Quadratic Equation Questions for IBPS PO PDF
Download Top-15 Banking Exams Quadratic Equation Questions PDF. Banking Exams Quadratic Equation questions based on asked questions in previous exam papers very important for the Banking exams.
Download Quadratic Equation Questions for IBPS PO PDF
105 IBPS PO Mocks for just Rs. 149
Download IBPS PO Previous Papers PDF
Take a IBPS PO free mock test
Instructions
In the following questions two equations numbered I and II are given. You have to solve both the equations and
Give answer a: if x > y
Give answer b: if x ≥ y
Give answer c: if x < y
Give answer d: if x ≤ y
Give answer e: if x = y or the relationship cannot be established.
Question 1: I. $x^{2}-3x-88=0$
II. $y^{2}+8y-48=0$
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.
Question 2: I. $5x^{2}+29x+20=0$
II. $25y^{2}+25y+6=0$
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.
Question 3: I. $2x^{2}-11x+12=0$
II. $2y^{2}-19y+44=0$
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.
Question 4: I. $3x^{2}+10x+8=0$
II. $3y^{2}+7y+4=0$
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.
Question 5: I. $2x^{2}+21x+10=0$
II. $3y^{2}+13y+14=0$
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.
Instructions
In the following questions two equations numbered I and II are given. You have to solve both equations and
Give answer If
a. x ˃ y
b. x ≥ y
c. x ˂ y
d. x ≤ y
e. x = y or the relationship cannot be established
Question 6: I. ${x^2}$ – 7x + 10 = 0
II. ${y^2}$ + 11y + 10 = 0
a) x ˃ y
b) x ≥ y
c) x ˂ y
d) x ≤ y
e) x = y or the relationship cannot be established
Take a free mock test for IBPS-PO
790 Mocks (cracku Pass) Just Rs. 299
Question 7: I. ${x^2}$ + 28x + 192 = 0
II. ${y^2}$ + 16y + 48 = 0
a) x ˃ y
b) x ≥ y
c) x ˂ y
d) x ≤ y
e) x = y or the relationship cannot be established
Question 8: I.2x – 3y = – 3.5
II. 3x + 2y = – 6.5
a) x ˃ y
b) x ≥ y
c) x ˂ y
d) x ≤ y
e) x = y or the relationship cannot be established
Question 9: I. ${x^2}$ + 8x + 15 = 0
II. ${y^2}$ + 11y + 30 = 0
a) x ˃ y
b) x ≥ y
c) x ˂ y
d) x ≤ y
e) x = y or the relationship cannot be established
Question 10: I. x = $\sqrt {3136} $
II.$ {y^2}$ = 3136
a) x ˃ y
b) x ≥ y
c) x ˂ y
d) x ≤ y
e) x = y or the relationship cannot be established
Take a free mock test for IBPS-PO
Instructions
In each of these questions two equations numbered I and II are given. You have to solve both the equations and –
Give answer a: if x < y
Give answer b: if x ≤ y
Give answer c: if x > y
Give answer d: if x ≥ y
Give answer e: if x = y or the relationship cannot be established.
Question 11: I. $x^{2}+13x+42=0$
II. $y^{2} +19y+90=0$
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.
Question 12: I. $x^{2}-15x+56=0$
II. $y^{2} -23y+132=0$
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.
Question 13: I. $x^{2}+7x+12=0$
II. $y^{2} +6y+8=0$
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.
Question 14: I. $x^{2}-22x+120=0$
II. $y^{2} -26y+168=0$
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.
Question 15: I.$x^{2}+12x+32=0$
II. $y^{2} +17y+72=0$
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.
Answers & Solutions:
1) Answer (E)
I.$x^{2} – 3x – 88 = 0$
=> $x^2 + 8x – 11x – 88 = 0$
=> $x (x + 8) – 11 (x + 8) = 0$
=> $(x + 8) (x – 11) = 0$
=> $x = -8 , 11$
II.$y^{2} + 8y – 48 = 0$
=> $y^2 + 12y – 4y – 48 = 0$
=> $y (y + 12) – 4 (y + 12) = 0$
=> $(y + 12) (y – 4) = 0$
=> $y = -12 , 4$
$\therefore$ No relation can be established.
2) Answer (C)
I.$5x^{2} + 29x + 20 = 0$
=> $5x^2 + 25x + 4x + 20 = 0$
=> $5x (x + 5) + 4 (x + 5) = 0$
=> $(x + 5) (5x + 4) = 0$
=> $x = -5 , \frac{-4}{5}$
II.$25y^{2} + 25y + 6 = 0$
=> $25y^2 + 10y + 15y + 6 = 0$
=> $5y (5y + 2) + 3 (5y + 2) = 0$
=> $(5y + 3) (5y + 2) = 0$
=> $y = \frac{-3}{5} , \frac{-2}{5}$
Therefore $x < y$
3) Answer (D)
I.$2x^{2} – 11x + 12 = 0$
=> $2x^2 – 8x – 3x + 12 = 0$
=> $2x (x – 4) – 3 (x – 4) = 0$
=> $(x – 4) (2x – 3) = 0$
=> $x = 4 , \frac{3}{2}$
II.$2y^{2} – 19y + 44 = 0$
=> $2y^2 – 8y – 11y + 44 = 0$
=> $2y (y – 4) – 11 (y – 4) = 0$
=> $(y – 4) (2y – 11) = 0$
=> $y = 4 , \frac{11}{2}$
$\therefore x \leq y$
4) Answer (D)
I.$3x^{2} + 10x + 8 = 0$
=> $3x^2 + 6x + 4x + 8 = 0$
=> $3x (x + 2) + 4 (x + 2) = 0$
=> $(x + 2) (3x + 4) = 0$
=> $x = -2 , \frac{-4}{3}$
II.$3y^{2} + 7y + 4 = 0$
=> $3y^2 + 3y + 4y + 4 = 0$
=> $3y (y + 1) + 4 (y + 1) = 0$
=> $(y + 1) (3y + 4) = 0$
=> $y = -1 , \frac{-4}{3}$
$\therefore x \leq y$
5) Answer (E)
I.$2x^{2} + 21x + 10 = 0$
=> $2x^2 + x + 20x + 10 = 0$
=> $x (2x + 1) + 10 (2x + 1) = 0$
=> $(x + 10) (2x + 1) = 0$
=> $x = -10 , \frac{-1}{2}$
II.$3y^{2} + 13y + 14 = 0$
=> $3y^2 + 6y + 7y + 14 = 0$
=> $3y (y + 2) + 7 (y + 2) = 0$
=> $(y + 2) (3y + 7) = 0$
=> $y = -2 , \frac{-7}{3}$
$\therefore$ No relation can be established.
6) Answer (A)
I.$x^{2} – 7x + 10 = 0$
=> $x^2 – 5x – 2x + 10 = 0$
=> $x (x – 5) – 2 (x – 5) = 0$
=> $(x – 5) (x – 2) = 0$
=> $x = 5 , 2$
II.$y^{2} + 11y + 10 = 0$
=> $y^2 + 10y + y + 10 = 0$
=> $y (y + 10) + 1 (y + 10) = 0$
=> $(y + 10) (y + 1) = 0$
=> $y = -10 , -1$
$\therefore x > y$
7) Answer (D)
I.$x^{2} + 28x + 192 = 0$
=> $x^2 + 16x + 12x + 192 = 0$
=> $x (x + 16) + 12 (x + 16) = 0$
=> $(x + 16) (x + 12) = 0$
=> $x = -16 , -12$
II.$y^{2} + 16y + 48 = 0$
=> $y^2 + 12y + 4y + 48 = 0$
=> $y (y + 12) + 4 (y + 12) = 0$
=> $(y + 12) (y + 4) = 0$
=> $y = -12 , -4$
$\therefore x \leq y$
8) Answer (C)
I : $2x – 3y = -3.5$
II : $3x + 2y = -6.5$
Multiplying eqn(I) by 2 and eqn(II) by 3, and then adding both equations, we get :
=> $(4x + 9x) + (-6y + 6y) = (-7 -19.5)$
=> $13x = -26.5$
=> $x = \frac{-26.5}{13} \approx -2$
=> $y = \frac{3x + 6.5}{2} = 0.25$
Hence $x < y$
9) Answer (B)
I.$x^{2} + 8x + 15 = 0$
=> $x^2 + 5x + 3x + 15 = 0$
=> $x (x + 5) + 3 (x + 5) = 0$
=> $(x + 5) (x + 3) = 0$
=> $x = -5 , -3$
II.$y^{2} + 11y + 30 = 0$
=> $y^2 + 5y + 6y + 30 = 0$
=> $y (y + 5) + 6 (y + 5) = 0$
=> $(y + 6) (y + 5) = 0$
=> $y = -6 , -5$
$\therefore x \geq y$
10) Answer (B)
I. $x = \sqrt {3136} $
=> $x = 56$
II.$ {y^2} = 3136$
=> $y = \sqrt{3136} = \pm 56$
$\therefore x \geq y$
11) Answer (C)
I.$x^{2} + 13x + 42 = 0$
=> $x^2 + 7x + 6x + 42 = 0$
=> $x (x + 7) + 6 (x + 7) = 0$
=> $(x + 7) (x + 6) = 0$
=> $x = -7 , -6$
II.$y^{2} + 19y + 90 = 0$
=> $y^2 + 9y + 10y + 90 = 0$
=> $y (y + 9) + 10 (y + 9) = 0$
=> $(y + 9) (y + 10) = 0$
=> $y = -9 , -10$
$\therefore x > y$
12) Answer (A)
I.$x^{2} – 15x + 56 = 0$
=> $x^2 – 8x – 7x + 56 = 0$
=> $x (x – 8) – 7 (x – 8) = 0$
=> $(x – 8) (x – 7) = 0$
=> $x = 8 , 7$
II.$y^{2} – 23y + 132 = 0$
=> $y^2 – 11y – 12y + 132 = 0$
=> $y (y – 11) – 12 (y – 11) = 0$
=> $(y – 11) (y – 12) = 0$
=> $y = 11 , 12$
$\therefore x < y$
13) Answer (E)
I.$x^{2} + 7x + 12 = 0$
=> $x^2 + 3x + 4x + 12 = 0$
=> $x (x + 3) + 4 (x + 3) = 0$
=> $(x + 3) (x + 4) = 0$
=> $x = -3 , -4$
II.$y^{2} + 6y + 8 = 0$
=> $y^2 + 4y + 2y + 8 = 0$
=> $y (y + 4) + 2 (y + 4) = 0$
=> $(y + 4) (y + 2) = 0$
=> $y = -4 , -2$
Because $-2 > -4$ and $-3 > -4$
Therefore, no relation can be established.
14) Answer (B)
I.$x^{2} – 22x + 120 = 0$
=> $x^2 – 10x – 12x + 120 = 0$
=> $x (x – 10) – 12 (x – 10) = 0$
=> $(x – 10) (x – 12) = 0$
=> $x = 10 , 12$
II.$y^{2} – 26y + 168 = 0$
=> $y^2 – 12y – 14y + 168 = 0$
=> $y (y – 12) – 14 (y – 12) = 0$
=> $(y – 12) (y – 14) = 0$
=> $y = 12 , 14$
$\therefore x \leq y$
15) Answer (D)
I.$x^{2} + 12x + 32 = 0$
=> $x^2 + 8x + 4x + 32 = 0$
=> $x (x + 8) + 4 (x + 8) = 0$
=> $(x + 8) (x + 4) = 0$
=> $x = -8 , -4$
II.$y^{2} + 17y + 72 = 0$
=> $y^2 + 9y + 8y + 72 = 0$
=> $y (y + 9) + 8 (y + 9) = 0$
=> $(y + 9) (y + 8) = 0$
=> $y = -9 , -8$
$\therefore x \geq y$
Top Rated Banking Study Material
DOWNLOAD APP FOR IBPS Clerk MOCKS
We hope this Quadratic Equation Questions for IBPS PO will be highly useful for your preparation.
