## 1 Jun 2016

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

SBI Clerk/UIIC Exam 2016- Practice Aptitude Questions (Quadratic Equations) Set-62:
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. 1-5) In the following questions, two equations I and II are given. You have to solve both the equations.
Give Answer
I.        If x > y
II.        If x > = y
III.        If x < y
IV.        If x < = y
V.        If x = y or the relation cannot be established
1). I. (25 / x2) – (12 / x) + (9 / x2) = (4 / x2)
II.       9.84 – 2.64 = 0.95 + y2
2). I . √(900)x + √(1296) = 0
II.       (256)1/4 y + (216)1/3 = 0
3).  I. [(3)5 + (7)3 / 3] = x3
II. 7y3 = - (15 × 2) + 17y3
4). I. (x1/4 / 16)2 = 144 / x3/2
II. y1/3  × y2/3  × 3014 = 16 × y2
5). I. 3x2  –  9x + 28 = 0
II. 5y2 – 18y + 16 = 0

Directions (Q. 6-10) In the following questions, two equations I and II are given. You have to solve both the equations.
Give Answer
I.        If x > y
II.        If x >= y
III.        If x < y
IV.        If x < = y
V.        If x = y or the relationship cannot be established
6). I. x2 – 1200 = 244
II. y + 122 = 159
7). I. 14x – 25 = 59 – 7x
II. √(y + 222) - √(36) = √(81)
8). I. 144x2 – 16 = 9
II. 12y + √4 = √(49)
9). I.  x2 – 9x + 20 = 0
II. y2 – 13y + 42 = 0
10). I. (√(x) / 5) + (3 √(x) / 10) = (1 / √(x))
II. (10 / √(y)) – (2 / √(y)) = 4√(y)

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

Solution:
1). I. (25/x2 ) + (9/x2 ) – (4/x2 )  = (12/x)
(25 + 9 - 4) / x2 = 12/x = 30/x2 = 12/x
12x = 30
x = 30 / 12 = 5/2 = 2.5
II. 9.84 -2.64 = 0.95 + y2
7.2 – 0.95 = y2
y = √(6.25) = ± (2.5)
clearly x ≥ y
Answer: b)
2). I. √(900)x + √(1296) =0
√(900)x = -√(1296)
30x = -36
x = -36 / 30 = -1.2
II. (256)1/4 y = (216)1/3
(44)1/4 y = - (63)1/3 à 4y = -6
Y = -(6/4) = -1.5
Clearly, x > y
Answer: a)
3). I.  [(3)5 + (7)3] / 3 = x3
(243 + 343) / 3 = x3
(586 / 3) = x3
II. 7y3 = -30 + 17y3 = 10y3 = 30
y3  = 30/10 = 3
clearly, x > y
Answer: a)
4). I. (x1/4 / 16)2 = (144 / x3/2 )  = (x1/2 / 256) = (144 / x3/2 )
(x1/2 )  × (x3/2 ) = 256 × 144
x2 = (256 × 144)
x = √(256 × 144)
x = ± (16 × 12) = ±192
II. y1/3 × y2/3 × 3104 = 16y2
y × 3104 = 16y2
3104 = 16y
Y = 3104 / 16 = 194
Clearly, x > y
Answer: c)
5). I. 3x2 – 9x + 28 = 0
3x2  - 12x – 7x + 28 = 0
3x (x - 4) – 7 (x - 4) = 0
(x - 4) (3x - 7) = 0
x = 4, 7/3
II. 5y2 – 18y + 16 = 0
5y2 – 10y – 8y + 16 = 0
5y (y - 2) – 8 (y - 2) = 0
(y - 2) (5y - 8) = 0
Y = 2, 8/5
Clearly, x > y
Answer: a)
6). I. x2 = 1200 + 244
x2 = 1444
x = √(1444) = ± 38
II. y = 159 – 122 = 37
Clearly, x > y or x < y
Hence, the relationship cannot be established.
Answer: e)
7). I. 14x + 7x = 59 + 25
21x = 84; x = 4
II. √(y + 222) = √(36) + √(81)
√(y + 222) = ± 6 ±9
√(y + 222) = ± 15
Taking (+ve) sign,
√(y + 222) = 15
y + 222 = 225
y = 225 – 222 = 3
Taking (-ve) sign,
√(y + 222) = -15
(y + 222) = 225
Y = 225 – 222 = 3
Clearly, x > y
Answer: a)
8). I. 144x2 = 16 + 9
144x2 = 25 àx2 = 25 / 144
x = ± 5 / 12
II. 12y = √49 - √4
12y = ± 7 – (± 2)
12y = ± 5
y = ± 5 / 12
clearly, x = y
Answer: e)
9). x2 – 9x + 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 – 13y + 42 = 0
y2 – 7y – 6y + 42 = 0
y(y - 7) – 6 (y - 7) = 0
(y - 7) (y - 6) = 0
Y = 6 (or) 7
Clearly, x < y
Answer: c)
10). I. (2√x + 3√x) / 10 = 1 / √x
2x + 3x = 10
5x = 10
x = 2
II. (10 - 2) / √y = 4 √y
8 = 4y
y = 8 / 4 = 2
Clearly, x = y
Answer: e)

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