# Frank ICSE Class 8 Solutions Chapter 12

## Frank ICSE Mathematics Class 8 Solutions Chapter 12 Linear Equations in One Variable

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Frank ICSE Mathematics Class 8 Math Book, Chapter 12, Linear Equations in One Variable. Here students can easily find step by step solutions of all the problems for Linear Equations in One Variable, Exercise 12.1, 12.2, 12.3 and 12.4 Also our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 12 solutions. Here in this post all the solutions are based on latest Syllabus.

Linear Equations in One Variable Exercise 12.1 Solution :

Question no – (1)

Solution :

(a) 4x- 11 = 49

or, 4x = 60

or, x = 60/4 = 15

(b) 3y/4 – 1 = 14

or, 3y/4 = 15

or, 3y = 15 × 4 = 60

y = 60/3 = 20

(c) 8x/5 + 26 = 50

or, 8x/5 = 24

or, x = 24 × 5/8 = 15

(d) 5a/2 – 7 = 3

or, 5a/2 = 10

or, a = 10 × 2/5

= 4

(e) 5x – 13 = 2x + 8

or, 5x – 2x = 8 + 13

or, 3x = 21

or, x = 21/3

= 7

(f) 17x + 12 = 13x + 24

or, 17x – 13x = 24 – 12

or, 4x = 12

or, x = 12/4 = 3

(g) 7 (x + 3) = 5 (2x – 3)

or, 7x + 21 = 10x – 15

or, 10x – 7x = 21 + 15

or, 3x = 36

∴ x = 36/3

= 12

(h) 7y/3 + 11 = 9y/4 + 15

or, 7y/3 – 9y/4 = 15 – 1`1

or, y/12 = 4

or, y = 12 × 4

= 48

(i) 2m – 2/9

= 4m – 4/3

or, 4m – 2m

= 4/3 – 2/9

= 2m = 1/9

or, m = 10/9/2

= 5/9

(j) 6 (4y – 2) = 3 (5 + 3y)

or, 24y – 12 = 15 + 9y

or, 5y = 27

or, y = 27/15

= 9/5

Question no – (2)

Solution :

(a) x – 1/4 = x – 3/5

or, x/4 – 1/4 = x/5 = x/5 – 3/5

or, x/4 – x/5 = 1/4 – 3/5

or, x/20 = – 7/20

or, x = – 7

(b) 2x/3x + 1 =x – 3/5

or, x/4 – 1/4 = x/5 – 3/5

or, x/4 – x/5 = 1/4

or, 1/4 – 3/5

or, x/20 = – 7/20

or, x = – 7

(c) 2n/3x + 1 = 3/5

or, 21x+ 3 = 125 – 40x

or, 61x = 122

or, x = 122

or, x 122/61

= 2

(d) 25 – 8x/7x + 1 = 3/5

= 9x – 9 = 8x

or, 9x – 8x = 9

or, x = 9

(e) 5x – 2/7 = 2x + 1/3

= 15x – 6

= 14x + 7

or, 15 – 14x = 7

or, 15x – 14x = 7 + 6

or, x = 13

(f) 2x + 3/5 = 4x + 1/3

20x – 5 = 6x + 9

or, 14x = 14

or, x = 14/14

= 1

Question no – (3)

Solution :

(a) (x + 2 (2x – 3) = (2x + 1) (x – 2)

or, 2x2 + x – 6

= 2x2 – 3x – 2

or, x – 6 = – 3x – 2

or, x + 3x = 6 – 2

or, 4x = 4

or, 4x = 4

or, x = 4/4

= 1

(b) (6x + 5) (2x + 3) = (4x + 7) (3x + 2)

or, 12x2 + 28x + 15

= 12x2 + 29x + 14

or, 29x – 28x = 15 – 14

or, x = 1

(c) 9x – 7/3x+ 5 = 3x – 4/x + 6

or, 9x2 + 47x – 42 = 9x2 + 3x – 20

or, 47x – 42 = 3x – 20

or, 44x = 42 – 20 = 22

or, 44x = 22

or, x = 22/24 = 1/2

(d) 6x + 7/3x + 4 = 4x + 5/2x + 3

or, 12x2 + 32x + 21

= 12x2 + 31x + 20

or, 32x – 31x = 20 – 21

or, x = – 1

(e) 5x + 4/7x + 3 = 5x- 2/7x – 2

or, 35x2 + 8x – 3

= 35x2 + 18x – 8

or, 8x – 18x = – 8 + 3

or, – 10x = – 5

or, x = – 5/- 10

= 1/2

(f) x + 3/x – 2 = x – 2/x + 3

= x2 + 6x + 9

= x2 – 4x + 4

or, 6x + 4x = 4 – 9

or, 10x = – 5

or, 10x = – 5

or, x = – 5/10

= – 1/2

Question no – (4)

Solution :

(a) 4x + 1/3 + 3x + 1/2 = 3x – 7/4 + 13

or, 4x/3 + 3x/2  3x/4

= 13 – 7/4 – 1/2 – 1/3

or, 16 + 18x – 9x/12

= 156 – 21 – 6 – 4/12

or, 25x = 125

or, x = 125/25

= 5

(b) x – 5/4 = 4x + 3/3 – x – 3/6

or, x/4 – 4x/3 + x/6

= 3/3 + 3/6 + 5/4

or, 3x – 6 + 2x/12

= 12 + 6 + 15/12

or, – 11x = 33

or, x 33/- 11

= – 3

(c) 5x – 4/7 – 3x – 1/14 = 1/2

or, 5x/7 – 3x/14

= 1/2  4/7 – 1/14

or, 10x – 3x/14

= 7 + 8 – 1/14

or, 7x/14 = 14/14

or, x/2 = 1

or, x = 2

(d) 3y – 9/4 + 2y + 1/3

= 7y – 1/6

or, 3y/4 + 2y/3 – 7y/6

= 9/4 – 1/3 -1/6

or, 9y + 8y – 14y/12

= 27 – 4 – 2/12

or, 3y = 21

or, y = 21/3

= 7

(e) x – 1/6 – 3x + 1/6 = x + 4/3

or, x/6 – 3x/6 – x

= 4/3 + 1/6 + 1/6

or, x – 3x – 6x/6

= 8 + 1 + 1/6

or, – 8x = 10

or, x = – 10/8

= – 5/4

(f) 2/3y + 1 = 4/5 (y -1/4)

or, 2/3y – 4/5y

= – 1/5 – 1

or, 10y – 12y/15

= – 1 – 5/5

or, – 2y/15 = – 6/5

or, y 6 × 15/5 × 2

= 9

(g) 5 (y + 12) – 17 (2 – y)/7y – 1 = 8

or, 5y + 60 – 34 + 17y = 56y – 6

or, – 12y + 26 = 56y – 8

or, 5y + 60 – 34 + 17y

= 56y – 8

(h) 5x + 2/6 – x – 5/3 = 3

or, 5x/6 – x/3 = 3 – 5 – 2/6

or, 5x – 2x/6 = 18 – 10 – 2/6

or, 3x = 6

or, x = 6/3

= 2

(i) 2 (x – 4)/(x – 7) = 4/5

or, 12x – 28 = 10x – 20

or, 2x = 8

or, x = 4

(l) 6x – 2/3 – 2x – 1/5 = 1/3

or, 6x/3 – 2x/5

= 1/3 – 2/3 – 1/5

or, 30x – 6x/15

= 5 – 10 – 3/15

or, – 3x = – 8

or, x = 8/3

(k) x – 2/4 + 1/3 = 2x – 1/3

or, x/4 – x + 2x/3

= 1/3 – 1/3 + 1/2

or, 3x – 1x+ 8x/12

= 4 – 4 + 6/12

or, – x = 36

or, x = – 6

(l) 1 – x/6 + 2x/3 – 1 – 7x/4  = 6 2/3

or, x/6 + 2x/6 = 20/3 – 1/6 + 1/4

or, – 2x + 4x + 2x/12 = 80 – 2 + 3/12

or, 23x = 81

or, x = 81/23

(m) 3(x – 1)/5 – 5 (x – 4)/7 = 1

or, 3x/5 – 5x/7 = 1 – 20/7 + 3/5

or, 21x – 25x/35

= 35 – 100 + 21/35

or, – 4x = – 44

or, x = 44/4 = 11

(n) 5(x + 2)/3 – 3(x – 2)/2 = 7

= 5x/3 – 3x/2

= 7 – 10/3 – 6/2

or, 10x – 9x/6

= 42 – 20 – 18/6

or, x = 4

Linear Equations in One Variable Exercise 12.2 Solution :

Question no – (1)

Solution :

Suppose the three numbers are,

= 7 (x – 1), 7x, 7 (x + 1)

= 7x + 7(x – 1) + 7 (x + 1) = 777

or, x + (x – 1) + (x + 1) = 111

or, 3x = 111 or, x = 111/3 = 37

or, 3x = 111

or, x = 111/3 = 37

The numbers are (7 × 36), (7 × 37) (7 × 38) = 252, 259, 266

Question no – (2)

Solution :

Suppose the multiples are 11 (x – 1), 11x 11 (x + 1)

2 × 11 (x – 1) + (3 × 11x) + 4 × 11 (x + 1) = 814

or, 11 [2x – 2 + 3x + 4x + 4] × 11 × [9x + 2] = 814

or, 9x + 2 = 814/14 = 74

or, 9x + 2 = 74

or, 9x = 74 – 2 = 72

9x = 72

or, x = 72/9

= 8

Question no – (3)

Solution :

Suppose the parts are x, (300 – x)

= 1/2 x = 1/3 (300 – x)

or, x/2 + x/3 = 300/3

or, 5x/6 = 100

or, x = 100 × 6/5 = 120

The parts are = 120 (300 – 120)

= 180

Question no – (4)

Solution :

Suppose the number is x

5x – 8 = 4x + 4

or, 5x – 4x = 4 + 8

or, x = 12

Question no – (5)

Solution :

Suppose the integers are 7x, 4x

7x – 4x = 75

or, 3x = 75

or, x = 75/3 = 25

The integers are 7x = 7 × 25

= 175,

= 4x = 4 × 25

= 100

Question no – (6)

Solution :

Suppose, the number are – 3x, 4x, 5x

(3x + 5x) – 4x

= 64

or, 8x – 4x = 64

or, 4x = 64

or, x = 64/4 = 16

The numbers are – 16 × 4 = 64

= 15 × 5

= 80

Question no – (7)

Solution :

Suppose she thinks of x

∴ 9 (x – 4) = 7x

or, 9x – 36 = 7x

or, 9x – 7x = 36

or, 2x = 36

∴ x = 36/2

= 18

Question no – (8)

Solution :

Suppose initially the fraction was = x/x + 5

= x + 11/x + 5 – 14 = 5

or, x + 11/x – 9 = 5

or, 5x – 45 =  x + 11

or, 5x – x = 11 + 45

or, 4x = 56

x = 56/4

= 14

Question no – (9)

Solution :

Suppose the number is x

= x/15 – x + 1 = 1/3

or, 3x = 15 – x + 1

or, 3x + x = 15 + 1

or, 4x = 16

or,  = 16/4 = 4

Question no – (10)

Solution :

Suppose 4 years ago son’s age x years

Father’s age was 4x year

= (4x + 4) + (x + 4) = 53

= 5x + 8 = 53

or, 5x = 45

or, x = 45/5 = 9

Son’s present age

= 9 + 4

= 13 years

Fathers present age

= 53 – 13

= 40 year

= 9 years ago

Question no – (11)

Solution :

Suppose Ankit’s son’s age was x years

Akint present age = 2(x + 9) years

Ankit present age

Ankit’s age after 4 years = 4 × x = 4x

4x = 2 (x + 9) + 4

or, 4x = 2x + 18 + 4

or, 4x – 2x = 22

or, 2x = 22

or, x = 11

Ankit’s son’s present age,

= 11 + 9

= 20 years

Ankit’s son’s age,

= 2 × 20

= 40 years

Question no – (12)

Solution :

Suppose, at the time of wedding the ages

of bride and groom was 3x, 4x years

= 3x + 8/4x + 8 = 4/5

or, 16x + 32 = 15x + 40

or, 16x – 15x

= 40 – 32

or, x = 8 years

The age of bride at the time of wedding

= 3 × 8

= 24 years

Therefore, age of the bride will be 24 years.

Question no – (13)

Solution :

Suppose Lalitha’s and Sarita’s present ages are 5x, 4x, years

5x + 3/4x + 3 = 11/9

or, 45x + 27 = 44x + 33

or 45x – 44x

= 33 – 27

or, x = 6 years

Saritha’s present age = 4 × 6

= 24 years

Question no – (14)

Solution :

Suppose the breath is x cm

= 2[x + (x + 14) = 180

or, 2x + 14 = 180/2

= 90

or, 2x = 90 – 14 = 76

or, x = 76/2 = 38 cm

Breath = 38 cm,

Length = 38 + 14 = 52 cm

Area = 38 × 52 = 1976 cm2

Question no – (15)

Solution :

Suppose breath, length are x cm (x + 7) cm

x (x + 7) =(x + 3) (x +7 – 4)

or, x2 + 7x = (x + 3) (x + 3)

or, x2 + 7x = x2 + 6x + 9

or, 7x – 6x = 9

or, x = 9

Breath = 9 cm,

Length = (9 + 7)

= 16 cm

Question no – (16)

Solution :

Suppose, the base and heights are

= 1/2 (4x) (3x)= 1/2

(4x + 4) (3x – 2)

or, 12x2 = 12x2 + 12x – 8x – 8

or, 4xx – 8 = 0

or, 4x = 8

or, x = 8/4= 2 cm

Base = 4 × 2= 8 cm

Height = 3 × 2 = 6 cm

Question no – (17)

Solution :

x+ 3x/2 + 3x/2 = 32

or, 2x + 3x + 3x/2 = 32

or, 8x = 32 × 2 = 64

or, x = 64/8 = 8 cm

Length of base = 8 cm

Length of others side 8 × 3/2 = 12 cm

Question no – (18)

Solution :

Suppose, cost of the chair = x rupees

x = + 116x/100 = 1620

or, 100x + 116x/100 = 1620

or, 216x = 1620 × 100

or, x = 1620 × 100/216

= 750 rupees

Cost of chair = 750 rupees

Cost of table = 750 × 116/100

= 870 rupees

Question number – (19)

Solution :

Suppose Srividya gets x rupees

So, she deposits in bank = x/2 rupees

She gave her daughter = x/4 + 6000

She gave her daughter son = x/8 + 3000

(x/8 + 3000) + (x/4 + 6000) = x/2

or, x/8 + x/4 – x/2 = – 6000 – 3000

or, x + 2x – 4x/8 = – 9000

or, – x/8 = – x/8 – 9000

or, x/8= 9000

x = 8 × 900 = 72000

Question no – (20)

Solution :

Suppose Anuj has x coins of 5 rupees

Suppose Anuj has 5x/8 coins of 2 rupees

(5 × x) + (5x/8 × 2) = 1000

or, 5x + 5x/4 = 1000

or, 20x + 5x/4 = 1000

or, 25x = 4 × 1000

or, x = 1000 × 4/25

= 160

= 160 coins of 5 rupees

= 160 × 5/8 = 1000 coins of 2 rupees

Question no – (21)

Solution :

Suppose the labourer came to work x days

He did not come to work = (30 – x) days

(80 × x) – 10 (30 – x) = 1860

or, 80x – 300 + 10x = 1860

or, 90x = 1860 + 300

or, 90x = 2160

or, x = 2160/90

or, = 24

He come to work 24 days

Question no – (22)

Solution :

Let, A’s share = x rupees B’s share = 5x/6

C’s share = 4/5 (5x/6) = 2x/3

x + 5x/6 + 2x/3 = 1500

or, 6x + 5x + 4x = 1500

or, 15x/6 = 1500

or, x = 1500 × 6/15 = 600 rupees

A’s Share = 600 × 5/6 = 500 rupees

C’s Share 500 × 4/5 = 400 rupees

Question no – (23)

Solution :

Suppose x adult tickets were sold

25x + 15 (210 – x) = 4300

or, 25x + 3150 – 15x = 4300

or, 10x = 4300 – 3150

or, 10x = 1150

or, x = 1150/10 = 115

Question no – (24)

Solution :

Suppose x no of guests attended the party

x – (x/3 + x/5 + x/6) = 27

or, x – (10x + 6x + 5x/30) = 27

or, x – 7x/10 = 27

or, 10x – 7x/10 = 27

or, 3x/10 = 27

or, x, 27 × 10/3

= 90

Question no – (25)

Solution :

Let, the speed of stream be x km/hr

2(18 + x) = 2 1/2 × (18 – x

or, 2x + 36 = 5/2 (18 – x)

or, 2x + 36 = 45 – 5x/2

or, 2x + 5x/2 = 45 – 36

or, 9x/2 = 9

or, x = 9 × 2/9

= 2 km/hr

Question no – (26)

Solution :

Suppose the speeds of the trains are x km/hr, (x + 10) km/hr respectively

Since they were morning towards each other, their relative speed = (x + x + 10) km/hr

= (2x + 10) km/hr

with (2x + 10) km/hr speed they travelled

= (500 – 45) = 455 km in 3 1/2 = 7/2 hours

455/7/2 = 2x + 10

or, 2x + 10 = 455 × 2/7 = 130

or, 2x = 120

or, x = 120/2

= 60

Speed of the trains = 60 km/hr

= 70 km/hr

Question no – (27)

Solution :

Suppose Devender bought x apples x bananas

Total cost price of apples = 5x rupees

Total cost price of bananas = 2x rupees

Profit from apples

= 5x × 20/100

= x rupees

Loss from apples

= 2x × 20/100

= 2x/5 rupees

Overall profit = x – 2x/5 = 360

or, 5x – 2x/5 = 360

or, 3x/5 = 360

or, x = 360 × 5/3 = 600

He bought 600 apples.

Linear Equations in One Variable Exercise 12.3 Solution :

Question no – (1)

Solution :

= 5x + 7y = – 9 (i)

= 4x + 11y = 26 – (ii)

= 20x + 28y – 36 – (iii)

+ 20x – 557 = + 130 – (iv)

(-)    (+)    (-)
————————————————–
83y = – 166

or, Y = – 2

Multiplying eqn (i) with 4 and eqn (ii) with 5 we get

Putting y = – 2 (i) we get

= 5x + 7 (- 2) = – 9

or, 5x – 14x = – 9

or, 5x – 14 = – 9 = 5

x = 5/5 = 1

The solution is = x = 1, y = – 2

Question no – (2)

Solution :

5x + 6y = 13 – (i)

6x  5y = 8 – (ii)

25x + 30y = 65 – (iii)

+ 42x + 3yy = + 48 – (iv)
(- )   (- )    (- )
———————————————————–
= – 17x = 17

Multiplying (i) with 5 and (ii) with 6 we get,

Putting x = – 1 in (i) we get,

= 5(1) + 6y = 13

or, 6y – 5 = 13

or, 6y – 5 = 13

or, 6y = 13 + 5 = 18

y = 18/6 = 3

The solution is x = – 1, y = 3

Question no – (3)

Solution :

= 3x – 5y = – 34 – (i)

= 8x – 7y = – 21 (ii)

= 21x – 35y = – 238

+ 40x – 35y = – 238 – (iii)
————————————————————–
= – 19x = – 133

or, x = – 133/ – 19 = 7

Multiplying (i) 7 and (ii) with 5 we get

Putting x = 7 in (i) we get

= 3 (7) – 5y = – 34

or, 21 – 5y = – 34

or, 5y = 21 + 34 = 55

or, y = 55/5 = 14

The solution is x = 7, y = 11

Question no – (4)

Solution :

= 3a + 8b = 28 – (i)

= 7a + 10b = 22 – (ii)

= 21a + 56b = 196 – (iii)

= 21a + 30b = 66

————————————————————–
= 26b = 130

= 130/26 = 5 Multiplying (i) with 7 and (ii) with 3, we get,

Putting b = 5 in (i) we get,

= 3a + 8(5) = 28

or, 3a = 28 – 40 = – 12

a = – 12/3 = – 4

The solution is a = – 4, b = 3

Question no – (5)

Solution :

= 7m + 8n = 52

= 9m- 11n = 3

= 63m + 72n = 468

= + 63m – 77n = 21

(-)     (+)    (-)
————————————————–
= 149 = 447

= n = 447/149 = 3

Multiplying (i) with 9 and (ii) with (7) we get

= 7 m + 24 = 52

or, m = 4

The solution is m = 4, n = 3

Question no – (6)

Solution :

= x – 5/3 = y + 2/7

= 7x – 35

= 3y + 6

= 7x – 3y = 47 (i)

= 2x – 3y = 1 (i)

(-)    (+)   (-)
————————————-
= 5x = 40

x = 40/5 = 8

Putting x = 8 (ii) we get,

= 2 (8) – 3y = 1

or, 3y = 16 – 1 = 15

or, y = 15/3 = 5

Question no – (7)

Solution :

= x/3 + y/2 = 8 (i)

= 3x/4 – y/4 = 7 (ii) × 2

= 3x/2 – y/2 = 14 (iii)

Now, (i) + (iiii) (x/3 + 3x/2) + (y/2 – y/20 = 8+ 14

or, 11x/6 = 22

or, x = 22 × 6/11 = 12

Putting x = 12 in (i) we get, 12/3 + y/2 = 8

or, y/2 = 8 – 4

or, y = 4 × 2= 8

The solution is x = 12, y = 8

Question no – (8)

Solution :

= x/3 – y2 = 1 – (i)

= x/5 + y/4 = 5 – (ii) × 2

= 2x/5 + y/2 = 10 (iii)

Now, (i) + (iii) – (x/3 + 2x/5) + (- y/2) + y/2) = 1 + 10

or, 11x/15 = 11

or, 11 × 15/11 = 15

Putting x = 15 in (i) 15/3 – y/2 = 1

or, y/2 = 5 – 1 = 4

or, y = 8

= [x = 15, y = 8]

Question number – (9)

Solution :

= 3x + 8y = 7 – (i)

= 2x + 9y = 1 (ii)

= 6x + 16y = 14 – (iii)

= 6x + 27y = 3 (iv)
—————————————
= – 11y = 11

or, y = 11/- 11 = – 1

Multiplying 2 with (i) and 3 with (ii) we get,

Putting y = – 1 or, (i) we get

= 3x – 8 = 7

or, 3x = 15

= x = 15/3 = 5

[x – 5, y = – 1]

Question no – (10)

Solution :

= 22x + 21y = 3

= 21x + 22y = – 3
—————————–

= 43x  + 43y = 0

or, x + y = 0 – (i)

= 22x + 21y = 3

= 21x + 22y = – 3

(-)  (-)   (+)
—————————————
= x – y = 6 – (ii)

= x + y = 0

= x – y = 6
—————————————
= 2x = 6

x = 3

Putting x = 3 in (ii) we get 3 – y = 6

or, y = – 3

= [x = 3, y = – 3]

Question no – (11)

Solution :

= 4x – 47y = 35

= 4x – 41y = 53
—————————————
= 88x – 88y = 88

or, x – y = 1 (i)

= 41x – 41y = 53
—————————————
6x + 6y = 18

or, x + y = 3

x – y = 1

x + y = 3
—————————————
= 2x = 4

x = 2

Putting x = 2 in (ii) 2 + y = 3

or, y = – 2 = 1

= [x = 2, y = 1]

= [(12) → Similar]

Question no – (13)

Solution :

= 2x – 3y = 4 – (i)

= – x + 3y = 1 – (ii)
—————————————
= x = 5

Putting x = 5 in (ii) we get,

= 3y = 1 + 5 = 6

= y = 6/3 = 2

[x = 5, y = 2]

Question number – (14)

Solution :

= 3x – 7y = 6 – (i)

= – 4x + 6y = 2 – (ii)

= 12x – 28y = 24 – (iii)

= – 12x + 18y = 6 – (iv)
—————————————
= – 10 = 30

= y = – 3

Multiplying (i) with 4 and (ii) with 3, we get

Putting y = – 3 in (i) we get

= 3x – 7 (- 3) = 6

or, 3x = 6 – 21 = – 15

= x – 15/3 = – 5

[x = – 5, y = – 3]

Question no – (15)

Solution :

= 2x + y = 7 – (i)

= 2x/5 + y/3 = 1 – (ii)

Subtracting (iii) from (i) we get,

= 2x + y = 7

= + 2x + 5y/3 = 5
—————————————
= – 2y/3 = 2

or, y = 2 × 3/ – 2 = – 3

Multiplying (ii) with 5 get 2x + 5y/3 = 5 – (iii)

Subtracting (iii) from (i) we get,

putting y = – 3 in (i) we get

= 2x – 3 = 7

or, 2x = 10

or, x = 10/2 = 5

= [x = 5, y = – 3]

Question no – (16)

Solution :

= 57x – 56y = – 169

= 56x – 57y = – 170
—————————————
= 113x – 113y – 339

or, x – y – 3 –(i)

= 57x – 56y = – 169

+ 56x – 57y = – 170
—————————————
= x + y = 1 – (ii)

= x – y = – 3

= x + y = 1
—————————————
or, x = – 2/2 = – 1

Putting x = – 1 in (i) we get,

= 1 – y = – 3

or, y = 3 – 1 = 2

Question no – (17)

Solution :

= 2x – 3y = 3 – (i)

= 2x/3 + 4y = 6 – (ii)

Subtracting (i) from (iii)

= 2x + 12y = 18

= 2x – 3y = 3
————————————
15y = 15

or, y = 1

Multiplying (ii) with 3 we get,

= 2x + 12y = 18 – (iii)

Subtracting y = 1 in (i) we get

= 2x – 3 (i) = 3

or, 2x = 6

or, x = 6/2 = 2

= [x = 3, y = 1]

Question no – (18)

Solution :

= x – 1/y + 1 = 3/4

= 4x – 4 = 3y + 3

= 4x – 3y = 7 (i)

= x + 2/y – 2 = 4/3

= 3x + 6

= 4y – 8

= 3x – 4y = – 14 – (ii)

Multiplying (i) with (3) and (ii) with we get

= 12x – 9y = 21 – (iii)

= + 12x – 16y = – 56 – (ii)
—————————————–
= 7y = 77

or, y = 11

Putting y = 11 in (i) we grt

= 4x – 3 (ii) = 7

or, 4x = 33 + 7 = 40

x = 40/4 = 10

or, y = 11

= [x = 10, y = 11]

Question no – (19)

Solution :

= 3x/2 + y/2 = – 1 – (i)

= x/2 + 3y/2 = 5 – (ii)

Multiplying (ii) with 3 we get

= 3x/2 + 9y/2 = 15 – (iii)

Subtracting (i) from (iii) we get

= (3x/2 – 3x/2) + (9y/2 – y/2) = 15 – (- 1)

or, 9y – y/2 = 16

or, 8y/2 = 16

or, 8y/2 = 16

= y = 2 × 16/8

or, y = 4

Putting y = 4 in (i) 3x/2 + 1 = 1

or, 3x/2 = – 3

or, x = – 3 × 2/3

= – 2

= [x = – 2, y = 4]

Question no – (20)

Solution :

= 371x + 197y = 454

= 197x + 371y = 23
—————————————–
= 568x + 567 = 568

or, x + y = 1 (i)

= x + y = 1

= x – y = 2
—————————————–
= 2x = 3

= 371x + 197y = 545

= 197x + 371y = 23
—————————————–
= 174x – 174y = 522

or, x – y = 2 – (ii)

[ x = 3/2, y = – 1/2]

Linear Equations in One Variable Exercise 12.4 Solution :

Question no – (1)

Solution :

Suppose the numbers are x, y

So, x + y = 187

= x – y = 23
—————————-
= 2x = 210

x = 105

Putting x = 105 in (i) we get

= 105 + y = 187

or, 187 – 105 = 82

∴ [x = 105, y = 82]

Question no – (2)

Solution :

= x + y = 53 – (i)

= 2x – 3y = 11 (ii)

Multiplying (i) by 2 we get

= 2x + 2y = 106 – (iii)

Subtracting (ii) from (iii) we get

= 2x + 2y = 106

= 2x – 3y = 11

Putting = y = 19 in (i) we get

= 19 + x = 53

or, x = 53 – 19 = 34

y = 19

= [x= 34, y = 19]

Question no – (3)

Solution :

= x + 3/y = 56

= 6x + 18 = 5y

= 6x – 5y – 18 (i)

= x + 5/y + 4 = 3/4

= 4x + 20 = 3y + 12

or, 4x – 3y – 8 – (ii)

Multiplying (i) with 3 and (ii) with 5 we, get

= 18x – 15y = – 54

= + 20x – 15y = – 40

(-)    (+)    (+)
—————————————
– 2x = – 14

x = – 14/- 2 = – 17

= [7/12]

Putting x = 7 in (i) we get

= 6 (7) – 5y = – 18

or, 5y = + 18 + 42 = 60

or, y = 60/5 = 12

Question no – (4)

Solution :

= x + 1/y – 1 = 2/3

= 3x + 3 = 2 + 2

= 3x – 2y – 5 (i)

= x + 2/y + 2 = 3/5

= 5x + 10

= 3y + 6

= 5x – 3y

= – 4 – (ii)

Multiplying (i) with 3 and (ii) with2, we get,

9x – 6y = – 15 – (iii)

+ 10x – 6y = – 8 (iv)
————————————-
– x = – 7

x = 7

Putting x = 7 in (i)

= 3 (7) – 2y = – 5

or, 2y = 5 + 21 = 26

or, y = 26/2 = 13

= [7/13]

Question no – (5)

Solution :

= x + 7/5x + 2 + 7 = 1/3

or, 5x + 9

= 3x + 21

or, 5x – 3x = 21 – 9

or, 2x = 12

or, x = 6

Daughter’s present age = 6 years, Anita’s present age

= 5(6) + 2

= 32 years

Question no – (6)

Solution :

Suppose son’s age = x years

= x + 3/66 – x + 3

= 1/3

or, 3x + 9 = 69 – x

or, 3x + x = 69 – 9

or, 4x = 60

or, x = 15

Son’s age = 15 years, fathers age

= 66 – 15

= 51 years

Question no – (7)

Solution :

Suppose Suresh’s age = years

= x/x + 16 + 20 = 1

or, 4x + = x = 36

or, 4x – x = 36

or, 3x = 36

or, x = 36/3 = 12

Suresh’s age = 12 years

Ramesh’s age = 12 + 16

= 28 years

Question no – (8)

Solution :

Suppose their incomes are 8x rupees 7x rupees

Suppose their expenditures are 5y rupees 4y rupees

8x – 5y = 21000 – (i)

= 7x – 4y = 21000 – (ii)

= 32x – 20y = 84000 – (iii)

= + 35x – 20y = 105000 – (iv)
——————————————————
= – 3x = – 21000

or, x = 7000

Multiplying (i) with 4 and (ii) with 5, we get

Putting x = 7000 in (i) we get

= 8 (7000) – 5y = 21000

or, 5y = 56000 – 21000

or, = 35000/5

= 7000

Their income,

= (8 × 7000) = 56000

= (7 × 7000) = 49000

Question no – (9)

Solution :

= 5x + 27/4x + 27 = 8/7

or, 35x + 189 = 32x + 216

or, 3x = 27

or, x = 27/3 = 9

The numbers are = (9 × 5) = 45

= (9 × 4) = 36

Question no – (10)

Solution :

Suppose costs of a pen = x rupees, cost of a pencil = y rupees

= 18x + 72y = 343 – (i)

= 11x + 96y = 237 – (ii)

Multiplying (i) with 4 and (ii)

with 3 we get

= 72x + 288y = 1296

+ 33x + 288y = 711
—————————————
= 39x = 585

= x = 585/39 = 15

Putting x = 15 in (i) we get

= 18(15) + 72y = 324

or, 12y = 324 – 270 = 54

or, y = 54/72

= 63/4 rupees

[Pen = 15 rupees, pencil = 3/4 rupees]

Question no – (11)

Solution :

Suppose cost of a chair = y rupees cost of a table = x rupees

= 8x + 15y = 5225 – (i)

= 10x + 17y = 6225 – (ii)

= 40x + 75y = 26125

= 4x + 63y = 24900
—————————————
= 7y = 1255

y = 1225/7 = 175

Multiplying (i) with 5 and (ii) with 4 we get

Putting y = 175 we, get

= 8x + 15 (175) – 5225

or, 8x = 5225 – 2625 = 2600

or, x = 2600/8 = 325

[ Table = 325, chair = 175]

Question no – (12)

Solution :

Suppose speed of bus = km/hr

Speed of  taxi = y km/hr

= 9x + 4y = 670 – (i)

= 8x + 7y = 785 – (ii)

= 63x + 28y = 4690 – (iii)

+ 32x + 28y = 3140 – (iv)
————————————————-
= 31x = 1550

x = 1550/31 = 50

Multiplying (i) with 7 and (ii) with 4 we get

Putting x = 50 in(i) we get

= 9 (50) + 4y = 670

or, 4y = 670 – 450 = 220

y = 220/4 = 55

[ Bus = 50 km/hr, Taxi = 55 km/hr]

Question no – (13)

Solution :

Speed in downstream

= 88/4

= 22 km/hr

Speed in upstream

= 40/4

= 10 km/hr

Suppose speed of the boat I still = x km/hr

Speed of the stream = y km/hr

x + y = 22

= x – y
——————————–
2x = 32

x = 16km/hr

= y = 6km/hr

[Boat = 16km/hr Stream/hr]

Question no – (14)

Solution :

Speed in upstream

= 45/4 1/2

= 45/9/2

= 10 km/hr

Speed in downstream

= 56/4

= 14 km/hr

Suppose : Speed of the boat in still water = x km/hr

x + y =14

x – y = 10
——————————–
= 2x = 24

x = + 2 km/hr

y = 2 km/hr

[Speed of the boat book = 12 km/hr : Speed of the stream = 2 km/hr]

Question no – (15)

Solution :

Speed in upstream = 30/31/3 = 30/10/3 = 9 km/hr

Speed in downstream = 30/2 = km/hr

x + y = 15

= x – y = 0
——————————–
= 2x = 24

x = 12

= y = 3

[Speed of the book = 12 km/hr : Speed of the stream = 3 km/hr]

Question no – (16)

Solution :

= 15 + 2y = 97

= 2x + 5y = 106
——————————–
7x + 7y = 203

or, x + y = 29 – (i)

= 5x + 2y = 97

= 2x + 5y = 106
——————————–
= 3x – 3y = – 9

= x – y = – 3 – (ii)

x + y = 29

x – y = – 3
——————————–
=  2x = 20

x = 26/2 = 13

= y = – 15

[ x = 13 y = 15]

Question no – (17)

Solution :

Suppose the tens digit is x

Suppose the tens unit digit is = 4x

(10 × x) + (4x × 1) + 54

= (4x × 10) + (x + 1)

or, 10x + 4x + 54 = 40x + x

or, 41x – 14x = 54

or, 27 x = 54

or, x = 54/27 = 2

The tens digit is 2 unit digit is 8

The number is 28

Question no – (18)

Solution :

Suppose the unit digit is

Suppose the unit tens is (9 – x)

10(9 – x) + (x × 1) + 27

= 10 (x) + 1 (9 – x)

or, 90 – 10x + x + 27

= 10x + 9 – x

or, 18x = 108

or, x = 6

Unit digit = 6

Tens digit = 3

The number is 36

Question no – (19)

Solution :

Suppose the unit digit is x

Suppose the unit digit = (2x + 1)

10x + 1 (2x + 1) + 36

= 10 (2x + 1) + (1 ∴ x)

or, 10x + 2x + 1 + 36

= 20x + 10 + x

or, 9x = 27

or, = 27/9 = 3

Tens digit = 3

Unit digit = (3 × 2) + 1 = 7

Question no – (20)

Solution :

Suppose Arun got x rupees Varun got y

x+ y = 76 – (i)

= x – 7 = y + 7

= x – y = 14 – (ii)

Adding (i) and (ii) (x + x) + (y – y)

= (76 + 14)

or, 2x = 90

or, x = 90/2 = 45

= y = 76 – 45 = 31

[Arun got = 45 rupees : Varun got = 31 rupees]

Revision Exercise Questions Solution :

Question no – (1)

Solution :

(a) 6x – 2/3 – 2x – 1/5 = 1/3

or, 6x/3 – 2/3- 2x/5 + 1/5 = 1/3

or, 2x – 2x/5 = 1/3 + 2/3 =  1/5

or, 2x/4 = 4/5

or, x/2 = 4/5

or, x = 8/5

(b) x/3 + 1/2 = x/4 + 1/3

or, x/3 – x/4

= 1/3 – 1/2

or, x/12

= – 1/6

or, x = – 12/6

or, x = – 2

Question no – (2)

Solution :

= 5 – x/7 – x = 2 + x/7 + x

or, x2 – 2x + 35

= x2 + 5x + 14

or, 7x = 21

or, x = 21/7

= 3

Question no – (3)

Solution :

= 3x + 5/2x + 1 = 4/3

or, 9x + 15

= 8x + 4

or, x = 4 – 15

= 11

Question no – (4)

Solution :

= x + 4/x + 6 = x – 2/x – 1

or, x2 + 4x – 12

= x2 + 3x – 4

or, x = 12 – 4

= 8

Question no – (5)

Solution :

= (x + 40)° + (2x – 10)° + (x + 50)° = 180°

or, (x + 2x + x)° + (40 + 50 – 10)° = 180°

or, 4x + 80° = 180°

or, 4x = 180° – 80° = 100°

or, x = 100°/4

= 25°

Question no – (6)

Solution :

Suppose the man’s age is – x years

His father age = 2x years his grandfather age = 3x years

x + 2x + 3x = 150

or, 6x = 150

or, x = 25 years

Question no – (7)

Solution :

Suppose now Sarita is x years old

4x + 5/x + 5 = 3/1

or, 4x + 5 = x + 15

or, 3x = 10

or, x = 10/3

Question no – (8)

Solution :

Suppose the numbers are 9x, 5x

9x + 5x = 56

or, 14x = 56

or, x = 56/14

= 4

Question no – (9)

Solution :

Suppose the unit’s digit is x ten’s digit = (7 – x)

10 (7 – x) + (x × 1) + 27

= (10 × x) + (7 – x)

or, 70x – 10x + x + 27

= 10x + 7 – x

or, 18x = 90

or, x = 90/18

= 5

∴ (7 – x) = 2

The number is 25

Question no – (10)

Solution :

Suppose the tens digit = x

units digit = (x + 2)

x + x + 2 = 16

or, 2x = 14

or, x = 14/2 = 7

(x + 2) = 9

The number is = 79

Question no – (11)

Solution :

Suppose the numerator = x denominator = x + 6

x + 3/x + 6 = 2/3

or, 3x + 9 = 2x + 12

or, 3x – 2x = 12 – 9

or, x = 3

The original fraction = 3/3 + 6

= 3/9

Question no – (12)

Solution :

Suppose he hit x times, missed y times

x+ y = 160 – (i)

4x – y = 440 – (ii)

——————————————-
5x = 600

x = 600/5 = 120

He hit 120 times

Question no – (13)

Solution :

Suppose speed of the stream = x km/hr

= 20/8 + x

= 12/8 – x

or, 96 + 12x = 160 – 20x

or, 12x + 20x – 160 – 96

or, 32x = 64

or, x = 64/32

= 2 km/hr

Question number – (14)

Solution :

Suppose speed of the streams = x km/hr

9 (x + 1) = 10(x – 1)

or, 10x – 9x

= 10 + 9

or, x = 19 km/hr

The total distance 2 × 9 × 919 + 1)

= 18 × 20

= 360 km

Question no – (15)

Solution :

Suppose three were x number of 50 rupees notes

50x + 100 (25 – x) = 200

or, 50x – 100x

= 2000 – 2500

or, 50x = 500

or, x = 500/50 = 10

Three were 10 notes of 50 rupees

Question no – (16)

Solution :

Suppose the least one is 2x

(4 × 2x) – 6

= (2x + 2) + (2x + 4) + (2x + 6)

or, 8x – 6 = 6x + 12

or, 8x – 6x

= 12 + 6

or, 2x = 18

x = 18/2 = 9

The numbers are (9 × 2) = 18

= (9 × 2 + 2) = 20

= 20 + 2 = 22

= 22 + 2 = 24

Next Chapter Solution :

Updated: June 24, 2023 — 11:37 am