# Quadratic Equation Quiz Set -2 IBPS PO | IBPS RRB| IBPS Clerk| SBI| RBI

Quadratic Equation Quiz. Welcome to Online Quantitative Aptitude Section in  letsstudytogether.co .Quadratic Equation is one of the most favourite topics of almost every banking exam. Here we are providing a Quiz Set on Quadratic Equation so that you can practice more and get good marks in the examination.

# Quadratic Equation Quiz Set -2

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Directions: In each of these questions two equations numbered I and II are given. You have to solve both the equations and give answer.

## II. 3y2-y-10=0

(a) if x < y

(b) if x ≤ y

(c) if x = y or Relationship between x and y cannot be established

(d) if x > y

(e) if x ≥ y

C. if x = y or Relationship between x and y cannot be established

From I,
9x2 + 15x – 21x – 35 = 0
3x(3x + 5)-7(3x + 5) = 0
(3x – 7)(3x + 5) = 0
x = 7/3 or -5/3
From II,
3y2– y – 10 = 0
3y2– 6y + 5y – 10 = 0
3y(y – 2)+5(y – 2) = 0
y = -5/3 or 2
So, no relationship can be established between
x and y.

## 2.I. 85x2-77x-30=0    II. 255y2– 231y-90=0

(a) if x < y

(b) if x ≤ y

(c) if x = y or Relationship between x and y cannot be established

(d) if x > y

(e) if x ≥ y

C. if x = y or Relationship between x and y cannot be established

From I,
85x2– 77x – 30 = 0
85x2– 102x + 25x – 30 = 0
17x(5x – 6) + 5(5x – 6) = 0
(17x + 5)(5x – 6) = 0
x = -5/17 or 6/5
From II,
255y2– 231y – 90 = 0
3(85y2– 77y – 30) = 0
85y2– 77y – 30 = 0
y = -5/17 or 6/5
So, no relationship can be established between
x and y.

## 3. I. 7x2– 45x + 50 = 0    II. 21y2– 44y + 20 = 0

(a) if x < y

(b) if x ≤ y

(c) if x = y or Relationship between x and y cannot be established

(d) if x > y

(e) if x ≥ y

E. if x ≥ y

From I,
7x2– 45x + 50 = 0
7x2– 35x – 10x + 50 = 0
7x(x – 5)-10(x – 5) = 0
(7x – 10)(x – 5) = 0
x = 10/7 or 5
From II,

21y2– 44y + 20 =0

21y2– 14y – 30y + 20 = 0
7y(3y – 2)-10(3y – 2) = 0
(7y – 10)(3y – 2) = 0
y = 10/7 or 2/3
So, x ≥ y.

## 4. I. 12x2-24x+12=0 II. 5y2-7y – 6=0

(a) if x < y

(b) if x ≤ y

(c) if x = y or Relationship between x and y cannot be established

(d) if x > y

(e) if x ≥ y

C. if x = y or Relationship between x and y cannot be established

From I,
12x2-12x – 12x + 12 = 0
12x(x – 1)-12(x – 1) = 0
(12x – 12)(x – 1) = 0
x = 1
From II,
5y2– 7y – 6 = 0
5y2– 10y – 3y – 6 = 0
5y(y – 2)-3(y – 2) = 0
y = 2/5 or 2
So, no relationship can be established between x and y.

## 5. I. 2x2-3x+1=0     II. 15y2-16y + 4=0

(a) if x < y

(b) if x ≤ y

(c) if x = y or Relationship between x and y cannot be established

(d) if x > y

(e) if x ≥ y

C. if x = y or Relationship between x and y cannot be established

From I,
2x2– 3x + 1 = 0
2x2– 2x – x + 1 = 0
2x(x – 1) – 1(x – 1) = 0
(2x – 1)(x – 1) = 0
x = 1/2 or 1

From II,
15y2– 16y + 4 = 0
15y2-10y – 6y + 4 = 0
5y(3y – 2)-3(3y – 2) = 0
(5y – 3)(3y – 2) = 0
y = 3/5 or 2/3
So, no relationship can be established between
x and y.

## 6. I. x2 +4x -5 = 0 II. y2-20y +19 = 0

(a) x < y

(b) x > y

(c) x ≤ y

(d) x ≥ y

(e) x = y or relationship cannot be established

C. x ≤ y

From equation I:
x2 +4x -5 = (x -1)(x + 5)= 0
=> x = 1, -5
From equation II:
y2-20y +19 = (y -1)(y -19) = 0
=> y = 1, 19

 X =1 X = -5 Y = 1 x = y x < y Y = 19 x < y x < y

So, x ≤ y

## 7. I. x2 +17x +52 = 0    II. 6y2 +y -40 = 0

(a) x < y

(b) x > y

(c) x ≤ y

(d) x ≥ y

(e) x = y or relationship cannot be established

A. x < y

From equation I:
x2 +17x +52 = (x + 4)(x + 13)= 0
=> x = -4, -13
From equation II:
6y2 +y -40 = (3y + 8)(2y -5) = 0
=> y = -8/3, 5/2

 X= -4 X = -13 Y=-8/3 x < y x < y Y = 5/2 x < y x < y

So, x < y

## 8. I. x2 + 12 = 7x II. y2 + 30 = 11y

a) x > y
b) x ≥ y
c) x < y
d) x ≤ y
e) x = y or the relationship cannot be established

C. x < y

From I,
x2– 7x + 12 = 0
=> x=3,4
From II,
y2– 11y + 30 = 0
=> y=5,6
Hence x<y

## 9. I.  x2– 5x – 14 = 0 II. 3y2 + 17y + 22 = 0

a) If x ≤ y

b) If x ≥ y

c) If x < y

d) If x > y

e) If x = y or no specific relation can be established.

B. If x ≥ y

First equation –
x2– 5x – 14 = 0
or, x2– 7x + 2x – 14 = 0
or, x (x – 7) + 2 (x – 7) = 0
or, (x – 7) (x + 2) = 0
or, x = – 2 or 7
Second equation –
3y2 + 17y + 22 = 0
or, 3y2 + 11y + 6y + 22 = 0
or, y (3y + 11) + 2 (3y + 11) = 0
or, (y + 2) (3y + 11) = 0
or, y = – 2 or – 11/3
So here x is always more than y or equals to y so the answer is 2

## 10. I. 12x2– 25x + 12 = 0 II. 2y2– 7y + 6 = 0

a) If x ≤ y

b) If x ≥ y

c) If x < y

d) If x > y

e) If x = y or no specific relation can be established.

C. If x < y

First equation –
12x2– 25x + 12 = 0
or, 12x2– 16x – 9x + 12 = 0
or, 4x (3x – 4) – 3 (3x – 4) = 0
or, (4x – 3) (3x – 4) = 0
or, x = ¾ or 4/3
Second equation –
2y2– 7y + 6 = 0
or, 2y2– 4y – 3y + 6 = 0
or, 2y (y – 2) – 3 (y – 2) = 0
or, (2y – 3) (y – 2) = 0
or, y = 3/2 or 2
So here we can see that y is always more than x. The answer is 3.