# Exercise 3.1

**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.**

**Answer:**

### Let the present age of Aftab and his daughter be x and y respectively.

Seven years ago, Age of Aftab = x – 7 and Age of his daughter = y – 7

According to the given condition,

Thus, the given conditions can be algebraically represented as:

x – 7y = –42

### And x – 3y = 6

### The graphical representation is as follows:

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

**Answer:<.**

### Let the cost of a bat and a ball be Rs x and Rs y respectively.

The given conditions can be algebraically represented as:

3x + 6y = 3900

x + 2y = 1300

### Three solutions of this equation can be written in a table as follows:

## x |
## 3900 |
## 1300 |
## -1300 |

## y |
## -1300 |
## 0 |
## 1300 |

### Three solutions of this equation can be written in a table as follows:

## x |
## 3900 |
## 1300 |
## -1300 |

## y |
## -1300 |
## 0 |
## 1300 |

### The graphical representation is as follows:

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

**Answer:**

### Let cost of 1 kg of apples = Rs x and let cost of 1 kg of grapes= Rs y

### According to given conditions, we have

### 2x + y = 160… (1)

### 4x + 2y = 300

### ⇒ 2x + y = 150… (2)

### So, we have equations (1) and (2), 2x + y = 160 and 2x + y = 150 which represent given situation algebraically.

### For equation 2x + y = 160, we have following points which lie on the line

### We plot the points for both of the equations and it is the graphical representation of the given situation.

# EXERCISE 3.2

**1. Form the pair of linear equations in the following problems, and find their solutions graphically.**

**(i) 10 students of class X took part in a 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 Rs 50, whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.**

**Answer:**

#### (i) Let number of boys = x

#### Let number of girls = y

#### According to given conditions, we have

#### x + y = 10

#### And, x = 10 – y

#### putting y=0,5,10,we get,

X=10-0=10

X=10-5=5,

X=10-10=0

## x |
## 10 |
## 5 |
## 0 |

## y |
## 0 |
## 5 |
## 10 |

#### Number of girls is 4 more than number of boys ……..Given,

so,

Y=x+4

putting x=-4,0,4 we get,

Y=-4+4=0

Y=0+4

Y=4+4=8

## x |
## -4 |
## 0 |
## 4 |

## y |
## 0 |
## 4 |
## 8 |

#### We plot the points for both of the equations to find the solution.

#### (ii)

#### Let the cost of one pencil=Rs.X

and Let the cost of one pen=Rs.Y

According to the given conditions, we have:

=5x + 7y = 50

=5x=50-7y

=x=10-7/5y

Three solutions of this equation can be written in a table as follows:

x | 3 | -4 | -11 |

y | 5 | 10 | 15 |

####

Three solutions of this equation can be written in a table as follows:

x | 0 | 2 | 4 |

y | 9.2 | 6.4 | 3.6 |

#### The graphical representation is as follows:

**2. On comparing the ratios a**_{1}/a_{2},b_{1}/b_{2} and c_{1}/c_{2} , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

_{1}/a

_{2},b

_{1}/b

_{2}and c

_{1}/c

_{2}, 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

#### (ii)9x + 3y + 12 = 0

#### 7x + 6y – 9 = 018x + 6y + 24 = 0

#### (iii) 6x − 3y + 10 = 0

#### 2x – y + 9 = 0

**Answer:**

#### (i) 5x − 4y + 8 = 0, 7x + 6y – 9 = 0

#### Comparing equation 5x − 4y + 8 = 0 with a_{1}x + b_{1}y + c_{1} = 0and 7x + 6y – 9 = 0 with a_{2}x + b_{2}y + c_{2} = 0,

#### We get,

#### Hence,

#### we find that,

#### (ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0

#### Comparing equation 9x + 3y + 12 = 0 with a_{1}x + b_{1}y + c_{1} = 0and 7x + 6y – 9 = 0 with a_{2}x + b_{2}y + c_{2} = 0,

#### We get,

#### Hence

#### We find that,

#### Hence, lines are coincident.

#### (iii) 6x − 3y + 10 = 0, 2x – y + 9 = 0

#### Comparing equation 6x − 3y + 10 = 0 with a_{1}x + b_{1}y + c_{1} = 0and 7x + 6y – 9 = 0 with a_{2}x + b_{2}y + c_{2} = 0,

#### We get,

#### Hence

#### We find that,

#### Hence,

lines are parallel to each other.

**3. On comparing the ratios a**_{1}/a_{2},b_{1}/b_{2} and c_{1}/c_{2} , find out whether the following pair of linear equations are consistent, or inconsistent.

_{1}/a

_{2},b

_{1}/b

_{2}and c

_{1}/c

_{2}, find out whether the following pair of linear equations are consistent, or inconsistent.

#### (i) 3x + 2y = 5, 2x − 3y = 8

#### (ii) 2x − 3y = 7, 4x − 6y = 9

#### (iii) 3x/2 + 5y/3 = 7, 9x − 10y = 14

#### (iv) 5x − 3y = 11, −10x + 6y = −22

**Answer:**

#### (i) 3x + 2y = 5, 2x − 3y = 7

#### Comparing equation 3x + 2y = 5 with a_{1}x + b_{1}y + c_{1} = 0and 7x + 6y – 9 = 0 with a_{2}x + b_{2}y + c_{2} = 0,

#### We get,

#### Hence,

#### Therefore,these linear equations will intersect at one point only and have only one possible solution.

And,pair of linear euations is consistent

#### (ii) 2x − 3y = 8, 4x − 6y = 9

#### Comparing equation 2x − 3y = 8 with a_{1}x + b_{1}y + c_{1} = 0and 7x + 6y – 9 = 0 with a_{2}x + b_{2}y + c_{2} = 0,

#### We get,

#### Hence,

#### Therefore,these linear equations are parallel to each other and have no possible solution.in

And,pair of linear euations is inconsistent

#### (iii) 9x − 10y = 14

#### We get,

#### Hence,

#### Therefore, these linear equations will intersect each other at one point and have only one possible solution.

And,pair of linear euations is consistent

#### (iv) 5x − 3y = 11, −10x + 6y = −22

#### Comparing equation 5x − 3y = 11 with a_{1}x + b_{1}y + c_{1} = 0and 7x + 6y – 9 = 0 with a_{2}x + b_{2}y + c_{2} = 0,

#### We get,

#### Hence,

#### Therefore these pair of lines have infinite number of solutions

And,pair of linear euations is consistent

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

**4. (i) x + y = 5, 2x + 2y = 10**

**(ii) x – y = 8, 3x − 3y = 16**

**(iii) 2x + y = 6, 4x − 2y = 4**

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

**Answer:**

#### (i) x + y = 5, 2x + 2y = 10

#### We get,

#### Hence,

#### (ii) x – y = 8, 3x − 3y = 16

#### We get,

#### Hence,

#### Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution.

Hence,the pair of linear equations is inconsistent.

#### (iii) 2x + y = 6, 4x − 2y = 4

#### We get,

#### Hence,

#### Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution.

Hence,the pair of linear equations is consistent

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

#### We get,

#### Hence,

#### Therefore, these linear equations are parallel to each other and have no possible solution,

Hence,the pair of linear equations is inconsistent.

**5. Half the perimeter of a rectangle garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.**

**Answer:**

#### Let width of rectangular garden = x metres

and length=y

#### So,

#### Hence, the graphic representation is as follows.

**6. 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

**Answer:**

#### (i) Let the second line be equal to a_{2}x + b_{2}y + c_{2} = 0,

#### Intersecting Lines:For this Condition,

The Second line such that it is intersecting the given line is

2x+4y-6=0

#### As,

#### (ii) Let the second line be equal to a_{2}x + b_{2}y + c_{2 }= 0,

#### parallel Lines:

For this Condition,

Hence,the second line can be 4x+6y-8=0

#### As,

#### (iii) Let the second line be equal to a_{2}x + b_{2}y + c_{2 }= 0,

#### Coincident lines: For coincident lines,

Hence,the second line can be 6x+9y-24=0

#### As,

**7. 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.**

**Answer:**

#### For equation x – y + 1 = 0, we have following points which lie on the line

#### For equation 3x + 2y – 12 = 0, we have following points which lie on the line

#### We can see from the graphs that points of intersection of the lines with the x–axis are (–1, 0), (2, 3) and (4, 0).

# EXERCISE 3.3

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

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

#### x – y = 4

#### (ii) s – t = 3

#### s/3 + t/2 = 6

#### (iii) 3x – y = 3

#### 9x − 3y = 9

#### (iv)0.2x + 0.3y = 1.3

#### 0.4x + 0.5y = 2.3

#### (v) √2x+ √3y = 0

√3x – √8y = 0

#### (vi) 3x/2 – 5y/3 = -2

#### x/3 + y/2 = 13/6

**Answer:**

#### (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

#### ⇒ 2y = 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

#### ⇒ 2t + 6 + 3t =36

#### ⇒ 5t + 6 = 36

#### ⇒ 5t = 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) 3x – y = 3 … (i)

#### 9x − 3y = 9 … (ii)

#### From equation (i),we get,

y =3x − 3 … (iii)

#### putting value of y from equation(iii) to equation(ii)

9x-3(3x-3)=9

=9x-9x+9=9

=9=9

#### This is always true,and pair of these equations have infinite possible solutions.

Therefore one possible solutions is x=1 and y=0

#### (iv) 0.2x + 0.3y = 1.3 … (1)

#### 0.4x + 0.5y = 2.3 … (2)

#### Using equation (1), we can say that

#### 0.2x = 1.3 − 0.3y

#### ⇒ x =

#### Putting this in equation (2), we get

#### 0.4x(6.5-1.5y)+ 0.5y = 2.3

#### ⇒ 2.6 − 0.6y + 0.5y = 2.3

#### ⇒ −0.1y = −0.3 ⇒ y = 3

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

#### 0.2x + 0.3 (3) = 1.3

#### ⇒ 0.2x + 0.9 = 1.3

#### ⇒ 0.2x = 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 =

#### 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

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

#### y = mx + 3.

**Answer:**

#### 2x + 3y = 11 … (1)

#### 2x − 4y = −24 … (2)

#### Using equation (2), we can say that

#### 2x = −24 + 4y

#### ⇒ x = −12 + 2y

#### Putting this in equation (1), we get

#### 2 (−12 + 2y) + 3y = 11

#### ⇒ −24 + 4y + 3y = 11

#### ⇒ 7y = 35 ⇒ y = 5

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

#### 2x + 3 (5) = 11

#### ⇒ 2x + 15 = 11

#### ⇒ 2x = 11 – 15 = −4⇒ x = −2

#### Therefore, x = −2 and y = 5

#### Putting values of x and y in y = mx + 3, we get

#### 5 = m (−2) + 3

#### ⇒ 5 = −2m + 3

#### ⇒ −2m = 2 ⇒ m = −1

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

**3. (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 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and denominator it becomes 5/6 . 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?**

**Answer:**

#### (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 = 3y(x > y)… (2)

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

#### 3y – y = 26

#### ⇒ 2y = 26

#### ⇒ y = 13

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

#### x = 3y =

#### 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, (Sum of supplementary angles) … (2)

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

#### x + x + 18 = 180

#### ⇒ 2x = 180 – 18 = 162

#### ⇒ x=162/2

x=81

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

#### y = x + 18 = 81 + 18 =

#### Therefore, two angles are .

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

#### According to given conditions, we have

#### 7x + 6y = 3800 … (1)

#### And,3x + 5y = 1750 … (2)

#### Using equation (1), we can say that 7x+6y=3800

= 6y = 3800-7x

#### And,

#### Putting value of y from equation(1) to equation (2)

#### 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 + 10y = 105… (1)

#### x + 15y = 155… (2)

#### Using equation (1), we can say that

#### x = 105 − 10y

#### Putting this in equation (2), we get

#### 105 − 10y + 15y = 155

#### ⇒ 5y = 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,

Fraction becomes 9/11 , if 2 is added in both numerator and denominator

#### =11x+22 =9y+18 (by cross multiplication)

=11x =9y – 4

#### And, if 3 is added in both numerator and denominator

#### =6x+18 =5y+15 …..(ii) (by cross multiplication)

#### putting value of x from equation (i) to equation (ii)

#### Putting value of y in equation (i), we get

#### (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 = 3y + 15

#### ⇒ x = 10 + 3y

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

#### 10 + 3y – 5 = 7y − 35

#### ⇒ −4y = −40

#### ⇒ y = 10 years

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

#### x + 5 = 3 (10 + 5)

#### ⇒ x = 45 – 5 = 40 years

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

# EXERCISE 3.4

**1. 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) 3x − 5y – 4 = 0, 9x = 2y + 7

#### (iv) x/2 + 2y/3 = -1, x – y/3 = 3

**Answer:**

#### (i) x + y = 5 … (1)

#### 2x – 3y = 4 … (2)

**Elimination method:**

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

#### 2x + 2y = 10 … (3)

#### 2x − 3y = 4 … (2)

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

#### 5y = 6⇒ y = 6/5

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

#### x + 6/5= 5

#### ⇒ x = 5 − 6/5= 19/5

#### Therefore, x = 19/5 and y = 6/5

**Substitution method:**

#### x + y = 5 … (1)

#### 2x − 3y = 4 … (2)

#### From equation (1), we get,

#### x = 5 − y

#### Putting this in equation (2), we get

#### 2 (5 − y) − 3y = 4

#### ⇒ 10 − 2y − 3y = 4

#### ⇒ 5y = 6 ⇒ y = 6/5

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

#### x = 5 − 6/5= 19/5

#### Therefore, x = 19/5 and y = 6/5

#### (ii) 3x + 4y = 10… (1)

#### 2x – 2y = 2… (2)

**Elimination method:**

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

#### 4x − 4y = 4 … (3)

#### 3x + 4y = 10 … (1)

#### Adding (3) and (1), we get

#### 7x = 14⇒ x = 2

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

#### 3 (2) + 4y = 10

#### ⇒ 4y = 10 – 6 = 4

#### ⇒ y = 1

#### Therefore, x = 2 and y = 1

**Substitution method:**

#### 3x + 4y = 10… (1)

#### 2x − 2y = 2… (2)

#### From equation (2), we get

#### 2x = 2 + 2y

#### ⇒ x = 1 + y … (3)

#### Putting this in equation (1), we get

#### 3 (1 + y) + 4y = 10

#### ⇒ 3 + 3y + 4y = 10

#### ⇒ 7y = 7⇒ y = 1

#### Putting value of y in (3), we get x = 1 + 1 = 2

#### Therefore, x = 2 and y = 1

#### (iii) 3x − 5y – 4 = 0 … (1)

#### 9x = 2y + 7… (2)

**Elimination method:**

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

#### 9x − 15y – 12 = 0… (3)

#### 9x − 2y – 7 = 0… (2)

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

#### −13y – 5 = 0

#### ⇒ −13y = 5

#### ⇒ y = −5/13

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

#### 3x – 5 (−5/13)− 4 = 0

#### ⇒ 3x = 4 −

#### ⇒ x =

#### Therefore, x = and y =

**Substitution Method:**

#### 3x − 5y – 4 = 0 … (1)

#### 9x = 2y + 7… (2)

#### From equation (1), we can say that

#### 3x = 4 + 5y⇒ x =

#### Putting this in equation (2), we get

#### 9 − 2y = 7

#### ⇒ 12 + 15y − 2y = 7

#### ⇒ 13y = −5 ⇒ y =

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

#### 3x – 5 = 4

#### ⇒ 3x = 4 −

#### ⇒ x =

#### Therefore, x = and y =

#### (iv) … (1)

#### … (2)

**Elimination method:**

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

#### … (3)

#### substracting equation (2)from (3) we get

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

#### =x-4=-2

=x=2

**Substitution method:**

#### … (1)

#### … (2)

#### From equation (2), we can say that

#### Putting this in equation (1), we get

#### Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

**2. **(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 togive 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.

**Answer:**

#### (i) Let numerator =x and let denominator =y

#### According to given condition, we have

#### ⇒ x + 1 = y – 1 and 2x = y + 1

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

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

#### 2x − 2y = −4… (3)

#### 2x – 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 = 3y – 15

#### ⇒ x − 3y = −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 = 2y + 20

#### ⇒ x − 2y = 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 (10x + y) = 2 (10y + x)

#### ⇒ 90x + 9y = 20y + 2x

#### ⇒ 88x = 11y

#### ⇒ 8x = y

#### ⇒ 8x – y = 0 … (2)

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

#### 9x = 9⇒ x = 1

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

#### 1 + y = 9

#### ⇒ y = 9 – 1 = 8

#### Therefore, number = 10x + 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 100x + 50y = 2000

#### ⇒ 2x + 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 + 4y = 27 … (1)

#### x + 2y = 21 … (2)

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

#### 2y = 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

# Exercise 3.5

**1. 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)2x + y = 5

#### 3x + 2y = 8

#### (iii) 3x − 5y = 20

#### 6x − 10y = 40

#### (iv) x − 3y – 7 = 0

#### 3x − 3y – 15 = 0

**Answer:**

#### (i) x − 3y – 3 = 0

#### 3x − 9y – 2 = 0

#### Comparing equation x − 3y – 3 = 0 with a1x +b1y + c1 = 0 and 3x − 9y – 2 = 0 with a2x +b2y + c2 = 0,

#### We get

#### Here

#### Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations.

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

#### 3x + 2y = 8

#### Comparing equation 2x + y -5= 0 with a1x +b1y + c1 = 0 and 3x + 2y -8= 0 with a2x +b2y + c2 = 0,

#### We get

#### Herethis means that there is unique solution for the given equations.

#### By cross-multiplication method,

#### ⇒

#### ⇒ x = 2 and y = 1

#### (iii) 3x − 5y = 20

#### 6x − 10y = 40

#### Comparing equation 3x − 5y = 20 with a1x +b1y + c1 = 0 and 6x − 10y = 40 with a2x +b2y + c2 = 0 ,

#### We get

#### Here

#### It means lines coincide with each other.

#### Hence, there are infinite many solutions.

#### (iv) x − 3y – 7 = 0

#### 3x − 3y – 15 = 0

#### Comparing equation x − 3y – 7 = 0 with a1x +b1y + c1 = 0 and 3x − 3y – 15 = 0 with a2x +b2y + c2 = 0,

#### We get

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

#### By cross-multiplication,

#### ⇒

#### ⇒

#### ⇒ x = 4 and y = –1

**2. (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

**Answer:**

#### (i) Comparing equation 2x + 3y – 7 = 0 with a1x +b1y + c1 = 0 and (a − b) x + (a + b)

#### y − 3a – b + 2 = 0 with a2x +b2y + c2 = 0

#### We get andand

#### Linear equations have infinite many solutions if

#### ⇒

#### ⇒

#### ⇒ 2a + 2b = 3a − 3b and 6 − 3b − 9a = −7a − 7b

#### ⇒ a = 5b… (1) and −2a = −4b – 6… (2)

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

#### −2 (5b) = −4b – 6

#### ⇒ −10b + 4b = −6

#### ⇒ −6b = –6 ⇒ b = 1

#### Putting value of b in (1), we get

#### a = 5b = 5 (1) = 5

#### Therefore, a = 5 and b = 1

#### (ii) Comparing (3x + y – 1 = 0) with a1x +b1y + c1 = 0 and (2k − 1)x + (k − 1)y −2k – 1 = 0) with a2x +b2y + c2 = 0,

#### We get andand c2 = −2k − 1

#### Linear equations have no solution if

#### ⇒

#### ⇒

#### ⇒ 3 (k − 1) = 2k – 1

#### ⇒ 3k – 3 = 2k − 1

#### ⇒ k = 2

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

#### 8x + 5y = 9

#### 3x + 2y = 4

**Answer.**

#### Substitution Method:

#### 8x + 5y = 9 … (i)

#### 3x + 2y = 4 … (ii)

#### From equation (ii),we get

#### x = … (iii)

#### Substituting this value in equation (i), we obtain

32-16y+15y =27

-y =-5

#### Y = 5 … (iv)

#### Substituting this value in equation (ii), we obtain

#### 3x+10=4

x=-2

Hence, x=-2,y=5

#### Cross multiplication method

#### 8x + 5y = 9 … (1)

#### 3x + 2y = 4 … (2)

#### ⇒

#### ⇒

#### ⇒ x = −2 and y = 5

#### Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

**4. (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 1/3 when 1 is subtracted from the numerator and it becomes 1/4 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.**

**Answer:**

#### (i)Let fixed monthly charge = Rs x and let charge of food for one day = Rs y

#### According to given conditions,

#### x + 20y = 1000 … (1),

#### and x + 26y = 1180 … (2)

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

#### 6y = 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,

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

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

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

#### 4x – y − (3x − 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

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

#### According to given conditions,

#### 3x – y = 40 … (1)

#### And, 4x − 2y = 50 … (2)

#### From equation (1), y = 3x − 40

#### Putting this in (2), we get

#### 4x – 2 (3x − 40) = 50

#### ⇒ 4x − 6x + 80 = 50

#### ⇒ −2x = −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,5(x-y) =100(Assuming x > y)

#### (Assuming x > y)

#### ⇒ 5x − 5y = 100

#### ⇒ x – y = 20 … (1)

#### And,1(x+y) =100

#### ⇒ x + y = 100 … (2)

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

#### 2x = 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 + 3x − 5y – 15

#### ⇒ 3x − 5y = 6 … (1)

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

#### ⇒ xy + 67 = xy + 2x + 3y + 6

#### ⇒ 2x + 3y = 61 … (2)

#### By cross-multiplication method,we obtain,

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

# EXERCISE 3.6

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

#### (i) 1/2x + 1/3y = 2

1/3x + 1/2y = 13/6

(ii) 2/√x +3/√y = 2

4/√x – 9/√y = -1

(iii) 4/x + 3y = 14

3/x – 4y = 23

(iv) 5/x-1 + 1/y-2 = 2

6/x-1 – 3/y-2 = 1

(v) 7x-2y/xy = 5

8x + 7y/xy = 15

(vi) 6x + 3y = 6xy

2x + 4y = 5xy

(vii) 10/x+y + 2/x-y = 4

15/x+y – 5/x-y = -2

(viii) 1/3x+y + 1/3x-y = 3/4

1/2(3x-y) – 1/2(3x-y) = -1/8

**Answer:**

#### (i) … (1)

#### … (2)

#### Let =p and =q

#### then the equations becomes.

#### Using cross-multiplication method,we obtain,

#### ⇒p = 2 andq = 3

#### (ii)

#### Let =p and =q

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

#### 2p + 3q = 2 … (i)

#### 4p − 9q = −1 … (ii)

#### Multiplying (i) by 3, we get

#### 6p + 9q =6 … (iii)

#### Adding equation (ii) and (iii),we obtain

#### 10p =5

#### p =

#### Putting value of p in (i), we get

#### Hence,x = 4 and y = 9.

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

#### − 4y = 23 and Let =p

#### Putting value of p in equation, we get

#### 4p + 3y = 14 … (i)

#### 3p − 4y = 23 … (ii)

#### By cross-multiplication,we get

#### Also,

#### (iv)

#### putting= p and=q,we get,

#### =5p+q =2 … (i)

=6p-3q =1 … (ii)

#### Now, multiplying equation (i) by 3,we get

#### 15p +3q = 6 … (iii)

#### Adding equation (ii) and (iii)

#### 21 p = 7

#### p =

#### putting value of p in equation (iii),we get,

#### = -3q = -1

#### q =

#### we know that,

p =

= 3 = x – 1

= x = 4

#### q =

#### =y – 2 = 3

y = 5

#### Hence,x=4 and y =5

#### (v) 7x − 2y = 5xy … (1)

#### 8x + 7y = 15xy … (2)

#### Dividing both the equations by xy, we get

#### … (i)

#### … (ii)

#### putting =p and =q

#### we get,

#### 7q − 2p = 5 … (iii)

#### 8q + 7p = 15 … (iv)

#### multiplying equation (iii) by 7 and equation (iv) by 2,we get,

#### 49q – 14p =5 … (v)

#### 16q + 14p = 30 … (vi)

#### After adding equation (v) and (vi),we get,

65q = 65

=q = 1

#### putting value of q in equation (iv), we get,

#### 8 + 7p = 15

= 7p = 15 -8 = 7

= p =1

#### Now,

p =

#### q =

#### Hence,x = 1 and y = 1

#### (vi) 6x + 3y − 6xy = 0 … (1)

#### 2x + 4y − 5xy = 0 … (2)

#### Dividing both the equations by xy, we get

#### Let =p and =q

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

#### 6q + 3p – 6 = 0 … (5)

#### 2q + 4p – 5 = 0 … (6)

#### From (5),

#### 3p = 6 − 6q

#### ⇒p = 2 − 2q

#### Putting this in (6), we get

#### 2q + 4 (2 − 2q) – 5 = 0

#### ⇒ 2q + 8 − 8q – 5 = 0

#### ⇒ −6q = −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 getx = 1 andy = 2

#### (vii) … (1)

#### …(2)

#### Let

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

#### 10p + 2q = 4 … (3)

#### 15p − 5q = −2 … (4)

#### From equation (3),

#### 2q = 4 − 10p

#### ⇒q = 2 − 5p … (5)

#### Putting this in (4), we get

#### 15p – 5 (2 − 5p) = −2

#### ⇒ 15p – 10 + 25p = −2

#### ⇒ 40p = 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) andx –y = 1 … (7)

#### Adding (6) and (7), we get

#### 2x = 6 ⇒x = 3

#### Puttingx = 3 in (7), we get

#### 3 –y = 1

#### ⇒y = 3 – 1 = 2

#### Therefore,x = 3 andy = 2

#### (viii)… (1)

#### … (2)

#### Let

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

#### p +q = and

#### ⇒ 4p + 4q = 3 … (3) and 4p − 4q = −1 … (4)

#### Adding (3) and (4), we get

#### 8p = 2 ⇒p = ¼

#### Putting value of p in (3), we get

#### 4 (¼) + 4q = 3

#### ⇒ 1 + 4q = 3

#### ⇒ 4q = 3 – 1 = 2

#### ⇒q = ½

#### Putting value of p and q,we get,

#### ⇒ 3x +y = 4 … (5) and 3x –y = 2 … (6)

#### Adding (5) and (6), we get

#### 6x = 6 ⇒x = 1

#### Puttingx = 1 in (5) , we get

#### 3 (1) +y = 4

#### ⇒y = 4 – 3 = 1

#### Therefore,x = 1 andy = 1

**2. 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.

**Answer:**

#### (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+y) = 20 and 2(x-y) = 4

#### ⇒ 2x + 2y = 20 and 2x − 2y = 4

#### ⇒x +y = 10 … (1) andx –y = 2 … (2)

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

#### 2x = 12⇒x = 6

#### Puttingx = 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

#### Therefore,work done by a woman in 1 day =

#### work done by a man in 1 day =

#### According to the question,

#### putting = p and = q, in these equations

we obtain,

#### 2p + 5q =

8p + 20q = 1

#### 3p + 6q =

= 9p + 18q = 1

#### By cross-multiplication,we obtain

#### x = 18, y = 36

Hence, number of days taken by a woman = 18 Number of days taken by a man = 36

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

#### According to given conditions,

#### putting in these equations,we obtain

#### Putting this in the above equations, we get

#### 60p + 240q = 4 … (3)

#### And 100p + 200q =

#### 600p + 1200q = 25 … (4)

#### Multiplying (3) by 10, we get

#### 600p +2400q = 40 … (5)

#### substracting equation (4) from (5),we obtain

1200q =15

q =… (6)

#### Substituting in equation(3),we obtain

60p + 3 = 4

60p = 1

#### u = 60km/h and v = 80km/h

Hence,speed of train = 60 km/h

speed of bus = 80km/h

# EXERCISE 3.7

**1. The age of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.**

**Answer:**

#### The difference between the ages of Biju and Ani is 3 years. Either Biju is 3 years older than Ani or Ani is 3 years older than Biju. However, it is obvious that in both cases, Ani’s father’s age will be 30 years more than that of Cathy’s age.

#### Let the age of Ani and Biju be x and y years respectively.

#### Therefore, age of Ani’s father, Dharam = 2 × x = 2x years

#### Biju’s sister Cathy And age of = years

#### By using the information given in the question,

#### Case (I) When Ani is older than Biju by 3 years, x – y = 3 (i)

#### 2x –= 30

#### 4x – y = 60 (ii)

#### Subtracting (i) from (ii), we obtain 3x = 60 – 3 = 57

#### x –= 19

#### Therefore, age of Ani = 19 years

#### And age of Biju = 19 – 3 = 16 years

#### Case (II) When Biju is older than Ani, y – x = 3 (i)

#### 2x –= 30

#### 4x – y = 60 (ii)

#### Adding (i) and (ii), we obtain 3x = 63

#### x = 21

#### Therefore, age of Ani = 21 years

#### And age of Biju = 21 + 3 = 24 years

**2. One says, “Give me a hundred, friend! I shall then become twice as rich as you.” The other replies, “If you give me ten, I shall be six times as rich as you.” Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II]**

**Answer:**

#### Let the money with the first person and second person be Rs x and Rs y respectively.

According to the question,

x + 100 = 2(y – 100)

x + 100 = 2y – 200

x – 2y = – 300 … (1)

6(x – 10) = (y + 10)

6x – 60 = y + 10

6x – y = 70 … (2)

Multiplying equation (2) by 2, we obtain:

12x – 2y = 140 … (3)

Subtracting equation (1) from equation (3), we obtain:

11x = 140 + 300

11x = 440

x = 40

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

40 – 2y = -300

40 + 300 = 2y

2y = 340

y = 170

Thus, the two friends had Rs 40 and Rs 170 with them.

**3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.**

**Answer:**

#### Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km.

#### We know that, Speed =

#### x =

#### Or, d = xt (i)

#### Using the information given in the question, we obtain

#### (x + 10) =

#### (x+10) (t-2) = d

#### Xt+10t-2x-20 = d

#### By using equation (i),

#### we obtain – 2x + 10t = 20 (ii)

#### (x-10) =

#### (x-10) (t+3) = d

#### Xt-10t+3x-30 = d

#### By using equation (i), we

#### obtain 3x – 10t = 30 (iii)

#### Adding equations (ii) and (iii),

#### we obtain x = 50

#### Using equation (ii), we

#### obtain ( – 2) × (50) + 10t = 20

#### – 100 + 10t =

#### 10t = 120

#### t = 12 hours

#### From equation (i), we obtain

#### Distance to travel = d = xt

#### = 50 × 12

#### = 600 km

#### Hence, the distance covered by the train is 600 km.

**4. 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.**

**Answer:**

#### Let the number of rows be x and number of students in a row be y.

Total number of students in the class = Number of rows x Number of students in a row

= xy

According to the question,

Total number of students = (x – 1) (y + 3)

xy = (x – 1) (y + 3)

= xy – y + 3x – 3

3x – y – 3 = 0

3x – y = 3 … (1)

Total number of students = (x + 2) (y – 3)

xy = xy + 2y – 3x – 6

3x – 2y = -6 … (2)

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

y = 9

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

3x – 9 = 3

3x = 9 + 3 = 12

x = 4

Number of rows = x = 4

Number of students in a row = y = 9

Total number of students in a class = xy = 4 x 9 = 36