CBSE Class 8 Mathematics Linear Equations in One Variable Practice Paper Assignment Set 12 with Answers (Old NCERT Syllabus)

This Class 8 Maths Linear Equations in One Variable practice paper carries 40 marks and has 17 questions in Sections A to E, all solved step by step. It follows the earlier (old) NCERT Class 8 Maths syllabus, so it suits state-board students and anyone revising this chapter. The paper tests transposing terms, opening brackets, clearing fractions with the LCM, cross-multiplication, checking a solution by substitution, and forming equations from a taxi-fare case study.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 7. 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 8. 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 9. 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 10. Solve: \( \frac{1}{4}x + \frac{1}{6}x = x - 7 \)
Answer:
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.
In simple words: Multiply through by 12 to clear fractions: 5x = 12x − 84, so 7x = 84 and x = 12.

Teacher's Note:
1. Multiply by the LCM of the denominators to clear all fractions at once.
2. Do not forget to multiply the constant 7 by 12 as well.

Question 11. Simplify and solve the linear equation: 15 (y − 4) − 2 (y − 9) + 5 (y + 6) = 0
Answer:
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: Open the brackets, collect terms to get 18y = 12, so y = 2/3.

Teacher's Note:
1. Watch the sign: −2 × (−9) = +18.
2. Reduce 12/18 to lowest terms.

Question 12. 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 13. 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 Questions (4 Marks Each)

Question 14. Solve: \( \frac{6x + 7}{3x + 2} = \frac{4x + 5}{2x + 3} \)
Answer:
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 = \( \frac{1}{5} \).
In simple words: Cross-multiply; the x² terms cancel, leaving 9x = −11, so x = −11/9.

Teacher's Note:
1. Cross-multiplication can create x² terms, but here they cancel, so the equation is linear.
2. Multiply out each pair of brackets carefully, term by term.

Question 15. Solve: \( \frac{x}{2} - \frac{1}{4}\left(x - \frac{1}{3}\right) = \frac{1}{6}(x + 1) + \frac{1}{12} \)
Answer:
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: Open the brackets, collect the x-terms on one side: x/12 = 1/6, so x = 2.

Teacher's Note:
1. −1/4 × (−1/3) = +1/12; watch the sign.
2. Equal terms on both sides can be cancelled straight away.

Question 16. 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.

Section E: Case Study Based Question (4 Marks)

Question 17. 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.