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# Maharashtra State Textbook Bureau Solution for Class 10 Mathematics Chapter 1 - Linear Equations in Two Variables

Exercise/Page

## Maharashtra State Textbook Bureau Solution for Class 10 Mathematics Chapter 1 - Linear Equations in Two Variables Ex. 1.1

Solution 1

5x + 3y = 9 … (I)

2x - 3y = 12 … (II)

Let's add equations (I) and (II).

Place x = 3 in equation (I).

Solution 2(1)

3a + 5b = 26 … (I)

a + 5b = 22 … (II)

Let's subtract equation (II) from (I).

Substituting a = 2 in equation (II)

a + 5b = 22

2 + 5b = 22

5b = 22 - 2

5b = 20

b = 4

∴ Solution is (x, y) = (2, 4).

Solution 2(2)

x + 7y = 10 … (I)

3x - 2y = 7 … (II)

Multiplying equation (I) by 3.

x × 3 + 7y × 3 = 10 × 3

3x + 21y = 30 … (III)

Subtracting equation (II) from equation (III).

Substituting y = 1 in equation (II).

3x - 2y = 7

3x - 2×1 = 7

3x - 2 = 7

3x = 9

x = 3

∴ Solution is (x, y) = (3, 1).

Solution 2(3)

2x - 3y = 9 … (I)

2x + y = 13 … (II)

Subtracting equation (I) from equation (II).

Substituting y = 1 in equation (I).

2x - 3y = 9

2x - 3×1 = 9

2x - 3 = 9

2x = 9 + 3

2x = 12

x = 6

∴ Solution is (x, y) = (6, 1).

Solution 2(4)

5m - 3n = 19 … (I)

m - 6n = -7 … (II)

Multiplying equation (II) by 5

m × 5 - 6n × 5 = -7 × 5

5m - 30n = -35 (III)

Subtracting equation (III) from equation (I)

Substituting n = 2 in equation (II)

m - 6n = -7

m - 6×2 = -7

m - 12 = -7

m = -7 + 12

m = 5

∴ Solution is (m, n) = (5, 2).

Solution 2(5)

5x + 2y = -3 … (I)

x + 5y = 4 … (II)

Multiplying equation (II) by 5

x × 5 + 5y × 5 = 4 × 5

5x + 25y = 20 (III)

Subtracting equation (I) from equation (III)

Substituting y = 1 in equation (II)

x + 5y = 4

x + 5×1 = 4

x + 5 = 4

x = 4 - 5

x = -1

∴ Solution is (x, y) = (-1, 1).

Solution 2(6)

Multiplying both sides of above equation by 3

x + 3y = 10 (I)

And,

Multiplying both sides of above equation by 4

8x + y = 11 (II)

Multiplying equation (I) by 8

x × 8 + 3y × 8 = 10 × 8

8x + 24y = 80 … (III)

Subtracting equation (II) from equation (III)

Substituting y = 3 in equation (II)

8x + y = 11

8x + 3 = 11

8x = 11 - 3

8x = 8

x = 1

∴ Solution is (x, y) = (1, 3).

Solution 2(7)

99x + 101y = 499 …(I)

101x + 99y = 501 …(II)

Let's add the two equations (I) and (II)

Dividing both sides of the above equation by 200

x + y = 5 (III)

Now, subtracting equation (I) from (II)

Dividing both sides of the above equation by 2

x - y = 1 (IV)

Adding the equations (III) and (IV)

2x = 6

x = 3

Substituting x = 3 in equation (III)

x + y = 5

3 + y = 5

y = 5 - 3

y = 2

∴ Solution is (x, y) = (3, 2).

Solution 2(8)

49x - 57y = 172 … (I)

57x - 49y = 252 … (II)

Let's add the two equations (I) and (II)

Dividing both sides of the above equation by 106

x - y = 4 … (III)

Now, subtracting equation (I) from (II)

Dividing both sides of the above equation by 8

x + y = 10 … (IV)

Adding the equations (III) and (IV)

x = 7

Substituting x = 7 in equation (IV)

x + y = 10

7 + y = 10

y = 10 - 7

y = 3

∴ Solution is (x, y) = (7, 3).

## Maharashtra State Textbook Bureau Solution for Class 10 Mathematics Chapter 1 - Linear Equations in Two Variables Ex. 1.2

Solution 1

(I)

x + y = 3

(II)

x - y = 4

Solution 2(1)

x + y = 6

 X 0 6 3 -1 Y 6 0 3 7 (x, y) (0, 6) (6, 0) (3, 3) (-1, 7)

x - y = 4

 X 0 4 8 2 Y -4 0 4 -2 (x, y) (0, -4) (4, 0) (6, 2) (2, -2)

Let's plot these two equations on the graph

The two lines intersect at (5, 1).

So, the ordered pair i.e. x = 5, y = 1 satisfies the two equations given in (I) and (II).

∴ Solution of the given equations is x = 5, y = 1.

Solution 2(2)

x + y = 5

 X 0 5 2 3 Y 5 0 3 2 (x, y) (0, 5) (5, 0) (2, 3) (3, 2)

x - y = 3

 X 0 3 6 1 Y -3 0 3 -1 (x, y) (0, -3) (3, 0) (6, 3) (2, -1)

Let's plot these two equations on the graph

The two lines intersect at (4, 1).

So, the ordered pair i.e. x = 4, y = 1 satisfies the two equations given in (I) and (II).

∴ Solution of the given equations is x = 4, y = 1.

Solution 2(3)

x + y = 0

 x 0 1 -1 2 Y 0 -1 1 -2 (x, y) (0, 0) (1, -1) (-1, 1) (2, -2)

2x - y = 9

 X 0 4.5 2 5 Y -9 0 -5 1 (x, y) (0, -9) (4.5, 0) (2, -5) (5, 1)

Let's plot these two equations on the graph

The two lines intersect at (3, -3).

So, the ordered pair i.e. x = 3, y = -3 satisfies the two equations given in (I) and (II).

∴ Solution of the given equations is x = 3, y = -3.

Solution 2(4)

3x - y = 2

 X 0 2/3 1 -1 Y -2 0 1 -5 (x, y) (0, -2) (2/3, 0) (1, 1) (-1, -5)

2x - y = 3

 X 0 3/2 2 1 Y -3 0 1 -1 (x, y) (0, -3) (3/2, 0) (2, 1) (1, -1)

Let's plot these two equations on the graph

The two lines intersect at (-1, -5).

So, the ordered pair i.e. x = -1, y = -5 satisfies the two equations given in (I) and (II).

∴ Solution of the given equations is x = -1, y = -5.

Solution 2(5)

3x - 4y = -7

 X 0 -7/3 -1 3 Y 7/4 0 1 4 (x, y) (0, 7/4) (2/3, 0) (-1, 1) (3, 4)

5x - 2y = 0

 X 0 1 2 -1 Y 0 2.5 5 -2.5 (x, y) (0, 0) (1, 2.5) (2, 5) (1, -2.5)

Let's plot these two equations on the graph

The two lines intersect at (1, 2.5).

So, the ordered pair i.e. x = 1, y = 2.5 satisfies the two equations given in (I) and (II).

∴ Solution of the given equations is x = 1, y = 2.5.

Solution 2(6)

2x - 3y = 4

 x 0 2 -1 5 y -4/3 0 -2 2 (x, y) (0, -4/3) (2, 0) (-1, -2) (5, 2)

3y - x = 4

 x -4 0 2 5 y 0 4/3 2 3 (x, y) (-4, 0) (0, 4/3) (2, 2) (5, 3)

Let's plot these two equations on the graph

The two lines intersect at (8, 4).

So, the ordered pair i.e. x = 8, y = 4 satisfies the two equations given in (I) and (II).

∴ Solution of the given equations is x = 8, y = 4.

## Maharashtra State Textbook Bureau Solution for Class 10 Mathematics Chapter 1 - Linear Equations in Two Variables Ex. 1.3

Solution 1

Solution 2(1)

Solution 2(2)

Solution 2(3)

Solution 3(1)

3x - 4y = 10 … (I)

4x + 3y = 5 … (II)

∴ The solution is (x, y) = (2, -1).

Solution 3(2)

4x + 3y - 4 = 0

4x + 3y = 4 (I)

6x = 8 - 5y

6x + 5y = 8 (II)

∴ The solution is (x, y) = (-2, 4).

Solution 3(3)

x + 2y = -1 … (I)

2x - 3y = 12 … (II)

∴ The solution is (x, y) = (3, -2).

Solution 3(4)

6x - 4y = -12 … (I)

8x - 3y = -2 … (II)

∴ The solution is (x, y) = (2, 6).

Solution 3(5)

4m + 6n = 54 … (I)

3m + 2n = 28 … (II)

∴ The solution is (m, n) = (6, 5).

Solution 3(6)

2x + 3y = 2 … (I)

Multiplying both sides of above equation by 2.

∴ 2x - y = 1 … (II)

Let's solve the equations (I) and (II) by Cramer's rule

## Maharashtra State Textbook Bureau Solution for Class 10 Mathematics Chapter 1 - Linear Equations in Two Variables Ex. 1.4

Solution 1(1)

Replacing by m and by n in equations (I) and (II), we get

2m - 3n = 15 … (III)

8m + 5n = 77 … (IV)

Multiplying equation (III) by 5 and equation (IV) by 3, we get

10m - 15n = 75 … (V)

24m + 15n = 231 … (VI)

34m = 306

m = 9

Substituting m = 9 in equation (III)

2×9 - 3n = 15

18 - 3n = 15

3n = 18 - 15

n = 1

Now,

And,

∴ Solution of the given simultaneous equations is

Solution 1(2)

Replacing by m and by n in equations (I) and (II), we get

10m + 2n = 4 … (III)

15m - 5n = -2 … (IV)

Multiplying equation (III) by 5 and equation (IV) by 2, we get

50m - 10n = 20 … (V)

30m - 10n = -4 … (VI)

∴ 80m = 16

Substituting in equation (III)

Now,

∴ x + y = 5 … (VII)

And,

∴ x - y = 1 … (VIII)

Solving equations (VII) and (VIII), we get

x = 3, y = 2

∴ Solution of the given simultaneous equations is (x,y)(3,2)

Solution 1(3)

Replacing by m and by n in equations (I) and (II), we get

27m + 31n = 85 … (III)

31m + 27n = 89 … (IV)

∴ 58m + 58n = 174

Dividing both sides of above equation by 58, we get,

m + n = 3 … (V)

Subtracting equation (III) from equation (IV)

∴ 4m - 4n = 4

Dividing both sides of above equation by 4, we get

m - n = 1 … (VI)

Solving equations (V) and (VI)

∴ m = 2 and n = 1

Now,

And,

∴ Solution of the given simultaneous equations is

Solution 1(4)

Replacing by m and by n in equations (I) and (II), we get

Multiplying both sides of above equation by 2

Substituting in equation (III), we get

Now,

… (VI)

And,

3x - y = 2 (VII)

Solving equations (VI) and (VII), we get

x = 1 and y = 1

Solution of the given simultaneous equations is (x, y) = (1, 1).

## Maharashtra State Textbook Bureau Solution for Class 10 Mathematics Chapter 1 - Linear Equations in Two Variables Ex. 1.5

Solution 1

Let the greater number be x and the smaller number be y.

From the first condition we get,

x - y = 3 … (I)

From the second condition we get,

3x + 2y = 19 … (II)

Multiplying equation (I) by 2, we get

2x - 2y = 6 … (III)

Adding equations (II) and (III), we get

∴ x = 5

Substituting x = 5 in equation (II)

3x + 2y = 19

∴ 3×5 + 2y = 19

∴ 15 + 2y = 19

∴ 2y = 4

∴ y = 2

∴ The required numbers are 5 and 2.

Solution 2

We know that, opposite sides of a rectangle are equal.

∴ 2y = x + 4

∴ x - 2y = -4 … (I)

Also, 2x + y + 8 = 4x - y

∴ 2x - 2y = 8

∴ x - y = 4… (II)

Subtracting equation (II) from equation (I), we get

∴ y = 8

Substituting y = 8 in equation (II)

x - y = 4

∴ x - 8 = 4

∴ x = 12

Length of the rectangle = 4x - y

= 4 × 12 - 8

= 48 - 8

= 40

Breadth of the rectangle = x + 4 = 12 + 4 = 16

Perimeter of the rectangle

= 2(40 + 16)

= 112 units

Area of rectangle = length × breadth = 40 × 16 = 640 sq. units

Solution 3

Let the present ages of father and son be x years and y years respectively.

From the first condition, we get

x + 2y = 70 … (I)

From the second condition, we get

2x + y = 95 … (II)

Multiplying equation (I) by 2, we get

2x + 4y = 140 … (III)

Subtracting equation (II) from (III), we get

∴ y = 15

Substituting y = 15 in equation (I)

x + 2y = 70

∴ x + 2×15 = 70

∴ x + 30 = 70

∴ x = 40

∴ The present ages of father and son are 40 years and 15 years respectively.

Solution 4

Let x and y be the numerator and the denominator of the fraction respectively.

From the first condition, we get

y = 2x + 4

2x - y = -4 (I)

From the second condition, we get

(y - 6)= 12(x - 6)

y - 6 = 12x - 72

12x - y = 72 - 6

12x - y = 66 (II)

Subtracting equation (I) from (II), we get

∴ x = 7

Substituting x = 7 in equation (I)

2x - y = -4

∴ 2×7 - y = -4

∴ y = 14 + 4

∴ y = 18

Solution 5

Let the weights of box of type A be x kg and that of box of type B be y kg.

Now, 1 ton = 1000 kg

10 tons = 10000 kg

From the first condition, we get

150x + 100y = 10000

Dividing both sides of the above equation by 50

3x + 2y = 200 (I)

From the second condition, we get

260x + 40y = 10000

Dividing both sides of the above equation by 20

13x + 2y = 500 (II)

Subtracting equation (I) from (II), we get

∴ x =30

Substituting x = 30 in equation (I)

3x + 2y = 200

∴ 3×30 + 2y = 200

∴ 2y = 200 - 90

∴ 2y = 110

∴ y =55

∴ Weights of box of type A is 30 kg and that of box of type B is 55 kg.

Solution 6

Let the distance Vishal travelled by bus be x km and by aeroplane be y km.

From the first condition, we get

x + y = 1900 … (I)

Time required to cover x km by bus

Time required to cover y km by aeroplane

From the second condition, we get

Dividing both sides of the above equation by 10

∴ 70x + 6y = 21000 … (II)

Multiplying both sides of equation (I) by 6

∴ 6x + 6y = 11400 … (III)

Subtracting equation (III) from equation (II)

∴ x = 150

∴ The distance Vishal travelled by bus is 150 km.

## Maharashtra State Textbook Bureau Solution for Class 10 Mathematics Chapter 1 - Linear Equations in Two Variables Problem Set 1

Solution 1(1)

4x + 5y = 19

When x = 1

4×1 + 5y = 19

∴ 4 + 5y = 19

∴ 5y = 15

∴ y = 3

Solution 1(2)

Solution 1(3)

Solution 1(4)

x + y = 3 … (I)

3x - 2y = 4 … (II)

Solution 1(5)

Since,

∴ The given simultaneous equations have only one common solution.

Solution 2

Solution 3(1)

2x + 3y = 12 …(I)

x - y = 1 …(II)

2x + 3y = 12

 X 0 6 3 -3 Y 4 0 2 6 (x, y) (0, 4) (6, 0) (3, 2) (-3, 6)

x - y = 1

 X 0 2 4 5 Y -1 1 3 4 (x, y) (0, -1) (2, 1) (4, 3) (5, 4)

Let's plot these two equations on the graph

The two lines intersect at (3, 2).

So, the ordered pair i.e. x = 3, y = 2 satisfies the two equations given in (I) and (II).

Solution of the given equations is x = 3, y = 2.

Solution 3(2)

x - 3y = 1 … (I)

3x - 2y + 4 = 0

3x - 2y = -4 … (II)

x - 3y = 1

 x 4 -2 -5 1 y 1 -1 -2 0 (x, y) (4, 1) (-2, -1) (-5, -2) (1, 0)

3x - 2y = -4

 x 0 2 -4 5 y 2 5 -4 4 (x, y) (0, 2) (2, 5) (4, 8) (5, 7)

Let's plot these two equations on the graph

The two lines intersect at (-2, -1).

So, the ordered pair i.e. x = -2, y = -1 satisfies the two equations given in (I) and (II).

Solution of the given equations is x = -2, y = -1.

Solution 3(3)

5x - 6y + 30 = 0

5x - 6y = -30 … (I)

5x + 4y - 20 = 0

5x + 4y = 20 … (II)

5x - 6y = -30

 x 0 -6 6 3 y 5 0 10 7.5 (x, y) (0, 5) (-6, 0) (6, 10) (3, 7.5)

5x + 4y = 20

 x 0 4 -4 2 y 5 0 10 2.5 (x, y) (0, 5) (4, 0) (-4, 10) (2, 2.5)

Let's plot these two equations on the graph

The two lines intersect at (0, 5).

So, the ordered pair i.e. x = 0, y = 5 satisfies the two equations given in (I) and (II).

Solution of the given equations is x = 0, y = 5.

Solution 3(4)

3x - y - 2 = 0

3x - y = 2 … (I)

2x + y = 8 … (II)

3x - y = 2

 x 0 1 3 -1 y -2 1 7 -5 (x, y) (0, -2) (1, 1) (3, 7) (-1, -5)

2x + y = 8

 x 0 4 1 3 y 8 0 6 2 (x, y) (0, 8) (4, 0) (1, 6) (3, 2)

Let's plot these two equations on the graph

The two lines intersect at (2, 4).

So, the ordered pair i.e. x = 2, y = 4 satisfies the two equations given in (I) and (II).

Solution of the given equations is x = 2, y = 4.

Solution 3(5)

3x + y = 10 … (I)

x - y = 2 … (II)

3x + y = 10

 x 2 3 4 5 y 4 1 -2 -5 (x, y) (2, 4) (3, 1) (4, -2) (5, -5)

x - y = 2

 x 0 2 4 5 y -2 0 2 3 (x, y) (0, -2) (2, 0) (4, 2) (5, 3)

Let's plot these two equations on the graph

The two lines intersect at (3, 1).

So, the ordered pair i.e. x = 3, y = 1 satisfies the two equations given in (I) and (II).

Solution of the given equations is x = 3, y = 1

Solution 4(1)

Solution 4(2)

Solution 4(3)

Solution 5

(1)

6x - 3y = -10 … (I)

3x + 5y - 8 = 0

3x + 5y = 8 (II)

(2)

4m - 2n = -4 … (I)

4m + 3n = 16 … (II)

By Cramer's rule, we get

The solution is (m, n) = (1, 4).

(3)

Multiplying both sides of the above equation by 3

… (II)

(4)

7x + 3y = 15 … (I)

12y - 5x = 39

-5x + 12y = 39 (II)

By Cramer's rule, we get

(5)

3(x + y - 8) = 2(x + 2y - 14)

3x + 3y - 24 = 2x + 4y - 28

x - y = -4 (II)

From (I), we have

4(x + y - 8) = 2(3x - y)

4x + 4y - 32 = 6x - 2y

2x - 6y = -32

x - 3y = -16 (III)

By Cramer's rule, we get

The solution is (x, y) = (2, 6).

Solution 6

(1)

Replacing by m and by n in equations (I) and (II), we get

Multiplying throughout by 3

3m + 2n = 0 … (IV)

Subtracting equation (IV) from (III)

Substituting in equation (IV), we get

Now,

x = 6

And,

y = -4

Solution of the given simultaneous equations is (x, y) = (6, -4).

(2)

… (I)

… (II)

Replacing by m and by n in equations (I) and (II), we get

7m + 13n = 27 … (III)

13m + 7n = 33 … (IV)

Dividing throughout by 20, we get

m + n = 3 … (V)

Subtracting equation (IV) from (III)

Dividing throughout by 6, we get

-m + n = -1 … (VI)

2n = 2

n = 1

Substituting n = 1 in (V)

m + n = 3

m + 1 = 3

m = 2

Now,

And,

y = -4

Solution of the given simultaneous equations is .

(3)

Multiplying throughout by xy, we get

148y + 231x = 527

231x + 148y = 527 … (I)

Multiplying throughout by xy, we get

231y + 148x = 610

148x + 231y = 610 … (II)

Dividing both sides of the equation by 379, we get

x + y = 3 … (III)

Subtracting equation (II) from (I)

Dividing both sides of the equation by 83, we get

x - y = -1 … (IV)

2x = 2

x = 1

Substituting x = 1 in (III)

x + y = 3

1 + y = 3

y = 2

Solution of the given simultaneous equations is (x, y) = (1, 2).

(4)

… (I)

… (II)

Replacing by m and by n in equations (I) and (II), we get

7n - 2m = 5

-2m + 7n = 5 … (III)

7m + 8n = 15 … (IV)

Multiplying both sides of equation (III) by 7

-14m + 49n = 35 … (V)

Multiplying both sides of equation (IV) by 2

14m + 16n = 30 … (VI)

m = 1

Substituting m = 1 in equation (IV)

7m + 8n = 15

7×1 + 8n = 15

8n = 8

n = 1

Now,

x = 1

And,

y = 1

Solution of the given simultaneous equations is(x, y) = (1, 1).

(5)

… (I)

… (II)

Replacing by m and by n in equations (I) and (II), we get

Multiplying both sides of the equation by 20

10m + 4n = 5 … (III)

Also,

Multiplying both sides of the above equation by 2

10m - 4n = -3 … (IV)

Substituting in equation (III)

Now,

3x + 4y = 10 … (V)

And,

2x - 3y = 1 … (VI)

Multiplying equation (V) by 3 and equation (VI) by 4

9x + 12y = 30 … (VII)

8x - 12y = 4 … (VIII)

Substituting x = 2 in the equation (V)

3x + 4y = 10

3×2 + 4y = 10

4y = 10 - 6

y = 1

Solution of the given simultaneous equations is (x, y) = (2, 1).

Solution 7(1)

Let the digit in unit's place is x and that in the ten's place is y.

∴ the number = 10y + x

The number obtained by interchanging the digits is .

According to first condition,

Two digit number + the number obtained by interchanging the digits = 143

10y + x + = 143

= 143

Dividing both the sides by 11

x + y = … (I)

From the second condition,

digit in unit's place = digit in the ten's place + 3

x = + 3

∴ x - y = 3 … (II)

2x =

∴ x = 8

Putting this value of x in equation (I)

x + y = 13

8 + = 13

y =

The original number is (10y + x)

= + 8

= 58

Solution 7(2)

Let the rate of tea be Rs x per kg and that of sugar be Rs y per kg.

From the first condition, we have

Cost of 1½ kg tea + Cost of 5 kg sugar + fare for rickshaw = Total expense

∴ 1½ x + 5y + 50 = 700

∴ 3/2 x + 5y = 650

∴ 3x + 10y = 1300 … (I)

From the second condition, we have

Cost of 2 kg tea Cost of 7 kg sugar = Total expense

∴ 2x + 7y = 880 … (II)

Multiplying equation (I) by 2

∴ 6x + 20y = 2600 … (III)

Multiplying equation (II) by 3

∴ 6x + 21y = 2640 … (IV)

Subtracting equation (III) from (IV)

Substituting y = 40 in equation (I)

3x + 10y = 1300

∴ 3x + 10×40 = 1300

∴ 3x = 1300 - 400

∴ 3x = 900

∴ x = 300

∴ The rate of tea is Rs 300 per kg and that of sugar is Rs. 40 per kg.

Solution 7(3)

Anushka had x notes of Rs 100 and y notes of Rs 50.

From the first condition,

100x + 50y = 2500

Dividing both sides of the equation by 50

2x + y = 50 (I)

From the second condition,

100y + 50x = 2000

Dividing both sides of the equation by 50

2y + x = 40

x + 2y = 40 … (II)

Multiplying both sides of equation (II) by 2

2x + 4y = 80 (III)

Subtracting equation (I) from (III), we get

∴ y = 10

Substituting y = 10 in equation (I)

2x + y = 50

∴ 2x + 10 = 50

∴ 2x = 40

∴ x = 20

∴ Anushka had 20 notes of Rs 100 and 10 notes of Rs 50.

Solution 7(4)

Let the present ages of Manish and Savita be x years and y years respectively.

From the first condition,

x + y = 31 … (I)

3 years ago,

Manish's age = (x - 3) years

Savita's age = (y - 3) years

From the second condition,

(x - 3) = 4 (y - 3)

x - 3 = 4y - 12

x - 4y = -12 + 3

x - 4y = -9 (II)

Subtracting equation (II) from (I), we get

∴ y = 8

Substituting y = 8 in equation (I), we get

x + 8 = 31

∴ x = 23

∴ The present ages of Manish and Savita are 23 years and 8 years respectively.

Solution 7(5)

Let the daily wages of skilled workers be Rs x and that of unskilled workers be Rs y.

From the first condition,

∴ 3x - 5y = 0 … (I)

From the second condition,

x + y = 720 … (II)

Multiplying equation (II) by 5

∴ 5x + 5y = 3600 … (III)

∴ x = 450

Substituting x = 450 in equation (I)

∴ 3×450 - 5y = 0

∴ 5y = 1350

∴ y = 270

∴ The daily wages of skilled workers is Rs 450 and that of unskilled workers is Rs 270.

Solution 7(6)

Let the speeds of Hamid and Joseph be x km/hr andy km/hr respectively.

Distance travelled by Hamid in 20 minutes i.e. hrs x km

Distance travelled by Joseph in 20 minutes i.e. hrs y km

From the first condition, we get

Multiplying both sides of the equation by 30

∴ x + y = 90 … (I)

From the second condition, we have

3x - 3y = 30

Dividing both sides of the equation by 3

∴x - y = 10 … (II)

Adding equations (I) and (II), we get

∴ x = 50

Substituting x = 50 in equation (I), we get

50 + y = 90

∴ y = 90 - 50

∴ y = 40

∴ The speeds of Hamid and Joseph are 50 km/hr and 40 km/hr respectively.

# Text Book Solutions

Maharashtra X - Mathematics

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