0
312

# Syllogism Questions For IBPS SO

Download important Syllogism PDF based on previously asked questions in IBPS Clerk and other Banking Exams. Practice Syllogism for IBPS Clerk Exam.

Instructions

In the following questions two equations numbered I and II are given. You have to solve both the equations and
Give answer a: if x > y
Give answer b: if x â‰¥ y
Give answer c: if x < y
Give answer d: if x â‰¤ y
Give answer e: if x = y or the relationship cannot be established.

Question 1:Â I.Â  $x^{2}-3x-88=0$
II.Â $y^{2}+8y-48=0$

a)Â if x > y

b)Â if x â‰¥ y

c)Â if x < y

d)Â if x â‰¤ y

e)Â if x = y or the relationship cannot be established.

Question 2:Â I.Â  $5x^{2}+29x+20=0$
II.Â $25y^{2}+25y+6=0$

a)Â if x > y

b)Â if x â‰¥ y

c)Â if x < y

d)Â if x â‰¤ y

e)Â if x = y or the relationship cannot be established.

Question 3:Â I.Â  $2x^{2}-11x+12=0$
II.Â $2y^{2}-19y+44=0$

a)Â if x > y

b)Â if x â‰¥ y

c)Â if x < y

d)Â if x â‰¤ y

e)Â if x = y or the relationship cannot be established.

Question 4:Â I.Â  $3x^{2}+10x+8=0$
II.Â $3y^{2}+7y+4=0$

a)Â if x > y

b)Â if x â‰¥ y

c)Â if x < y

d)Â if x â‰¤ y

e)Â if x = y or the relationship cannot be established.

Question 5:Â I.Â  $2x^{2}+21x+10=0$
II.Â $3y^{2}+13y+14=0$

a)Â if x > y

b)Â if x â‰¥ y

c)Â if x < y

d)Â if x â‰¤ y

e)Â if x = y or the relationship cannot be established.

Instructions

In each of these questions two equations numbered I and II are given. You have to solve both the equations and â€“
Give answer a: if x < y
Give answer b: if x â‰¤ y
Give answer c: if x > y
Give answer d: if x â‰¥ y
Give answer e: if x = y or the relationship cannot be established.

Question 6:Â I.Â  $x^{2}+13x+42=0$
II.Â $y^{2} +19y+90=0$

a)Â if x < y

b)Â if x â‰¤ y

c)Â if x > y

d)Â if x â‰¥ y

e)Â if x = y or the relationship cannot be established.

Question 7:Â I.Â  Â $x^{2}-15x+56=0$
II.Â $y^{2} -23y+132=0$

a)Â if x < y

b)Â if x â‰¤ y

c)Â if x > y

d)Â if x â‰¥ y

e)Â if x = y or the relationship cannot be established.

Question 8:Â I.Â $x^{2}+7x+12=0$
II.Â $y^{2} +6y+8=0$

a)Â if x < y

b)Â if x â‰¤ y

c)Â if x > y

d)Â if x â‰¥ y

e)Â if x = y or the relationship cannot be established.

Question 9:Â I.Â $x^{2}-22x+120=0$
II.Â $y^{2} -26y+168=0$

a)Â if x < y

b)Â if x â‰¤ y

c)Â if x > y

d)Â if x â‰¥ y

e)Â if x = y or the relationship cannot be established.

Question 10:Â I.$x^{2}+12x+32=0$
II.Â $y^{2} +17y+72=0$

a)Â if x < y

b)Â if x â‰¤ y

c)Â if x > y

d)Â if x â‰¥ y

e)Â if x = y or the relationship cannot be established.

Instructions

In each of the following questions, two equations I and II have been given.

(1)if x < y
(2) if x â‰¤ y

(3) if x = y or the relation cannot be established

(4) if â‰¥ y

(5) if x > y

Question 11:Â I. $30 x^{2} + 11x + 1 = 0$
II. $42 y^{2} + 13y + 1 = 0$

a)Â if x < y

b)Â if x â‰¤ y

c)Â if x = y or the relation cannot be established

d)Â  if â‰¥ y

e)Â if x > y

Question 12:Â I. $x^{2}- x – \sqrt{2}x + \sqrt{2}=0$
II.$y^{2}-3y+2=0$

a)Â if x < y

b)Â  if x â‰¤ y

c)Â if x = y or the relation cannot be established

d)Â if â‰¥ y

e)Â if x > y

Question 13:Â I.$x^{2}-2x-\sqrt{5}x+2\sqrt{5}=0$
II.$y^{2}-\sqrt{3}y-\sqrt{2}y+\sqrt{6}=0$

a)Â if x < y

b)Â  if x â‰¤ y

c)Â if x = y or the relation cannot be established

d)Â if â‰¥ y

e)Â if x > y

Question 14:Â I.$x^{2}+12x+36=0$
II.$y^{2}=16$

a)Â if x < y

b)Â if x â‰¤ y

c)Â if x = y or the relation cannot be established

d)Â  if â‰¥ y

e)Â if x > y

Question 15:Â I.$9x^{2}+3x-2=0$
II.$8y^{2}+6y+1=0$

a)Â if x < y

b)Â  if x â‰¤ y

c)Â if x = y or the relation cannot be established

d)Â if â‰¥ y

e)Â if x > y

I.$x^{2} – 3x – 88 = 0$

=> $x^2 + 8x – 11x – 88 = 0$

=> $x (x + 8) – 11 (x + 8) = 0$

=> $(x + 8) (x – 11) = 0$

=> $x = -8 , 11$

II.$y^{2} + 8y – 48 = 0$

=> $y^2 + 12y – 4y – 48 = 0$

=> $y (y + 12) – 4 (y + 12) = 0$

=> $(y + 12) (y – 4) = 0$

=> $y = -12 , 4$

$\therefore$Â No relation can be established.

I.$5x^{2} + 29x + 20 = 0$

=> $5x^2 + 25x + 4x + 20 = 0$

=> $5x (x + 5) + 4 (x + 5) = 0$

=> $(x + 5) (5x + 4) = 0$

=> $x = -5 , \frac{-4}{5}$

II.$25y^{2} + 25y + 6 = 0$

=> $25y^2 + 10y + 15y + 6 = 0$

=> $5y (5y + 2) + 3 (5y + 2) = 0$

=> $(5y + 3) (5y + 2) = 0$

=> $y = \frac{-3}{5} , \frac{-2}{5}$

Therefore $x < y$

I.$2x^{2} – 11x + 12 = 0$

=> $2x^2 – 8x – 3x + 12 = 0$

=> $2x (x – 4) – 3 (x – 4) = 0$

=> $(x – 4) (2x – 3) = 0$

=> $x = 4 , \frac{3}{2}$

II.$2y^{2} – 19y + 44 = 0$

=> $2y^2 – 8y – 11y + 44 = 0$

=> $2y (y – 4) – 11 (y – 4) = 0$

=> $(y – 4) (2y – 11) = 0$

=> $y = 4 , \frac{11}{2}$

$\therefore x \leq y$

I.$3x^{2} + 10x + 8 = 0$

=> $3x^2 + 6x + 4x + 8 = 0$

=> $3x (x + 2) + 4 (x + 2) = 0$

=> $(x + 2) (3x + 4) = 0$

=> $x = -2 , \frac{-4}{3}$

II.$3y^{2} + 7y + 4 = 0$

=> $3y^2 + 3y + 4y + 4 = 0$

=> $3y (y + 1) + 4 (y + 1) = 0$

=> $(y + 1) (3y + 4) = 0$

=> $y = -1 , \frac{-4}{3}$

$\therefore x \leq y$

I.$2x^{2} + 21x + 10 = 0$

=> $2x^2 + x + 20x + 10 = 0$

=> $x (2x + 1) + 10 (2x + 1) = 0$

=> $(x + 10) (2x + 1) = 0$

=> $x = -10 , \frac{-1}{2}$

II.$3y^{2} + 13y + 14 = 0$

=> $3y^2 + 6y + 7y + 14 = 0$

=> $3y (y + 2) + 7 (y + 2) = 0$

=> $(y + 2) (3y + 7) = 0$

=> $y = -2 , \frac{-7}{3}$

$\therefore$Â No relation can be established.

I.$x^{2} + 13x + 42 = 0$

=> $x^2 + 7x + 6x + 42 = 0$

=> $x (x + 7) + 6 (x + 7) = 0$

=> $(x + 7) (x + 6) = 0$

=> $x = -7 , -6$

II.$y^{2} + 19y + 90 = 0$

=> $y^2 + 9y + 10y + 90 = 0$

=> $y (y + 9) + 10 (y + 9) = 0$

=> $(y + 9) (y + 10) = 0$

=> $y = -9 , -10$

$\therefore x > y$

I.$x^{2} – 15x + 56 = 0$

=> $x^2 – 8x – 7x + 56 = 0$

=> $x (x – 8) – 7 (x – 8) = 0$

=> $(x – 8) (x – 7) = 0$

=> $x = 8 , 7$

II.$y^{2} – 23y + 132 = 0$

=> $y^2 – 11y – 12y + 132 = 0$

=> $y (y – 11) – 12 (y – 11) = 0$

=> $(y – 11) (y – 12) = 0$

=> $y = 11 , 12$

$\therefore x < y$

I.$x^{2} + 7x + 12 = 0$

=> $x^2 + 3x + 4x + 12 = 0$

=> $x (x + 3) + 4 (x + 3) = 0$

=> $(x + 3) (x + 4) = 0$

=> $x = -3 , -4$

II.$y^{2} + 6y + 8 = 0$

=> $y^2 + 4y + 2y + 8 = 0$

=> $y (y + 4) + 2 (y + 4) = 0$

=> $(y + 4) (y + 2) = 0$

=> $y = -4 , -2$

Because $-2 > -4$ and $-3 > -4$

Therefore, no relation can be established.

I.$x^{2} – 22x + 120 = 0$

=> $x^2 – 10x – 12x + 120 = 0$

=> $x (x – 10) – 12 (x – 10) = 0$

=> $(x – 10) (x – 12) = 0$

=> $x = 10 , 12$

II.$y^{2} – 26y + 168 = 0$

=> $y^2 – 12y – 14y + 168 = 0$

=> $y (y – 12) – 14 (y – 12) = 0$

=> $(y – 12) (y – 14) = 0$

=> $y = 12 , 14$

$\therefore x \leq y$

I.$x^{2} + 12x + 32 = 0$

=> $x^2 + 8x + 4x + 32 = 0$

=> $x (x + 8) + 4 (x + 8) = 0$

=> $(x + 8) (x + 4) = 0$

=> $x = -8 , -4$

II.$y^{2} + 17y + 72 = 0$

=> $y^2 + 9y + 8y + 72 = 0$

=> $y (y + 9) + 8 (y + 9) = 0$

=> $(y + 9) (y + 8) = 0$

=> $y = -9 , -8$

$\therefore x \geq y$

Statement I : $30 x^{2} + 11x + 1 = 0$

=> $30x^2 + 6x + 5x + 1 = 0$

=> $6x (5x + 1) + 1 (5x + 1) = 0$

=> $(6x + 1) (5x + 1) = 0$

=> $x = \frac{-1}{6} , \frac{-1}{5}$

Statement II : $42 y^{2} + 13y + 1 = 0$

=> $42y^2 + 7y + 6y + 1 = 0$

=> $7y (6y + 1) + 1 (6y + 1) = 0$

=> $(7y + 1) (6y + 1) = 0$

=> $y = \frac{-1}{7} , \frac{-1}{6}$

$\therefore$ $x \leq y$

I. $x^{2}- x – \sqrt{2}x + \sqrt{2}=0$

=> $x (x – 1) – \sqrt{2} (x – 1) = 0$

=> $(x – \sqrt{2}) (x – 1) = 0$

=> $x = \sqrt{2} , 1$

II. $y^{2}-3y+2=0$

=> $y^2 – 2y – y + 2 = 0$

=> $y (y – 2) – 1 (y – 2) = 0$

=> $(y – 2) (y – 1) = 0$

=> $y = 1 , 2$

$\therefore$ No relation established.

Statement I : $x^{2}-2x-\sqrt{5}x+2\sqrt{5}=0$

=> $x (x – 2) – \sqrt{5} (x – 2) = 0$

=> $(x – \sqrt{5}) (x – 2) = 0$

=> $x = \sqrt{5} , 2$

Statement II : $y^{2}-\sqrt{3}y-\sqrt{2}y+\sqrt{6}=0$

=> $y (y – \sqrt{3}) – \sqrt{2} (y – \sqrt{3}) = 0$

=> $(y – \sqrt{2}) (y – \sqrt{3}) = 0$

=> $y = \sqrt{2} , \sqrt{3}$

$\therefore$ $x > y$

StatementÂ 1Â : $x^2 + 12x + 36 = 0$

=> $x^2 + 2.x.6 + 6^2 = 0$

=> $(x + 6)^2 = 0$

=> $x = -6$

Statement II : $y^2 = 16$

=> $(y)^2 = (\pm 4)^2$

=> $y = \pm 4$

$\therefore$ $x < y$

Statement I : $9x^{2}+3x-2=0$

=> $9x^2 + 6x – 3x – 2 = 0$

=> $3x (3x + 2) – 1 (3x + 2) = 0$

=> $(3x – 1) (3x + 2) = 0$

=> $x = \frac{1}{3} , \frac{-2}{3}$

Statement II : $8y^{2}+6y+1=0$

=> $8y^2 + 4y + 2y + 1 = 0$

=> $4y (2y + 1) + 1 (2y + 1) = 0$

=> $(4y + 1) (2y + 1) = 0$

=> $y = \frac{-1}{4} , \frac{-1}{2}$

$\therefore$ No relation can be established.