8 Jun 2016

SBI Clerk/UIIC Exam 2016- Practice Aptitude Questions (Quadratic Equations)

SBI Clerk Prelims 2016- Practice Aptitude Questions (Data Interpretation)
SBI Clerk/UIIC Exam 2016- Practice Aptitude Questions (Quadratic Equations) Set-70:
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. 1-9) In each question, two equations numbered I and II have been given. You have to solve both the equations and mark the appropriate option.

1). I. 2x2 + 19x + 45 = 0
II. 2y2 + 11y + 12 = 0
a)   x > y
b)   x ≥ y
c)   x < y
d)   Relationship between x and y cannot be determined
e)   x ≤ y

2). I. 3x2 - 13x + 12 = 0
II. 2y2 - 15y + 28 = 0
a)   x ≥ y
b)   Relationship between x and y cannot be determined
c)   x > y
d)   x < y
e)   x ≤ y

3). I. x2 = 16
II. 2y2 - 17y + 36 = 0
a)   x < y
b)   Relationship between x and y cannot be determined
c)   x > y
d)   x ≥ y
e)   x ≤ y

4).I. 6x2 + 19x + 15 = 0
II. 3y2 + 11y + 10 = 0
a)   x ≤ y
b)   x < y
c)   x ≥ y
d)   x > y
e)   Relationship between x and y cannot be determined

5).I. 2x2 - 11x + 15 = 0
II. 2y2 - 11y + 14 = 0
a)   x ≤ y
b)   x > y
c)   x ≥ y
d)   x < y
e)   Relationship between x and y cannot be determined

6).I. x2 + 9x - 36 = 0
II. y2 - 2y - 24 = 0
a)   x = y or relationship cannot be established
b)   x < y
c)   x ≥ y
d)   x ≤ y
e)   x > y

7).I. 6x – 5y = 17
II. 3x – 4y = 13
a)   x > y
b)   x ≤ y
c)   x = y or relationship cannot be established
d)   x ≥ y
e)   x < y

8).I. x = (- 3)2
II. y2/3 = 64
a)   x > y
b)   x < y
c)   x ≥ y
d)   x ≤ y
e)   x = y or relationship cannot be established

9).I. x2 - 20x + 99 = 0
II. y2 - 18y + 72 = 0
a)   x < y
b)   x > y
c)   x = y or relationship cannot be established
d)   x ≤ y
e)   x ≥ y

Answers:                    

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

Solution:

1). I. 2x2 + 19x + 45 = 0
2 × 45 = 90 = (10 × 9)
(10 + 9 = 19)  x = (-10/2), (-9/2) (dividing by co efficient of x2 and changing signs)
x = -5, -4.5
II. 2y2 + 11y + 12 = 0
2 × 12 = 24 (8 × 3 = 24) (8 + 3 = 11)
y = (-8/2), (-3/2) ) (dividing by co efficient of y2 and changing signs)
y = -4, -1.5
Hence x < y
Answer: c)

2). I. 3x2 - 13x + 12 = 0
12 × 3 = 36 ( -9 × -4 = 36) (-9 – 4 = -13)
x = 9/3, 4/3 (dividing by co efficient of x2 and changing signs)
x = 3, 4/3
II. 2y2 - 15y + 28 = 0
2 × 28 = 56 (-8 × -7 = 56)
(-8 -7 = -15) (dividing by co efficient of y2 and changing signs)
y = 4, 3.5
Hence = x < y
Answer: d)

3). I. x2 = 16
x = ±4
II. 2y2 - 17y + 36 = 0
2 × 36 = 72 (-9 × -8 = 72)
(-9 -8 = -17) y = 9/2, 8/2
(dividing by co efficient of y2 and changing signs)
y = 4.5, 4
Hence = x ≤ y
Answer: e)

4).I. 6x2 + 19x + 15 = 0
6 × 15 = 90 (10 × 9 = 90) (10 + 9 = 19)
x = -10/6, -9/6 (dividing by co efficient of x2 and changing signs)
x = -5/3, -3/2 = -1.66, -1.5
II. 3y2 + 11y + 10 = 0
3 × 10 = 30 (6 × 5 = 30) (6 + 5 = 11)
y = -6/3, -5/2  (dividing by co efficient of y2 and changing signs)
y = -2, -2.5
Hence x > y
Answer: d)

5).I. 2x2 - 11x + 15 = 0
2 × 15 = 30 (-6 × -5 = 30) (-6 – 5 = -11)
x = -6/2, -5/2 (dividing by co efficient of x2 and changing signs)
x = 3, 2.5
II. 2y2 - 11y + 14 = 0
2 × 14 = 28 (-7 × -4 = 28) (-7 -4 = -11)
y = 7/2, 4/2   (dividing by co efficient of y2 and changing signs)
y = 3.5, 2
Hence relationship cannot be established.
Answer: e)

6).I. x2 + 9x - 36 = 0
x2 + 12x -3x -36 = 0
 x (x + 12) – 3 (x +12) = 0
(x - 3) (x + 12) = 0
x = 3, -12
II. y2 - 2y - 24 = 0
y2  -  6y + 4y – 24 = 0
y (y - 6) + 4 (y - 6) = 0
(y + 4) (y - 6) = 0
Y = - 4, 6
Hence relationship cannot be established
Answer: a)

7).I. 6x – 5y = 17                ..(i)
II. 3x – 4y = 13                    ..(ii)     
Solving (i) × 3 – (ii) × 6, we get
18x – 15y = 51
-18x + 24y = -78
We get 9y = -27
y = -3
Putting the value of y in equation (i), we get
6x = 2,  x = 1/3
Hence x >y
Answer: a)

8).I. x = (- 3)2  = 9
II. y2/3 = 64
y3  = 82 à y = 8 2×(3 / 2) = 512
Hence x < y
Answer: b)

9).I. x2 - 20x + 99 = 0
x2 – 11x – 9x + 99 = 0
x (x - 11) -9 (x - 11) = 0
(x - 9) (x - 11) = 0
x = 9,11
II. y2 - 18y + 72 = 0
y2 – 12y – 6y +72 = 0
y (y - 12) -6 (y - 12) = 0
(y - 6) (y - 12) = 0
Y = 6, 12
Hence relationship cannot be established.
Answer: c)




 

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