## CBSE Class 10 Maths Chapter-3 Pair of Linear Equations in Two Variables – Free PDF Download

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CBSE Class 10 Maths Chapter-3 Pair of Linear Equations in Two Variables Important Questions

**CBSE Class 10 Maths Important Questions Chapter 3 – Pair of Linear Equations in Two Variables**

**4 Mark Questions**

**1. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:**

**(i)**

(ii)

(iii) 2

(iv) 2

*x*+*y*=5, 2*x*+2*y*=10(ii)

*x*–*y*= 8, 3*x*− 3*y*= 16(iii) 2

*x*+*y*= 6, 4*x*− 2*y*= 4(iv) 2

*x*− 2*y*– 2 = 0, 4*x*− 4*y*– 5 = 0**Ans. (i)**

*x*+

*y*= 5, 2

*x*+ 2

*y*= 10

For equation x + y – 5 = 0, we have following points which lie on the line

x | 0 | 5 |

y | 5 | 0 |

For equation 2x + 2y – 10 = 0, we have following points which lie on the line

x | 1 | 2 |

y | 4 | 3 |

We can see that both of the lines coincide. Hence, there are infinite many solutions. Any point which lies on one line also lies on the other. Hence, by using equation (*x*+*y*– 5=0), we can say that*x*=5−*y*

We can assume any random values for y and can find the corresponding value of x using the above equation. All such points will lie on both lines and there will be infinite number of such points.

**(ii)** *x*– *y *= 8, 3*x *− 3*y *= 16

For x – y = 8, the coordinates are:

x | 0 | 8 |

y | -8 | 0 |

And for 3x – 3y = 16, the coordinates

x | 0 | |

y | 0 |

Plotting these points on the graph,it is clear that both lines are parallel. So the two lines have no common point. Hence the given equations have no solution and lines are inconsistent.

**(iii)** 2*x *+ *y *= 6, 4*x *− 2*y *= 4

For equation 2x + y – 6 = 0, we have following points which lie on the line

x | 0 | 3 |

y | 6 | 0 |

For equation 4x – 2y – 4 = 0, we have following points which lie on the line

x | 0 | 1 |

y | -2 | 0 |

We can clearly see that lines are intersecting at (2, 2) which is the solution.

Hence x = 2 and y = 2 and lines are consistent.

**(iv)** 2*x *− 2*y *– 2 = 0, 4*x *− 4*y *– 5 = 0

For 2x – 2y – 2 = 0, the coordinates are:

x | 2 | 0 |

y | 0 | -2 |

And for 4x – 4y – 5 = 0, the coordinates

x | 0 | |

y | 0 |

Plotting these points on the graph, it is clear that both lines are parallel. So the two lines have no common point. Hence the given equations have no solution and lines are inconsistent.

**2. Solve the following pair of linear equations by the substitution method.**

**(i) x+ y = 14**

*x*–*y*= 4**(ii)**

*s*–*t*= 3**(iii)**

*3x – y = 3*

*9x − 3y = 9***(iv)**

*0.2x + 0.3y = 1.3*

*0.4x + 0.5y = 2.3***(v)**

**(vi)**

**Ans. (i)**

*x*+

*y*=14 … (1)

*x*–

*y*=4 … (2)

*x*=4+

*y*from equation (2)

Putting this in equation (1), we get

4+

*y*+

*y*=14

⇒ 2

*y*=10

⇒

*y*=5

Putting value of y in equation (1), we get

*x*+5=14

⇒

*x*=14 – 5=9

Therefore,

*x*=9 and

*y*=5

**(ii)**

*s*–

*t*=3 … (1)

*…*(2)

Using equation (1), we can say that

*s*=3+

*t*

Putting this in equation (2), we get

⇒

⇒ 5

*t*+6=36

⇒ 5

*t*=30

⇒

*t*=6

Putting value of t in equation (1), we get

*s*– 6=3

⇒

*s*=3+6=9

Therefore,

*t*=6 and

*s*=9

**(iii)**3

*x*–

*y*=3 … (1)

9

*x*−3

*y*=9 … (2)

Comparing equation 3

*x*–

*y*=3 with

*a*

_{1}

*x*+

*b*

_{1}

*y*+

*c*

_{1}=0 and equation 9

*x*−3

*y*=9 with

*a*

_{2}

*x*+

*b*

_{2}

*y*+

*c*

_{2}=0,

We get

*a*

_{1}=3,

*b*

_{1}=−1,

*c*

_{1}=−3,

*a*

_{2}=9,

*b*

_{2}=−3 and

*c*

_{2}=−9

Here

Therefore, we have infinite many solutions for x and y

**(iv)**0.2

*x*+0.3

*y*=1.3 … (1)

0.4

*x*+0.5

*y*=2.3 … (2)

Using equation (1), we can say that

0.2

*x*=1.3−0.3

*y*

⇒

*x*=

Putting this in equation (2), we get

0.4+0.5

*y*=2.3

⇒ 2.6−0.6

*y*+0.5

*y*=2.3

⇒ −0.1

*y*=−0.3

⇒

*y*=3

Putting value of y in (1), we get

0.2

*x*+0.3(3)=1.3

⇒ 0.2

*x*+0.9=1.3

⇒ 0.2

*x*=0.4

⇒

*x*=2

Therefore,

*x*=2 and

*y*=3

**(v)**……….(1)

*……….(2)*

Using equation (1), we can say that

*x*=

Putting this in equation (2), we get

⇒

⇒

⇒

*y*=0

Putting value of y in (1), we get

*x*=0

Therefore,

*x*=0 and

*y*=0

**(vi)**… (1)

… (2

*)*

Using equation (2), we can say that

*x*=

⇒

*x*=

Putting this in equation (1), we get

⇒

⇒

⇒

⇒

⇒

*y*=3

Putting value of y in equation (2), we get

⇒

⇒

⇒ x = 2

Therefore,

*x*=2 and

*y*=3

**3. Form a pair of linear equations for the following problems and find their solution by substitution method.**

**(i) The difference between 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 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 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 Rs 105 and for a journey of 15 km, the charge paid is Rs 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 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and denominator it becomes . 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 first number be x and second number be y.

According to given conditions, we have

*x*– *y*=26 (assuming x>y) … (1)

*x*=3*y *(x>y) … (2)

Putting equation (2) in (1), we get

3*y*– *y*=26 ⇒ 2*y*=26 ⇒ *y*=13

Putting value of y in equation (2), we get

*x*=3*y*=3×13=39

Therefore, two numbers are 13 and 39.

**(ii)** Let smaller angle =*x*and let larger angle =*y*

According to given conditions, we have

*y*=*x*+18 … (1)

Also, *x*+*y*=180^{0 }(Sum of supplementary angles) … (2)

Putting (1) in equation (2), we get

*x*+*x*+18=180

⇒ 2*x*=180 – 18=162

⇒ *x*=81^{0}

Putting value of x in equation (1), we get

*y*=*x*+18=81+18=99^{0}

Therefore, two angles are 81^{0} and 99^{0}.

**(iii)** Let cost of each bat = Rs x and let cost of each ball = Rs y

According to given conditions, we have

7*x*+6*y*=3800 … (1)

And, 3*x*+5*y*=1750 … (2)

Using equation (1), we can say that

7*x*=3800−6

⇒ *x*=

Putting this in equation (2), we get

3+5*y*=1750

⇒ +5*y*=1750

⇒

⇒

⇒ 17*y*=850

⇒ *y*=50

Putting value of y in (2), we get

3*x*+250=1750

⇒ 3*x*=1500

⇒ *x*=500

Therefore, cost of each bat = Rs 500 and cost of each ball = Rs 50

**(iv) **Let fixed charge = Rs x and let charge for every km = Rs y

According to given conditions, we have

*x*+10*y*=105 … (1)

*x*+15*y*=155 … (2)

Using equation (1), we can say that

*x*=105−10*y*

Putting this in equation (2), we get

105−10*y*+15*y*=155

⇒ 5*y*=50

⇒ *y*=10

Putting value of y in equation (1), we get

*x*+10(10)=105

⇒ *x*=105 – 100=5

Therefore, fixed charge = Rs 5 and charge per km = Rs 10

To travel distance of 25 Km, person will have to pay = Rs (x+25y) = Rs (5+25 × 10) = Rs(5+250) = Rs 255

**(v) **Let numerator = x and let denominator = y

According to given conditions, we have

* *… (1)

* *… (2)

Using equation (1), we can say that

11(*x*+2)=9*y*+18

⇒ 11*x*+22=9*y*+18

⇒ 11*x*=9*y*– 4

⇒ *x*=

Putting value of x in equation (2), we get

6 =5(*y*+3)

⇒

⇒

⇒

⇒ *y*=9

Putting value of y in (1), we get

* *⇒ *x*+2=9 ⇒ *x*=7

Therefore, fraction =

**(vi)** Let present age of Jacob = x years

Let present age of Jacob’s son = y years

According to given conditions, we have

(*x*+5)=3(*y*+5) … (1)

And, (*x*−5)=7(*y*−5) … (2)

From equation (1), we can say that

*x*+5=3*y*+15

⇒ *x*=10+3*y*

Putting value of x in equation (2) we get

10+3*y*– 5=7*y*−35

⇒ −4*y*=−40

⇒ *y*=10 years

Putting value of y in equation (1), we get

*x*+5=3(10+5)=3×15=45

⇒ *x*=45 – 5=40 years

Therefore, present age of Jacob = 40 years and, present age of Jacob’s son = 10 years

**4. Solve the following pair of linear equations by the elimination method and the substitution method:**

**(i) x + y = 5, 2x – 3y = 4**

**(ii) 3x + 4y = 10, 2x – 2y = 2**

**(iii) 3 x − 5y – 4 = 0, 9x = 2y + 7**

**(iv)**

**Ans. (i)**x+y=5 … (1)

2x – 3y=4 … (2)

**Elimination method:**

Multiplying equation (1) by 2, we get equation (3)

2

*x*+2

*y*=10 … (3)

2

*x*−3

*y*=4 … (2)

Subtracting equation (2) from (3), we get

5

*y*=6

⇒

*y*=

Putting value of y in (1), we get

*x*+=5 ⇒

*x*=5−=

Therefore,

*x*= and

*y*=

**Substitution method:**

*x*+

*y*=5 … (1)

2

*x*−3

*y*=4 … (2)

From equation (1), we get,

*x*=5−

*y*

Putting this in equation (2), we get

2(5−

*y*)−3

*y*=4 ⇒ 10−2

*y*−3

*y*=4

⇒ 5

*y*=6

⇒

*y*=

Putting value of y in (1), we get

*x*=5−=

Therefore,

*x*= and

*y*=

**(ii)**3x+4y=10 … (1)

2x – 2y=2 … (2)

**Elimination method:**

Multiplying equation (2) by 2, we get (3)

4

*x*−4

*y*=4 … (3)

3

*x*+4

*y*=10 … (1)

Adding (3) and (1), we get

7

*x*=14

⇒

*x*=2

Putting value of x in (1), we get

3(2)+4

*y*=10

⇒ 4

*y*=10 – 6=4

⇒

*y*=1

Therefore,

*x*=2 and

*y*=1

**Substitution method:**

3

*x*+4

*y*=10 … (1)

2

*x*−2

*y*=2 … (2)

From equation (2), we get

2

*x*=2+2

*y*

⇒

*x*=1+

*y …*(3)

Putting this in equation (1), we get

3(1+

*y*)+4

*y*=10

⇒ 3+3

*y*+4

*y*=10

⇒ 7

*y*=7

⇒

*y*=1

Putting value of y in (3), we get

*x*=1+1=2

Therefore,

*x*=2 and

*y*=1

**(iii)**3

*x*−5

*y*– 4=0 … (1)

9

*x*=2

*y*+7 … (2)

**Elimination method:**

Multiplying (1) by 3, we get (3)

9

*x*−15

*y*– 12=0 … (3)

9

*x*−2

*y*– 7=0 … (2)

Subtracting (2) from (3), we get

−13

*y*– 5=0

⇒ −13

*y*=5

⇒

*y*=

Putting value of y in (1), we get

3

*x*– 5−4=0

⇒ 3

*x*=4−=

⇒

*x*=

Therefore,

*x*= and

*y*=

**Substitution Method:**

3

*x*−5

*y*– 4=0 … (1)

9

*x*=2

*y*+7 … (2)

From equation (1), we can say that

3

*x*=4+5

*y*

⇒

*x*=

Putting this in equation (2), we get

9−2

*y*=7

⇒ 12+15

*y*−2

*y*=7

⇒ 13

*y*=−5

⇒

*y*=

Putting value of y in (1), we get

3

*x*– 5=4

⇒ 3

*x*=4−=

⇒

*x*=

Therefore,

*x*= and

*y*=

**(iv)**…(1)

… (2)

**Elimination method:**

Multiplying equation (2) by 2, we get (3)

… (3)

… (1)

Adding (3) and (1), we get

⇒

*x*=2

Putting value of x in (2), we get

2−= 3

⇒

*y*=−3

Therefore,

*x*=2 and

*y*=−3

**Substitution method:**

… (1)

… (2)

From equation (2), we can say that

Putting this in equation (1), we get

⇒

⇒ 5

*y*+9=−6

⇒ 5

*y*=−15

⇒

*y*=−3

Putting value of y in (1), we get

⇒

*x*= 2

Therefore,

*x*=2 and

*y*=−3

**5. 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 Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 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 Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.**

**Ans. (i)** Let numerator =*x*and let denominator =*y*

According to given condition, we have

⇒ *x*+1=*y*– 1 and 2*x*=*y*+1

⇒ *x*– *y*=−2 … **(1)** and 2*x*– *y*=1 … **(2)**

So, we have equations **(1)** and **(2)**, multiplying equation **(1)** by 2 we get **(3)**

2*x*−2*y*=−4 … **(3)**

2*x*– *y*=1 … **(2)**

Subtracting equation **(2)** from **(3)**, we get

−*y*=−5

⇒ *y*=5

Putting value of y in **(1)**, we get

*x*– 5=−2

⇒ *x*=−2+5=3

Therefore, fraction =

**(ii) **Let present age of Nuri =*x* years and let present age of Sonu =*y* years

5 years ago, age of Nuri = (x – 5) years

5 years ago, age of Sonu = (y – 5) years

According to given condition, we have

(*x*−5)=3(*y*−5)

⇒ *x*– 5=3*y*– 15

⇒ *x*−3*y*=−10 … **(1)**

10 years later from present, age of Nuri =(*x*+10) years

10 years later from present, age of Sonu =(*y*+10) years

According to given condition, we have

(*x*+10)=2(*y*+10)

⇒ *x*+10=2*y*+20

⇒ *x*−2*y*=10 … **(2)**

Subtracting equation **(1)** from **(2)**, we get

*y*=10−(−10)=20 years

Putting value of **y** in **(1)**, we get

*x*– 3(20)=−10

⇒ *x*– 60=−10

⇒ *x*=50 years

Therefore, present age of Nuri =50 years and present age of Sonu =20 years

**(iii) **Let digit at ten’s place =*x*and Let digit at one’s place =*y*

According to given condition, we have

*x*+*y*=9 … **(1)**

And 9(10*x*+*y*)=2(10*y*+*x*)

⇒ 90*x*+9*y*=20*y*+2*x*

⇒ 88*x*=11*y*

⇒ 8*x*=*y*

⇒ 8*x*– *y*=0 … **(2)**

Adding **(1)** and **(2)**, we get

9*x*=9

⇒ *x*=1

Putting value of **x** in **(1)**, we get

1+*y*=9

⇒ *y*=9 – 1=8

Therefore, number =10*x*+*y*=10(1)+8=10+8=18

**(iv) **Let number of Rs 100 notes =*x*and let number of Rs 50 notes =*y*

According to given conditions, we have

*x*+*y*=25 … **(1)**

and 100*x*+50*y*=2000

⇒ 2*x*+*y*=40 … **(2)**

Subtracting **(2)** from **(1)**, we get

−*x*=−15

⇒ *x*=15

Putting value of **x** in **(1)**, we get

15+*y*=25

⇒ *y*=25 – 15=10

Therefore, number of Rs 100 notes =15 and number of Rs 50 notes =10

**(v) **Let fixed charge for 3 days = Rs *x*

Let additional charge for each day thereafter = Rs *y*

According to given condition, we have

*x*+4*y*=27 … **(1)**

*x*+2*y*=21 … **(2)**

Subtracting **(2)** from **(1)**, we get

2*y*=6

⇒ *y*= 3

Putting value of **y** in **(1)**, we get

*x*+4(3)=27

⇒ *x*=27 – 12=15

Therefore, fixed charge for 3 days = Rs 15 and additional charge for each day after 3 days = Rs 3

**6. C of linear equations has unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method.**

**(i) x – 3y – 3 = 0**

**3x – 9y – 2 = 0**

**(ii) 2 x + y = 5**

**3**

*x*+ 2*y*= 8**(iii) 3**

*x*− 5*y*= 20**6**

*x*− 10*y*= 40**(iv)**

*x*− 3*y*– 7 = 0**3**

*x*− 3*y*– 15 = 0**Ans. (i)**

*x*−3

*y*– 3=0

3

*x*−9

*y*– 2=0

Comparing equation

*x*−3

*y*– 3=0 with

*a*

_{1}

*x*+

*b*

_{1}

*y*+

*c*

_{1}=0 and 3

*x*−9

*y*– 2=0 with

*a*

_{2}

*x*+

*b*

_{2}

*y*+

*c*

_{2}=0,

We get

*a*

_{1}=1,

*b*

_{1}= −3,

*c*

_{1}=−3,

*a*

_{2}=3,

*b*

_{2}=−9,

*c*

_{2}=−2

Here this means that the two lines are parallel.

Therefore, there is no solution for the given equations i.e. it is inconsistent.

**(ii)**2

*x*+

*y*=5

3

*x*+2

*y*=8

Comparing equation 2

*x*+

*y*=5 with

*a*

_{1}

*x*+

*b*

_{1}

*y*+

*c*

_{1}=0 and 3

*x*+2

*y*=8 with

*a*

_{2}

*x*+

*b*

_{2}

*y*+

*c*

_{2}=0,

We get

*a*

_{1}= 2,

*b*

_{1}= 1,

*c*

_{1}=−5,

*a*

_{2}=3,

*b*

_{2}= 2,

*c*

_{2}=−8

Here this means that there is unique solution for the given equations.

⇒

⇒

*x*=2 and

*y*=1

**(iii)**3

*x*−5

*y*=20

6

*x*−10

*y*=40

Comparing equation 3

*x*−5

*y*=20 with

*a*

_{1}

*x*+

*b*

_{1}

*y*+

*c*

_{1}=0 and 6

*x*−10

*y*=40 with

*a*

_{2}

*x*+

*b*

_{2}

*y*+

*c*

_{2}=0,

We get

*a*

_{1}= 3,

*b*

_{1}= –5,

*c*

_{1}=−20,

*a*

_{2}= 6,

*b*

_{2}= –10,

*c*

_{2}=−40

Here

It means lines coincide with each other.

Hence, there are infinite many solutions.

**(iv)**

*x*−3

*y*– 7=0

3

*x*−3

*y*– 15=0

Comparing equation

*x*−3

*y*– 7=0 with

*a*

_{1}

*x*+

*b*

_{1}

*y*+

*c*

_{1}=0 and 3

*x*−3

*y*– 15=0 with

*a*

_{2}

*x*+

*b*

_{2}

*y*+

*c*

_{2}=0,

We get

*a*

_{1}= 1,

*b*

_{1}= –3,

*c*

_{1}=−7,

*a*

_{2}= 3,

*b*

_{2}= –3,

*c*

_{2}=−15

Here this means that we have unique solution for these equations.

⇒

⇒

⇒

*x*= 4 and

*y*= –1

**7. Form the pair of linear equations in the following problems 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 Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.**

**(ii) A fraction becomes when 1 is subtracted from the numerator and it becomes 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 fixed monthly charge = Rs x and let charge of food for one day = Rs y

According to given conditions,

*x *+ 20*y *= 1000 … (1),

and *x *+ 26*y *= 1180 … (2)

Subtracting equation (1) from equation (2), we get

6*y *= 180

⇒ *y *= 30

Putting value of y in (1), we get

*x *+ 20 (30) = 1000

⇒ *x *= 1000 – 600 = 400

Therefore, fixed monthly charges = Rs 400 and, charges of food for one day = Rs 30

**(ii)** Let numerator = *x*and let denominator = *y*

According to given conditions,

⇒ 3*x *– 3 = *y … *(1) 4*x *= *y *+ 8 … (1)

⇒ 3*x *– *y *= 3 … (1) 4*x *– *y *= 8 … (2)

Subtracting equation (1) from (2), we get

4*x *– *y *− (3*x *− *y*) = 8 – 3

⇒ *x *= 5

Putting value of x in (1), we get

3 (5) – *y *= 3

⇒ 15 – *y *= 3

⇒ *y *= 12

Therefore, numerator = 5 and, denominator = 12

It means fraction =

**(iii)** Let number of correct answers = *x* and let number of wrong answers = *y*

According to given conditions,

3*x *– *y *= 40 … (1)

And, 4*x *− 2*y *= 50 … (2)

From equation (1), *y *= 3*x *– 40

Putting this in (2), we get

4*x *– 2 (3*x *− 40) = 50

⇒ 4*x *− 6*x *+ 80 = 50

⇒ −2*x *= −30

⇒ *x *= 15

Putting value of x in (1), we get

3 (15) – *y *= 40

⇒ 45 – *y *= 40

⇒ *y *= 45 – 40 = 5

Therefore, number of correct answers = *x *= 15and number of wrong answers = *y *= 5

Total questions = *x *+ *y *= 15 + 5 = 20

**(iv)** Let speed of car which starts from part A = x km/hr

Let speed of car which starts from part B = y km/hr

According to given conditions,

(Assuming x > y)

⇒ 5*x *− 5*y *= 100

⇒ *x *– *y *= 20 … (1)

And,

⇒ *x *+ *y *= 100 … (2)

Adding (1) and (2), we get

2*x *= 120

⇒ *x *= 60 km/hr

Putting value of x in (1), we get

60 – *y*= 20

⇒ *y *= 60 – 20 = 40 km/hr

Therefore, speed of car starting from point A = 60 km/hr

And, Speed of car starting from point B = 40 km/hr

**(v)** Let length of rectangle = x units and Let breadth of rectangle = y units

Area =*xy* square *units. *According to given conditions,

*xy*– 9 = (*x *− 5) (*y *+ 3)

⇒ *xy*– 9 = *xy *+ 3*x *− 5*y *– 15

⇒ 3*x *− 5*y *= 6 … (1)

And, *xy *+ 67 = (*x *+ 3) (*y *+ 2)

⇒ *xy*+ 67 = *xy *+ 2*x *+ 3*y *+ 6

⇒ 2*x *+ 3*y *= 61 … (2)

From equation (1),

3*x *= 6 + 5*y*

⇒ *x *=

Putting this in (2), we get

2 + 3*y *= 61

⇒ 12 + 10y + 9y = 183

⇒ 19*y *= 171

⇒ *y *= 9 units

Putting value of y in (2), we get

2*x *+ 3 (9) = 61

⇒ 2*x *= 61 – 27 = 34

⇒ *x *= 17 units

Therefore, length = 17 units and, breadth = 9 units

**8. Solve the following pairs of equations by reducing them to a pair of linear equations:**

**(i) **

**(ii) **

**(iii) +3 y=14**

**−4**

*y*=23**(iv)**

**(v) 7**

*x*−2*y*=5*xy***8**

*x*+7*y*=15*xy***(vi) 6**

*x*+3*y*−6*xy*=0**2**

*x*+4*y*−5*xy*=0**(vii)**

**(viii)**

**Ans. (i)**… (1)

… (2)

Let =

*p*and =

*q*

Putting this in equation (1) and (2), we get

⇒ 3

*p*+2

*q*=12 and 6(2

*p*+3

*q*)=13(6)

⇒ 3

*p*+2

*q*=12 and 2p + 3q =13

⇒ 3

*p*+2

*q*– 12=0 … (3) and 2

*p*+3

*q*– 13=0 … (4)

⇒

⇒

⇒

⇒

*p*=2 and

*q*=3

But =

*p*and =

*q*

Putting value of p and q in this we get

*x*= and

*y*=

**(ii)**… (1)

… (2)

Let =

*p*and =

*q*

Putting this in (1) and (2), we get

2

*p*+3

*q*=2 … (3)

4

*p*−9

*q*=−1 … (4)

Multiplying (3) by 2 and subtracting it from (4), we get

4

*p*−9

*q*+1 – 2(2

*p*+3

*q*−2)=0

⇒ 4

*p*−9

*q*+1−4

*p*−6

*q*+4=0

⇒ −15

*q*+5=0

⇒

*q*=

Putting value of q in (3), we get

2

*p*+1=2

⇒ 2

*p*=1 ⇒

*p*= ½

Putting values of p and q in (=

*p*and =

*q*), we get

and

⇒

⇒

*x*=4 and

*y*=9

**(iii)**+3

*y*=14 … (1)

−4

*y*=23 … (2) and Let =

*p …*(3)

Putting (3) in (1) and (2), we get

4

*p*+3

*y*=14 … (4)

3

*p*−4

*y*=23 … (5)

Multiplying (4) by 3 and (5) by 4, we get

3(4

*p*+3

*y*– 14= 0) and, 4(3

*p*−4

*y*– 23= 0)

⇒ 12

*p*+9

*y*– 42=0 … (6) 12

*p*−16

*y*– 92=0 … (7)

Subtracting (7) from (6), we get

9

*y*−(−16

*y*) – 42−(−92)=0

⇒ 25

*y*+50=0

⇒

*y*=50 – 25=−2

Putting value of y in (4), we get

4

*p*+3(−2)=14

⇒ 4

*p*– 6=14

⇒ 4

*p*=20

⇒

*p*=5

Putting value of p in (3), we get

=5

⇒

*x*=

Therefore,

*x*= and

*y*=−2

**(iv)**… (1)

… (2)

Let

Putting this in (1) and (2), we get

5

*p*+

*q*=2

⇒ 5

*p*+

*q*– 2=0 … (3)

And, 6

*p*−3

*q*=1

⇒ 6

*p*−3

*q*– 1=0 … (4)

Multiplying (3) by 3 and adding it to (4), we get

3(5

*p*+

*q*−2)+6

*p*−3

*q*– 1=0

⇒ 15

*p*+3

*q*– 6+6

*p*−3

*q*– 1=0

⇒ 21

*p*– 7=0

⇒

*p*=

Putting this in (3), we get

5()+

*q*– 2=0

⇒ 5+ 3

*q*= 6

⇒ 3q = 6 – 5 = 1

⇒

*q*=

Putting values of p and q in (), we get

⇒ 3=

*x*−1 and 3=

*y*– 2

⇒

*x*=4 and

*y*=5

**(v)**7

*x*−2

*y*=5

*xy …*(1)

8

*x*+7

*y*=15

*xy …*(2)

Dividing both the equations by xy, we get

Let =

*p*and =

*q*

Putting these in (3) and (4), we get

7

*q*−2

*p*=5 … (5)

8

*q*+7

*p*=15 … (6)

From equation (5),

2

*p*=7

*q*– 5

⇒

*p*=

Putting value of p in (6), we get

8

*q*+7()=15

⇒ 16

*q*+49

*q*– 35= 30

⇒ 65

*q*= 30 +35=65

⇒

*q*=1

Putting value of q in (5), we get

7(1)−2

*p*=5

⇒ 2

*p*=2 ⇒

*p*=1

Putting value of p and q in (=

*p*and =

*q*), we get

*x*=1 and

*y*=1

**(vi)**6

*x*+3

*y*−6

*xy*=0 … (1)

2

*x*+4

*y*−5

*xy*=0 … (2)

Dividing both the equations by xy, we get

Let =

*p*and =

*q*

Putting these in (3) and (4), we get

6

*q*+3

*p*– 6=0 … (5)

2

*q*+4

*p*– 5=0 … (6)

From (5),

3

*p*=6−6

*q*

⇒

*p*=2−2

*q*

Putting this in (6), we get

2

*q*+4(2−2

*q*) – 5=0

⇒ 2

*q*+8−8

*q*– 5=0

⇒ −6

*q*=−3

⇒

*q*= ½

Putting value of q in (p=2 – 2q), we get

*p*=2 – 2 (½)=2 – 1=1

Putting values of p and q in (=

*p*and =

*q*), we get

*x*=1 and

*y*=2

**(vii)**… (1)

… (2)

Let

Putting this in (1) and (2), we get

10

*p*+2

*q*=4 … (3)

15

*p*−5

*q*=−2 … (4)

From equation (3),

2

*q*=4−10

*p*

⇒

*q*=2−5

*p …*(5)

Putting this in (4), we get

15

*p*– 5(2−5

*p*)=−2

⇒ 15

*p*– 10+25

*p*=−2

⇒ 40

*p*=8

⇒

*p*=

Putting value of p in (5), we get

*q*=2 – 5()=2 – 1=1

Putting values of p and q in (),we get

⇒

*x*+

*y*=5 … (6) and

*x*–

*y*=1 … (7)

Adding (6) and (7), we get

2

*x*=6

⇒

*x*=3

Putting

*x*=3 in (7), we get

3 –

*y*=1

⇒

*y*=3 – 1=2

Therefore,

*x*=3 and

*y*=2

**(viii)**… (1)

… (2)

Let

Putting this in (1) and (2), we get

*p*+

*q*= and

⇒ 4

*p*+4

*q*=3 … (3) and 4

*p*−4

*q*=−1 … (4)

Adding (3) and (4), we get

8

*p*=2

⇒

*p*= ¼

Putting value of p in (3), we get

4 (¼)+4

*q*=3

⇒ 1+4

*q*=3

⇒ 4

*q*=3 – 1=2

⇒

*q*= ½

Putting value of p and q in , we get

⇒ 3

*x*+

*y*=4 … (5) and 3

*x*–

*y*=2 … (6)

Adding (5) and (6), we get

6

*x*=6

⇒

*x*=1

Putting

*x*=1 in (5) , we get

3(1)+

*y*=4

⇒

*y*=4 – 3=1

Therefore,

*x*=1 and

*y*=1

**9. Formulate the following problems as a part of equations, and hence find their solutions.**

**(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 the 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 rowing in still water =*x* km/h

Let speed of current =*y* km/h

So, speed of rowing downstream =(*x*+*y*) km/h

And, speed of rowing upstream =(*x*−*y*) km/h

According to given conditions,

⇒ 2*x*+2*y*=20 and 2*x*−2*y*=4

⇒ *x*+*y*=10 … (1) and *x*– *y*=2 … (2)

Adding (1) and (2), we get

2*x*=12

⇒ *x*=6

Putting *x*=6 in (1), we get

6+*y*=10

⇒ *y*=10 – 6=4

Therefore, speed of rowing in still water =6 km/h

Speed of current =4 km/h

**(ii)** Let time taken by 1 woman alone to finish the work =*x* days

Let time taken by 1 man alone to finish the work =*y* days

So, 1 woman’s 1 day work =()*th* part of the work

And, 1 man’s 1 day work =()*th* part of the work

So, 2 women’s 1 day work =()*th* part of the work

And, 5 men’s 1 day work =()*th* part of the work

Therefore, 2 women and 5 men’s 1 day work =(+)*th* part of the work… (1)

It is given that 2 women and 5 men complete work in = 4 days

It means that in 1 day, they will be completing *th* part of the work … (2)

Clearly, we can see that (1) = (2)

⇒ … (3)

Similarly, … (4)

Let

Putting this in (3) and (4), we get

2*p*+5*q*= and 3*p*+6*q*=

⇒ 8*p*+20*q*=1 … (5) and 9*p*+18*q*=1 … (6)

Multiplying (5) by 9 and (6) by 8, we get

72*p*+180*q*=9 … (7)

72*p*+144*q*=8 … (8)

Subtracting (8) from (7), we get

36*q*=1

⇒ *q*=

Putting this in (6), we get

9*p*+18()=1

⇒ 9*p*= ½

⇒ p =

Putting values of p and q in , we get *x*=18 and *y*=36

Therefore, 1 woman completes work in =18 days

And, 1 man completes work in =36 days

**(iii)** Let speed of train =*x* km/h and let speed of bus =*y* km/h

According to given conditions,

Let

Putting this in the above equations, we get

60*p*+240*q*=4 … (1)

And 100*p*+200*q*= … (2)

Multiplying (1) by 5 and (2) by 3, we get

300*p*+1200*q*=20 … (3)

300*p*+600*q*= … (4)

Subtracting (4) from (3), we get

600*q*=20−=7.5

⇒ *q*=

Putting value of q in (2), we get

100*p*+200()=

⇒ 100*p*+ 2.5 =

⇒ 100*p*=– 2.5

⇒ *p*=

But

Therefore, *x*=km/h and *y*=km/h

Therefore, speed of train =60 km/h

And, speed of bus =80 km/h

**10. Solve the following pair of linear equations:**

**(i) **

**(ii) **

**(iii) **

**(iv) **

**(v) **

**Ans. (i) ** … (1)

… (2)

Multiplying equation (1) by p and equation (2) by q, we obtain:

… (3)

… (4)

Adding equations (3) and (4), we obtain:

= 1

Substituting the value of in equation (1), we obtain:

Hence the required solution is x = 1 and y = –1.

**(ii)** … (1)

… (2)

Multiplying equation (1) by a and equation (2) by b, we obtain:

… (3)

… (4)

Subtracting equation (4) from equation (3),

Substituting the value of x in equation (1), we obtain:

**(iii)** ……..(1)

……..(2)

Multiplying equation (1) and (2) by b and a respectively, we obtain:

……..(3)

……..(4)

Adding equations (3) and (4), we obtain:

Substituting the value of in equation (1), we obtain:

**(iv)** … (1)

……..(2)

Subtracting equation (2) from (1), we obtain:

Substituting the value of in equation (1), we obtain:

**(v) **152x – 378y = –74 … (1)

–378x + 152y = –604 … (2)

Adding the equations (1) and (2), we obtain:

–226x – 226y = –678

x + y = 3 ………(3)

Subtracting the equation (2) from equation (1), we obtain:

530x – 530y = 530

x – y = 1 ……..(4)

Adding equations (3) and (4), we obtain:

2x = 4

x = 2

Substituting the value of x in equation (3), we obtain:

y = 1

**11. Abdul travelled 300 km by train and 200 km by taxi, it took him 5 hours 30 minutes. But if he travels 260 km by train and 240 km by taxi he takes 6 minute longer. Find the speed of the train and that of the taxi**

**Ans. **

According to question,

**12. If in a rectangle the length is increased and breadth is decreased by 2 units each, the area is reduced by 28 square units, and if the length is reduced by 1 unit and breadth is increased by 2 units, the area increased by 33 square units. Find the dimensions of the rectangle.**

**Ans. **

**13. On comparing the ratios , find out whether the following pair of linear equations are consistent, or inconsistent.**

**(i) 3 x + 2y = 5**

**2**

*x*− 3*y*= 7**(ii) 2**

*x*− 3*y*= 8**4**

*x*− 6*y*= 9**(iii)**

**9**

*x*− 10*y*= 14**(iv) 5**

*x*− 3*y*= 11**−10**

*x*+ 6*y*= −2**Ans. (i)**3

*x*+2

*y*=5, 2

*x*−3

*y*=7

Comparing equation 3

*x*+2

*y*=5 with

*a*

_{1}

*x*+

*b*

_{1}

*y*+

*c*

_{1}=0and 2

*x*−3

*y*=7 with

*a*

_{2}

*x*+

*b*

_{2}

*y*+

*c*

_{2}=0,

We get,

*a*

_{1}= 3,

*b*

_{1}= 2,

*c*

_{1}= 5,

*a*

_{2}= 2,

*b*

_{2}= –3,

*c*

_{2}= –7

and

Here which means equations have unique solution.

Hence they are consistent.

**(ii)**2

*x*−3

*y*=8, 4

*x*−6

*y*=9

Comparing equation 2

*x*−3

*y*=8 with

*a*

_{1}

*x*+

*b*

_{1}

*y*+

*c*

_{1}=0and 4

*x*−6

*y*=9 with

*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

Here because

Therefore, equations have no solution because they are parallel.

Hence, they are inconsistent.

**(iii)**9

*x*− 10

*y*= 14

Comparing equation with

*a*

_{1}

*x*+

*b*

_{1}

*y*+

*c*

_{1}=0and 9

*x*− 10

*y*= 14 with

*a*

_{2}

*x*+

*b*

_{2}

*y*+

*c*

_{2}=0,

We get,

*a*

_{1}=,

*b*

_{1}=,

*c*

_{1}= –14,

*a*

_{2}= 9,

*b*

_{2}= –10,

*c*

_{2}= –14

and

Here

Therefore, equations have unique solution.

Hence, they are consistent.

**(iv)**5

*x*−3

*y*=11, −10

*x*+6

*y*=−22

Comparing equation 5

*x*−3

*y*=11 with

*a*

_{1}

*x*+

*b*

_{1}

*y*+

*c*

_{1}=0and −10

*x*+6

*y*=−22 with

*a*

_{2}

*x*+

*b*

_{2}

*y*+

*c*

_{2}=0,

We get,

*a*

_{1}= 5,

*b*

_{1}= –3,

*c*

_{1}= –11,

*a*

_{2}= –10,

*b*

_{2}= 6,

*c*

_{2}= 22

and

Here

Therefore, the lines have infinite many solutions.

Hence, they are consistent.

**14. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method.**

**(i) x – 3y – 3 = 0**

**3x – 9y – 2 = 0**

**(ii) 2 x + y = 5**

**3**

*x*+ 2*y*= 8**(iii) 3**

*x*− 5*y*= 20**6**

*x*− 10*y*= 40**(iv)**

*x*− 3*y*– 7 = 0**3**

*x*− 3*y*– 15 = 0**Ans**.

**(i)**

*x*−3

*y*– 3=0

3

*x*−9

*y*– 2=0

Comparing equation

*x*−3

*y*– 3=0 with

*a*

_{1}

*x*+

*b*

_{1}

*y*+

*c*

_{1}=0 and 3

*x*−9

*y*– 2=0 with

*a*

_{2}

*x*+

*b*

_{2}

*y*+

*c*

_{2}=0,

We get

*a*

_{1}=1,

*b*

_{1}= −3,

*c*

_{1}=−3,

*a*

_{2}=3,

*b*

_{2}=−9,

*c*

_{2}=−2

Here this means that the two lines are parallel.

Therefore, there is no solution for the given equations i.e. it is inconsistent.

**(ii)**2

*x*+

*y*=5

3

*x*+2

*y*=8

Comparing equation 2

*x*+

*y*=5 with

*a*

_{1}

*x*+

*b*

_{1}

*y*+

*c*

_{1}=0 and 3

*x*+2

*y*=8 with

*a*

_{2}

*x*+

*b*

_{2}

*y*+

*c*

_{2}=0,

We get

*a*

_{1}= 2,

*b*

_{1}= 1,

*c*

_{1}=−5,

*a*

_{2}=3,

*b*

_{2}= 2,

*c*

_{2}=−8

Here this means that there is unique solution for the given equations.

⇒

⇒

*x*=2 and

*y*=1

**(iii)**3

*x*−5

*y*=20

6

*x*−10

*y*=40

Comparing equation 3

*x*−5

*y*=20 with

*a*

_{1}

*x*+

*b*

_{1}

*y*+

*c*

_{1}=0 and 6

*x*−10

*y*=40 with

*a*

_{2}

*x*+

*b*

_{2}

*y*+

*c*

_{2}=0,

We get

*a*

_{1}= 3,

*b*

_{1}= –5,

*c*

_{1}=−20,

*a*

_{2}= 6,

*b*

_{2}= –10,

*c*

_{2}=−40

Here

It means lines coincide with each other.

Hence, there are infinite many solutions.

**(iv)**

*x*−3

*y*– 7=0

3

*x*−3

*y*– 15=0

Comparing equation

*x*−3

*y*– 7=0 with

*a*

_{1}

*x*+

*b*

_{1}

*y*+

*c*

_{1}=0 and 3

*x*−3

*y*– 15=0 with

*a*

_{2}

*x*+

*b*

_{2}

*y*+

*c*

_{2}=0,

We get

*a*

_{1}= 1,

*b*

_{1}= –3,

*c*

_{1}=−7,

*a*

_{2}= 3,

*b*

_{2}= –3,

*c*

_{2}=−15

Here this means that we have unique solution for these equations.

⇒

⇒

⇒

*x*= 4 and

*y*= –1

**15. Solve the following pair of linear equations by the substitution and cross-multiplication methods:**

**8 x+5y=9**

**3**

*x*+2*y*=4**Ans. Substitution Method**

8

*x*+5

*y*=9 …

**(1)**

3

*x*+2

*y*=4 …

**(2)**

From equation

**(1)**,

5

*y*=9−8

*x*

⇒

*y*=

Putting this in equation

**(2)**, we get

3

*x*+2=4

⇒ 3

*x*+=4

⇒ 3

*x*−

⇒ 15

*x*−16

*x*=20 – 18

⇒

*x*=−2

Putting value of

**x**in

**(1)**, we get

8(−2)+5

*y*=9

⇒ 5

*y*=9+16=25

⇒

*y*=5

Therefore,

*x*=−2 and

*y*=5

**Cross multiplication method**

8

*x*+5

*y*=9 …

**(1)**

3

*x*+2

*y*=4 …

**(2)**

⇒

⇒

⇒

*x*=−2 and

*y*=5

**16. In aABC, C = 3B = 2(A + B). Find three angles**

**Ans. **C = 3B = 2(A + B)

Taking 3B = 2(A + B)

B = 2A

2A – B = 0 …….(1)

We know that the sum of the measures of all angles of a triangle is 180°.

A + B + C = 180°

A + B + 3B = 180°

A + 4B = 180° …….(2)

Multiplying equation (1) by 4, we obtain:

8A – 4B = 0 …….(3)

Adding equations (2) and (3), we get

9A = 180°

A = 20°

From eq. (2), we get,

20° + 4B = 180°

B = 40°

And C = 3 x 40°= 120°

Hence the measures of A, B and C are 20°, 40° and 120° respectively.