In this article, you can download the standard Quadratic Equation for Bank PO Mains PDF provided. The quadratic equation is one of the easiest and tricky parts of the quantitative aptitude section. This section requires good observation to avoid silly mistakes during the examination. For the bank PO Mains examination, the quadratic equation will be a complex one. But with good practice of ample questions in the Quadratic Equation For bank PO Mains examination PDF provided here. An algebraic statement of the second degree in x is called a quadratic equation. Bank exams require accuracy in the allotted time during the main exam so shortcut tricks are important for an aspirant to quickly solve a particular bank exam question. There are some tricks and strategies which are essential to solve a quadratic equation, which is provided in the Quadratic Equation For bank PO Mains examination PDF with solutions.
Most questions from this topic are asked in the form of inequalities in the first phase of the exam and also help in various other word problems which can be solved by forming equations. Candidates must dedicate quality time in the Quadratic Equation For bank PO Mains examination PDF practice to ace this topic. Solving questions in the numerical ability section requires good accuracy. Candidates can expect 3-5 questions in the quadratic equation in the bank PO Mains examination. Utilize the Quadratic Equation For bank PO Mains examination PDF provided and ace the concepts with the help of solutions provided with detailed explanations.
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Directions (1 –10): In each of these questions, two equations numbered I and II with variables x and y are given. You have to solve both the equations to find the relation between x and y.
1) I) 6x2 – 29x + 34 = 0
II) 4y2– 13y + 9 = 0
A. x > y
B. x ≥ y
C. x < y
D. x ≤ y
E. x = y or relationship between x and y cannot be determined
2) I) 2x2Â + 17x +33 = 0
II) 17y2+ 48y – 9 = 0
A. x > y
B. x ≥ y
C. x < y
D. x ≤ y
E. x = y or relationship between x and y cannot be determined
3) I) 9x2Â + 81x + 162 = 0
II) 14y2– 132y + 238 = 0
A. x > y
B. x ≥ y
C. x < y
D. x ≤ y
E. x = y or relationship between x and y cannot be determined
4) I) 14x2Â + 210x + 196 = 0
II) 6y2+ 25y + 21 = 0
A. x > y
B. x ≥ y
C. x < y
D. x ≤ y
E. x = y or relationship between x and y cannot be determined
5) I) 6x2 – 32x + 42 = 0
II) y2Â + 7y + 12 = 0
A. x > y
B. x ≥ y
C. x < y
D. x ≤ y
E. x = y or relationship between x and y cannot be determined
6) I)Â 2x2Â + 32x + 120 = 0
II)y2+ 16y + 48 = 0
A. x > y
B. x ≥ y
C. x < y
D. x ≤ y
E. x = y or relationship between x and y cannot be determined
7) I) x2 – 23x + 102 = 0
II) y2– 35y + 306 = 0
A. x > y
B. x ≥ y
C. x < y
D. x ≤ y
E. x = y or relationship between x and y cannot be determined
8) I) 3x2 – 15x + 18 = 0
II) 2y2– 40y + 198 = 0
A. x > y
B. x ≥ y
C. x < y
D. x ≤ y
E. x = y or relationship between x and y cannot be determined
9) I)Â x2Â + 31x + 238 = 0
II)2y2+ 70y + 612 = 0
A. x > y
B. x ≥ y
C. x < y
D. x ≤ y
E. x = y or relationship between x and y cannot be determined
10) I) 7x2Â + 13x + 6 = 0
II)Â 3y2Â + 8y + 4 = 0
A. x > y
B. x ≥ y
C. x < y
D. x ≤ y
E. x = y or relationship between x and y cannot be determined
Answers :
1) Answer: E
I) 6x2– 29x + 34 = 0
=> (x-2)(6x – 17) = 0
=> x = 2, 17/6
II) 4y2– 13y + 9 = 0
=> (y-1)(4y-9) = 0
=> y = 1, 9/4
Hence, relationship between x and y cannot be determined.
2) Answer: D
I) 2x2+ 17x +33 = 0
=> (x+3) (2x+11) = 0
=> x = -3, -11/2
II) 17y2+ 48y – 9 = 0
=> (17y-3) (y+3)
=> y = 3/17, -3
Hence, x ≤ y
3) Answer: C
I) 9x2+ 81x + 162 = 0
=> x2Â + 9x + 18 = 0
=> (x + 6)(x + 3) = 0
=> x = -6, -3
II) 14y2– 132y + 238 = 0
=> 7y2 – 66y + 119 = 0
=> (y – 7)(7y – 17) = 0
=> y = 7, 17/7
Hence, x<y
4) Answer: E
I) 14x2+ 210x + 196 = 0
=> X= 196/14 , 14/14
X=-14, -1
II) 6y2+ 25y + 21 = 0
=> (6y + 7)(y + 3) = 0
=> y = -7/6, -3
Hence, Relationship cannot be established
5) Answer: A
I) 6x2– 32x + 42 = 0
=> (6x – 14)(x – 3) = 0
=> x = 3, 7/3
II) y2Â + 7y + 12 = 0
=> (y + 3)(y + 4) = 0
=> y = -3, -4
Hence, x > y
6) Answer: E
2x2Â + 32x + 120 = 0
2x2Â + 20x + 12x + 120 = 0
2x(x + 10) + 12(x + 10) = 0
(x + 10)(2x + 12) = 0
x = -10, -6
y2Â + 16y + 48 = 0
y2Â + 12y + 4y + 48 = 0
y(y + 12) + 4(y + 12) = 0
(y + 12)(y + 4) = 0
y = -12, -4
Relationship between x and y cannot be established.
7) Answer: D
x2 – 23x + 102 = 0
x2 – 17x – 6x + 102 = 0
x(x – 17) – 6(x – 17) = 0
(x – 6)(x – 17) = 0
x = 6, 17
y2 – 35y + 306 = 0
y2 – 17y – 18y + 306 = 0
y(y – 17) – 18(y – 17) = 0
(y – 18)(y – 17) = 0
y = 18, 17
x ≤ y
8) Answer: C
3x2 – 15x + 18 = 0
3x2 – 9x – 6x + 18 = 0
3x(x – 3) – 6(x – 3) = 0
(3x – 6)(x – 3) = 0
x = 2, 3
2y2 – 40y + 198 = 0
2y2 – 22y – 18y + 198 = 0
2y(y – 11) – 18(y – 11) = 0
(2y – 18)(y – 11) = 0
y = 9, 11
x < y
9) Answer: B
x2Â + 31x + 238 = 0
x2Â + 17x + 14x + 238 = 0
x(x + 17) + 14(x + 17) = 0
(x + 14) (x + 17) = 0
x = -14, -17
2y2Â + 70y + 612 = 0
2y2Â +36y + 34y + 612 = 0
2y(y + 18) + 34(y + 18) = 0
(2y + 34) (y + 18) = 0
y = -17, -18
x ≥ y
10) Answer: E
I) 7x2Â + 13x + 6 = 0
7x2Â + 7x + 6x + 6 = 0
7x (x + 1) + 6 (x + 1) = 0
= > (7x + 6) (x + 1) = 0
= > x = -1, -6/7
II) 3y2Â + 8y + 4 = 0
3y2Â + 6y + 2y + 4 = 0
3y (y + 2) + 2 (y + 2) = 0
(3y + 2) (y + 2) = 0
= > y = -2, -2/3
Hence no relation can be established between x and y
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