Linear Inequations - Class 10 MCQ

Add 4 to both sides: 3x < 15.
Divide by 3: x < 5.
Since x is a natural number (N), the solution set is {1, 2, 3, 4}.

Subtract 2: 5x ≥ 10.
Divide by 5: x ≥ 2.
Since x is a whole number (W), the solution set is {2, 3, 4}.

Subtract 7: -2x > -6.
Divide by -2 (reverse sign): x < 3.
Since x is an integer (Z), the solution set is {x : x < 3}.

Add 9: 6x ≤ 9.
Divide by 6: x ≤ 1.5.
Since x is a natural number (N), no values satisfy, so {}.

Subtract 3: 4x > 12.
Divide by 4: x > 3.
Since x is an integer (Z), the solution set is {x : x > 3}.

Subtract 5: -x < 3.
Multiply by -1 (reverse sign): x > -3.
Since x is real (R), the solution set is {x : x > -3}.

Add 5: 2x ≤ 6.
Divide by 2: x ≤ 3.
Since x is a whole number (W), the solution set is {0, 1, 2, 3}, but check shows {0, 1, 2} fits strictly. Wait, error – x ≤ 3 includes 3, so {0, 1, 2, 3}. Adjust to C.

Subtract 7: -3x > -6.
Divide by -3 (reverse sign): x < 2.
Since x is a natural number (N), the solution set is {1}. Wait, x < 2 gives {1}, so B incorrect, adjust to {1}.

Subtract 1: 3x < 9.
Divide by 3: x < 3.
Since x is an integer (Z), the solution set is {x : x < 3}.

Subtract 6: -4x ≥ -8.
Divide by -4 (reverse sign): x ≤ 2.
Since x is real (R), the solution set is {x : x ≤ 2}.

Add 8: 5x < 10.
Divide by 5: x < 2.
Since x is a natural number (N), the solution set is {1}. Wait, {1, 2} incorrect, adjust to {1}. Error, recheck: x < 2 gives {1}, so A.

Subtract 4: 2x ≥ 4.
Divide by 2: x ≥ 2.
Since x is an integer (Z), the solution set is {2, 3, 4, ...}.

Subtract 9: -3x > -9.
Divide by -3 (reverse sign): x < 3.
Since x is real (R), the solution set is {x : x < 3}.

Subtract 4: -2x ≤ -4.
Divide by -2 (reverse sign): x ≥ 2.
Since x is a whole number (W), the solution set is {2}. Wait, error – 4 - 2x ≤ 0 gives x ≥ 2, so {2}. Adjust to C.

Add 3: 6x < 12.
Divide by 6: x < 2.
Since x is a natural number (N), the solution set is {1}. Wait, x < 2 gives {1}, so A incorrect, adjust to {1}.

Subtract 10: -5x ≥ -10.
Divide by -5 (reverse sign): x ≤ 2.
Since x is an integer (Z), the solution set is {x : x ≤ 2}.

Subtract 2: 3x > 6.
Divide by 3: x > 2.
Since x is real (R), the solution set is {x : x > 2}.

Subtract 7: -4x ≥ -4.
Divide by -4 (reverse sign): x ≤ 1.
Since x is a whole number (W), the solution set is {0, 1}. Wait, check: x ≤ 1 gives {0, 1}, so C.

Add 1: 2x < 6.
Divide by 2: x < 3.
Since x is a natural number (N), the solution set is {1, 2}.

Subtract 12: -6x ≤ -12.
Divide by -6 (reverse sign): x ≥ 2. Wait, error: -6x ≤ -12, x ≥ 2 is wrong, correct: x ≤ 2.