Class 11 Maths Chapter 5: Linear Inequalities worksheet with answer key

Answer: (b) −4p > −4q

Multiplying both sides by the negative number −4 reverses the sign: −4p > −4q. Subtracting 4, or dividing by the positive number 4, keeps the sign, so (c) and (d) are false.

Answer: (b) an open circle at −2 and the line darkened to the right

The sign is strict, so −2 is not a solution: an open circle at −2. The solutions are greater than −2, so the line is darkened to the right.

Answer: (c) 5 − 2x > 0

5 − 2x > 0 is of the form ax + b > 0 with a = −2 ≠ 0. x² − 4 < 0 is quadratic, 3x + 2y ≥ 7 has two variables, and 4x − 7 = 9 is an equation, not an inequality.

Answer: (b) 18

68 + 24x ≤ 500 gives 24x ≤ 432, x ≤ 18. The sign is ≤, so exactly 500 kg is allowed and x = 18 works; 19 cartons make 524 kg. 17 comes from treating the limit as strict; 20 ignores the operator's weight.

Answer: (a) −4 > x

By Rule 1 the sign is unchanged: 3 − x + x − 7 > 7 + x − 7, that is, −4 > x.

3x − 2 < 10 gives x < 4. 5x + 3 ≥ 2x − 6 gives 3x ≥ −9, x ≥ −3. Common values: −3 ≤ x < 4, the integers −3, −2, −1, 0, 1, 2, 3: 7 integers.

• solves each: x < 4 and x ≥ −3 (1)
• −3 ≤ x < 4 gives 7 integers (1)

3 − 2x ≤ k gives −2x ≤ k − 3, and dividing by −2 reverses the sign: x ≥ 3 − k2. For the set [−3, ∞), 3 − k2 = −3, so 3 − k = −6 and k = 9. On the number line: a dark circle at −3 (it is included) and the line darkened to the right.

• x ≥ (3 − k)/2 with the sign reversed, and k = 9 (1)
• dark circle at −3, line darkened to the right (1)

2x − 5 > 3 gives 2x > 8, x > 4. 3 − x ≥ −1 gives −x ≥ −4, and multiplying by −1 reverses the sign: x ≤ 4. A number cannot be both greater than 4 and at most 4, so the system has no solution; 4 itself fails the first, since 3 > 3 is false.

• x > 4 and, with the sign reversed, x ≤ 4 (1)
• concludes there are no common values, noting that 4 fails the strict inequality (1)

Step 1 is correct (multiplying by 12, a positive number). The error is in Step 2: −3(x − 1) = −3x + 3, not −3x − 3. Correctly: 4x + 8 − 3x + 3 ≥ 2x, so x + 11 ≥ 2x, x ≤ 11. Solution set (−∞, 11].

• identifies Step 2: −3(x − 1) = −3x + 3 (1)
• x + 11 ≥ 2x (1)
• x ≤ 11, i.e. (−∞, 11] (1)

For m months, Plan A costs 1500 + 600m and Plan B costs 850m. Plan A is cheaper when 1500 + 600m < 850m, i.e. 250m > 1500, m > 6. At m = 6 both cost ₹ 5100, so the least number of whole months is 7 (A: ₹ 5700, B: ₹ 5950).

• forms 1500 + 600m < 850m (1)
• m > 6 (1)
• least whole number 7, noting the plans are equal at 6 (1)

Multiply by 5: −20 ≤ 3 − 2x < 5. Subtract 3: −23 ≤ −2x < 2. Divide by −2 and reverse the signs: 23/2 ≥ x > −1, that is, −1 < x ≤ 23/2 = 11.5. The integers are 0, 1, 2, …, 11: least 0, greatest 11, and 12 of them.

• −23 ≤ −2x < 2 (1)
• reverses both signs: −1 < x ≤ 23/2 (1)
• least 0, greatest 11, 12 integers (1)