# Bank Exams :: Quantitative Aptitude :: Quadratic equations IBPS Recruitment Latest Govt Jobs

## Home Bank Exams / Quantitative Aptitude Quadratic equations Questions and Answers

1 . 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 $\leq$ y
Give answer (c) if x > y
Give answer (d) if x $\geq$ y
Give answer (e) if x = y or the relationship cannot be established.
I. $x^2$ + 15$x$ + 56 = 0
II. $y^2$ - 23y + 132 = 0
a
b
c
d
e View Answer Discuss in Forum
2 . 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 $\leq$ y
Give answer (c) if x > y
Give answer (d) if x $\geq$ y
Give answer (e) if x = y or the relationship cannot be established.
I. $x^2$ - 22$x$ + 120 = 0
II. $y^2$ - 26y + 168 = 0
a
b
c
d
e View Answer Discuss in Forum
3 . 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 $\leq$ y
Give answer (c) if x > y
Give answer (d) if x $\geq$ y
Give answer (e) if x = y or the relationship cannot be established.
I. $x^2$ + 12$x$ + 32 = 0
II. $y^2$ + 17y + 72 = 0
a
b
c
d
e View Answer Discuss in Forum
4 . 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 $\geq$ y
Give answer (c) If x < y
Give answer (d) If x $\leq$ y
Give answer (e) If x = y or the relationship cannot be established.
I. $x^2$ - 32 = 112
II. y - $\sqrt {169}$ = 0
a
b
c
d
e View Answer Discuss in Forum
5 . 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 $\geq$ y
Give answer (c) If x < y
Give answer (d) If x $\leq$ y
Give answer (e) If x = y or the relationship cannot be established.
I. $x$ - $\sqrt {121}$ = 0
II.$y^2$ - 121 = 0
a
b
c
d
e View Answer Discuss in Forum
6 . 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 $\geq$ y
Give answer (c) If x < y
Give answer (d) If x $\leq$ y
Give answer (e) If x = y or the relationship cannot be established.
I. $3 \over {\sqrt {x}}$ + $4 \over {\sqrt {x}}$ = $\sqrt {x}$
II. $y^2$ = ${(7)^{5 \over 2}} \over {\sqrt {y}}$ = 0
a
b
c
d
e View Answer Discuss in Forum
7 . In each of these questions, two are given. You have to solve these equations and find out the values of x and y and-
Give answer
(a) If x < y
(b) If x > y
(c) If x $\leq$ y
(d) If x $\geq$ y
(e) If x = y
I. 16$x^2$ + 20$x$ + 6 = 0
II. 10$y^2$ + 38y + 24 = 0
a
b
c
d
e View Answer Discuss in Forum
8 . In each of these questions, two are given. You have to solve these equations and find out the values of x and y and-
Give answer
(a) If x < y
(b) If x > y
(c) If x $\leq$ y
(d) If x $\geq$ y
(e) If x = y
I. 18$x^2$ + 18$x$ + 4 = 0
II. 12$y^2$ + 29y + 14 = 0
a
b
c
d
e View Answer Discuss in Forum
9 . In each of these questions, two are given. You have to solve these equations and find out the values of x and y and-
Give answer
(a) If x < y
(b) If x > y
(c) If x $\leq$ y
(d) If x $\geq$ y
(e) If x = y
I. 8$x^2$ + 6$x$ = 5
II. 12$y^2$ - 22y + 8 = 0
a
b
c
d
e View Answer Discuss in Forum
10 . In each of these questions, two are given. You have to solve these equations and find out the values of x and y and-
Give answer
(a) If x < y
(b) If x > y
(c) If x $\leq$ y
(d) If x $\geq$ y
(e) If x = y
I. 17$x^2$ + 48$x$ = 9
II. 13$y^2$ - 32y - 12 = 0
a
b
c
d
e View Answer Discuss in Forum
11 . 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
(b) If x $\geq$ y
(c) If x < y
(d) If x $\leq$ y
(e) If x = y or the relationship cannot be established
I. $\sqrt {25x^2}$ - 125 = 0
II.$\sqrt {361}y$ + 95 = 0
a
b
c
d
e View Answer Discuss in Forum
12 . 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
(b) If x $\geq$ y
(c) If x < y
(d) If x $\leq$ y
(e) If x = y or the relationship cannot be established
I. $5 \over 7$ - $5 \over 21$ = ${\sqrt {x}} \over 42$
II. ${\sqrt {y}} \over 4$ + ${\sqrt {y}} \over 16$ = $250 \over {\sqrt {y}}$
a
b
c
d
e View Answer Discuss in Forum
13 . 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
(b) If x $\geq$ y
(c) If x < y
(d) If x $\leq$ y
(e) If x = y or the relationship cannot be established
I. 5$x^2$ - 18$x$ + 9 = 0
II. 3$y^2$ + 5y - 2 = 0
a
b
c
d
e View Answer Discuss in Forum
14 . In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answers.
(a) if x > y
(b) if x $\geq$ y
(c) if x < y
(d) if x $\leq$ y
(e) if x = y or the relationship cannot be established
I. 12$x^2$ + 11$x$ + 12 = 10$x^2$ + 22$x$
II. 13$y^2$ - 18y + 3 = 9$y^2$ - 10y
a
b
c
d
e View Answer Discuss in Forum
15 . In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answers.
(a) if x > y
(b) if x $\geq$ y
(c) if x < y
(d) if x $\leq$ y
(e) if x = y or the relationship cannot be established
I. $18 \over {x^2}$ + $6 \over x$ - $12 \over {x^2}$ = $8 \over {x^2}$
II. $y^3$ + 9.68 + 5.64 = 16.95
a
b
c
d
e View Answer Discuss in Forum
16 . In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answers.
(a) if x > y
(b) if x $\geq$ y
(c) if x < y
(d) if x $\leq$ y
(e) if x = y or the relationship cannot be established
I. $\sqrt {1225x}$ + $\sqrt {4900}$ = 0
II. $(81)^{1 \over 4}$ y + ${343}^{1 \over 3}$ = 0
a
b
c
d
e View Answer Discuss in Forum
17 . In each of the following questions two equations are given. You have to solve them and give answer accordingly.
(a) If x > y
(b) If x < y
(c) If x = y, or the relationship cannot be established
(d) If x $\geq$ y
(e) If x $\leq$ y
I. ${x^2} \over 2$ + $x$ - $1 \over 2$ = 1
II. 3$y^2$ - 10y + 8 = $y^2$ + 2y - 10
a
b
c
d
e View Answer Discuss in Forum
18 . In each of the following questions two equations are given. You have to solve them and give answer accordingly.
(a) If x > y
(b) If x < y
(c) If x = y, or the relationship cannot be established
(d) If x $\geq$ y
(e) If x $\leq$ y
I. 4$x^2$ - 20x + 19 = 4x - 1
II. 2$y^2$ = 26y - 84
a
b
c
d
e View Answer Discuss in Forum
19 . In each of the following questions two equations are given. You have to solve them and give answer accordingly.
(a) If x > y
(b) If x < y
(c) If x = y, or the relationship cannot be established
(d) If x $\geq$ y
(e) If x $\leq$ y
I. $y^2$ + y - 1 = 4 - 2y - $y^2$
II. ${x^2} \over 2$ - $3 \over 2$$x$ = $x$ - 3
a
b
c
d
e View Answer Discuss in Forum
20 . In each of the following questions two equations are given. You have to solve them and give answer accordingly.
(a) If x > y
(b) If x < y
(c) If x = y, or the relationship cannot be established
(d) If x $\geq$ y
(e) If x $\leq$ y
I. 6$x^2$ + 13$x$ = 12 - $x$
II. 1 + 2y2 = 2y + $5y \over 6$
a
b
c
d
e View Answer Discuss in Forum
21 . In each of the following questions equation I and equation II have been given. You have to solve both of these equations and Given answer
If (a) x < y
(b) x > y
(c) x $\leq$ y
(d) x $\geq$ y
(e) x = y or no relation between two can be established.
I. 4$x^2$ - 32$x$ + 63 = 0
II. 2$y^2$ - 11y + 15 = 0
a
b
c
d
e View Answer Discuss in Forum
22 . In each of the following questions equation I and equation II have been given. You have to solve both of these equations and Given answer
If (a) x < y
(b) x > y
(c) x $\leq$ y
(d) x $\geq$ y
(e) x = y or no relation between two can be established.
I. $x^3$ = $(\sqrt {216})^3$
II. 6$y^2$ = 150
a
b
c
d
e View Answer Discuss in Forum
23 . In each of the following questions equation I and equation II have been given. You have to solve both of these equations and Given answer
If (a) x < y
(b) x > y
(c) x $\leq$ y
(d) x $\geq$ y
(e) x = y or no relation between two can be established.
I. 12$x^2$ + 17$x$ + 6 = 0
II. 6$y^2$ + 5y + 1 = 0
a
b
c
d
e View Answer Discuss in Forum
24 . In each of the following questions equation I and equation II have been given. You have to solve both of these equations and Given answer
If (a) x < y
(b) x > y
(c) x $\leq$ y
(d) x $\geq$ y
(e) x = y or no relation between two can be established.
I. 20$x^2$ + 9$x$ + 1 = 0
II. 30$y^2$ + 11y + 1 = 0
a
b
c
d
e View Answer Discuss in Forum
25 . In the following questions two equations numbered I and II are given. You have to solve both the equations and Give Answer
if (a) x > y
(b) x $\geq$ y
(c) x < y
(d) x $\leq$ y
(e) x = y or the relationship cannot be established.
I. 20$x^2$ - $x$ - 12 = 0
II. 20$y^2$ + 27y + 9 = 0
a
b
c
d
e View Answer Discuss in Forum
26 . In the following questions two equations numbered I and II are given. You have to solve both the equations and Give Answer
if (a) x > y
(b) x $\geq$ y
(c) x < y
(d) x $\leq$ y
(e) x = y or the relationship cannot be established.
I. $x^2$ - 218 = 106
II. $y^2$ - 37y + 342 = 0
a
b
c
d
e View Answer Discuss in Forum
27 . In the following questions two equations numbered I and II are given. You have to solve both the equations and Give Answer
if (a) x > y
(b) x $\geq$ y
(c) x < y
(d) x $\leq$ y
(e) x = y or the relationship cannot be established.
I. $7 \over {\sqrt {x}}$ + $5 \over {\sqrt {x}}$ = $\sqrt {x}$
II. $y^2$ - ${P(12)^{5 \over 2}} \over {\sqrt P{y}}$ = 0
a
b
c
d
e View Answer Discuss in Forum
28 . In the following questions two equations numbered I and II are given. You have to solve both the equations and Give Answer
if (a) x > y
(b) x $\geq$ y
(c) x < y
(d) x $\leq$ y
(e) x = y or the relationship cannot be established.
I. $\sqrt {361x}$ + $\sqrt {16}$ = 0
II. $\sqrt {441y}$ + 4 = 0
a
b
c
d
e View Answer Discuss in Forum

Profit loss and discount Ratio and proportion 1

Sponsored Links

### Newest Comment Oldest Comment Best Comment

Write your Comment or Discuss..
Log in | Sign up
Advertisements

Copyright 2019-2020 | Privacy Policy | Terms and Conditions | Contact us | Advertise