# Maths Ace Class 8 Solutions Chapter 8

## Maths Ace Class 8 Solutions Chapter 8 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 Maths Ace Prime Class 8 Math Book, Chapter 8, 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 8.1 and 8.2 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. Here in this post students will get chapter 8 solutions.

Linear Equations in One Variable Exercise 8.1 Solution :

Question no – (1)

Solution :

(a) -s/24 = 7/6

= – 6s = 7 × 24

= s = 7 × 24/- 6

= – 28

Therefore, the value of unknown variable is – 28

(c) 6/3x + 10/3x = 256

= 6/3x + 10/3x = 256

= 6 + 10/3x = 256

= 3x × 256 = 16

= x = 16/256 × 3

= 1/48

Thus, the value of unknown variable is 1/48

(d) 8/y + 3 = 10

= 10y + 30 = 8

= 10y = 8 – 30

= y = – 22/10

= – 2.2

Thus, the value of unknown variable is – 2.2

(e) a + 7/2 + 3a/5 = 1

= 5a + 35 + 6a/10 = 1

= 11a + 35 = 10

= 11a = 10 – 35

= a =- 25/11

= – 2 3/11

Hence, the value of unknown variable -2 3/11.

(f) 3a + 2/5 = 5

= 3a = 25 – 2

= a = 23/3

= 7 2/3

Thus, the value of unknown variable 7 2/3

Question no – (2)

Solution :

(a) 2(0.25y – 1) + 1.5y = 8

2(0.25y – 1) + 1.5y = 8

= 0.5y – 2 + 1.5y = 8

= 2y = 8 + 1

= y = 10/2

= 5

(b) x -2/3 – [3/4 – x – 3/2] – x – 5/6 – x = 6/5

Now, x -2/3 – [3/4 – x – 3/2] – x – 5/6 – x = 6/5

= x – 2/3 – [3 – 2x + 6/4] – x – 5/6 – x = 6/5

= x – 2/3 – 9 – 2x/4 – x – 5/6 – x/1 = 6/5

= 4x – 8 – 27 + 6x – 2×2 + 10 – 12x/12 = 6/5

= – 4x – 25/12 = 6/5

= – 20x = 72 + 125

= x = – 197/20

= – 9 17/20

(e) -7p + 13/6 = 2p – 1

Now, -7p + 13/6 = 2p – 1

= -7p + 13 = 12p – 6

= -7p – 12p = -6 -13

= -19p = -19

= p = 1

(f) x + 1/4 + x + 2/2 = x – 3/2 – x – 1/3

Now, x + 1/4 + x + 2/2 = x – 3/2 – x – 1/3

= x + 1 + 2x + 4/4 = 3x – 9 – 2x + 2/6

= 3x + 5/4 = x – 7/6

= 18x + 30 = 4x – 28

= 18x – 4x = -28 – 30

= 14x = -58

= x = -58/14

= -4 2/14

= -4 1/7

(g) 2(1/a + 2) = 5 – 2/a

Now, 2(1/a + 2) = 5 – 2/a

= 2(1 + 2a/a) = 5a – 2/a

= 2 + 4a = 5a – 2

= 4a – 5a = -2 -2

= -a = -4

= a = 4

(h) 4/3x – 2/x = 5/3x – 5

Now, 4/3x – 2/x = 5/3x – 5

= 4 – 6/3x = 5/3x – 5

= -2/3x = 5/3x – 5

= -6x + 10 = 15x

= -21x = -10

= x = 10/21

(i) -1/x = – 7/2 + 2/x

So now, – 1/x = – 7/2 + 2/x

= – 1/x – 2/x = – 7/2

= -1 – 2/x = -7/2

= – 3/x = – 7/2

= 7x = 6

= x = 6/7

Linear Equations in One Variable Exercise 8.2 Solution :

Question no – (1)

Solution :

Let, number is x

3x – 1/2x = 2500

= 6x – x/2 = 2500

= 5x = 5000

= x = 5000/5

= 1000

Therefore, the number will be 1000.

Question no – (2)

Solution :

Let, the number is x

x/5 + 20 = 120

= x + 100/5 = 120

= x = 600 – 100

= x = 500

Thus, the required number is 500.

Question no – (3)

Solution :

Let, the length is = x cm

x + x + 72 = 252

= 2x = 252 – 72

= 180

= x = 180/2

= 90 cm

Therefore, the length of its equal sides will be 90 cm.

Question no – (4)

Solution :

Let, the number is = ‘x’

x + 3/4 × 3(-6/7) = 17/13

= x + 3/4 × -18/7 = 17/13

= x – 54/28 = 17/13

= x = 17/13 + 54/28

= 238 + 351/182

= 583/182

= 1184/364

Hence, 1184/364 should be added to the sum.

Question no – (5)

Solution :

Let, the required number is = x

x – 1/2 × 2/3 = 1/6

= x – 2/6 = 1/6

= x = 1/6 + 2/6

= 3/6

= 1/2

Therefore, the required number will be 1/2.

Question no – (6)

Solution :

Length is x × 2 3/4

= 11x/4

Now, 2 (11x/4 + x) = 75

= 2 (11x + 4x/4) = 75

= 2 × 15x/4 = 75

= 30x = 300

= x = 300/30

= 10 cm

Length = 11 × 10/4

= 55/2

= 27.5 cm

Therefore, the length is 10 cm and breadth will be 27.5 cm.

Question no – (7)

Solution :

Let, present age = x

3/5 × x – (x – 4) = 0

= 3x/5 – x + 4 = 0

= 3x/5 – x = -4

= 3x – 5x/5 = -4

= -2x/5 = -4

= x = 4 × 5/2 = 10

After 5 years, Ramani’s age

= (10 + 5)

= 15 years

Therefore, after 5 years his age will be 15 years.

Question no – (8)

Solution :

As per the given question,

The ratio of the ages of Ram and Gopal is 3 : 4.

The difference between their ages after two years will be 10.

Now, Let, Ram’s and Gopal’s age 3x and 4x

(4x + 2) – (3x + 2) = 10

= 4x + 2 – 3x – 2 = 10

= x = 10

Ram’s age,

= (3 × 10)

= 30 years

Gopal’s age,

= (4 × 10)

= 40 years

Thus, Ram ‘s age will be 30 years and Gopal’s age will be 40 years

Question no – (9)

Solution :

The breadth of a rectangular park is 5/6 of its length.

Suresh takes two rounds around the park and covers a distance of 1540 m

Now, Let, Length x

Perimeter,

= 2 (x + 5x/6)

= 2 × (6x + 5x/6)

= 2 × 11x/6

= 22x/6

In two turn

= 2 × 22x/6

= 22x/3

22x/3 = 1540

= x = 3 × 1540/22

= 210 m (Length)

= (210 × 5/6)

= 175 m

Therefore, the length of the park is 210 m and breadth of the park is 175 m

Question no – (10)

Solution :

The perimeter of an isosceles triangle is 90 m.

The length of the two equal sides is 3/4 of the length of unequal side.

Therefore,

Let, length of unequal side = x

x + 3x/4 + 3x/4 = 40

= 4x + 3x + 3x/4 = 90

= 10x = 4 × 90

= x = 4 ×90/10

= 36 m (unequal side)

3x/4 = 3 × 36/4

= 27 m, {Equal side}

Therefore, the dimensions of the triangle will be 36 m and 27 m.

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Updated: June 16, 2023 — 6:52 am