# SBI Clerk Mains 2016- Practice Aptitude Questions (Inequality)

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. 3x2 + 8x + 4 = 0  II. 4y2– 19y + 12 = 0
a)  x> y
b)  x ≥ y
c)  x< y
d)  x ≤ y
e)  The relationship cannot be established

2).I. x2 + x – 20 = 0        II. y2– y – 30 = 0
a)  x> y
b)  x ≥ y
c)  x< y
d)  x ≤ y
e)  The relationship cannot be established

3).I. x2 – 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. y2– [ (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. 225x2 – 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.2a2 + a – 1 = 0                  II. 12b2 – 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.a2 – 5a + 6 = 0                  II. 2b2 – 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.a2 + 5a + 6 = 0                 II. b2 + 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.16a2  = 1                II. 3b2 + 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.a2 + 2a + 1 = 0     II. b2  = ± 4
a)  a< b
b)  a ≤ b
c)  The relationship between a and b cannot be established
d)  a> b
e)  a ≥ b

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

Solution:

1).I. 3x2 + 8x + 4 = 0
3x2+ 6x + 2x + 4 = 0
3x(x + 2) + 2(x + 2) = 0
(x + 2) (3x + 2) = 0
x = -2 or -2 / 3
II. 4y2 – 19y + 12 = 0
4y2– 16y – 3y + 12 = 0
4y (y – 4) – 3 (y – 4) = 0
(y – 4) (4y – 3) = 0
y = 4 or 3/4
clearly, x< y

2).I. x2 + x – 20 = 0
x2+ 5x – 4x – 20 = 0
x(x + 5) – 4 (x + 5) = 0
(x + 5) (x – 4) = 0
x = -5 or 4
II. y2 – y – 30 = 0
y2– 6y + 5y – 30 = 0
y(y – 6) + 5(y – 6) = 0
(y – 6) (y + 5) = 0
y = 6 or -5
clearly x ≤ y

3).I. x2 – 365 = 364
x2= 365 + 364
x2= 729
x  = √729  = ±27
II. y – √324 = √81
y – 18 = 9
y = 9 + 18
y = 27
Clearly, x ≤ y.

4).I. (4 / √x )  + (7 / √x) = √x
(4 + 7) / √x = √x
x = 11
II. y2 – [ (11)5 / 2 / √y ] = 0
y5/2= (11)5 / 2
y = 11
Clearly, x = y

5).I. 225x2 – 4 = 0
225x2 =  4
x2= 4 / 225
x = √(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

6).I.2a2 + a – 1 = 0
2a2+ 2a – a – 1 = 0
2a(a + 1) – 1(a + 1) = 0
(a + 1) (2a – 1) = 0
a = -1 or 1 / 2
II. 12b2 – 17b + 6 = 0
12b2– 9b – 8b + 6 = 0
3b(4b – 3) – 2(4b – 3) = 0
(4b – 3)(2b – 3) = 0
b = 3 / 4 or 3 / 2
a< b

7).I.a2 – 5a + 6 = 0
a2 – 3a – 2a + 6 = 0
a(a – 3) – 2(a – 3) = 0
(a – 3) (a – 2) = 0
a = 3 or 2
II. 2b2 – 13b + 21 = 0
2b2– 7b  – 6b + 21 = 0
(2b – 7) -3(2b – 7) = 0
(2b – 7)(b – 3) = 0
b = 7 / 2 or 3
a ≤ b

8).I.a2 + 5a + 6 = 0
a2+ 3a + 2a + 6 = 0
a(a + 3) + 2 (a + 3) = 0
(a + 3)(a + 2) = 0
a = -3 or -2
II. b2 + 7b + 12 = 0
b2+ 4b + 3b + 12 = 0
b(b + 4) + 3(b + 4) = 0
(b + 4) (b + 3) = 0
b = -4 or -3
a ≥ b

9).I.16a2  = 1
a2= 1 / 16
a = √(1 / 16)
a = ±1 / 4
II. 3b2 + 7b + 2 = 0
3b2+ 6b + b + 2 = 0
3b(b + 2) + 1 (b + 2) = 0
(b + 2) (3b + 1) = 0
b = -2 or -1 / 3
a> b

10).I.a2 + 2a + 1 = 0
a2+ a + a + 1 = 0
a(a + 1) + 1(a + 1) = 0
(a + 1) (a + 1) = 0
(a + 1)2 = 0
a = – 1
II. b2  = ± 4
b = ± 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