- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.1
- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.2
- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.3
- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.4
- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.5
- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.6
- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.7
- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.8
- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.9
- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.10
- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-VSAQS

RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.1 |
RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.2 |
RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.3 |
RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.4 |
RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.5 |
RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.6 |
RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.7 |
RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.8 |
RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.9 |
RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.10 |
RD Chapter 3- Pair of Linear Equations in Two Variables Ex-VSAQS |

**Answer
1** :

Let the length of rectangle = x units

and breadth = y units

Area = Length x breadth = x x y = xy sq. units

According to the condition given,

(x + 2) (y – 2) = xy – 28

=> xy – 2x + 2y – 4 = xy – 28

=> -2x + 2y = -28 + 4 = -24

=> x – y = 12 ….(i)

(Dividing by -2)

and (x – 1) (y + 2) = xy + 33

=> xy + 2x – y – 2 = xy + 33

=> 2x – y = 33 + 2

=> 2x – y = 35 ….(ii)

Subtracting (i), from (ii)

x = 23

Substituting the value of x in (i)

23 – y = 12

=> -y = 12 – 23 = -11

=> y = 11

Area of the rectangle = xy = 23 x 11 = 253 sq. units

**Answer
2** :

Let the length of a rectangle = x m

and breadth = y m

Area = Length x breadth = xy sq. m

According to the condition given,

(x + 7)(y – 3) = xy

=> xy – 3x + 7y – 21 = xy

-3x + 7y = 21 ….(i)

and (x – 7) (y + 5) = xy

=>xy + 5x – 7y = xy

5x – 7y = 35 …(ii)

(i) from (ii)

2x = 56

=> x = 28

Substituting the value of x in (i)

-3 x 28 + 7y = 21

– 84 +7y = 21

=> 7y = 21 + 84 = 105

y = 15

Length of the rectangle = 28 m and breadth = 15 m

**Answer
3** :

Let length of rectangle = x m

and breadth = y m

Area = Length x breadth = xy m2

According to the given condition,

(x + 3) (y – 4) = xy – 67

=> xy – 4x + 3y – 12 = xy – 67

=>-4x + 3y = -67+ 12 = -55

=> 4x – 3y = 55 ….(i)

and (x -1) (y + 4) = xy + 89

=> xy + 4x – y – 4 = xy + 89

=> 4x – y = 89 + 4 = 93 ….(ii)

Subtracting (i) from (ii)

2y = 38

=> y = 19

Substituting the value of y in (i)

4x – 3 x 19 = 55

=> 4x – 57 = 55

=> 4x = 55 + 57 = 112

=> x = 28

Length of the rectangle = 28 m and breadth = 19 m

**Answer
4** :

Incomes of X and Y are in the ratio 8 : 7

and their expenditures = 19 : 16

Let income of X = ₹ 8x

and income of Y = ₹ 7x

and let expenditures of X = 19y

and expenditure of Y = 16y

Saving in each case is save i.e. ₹ 1250

8x – 19y = 1250 ….(i)

and 7x – 16y = 1250 ….(ii)

Multiplying (i) by 7 and (ii) by 8, we get

**Answer
5** :

Let A’s money = 7 x

and B’s money = 7 y

According to the given conditions,

2(x – 30) = 7 + 30

=> 2x – 60 = y + 30

=> 2x – y = 30 + 60 = 90 ….(i)

and x + 10 = 3 (y – 10)

=> x + 10 = 3y – 30

=> x – 3y = -30 – 10 = -40 ………(ii)

Multiplying (i) by 3 and (ii) by 1,

**Answer
6** :

**Answer
7** :
Let one man can do a work is = x days

**Answer
8** :

We know that sum of three angles of a triangle = 180°

∠A + ∠B + ∠C = 180°

=> x° + (3x – 2)° + y° = 180°

=> x + 3x – 2 + y = 180

=> 4x + y = 180 + 2 = 182 …(i)

∠C – ∠B = 9

y – (3x – 2) = 9

=> y – 3x + 2 = 9

=> y – 3x = 9 – 2

=> -3x + y = 7 ….(ii)

Subtracting (ii) from (i)

7x = 175 => x = 25

Substituting the value of x in (ii)

-3 x 25 + y = 7

=> -75 + y = 7

=> y = 7 + 75 = 82

∠A = x° = 25°

∠B = 3x – 2 = 3 x 25 – 2 = 73°

∠C = 7 = 82°

**Answer
9** :

ABCD is a cyclic quadrilateral

and ∠A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10)°, ∠D = (4x – 5)°

The sum of opposite angles = 180°

∠A + ∠C=180° and ∠B + ∠D = 180°

=> 2x + 4 + 2y + 10 = 180°

=> 2x + 2y + 14 = 180°

=> 2x + 2y = 180 – 14 = 166

=> x + y = 83 ….(i)

(Dividing by 2)

and y + 3 + 4x – 5 = 180°

=> 4x + 7 – 2 = 180°

4x + 7 = 180°+ 2= 182° ….(ii)

Subtracting (i) from (ii)

3x = 99

=> x = 33

Substituting the value of x in (i)

33 + y = 83

=> y = 83 – 33 = 50

∠A = 2x + 4 = 2 x 33 + 4 = 66 + 4 = 70°

∠B = y + 3 = 50 + 3 = 53°

∠C = 2y + 10 = 2 x 50 + 10 = 100 + 10 = 110°

∠D = 4x – 5 = 4 x 33 – 5 = 132 – 5 = 127°

**Answer
10** :

Let number of right answers questions = x

and number of wrong answers questions = y

Now according to the given conditions,

3x – y = 40 ….(i)

and 4x – 2y = 50

=> 2x – y = 25 ….(ii)

(Dividing by 2)

Subtracting (ii) from (i), x = 15

Substituting the value of x in (i)

3 x 15 – y = 40

=> 45 – y = 40

=> y = 45 – 40 = 5

x = 15 and y = 5

Total number of questions = 15 + 5 = 20

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