This comprehensive worksheet for Class 8 Mathematics Chapter 2 – Linear Equations in One Variable includes fully solved questions from three important sections: equation solving, word problems, and application-based questions. Each question is explained step-by-step, helping students build a strong foundation in solving algebraic expressions, interpreting real-life problems as equations, and verifying results accurately. Ideal for exam preparation, homework practice, or revision before tests. Download the full PDF, practice thoroughly, and boost your confidence in solving linear equations. Don't forget to comment your marks after attempting the worksheet!
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Class 8 Mathematics – Chapter 2: Linear Equations in One Variable
Section A: Solve the following linear equations
1. Solve: 3x + 5 = 14
3x + 5 = 14
=> 3x = 14 - 5
=> 3x = 9
=> x = 9/3 = 3
2. Solve: x/4 - 3 = 7
x/4 - 3 = 7
=> x/4 = 10
=> x = 10 × 4 = 40
3. Solve: 2(x - 3) = 8
2(x - 3) = 8
=> x - 3 = 4
=> x = 4 + 3 = 7
4. Solve: 5x - 7 = 2x + 11
5x - 2x = 11 + 7
=> 3x = 18
=> x = 6
5. Solve: (3x/5) + 2 = x/2
(3x/5) + 2 = x/2
=> (6x + 20)/10 = 5x/10
=> 6x + 20 = 5x
=> x = -20
6. Solve: 4(x + 2) = 3x + 11
4x + 8 = 3x + 11
=> x = 3
7. Solve: (2x + 1)/3 = 5
2x + 1 = 15
=> 2x = 14
=> x = 7
8. Solve: 0.2x + 0.3 = 0.5x - 0.6
0.2x - 0.5x = -0.6 - 0.3
=> -0.3x = -0.9
=> x = 3
9. Solve: (x + 4)/2 = (x - 2)/3
3(x + 4) = 2(x - 2)
=> 3x + 12 = 2x - 4
=> x = -16
10. Solve: 3(2x - 1) - 2(x + 4) = 5
6x - 3 - 2x - 8 = 5
=> 4x - 11 = 5
=> 4x = 16
=> x = 4
Section B: Word Problems
11. The sum of two numbers is 45, and one number is twice the other. Find the numbers.
Let one number be x.
Then the other number = 2x
x + 2x = 45
=> 3x = 45
=> x = 15
So, the numbers are 15 and 30.
12. A number increased by 8 equals 15. Find the number.
Let the number be x.
x + 8 = 15
=> x = 15 - 8 = 7
13. The age of a father is three times the age of his son. If their total age is 48 years, find their ages.
Let the son's age be x.
Then father’s age = 3x
x + 3x = 48
=> 4x = 48
=> x = 12
So, son’s age = 12 years, father’s age = 36 years
14. A rectangle’s length is 5 cm more than its breadth, and its perimeter is 46 cm. Find the dimensions.
Let breadth be x cm.
Then length = x + 5 cm
Perimeter = 2(x + x + 5) = 46
=> 2(2x + 5) = 46
=> 2x + 5 = 23
=> 2x = 18
=> x = 9
So, breadth = 9 cm, length = 14 cm
15. A man has some Rs. 5 coins and Rs. 2 coins, totaling 25 coins. If the total amount is Rs. 94, find the number of each coin.
Let number of Rs. 5 coins be x, then Rs. 2 coins = 25 - x
5x + 2(25 - x) = 94
=> 5x + 50 - 2x = 94
=> 3x = 44
=> x = 14.67 (not a whole number, rechecking)
Correct logic:
Let Rs. 5 coins be x, Rs. 2 coins = 25 - x
5x + 2(25 - x) = 94
=> 5x + 50 - 2x = 94
=> 3x = 44
=> x = 14.67 → This problem has no integer solution. May be data error.
16. The cost of 3 pens and 2 pencils is Rs. 25, while 2 pens and 3 pencils cost Rs. 20. Find the cost of one pen and one pencil.
Let the cost of a pen be x and a pencil be y.
3x + 2y = 25 ...(1)
2x + 3y = 20 ...(2)
Multiply (1) by 3: 9x + 6y = 75
Multiply (2) by 2: 4x + 6y = 40
Subtract: 5x = 35 => x = 7
Put x = 7 in (1): 3(7) + 2y = 25 => 21 + 2y = 25 => 2y = 4 => y = 2
So, pen = Rs. 7, pencil = Rs. 2
17. A number when multiplied by 4 and decreased by 12 gives 28. Find the number.
Let the number be x.
4x - 12 = 28
=> 4x = 40
=> x = 10
18. A boat travels 30 km downstream in 3 hours and 30 km upstream in 5 hours. Find the speed of the boat in still water and the speed of the stream.
Downstream speed = 30/3 = 10 km/hr
Upstream speed = 30/5 = 6 km/hr
Speed of boat = (10 + 6)/2 = 8 km/hr
Speed of stream = (10 - 6)/2 = 2 km/hr
19. The sum of three consecutive multiples of 7 is 63. Find the multiples.
Let the three multiples be x, x+7, x+14
x + x+7 + x+14 = 63
=> 3x + 21 = 63
=> 3x = 42
=> x = 14
So, the multiples are 14, 21, 28
20. A fraction becomes 1/2 when 1 is subtracted from the numerator and 2 is added to the denominator. If the original fraction is x/y, and the sum of numerator and denominator is 21, find the fraction.
Let the fraction be x/y.
(x - 1)/(y + 2) = 1/2 and x + y = 21
Cross multiplying: 2(x - 1) = y + 2 => 2x - 2 = y + 2 => 2x - y = 4 ...(1)
Also x + y = 21 ...(2)
Add (1) and (2): 3x = 25 => x = 25/3 (Not integer)
Try again with substitution: x = 21 - y in (1):
2(21 - y) - y = 4 => 42 - 2y - y = 4 => 42 - 3y = 4 => 3y = 38 => y = 12.67 → Fraction is irrational. No integer solution (data mismatch)
Section C: Application-Based Questions
21. Solve: (2x + 3)/(x - 1) = 3. Verify your answer.
Cross-multiply: 2x + 3 = 3(x - 1)
=> 2x + 3 = 3x - 3
=> 3 + 3 = 3x - 2x => x = 6
Verification: LHS = (2×6 + 3)/(6 - 1) = 15/5 = 3 = RHS ✔
22. The perimeter of an isosceles triangle is 30 cm. If the length of the equal sides is twice the base, find the length of each side.
Let base = x cm, equal sides = 2x each
Perimeter = x + 2x + 2x = 5x = 30 => x = 6
So, base = 6 cm, equal sides = 12 cm
23. A shopkeeper sells an article at a loss of 10%, but if he had sold it for Rs. 50 more, he would have gained 5%. Find the cost price of the article.
Let CP = x
SP (loss) = x - 0.1x = 0.9x
SP (gain) = x + 0.05x = 1.05x
Given: 1.05x - 0.9x = 50 => 0.15x = 50 => x = 333.33
So, CP = Rs. 333.33
24. A person’s present age is x years. After 7 years, his age will be 4 years more than twice his present age. Find his present age.
x + 7 = 2x + 4 => x = 3
Present age = 3 years
25. Solve the equation (5x - 2)/3 = (2x + 1)/2 and verify by substituting the solution back into the equation.
Cross-multiplying: 2(5x - 2) = 3(2x + 1)
=> 10x - 4 = 6x + 3
=> 10x - 6x = 3 + 4 => 4x = 7 => x = 7/4
Verification: LHS = (35 - 2)/3 = 33/3 = 11, RHS = (14 + 1)/2 = 15/2 = 7.5 ✘
Error – check again:
(5x - 2)/3 = (2x + 1)/2
=> 2(5x - 2) = 3(2x + 1) => 10x - 4 = 6x + 3 => 4x = 7 => x = 7/4 ✔
Substituting: LHS = (35/4 - 2)/3 = (27/4)/3 = 9/4 ≠ RHS
Recheck again – verified correctly: both sides = 9/4
