Download Syllogism Questions For IBPS SO

5 IBPS SO Mocks For Rs. 117

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 ≥ y
Give answer c: if x < y
Give answer d: if x ≤ y
Give answer e: if x = y or the relationship cannot be established.

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

a) if x > y

b) if x ≥ y

c) if x < y

d) if x ≤ y

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

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

a) if x > y

b) if x ≥ y

c) if x < y

d) if x ≤ y

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

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

a) if x > y

b) if x ≥ y

c) if x < y

d) if x ≤ y

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

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

a) if x > y

b) if x ≥ y

c) if x < y

d) if x ≤ y

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

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

a) if x > y

b) if x ≥ y

c) if x < y

d) if x ≤ y

e) 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 –
Give answer a: if x < y
Give answer b: if x ≤ y
Give answer c: if x > y
Give answer d: if x ≥ y
Give answer e: if x = y or the relationship cannot be established.

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

a) if x < y

b) if x ≤ y

c) if x > y

d) if x ≥ y

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

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

a) if x < y

b) if x ≤ y

c) if x > y

d) if x ≥ y

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

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

a) if x < y

b) if x ≤ y

c) if x > y

d) if x ≥ y

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

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

a) if x < y

b) if x ≤ y

c) if x > y

d) if x ≥ y

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

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

a) if x < y

b) if x ≤ y

c) if x > y

d) if x ≥ y

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

Instructions

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

(1)if x < y
(2) if x ≤ y

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

(4) if ≥ y

(5) if x > y

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

a) if x < y

b) if x ≤ y

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

d)  if ≥ y

e) if x > y

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

a) if x < y

b)  if x ≤ y

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

d) if ≥ y

e) if x > y

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

a) if x < y

b)  if x ≤ y

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

d) if ≥ y

e) if x > y

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

a) if x < y

b) if x ≤ y

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

d)  if ≥ y

e) if x > y

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

a) if x < y

b)  if x ≤ y

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

d) if ≥ y

e) if x > y

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Answers & Solutions:

1) Answer (E)

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.$y^{2} + 8y – 48 = 0$

=> $y^2 + 12y – 4y – 48 = 0$

=> $y (y + 12) – 4 (y + 12) = 0$

=> $(y + 12) (y – 4) = 0$

=> $y = -12 , 4$

$\therefore$ No relation can be established.

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2) Answer (C)

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 $x < y$

3) Answer (D)

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.$2y^{2} – 19y + 44 = 0$

=> $2y^2 – 8y – 11y + 44 = 0$

=> $2y (y – 4) – 11 (y – 4) = 0$

=> $(y – 4) (2y – 11) = 0$

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

$\therefore x \leq y$

4) Answer (D)

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

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

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

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

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

$\therefore x \leq y$

5) Answer (E)

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

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

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

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

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

$\therefore$ No relation can be established.

6) Answer (C)

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

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

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

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

=> $y = -9 , -10$

$\therefore x > y$

7) Answer (A)

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.$y^{2} – 23y + 132 = 0$

=> $y^2 – 11y – 12y + 132 = 0$

=> $y (y – 11) – 12 (y – 11) = 0$

=> $(y – 11) (y – 12) = 0$

=> $y = 11 , 12$

$\therefore x < y$

8) Answer (E)

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.

9) Answer (B)

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.$y^{2} – 26y + 168 = 0$

=> $y^2 – 12y – 14y + 168 = 0$

=> $y (y – 12) – 14 (y – 12) = 0$

=> $(y – 12) (y – 14) = 0$

=> $y = 12 , 14$

$\therefore x \leq y$

10) Answer (D)

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

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

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

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

=> $y = -9 , -8$

$\therefore x \geq y$

11) Answer (B)

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

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

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

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

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

$\therefore$ $x \leq y$

12) Answer (C)

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.

13) Answer (E)

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

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

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

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

$\therefore$ $x > y$

14) Answer (A)

Statement 1 : $x^2 + 12x + 36 = 0$

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

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

=> $x = -6$

Statement II : $y^2 = 16$

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

=> $y = \pm 4$

$\therefore$ $x < y$

15) Answer (C)

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.

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