Chapter-wise Worksheets for Class 8 Mathematics: Chapter 02 Linear Equations in One Variable
Explore structured practice materials through the CBSE Class 8 Mathematics Linear Equations in One Variable Practice Paper Worksheet Set 08 with Answers (Old NCERT Syllabus). Tailored for Class 8 learners, utilizing these Mathematics worksheets ensures thorough preparation and strengthens problem-solving accuracy before final school evaluations.
Practice Class 8 Mathematics Worksheets: Chapter 02 Linear Equations in One Variable
Access the complete worksheet PDF for Class 8 Mathematics below. Regular practice with these targeted academic tasks builds familiarity with standard question patterns and helps secure higher marks in final school examinations.
This Class 8 Maths Linear Equations in One Variable practice paper (40 marks) has 20 questions: ten 1-mark MCQs including two Assertion–Reason items, short and long answer questions, and two case studies, each with a step-by-step answer. It follows the earlier (old) NCERT Class 8 Maths syllabus, which makes it handy for state-board students and for revising this topic. It checks equations with no solution, solving equations with brackets and fractions, cross-multiplication, word problems on numbers and marks, and fare equations.
Section A: Multiple Choice Questions (1 Mark Each)
(a) remains unchanged
(b) becomes double
(c) becomes zero
(d) becomes undefined
Show Answer & Explanation
Answer: (a) remains unchanged
Explanation:
1. Multiplying both sides of an equation by the same non-zero number keeps the two sides equal.
2. For example, x + 1 = 4 has solution x = 3; multiplying by 2 gives 2x + 2 = 8, whose solution is still x = 3.
3. So the solution remains unchanged.
Teacher's Note:
1. This rule is what lets us clear fractions by multiplying by the LCM.
2. The number must be non-zero: multiplying by 0 makes every value of x satisfy 0 = 0.
(a) 3x + 2 = 5x + 2
(b) 4x − 18 = 2
(c) 4x + 7 = x + 2
(d) 5x − 8 = x + 4
Show Answer & Explanation
Answer: (c) 4x + 7 = x + 2
Explanation:
1. (a) 3x + 2 = 5x + 2 gives 0 = 2x, so x = 0 (an integer).
2. (b) 4x − 18 = 2 gives 4x = 20, so x = 5 (an integer).
3. (c) 4x + 7 = x + 2 gives 3x = −5, so x = \( \frac{-5}{3} \) (a negative fraction).
4. (d) 5x − 8 = x + 4 gives 4x = 12, so x = 3 (an integer).
5. So the answer is 4x + 7 = x + 2.
Teacher's Note:
1. Solve every option quickly; only one solution is not a whole number.
2. Transpose the x-terms to one side and the numbers to the other.
(a) −2
(b) 2
(c) −3
(d) 3
Show Answer & Explanation
Answer: (a) −2
Explanation:
1. Open the brackets: 3t − 9 = 10t + 5.
2. 3t − 10t = 5 + 9, so −7t = 14.
3. t = 14 ÷ (−7) = −2.
4. Check: 3(−2 − 3) = −15 and 5(2 × (−2) + 1) = 5 × (−3) = −15.
Teacher's Note:
1. Multiply every term inside the bracket by the number outside.
2. Dividing a positive number by a negative number gives a negative answer.
(a) 1
(b) \( \frac{2}{3} \)
(c) \( \frac{6}{5} \)
(d) \( \frac{7}{3} \)
Show Answer & Explanation
Answer: (d) \( \frac{7}{3} \)
Explanation:
1. Bring the y-terms to the left and the numbers to the right: \( 2y + y = \frac{26}{3} - \frac{5}{3} \).
2. \( 3y = \frac{21}{3} = 7 \).
3. y = \( \frac{7}{3} \). So the answer is \( \frac{7}{3} \).
Teacher's Note:
1. The two fractions already have the same denominator, so subtract the numerators.
2. Check: 14/3 + 5/3 = 19/3 and 26/3 − 7/3 = 19/3.
(a) 6
(b) −6
(c) 5
(d) −5
Show Answer & Explanation
Answer: (b) −6
Explanation:
1. Simplify each side: 3x − 5 = 5x + 7.
2. 3x − 5x = 7 + 5, so −2x = 12.
3. x = 12 ÷ (−2) = −6.
4. Check: LHS = −60 − 5 + 42 = −23; RHS = −30 + 15 − 8 = −23.
Teacher's Note:
1. Combine like terms on each side before transposing.
2. When a term changes sides, its sign changes.
(a) 2
(b) −2
(c) 3
(d) none of these
Show Answer & Explanation
Answer: (b) −2
Explanation:
1. Cross-multiply: 3(3x + 5) = 1 × (2x + 1).
2. 9x + 15 = 2x + 1, so 7x = −14.
3. x = −2.
4. Check: (3(−2) + 5) ÷ (2(−2) + 1) = (−1) ÷ (−3) = \( \frac{1}{3} \).
Teacher's Note:
1. Cross-multiplication removes both denominators in one step.
2. Keep the brackets while cross-multiplying so that every term is multiplied.
(a) \( \frac{3}{4} \)
(b) \( -\frac{3}{4} \)
(c) 1
(d) none of these
Show Answer & Explanation
Answer: (b) \( -\frac{3}{4} \)
Explanation:
1. \( -7x = 9 - \frac{15}{4} = \frac{36 - 15}{4} = \frac{21}{4} \).
2. \( x = \frac{21}{4} \div (-7) = -\frac{3}{4} \).
3. So the answer is \( -\frac{3}{4} \).
4. Check: \( \frac{15}{4} - 7 \times \left(-\frac{3}{4}\right) = \frac{15}{4} + \frac{21}{4} = \frac{36}{4} = 9 \).
Teacher's Note:
1. Move the constant first, then divide by the coefficient of x.
2. The coefficient is −7, so the sign of the answer flips.
(a) 10.5
(b) 10.2
(c) 21
(d) 9
Show Answer & Explanation
Answer: (a) 10.5
Explanation:
1. Let the number be x. Then \( x - 7 = \frac{x}{3} \).
2. Multiply by 3: 3x − 21 = x, so 2x = 21.
3. x = 21 ÷ 2 = 10.5.
4. Check: 10.5 − 7 = 3.5 and 10.5 ÷ 3 = 3.5.
Teacher's Note:
1. Translate the words into an equation first: 'decreased by 7' means x − 7.
2. The answer need not be a whole number.
Assertion (A): The equation 2x + 3 = 2x + 5 has no solution.
Reason (R): If both sides of an equation are exactly equal, then the equation has infinite solutions.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. Subtracting 2x from both sides of 2x + 3 = 2x + 5 gives 3 = 5, which is never true. So no value of x works and A is true.
2. If both sides are exactly the same expression (for example 2x + 3 = 2x + 3), every value of x satisfies it, so it has infinitely many solutions. R is true.
3. R talks about identical sides, but A is about sides that can never be equal, so R does not explain A.
4. So the answer is (b): both A and R are true, but R is not the correct explanation of A.
Teacher's Note:
1. Same x-terms with different constants: no solution.
2. Same x-terms with the same constants: infinitely many solutions (an identity).
Assertion (A): The equation x2 − 2x + 1 is a linear equation in one variable.
Reason (R): Standard form for linear equation in one variable is ax + b = 0, where x is variable and a, b are arbitrary constants.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. x2 − 2x + 1 contains x2, so its highest power of the variable is 2. It is not linear (it is not even an equation, as there is no = sign). A is false.
2. A linear equation in one variable can be written as ax + b = 0 (with a ≠ 0), so R is true.
3. So the answer is (d): A is false but R is true.
Teacher's Note:
1. Linear means the highest power of the variable is 1.
2. Strictly, a must not be zero in ax + b = 0; otherwise there is no x-term.
Section B: Short Answer Questions (2 Marks Each)
Question 11. Solve: 8x + 4 = 3 (x − 1) + 7
Answer:
8x + 4 = 3x − 3 + 7
8x + 4 = 3x + 4
8x − 3x = 4 − 4
5x = 0, so x = 0.
Check: LHS = 0 + 4 = 4; RHS = 3(0 − 1) + 7 = 4.
In simple words: After opening the bracket both sides have + 4, which leaves 5x = 0, so x = 0.
Teacher's Note:
1. x = 0 is a perfectly valid solution.
2. Simplify the right side fully before transposing.
Question 12. Solve: \( 2y + \frac{5}{3} = \frac{26}{3} - y \)
Answer:
\( 2y + y = \frac{26}{3} - \frac{5}{3} \)
\( 3y = \frac{21}{3} = 7 \)
\( y = \frac{7}{3} \), so \( y = \frac{7}{3} \).
Check: LHS = \( \frac{14}{3} + \frac{5}{3} = \frac{19}{3} \); RHS = \( \frac{26}{3} - \frac{7}{3} = \frac{19}{3} \).
In simple words: Move y to the left and 5/3 to the right: 3y = 7, so y = 7/3.
Teacher's Note:
1. Group variable terms on one side and constants on the other.
2. Substituting the answer back is the surest check.
Question 13. Solve: \( \frac{n}{2} - \frac{3n}{4} + \frac{5n}{6} = 21 \)
Answer:
LCM of 2, 4 and 6 is 12. Multiply every term by 12: 6n − 9n + 10n = 252.
7n = 252
n = 252 ÷ 7, so n = 36.
Check: 18 − 27 + 30 = 21.
In simple words: Clear the fractions with 12: 7n = 252, so n = 36.
Teacher's Note:
1. Also multiply the right-hand side 21 by 12.
2. Check by substituting: 36/2 − 108/4 + 180/6 = 18 − 27 + 30 = 21.
Question 14. Solve: \( x = \frac{4}{5}(x + 10) \)
Answer:
Multiply both sides by 5: 5x = 4(x + 10).
5x = 4x + 40
5x − 4x = 40, so x = 40.
Check: \( \frac{4}{5} \times (40 + 10) = \frac{4}{5} \times 50 = 40 \).
In simple words: Clear the fraction by multiplying by 5, then 5x = 4x + 40 gives x = 40.
Teacher's Note:
1. Multiplying both sides by the denominator removes the fraction.
2. Remember to multiply every term on both sides.
Section C: Short Answer Questions (3 Marks Each)
Question 15. Simplify and solve the linear equation: 4(3p + 2) − 5(6p − 1) = 2(p − 8) − 6(7p − 4)
Answer:
LHS: 12p + 8 − 30p + 5 = −18p + 13.
RHS: 2p − 16 − 42p + 24 = −40p + 8.
−18p + 13 = −40p + 8
40p − 18p = 8 − 13, so 22p = −5.
\( p = -\frac{5}{22} \).
In simple words: Both sides simplify to −18p + 13 and −40p + 8, giving 22p = −5 and p = −5/22.
Teacher's Note:
1. −5 × (−1) = +5 and −6 × (−4) = +24.
2. A fractional answer is fine; the equation is still linear.
Question 16. Solve: \( \frac{1}{4}x + \frac{1}{6}x = x - 7 \)
OR
Simplify and solve the linear equation: 15 (y − 4) − 2 (y − 9) + 5 (y + 6) = 0
Answer:
First question:
LCM of 4 and 6 is 12. Multiply every term by 12: 3x + 2x = 12x − 84.
5x = 12x − 84
84 = 12x − 5x = 7x, so x = 12.
Check: LHS = 3 + 2 = 5; RHS = 12 − 7 = 5.
OR question:
Open the brackets: 15y − 60 − 2y + 18 + 5y + 30 = 0.
Collect like terms: (15 − 2 + 5)y + (−60 + 18 + 30) = 0, so 18y − 12 = 0.
18y = 12, so y = \( \frac{12}{18} \) = \( \frac{2}{3} \).
So \( y = \frac{2}{3} \).
Check: 15(2/3 − 4) − 2(2/3 − 9) + 5(2/3 + 6) = −50 + \( \frac{50}{3} \) + \( \frac{100}{3} \) = −50 + 50 = 0.
In simple words: Clear fractions with the LCM 12 to get x = 12; in the OR question open the brackets to get 18y = 12, so y = 2/3.
Teacher's Note:
1. Multiply by the LCM of the denominators to clear all fractions at once.
2. Watch the sign: −2 × (−9) = +18.
Question 17. Simplify and solve the linear equation: 3 (5z − 7) − 2 (9z − 11) = 4 (8z − 13) − 17
Answer:
LHS: 15z − 21 − 18z + 22 = −3z + 1.
RHS: 32z − 52 − 17 = 32z − 69.
−3z + 1 = 32z − 69
1 + 69 = 32z + 3z, so 35z = 70.
z = 2.
Check: LHS = 3(3) − 2(7) = −5; RHS = 4(3) − 17 = −5.
In simple words: Each side simplifies to −3z + 1 and 32z − 69; solving gives 35z = 70, z = 2.
Teacher's Note:
1. Simplify each side separately before moving terms.
2. −2 × (−11) = +22, a common sign slip.
Section D: Long Answer Question (5 Marks)
Question 18. Solve: (a) \( \frac{6x + 7}{3x + 2} = \frac{4x + 5}{2x + 3} \)
(b) \( \frac{x}{2} - \frac{1}{4}\left(x - \frac{1}{3}\right) = \frac{1}{6}(x + 1) + \frac{1}{12} \)
Answer:
(a)
Cross-multiply: (6x + 7)(2x + 3) = (4x + 5)(3x + 2).
LHS = 12x2 + 18x + 14x + 21 = 12x2 + 32x + 21.
RHS = 12x2 + 8x + 15x + 10 = 12x2 + 23x + 10.
The 12x2 terms cancel: 32x + 21 = 23x + 10.
9x = −11, so \( x = -\frac{11}{9} \).
Check: 6x + 7 = \( \frac{-1}{3} \) and 3x + 2 = \( \frac{-5}{3} \), ratio = \( \frac{1}{5} \); 4x + 5 = \( \frac{1}{9} \) and 2x + 3 = \( \frac{5}{9} \), ratio = \( \frac{1}{5} \).
(b)
Open the brackets: \( \frac{x}{2} - \frac{x}{4} + \frac{1}{12} = \frac{x}{6} + \frac{1}{6} + \frac{1}{12} \).
The \( \frac{1}{12} \) on both sides cancels: \( \frac{x}{2} - \frac{x}{4} - \frac{x}{6} = \frac{1}{6} \).
LCM 12: \( \frac{6x - 3x - 2x}{12} = \frac{1}{6} \), so \( \frac{x}{12} = \frac{1}{6} \).
x = \( \frac{12}{6} \) = 2, so x = 2.
Check: LHS = 1 − \( \frac{1}{4} \times \frac{5}{3} \) = 1 − \( \frac{5}{12} \) = \( \frac{7}{12} \); RHS = \( \frac{3}{6} \) + \( \frac{1}{12} \) = \( \frac{7}{12} \).
In simple words: In (a) cross-multiply and the x² terms cancel, giving x = −11/9; in (b) open the brackets and clear fractions, giving x = 2.
Teacher's Note:
1. Cross-multiplication can create x² terms, but here they cancel, so the equation is linear.
2. −1/4 × (−1/3) = +1/12; watch the sign.
3. Equal terms on both sides can be cancelled straight away.
Section E: Case Study Based Questions (4 Marks Each)
Question 19. The teacher tells the class that the lowest marks obtained by a student in his class is half the highest marks plus 5. The lowest score is 45. Just to pass in the examination, the borderline marks are just 12 less than the lowest marks obtained by the student in the class.
(a) Find the highest marks. (2)
(b) Find the required passing marks. (1)
(c) If one student Aditya of another class scored 9 marks more than double the lowest marks of this class, find Aditya's marks. (1)
Answer:
(a) Let the highest marks be x. Lowest marks = \( \frac{1}{2}x + 5 \).
So \( \frac{x}{2} + 5 = 45 \), which gives \( \frac{x}{2} = 40 \) and x = 80.
Highest marks = 80.
(b) Passing marks = lowest marks − 12 = 45 − 12 = 33.
(c) Aditya's marks = 2 × lowest marks + 9 = 2 × 45 + 9 = 99.
In simple words: Half of the highest plus 5 is 45, so the highest is 80; pass marks are 45 − 12 = 33; Aditya got 2 × 45 + 9 = 99.
Teacher's Note:
1. Write the sentence as an equation: lowest = highest ÷ 2 + 5.
2. Check (a): 80 ÷ 2 + 5 = 45.
Question 20. It is common that government revises fares from time to time based on various factors such as taxes, economy and inflation, for various vehicles like auto-rickshaw, taxis and radio cab etc. The auto and Taxi charge in a city comprise of fixed charge and the charge for the distance covered. Few situations are given below in the form of questions.
(a) If the fixed charge in a city is ₹ x and charge per km is ₹ 5 and total fare is ₹ 60 then find the linear equation for the journey of 10 km. (1)
(b) In the above question, what is the value of fixed charge? (1)
(c) If in a city a person has to pay ₹ 110 for a journey of 15 km and fixed charge is ₹ 20 then what is the charge per km? (1)
(d) If in a city fixed charge is double of the charge per km and a person paid ₹ 75 for a journey of 1 km, then what is the linear equation for this situation? (1)
Answer:
(a) Fixed charge = ₹ x; charge for 10 km = 10 × ₹ 5 = ₹ 50.
Total fare = fixed charge + distance charge, so the equation is x + 50 = 60.
(b) x + 50 = 60 gives x = 60 − 50 = 10. Fixed charge = ₹ 10.
(c) Let the charge per km be ₹ y. Then 20 + 15y = 110.
15y = 90, so y = 6. Charge per km = ₹ 6.
(d) Let the charge per km be ₹ x. Then the fixed charge = ₹ 2x.
For 1 km: 2x + 1 × x = 75, so the equation is 2x + x = 75 (that is, 3x = 75).
(Solving it, x = 25: charge per km = ₹ 25 and fixed charge = ₹ 50.)
In simple words: Fare = fixed charge + (rate per km × km); write that as an equation and solve.
Teacher's Note:
1. Name the unknown clearly (fixed charge or rate per km) before writing the equation.
2. In (d), 'double the charge per km' means fixed charge = 2x.
3. Check (c): 20 + 15 × 6 = 110.
Free study material for Mathematics
Chapter 02 Linear Equations in One Variable Printable Worksheets and Exercises for Class 8 Mathematics
Mastering Chapter 02 Linear Equations in One Variable with Printable Worksheets
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