Brilliant’s Composite Mathematics Class 8 Solutions Chapter 6

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Brilliant’s Composite Mathematics Class 8 Solutions Chapter 6 Linear Equations

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Brilliant’s Composite Mathematics Class 8 Math Book, Chapter 6, Linear Equations. Here students can easily find step by step solutions of all the problems for Linear Equations, Exercise 6.1 and 6.2 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.

Linear Equations Exercise 6.1 Solution

Question no – (1)

Solution : 

Given, 2x + 1/3x – 5 = 0

Or, 2x + 1 = 0 (3x – 5)

Or, 2x + 1 = 0

Or, 2x = – 1

∴ x = – 1/2

Question no – (2)

Solution : 

Given, 3x – 9/3x – 6 = – 2/3

 Or, 3(3x – 9) = – 2(3x – 5)

Or, 9x – 27 = -6x + 10

Or, 9x + 6x = 10 + 27

Or, 15x = 37

Or, x = 37/15

Question no – (3)

Solution : 

2x – 3/3x + 2 = – 2/3

Or, 3(2x – 30 = – 2(3x + 2)

Or, 6x – 9 = – 6x – 4

Or, 6x + 6x = 9 – 4

Or, 12x = 5

Or , x = 5/12

Question – (4)

Solution : 

y – (7 – 8y) / 9y – (3 + 4y)

Printing deficiency.

Question no – (5)

Solution : 

Given, 2/x – 7 = 3/x – 4

Or, 3(x – 7) = 2(x – 4)

Or, 3x – 21 = 2x – 8

Or, 3x – 2x = 21 – 8

Or, x = 13.

Question no – (6)

Solution : 

Given, 2x + 3/ 3x + 4 = 5

Or, 5(3x + 4) = 2x + 4

Or, 15x + 20 = 2x + 4

Or 15x – 2x = 4 – 20

Or, 13x = – 16

Or, x = – 16/13

Question no – (7)

Solution : 

Given, 5x – 7/ 3x = 2

Or, 2(3x) = 5x – 7

Or, 6x = 5x – 7

Or, 6x – 5x = – 7

Or, x = – 7

Question no – (8)

Solution : 

Given,  -4x + 3/2x + 6 = – 3

Or, – 3 (2x + 3) = (- 4x + 3)

Or, – 6x – 9 = – 4x – 3

Or, – 6x + 4x = 9 – 3

Or, – 2x = 6

Or, x = 6/- 2 = – 3

Question no – (9)

Solution : 

Given, 8/x = 9/2x – 1

Or, 8(2x – 1) = 9x

Or,  16x – 8 = 9x

Or, 16x – 9x = 8

Or, 7x = 8

Or, x = 8/7

Question no – (10)

Solution : 

Given, x – (6 – x)/9x – (7 – x) = 8

Or, x – 6  x/9x – 7 + x = 8

Or, 2x – 6/10x – 7 = 8

Or, 8(10x – 7) = 2x – 6

Or, 80x – 56 = 2x – 6

Or, 80x – 2x = 56 – 6

Or, 78x = 50

Or, x = 50/78

Or, x = 25/39

Linear Equations Exercise 6.2 Solution

Question no – (1) 

Solution : 

Suppose, the number is x.

∴ 4x – 15 = 9

Or, 4x = 9 + 15

Or, 4x = 24

Or, x = 24/4

Or, x = 6

Question no – (2) 

Solution : 

Suppose, the age of Ram and Puneet are 5x, 7x years.

∴ 5x + 9/7x – 9 = 2/1

Or, 2(7x – 9) = 1(5x + 9)

Or, 14x – 18 = 15x + 9

Or, 14x – 5x = 9 + 18

Or, 9x = 27

Or, x = 27/9

Or, x = 3

∴ Ages of Ram and Puneet are,

= (5 × 3)

= 15 years,

= (7 × 3) = 21 years.

Question no – (3) 

Solution : 

Suppose, the number for = x.

∴ x + 11/(x + 4) – 1 = 7/3

Or, x + 11/x + 3 = 7/3

Or, 7(x + 3) = 3(x + 11)

Or, 7x – 3x = 33 – 21

Or, 4x = 12

Or, x = 12/4

∴ x = 3

Hence, the original number is 3.

Question no – (4) 

Solution :  

Let the number be x.

∴ 3x – 5 = – 20

Or, 3x = – 20 + 5

Or, 3x = – 15

Or, x = – 15/3

Or, x = – 5

There fore, the number will be -5

Question no – (5) 

Solution :  

Let the age of the son be x years.

∴ The age of the father will be 5x years.

Now, Their Sum = 5x + x = 48

Or, 6x = 48

Or, x = 48/6

Or, x = 8

∴ The age of the son = 8 years and the age of the father = 5 × 8 = 40 years.

Question no – (6) 

Solution :  

Let the salary of B be x rupees

∴ Then the salary of a will be 2x rupees.

∴ 2x + x = 2754

Or, 3x = 2754

Or, x = 2754/3

Or, x = 918 rupees.

∴ Salary of B = 918 rupees,

Salary of a = 918 × 2 = 1836 rupees.

Question no – (7) 

Solution :  

Let the weights of silver and gold be x gm and (30 – x) gm in air.

∴ loss of eights  x/10 + (30 – x)/19 = (30 – 28)

Or, 19x + 10(3 – x)/190 = 2

Or, 19x + 300 – 10x/90 = 2

Or, 300 – 9x = 90 × 2

Or, – 9x = 180 – 300

Or, – 9x = – 120

Or, 9x = 120

Or, x = 120/9

Or, x = 40/3gm

∴ Weight of silver = 40/3g

= 13 1/3g.

∴ Weight of gold,

= 30 – 40/3

= 30 – 13 1/3

= 16 2/3g

Question no – (8) 

Solution :  

Let the numbers be x, 50 + x

∴ Clearly, x/50 + x  =  3/8

Or, 8x = 3 (50+ x)

Or, 8x = 150 + 3x

Or, 8x – 3x = 150

Or, 5x = 105

Or, x = 150/5

∴ x = 30

Hence, the smaller number is 30 and bigger number is 30 + 50 = 80

Question no – (9) 

Solution :  

Let the breath be x cm. then length = (x + 7)cm

∴ The area = x(x + 7)cm²

Now, the reduced length

= x + 7 – 4

= (x + 3) cm

And the increased breath = ( x + 3) cm

∴ The new area = (x + 3)(x + 3) cm²

∴ Since both areas are equal , we have

X(x + 7) = (x + 3)(x + 3)

Or, x2 + 7x = x2 + 6x + 9

Or, 7x = 6x + 9

Or, 7x – 6x = 9

Or, x = 9 cm

∴ The original breadth = 9 cm, original length = 9 + 7

= 16 cm

Question no – (10) 

Solution :  

Let the shortest side be = x cm

∴ The length of the second side = (x + 7)cm

∴ The length of the second side = (x + 7) + 5 cm

= (x + 12)cm

∴ x + (x + 7) + (x + 12) = 49

Or, (x + x + x) + (7 + 12) = 49

Or, 3x + 19 = 49

Or, 3x = 49 – 19 = 30

Or, x = 30/3

Or, x = 10

∴ The length of sides = 10 cm,

∴  (10 + 7) = 17 cm,

∴  (10 + 7 + 5) = 22 cm

Question no – (11) 

Solution :  

Let Susheel’s son’s age be x years, 2 years age.

∴ Then susheel’s age was = 3x years.

∴ 3x + 4/x + 4= 5/2

Or, 2(3x + 4) = 5(x + 4)

Or, 6x + 8 = 5x + 20

Or, 6x – 5x = 20 – 8

∴ x = 12 years.

∴ Susheel’s sons present age,

= 12 + 2

= 14 years.

∴ Susheel present age,

= (12 × 3) + 2

= 36 + 2

= 38 years.

Question no – (12) 

Solution :  

Let the speed of boat in still water be x km/hr

∴ (x + 2) × 4 = (x – 2) × 5

Or, 4x + 8 = 5x – 10

Or, 5x – 4x = 10 + 8

Or, x = 18 km/hr

L.H.S :- (18 + 2) × 4

= 20× 4

= 80

R.H.S :- (18 – 2) × 5

= 16 × 5

= 80

Since , L.H.S = R.H.S

∴ The solution is correct.

Question no – (13) 

Solution :  

Let the breadth be x cm, Length = (x + 9) cm.

∴ area = x(x + 9)cm²

The increased breadth and length = (x + 3), (x + 12) cm.

∴ new area = (x + 3)(x + 12)

∴ we have, (x + 3)(x + 12) – x(x + 9) = 48

Or, x² + 15x + 36 – x² – 9x = 84

Or, 15x – 9x = 84 – 36

Or, 6x = 48

Or, x = 48/6

Or, x = 8

∴ Breadth = 8 cm,

∴ length = (8 + 9) = 17 cm.

Question no – (14) 

Solution : 

Let’s tens digit = x, unit’s digit = (15-x)

∴ The number is =  10 x + (15 – x). When the digits are reversed, then the number will be 10 (15-x) + x = 150 – 10x + x = 150 – 9x.

∴ We have {10x + (15 – x )} – {150 – 9x} =27

Or, {10x + 15 – x } – 150 + 9x = 27

Or, 9x + 15 – 150 + 9x = 27

Or, 18x – 135 + 27

Or, 18x = 135 + 27 = 162

Or,  x = 162/18 = 9

∴ Ten’s digit = 9,

∴ unit’s digit  = (15 – 9) = 6

Therefore, The number is 96.

Question no – (15) 

Solution : 

Let the speed of the slower train be x km/hr

∴ The speed of the faster train,

= (x + x + 5)

= (2x + 5) km/hr .

In 3 hours together they crossed

= 425 – 20

= 405 km.

∴ We have, 3 (2x + 5 )

= 405 or, 6x + 15 = 405

or, 6x = 405 – 15

or, 6x = 390

or, x = 390/6

= 65 km/hr.

∴ Speed of the slower train = 65 km/hr.

Question no – (16) 

Solution : 

Let the altitude be 2x cm and the corresponding base be 3x cm.

∴ Area = 1/2 × 2x × 3x.

Now, 1/2 ×2x × 3x

= 1/2 x (2x – 1) × ( 3x + 4 )

Or, 6x² = 6x² + 8x – 3x – 4

Or, 8x – 3x = 4

Or, 5x = 4

Or, x = 4/5

Question no – (17) 

Solution : 

Let the volume of the gold piece be x cm^3, and the volume of the silver piece be 3x cm³

∴ (10.5) (x) +   (8.5) (3x) = 1350

Or, 10.5x = 25.5x = 1350

Or, 36x = 1350

Or, x = 1350/36

Or, x = 75/2

∴ x = 37.5 cm³

∴ The volume of gold piece = 37.5 cm³ and the volume of silver piece = 37.5 × 3 = 112.5 cm³

Question no – (18) 

Solution : 

Let the area be x hectors, Which can be ploughed daily.

∴ We have, 14x = (x+20) c 10

Or, 14x = 10x + 200

Or, 14x – 10x = 200

Or, 4x = 200

Or, x = 200/4 = 50 hectors 0

∴ 50 hectors can be ploughed daily.

The area of the field = 50 × 14 = 700 hectors.

Question no – (25) 

Solution : 

Let the numbers be x, (2490 – x ).

= x × 6.5/100 =(2490 –x) × 8.5/100

Or, x × 65/1000 = (2490 – x ) × 85/1000

Or, 65x  = (2490 – x ) × 85

Or, 13x = (2490 – x ) × 17

Or, 13x = 2490 × 17 – 17x

Or, 13x + 17x = 2490 × 17

Or, 30x = 2490 × 17

Or, x = 2490 × 17/30

∴ x = 83 × 17 = 1431

∴ The numbers are 1431, (2490-1431) = 1059

Question no – (27) 

Solution : 

Let the angles be (x – 2), x, (x + 2)

∴ We have, (x – 2) + x + (x + 2) = 57

Or, 3x = 57

Or, x = 57/3 = 19

∴ The numbers are 17, 19, 21

Question no – (28) 

Solution : 

Let the angles be 4x,5x,9x

∴ 4x + 5x + 9x = 180

Or, 18x = 180

Or, x = 180/18 = 10°

∴ (4×10) = 40°,

∴ (5×10) = 50°,

∴ (9×10°) = 90°

Hence, the angles are 40°, 50° and 90°

Previous Chapter Solution : 

👉 Chapter 5