# Quadratic equation Questions For SBI Clerk PDF

Download SBI Clerk Quadratic Equation Questions & Answers PDF for SBI Clerk Prelims and Mains exam. Very Important SBI Clerk Quadratic Equation questions on with solutions.

Download quadratic equation questions for sbi clerk pdf

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**Instructions**

In the following question, two equations are given. You have to solve both the equations & find out the relationship between the variables:

**Question 1:Â **$10x^2-27x-28=0$

$6y^2-17y-14=0$

a)Â $x > y$

b)Â $x \geq y$

c)Â $x < y$

d)Â $x \leq y$

e)Â x = y or No relationship can be established

**Question 2:Â **$6x^2-5x+1=0$

$y^2-7y+12=0$

a)Â $x > y$

b)Â $x \geq y$

c)Â $x < y$

d)Â $x \leq y$

e)Â x = y or No relationship can be established

**Instructions**

In the following question, two equations numbered I and II are given. You have to solve both the equations & find out the relationship between the variables:

**Question 3:Â **$8x^2-10x+3=0$

$6y^2-23y+20=0$

a)Â $x > y$

b)Â $x \geq y$

c)Â $x < y$

d)Â $x \leq y$

e)Â x = y or No relationship can be established

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**Question 4:Â **$2x^2-17x+35=0$

$12y^2-11y-5=0$

a)Â $x > y$

b)Â $x \geq y$

c)Â $x < y$

d)Â $x \leq y$

e)Â x = y or No relationship can be established

**Question 5:Â **$x^2-40x+391=0$

$4y^2-180y+2021=0$

a)Â $x > y$

b)Â $x \geq y$

c)Â $x < y$

d)Â $x \leq y$

e)Â x = y or No relationship can be established

**Question 6:Â **$4x^2-11x-3=0$

$6y^2-29y+35=0$

a)Â $x > y$

b)Â $x \geq y$

c)Â $x < y$

d)Â $x \leq y$

e)Â x = y or No relationship can be established

**Question 7:Â **$18x^2+3x-28=0$

$30y^2-47y+14=0$

a)Â $x > y$

b)Â $x \geq y$

c)Â $x < y$

d)Â $x \leq y$

e)Â x = y or No relationship can be established

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**Instructions**

In the following question, two equations numbered I and II are given. You have to solve both the equations & find out the relationship between the variables:

**Question 8:Â **$ 2x^2-11x+15 = 0 $

$ 2y^2-9y+10 = 0 $

a)Â $x > y$

b)Â $x \geq y$

c)Â $x < y$

d)Â $x \leq y$

e)Â x = y or No relationship can be established

**Question 9:Â **$ 15x^2+x-2 = 0 $

$ 20y^2-23y+6 = 0 $

a)Â $x > y$

b)Â $x \geq y$

c)Â $x < y$

d)Â $x \leq y$

e)Â x = y or No relationship can be established

**Question 10:Â **$ x^2-7x+12 = 0 $

$ 8y^2-70y+153 = 0 $

a)Â $x > y$

b)Â $x \geq y$

c)Â $x < y$

d)Â $x \leq y$

e)Â x = y or No relationship can be established

**Question 11:Â **$ 6x^2-11x+3 = 0 $

$ 3y^2-16y+5 = 0 $

a)Â $x > y$

b)Â $x \geq y$

c)Â $x < y$

d)Â $x \leq y$

e)Â x = y or No relationship can be established

**Question 12:Â **$ 15x^2-14x-8 =0 $

$ 10y^2-17y+3 = 0 $

a)Â $x > y$

b)Â $x \geq y$

c)Â $x < y$

d)Â $x \leq y$

e)Â x = y or No relationship can be established

**Instructions**

In the following question, two equations numbered I and II are given. You have to solve both the equations & find out the relationship between the variables:

**Question 13:Â **$2x^2-11x+15 = 0$

$2y^2-7y+6 = 0$

a)Â $x > y$

b)Â $x \geq y$

c)Â $x < y$

d)Â $x \leq y$

e)Â x = y or No relationship can be established

**Question 14:Â **$6x^2-5x-4 = 0$

$6y^2-11y+4 = 0$

a)Â $x > y$

b)Â $x \geq y$

c)Â $x < y$

d)Â $x \leq y$

e)Â x = y or No relationship can be established

**Question 15:Â **$12x^2-25x+12 = 0$

$12y^2-11y-5 = 0$

a)Â $x > y$

b)Â $x \geq y$

c)Â $x < y$

d)Â $x \leq y$

e)Â x = y or No relationship can be established

**Question 16:Â **$12x^2-41x+35 = 0$

$8y^2-14y+5 = 0$

a)Â $x > y$

b)Â $x \geq y$

c)Â $x < y$

d)Â $x \leq y$

e)Â x = y or No relationship can be established

**Question 17:Â **$9x^2-18x+5 = 0$

$12y^2-19y+5 = 0$

a)Â $x > y$

b)Â $x \geq y$

c)Â $x < y$

d)Â $x \leq y$

e)Â x = y or No relationship can be established

**Instructions**

**Question 18:Â **$12x^2-19x+5 = 0$

$20y^2-57y+40 = 0$

a)Â $x > y$

b)Â $x \geq y$

c)Â $x < y$

d)Â $x \leq y$

e)Â x = y or No relationship can be established

**Question 19:Â **$9x^2-15x+4 = 0$

$12y^2-19y+5 = 0$

a)Â $x > y$

b)Â $x \geq y$

c)Â $x < y$

d)Â $x \leq y$

e)Â x = y or No relationship can be established

**Question 20:Â **$5x^2-11x+2 = 0$

$21y^2+4y-1 = 0$

a)Â $x > y$

b)Â $x \geq y$

c)Â $x < y$

d)Â $x \leq y$

e)Â x = y or No relationship can be established

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**Answers & Solutions:**

**1)Â AnswerÂ (E)**

$10x^2-27x-28=0$ can be written as,

$(x-\frac{7}{2})(x+\frac{4}{5})=0$

So $x = \frac{7}{2}$ or $x = -\frac{4}{5}$

$6y^2-17y-14=0$ can be written as,

$(y+\frac{2}{3})(y-\frac{7}{2})=0$

So $y = -\frac{2}{3}$ or $y = \frac{7}{2}$

x = y or No relationship can be established

Hence, option E is the correct choice.

**2)Â AnswerÂ (C)**

$6x^2-5x+1=0$ can be written as,

$(x-\frac{1}{2})(x-\frac{1}{3})=0$

So $x = \frac{1}{2}$ or $x = \frac{1}{3}$

$y^2-7y+12=0$ can be written as,

$(y-3)(y-4)=0$

So $y = 3$ or $y = 4$

c: $x < y$

Hence, option C is the correct choice.

**3)Â AnswerÂ (C)**

$8x^2-10x+3=0$ can be written as,

$(x-\frac{1}{2})(x-\frac{3}{4})=0$

So $x = \frac{1}{2}$ or $x = \frac{3}{4}$

$6y^2-23y+20=0$ can be written as,

$(y-\frac{4}{3})(y-\frac{5}{2})=0$

So $y = \frac{4}{3}$ or $y = \frac{5}{2}$

So $x < y$

Hence, option C is the correct choice.

**4)Â AnswerÂ (A)**

$2x^2-17x+35=0$ can be written as,

$(x-5)(x-\frac{7}{2})=0$

So $x = 5$ or $x = \frac{7}{2}$

$12y^2-11y-5=0$ can be written as,

$(y-\frac{5}{4})(y+\frac{1}{3})=0$

So $y = \frac{5}{4}$ or $y = -\frac{1}{3}$

So $x > y$

Hence, option A is the correct choice.

**5)Â AnswerÂ (E)**

$x^2-40x+391=0$ can be written as,

$(x-23)(x-17)=0$

So $x = 23$ or $x = 17$

$4y^2-180y+2021=0$ can be written as,

$(y-\frac{47}{2})(y-\frac{43}{2})=0$

So $y = \frac{47}{2}$ or $y = \frac{43}{2}$

So No relation can be established.

Hence, option E is the correct choice.

**6)Â AnswerÂ (E)**

$4x^2-11x-3=0$ can be written as,

$(x-3)(x+\frac{1}{4})=0$

So $x = 3$ or $x = -\frac{1}{4}$

$6y^2-29y+35=0$ can be written as,

$(y-\frac{5}{2})(y-\frac{7}{3})=0$

So $y = \frac{5}{2}$ or $y = \frac{7}{2}$

So, No relationship can be established

Hence, option E is the correct choice.

**7)Â AnswerÂ (E)**

$18x^2+3x-28=0$ can be written as,

$(x+\frac{4}{3})(x-\frac{7}{6})=0$

So $x = -\frac{4}{3}$ or $x = \frac{7}{6}$

$30y^2-47y+14=0$ can be written as,

$(y-\frac{7}{6})(y-\frac{2}{5})=0$

So $y = \frac{7}{6}$ or $y = \frac{2}{5}$

So No relation can be established

Hence, option E is the correct choice.

**8)Â AnswerÂ (B)**

$ 2x^2-11x+15 = 0 $ can be written as $(x-\frac{5}{2})(x-3)= 0 $ therefore, x = $\frac{5}{2}$ or $ 3 $

$ 2y^2-9y+10 = 0 $ can be written as $ (y-\frac{5}{2})(y-2) =0 $

Therefore, y = $\frac{5}{2} $ or $ 2 $

Therefore, $x \geq y$

Hence, option B is the correct answer.

**9)Â AnswerÂ (C)**

$ 15x^2+x-2 = 0 $ can be written as $(x-\frac{1}{3})(x+\frac{2}{5})= 0 $ therefore, x = $\frac{1}{3}$ or $ \frac{-2}{5} $

$ 20y^2-23y+6 = 0 $ can be written as $ (y-\frac{2}{5})(y-\frac{3}{4}) =0 $

Therefore, y = $\frac{2}{5}$ or $ \frac{3}{4} $

Therefore, $x < y$

Hence, option C is the correct answer.

**10)Â AnswerÂ (C)**

$ x^2-7x+12 = 0 $ can be written as $(x-4)(x-3)= 0 $ therefore, x = $ 4 $ or $ 3 $

$ 8y^2-70y+153 = 0 $ can be written as $ (y-\frac{9}{2})(y-\frac{17}{4}) =0 $

Therefore, y = $\frac{9}{2} $or $ \frac{17}{4} $

Therefore, $x < y$

Hence, option C is the correct answer.

**11)Â AnswerÂ (E)**

$ 6x^2-11x+3 = 0 $ can be written as $(x-\frac{3}{2})(x-\frac{1}{3})= 0 $ therefore, x = $\frac{3}{2}$ or $ \frac{1}{3} $

$ 3y^2-16y+5 = 0 $ can be written as $ (y-\frac{1}{3})(y-5) =0 $

Therefore, y = $\frac{1}{3}$ or $ 5 $

Therefore, No relation can be established

Hence, option E is the correct answer.

**12)Â AnswerÂ (E)**

$ 15x^2-14x-8 =0 $ can be written as $(x-\frac{4}{3})(x+\frac{2}{5})= 0 $ therefore, x = $\frac{4}{3}$ or $ \frac{-2}{5} $

$ 10y^2-17y+3 = 0 $ can be written as $ (y-\frac{1}{5})(y-\frac{3}{2}) =0 $

Therefore, y = $\frac{3}{2}$ or $ \frac{1}{5} $

Therefore, no relation can be established

Hence, option B is the correct answer.

**13)Â AnswerÂ (A)**

$2x^2-11x+15 = 0$ can be written as $(x-\frac{5}{2})(x-3) = 0$ i.e. x = $\frac{5}{2}$ or $3$

$2y^2-7y+6 = 0$ can be written as $(y-\frac{3}{2})(y-2) = 0$

i.e. y = $\frac{3}{2}$ or $2$

Hence, $x>y$

Hence, option A is the correct answer.

**14)Â AnswerÂ (E)**

$6x^2-5x-4 = 0$ can be written as $(x+\frac{1}{2})(x-\frac{4}{3}) = 0$ i.e. x = -$\frac{1}{2} or \frac{4}{3}$

$6y^2-11y+4 = 0$ can be written as $(y-\frac{4}{3})(y-\frac{1}{2}) = 0$

i.e. y = $\frac{4}{3} or \frac{1}{2}$

Hence, no relation can be established

Hence, option E is the correct answer.

**15)Â AnswerÂ (E)**

$12x^2-25x+12 = 0$ can be written as $(x-\frac{3}{4})(x-\frac{4}{3}) = 0$ i.e. x = $\frac{3}{4} or \frac{4}{3}$

$12y^2-11y-5 = 0$ can be written as $(y-\frac{5}{4})(y+\frac{1}{3}) = 0$

i.e. y = $\frac{5}{4} or -\frac{1}{3}$

Hence, no relation can be established.

Hence, option E is the correct answer.

**16)Â AnswerÂ (A)**

$12x^2-41x+35 = 0$ can be written as $(x-\frac{7}{4})(x-\frac{5}{3}) = 0$ i.e. x = $\frac{7}{4} or \frac{5}{3}$

$8y^2-14y+5 = 0$ can be written as $(y-\frac{5}{4})(y-\frac{1}{2}) = 0$

i.e. y = $\frac{5}{4} or \frac{1}{2}$

Hence, $x > y$

Hence, option A is the correct answer.

**17)Â AnswerÂ (E)**

$9x^2-18x+5 = 0$ can be written as $(x-\frac{5}{3})(x-\frac{1}{3}) = 0$ i.e. x = $\frac{5}{3} or \frac{1}{3}$

$12y^2-19y+5 = 0$ can be written as $(y-\frac{1}{3})(y-\frac{5}{4}) = 0$

i.e. y = $\frac{1}{3} or \frac{5}{4}$

Hence, No relation can be established

Hence, option E is the correct answer.

**18)Â AnswerÂ (D)**

$12x^2-19x+5 = 0$ can be written as $(x-\frac{5}{4})(x-\frac{1}{3}) = 0$

i.e. x = $\frac{5}{4} or \frac{1}{3}$

$20y^2-57y+40 = 0$ can be written as $(y-\frac{5}{4})(y-\frac{8}{5}) = 0$

i.e. y = $\frac{5}{4} or \frac{8}{5}$

Hence, $x\leq y$

Hence, option D is the correct answer.

**19)Â AnswerÂ (E)**

$9x^2-15x+4 = 0$ can be written as $(x-\frac{1}{3})(x-\frac{4}{3}) = 0$ i.e. x = $\frac{1}{3} or \frac{4}{3}$

$12y^2-19y+5 = 0$ can be written as $(y-\frac{1}{3})(y-\frac{5}{4}) = 0$

i.e. y = $\frac{1}{3} or \frac{5}{4}$

Hence, no relation can be established.

Hence, option E is the correct answer.

**20)Â AnswerÂ (A)**

$5x^2-11x+2 = 0$ can be written as $(x-\frac{1}{5})(x-2) = 0$

i.e. x = $\frac{1}{5} or 2$

$21y^2+4y-1 = 0$ can be written as $(y-\frac{1}{7})(y+\frac{1}{3}) = 0$

i.e. y = $\frac{1}{7} or -\frac{1}{3}$

Hence, $x>y$

Hence, option A is the correct answer.

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