NCERT Solutions for Linear equation in One Variable Class 8 Maths Chapter 2 Exercise 2.5

NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.5

Question 1
Solve the linear equation
Answer

L.C.M. of the denominators, 2, 3, 4, and 5, is 60.
So Multiplying both sides by 60, we obtain
$30x- 12 = 20x+ 15$
Transposing 20x to R.H.S and 12 to L.H.S, we obtain
$30x- 20x= 15 + 12$
10x= 27
X=27/10=2.7

Question 2
Solve the linear equation
Answer

L.C.M. of the denominators, 2, 4, and 6, is 12.
Multiplying both sides by 12, we obtain
$6n- 9n+ 10n= 252$
7n= 252
Dividing by 7 on both the sides
n=36

Question 3
Solve the linear equation
Answer

L.C.M. of the denominators, 2, 3, and 6, is 6.
Multiplying both sides by 6, we obtain
6x+ 42 - 16x= 17 - 15x
Transposing 15x to RHS and 42 to LHS
6x- 16x+ 15x= 17 - 42
5x= -25
Dividing by 5 on both the sides
x=-5

Question 4
Solve the linear equation
Answer

L.C.M. of the denominators, 3 and 5, is 15.
Multiplying both sides by 15, we obtain
$5(x- 5) = 3(x- 3)$
$5x- 25 = 3x- 9$
Transposing 3x to RHS and 25 to LHS
5x- 3x= 25 - 9
2x= 16
Dividing both the sides by 2
x=8

Question 5
Solve the linear equation
Answer

L.C.M. of the denominators, 3 and 4, is 12.
Multiplying both sides by 12, we obtain
$3(3t- 2) - 4(2t+ 3) = 8 - 12t$
$9t- 6 - 8t- 12 = 8 - 12t$
Transposing 12t to RHS and 6 and 12 to LHS
9t- 8t+ 12t= 8 + 6 + 12
13t= 26
Dividing by 13 on both the sides
t=2

Question 6
Solve the linear equation
Answer

L.C.M. of the denominators, 2 and 3, is 6.
Multiplying both sides by 6, we obtain
$6m- 3(m- 1) = 6 - 2(m- 2)$
$6m- 3m+ 3 = 6 - 2m+ 4$
Transposing 2m to RHS and 3 to LHS
6m- 3m+ 2m= 6 + 4 - 3
5m= 7
Dividing by 5 on both the sides
m=7/5



Question 7
Simplify and solve the linear equation $3(t- 3) = 5(2t+ 1)$

Answer
$3(t- 3) = 5(2t+ 1)$
3t- 9 = 10t+ 5
-9 - 5 = 10t- 3t
-14 = 7t
t=-2
Question 8
Simplify and solve the linear equation $15(y- 4) - 2(y- 9) + 5(y+ 6) = 0$
Answer
$15(y- 4) - 2(y- 9) + 5(y+ 6) = 0$
15y- 60 - 2y+ 18 + 5y+ 30 = 0
18y- 12 = 0
18y= 12 y=12/18=2/3
Question 9
Simplify and solve the linear equation
$3(5z- 7) - 2(9z- 11) = 4(8z- 13)-17$
Answer
$3(5z- 7) - 2(9z- 11) = 4(8z- 13)-17$
15z- 21 - 18z+ 22 = 32z- 52 - 17
-3z+ 1 = 32z- 69
-3z- 32z= -69 - 1
-35z= -70
z=2

Question 10
Simplify and solve the linear equation $0.25(4f- 3) = 0.05(10f- 9)$
Answer
$0.25(4f- 3) = 0.05(10f- 9)$
Multiplying both sides by 20, we obtain
$5(4f- 3) = 10f- 9$
20f- 15 = 10f- 9
20f- 10f= - 9 + 15
10f= 6
f==6/10=3/5


Download NCERT Solutions Exercise 2.5 as pdf
link to this page by copying the following text
<a href="https://physicscatalyst.com/class8/ncert_solutions_class8_maths_linear_equation_5.php">NCERT Solutions for Linear equation in One Variable Class 8 Maths Chapter 2 Exercise 2.5</a> Also Read

---

• Notes
  • Linear equations in One variable Class 8 notes
• Worksheets
  • Linear equations in One variable class 8 important questions
  • linear equations in one variable class 8 worksheets
  • Practice Questions for Linear equations Class 8 maths
  • Linear equations in One variable Test Paper
  • Linear equations in One variable Problems with Answers
• Ncert Solutions
  • NCERT Solution Linear equations in One variable Exercise 2.1
  • NCERT Solution Linear equations in One variable Exercise 2.2
  • NCERT Solution Linear equations in One variable Exercise 2.3
  • NCERT Solution Linear equations in One variable Exercise 2.4
  • NCERT Solution Linear equations in One variable Exercise 2.5
  • NCERT Solution Linear equations in One variable Exercise 2.6

---

Class 8 Maths Class 8 Science