 High Level Quadratic Equation Quiz For SBI Clerk | NABARD 2018 Set – 11

## Quadratic Equation Quiz For SBI Clerk 2018

High-Level Quadratic Equation Quiz for SBI Clerk 2018. Welcome to the www.letsstudytogether.co online Quant section. If you are preparing for SBI PO/Clerk, NABARD Exams 2018, you will come across a section of Quant Section.Here we are providing you with Quant quiz on “High-Level Quadratic Equation Quiz for SBI Clerk 2018″based on the latest pattern for your daily practice.

This “High-Level Quadratic Equation Quiz” is also important for other banking exams such as SBI PO/Clerk, NABARD, IBPS PO, IBPS Clerk, SBI Clerk, IBPS RRB Officer, IBPS RRB Office Assistant and other competitive exams.

## High-Level Quadratic Equation Quiz: Set-11

Directions:(1-5) 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

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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

2.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

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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

3. 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

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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

4.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

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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

5. 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

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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

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. 45x – 14 = 25x2
II. 6y2+ 15 = 19y
A  x > y
B  x < y
C  x = y or the relationship cannot be established
D  x ≥ y
E  x ≤ y

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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. 10x– 33x + 20 = 0
II. 9y2 = 14 – 15y
A  x > y
B  x < y
C  x = y or the relationship cannot be established
D  x ≥ y
E  x ≤ y

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8. A. x > y
B. x < y
C. x = y or the relationship cannot be established
D. x ≥ y
E. x ≤ y

9. I. 9x – 2y = 4
II. 4x + 3y = 29
A  x > y
B  x < y
C  x = y or the relationship cannot be established
D  x ≥ y
E  x ≤ y

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10. I. x2 = 2x + 15
II. y2 – 12y + 35 = 0
A  x > y
B  x < y
C  x = y or the relationship cannot be established
D  x ≥ y
E  x ≤ y

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E. x ≤ y

From I:
x2 = 2x + 15
=>x2 – 2x – 15 = 0
=>x2 + 3x – 5x – 15 = 0
=>x(x + 3) – 5(x + 3) = 0
=>(x + 3)(x – 5) = 0
=>x = -3, 5
From II:
y2– 12y + 35 = 0
=>y2– 7y – 5y + 35 = 0
=>y(y – 7) – 5(y – 7) = 0
=>(y – 7)(y – 5) = 0
=>y = 7, 5
So, x ≤ y

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