__SBI Clerk Mains Exam 2016- Practice Aptitude Questions (__**:**

__Inequality) Set-76__Dear Readers, Important Practice Aptitude Questions with solution for Upcoming SBI Clerk Exam, candidates those who are preparing for those exams can use this practice questions.

**Directions (Q. No 1 – 5)**In each question, two Equations I and II are given. You have to solved both the equations and find out whether.

**1).**I. 3x

^{2}+ 8x + 4 = 0 II. 4y

^{2}– 19y + 12 = 0

a) x> y

b) x ≥ y

c) x< y

d) x ≤ y

e) The relationship cannot be established

**2).**I. x

^{2}+ x – 20 = 0 II. y

^{2}– y – 30 = 0

a) x> y

b) x ≥ y

c) x< y

d) x ≤ y

e) The relationship cannot be established

**3).**I. x

^{2}– 365 = 364 II. y – √324 = √81

a) x> y

b) x ≥ y

c) x< y

d) x ≤ y

e) The relationship cannot be established

**4).**I. (4 / √x ) + (7 / √x) = √x II. y

^{2}– [ (11)

^{5 / 2}/ √y ] = 0

a) x> y

b) x ≥ y

c) x< y

d) x ≤ y

e) The relationship cannot be established

**5).**I. 225x

^{2}– 4 = 0 II. √(225y) + 2 = 0

a) x> y

b) x ≥ y

c) x< y

d) x ≤ y

e) The relationship cannot be established

**Directions (Q. No 6 – 10)**In each question, two Equations I and II are given. You have to solved both the equations and find out values of a and b

**6).**I.2a

^{2}+ a – 1 = 0 II. 12b

^{2}– 17b + 6 = 0

a) a< b

b) a ≤ b

c) The relationship between a and b cannot be established

d) a> b

e) a ≥ b

**7).**I.a

^{2}– 5a + 6 = 0 II. 2b

^{2}– 13b + 21 = 0

a) a< b

b) a ≤ b

c) The relationship between a and b cannot be established

d) a> b

e) a ≥ b

**8).**I.a

^{2}+ 5a + 6 = 0 II. b

^{2}+ 7b + 12 = 0

a) a< b

b) a ≤ b

c) The relationship between a and b cannot be established

d) a> b

e) a ≥ b

**9).**I.16a

^{2}= 1 II. 3b

^{2}+ 7b + 2 = 0

a) a< b

b) a ≤ b

c) The relationship between a and b cannot be established

d) a> b

e) a ≥ b

**10).**I.a

^{2}+ 2a + 1 = 0 II. b

^{2}= ± 4

a) a< b

b) a ≤ b

c) The relationship between a and b cannot be established

d) a> b

e) a ≥ b

__Answers__**:**

**1). c) 2). d) 3). d) 4). e) 5). e) 6). a) 7). b) 8). e) 9). d) 10). c)**

__Solution:__

**1).**I. 3x

^{2}+ 8x + 4 = 0

3x

^{2}+ 6x + 2x + 4 = 03x(x + 2) + 2(x + 2) = 0

(x + 2) (3x + 2) = 0

x = -2 or -2 / 3

II. 4y

^{2}– 19y + 12 = 04y

^{2}– 16y – 3y + 12 = 04y (y – 4) – 3 (y – 4) = 0

(y – 4) (4y – 3) = 0

y = 4 or 3/4

clearly, x< y

**Answer: c)**

**2).**I. x

^{2}+ x – 20 = 0

x

^{2}+ 5x – 4x – 20 = 0x(x + 5) – 4 (x + 5) = 0

(x + 5) (x – 4) = 0

x = -5 or 4

II. y

^{2}– y – 30 = 0y

^{2}– 6y + 5y – 30 = 0y(y – 6) + 5(y – 6) = 0

(y – 6) (y + 5) = 0

y = 6 or -5

clearly x ≤ y

**Answer: d)**

**3).**I. x

^{2}– 365 = 364

x

^{2}= 365 + 364x

^{2}= 729x = √729 = ±27

II. y – √324 = √81

y – 18 = 9

y = 9 + 18

y = 27

Clearly, x ≤ y.

**Answer: d)**

**4).**I. (4 / √x ) + (7 / √x) = √x

(4 + 7) / √x = √x

x = 11

II. y

^{2}– [ (11)^{5 / 2}/ √y ] = 0y

^{5/2}= (11)^{5 / 2}y = 11

Clearly, x = y

**Answer: e)**

**5).**I. 225x

^{2}– 4 = 0

225x

^{2}= 4x

^{2}= 4 / 225x = √(4 / 225) = ± 2 / 15

II. √225y + 2 = 0

√225y = – 2

Squaring on both sides, we get 225 = y; y = 4 / 225

Hence, the relationship cannot be established

**Answer: e)**

**6).**I.2a

^{2}+ a – 1 = 0

2a

^{2}+ 2a – a – 1 = 02a(a + 1) – 1(a + 1) = 0

(a + 1) (2a – 1) = 0

a = -1 or 1 / 2

II. 12b

^{2}– 17b + 6 = 012b

^{2}– 9b – 8b + 6 = 03b(4b – 3) – 2(4b – 3) = 0

(4b – 3)(2b – 3) = 0

b = 3 / 4 or 3 / 2

a< b

**Answer: a)**

**7).**I.a

^{2}– 5a + 6 = 0

**a**

^{2}– 3a – 2a + 6 = 0

a(a – 3) – 2(a – 3) = 0

(a – 3) (a – 2) = 0

a = 3 or 2

II. 2b

^{2}– 13b + 21 = 02b

^{2}– 7b – 6b + 21 = 0(2b – 7) -3(2b – 7) = 0

(2b – 7)(b – 3) = 0

b = 7 / 2 or 3

a ≤ b

**Answer: b)**

**8).**I.a

^{2}+ 5a + 6 = 0

a

^{2}+ 3a + 2a + 6 = 0a(a + 3) + 2 (a + 3) = 0

(a + 3)(a + 2) = 0

a = -3 or -2

II. b

^{2}+ 7b + 12 = 0b

^{2}+ 4b + 3b + 12 = 0b(b + 4) + 3(b + 4) = 0

(b + 4) (b + 3) = 0

b = -4 or -3

a ≥ b

**Answer: e)**

**9).**I.16a

^{2}= 1

a

^{2}= 1 / 16a = √(1 / 16)

a = ±1 / 4

II. 3b

^{2}+ 7b + 2 = 03b

^{2}+ 6b + b + 2 = 03b(b + 2) + 1 (b + 2) = 0

(b + 2) (3b + 1) = 0

b = -2 or -1 / 3

a> b

**Answer: d)**

**10).**I.a

^{2}+ 2a + 1 = 0

a

^{2}+ a + a + 1 = 0a(a + 1) + 1(a + 1) = 0

(a + 1) (a + 1) = 0

(a + 1)

^{2}= 0a = – 1

II. b

^{2}= ± 4b = ± 2 , ± √-4

±√-4 is imaginary number

Now compare, a = -1 and b = ± 2

As -1> – 2 but – 1< + 2, so relationship between a and b cannot be established

**Answer: c)**