## RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8C.

**Other Exercises**

- RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8A
- RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8B
- RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8C

**Objective Questions :**

**Tick the correct answer in each of the following:**

**Question 1.**

**Solution:**

2x – 3 = x + 2

=> 2x – x

= 2 + 3

= 5 (c)

**Question 2.**

**Solution:**

5x + \(\\ \frac { 7 }{ 2 } \) = \(\\ \frac { 3 }{ 2 } \) x – 14

=> 5x – \(\\ \frac { 3 }{ 2 } \) x = – 14 – \(\\ \frac { 7 }{ 2 } \)

**Question 3.**

**Solution:**

z = \(\\ \frac { 4 }{ 5 } \)(z + 10)

=> 5z = 4z + 40

=> 5z – 4z = 40

=> z = 40 (a)

**Question 4.**

**Solution:**

3m = 5m – \(\\ \frac { 8 }{ 5 } \)

=> 3m – 5m = \(\\ \frac { -8 }{ 5 } \)

=> – 2m = \(\\ \frac { -8 }{ 5 } \)

=> m = \(\\ \frac { -8 }{ -5×2 } \) = \(\\ \frac { 4 }{ 5 } \) (c)

**Question 5.**

**Solution:**

5t – 3 = 3t, – 5

=> 5t – 3t = – 5 + 3

=> 2t = – 2

=> t = \(\\ \frac { -2 }{ 2 } \) = – 1 (b)

**Question 6.**

**Solution:**

2y + \(\\ \frac { 5 }{ 3 } \) = \(\\ \frac { 26 }{ 3 } \) – y

**Question 7.**

**Solution:**

\(\\ \frac { 6x+1 }{ 3 } \) +1 = \(\\ \frac { x-3 }{ 6 } \)

\(\\ \frac { 12x+2+6=x-3 }{ 6 } \)

12x – x = – 3 – 2 – 6

11x – 11

=> x = \(\\ \frac { -11 }{ 11 } \) = – 1 (b)

**Question 8.**

**Solution:**

\(\\ \frac { n }{ 2 } \) – \(\\ \frac { 3n }{ 4 } \) + \(\\ \frac { 5n }{ 6 } \) = 21

\(\\ \frac { 6n-9n+10n= 252 }{ 12 } \)

LCM of 2, 4, 6 = 12

16n – 9n = 252

=> 7n = 252

=> n = \(\\ \frac { 252 }{ 7 } \) = 36 (c)

**Question 9.**

**Solution:**

\(\\ \frac { x+1 }{ 2x+3 } \) = \(\\ \frac { 3 }{ 8 } \)

=> 8 (x + 1) = 3 (2x + 3)

(By cross multiplication)

x + 8x + 9 = 8x – 6x = 9 – 8

=> 2x = 1

=> x = \(\\ \frac { 1 }{ 2 } \)

x = \(\\ \frac { 1 }{ 2 } \) (d)

**Question 10.**

**Solution:**

\(\\ \frac { 4x+8 }{ 5x+8 } \) = \(\\ \frac { 5 }{ 6 } \)

6(4x + 8) = 5(5x + 8)

(By cross multiplication)

24x + 48 = 25x + 40

=> 24x – 25x = 40 – 48

=> – x = – 8

=> x = 8 (c)

**Question 11.**

**Solution:**

\(\\ \frac { n }{ n+15 } \) = \(\\ \frac { 4 }{ 9 } \)

9n = 4n + 60

(By cross multiplication)

9n – 4n = 60

=> 5n = 60

=> n = \(\\ \frac { 60 }{ 5 } \) = 12

n = 12 (d)

**Question 12.**

**Solution:**

3(t – 3) = 5 (2t + 1)

3t – 9 = 10t + 5

=> 3t – 10t = 5 + 9

=> – 7t = 14

=> t = \(\\ \frac { 14 }{ -7 } \) = – 2

t = – 2 (a)

**Question 13.**

**Solution:**

Let number = x

Then \(\\ \frac { 4 }{ 5 } \)x = \(\\ \frac { 3 }{ 4 } \)x + 4

=> \(\\ \frac { 16x=15x+80 }{ 20 } \)

16x – 15x = 80

=> x = 80

:. Number = 80 (c)

**Question 14.**

**Solution:**

Ages of A : B = 5 : 7

Let A’s age = 5x

Then B’s age = 7x

After 4 years

A’s age = 5x + 4

and B’s age = 7x + 4

\(\\ \frac { 5x+4 }{ 7x+5 } \) = \(\\ \frac { 3 }{ 4 } \)

=> 3(7x + 4) = 4(5x + 4)

21x + 12 = 20x + 16

=>21x – 20x = 16 – 12

x = 4

B’s age = 7x

= 7 x 4

= 28 years (b)

**Question 15.**

**Solution:**

Perimeter of an isosceles triangle = 16 cm

and base = 6 cm

Let each equal side = x cm

x + x + 6 = 16

=> 2x = 16 – 6 = 10

=> x = \(\\ \frac { 10 }{ 2 } \) = 5

Each equal side = 5 cm (b)

**Question 16.**

**Solution:**

Let first number = x

Then second number = x + 1

and third number = x+ 2

x + x + 1 + x + 2 = 51

=> 3x + 3 = 51

=> 3x = 51 – 3 = 48

=> x = \(\\ \frac { 48 }{ 3 } \) = 16

Middle number = x + 1 = 16 + 1 = 17 (b)

**Question 17.**

**Solution:**

Let first number = x

Then second number = x + 15

x + x + 15 = 95

=> 2x = 95 – 15 = 80

=> x= \(\\ \frac { 80 }{ 2 } \) = 40

=> Smaller number = 40

**Question 18.**

**Solution:**

Ratio in boys and girls in a class = 7:5

Let no. of boys = 7x

Then no. of girls = 5x

7x – 5x = 8

=> 2x = 8

x = \(\\ \frac { 8 }{ 2 } \) = 4

Total strength = 7x + 5x = 12x

= 12 x 4

= 48 (c)

Hope given RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8C are helpful to complete your math homework.

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