Aptitude Questions (Inequality) for Upcoming AAO/SO Exams

Aptitude Questions (Inequality) for Upcoming AAO/SO Exams
Aptitude Questions (Inequality) for Upcoming AAO/SO Exams Set-41:
Dear Readers, Important Practice Aptitude Questions for Upcoming AAO/SO Exams was given here with Solutions. Aspirants those who are preparing for the examination can use this.

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Directions (Q. 1 – 10): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer

1).I. 9x2 – 9x + 2 = 0                        II. y2 – 3y + 2 = 0
a)   x> y
b)   x ≥ y
c)   x< y
d)   x ≤ y
e)   x = y or the relationship cannot be established

2).I. 4x2 – 8x – 5 = 0             II. 2y2 + 5y + 3 = 0
a)   x> y
b)   x ≥ y
c)   x< y
d)   x ≤ y
e)   x = y or the relationship cannot be established

3).I. x2 – x – 12 = 0                II. y2 + 8y + 15  = 0
a)   x> y
b)   x ≥ y
c)   x< y
d)   x ≤ y
e)   x = y or the relationship cannot be established

4).I. 6x2 + 7x + 2 = 0                        II. 2y2 – 3y – 2 = 0
a)   x> y
b)   x ≥ y
c)   x< y
d)   x ≤ y
e)   x = y or the relationship cannot be established

5).I. x2 – 8x + 15 = 0             II. y2 – 6y + 8 = 0
a)   x> y
b)   x ≥ y
c)   x< y
d)   x ≤ y
e)   x = y or the relationship cannot be established

 6).I. 2x2+ 7x + 6 = 0                      II. 2y2+ y – 3 = 0
f)    x> y
g)   x ≥ y
h)   x< y
i)    x ≤ y
j)    x = y or the relationship cannot be established

7).I. 4x2 – 1 = 0                     II. 2y2 + 5y + 2 = 0
f)    x> y
g)   x ≥ y
h)   x< y
i)    x ≤ y
j)    x = y or the relationship cannot be established

8).I. x2 – 5x + 6 = 0               II. y2 + y  = 0
f)    x> y
g)   x ≥ y
h)   x< y
i)    x ≤ y
j)    x = y or the relationship cannot be established

9).I. 4x2 + 7x – 36 = 0                      II. y2 – 7y + 12 = 0
f)    x> y
g)   x ≥ y
h)   x< y
i)    x ≤ y
j)    x = y or the relationship cannot be established

10).I. x2 – (1 + √2)x + √2 = 0                      II. y2 – y – 2 = 0
f)    x> y
g)   x ≥ y
h)   x< y
i)    x ≤ y
j)    x = y or the relationship cannot be established

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

Solution:

1).  I. 9x2 – 9x + 2 = 0          
Or, 9x2 – 6x  – 3x + 2 = 0
Or, 3x(3x – 2) – 1(3x – 2) = 0
Or, (3x – 1) (3x – 2) = 0
x = (1 / 3), (2 / 3)
II. y2 – 3y + 2 = 0
Or, y2 – 2y – y + 2 = 0
Or, y(y – 2) – 1(y – 2) = 0
Or, (y – 1) (y – 2) = 0
Or, y = 1, 2
Hence x< y
Answer:  c)

2).I. 4x2 – 8x – 5 = 0
Or, 4x2 – 10x + 2x – 5 = 0    
Or, 2x(2x – 5) + 1(2x – 5) = 0
Or, (2x – 5) (2x + 1) = 0
x = (5 / 2), -(1 / 2)
II. 2y2 + 5y + 3 = 0
Or, 2y2 + 3y + 2y + 3 = 0
Or, y(2y + 3) + 1(2y + 3) = 0
Or, (2y + 3) (y + 1) = 0
y = – (3 / 2), -1
Hence x> y
Answer: a)

3).I. x2 – x – 12 = 0
Or, x2 – 4x + 3x – 12 = 0
Or, x(x – 4) + 3(x – 4) = 0
Or, (x – 4) (x + 3) = 0
Or, x = 4, -3
II. y2 + 8y + 15  = 0
Or, y2 + 5y + 3y + 15  = 0
Or, y(y + 5) + 3(y + 5) = 0
Or, (y + 5) (y + 3) = 0
Or, y = -3, -5
Hence x ≥ y
Answer: b)

4).I. 6x2 + 7x + 2 = 0
Or, 6x2 + 4x + 3x + 2 = 0
Or, 2x(3x + 2) + (3x + 2) = 0
Or, (3x + 2) (2x + 1) = 0
x = – (2 / 3), -(1 / 2)   
II. 2y2 – 3y – 2 = 0
Or, 2y2 – 4y + y – 2 = 0
Or, 2y(y – 2) + 1(y – 2) = 0
Or, (y – 2) (2y + 1) = 0
Or, y = 2, – (1 / 2)
Hence x ≤ y
Answer: d)

5).I. x2 – 8x + 15 = 0
Or, x2 – 5x – 3x + 15 = 0
Or, x(x – 5) – 3(x – 5) = 0
Or, (x – 5) (x  – 3) = 0
x = 3, 5
II. y2 – 6y + 8 = 0
Or, y2 – 4y – 2y + 8 = 0
Or, y(y – 4) – (y – 4) = 0
Or, (y – 4) (y – 2) = 0
y = 2, 4
Hence the relationship cannot be established.
Answer: e)

6).I. 2x2 + 7x + 6 = 0
Or, 2x2 + 4x + 3x + 6 = 0     
Or, 2x(x + 2) + 3(x + 2) = 0
Or, (x + 2) (2x + 3) = 0
x = -2, – (3 / 2)
II. 2y2 + y – 3 = 0
Or, 2y2 + 3y – 2y – 3 = 0
Or, y(2y + 3) – 1(2y + 3) = 0
Or, (2y + 3) (y – 1) = 0
y = 1, – (3 / 2)
Hence x ≤ y
Answer: d)

7).I. 4x2 – 1 = 0        
x2 = (1 / 4)
x =+ (1 / 2)
or, x = (1 / 2), (-1 / 2)
II. 2y2 + 5y + 2 = 0
Or, 2y2 + 4y + y + 2 = 0
Or, 2y(y + 2) + 1(y + 2) = 0
Or, (2y + 1) (y + 2) = 0
y = – (1 / 2), – 2
Hence, x ≥ y
Answer: b)

8).I. x2 – 5x + 6 = 0  
Or, x2 – 3x – 2x + 6 = 0
Or, x(x – 3) – 2(x – 3) = 0
Or, (x – 2) (x – 3) = 0
x = 2, 3                     
II. y2 + y  = 0
Or, y(y + 1) = 0
y = 0, – 1
Hence x> y
Answer: a)

9).I. 4x2 + 7x – 36 = 0
Or, 4x2 + 16x – 9x – 36 = 0  
Or, 4x(x + 4) – 9(x + 4) = 0
Or, (x + 4) (4x – 9)  = 0
x = – 4, (9 / 4)
II. y2 – 7y + 12 = 0
Or, y2 – 4y – 3y + 12 = 0
Or, y(y – 4) – 3(y – 4) = 0
Or, (y – 3) (y – 4) = 0
Or, y = 3, 4
Hence x< y
Answer: c)

10).I. x2 – (1 + √2) x + √2 = 0
x2 – x – √2x + √2 = 0           
x(x – 1) –  √2(x – 1) = 0
or, (x – 1) (x – √2) = 0
x = 1, √2
II. y2 – y – 2 = 0
Or, y2 – 2y + y – 2 = 0
Or, y(y – 2) + 1(y – 2) = 0
Or, (y – 2) (y + 1) = 0
y = 2, – 1
Hence relationship cannot be established.
Answer: e)

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