# SBI Clerk Mains 2016- Practice Aptitude Questions (Quadratic Equation)

SBI Clerk Mains Exam 2016- Practice Aptitude Questions (Quadratic Equation) Set-85:
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. 1-10): In the following questions, two Equations I and II are given. You have to solve both the equation.
a)  If x> y
b)  If x ≥ y
c)  If x< y
d)  If x ≤ y
e)  If x = y or the relationship cannot be established

1).I. 5x2 – 18x + 9 = 0
II. 20y2 – 13y + 2 = 0

2).I. x3 – 878 = 453
II. y2 – 82 = 39

3).I. 3/√x + 4/√x = √x
II. y3 – (7)7/2 /√y = 0

4).I. 9x – 15.45 = 54.55 + 4x
II. √(y + 155) – √36 = √49

5).I. x2 + 11x + 30 = 0
II. y2 + 7y + 12 = 0

6).I. x2 – 19x + 84 = 0
II. y2 – 25y + 156 = 0

7).I. x3 – 468 = 1729
II. y2 – 1733 + 1564 = 0

8).I. 9/√x + 19/√x = √x
II. y5 – (2 × 14)11/2 /√y = 0

9).I. √(784)x + 1234 = 1486
II. √(1089)y + 2081 = 2345

10).I. 12/√x – 23/√x = 5√x
II. √y/12 – 5√y/12 = 1/√y

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

Solution:

1).I. 5x2 – 18x + 9 = 0
5x2– 15x – 3x + 9 = 0
5x (x -3) – 3(x – 3) = 0
(x – 3) (5x – 3) = 0
x = 3 or 3/5
II. 20y2 – 13y + 2 = 0
20y2– 8y – 5y + 2 = 0
4y(5y – 2) -1(5y – 2) = 0
(4y – 1) (5y – 2) = 0
y = 1/4 or 2/5
Clearly x> y

2).I. x3 – 878 = 453
x =3√1331 = 11àx = 11
II. y2 – 82 = 39
y2= 82 + 39 = 121
y = ±11
x ≥ y

3).I. 3/√x + 4/√x = √x
3 +4 = x
x = 7
II. y3 – (7)7/2 /√y = 0
y3+1/2– (7)7/2 = 0
y7/2= 77/2ày = 7
Clearly, x = y

4).I. 9x – 15.45 = 54.55 + 4x
9x – 4x = 70
5x = 70à x = 14
II. √(y + 155) – √36 = √49
√(y + 155) = 6 + 7à√(y + 155) = 13
y + 155 = 169
y = 169 – 155 = 14
Clearly, x = y

5).I. x2 + 11x + 30 = 0
x2+ 6x + 5x + 30 = 0
x(x +6) + 5(x + 6) = 0
(x + 5) (x + 6) = 0
x = -5 (or) -6
II. y2 + 7y + 12 = 0
y2+ 4y + 3y + 12 = 0
y(y+4) + 3(y+4) = 0
(y+3) (y+4) = 0
y = -3 (or) -4
x

6).I. x2 – 19x + 84 = 0
x2– 12x – 7x + 84 = 0
x(x – 12) -7(x – 12) = 0
x = 7 (or) 12
II. y2 – 25y + 156 = 0
y2– 12y – 13y + 156 = 0
y(y – 12) – 13(y – 12) = 0
(y – 12) (y – 13) = 0
Y = 12 (or) 13
Clearly, x ≤ y

7).I. x3 – 468 = 1729
x3= 1729 + 468 = 2197
x =3√2197 = 13
II. y2 – 1733 + 1564 = 0
y2– 169 = 0
y = ±13
x ≥ y

8).I. 9/√x + 19/√x = √x
(9 + 19) / √x = √x
28 = √x ×√xàx = 28
II. y5 – (2 × 14)11/2 /√y = 0
[y5√y – (28)11/2] / √y = 0ày11/2 – (28)11/2 = 0
y = 28
Clearly, x = y

9).I. √(784)x + 1234 = 1486
√(784)x = 1486 – 1234à√(784)x = 252
x = 252 / 28àx = 9
II. √(1089)y + 2081 = 2345
√(1089)y = 2345 – 2081à√(1089)y = 264
y = 264 / 33ày = 8
x> y

10).I. 12/√x – 23/√x = 5√x
(12 – 23) / √x = 5xà -11
x = -11/5
II. √y/12 – 5√y/12 = 1/√y
(√y-5√y) / 12 = 1/√yà-4√y ×√y = 12
-4y = 12à y = -3
Clearly, x>y