15 Jun 2016

SBI Clerk Mains 2016- Practice Aptitude Questions (Inequality)

SBI Clerk Prelims 2016- Practice Aptitude Questions (Data Interpretation)
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

Answers:                    
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
Answer:  c)

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
Answer: d)

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.
Answer: d)

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
Answer: e)

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
Answer: e)

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
Answer: a)

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
Answer: 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
Answer: e)

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
Answer: d)

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

Answer: c)





 

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