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# Quadratic equation Questions For SBI Clerk PDF

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

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

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

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 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

$10x^2-27x-28=0$ 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.

$6x^2-5x+1=0$ 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.

$8x^2-10x+3=0$ 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.

$2x^2-17x+35=0$ 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.

$x^2-40x+391=0$ 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.

$4x^2-11x-3=0$ 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.

$18x^2+3x-28=0$ 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.

$2x^2-11x+15 = 0$ 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.

$15x^2+x-2 = 0$ 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.

$x^2-7x+12 = 0$ 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.

$6x^2-11x+3 = 0$ 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.

$15x^2-14x-8 =0$ 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.

$2x^2-11x+15 = 0$ 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.

$6x^2-5x-4 = 0$ 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.

$12x^2-25x+12 = 0$ 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.

$12x^2-41x+35 = 0$ 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.

$9x^2-18x+5 = 0$ 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.

$12x^2-19x+5 = 0$ 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.

$9x^2-15x+4 = 0$ 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.

$5x^2-11x+2 = 0$ 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, $x>y$