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SBI Clerk Previous Year Simplification Questions

Download SBI Clerk Simplification Previous Year Questions & Answers PDF for SBI Clerk Prelims and Mains exam. Very Important SBI Clerk Simplification questions on with solutions.

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: $10x^2-27x-28=0$
$6y^2-17y-14=0$

a) $x > y$

b) $x \geq y$

c) $x < y$

d) $x \leq y$

e) x = y or No relationship can be established

Question 2: $6x^2-5x+1=0$
$y^2-7y+12=0$

a) $x > y$

b) $x \geq y$

c) $x < y$

d) $x \leq y$

e) 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: $8x^2-10x+3=0$
$6y^2-23y+20=0$

a) $x > y$

b) $x \geq y$

c) $x < y$

d) $x \leq y$

e) x = y or No relationship can be established

Question 4: $2x^2-17x+35=0$
$12y^2-11y-5=0$

a) $x > y$

b) $x \geq y$

c) $x < y$

d) $x \leq y$

e) x = y or No relationship can be established

Question 5: $x^2-40x+391=0$
$4y^2-180y+2021=0$

a) $x > y$

b) $x \geq y$

c) $x < y$

d) $x \leq y$

e) x = y or No relationship can be established

Question 6: $4x^2-11x-3=0$
$6y^2-29y+35=0$

a) $x > y$

b) $x \geq y$

c) $x < y$

d) $x \leq y$

e) x = y or No relationship can be established

Question 7: $18x^2+3x-28=0$
$30y^2-47y+14=0$

a) $x > y$

b) $x \geq y$

c) $x < y$

d) $x \leq y$

e) 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: $2x^2-11x+15 = 0$
$2y^2-9y+10 = 0$

a) $x > y$

b) $x \geq y$

c) $x < y$

d) $x \leq y$

e) x = y or No relationship can be established

Question 9: $15x^2+x-2 = 0$
$20y^2-23y+6 = 0$

a) $x > y$

b) $x \geq y$

c) $x < y$

d) $x \leq y$

e) x = y or No relationship can be established

Question 10: $x^2-7x+12 = 0$
$8y^2-70y+153 = 0$

a) $x > y$

b) $x \geq y$

c) $x < y$

d) $x \leq y$

e) x = y or No relationship can be established

Question 11: $6x^2-11x+3 = 0$
$3y^2-16y+5 = 0$

a) $x > y$

b) $x \geq y$

c) $x < y$

d) $x \leq y$

e) x = y or No relationship can be established

Question 12: $15x^2-14x-8 =0$
$10y^2-17y+3 = 0$

a) $x > y$

b) $x \geq y$

c) $x < y$

d) $x \leq y$

e) 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: $2x^2-11x+15 = 0$
$2y^2-7y+6 = 0$

a) $x > y$

b) $x \geq y$

c) $x < y$

d) $x \leq y$

e) x = y or No relationship can be established

Question 14: $6x^2-5x-4 = 0$
$6y^2-11y+4 = 0$

a) $x > y$

b) $x \geq y$

c) $x < y$

d) $x \leq y$

e) x = y or No relationship can be established

Question 15: $12x^2-25x+12 = 0$
$12y^2-11y-5 = 0$

a) $x > y$

b) $x \geq y$

c) $x < y$

d) $x \leq y$

e) x = y or No relationship can be established

Question 16: $12x^2-41x+35 = 0$
$8y^2-14y+5 = 0$

a) $x > y$

b) $x \geq y$

c) $x < y$

d) $x \leq y$

e) x = y or No relationship can be established

Question 17: $9x^2-18x+5 = 0$
$12y^2-19y+5 = 0$

a) $x > y$

b) $x \geq y$

c) $x < y$

d) $x \leq y$

e) 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: $12x^2-19x+5 = 0$
$20y^2-57y+40 = 0$

a) $x > y$

b) $x \geq y$

c) $x < y$

d) $x \leq y$

e) x = y or No relationship can be established

Question 19: $9x^2-15x+4 = 0$
$12y^2-19y+5 = 0$

a) $x > y$

b) $x \geq y$

c) $x < y$

d) $x \leq y$

e) x = y or No relationship can be established

Question 20: $5x^2-11x+2 = 0$
$21y^2+4y-1 = 0$

a) $x > y$

b) $x \geq y$

c) $x < y$

d) $x \leq y$

e) 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, $x>y$