4)2y=10

3y42y=10

y=14 

Now, putting the value of \[y\] in equation (i)

x=3y4

 =3×144

 =38

Thus, Salim and his daughter’s age are 38 year and 14 year, respectively.


17. The age of the father is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father.

Ans: Let the present age (in year) of father and his two children be x, y and z year, respectively.
According to the question

x=2(y+z)(i)

And after \[20\] year

(x+20)=(y+20)+(z+20)

y+z+40=x+20

y+z=x20              
On putting the value of (y+z) in the equation (i) and get the present age of the father.

x=2(x 20)

x=2x40

x=40

x=40
Thus, the father’s age is 40 year.


18. Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4:5. Find the numbers.

Ans: Let the two numbers be x and y.

Two numbers are in the ratio 5 : 6 (given)

xy=56

x=5y6

If 8 is subtracted from each of the numbers, the ratio becomes 4:5 (given)

x8y8=45

5(x8)=4(y8)

5x40=4y32

5x4y=8

5(5y6)4y=8(from equation (i))

25y24y6=8

y=48

Putting the value of y in equation (i)

x=5y6

x=5×486

 =5×8

 =40

The value of x and y\]is\[40 and 48, respectively.


19. There are some students in the two examination halls AandB. To make the number of students equal in each hall, 10 students are sent from AtoB. But if 20 students are sent from BtoA , the number of students in A becomes double the number of students in B. Find the number of students in the two halls.

Ans: Let the number of students in halls A and B\]are\[x and y, respectively.
Now, by according to the question,                

x10=y+10

 x  y =20 (i)

and(x + 20) = 2 (y20)

x2y=60(ii)
On subtracting equation (ii) from equation (i), we get

(xy)(x 2 y)=20+60

xyx+2y=80 

y=8
On putting y=80 in equation (i), we get:

x80=20 

x=100 

and\; y=80
Hence, 100 students are in hall A and 80 students are in hall B.

  

20. A shopkeeper gives books on rent for reading. She takes a fixed charge for the
first two days, and an additional charge for each day thereafter. Latika paid Rs 22 for a book kept for six days, while Anand paid Rs 164 for the book kept for four days. Find the fixed charges and the charge for each extra day.

Ans: Let Latika take a fixed charge for the first two day is Rs x and additional charge for each day thereafter is Rs y.
Latika paid Rs 22 for a book kept for six days i.e.,

For Latika, Number of days = 6 = 2 (fixed days) + 4 (additional days).
x+4y=22...(i)
Anand paid Rs 16 for a book kept for four days i.e.,

For Anand, Number of days = 4 = 2 (fixed days) + 2 (additional days).
x+2y=16...(ii)
Now, subtracting equation (ii) from equation (i), we get

2y=6

y=3
On putting the value of y in equation  \[(ii)\], we get

x+2×3 =16

x=166

 =10
Hence,   the fixed charge =Rs 10
and the charge for each extra day =Rs 3


21. In a competitive examination, one mark is awarded for each correct answer while 12 mark is deducted for every wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly?

Ans: Total number of questions =120.

Let x be the number of correct answers.

The number of wrong answers =(120x).

The marks scored by Jayanti is 90.

Now, according to the given question

x×1(120x)×12=90

2x120+x=180

3x=180+120

3x=300

x=100

Thus, Jayanti answered correctly 100 questions.


22. The angles of a cyclic quadrilateral ABCD are
A=(6x+10),B=(5x)
C=(x+y),\angleD=(3y10)
Find xandy, and hence the values of the four angles.

Ans: Since, the sum of two opposite angles in a cyclic quadrilateral is 180

A+C=180and B+D=180

A+C=180

(6x+10)+(x+y)=180

(7x)+y=170

y=170(7x)...(i)

B+D=180

(5x)+(3y10)=180

(5x)+(3y)=190...(ii)

Now, equating equation (i) and (ii)

(5x)+3(170(7x))=190

(5x)+510(21x)=190

(16x)=320

Negative sign will get eliminated

x=20

Now, putting value of x in equation (i)

y=170(7x)

 = 170(7×20)

 =170140

 =30

Thus, four angle of a cyclic quadrilateral are:

A=(6x+10)

 =(6×20+10)

 =(130) 

B =(5x)

 =(5×20)

 =(100) 

C=(x+y)

 =(20+30)

 =(50)

D=(3y10)

 =(3×3010)

 =(9010)

 =80 


SAMPLE QUESTIONS

1. Draw the graphs of the lines x=2 and y=3. Write the vertices of the figure formed by these lines, the xaxis and the yaxis. Also, find the area of the figure.

Ans: The graph of x=2 is a line parallel to yaxis present at 2 units distance to the left of it.

Thus, the line l is the graph of x=2

The graph of y=3 is a line parallel to the xaxis at 3 units distance above it.

Thus, the line m is the graph of y=3

seo images


The figure enclosed by x=2,y=3, xaxis and yaxis is OABC.

The vertices of OABC are O(0,0), A(0,3), B(2,3), C(2,0).

The length of OABCis 2 units.

The breadth of OABCis 3units.

Area of a rectangle =l×b

Therefore, area of rectangle OABC=2×3=6 sq. units.


2. Determine, algebraically, the vertices of the triangle formed by the lines

5xy=5, x+2y=1 and6x+y=17.

Ans: The common solution of the two equations forming its two sides is the vertex of the triangle. So, solving the given equations pairwise gives us the required vertices of the triangle. 

The three pairs of equations are:

5xy=5 and x+2y=1

x+2y=1 and 6x+y=17

5xy=5 and 6x+y=17

Solving the first pair of equations 5xy=5 andx+2y=1:

x=1, y=0.

Thus, one vertex of the triangle is (1,0)

Solving the second pair of equationsx+2y=1 and 6x+y=17:

x=3, y=1.

Thus, other vertex of the triangle is (3,1) 

Solving the third pair of equations 5xy=5 and 6x+y=17:

x=2, y=5.

Thus, the third vertex of triangle is (2,5). 

Therefore, the three vertices of the triangle are (1,0), (3,1) and (2,5).


3. Jamila sold a table and a chair for Rs1050, thereby making a profit of 10% on the table and 25% on the chair. If she had taken a profit of 25% on the table and 10% on the chair she would have got Rs1065. Find the cost price of each. 

Ans: Assume that the cost price of the table isRsx and the cost price of the chair is Rs y.

The selling price of table when sold at a profit of 10%=Rs x+10100x=Rs 110100x.

The selling price of chair when sold at a profit of 25%=Rs y+25100y=Rs 125100y.

110100x+125100y=1050 ...(1)

When the table is sold at a profit of 25%, then selling price =Rs (x+25100x)=Rs 125100x.

When the chair is sold at a profit of 10%, then selling price =Rs (y+10100y)=Rs 110100y.

125100x+110100y=1065 ...(2)

From(1) and (2),

110x+125y=105000 and 125x+110y=106500

By adding and subtracting these equations,

235x+235y=211500 and 15x15y=1500

i.e. x+y=900 ...(3)

and xy=100 ...(4)

From(3) and (4),

x=500, y=400.

Therefore, the cost price of a table is Rs 500and the cost price of chair is Rs 400.


4. It can take 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for 4hours and the pipe of smaller diameter for 9 hours, only half the pool can be filled. How long would it take for each pipe to fill the pool separately?

Ans: Assume the time taken by the pipe of larger diameter to fill the pool as x hours and that taken by the pipe of smaller diameter pipe asy hours.

In x hours, the pipe having a larger diameter fills the pool.

Thus, in 1 hour the pipe of larger diameter fills the 1x part of the pool.

In 4 hours, the pipe of larger diameter fills the 4x part of the pool.

In 9 hours, the pipe of smaller diameter fills the 9y part of the pool.

4x+9y=12 ...(1)

Using both pipes, the pool can be filled in 12 hours.

12x+12y=1 ...(2)

Let 1x=u and 1y=v.

Then (1) and (2)will become:

4u+9v=12 ...(3)

And 12u+12v=1 ...(4)

Multiply equation(3) by 3 and subtract equation(4)from it.

15v=12

v=130

Substituting the value of v in equation(4)

u=120, v=130

1x=120, 1y=130

x=20,y=30

Therefore, the pipe of larger diameter alone can fill the pool in20 hours and the pipe of smaller diameter alone can fill the pool in 30 hours.


EXERCISE 3.4:

1. Graphically, solve the following pair of equations:

2x+y=6 and 2xy+2=0

Find the ratio of the areas of the two triangles formed by the lines representing these equations with the xaxis and the lines with the yaxis. 

Ans: Table for the line 2x+y=6is as follows:

x

0

3

y

6

0


Table for the line 2xy+2=0is as follows:

x

0

1

y

2

0


Graph of the lines is as follows:

seo images


So, the pair of equations intersect graphically at E(1,4) i.e. x=1 andy=4.

Let A1 and A2 represent the areas of ΔACE andΔBDE.

A1= Area of ΔACE=12×AC×PE=12×4×4=82.

And A2= Area of ΔBDE=12×BD×QE=12×4×1=2.

Therefore, the required ratio =A1:A2=8:2=4:1.


2. Determine, graphically, the vertices of the triangle formed by the lines y=x,3y=x and x+y=8. 

Ans: Table for the line y=x is as follows:

x

0

1

2

y

0

1

2

Table for the line 3y=x is as follows:

x

0

3

6

y

0

1

2


Table for the line x+y=8 is as follows:

x

0

4

8

y

8

4

0


Graph of the lines is as follows:

seo images


The lines y=x and 3y=x intersect the line x+y=8 at Q and D.

Thus, ΔOQD is formed by these lines. 

Therefore, the vertices of OQD are O(0,0),Q(4,4)and(6,2).


3. Draw the graphs of the equations x=3,x=5 and 2xy4=0. Also find the area of the quadrilateral formed by the lines and the x-axis.

Ans: The given equation of lines are 2xy4=0,x=3andx=5.

Table for the line2xy4=0  is as follows:

X

0

2

Y

-4

0


Plot (0, -4) and (2, 0) and join these points to get the graph of 2xy4=0. Draw x = 3 and x = 5.

seo images


From the graph, ABCD is the quadrilateral formed by the given lines and x-axis.

AB = OB - OA = 5 - 3 = 2,AD = 2 and BC = 6

Area of quadrilateral \ABCD=12x distance between parallel lines x sum of parallel sides 12×AB×(AD+BC)=12×2×(2+6)=8sq.units=8 sq. units. 

Therefore, the area of the quadrilateral is 8 sq units.


4. The cost of 4 pens and 4 pencils boxes is 100. Three times the cost of a pen is 15more than the cost of a pencil box. Form the pair of linear equations for the above situation. Find the cost of a pen and a pencil box.

Ans: Let the cost of one pen be x and the cost of one pencil box be y .

4x+4y=100

x+y=25.....(1)

and 3x=y+15

3xy=15......(2)

Add equation (1) and equation (2),

4x=40

x=10

Substitute x=10 in equation (1), 

y=2510=15

Therefore, the cost of one pen and one pencil box are ₹10 and ₹15, respectively.


5. Determine, algebraically, the vertices of the triangle formed by the lines 3xy=3,2x3y=2 and x+2y=8.

Ans: The common solution of the two equations forming its two sides is the vertex of the triangle.

Consider, the two lines AB:3xy=3,BC:2x3y=2

Multiply 3xy=3 by 3 and subtract 2x3y=2  from it.

(9x3y)(2x3y)=92

7x=7

x=1

Put x in 3xy=3:

3×1y=3

y=0

Thus, vertex Bis (1,0).

Now, consider the two lines AC:x+2y=8 and BC:2x3y=2.

Multiply x+2y=8 by 2 and subtract 2x3y=2 from it.

(2x+4y)(2x3y)=162

7y=14

y=2

Put y in x+2y=8 :

x+2×2=8

x=84

x=4

Thus, vertex C is (4,2).

Now, consider the lines AB:3xy=3 and AC:x+2y=8

Multiply 3xy=3 by 2 and add x+2y=8

(6x2y)+(x+2y)=6+8

7x=14

x=2

Put x in 3xy=3:

3×2y=3

y=63

y=3

Thus, vertex A is (2,3).

Therefore, the vertices of the ABC are A(2,3),B(1,0) and C(4,2)


6. Ankita travels 14km to her home partly by rickshaw and partly by bus. She takes half an hour, if she travels 2km by rickshaw, and the remaining distance by bus. On the other hand, if she travels 4km by rickshaw and the remaining distance by bus, she takes 9 minutes longer. Find the speed of the rickshaw and of the bus.

Ans: Let the speed of rickshaw and bus be x km/h and y km/h, respectively.

The time taken to travel 2 km by rickshaw:

As speed = distance  time 

$\Rightarrow t_{1}=\frac{2}{x}hours

And the time taken to travel the remaining distance i.e. (142)=12 km by bus, t2=12y hours.

Given that t1+t2=12

2x+12y=12 (i) The time taken to travel 4 km by rickshaw is t3=4x hours.

And the time taken to travel the remaining distance i.e. (144)=10 km by bus is t4=10y hours.

Given that t3+t4=12+960=12+320

4x+10y=1320

(ii)Let 1x=u and 1y=v, then equation (i) and equation(ii) will become:

2u+12v=12...(iii)

And 4u+10v=1320 (iv)

Multiply equation(iii) by 2 and subtract equation(iv) from it.

(4u+24v)(4u+10v)=11320

14v=720

2v=120

v=140

Put v in equation(iii), we get:

2u+12(140)=12

2u=12310=5310

2u=210

u=110

As, 1x=110

x=10

1y=140

y=40

Therefore, the speeds of rickshaw and bus are 10 km/h and 40 km/h, respectively.


7. A person, rowing at the rate of 5 km/h in still water, takes thrice as much time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream. 

Ans: Let the speed of the stream be v km/h.

Speed of a person rowing in still water =5 km/h.

Speed of a person rowing downstream =(5+v)km/h

Speed of person rowing upstream =(5v)km/h.

The time taken by the person to cover 40 km downstream is:

t1=405+v hours ........ speed = distance  time 

The time taken by the person to cover 40 km upstream is:

t2=405v hours

As t2=t1×3

405v=405+v×3 15v=35+v 5+v=153v 4v=10 v=104=2.5 km/h

Therefore, the speed of the stream is 2.5 km/h.


8. A motorboat can travel 30kmupstream and 28kmdownstream in 7 hours. It can travel 21kmupstream and return in 5 hours. Find the speed of the boat in still water and the speed of the stream.

Ans: Let the speed of the motorboat in still water and speed of stream be u km/h and v km/h, respectively.

The downstream speed of the motorboat =(u+v)km/h.

And the upstream speed of the motorboat =(uv)km/h.

The time taken to travel 30 km upstream is:

t1=30uv hours

And time taken to travel 28 km downstream is: t2=28u+v hours

Given that t1+t2=7 hours.

30uv+28u+v=7

Time taken to travel 21 km upstream is:

t3=21uv hours 

And time taken to travel 21 km downstream is:

t4=21u+v hours 

Given that t4+t3=5 hours.

21u+v+21uv=5

 Let x=1u+v and y=1uv 

Then (1) and (2) will become:

30x+28y=7 (3) 

And 21x+21y=5

x+y=521

Multiply (4) by 28 and subtract from (3), we get:

(30x28y)(28x+28y)=714021

2x=7203

2x=13

x=16

Put x in equation (4):

16+y=521

y=52116=10742=342

y=114

Now, x=1u+v=16

u+v=6

And y=1uv=114

uv=14

Add (5) and (6) :

2u=20

u=10

Put u in equation (5):

10+v=6

v=4

Therefore, the speed of the boat in still water is 10 km/h and speed of the stream is 4 km/h.


9. A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. Find the number. 

Ans: Let the two-digit number be 10x+y.

Case I: Multiplying the sum of digits by 8 and then subtracting 5= two-digit number 8(x+y)5=10x+y 8x+8y5=10x+y 2x7y=5 (i)

Case II: Multiplying the difference of digits by 16 and then adding 3= two-digit number 16x(xy)+3=10x+y 16x16y+3=10x+y 6x17y=3 (ii)

Multiply (i) by 3 and subtract from (ii):

(6x17y)(6x21y)=3(15)

4y=12

y=3

Put y in equation (i):

2x7×3=5

2x=215=16

x=8

Therefore, the required two-digit number =10x+y=10×8+3=80+3=83


10. A railway half ticket costs half the full fare, but the reservation charges are the same on a half ticket as on a full ticket. One reserved first class ticket from the stations A to B costs ₹2530. Also, one reserved first class ticket and one reserved first class half ticket from A to 8 costs 3810. Find the full first class fare from station A to B  and also the reservation charges for a ticket.

Ans: Let the cost of full and half first class fare be x and x2 respectively.

Let the reservation charges be ₹ y per ticket.

Case I: The cost of one reserved first class ticket from the stations A to Math input error x+y=2530...(1)

Case II: The cost of one reserved first class ticket and one reserved first class half ticket from stations A to Math input error

x+y+x2+y=3810

3x2+2y=3810

3x+4y=7620

Multiply (1) by 4 and subtract from (2),

(3x+4y)(4x+4y)=762010120

x=2500

x=2500

Put x in equation (1),

2500+y=2530

y=25302500

y=30

Therefore, full first class fare from stations A to B is Math input error and the reservation charge for the ticket is Math input error.


11. A shopkeeper sells a saree at 8% profit and a sweater at 10%discount, thereby, getting a sum 1008. If she had sold the saree at 10% profit and the sweater at 8% discount, she would have got 1028 . Find the cost price of the saree and the list price (price before discount) of the sweater.

Ans: Let the cost price of saree and list price of sweater be Math input error and Math input error, respectively. Case I: He sells a saree at 8% profit + He sells a sweater at 10% discount Math input error (100+8)% of x+(10010)% of y=1008

108% of x+90% of y=1008

1.08x+0.9y=1008... (i) 

Case II: He sells the saree at 10% profit + He sells the sweater at 8% discount Math input error (100+10)% of x+(1008)% of y=1028 110% of x+92% of y=1028

1.1x+0.92y=1028 (ii) 

Put y from equation (i) into equation (ii),

1.1×0.92(10081.08x0.9)=1028

1.1×0.9x+927.360.9936x=1028×0.9

0.99x0.9936x=925.2927.36

0.0036x=2.16

x=2.160.0036=600

Put x in equation (i),

1.08×600+0.9y=1008

648+0.9y=1008

0.9y=1008648

0.9y=360

y=3600.9=400

Therefore, the cost price of a saree and the list price of a sweater are Math input error and Math input error, respectively.


12. Susan invested a certain amount of money in two schemes A andB, which offer interest at the rate of 8% per annum and 9% per annum, respectively. She received 1860 as annual interest. However, had she interchanged the amount of investments in the two schemes, she would have received 20 more as annual interest. How much money did she invest in each scheme?

Ans: Let the amount of investments in schemes A and B be Math input error and Math input error, respectively. 

Case I: Interest at the rate of 8% per annum on scheme A+ Interest at the rate of 9% per annum on scheme B = ₹1860.

x×8×1100+y×9×1100=1860.. simple interest = principal × rate × time 100

8x+9y=186000( i )

Case II: Interest at the rate of 9% per annum on scheme A+ Interest at the rate of 8% per annum on scheme Math input error

x×9×1100+y×8×1100=20+1860

9x100+8y100=1880

9x+8y=188000...(ii) 

Multiply (i) by 9 and (ii) by 8 and subtract them.

(72x+81y)(72x+64y)=9×1860008×188000 17y=1000[(9×186)(8×188)] 17y=1000(16741504)=1000×170 17y=1000×170 17y=170000 y=10000 Put y in equation (i), 8x+9×10000=186000 8x=18600090000

8x=96000

x=12000

Therefore, she invested ₹ 12000 and ₹ 10000 in schemes A and B respectively.


13. Vijay had some bananas, and he divided them into two lots A and B. He sold the first lot at the rate of 2 for 3 bananas and the second lot at the rate of 1 per banana, and got a total of 400. If he had sold the first lot at the rate of 1 per banana and the second lot at the rate of 4 for 5 bananas, his total collection would have been ₹460. Find the total number of bananas he had.

Ans: Let the number of bananas in lots A and B be x and y respectively

Case I: Cost of first lot at the rate of Math input error for 3 bananas + Cost of second lot at the rate of Math input error per banana Math input error

23x+y=400

2x+3y=1200

Case II: Cost of first lot at the rate ₹ 1 per banana + Cost of second lot at the rate of ₹ 4 for 5 bananas = amount received

x+45y=460

5x+4y=2300

Multiply (1) by 4 and (2) by 3 and subtract them.

(8x+12y)(15x+12y)=48006900

7x=2100

x=300

Put x in equation (1),

2×300+3y=1200

600+3y=1200

3y=1200600

3y=600

y=200

The total number of bananas = Number of bananas in lot A+ Number of bananas in lotB=x+y=300+200=500

Therefore, he had 500 bananas.


Why NCERT Exemplar for Class 10 Maths - Pair of Linear Equations in Two Variables

The NCERT Maths Exemplar for Class 10 is useful for students who want to do well in their Class 10 board exams. Aside from that, you can use reference books to help you answer the questions, however, these reference books can occasionally lead you astray and confuse you. To achieve well in the Class 10 board exams, it is best to stick to the NCERT books for Class 10 and practice various tasks.


Refer to the NCERT Exemplar Solutions for Class 10 for the answers to the NCERT Maths exemplar questions. For a better comprehension of the subject's subjects, these solutions are offered in a concise, quick, and correct manner. These solutions can be used to effectively prepare for the Maths topic of the Class 10th board exams.


Concepts Included in Class 10 Maths - Pair of Linear Equations in Two Variables

  • Equations

  • Linear Equations

  • General Form of Linear Equation in two Variables

  • Geometrical Representation of a linear equation

  • Comparing the ratios of coefficients of a linear equation

  • Algebraic Solution

  • Substitution Method

  • Elimination Method

  • Cross-Multiplication Method


Number of Exercises in NCERT Exemplar for Class 10 Maths - Pair of Linear Equations in Two variables

  • Exercise 3.1 (13 Questions)

  • Exercise 3.2 (5 Questions)

  • Exercise 3.3 (22 Questions)

  • Exercise 3.4 (13 Questions)


Benefits of solving Problems from NCERT Exemplar for Class 10 Maths 

Mathematics is a subject that demands not only a thorough comprehension of the material but also many practice sessions to master it and get high grades. As a result, just finishing NCERT isn't enough to learn this subject; you'll also need the necessary study tools. As a result, to grasp the material quickly, you need to use NCERT Exemplar for Class 10 Maths. Take a look at the advantages of using these NCERT Maths examples to help you prepare for your board test. Some of these advantages are as follows:

  • NCERT exemplars adhere to the CBSE curriculum and cover the entire syllabus for the Class 10 board exam. This study material will also assist you in forming a foundation for studying for board examinations, as the questions in this textbook are based on the subject's prescribed syllabus.

  • Aids in board preparation- The NCERT Maths example contains advanced-level questions that appear in board exams. Solving NCERT example problems can assist you in honing your math skills. This will undoubtedly assist you in completing difficult questions and MCQs.

  • Clears topic fundamentals- The NCERT books not only cover the whole test syllabus but also offer all of the fundamentals on all topics in plain language. To avoid rehashing the same themes, make sure your concepts are crystal clear. To create a clear concept, you must refer to the NCERT Maths Exemplar for a complete and extensive study. As a result, you will be required to update the formulas, terms, and execution during the revision.

  • NCERT Exemplars are essentially practice books that contain additional key questions at a higher level and allow you to learn in greater depth. Conceptual sums are included in these examples, which are important for CBSE and competitive exams. These NCERT Exemplar Class 10 Maths Solutions will provide you with MCQs, skill-development questions, and new topics. All of the themes under each subject are covered in NCERT Exemplars.

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FAQs on NCERT Exemplar for Class 10 Maths Chapter 3 - Pair of Linear Equations in Two Variables (Book Solutions)

1. Will solving Exemplar exercises for Class 10 Maths benefit me?

These NCERT exemplars are written by subject experts following extensive research into the difficulties surrounding the subject. They offer in-depth information. You'll be able to get accurate and trustworthy information on the topics. Each topic in the exemplar is described in detail. By answering questions from the NCERT Exemplar, you will check your basic grasp and thoroughly understand the in-depth features of the material. Students can solve the Exemplar questions to practice for the final board exams and score high marks in the exams.