Rd Sharma Solutions Class 8 Chapter 9 Linear Equation in One Variable

Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of RD Sharma class 8 Mathematics, Chapter 9, Linear Equation in One Variable. Here students can easily find Exercise wise solution for chapter 9, Linear Equation in One Variable. Students will find proper solutions for Exercise 9.1, 9.2, 9.3 and 9.4 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily.

Linear Equation in One Variable Exercise 9.1 Solution :

Question no – (1)

Solution :

Give, 9 1/4 = y – 1 1/3

or, 37/4 = y – 4/3

or, y = 37/4 + 4/3

or, y = 111 + 16/12

∴ y = 127/12

Question no – (2)

Solution :

Given, 5x/3 + 2/5 = 1

or, 25x + 6/15 = 1

or, 25x = 15 – 6 = 15

or, 25x = 15 – 6 = 9

∴ x = 9/25

Question no – (3)

Solution :

Given, x/2 + x/3 + x/4 = 13

= 6x + 4x + 3x /12 = 13

or, 13x /12 = 13

or, x = 13 × 12/13

∴ x = 12

Question no – (4)

Solution :

Given, x/2 + x/8 = 1/8

or, 4x + x/8 = 1/8

or, 5x = 1

∴ x = 1/5

Question no – (5)

Solution :

Given,  2x/3 – 3x/8 = 7/12

or, 16x – 9x/24 = 7/12

or, 16x – 9x/2 = 7

or, 16x – 9x = 14

or, 7x = 14

or, x = 14/7

∴ x = 2

Question no – (6)

Solution :

Given, (x + 2) (x + 3) + (x – 3) (x – 2) – 2x (x + 1) = 0

or, x² + 3x + 2x + 6 + x² – 2x – 3x + 6 – 2x² – 2x = 0

or, 2x² – 2x² + 6x – 6x + 6 + -2x = 0

or, 12 – 2x = 0

or, 2x = 12

or, x = 12/2

∴ x = 6

Question no – (7)

Solution :

Given, x/2 – 4/5 + x/5 + 3x/10 = 1/5

or, (x/2 + x/5 + 3x/10) – 4/5 = 1/5

or, (5x + 2x + 3x/10) = 1/5 + 4/5 = 1 + 4/5

or, 10x/10 = 5/8

∴ x = 1

Question no – (8)

Solution :

Given, 7/x + 35 = 1/10

or, 7/x = 1/10 – 35

or, x/7 = 1 – 350/10

or, 70/x – 349

∴ x = -70/349

Question no – (9)

Solution :

Given, 2x – 1/3 – 6x – 2 /5 = 1/3

= 10x – 5 – 18x + 6/15 = 1/3

= -8x + 1/15 = 1/3

= -24x + 3 = 15

= -24x = 15-3

= x = -12/24 = -1/2

Substituting x = -1/2

L.H.S, 2x-1/3 – 6x -2/5

= 2 × (-1/2) – 1/3 – 6 × (-1/2) -2/5

= -1-1/3 – -3-2/5

= -2/3 + 5/5

= -2/3 +1

= -2 + 3/3

= 1/3 = R.H.S

Therefore, L.H.S = R.H.S

Question no – (10)

Solution :

Given, 13 (y – 4) – 3 (y – 9) – 5 (y + 4) = 0

or, 13y – 52 – 3y + 27 – 5y – 20 = 0

or, 13y – 3y – 5y = 52 + 20 – 27

or, 5y = 72 – 27 = 45

or, y = 45/5

∴ y = 9

Question no – (11)

Solution :

Given, 2/3 (x – 5) – 1/4 (x – 2) = 9/2

or, 2x/3 – 10/3 – x/4 + 1/2= 9/2

or, 2x/3 – x/4 = 9/2 + 10/3 – 1/2

or, 8x – 3x/12 = 27 + 20 – 3/6

or, 5x/2 = 44

or, x = 44 × 2/5

∴ x = 88/5

Linear Equation in One Variable Exercise 9.2 Solution :

Question no – (1)

Solution :

Given,  2x + 5/3 = 3x – 10

or, 2x + 5 = 9x – 30

or, 9x – 2x = 30 + 5

or, 7x = 35

or, x = 35/7

∴ x = 5

Question no – (2)

Solution :

Given, a – 8/3 = a – 3/2

or, 2a – 16 = 3a – 9

or, 3a – 2a = 9 = – 16

∴ a = – 7

Question no – (3)

Solution :

Given, 7y + 2/5 = 6y – 5 /11

or, 77y + 22 = 30y – 25

or, 77y – 30 = – 22 – 25

or, 47y = – 47

∴ y = – 1

Question no – (5)

Solution :

Given, 1/2x + 7x – 6 = 7x + 1/4

or, 1/2 x – 6 = 1/4

or, 1/2x = 1/4 + 6 = 1 + 24/4

x/2 = 25/4

or, 25 × 2/4

∴ = 25/2

Question no – (6)

Solution :

A per the question, 3/4x + 4x= 7/8+6x – 6

or, 3x/4 + 4x – 6x = 7/8 – 6

or, 3x/4 – 2x = 7 -48/8 = – 41/8

or, 3x – 8x /4 = – 41/8

or, – 5x = – 41/2

or, x = 41/2 × 5

∴ x = 41/10

Question no – (7)

Solution :

Given, 7/2x – 5/2x = 20/3x + 10

or, 7x/2 – 5x/2 – 20x/3 = 10

or, 21x – 15x – 40x/6 = 10

or, 21x – 55x = 60

or, – 34x = 60

or, x = – 60/34

∴ x = -30/17

Question no – (8)

Solution :

Given,  6x + 1/2 + 1 = 7x – 3/3

or, 7x – 3/3 – 6x + 1/2 = 1

or, 14x – 6 – 18x – 3 /6 = 1

or, – 4 x = 6 + 9 = 15

∴ x = – 15/4

Question no – (9)

Solution :

Given,  3a – 2/3 2a + 3/2 = a + 7/6

or, 3a/3 – 2/3 + 2a/2 + 3/2 = a + 7/6

or, a – 2/3 + a + 3/2 = a + 7/6

or, + a – (2/3 – 3/2) = 7/6

or, + a + (4 – 9/6) = 7/6

or, + a + (-5/6) = 7/6

or, + a – 5/6 = 7/6

or, + a = 7/6 + 5/6 = 7 – 5/6

or, a = 2/6

∴ a = 1/3

Question no – (10)

Solution :

Given, x – (x – 1)/2 = 1 – (x – 2)/3

or, x – x/2 + 1/2 = 1 – x/3 + 2/3

or, x – x/2 + x/3 = 1 + 2/3 – 1/2

or, 6x – 3x + 2x/6 = 6 + 4 – 3/6 = 7/6

or, 5x = 7

∴ x = 7/5

Question no – (11)

Solution :

Given, 3x/4 – (x – 1)/2 = (x – 2)/3

or, 3x – 2x + 2/4 = x-2/3

or, x + 2/4 = x – 2/3

or, 3x + 6 = 4x – 8

or, 4x – 3x = 6 + 8 = 14

∴ x = 14

Question no – (12)

Solution :

Given, 5x/3 – (x – 1)/4 = (x – 3)/5

or, 20x – 3x + 3 /12 = x – 3/5

or, 17x + 3 /12 = x – 3/5

or, 85x + 15 = 12x – 36

or, 85x – 12x = – 36 – 15

or, 73x = 51

∴ x = – 51/73

Question no – (12)

Solution :

Given, 5x/3 – (x – 1)/4 = (x – 3)/5

or, 20x – 3x + 3 /12 = x – 3/5

or, 17x + 3 /12 = x – 3/5

or, 85x + 15 = 12x – 36

or, 85x – 12x = – 36 – 15

or, 73x = 51

∴ x = – 51/73

Question no – (13)

Solution :

Given, (3x + 1)/16 + (2x – 3)/7 = (x+3)/8 +(3x – 1)/14

or, 21x +7 + 32x – 48/112 = 14x + 42 + 24x – 8 /112

or, 53x – 41 = 38x + 34

or, 53x – 38x = 41 + 34

or, 15x = 75

or, x = 75/15

∴ x = 5

Question no – (15)

Solution :

Given, 9x + 7/2 – (x – x – 2/7) = 36

∴ 9x + 7/2 – (7x – x – 2/7) = 36

= 9x + 7/2 – (6x – 2/7)

= 63 + 49 – 12x + 4/14

= 51x + 53 = 14 × 36

= 51x = 504 – 53

= x = 459/51

∴ x = 9

Question no – (16)

Solution :

Given, 0.18(5x – 4) = 0.5x + 0.8

= 90x – 72 = 50x + 80

= 40x = 152

= x= 152/40 = 19/5 = 3.8

Substituting x = 3.8,

L.H.S, 0.18 (5x – 4)

= 0.18 (5 × 3.8 – 4)

= 0.18 × 105

= 2.7

R.H.S, 0.5x + 0.8

= 0.5 × 2.7 + 0.8

= 1.9 + 0.8

= 2.7

Therefore, L.H.S = R.H.S

Question no – (17)

Solution :

Given, 2/3x – 3/2x = 1/12

or, 1/x(2/3 – 3/2) = 1/2

or, 1/x(4 – 9/6) = 1/2

or, -5/6 = x/12

∴ x = – 10

Question no – (18)

Solution :

Given, 4x/9 + 1/3 + 13/100x = 8x + 19/18

or, 4/9x + 1/3 + 13/108x = 8x + 19/18

or, 4/9x + 1/3 + 13x/108 = 8x/18 + 19/18

or, 4/9x + 1/3 + 13x/108 = 4x/9 + 19/18

or, 13x/108 = 19/18 – 1/3 = 19 – 6/18 = 13/18

or, x = 108/18

∴ x = 6

Question no – (20)

Solution :

As per the given question,

= 5 (7x + 5/3) – 23/3 = 13 – 4x – 2/3

∴ 35x + 25/3 – 23/3 = 39 – 4x + 41/3

or, 35x + 25 – 23/3 = -4x + 41 /3

or, 35 x + 2 = 4x + 41

or, 35x + 4x = 41 – 2

or, 39x = 39

or, x = 39/3

∴ x = 1

Question no – (21)

Solution :

Given, 7x – 1 /4 – 1/3 (2x – 1- x/2) = 10/3

or, or, 7x – 1/4 – 1/3 (44 – 1 + x /2) = 10/3

or, 7x – 1 /4 – 1/3(5x – 1/2) = 10/3

or, 7x – 1/4 = 5x – 1/6 = 10/3

or, 42x – 6 – 20x + 4 /24 = 10/3

or, 22x – 2 /8 = 10

or, 22x – 2 = 10 × 8 = 80

or, 22x = 80 + 2 = 82

or, x = 82/22

∴ x = 41/11

Question no – (22)

Solution :

Given, 0.5(x – 0.4)/0.35 – 0.6(x – 2.71)/0.42 = x + 6.1

or, 0.5/0.35( x – 0,4) – 0.6/0.4 (x – 2.71) = x + 6.1

or, 1/0.7(x – 0.4 – x + 2.71) = x + 6.1

or, 2.31 = 0.7x + 4.27

or, 0.7x = 2.31 – 4.27 = 1.96

or, x = 1.96/0.7

∴ x = – 2.8

Question no – (23)

Solution :

Given, 6.5x + 19.5x – 32. 5 /2 = 6.5x + 13 + (13x – 26/2)

or, 19.5x – 32. 5 /2 = 26 + 13x – 26/2

or, 19.5x – 13x = 13x

or, 19. 5x = 13x = 32. 5

or, 6.5x = 32.5

or, x = 325/6.5

∴ x = 5

Question no – (24)

Solution :

Given, (3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)

or, 9x² + 6x – 24x – 16 – (8x² + 4x – 22- 11) = x² + 7x – 3x – 21

or, 9x² + 18x – 16 – 8x² + 18x + 11 = x² + 4x – 21

or, x² – 5 = x² + 4x – 21

or, 4x = 21 – 5 = 16

or, x = 16/4

∴ x = 4

Linear Equation in One Variable Exercise 9.3 Solution : 

Question no – (1) 

Solution : 

Given, 2x – 3/3x + 2 = – 2/3

or, 6x – 9 = – 6x – 4

or, 6x + 6x = 9 – 4 = 5

or, 12x = 5

∴ x = 5/12

Question no – (2)

Solution : 

Given, 2 – y/y + 7 = 3/5

or, 10 – 5y = 3y + 21

or, 3y + 5y = – 21 + 10 = – 11

or, 8y = – 11

∴ y = – 11/8

Question no – (3) 

Solution :

Given, 5x – 7 /3x = 2

or, 5x – 7 = 6x

or, 6x – 5x = – 7

∴ x = – 7

Question no – (4) 

Solution :

Given, 3x + 5/2x + 7 = 4

or, 3x + 5 = 8x + 28

or, 8x – 3x = 5 – 28

or, 5x = – 23

∴ x = – 23/5

Question no – (5) 

Solution :

Given, 2y + 5/y + 4 = 1

or, 2y + 5 = y + 4

or, 2y – y = – 5 + 4 = -1

∴ y = – 1

Question no – (6) 

Solution :

Given, 2x + 1/3x – 2 = 5/9

or, 18x + 9 = 15 – 10

or, 18x – 15x = – 10 – 9

or, 3x = – 19

∴ x = – 19/3

Question no – (7) 

Solution :

Given, 1 – 9y/19 – 3y = 5/8

or, 8 – 72y = 95 – 15y

or, 72y – 15y = – 95 + 8

or, 57y = – 87

or, y = – 87/57

∴ y = – 29/19

Question no – (8) 

Solution :

Given, 2x/3x + 1 = – 3

or, 2x = – 9x – 3

or, 2x + 9x = -3

or, 11x = – 3

∴ x = – 3/11

Question no – (9) 

Solution :

Given in the question,

= y – (7 – 8y)/9y – (3 + 4y) = 2/3

or, y – 7 + 8y / 9y – 3 – 4y = 2/3

or, 9y – 7/5y – 3 = 2/3

or, 27y – 21 = 10y – 6

or, 27y – 10y = 21 – 6 = 15

or, 17y = 15

∴ y = 15/17

Question no – (10)

Solution : 

Given, 6/2x – (3 – 4x) = 2/3

or, 6/2x – 3 + 4x = 2/3

or, 6/6x – 3 = 2/3

or, 6/3 (2x – 1) = 2/3

or, 6/2x – 1 = 2

or, 4x – 2 = 6

or, 4x = 6 + 2 = 8

or, x = 8/4

∴ x = 2

Question no – (11)

Solution : 

2/3x – 3/2x = 1/12

or, 1/x(2/3 – 3/2) = 1/2

or, 1/x(4 – 9/6) = 1/2

or, -5/6 = x/12

∴ x = – 10

Question no – (12)

Solution : 

Given, 3x + 5 / 4x + 2 = 3x + 4 / 4x + 7

or, 12x + 21x + 20x + 35 = 12x + 6x + 16x + 8

or, 41x – 22x = 22x + 8

or, 41x – 22x = – 35 + 8 = – 27

or, 19x = – 27

∴ x = – 27/19

Question no – (13)

Solution :

Given, 7x – 2 / 5x – 1 = 7x + 3/5x + 4

= or, 35x² + 28x – 10x – 8 = 35x² – 7x + 15x – 3

or, 18x – 8 = 8x – 3

or, 18x – 8x = 8 – 3 = 5

or, 10x = 5

or, x = 5/10

∴ x = 1/2

Question no – (14)

Solution :

Given, (x + 1/x – 4)² = x + 8/x – 2

or, (x + 1)²/(x – 4)² = x + 8/x -2

or, x² + 2x + 1²/x² – 8x + 4² = x + 8/x – 2

or, x² + 2x + 1/x² – 8x + 16 = x +8/x – 2

or, x³ + 2x² + x – 2x² – 4x – 2 = x³ – 8x² + 16x + 8x² – 64x + 128

or, + x – 4x – 2 = 16x – 64x + 128

or, – 3x – 2 = – 48x + 128

or, 48x – 3x = 128 + 2 = 130

or, 45x = 130

or, x = 130/45

∴ x = 26/9

Question no – (15)

Solution :

Given, (x + 1/x – 4)² = x + 8/x – 2

or, (x + 1)²/(x – 4)² = x + 8/x -2

or, x² + 2x + 1²/x² – 8x + 4² = x + 8/x – 2

or, x² + 2x + 1/x² – 8x + 16 = x +8/x – 2

or, x³ + 2x² + x – 2x² – 4x – 2 = x³ – 8x² + 16x + 8x² – 64x + 128

or, + x – 4x – 2 = 16x – 64x + 128

or, – 3x – 2 = – 48x + 128

or, 48x – 3x = 128 + 2 = 130

or, 45x = 130

or, x = 130/45

∴ x = 26/9

Question no – (16)

Solution :

Given, 9x – 7/3x + 5 = 3x – 4/x + 6

= 9x² – 7x + 54x – 42 = 9x² – 12x + 15x – 20

= 47x – 3x = – 20 + 42

= 44x = 22

= x = 22/42

∴ x = 1/2

Question no – (18)

Solution :

Given, 2x – (7- 5x)/9x – (3 + 4x) = 7/6

= 2x – 7 + 5x/9x – 3 – 4x = 7/6

= 7x – 7/5x – 3 = 7/6

= 42x – 42 = 35x – 21

= 42x – 35x = – 21 + 42

= 7x = 21

∴ x = 3

Question no – (19)

Solution :

Given, 15 (2 – x) – 5(x + 6)/1 – 3x = 10

or, 30x – 15x – 5 – 30 = 10 – 30x

or, 15x + 30x = 10 + 5 + 30 = 45

or, 45x = 45

or, x = 45/45

∴ x = 1

Question no – (20)

Solution : 

As per the question,

= x + 3/x – 3 + x + 2/x – 2 = 2

or, (x + 3) (x – 2) + (x – 3) (x + 2)/(x – 3) (x – 2) = 2

or, x² – 2x + 3x – 6 + x² + 2x – 3x – 6/x² – 2x – 3x + 6 = 2

or, 2x² – 12 / x² – 5x + 6 = 2

or, 2x² – 12 = 2x² – 10 + 12

or, 10x = 12 + 12 = 24

or, x = 24/12

∴ x = 12/5

Question no – (21)

Solution :

Given, (x + 2) (2x – 3) – 2x² + 6/x – 5 = 2

or, 2x² – 3x + 4x – 6 – 2x² + 6 = 2x – 10

or, x = 2x – 10

or, 2x – x = 10

∴ x = 10

Question no – (22)

Solution :

Given, x² – (x + 1) (x + 2)/ 5x + 1 = 6

or, x² – (x² + 2x + x + 2) = 30x + 6

or, x² – x² – 3x – 2 = 30x + 6

or, 30x + 3x = – 6 – 2 = – 8

or, 33x = – 8

∴ x = – 8/33

Question no – (23)

Solution :

Given, (2x + 3) – (5x – 7) / 6x + 11 = – 8/3

or, 2x + 3 – 5x + 7 / 6x + 11 = – 8/3

or, – 3x + 10 / 6x + 11 = – 8/3

or, 9x + 30 = – 48x – 88

or, 9x + 48x = – 30 – 88

or, 39x = – 118

∴ x = – 118/39

Question no – (24) 

Solution :

(i) x² – 9/5 + x² = – 5/9

or, 9x² – 81 = – 25 – 5x²

or, 9x² + 5x² = 81 – 25 = 56

or, 14x² = 56

or, x² = 56/14 = 4

or, x = √4 = ± 2

∴ The positive value is 2.

(ii) y² + 4/3y² + 7 = 1/2

or, 2y² + 8 = 3y² + 7

or, 3y² – 2y² = 8 – 7 = 1

or, y² = 1

or, y = √1 = 1

∴ The positive value is 1.

Linear Equation in One Variable Exercise 9.4 Solution : 

Question no – (1) 

Solution : 

Let, the number = x

According to question,

4/5 x = 3x/4 + 4

or, 4x/5 = 3x + 16/4

or, 16x = 15x + 80

or, 16x – 15x = 80

or, x = 80

Therefore, the required number will be 80.

Question no – (2)

Solution : 

Let, 1st consecutive number = x

2nd consecutive number = x +1

Now, according to question,

= (x + 1) ² – x² = 31

or, x² + 2.x .1 + 1² – x² = 31

or, x² + 2x + 1 – x² = 31

or, 2x = 31 – 1 = 30

or, x = 30/2 = 15

∴ 1st consecutive number = 15

∴ 2nd consecutive number

= 15 + 1

= 16

Therefore, the required numbers are 15 and 16.

Question no – (3)

Solution : 

Let the number x.

According to question,

2x – x/2 = 45

or, 4x – x/5 = 45

or, 3x = 45 × 2

or, x = 45 × 2/3

∴ x = 30

Therefore, the required number will be 30.

Question no – (4)

Solution : 

Let, a number is x.

According to question

(5x – 5) = 2x + 4

or, 5x – 5 = 2x + 4

or, 5x – 2x = 5v + 4 = 9

or, 3x = 9

or, x =9/3

or, x = 3

Therefore, the required number will be 3.

Question no – (5)

Solution :

Let, the number = x

∴ 5th part of x = 1/5 × x = x/5

4th part of x = 1/4 × x = x/4

Now, according to question,

= x/5 + x/4 – 5

or, x + 25/5 = x – 20/4

or, 4x + 100 = 5x – 100

or, 5x – 4x = 100 + 100 = 200

or, x = 200

Hence, the required number will be 200.

Question no – (6)

Solution :

Let the original number’s xy = 10x+y

x is ten’s number and y is one number.

Interchanges of the digits = yx = 10y+x

∴ x + y = 9

y = 9 – x

and 10x + y – 27 = 10y + x

= 10x + y – 10y – x = 27

= 9x-9y = 27

= 9x – 9(9 – x) = 27

= 9x – 81 + 9x = 27

= 18x = 27 + 81

= x = 108/18

= x = 6

∴ y = 9 – 6

= y = 3

The original number

= (10 × 6) + 3

= 60 + 3

= 63

Therefore, the original number will be 63.

Question no – (8)

Solution :

Let, the denominator = x

The numerator = x – 6

According to question,

x – 6 + 3/x = 2/3

or, x – 3/x = 2/3

or, 3x – 9 = 2x

or, 3x – 2x = 9

or, x = 9

∴ Denominator = 9

∴ Numerator = 9 – 6 = 3

∴ Fraction = 3/9 = 1/3

Question no – (10)

Solution :

As we know,

9 rupees = 900 paise (∵ 1 rupee =  100 paise)

Let, Seeta Devi has 50 paise = x

∴ Seeta Devi has 25 = 2x

According to question,

50x + 2x × 25 = 900

or, 50x + 50x = 900

or, 100x = 900

or, x = 900/100 = 9

∴ Seeta Devi has 50 paise = 9

∴ Seeta Devi has 25 = 2 × 9 = 18

Question no – (11)

Solution :

Let, Ashima’s age = x

Sunita’s age = 2x

According to question,

4(x – 6) = (2x + 4)

or, 4x – (2x + 4)

or, 4x – 2x = 24 + 4 = 28

or, 2x = 28

or, x = 28/2 = 14

∴ Ashima’s age = 14 years

∴ Sunita’s age

= 2 × 14

= 28 years

∴ Ashima age before 2 years,

= 14 – 2

= 12 years

∴ Sunita’s age before 2 years,

= 18 – 2

= 26 years.

Question no – (12) 

Solution :

Sonu : Monu = 7 : 5

Let common factor = x

∴ Present age of Sonu = 7x

∴ Present age of Monu = 5x

After 1o year ages of Monu = 7x + 10

After 1o year ages of Monu = 5x + 10

Now according to question,

7x + 10/5x + 10 = 9/7

or, 49,x + 70 = 45x + 90

or, 49x – 5x = 90 – 70 = 20

or, 4x = 20

or, x = 20/4 = 5

∴ Present age of Sonu,

= 5 × 7

= 35 years

∴ Present age of Monu,

= 5 × 5

= 25 years.

Question no – (13)

Solution :

Let, 5 years ago age of Son = x

5 years ago age of Father = 7x

Present age of Son = x + 5

After 5 year age of Son,

= (x + 5) + 5

= x + 10

After 5 year age of Father,

= (7x + 5) + 5

= 7x + 10

Now, according to question,

7x + 10 = 3(x + 10)

or, 7x + 10 = 3x + 30

or, 7x – 3x = 30 – 10 = 20

or, 4x = 20

or, x = 20/4 = 5

∴ Present age of son,

= 5 + 5

= 10 years

∴ Present age of father,

= (7 × 5) + 5

= 35 + 5

= 40 year.

Question no – (15)

Solution :

Let, no of 5 rupees note = x

no of 5 rupees note = x + 10

According to question,

= 5x + 10 (x + 10) = 1000

or, 5x + 10x + 100 = 1000

or, 15x = 1000 – 100 = 900

or, x = 900/15

∴ Number of 5 rupees note = 60

∴ number of 5 rupees note,

= 60 + 10

= 70

Question no – (16)

Solution :

Let, speed of steamer in still water x km/h

Downstream speed = (x + 1) km/h

Upstream speed = (x – 1) km/h

In downstream, distance covered in 9h,

= 9 (x + 1) km and upstream = 10 (x – 1) km

∴ 9 (x + 1) = 10 (x – 1)

= 9x + 9 = 10x – 10

= 9x – 10x = – 10 – 9

= – x = + 19

∴ x = 19.

Therefore, Speed of steamer in still water 19 km/h.

Question no – (16)

Solution :

Let, Total no of guest = x

Colas drank = 1/4 × x = x/4

squash drank = 1/3 × x = x/3

Fruit juice drank = 2/5 × x = 2x/5

No of guest not drank = 3

According to question,

= x – (x/4 + x/3 + 2/5) = 3

or, x – (15x + 20x + 24x/60) = 3

or, x – 59x/60 = 3

or, 60x – 59x/60 = 3

or, x = 3 × 60

= x = 180

Question no – (17) 

Solution : 

Total question = 180

Total marks of question

= 180 × 4

= 720

A candidate scored = 450

Let, he answer currently = x

he answer wrongly = y

According to question,

x + y = 180 – (i) × 1

4x – y = 450 – (i) × 1

(i) × 1 4 (ii) 2 we get,

x + y = 180

4x – y = 450
——————————
5x = 630

or, x = 630/5 = 126

correctly = 126

Now, putting the value x in (i) question

x + y = 180

or, y = 180 – x

= 180 – 126

= 54

Therefore, he will answer correctly 54 questions.

Question no – (18)

Solution :

Let, laboure worked for = x days

laboure worked absent = (20 – x)

60x – m (20 – x) × 5 = 745

or, 60x – 1000 + 5x = 745

or, 65x = 745 + 100 = 845

or, x = 845/65 = 13

∴ laboure absent,

= (20 – 13)

= 7 days

Therefore, he was absent for 7 days.

Question no – (19)

Solution :

Let weight of box A = x kg

The weight of box B,

= (x + 3 1/2 )

= (x + 7/2) kg

The weight of box C,

= x + 7/2 + 5 1/3

= (x + 7/2 + 1/3)kg

Total weight of box,

= 601/2

= 121/2 kg

Now, according to the question,

x + (x + 7/2) + (x + 7/2 + 16/3) = 121/2

or, x + x + 7/2+ x + 7/2 + 16/3 = 121/2

or, 3x + (2 × 7/2) + 16/3 = 21/2

or, 3x + 7 + 16/3 = 121/2

or, 3x + 21 + 16/3 = 121/2

or, 3x + 37/3 = 121/2

or, 3x = 121/2 – 37/3 = 363 – 74/6 = 289/6

or, x = 289/6 × 3 = 289/18 kg

Therefore, the weight of box A will be 289/18 kg.

Question no – (20)

Solution :

Let, denominator = x

Numerator = x – 3

∴ The ration number = x – 3/x

According to question,

x-3 + 2/x + 5 = 1/2

or, x – 1/x + 5 = 1/2

or, 2x – 2 = x + 5

or, 2x – x = 5 + 2 = 7

or, x = 7

∴ Denominator,

= x = 7

∴ Numerator,

= x – 3

= 7 – 3

= 4

Therefore, the rational number 4/7.

Question no – (21)

Solution :

Let, Denominator = 2x – 2

And, Numerator = x

∴ Original number = x/2x – 2

Now, according to question,

x + 3/2x – 2 + 3= 2/3

or, x + 3 / 2x + 1 = 2/3

or, 3x + 9 = 4x + 2

or, 4x – 3x = 9 – 2

or, x = 7

∴ Numerator = 7

∴ Denominator,

= 2.7 – 2

= 14 – 2

= 12

Therefore, the Original number will be 7/12

Question no – (22)

Solution :

Let, 1st train speed = x

2nd trains speed = (x + 5)

In 2 hours train speed distance = 2x

In 2 hours train speed distance,

= (x + 5) × 2

= (2x + 10)

After 2 hrs distance between two train,

= (340 – 30)

= 300 km

Now, according to question,

(x + x ) + (2x + 10) = 310

or, 2x + 2x + 10 = 310

or, 4x = 310 – 10 = 310

or, x = 300/4 = 75 km/hr

∴ 1st trains speed = 75 km/hr

∴ 2nd train speed,

= (75 + 5)

= 80 km/hr

Therefore, the speed of 1st train will be 75 km/hr and 2nd train will be 80km/hr.

Question no – (24)

Solution :

Total money = 12000. 00

Let 1st investment = x

∴ 2nd investment = (12000 – x)

1st income = x × 10% = 10x/100

2nd income (2000 – x) × 12%

= (12000 – x) × 12/100

Now according to question,

10x/100 + 12/100 (12000 – x) = 1280

or, 1/100 (10x + 144000 – 12x) = 1280

or, – 2x + 14400 = 1280 × 100 = 128000

or, 2x = 144000 – 128000

or, 2x = 16000

or, 16000/2 = 8000

∴ 1st investment = 8000 Rs.

∴ 2nd investment,

= (12000 – 800)

= 4000 Rs.

Therefore, Bhagwanti 1st invested 8000 Rs and 2nd investment is 4000 Rs.

Question no – (25)

Solution :

Let, Breadth of a rectangle = B

And, Length of a rectangle = (B + 9)

∴ Area = B (B + 9)

New Length = (B + 9) + 3 = (B + 12)

New Breadth = (B + 3)

According to area,

[(B + 12 ) ( B + 3)] – [B (B + 9)] = 84

or, (B ² + 3B + 12B + 36) – (B² + 9B) = 84

OR, B ² + 15B + 36 – B² 9 = 84

OR, B = 84 – 36 = 48

OR, B = 48/6 = 8

∴ Breadth = 8 cm

∴ Length = (8 + 9) = 17 cm

Therefore, the length of the rectangle will be 8 cm and breadth will be 17 cm.

Question no – (26)

Solution :

Let, now Anup’s age = x

Now Anup’ Father age = (100 – x)

Now Anuj’s age,

= 1/5 × (100 – x)

= 100 – x/5

When Anup is as old as his father than his age = (100 – x)

∴ Difference between Anup’s age = (100 – x) – x

After 8 years Anuj’s age x + 8

According to question,

(100 – x/5) + (100 – 2x) = x + 8

or, (100 – x + 500 – 10x/5) = x + 8

or, 600 – 11x = 5x + 40

or, 11x + 5x = 600 – 40 = 650

or , 16x = 560

or, x = 060/16

∴ Now, Anup’s age = 35 years

∴ Anup father age,

= 100 – 35

= 65 years

∴ Anuj’s age,

= 100 – 35/5

= 65/5

= 13 years

Question no – (27)

Solution :

Let, she start with = x

she spent – on hankies = x/2

for Begger = 1

∴ Remaining amount = x – (x/2 +1)

= x – (x + 2/2) = 2x – x – 2/2 = x – 2/4

∴ Expenses for lunch

= (x – 2/2) × 1/2

= x – 2/4

Amount of tip = 2

∴ Remaining amount after lunch,

= (x – 2/2) – (x – 2/4) – 2

= 2x – 4 – x + 2 – 8/4 = x – 10/4

∴ Spent for books = (x – 10/4) × 1/2 (x – 10/8)

= 2x – 20 – x – 10 – 24 /8

= x – 34/8

for Bus fare = 3

∴ Remaining amount after this,

= (x – 10/4) – (x – 1 – 0/8) – 3

= 2x – 20 – x + 10 – 24 /8 = x – 34/8

∴ According to question,

x – 34/8 = 1

or, x – 34 = 8

or, x = 8 + 34 = 42

Therefore, She start with 42 Rs.

Next Chapter Solution : 

👉 Chapter 10 👈