# Inequality – Quantitative Aptitude Problems

(01) Find the range of Inequality
3x2 – 7x + 4 ≤ 0

(a) x > 0                                            (b) x < 0

(c) All x                                             (d) None of these.

Sol. 3x2 – 7x + 4 ≤ 0

3x2 – 4x – 3x + 4 ≤ 0

x (3x – 4) – 1 (3x – 4) ≤ 0

(x – 1) (3x – 4) ≤ 0

Case I – When (x – 1) ≤ 0 then (3x – 4) ≥ 0

x ≤ 1 and x ≥ 4/3

Hence, x belongs to [-∞, 1] and [4/3, ∞] which is not possible

Case II – When (x – 1) ≥ 0 then (3x – 4) ≤ 0

x ≥ 1 and x ≤ 4/3

Hence, x belongs to [1, 4/3]. It is possible but not in option.

Hence, option (d) is the correct answer.

(02) Solve the following inequality
3x2 – 7x – 6 < 0

(a) –0.66 < x < 3                                   (b) x < – 0.66 or x > 3

(c) 3 < x < 7                                          (d) –2 < x < 2

Sol.  3x2 – 7x – 6 < 0

3x2 – 9x + 2x – 6 < 0

3x (x –3) + 2 (x –3) < 0

(3x + 2) (x – 3) < 0

Case I: When (3x + 2) < 0 then (x – 3) > 0

x < -2/3 and x > 3

Hence, x belongs to [-∞, -2/3] and [3, ∞] which is not possible.

Case II: When (3x + 2) > 0 then (x – 3) < 0

x > -2/3 and x < 3

Hence, -2/3 < x < 3 or –0.66 < x < 3

Hence, option (a) is the correct answer.

(03) Find the right answer for given inequality
3x2 – 7x + 6 < 0

(a) 0.66 < x < 3                                    (b) –0.66 < x < 3

(c) –1 < x < 3                                       (d) None of these

Sol. 3x2 – 7x + 6 < 0

Putting values,

x = {- (-7) ± √(-7)2 – 4(3)(6)}/2(3)

Here, value of “d” = √49 – 72 = √-23 is negative.

Hence, the value of x will be an imaginary number (i.e., not real)

Hence, option (d) is the correct answer.

(04) Find the right value for given inequality
x2 – 14x – 15 > 0

(a) x < –1                                             (b) 15 < x

(c) Both (a) and (b)                            (d) –1 < x < 15

Sol. x2 – 14x – 15 > 0

x2 – 15x + x – 15 > 0

X (x – 15) + 1 (x – 15) > 0

(x + 1) (x – 15) > 0

Case I: When (x + 1) > 0 and (x – 15) > 0

x > -1 and x > 15

Hence, x > 15

Case II: When (x + 1) < 0 and (x – 15) < 0

x < -1 and x < 15

Hence, x < -1

Hence, option (c) is the correct answer.

(05) Select the right answer for the given inequality
2 – x – x2 ≥ 0

(a) –2 ≤ x ≤ 1                                       (b) –2 < x < 1

(c) x < –2                                             (d) x > 1

Sol. 2 – x – x2 ≥ 0

Multiplying by (-1)

x2 + x – 2 ≤ 0

x2 + 2x –x – 2 ≤ 0

x (x + 2) -1 (x + 2) ≤ 0

(x – 1) (x + 2) ≤ 0

Case I – When (x – 1) ≥ 0 and (x + 2) ≤ 0

x ≥ 1 and x ≤ -2

It is not possible.

Case II – When (x – 1) ≤ 0 then (x + 2) ≥ 0

x ≤ 1 and x ≥ -2

Hence, -2 ≤ x ≤ 1

Hence, option (a) is the correct answer.