CBSE Class 8 Mathematics Linear Equations in One Variable MCQs Set 18

Answer: (c) 125

Explanation:
1. Let the number be x: \( x-\frac35x=50 \).
2. \( \frac25x=50 \), so x = 125.

Teacher's Note:
1. Translate the words into an equation, using x for the unknown number.
2. Always substitute the answer back into the statement to check it.

Answer: (b) 10

Explanation:
1. Let the number be x: (x − 4)/6 = 8, so x = 52.
2. Now (52 − 2)/5 = 50/5 = 10.

Teacher's Note:
1. Write the condition as a single linear equation and solve step by step.
2. Check your result against the original wording, not only against the equation.

Answer: (d) 54

Explanation:
1. Let the number be x: \( \frac13\cdot\frac14x=15 \), so x = 180.
2. Three-tenths of 180 = 54.

Teacher's Note:
1. Translate the words into an equation, using x for the unknown number.
2. Always substitute the answer back into the statement to check it.

Answer: (c) 8

Explanation:
1. Let the number be x: 3(2x + 9) = 75.
2. 2x + 9 = 25, so x = 8.

Teacher's Note:
1. Write the condition as a single linear equation and solve step by step.
2. Check your result against the original wording, not only against the equation.

Answer: (c) 144

Explanation:
1. Let the number be x: \( \frac34x-\frac13x=60 \).
2. \( \frac{5}{12}x=60 \), so x = 144.

Teacher's Note:
1. Translate the words into an equation, using x for the unknown number.
2. Always substitute the answer back into the statement to check it.

Answer: (c) 11

Explanation:
1. Let the number be x: \( x-24=\frac47x \), so \( \frac37x=24 \) and x = 56.
2. Sum of digits = 5 + 6 = 11.

Teacher's Note:
1. Write the condition as a single linear equation and solve step by step.
2. Check your result against the original wording, not only against the equation.

Answer: (a) 14

Explanation:
1. Let the number be x: 15x = x + 196.
2. 14x = 196, so x = 14.

Teacher's Note:
1. Translate the words into an equation, using x for the unknown number.
2. Always substitute the answer back into the statement to check it.

Answer: (c) 28

Explanation:
1. Let the number be x: x − x/4 = 21.
2. \( \frac34x=21 \), so x = 28.

Teacher's Note:
1. Write the condition as a single linear equation and solve step by step.
2. Check your result against the original wording, not only against the equation.

Answer: (d) 280

Explanation:
1. Let the number be x: x/5 + 4 = x/4 − 10.
2. 14 = x/4 − x/5 = x/20, so x = 280.

Teacher's Note:
1. Translate the words into an equation, using x for the unknown number.
2. Always substitute the answer back into the statement to check it.

Answer: (a) 15

Explanation:
1. Let the larger number be x: x − 12 = 0.2x.
2. 0.8x = 12, so x = 15.

Teacher's Note:
1. Write the condition as a single linear equation and solve step by step.
2. Check your result against the original wording, not only against the equation.

Answer: (d) 1925

Explanation:
1. Let the number be x: x/7 − x/11 = 100.
2. \( \frac{4x}{77}=100 \), so x = 1925.

Teacher's Note:
1. Translate the words into an equation, using x for the unknown number.
2. Always substitute the answer back into the statement to check it.

Answer: (c) 20

Explanation:
1. Let the number be x: \( \frac x2+\frac x5-\frac x3=\frac{22}{3} \).
2. \( \frac{11x}{30}=\frac{22}{3} \), so x = 20.

Teacher's Note:
1. Write the condition as a single linear equation and solve step by step.
2. Check your result against the original wording, not only against the equation.

Answer: (c) 4

Explanation:
1. Let the number be x: 2x + 20 = 8x − 4.
2. 24 = 6x, so x = 4.

Teacher's Note:
1. Translate the words into an equation, using x for the unknown number.
2. Always substitute the answer back into the statement to check it.

Answer: (b) 216

Explanation:
1. Let the number be x: \( \frac23x-50=40+\frac14x \).
2. \( \frac{5}{12}x=90 \), so x = 216.

Teacher's Note:
1. Write the condition as a single linear equation and solve step by step.
2. Check your result against the original wording, not only against the equation.

Answer: (d) None of these

Explanation:
1. x + x² = 182, so x² + x − 182 = 0, i.e. (x − 13)(x + 14) = 0.
2. The number is 13 (or −14), which is not among options (a), (b), (c).

Teacher's Note:
1. Translate the words into an equation, using x for the unknown number.
2. Always substitute the answer back into the statement to check it.

Answer: (b) 24

Explanation:
1. Let the integer be x: x² − 20x = 96, so x² − 20x − 96 = 0.
2. (x − 24)(x + 4) = 0, and the positive root is x = 24.

Teacher's Note:
1. Write the condition as a single linear equation and solve step by step.
2. Check your result against the original wording, not only against the equation.

Answer: (b) 5

Explanation:
1. Let the number be x: 3x² − 4x = x + 50, so 3x² − 5x − 50 = 0.
2. (3x + 10)(x − 5) = 0, and the natural number is x = 5.

Teacher's Note:
1. Translate the words into an equation, using x for the unknown number.
2. Always substitute the answer back into the statement to check it.

Answer: (a) 8

Explanation:
1. x + 1/x = 34/8 = 17/4, so 4x² − 17x + 4 = 0, i.e. (4x − 1)(x − 4) = 0.
2. Taking x = 4 (the natural number): product of the number and its square root = 4 × 2 = 8.

Teacher's Note:
1. Write the condition as a single linear equation and solve step by step.
2. Check your result against the original wording, not only against the equation.

Answer: (a) \( \frac{5}{12} \)

Explanation:
1. Let the number be x: \( \frac23x=\frac{25}{216x} \).
2. \( x^2=\frac{25\times3}{216\times2}=\frac{25}{144} \), so \( x=\frac5{12} \).

Teacher's Note:
1. Translate the words into an equation, using x for the unknown number.
2. Always substitute the answer back into the statement to check it.

Answer: (a) 3

Explanation:
1. x + 17 = 60/x, so x² + 17x − 60 = 0.
2. (x + 20)(x − 3) = 0, and the positive root is x = 3. The question asks for that number.

Teacher's Note:
1. Write the condition as a single linear equation and solve step by step.
2. Check your result against the original wording, not only against the equation.