# High Level Quadratic Equation Quiz For IBPS Clerk | Syndicate PO Set – 9

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## High Level Quadratic Equation Quiz: Set-9

Directions:(1-5) In each of the questions, two equations numbered I and II with variables x and y are given. You have to solve both the equations to find the value of x and y and give answer.

1. I. 9x2– 6x – 35 = 0
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

D. 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.In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. x2– 16x + 63 = 0
II. y2– 8y + 15 = 0
A. x > y
B. x < y
C. x = y or the relationship cannot be established
D. x ≥ y
E. x ≤ y

A. x > y

From I,

x2– 16x + 63 = 0
x2– 7x – 9x + 63 = 0
x(x – 7) – 9(x – 7) = 0
(x – 7)(x – 9) = 0
x = 7, 9
From II,

y2– 8y + 15 = 0
y2– 3y – 5y + 15 = 0
y(y – 3) – 5(y – 3) = 0
(y – 3)(y – 5) = 0
y = 3, 5
Therefore, x > y

7.In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. x2– 13x + 42 = 0
II. y2– 17y + 70 = 0
A. x > y
B. x < y
C. x = y or the relationship cannot be established
D. x ≥ y
E. x ≤ y

E. x ≤ y

From I,

x2– 13x + 42 = 0
x2– 6x – 7x + 42 = 0
x(x – 6) – 7(x – 6) = 0
(x – 6)(x – 7) = 0
x = 6, 7
From II,

y2– 17y + 70 = 0
y2– 7y – 10y + 70 = 0
y(y – 7) – 10(y – 7) = 0
(y – 7)(y – 10) = 0
y = 7, 10
Therefore, x ≤ y

8. In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. x3 = 133– 866
II. y3 = 431 + 92
A. x > y
B. x < y
C. x = y or the relationship cannot be established
D. x ≥ y
E. x ≤ y

A. x > y

From I,

x3= 133- 866
x3= 2197 – 866 = 1331
x = 11
From II,

y3 = 431 + 92
y3 = 431 + 81 = 512
y = 8
Therefore, x > y

9.In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. x2– 15x + 54 = 0
II. y2– 9y + 18 = 0
A. x > y
B. x < y
C. x = y or the relationship cannot be established
D. x ≥ y
E. x ≤ y

D. x ≥ y

From I,

x2– 15x + 54 = 0
x2– 6x – 9x + 54 = 0
x(x – 6) – 9(x – 6) = 0
(x – 6)(x – 9) = 0
x = 6, 9
From II,

y2– 9y + 18 = 0
y2– 3y – 6y + 18 = 0
y(y – 3) – 6(y – 3) = 0
(y – 3)(y – 6) = 0
y = 3, 6
Therefore, x ≥ y

10.In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. 7x + 4y = 63
II. 9x + y = 52
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 II, we get y = 52 – 9x
So,
7x + 4 × (52 – 9x) = 63
7x + 208 – 36x = 63
x = 5
So, y = 52 – 9 × 5 = 7
Therefore, x < y

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