{"id":38338,"date":"2019-11-27T17:57:54","date_gmt":"2019-11-27T12:27:54","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=38338"},"modified":"2019-11-27T17:57:54","modified_gmt":"2019-11-27T12:27:54","slug":"syllogism-questions-for-ibps-so","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/syllogism-questions-for-ibps-so\/","title":{"rendered":"Syllogism Questions For IBPS SO"},"content":{"rendered":"
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Syllogism Questions For IBPS SO<\/h1>\n
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Download important Syllogism PDF based on previously asked questions in IBPS Clerk and other Banking Exams. Practice Syllogism for IBPS Clerk Exam.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

Download Syllogism Questions For IBPS SO<\/a><\/p>\n

5 IBPS SO Mocks For Rs. 117<\/a><\/p>\n

Instructions<\/b><\/p>\n

In the following questions two equations numbered I and II are given. You have to solve both the equations and
\nGive answer a: if x > y
\nGive answer b: if x \u2265 y
\nGive answer c: if x < y
\nGive answer d: if x \u2264 y
\nGive answer e: if x = y or the relationship cannot be established.<\/p>\n

Question 1:\u00a0<\/b>I.\u00a0 $x^{2}-3x-88=0$
\nII.\u00a0$y^{2}+8y-48=0$<\/p>\n

a)\u00a0if x > y<\/p>\n

b)\u00a0if x \u2265 y<\/p>\n

c)\u00a0if x < y<\/p>\n

d)\u00a0if x \u2264 y<\/p>\n

e)\u00a0if x = y or the relationship cannot be established.<\/p>\n

Question 2:\u00a0<\/b>I.\u00a0 $5x^{2}+29x+20=0$
\nII.\u00a0$25y^{2}+25y+6=0$<\/p>\n

a)\u00a0if x > y<\/p>\n

b)\u00a0if x \u2265 y<\/p>\n

c)\u00a0if x < y<\/p>\n

d)\u00a0if x \u2264 y<\/p>\n

e)\u00a0if x = y or the relationship cannot be established.<\/p>\n

Question 3:\u00a0<\/b>I.\u00a0 $2x^{2}-11x+12=0$
\nII.\u00a0$2y^{2}-19y+44=0$<\/p>\n

a)\u00a0if x > y<\/p>\n

b)\u00a0if x \u2265 y<\/p>\n

c)\u00a0if x < y<\/p>\n

d)\u00a0if x \u2264 y<\/p>\n

e)\u00a0if x = y or the relationship cannot be established.<\/p>\n

Take IBPS SO Mock Test<\/a><\/p>\n

Free Banking Study Material – 15000 Solved Questions<\/a><\/p>\n

Question 4:\u00a0<\/b>I.\u00a0 $3x^{2}+10x+8=0$
\nII.\u00a0$3y^{2}+7y+4=0$<\/p>\n

a)\u00a0if x > y<\/p>\n

b)\u00a0if x \u2265 y<\/p>\n

c)\u00a0if x < y<\/p>\n

d)\u00a0if x \u2264 y<\/p>\n

e)\u00a0if x = y or the relationship cannot be established.<\/p>\n

Question 5:\u00a0<\/b>I.\u00a0 $2x^{2}+21x+10=0$
\nII.\u00a0$3y^{2}+13y+14=0$<\/p>\n

a)\u00a0if x > y<\/p>\n

b)\u00a0if x \u2265 y<\/p>\n

c)\u00a0if x < y<\/p>\n

d)\u00a0if x \u2264 y<\/p>\n

e)\u00a0if x = y or the relationship cannot be established.<\/p>\n

Instructions<\/b><\/p>\n

In each of these questions two equations numbered I and II are given. You have to solve both the equations and \u2013
\nGive answer a: if x < y
\nGive answer b: if x \u2264 y
\nGive answer c: if x > y
\nGive answer d: if x \u2265 y
\nGive answer e: if x = y or the relationship cannot be established.<\/p>\n

Question 6:\u00a0<\/b>I.\u00a0 $x^{2}+13x+42=0$
\nII.\u00a0$y^{2} +19y+90=0$<\/p>\n

a)\u00a0if x < y<\/p>\n

b)\u00a0if x \u2264 y<\/p>\n

c)\u00a0if x > y<\/p>\n

d)\u00a0if x \u2265 y<\/p>\n

e)\u00a0if x = y or the relationship cannot be established.<\/p>\n

Question 7:\u00a0<\/b>I.\u00a0 \u00a0$x^{2}-15x+56=0$
\nII.\u00a0$y^{2} -23y+132=0$<\/p>\n

a)\u00a0if x < y<\/p>\n

b)\u00a0if x \u2264 y<\/p>\n

c)\u00a0if x > y<\/p>\n

d)\u00a0if x \u2265 y<\/p>\n

e)\u00a0if x = y or the relationship cannot be established.<\/p>\n

Question 8:\u00a0<\/b>I.\u00a0$x^{2}+7x+12=0$
\nII.\u00a0$y^{2} +6y+8=0$<\/p>\n

a)\u00a0if x < y<\/p>\n

b)\u00a0if x \u2264 y<\/p>\n

c)\u00a0if x > y<\/p>\n

d)\u00a0if x \u2265 y<\/p>\n

e)\u00a0if x = y or the relationship cannot be established.<\/p>\n

Question 9:\u00a0<\/b>I.\u00a0$x^{2}-22x+120=0$
\nII.\u00a0$y^{2} -26y+168=0$<\/p>\n

a)\u00a0if x < y<\/p>\n

b)\u00a0if x \u2264 y<\/p>\n

c)\u00a0if x > y<\/p>\n

d)\u00a0if x \u2265 y<\/p>\n

e)\u00a0if x = y or the relationship cannot be established.<\/p>\n

Get 790+ Mocks For Rs,100<\/a><\/p>\n

Question 10:\u00a0<\/b>I.$x^{2}+12x+32=0$
\nII.\u00a0$y^{2} +17y+72=0$<\/p>\n

a)\u00a0if x < y<\/p>\n

b)\u00a0if x \u2264 y<\/p>\n

c)\u00a0if x > y<\/p>\n

d)\u00a0if x \u2265 y<\/p>\n

e)\u00a0if x = y or the relationship cannot be established.<\/p>\n

Instructions<\/b><\/p>\n

In each of the following questions, two equations I and II have been given.
\nSolve these questions and answer<\/p>\n

(1)if x < y
\n(2) if x \u2264 y<\/p>\n

(3) if x = y or the relation cannot be established<\/p>\n

(4) if \u2265 y<\/p>\n

(5) if x > y<\/p>\n

Question 11:\u00a0<\/b>I. $30 x^{2} + 11x + 1 = 0$
\nII. $42 y^{2} + 13y + 1 = 0$<\/p>\n

a)\u00a0if x < y<\/p>\n

b)\u00a0if x \u2264 y<\/p>\n

c)\u00a0if x = y or the relation cannot be established<\/p>\n

d)\u00a0 if \u2265 y<\/p>\n

e)\u00a0if x > y<\/p>\n

Question 12:\u00a0<\/b>I. $x^{2}- x – \\sqrt{2}x + \\sqrt{2}=0$
\nII.$y^{2}-3y+2=0$<\/p>\n

a)\u00a0if x < y<\/p>\n

b)\u00a0 if x \u2264 y<\/p>\n

c)\u00a0if x = y or the relation cannot be established<\/p>\n

d)\u00a0if \u2265 y<\/p>\n

e)\u00a0if x > y<\/p>\n

200+ Banking Previous Papers (Download PDF)<\/a><\/p>\n

Question 13:\u00a0<\/b>I.$x^{2}-2x-\\sqrt{5}x+2\\sqrt{5}=0$
\nII.$y^{2}-\\sqrt{3}y-\\sqrt{2}y+\\sqrt{6}=0$<\/p>\n

a)\u00a0if x < y<\/p>\n

b)\u00a0 if x \u2264 y<\/p>\n

c)\u00a0if x = y or the relation cannot be established<\/p>\n

d)\u00a0if \u2265 y<\/p>\n

e)\u00a0if x > y<\/p>\n

Question 14:\u00a0<\/b>I.$ x^{2}+12x+36=0$
\nII.$y^{2}=16$<\/p>\n

a)\u00a0if x < y<\/p>\n

b)\u00a0if x \u2264 y<\/p>\n

c)\u00a0if x = y or the relation cannot be established<\/p>\n

d)\u00a0 if \u2265 y<\/p>\n

e)\u00a0if x > y<\/p>\n

Question 15:\u00a0<\/b>I.$9x^{2}+3x-2=0$
\nII.$8y^{2}+6y+1=0$<\/p>\n

a)\u00a0if x < y<\/p>\n

b)\u00a0 if x \u2264 y<\/p>\n

c)\u00a0if x = y or the relation cannot be established<\/p>\n

d)\u00a0if \u2265 y<\/p>\n

e)\u00a0if x > y<\/p>\n

Free Banking Study Material (15000 Solved Questions)<\/a><\/p>\n

Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

I.$x^{2} – 3x – 88 = 0$<\/p>\n

=> $x^2 + 8x – 11x – 88 = 0$<\/p>\n

=> $x (x + 8) – 11 (x + 8) = 0$<\/p>\n

=> $(x + 8) (x – 11) = 0$<\/p>\n

=> $x = -8 , 11$<\/p>\n

II.$y^{2} + 8y – 48 = 0$<\/p>\n

=> $y^2 + 12y – 4y – 48 = 0$<\/p>\n

=> $y (y + 12) – 4 (y + 12) = 0$<\/p>\n

=> $(y + 12) (y – 4) = 0$<\/p>\n

=> $y = -12 , 4$<\/p>\n

$\\therefore$\u00a0No relation can be established.<\/p>\n

5 IBPS SO Mocks For Rs. 117<\/a><\/p>\n

2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

I.$5x^{2} + 29x + 20 = 0$<\/p>\n

=> $5x^2 + 25x + 4x + 20 = 0$<\/p>\n

=> $5x (x + 5) + 4 (x + 5) = 0$<\/p>\n

=> $(x + 5) (5x + 4) = 0$<\/p>\n

=> $x = -5 , \\frac{-4}{5}$<\/p>\n

II.$25y^{2} + 25y + 6 = 0$<\/p>\n

=> $25y^2 + 10y + 15y + 6 = 0$<\/p>\n

=> $5y (5y + 2) + 3 (5y + 2) = 0$<\/p>\n

=> $(5y + 3) (5y + 2) = 0$<\/p>\n

=> $y = \\frac{-3}{5} , \\frac{-2}{5}$<\/p>\n

Therefore $x < y$<\/p>\n

3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

I.$2x^{2} – 11x + 12 = 0$<\/p>\n

=> $2x^2 – 8x – 3x + 12 = 0$<\/p>\n

=> $2x (x – 4) – 3 (x – 4) = 0$<\/p>\n

=> $(x – 4) (2x – 3) = 0$<\/p>\n

=> $x = 4 , \\frac{3}{2}$<\/p>\n

II.$2y^{2} – 19y + 44 = 0$<\/p>\n

=> $2y^2 – 8y – 11y + 44 = 0$<\/p>\n

=> $2y (y – 4) – 11 (y – 4) = 0$<\/p>\n

=> $(y – 4) (2y – 11) = 0$<\/p>\n

=> $y = 4 , \\frac{11}{2}$<\/p>\n

$\\therefore x \\leq y$<\/p>\n

4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

I.$3x^{2} + 10x + 8 = 0$<\/p>\n

=> $3x^2 + 6x + 4x + 8 = 0$<\/p>\n

=> $3x (x + 2) + 4 (x + 2) = 0$<\/p>\n

=> $(x + 2) (3x + 4) = 0$<\/p>\n

=> $x = -2 , \\frac{-4}{3}$<\/p>\n

II.$3y^{2} + 7y + 4 = 0$<\/p>\n

=> $3y^2 + 3y + 4y + 4 = 0$<\/p>\n

=> $3y (y + 1) + 4 (y + 1) = 0$<\/p>\n

=> $(y + 1) (3y + 4) = 0$<\/p>\n

=> $y = -1 , \\frac{-4}{3}$<\/p>\n

$\\therefore x \\leq y$<\/p>\n

5)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

I.$2x^{2} + 21x + 10 = 0$<\/p>\n

=> $2x^2 + x + 20x + 10 = 0$<\/p>\n

=> $x (2x + 1) + 10 (2x + 1) = 0$<\/p>\n

=> $(x + 10) (2x + 1) = 0$<\/p>\n

=> $x = -10 , \\frac{-1}{2}$<\/p>\n

II.$3y^{2} + 13y + 14 = 0$<\/p>\n

=> $3y^2 + 6y + 7y + 14 = 0$<\/p>\n

=> $3y (y + 2) + 7 (y + 2) = 0$<\/p>\n

=> $(y + 2) (3y + 7) = 0$<\/p>\n

=> $y = -2 , \\frac{-7}{3}$<\/p>\n

$\\therefore$\u00a0No relation can be established.<\/p>\n

6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

I.$x^{2} + 13x + 42 = 0$<\/p>\n

=> $x^2 + 7x + 6x + 42 = 0$<\/p>\n

=> $x (x + 7) + 6 (x + 7) = 0$<\/p>\n

=> $(x + 7) (x + 6) = 0$<\/p>\n

=> $x = -7 , -6$<\/p>\n

II.$y^{2} + 19y + 90 = 0$<\/p>\n

=> $y^2 + 9y + 10y + 90 = 0$<\/p>\n

=> $y (y + 9) + 10 (y + 9) = 0$<\/p>\n

=> $(y + 9) (y + 10) = 0$<\/p>\n

=> $y = -9 , -10$<\/p>\n

$\\therefore x > y$<\/p>\n

7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

I.$x^{2} – 15x + 56 = 0$<\/p>\n

=> $x^2 – 8x – 7x + 56 = 0$<\/p>\n

=> $x (x – 8) – 7 (x – 8) = 0$<\/p>\n

=> $(x – 8) (x – 7) = 0$<\/p>\n

=> $x = 8 , 7$<\/p>\n

II.$y^{2} – 23y + 132 = 0$<\/p>\n

=> $y^2 – 11y – 12y + 132 = 0$<\/p>\n

=> $y (y – 11) – 12 (y – 11) = 0$<\/p>\n

=> $(y – 11) (y – 12) = 0$<\/p>\n

=> $y = 11 , 12$<\/p>\n

$\\therefore x < y$<\/p>\n

8)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

I.$x^{2} + 7x + 12 = 0$<\/p>\n

=> $x^2 + 3x + 4x + 12 = 0$<\/p>\n

=> $x (x + 3) + 4 (x + 3) = 0$<\/p>\n

=> $(x + 3) (x + 4) = 0$<\/p>\n

=> $x = -3 , -4$<\/p>\n

II.$y^{2} + 6y + 8 = 0$<\/p>\n

=> $y^2 + 4y + 2y + 8 = 0$<\/p>\n

=> $y (y + 4) + 2 (y + 4) = 0$<\/p>\n

=> $(y + 4) (y + 2) = 0$<\/p>\n

=> $y = -4 , -2$<\/p>\n

Because $-2 > -4$ and $-3 > -4$<\/p>\n

Therefore, no relation can be established.<\/p>\n

9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

I.$x^{2} – 22x + 120 = 0$<\/p>\n

=> $x^2 – 10x – 12x + 120 = 0$<\/p>\n

=> $x (x – 10) – 12 (x – 10) = 0$<\/p>\n

=> $(x – 10) (x – 12) = 0$<\/p>\n

=> $x = 10 , 12$<\/p>\n

II.$y^{2} – 26y + 168 = 0$<\/p>\n

=> $y^2 – 12y – 14y + 168 = 0$<\/p>\n

=> $y (y – 12) – 14 (y – 12) = 0$<\/p>\n

=> $(y – 12) (y – 14) = 0$<\/p>\n

=> $y = 12 , 14$<\/p>\n

$\\therefore x \\leq y$<\/p>\n

10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

I.$x^{2} + 12x + 32 = 0$<\/p>\n

=> $x^2 + 8x + 4x + 32 = 0$<\/p>\n

=> $x (x + 8) + 4 (x + 8) = 0$<\/p>\n

=> $(x + 8) (x + 4) = 0$<\/p>\n

=> $x = -8 , -4$<\/p>\n

II.$y^{2} + 17y + 72 = 0$<\/p>\n

=> $y^2 + 9y + 8y + 72 = 0$<\/p>\n

=> $y (y + 9) + 8 (y + 9) = 0$<\/p>\n

=> $(y + 9) (y + 8) = 0$<\/p>\n

=> $y = -9 , -8$<\/p>\n

$\\therefore x \\geq y$<\/p>\n

11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Statement I : $30 x^{2} + 11x + 1 = 0$<\/p>\n

=> $30x^2 + 6x + 5x + 1 = 0$<\/p>\n

=> $6x (5x + 1) + 1 (5x + 1) = 0$<\/p>\n

=> $(6x + 1) (5x + 1) = 0$<\/p>\n

=> $x = \\frac{-1}{6} , \\frac{-1}{5}$<\/p>\n

Statement II : $42 y^{2} + 13y + 1 = 0$<\/p>\n

=> $42y^2 + 7y + 6y + 1 = 0$<\/p>\n

=> $7y (6y + 1) + 1 (6y + 1) = 0$<\/p>\n

=> $(7y + 1) (6y + 1) = 0$<\/p>\n

=> $y = \\frac{-1}{7} , \\frac{-1}{6}$<\/p>\n

$\\therefore$ $x \\leq y$<\/p>\n

12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

I. $x^{2}- x – \\sqrt{2}x + \\sqrt{2}=0$<\/p>\n

=> $x (x – 1) – \\sqrt{2} (x – 1) = 0$<\/p>\n

=> $(x – \\sqrt{2}) (x – 1) = 0$<\/p>\n

=> $x = \\sqrt{2} , 1$<\/p>\n

II. $y^{2}-3y+2=0$<\/p>\n

=> $y^2 – 2y – y + 2 = 0$<\/p>\n

=> $y (y – 2) – 1 (y – 2) = 0$<\/p>\n

=> $(y – 2) (y – 1) = 0$<\/p>\n

=> $y = 1 , 2$<\/p>\n

$\\therefore$ No relation established.<\/p>\n

13)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

Statement I : $x^{2}-2x-\\sqrt{5}x+2\\sqrt{5}=0$<\/p>\n

=> $x (x – 2) – \\sqrt{5} (x – 2) = 0$<\/p>\n

=> $(x – \\sqrt{5}) (x – 2) = 0$<\/p>\n

=> $x = \\sqrt{5} , 2$<\/p>\n

Statement II : $y^{2}-\\sqrt{3}y-\\sqrt{2}y+\\sqrt{6}=0$<\/p>\n

=> $y (y – \\sqrt{3}) – \\sqrt{2} (y – \\sqrt{3}) = 0$<\/p>\n

=> $(y – \\sqrt{2}) (y – \\sqrt{3}) = 0$<\/p>\n

=> $y = \\sqrt{2} , \\sqrt{3}$<\/p>\n

$\\therefore$ $x > y$<\/p>\n

14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Statement\u00a01\u00a0: $x^2 + 12x + 36 = 0$<\/p>\n

=> $x^2 + 2.x.6 + 6^2 = 0$<\/p>\n

=> $(x + 6)^2 = 0$<\/p>\n

=> $x = -6$<\/p>\n

Statement II : $y^2 = 16$<\/p>\n

=> $(y)^2 = (\\pm 4)^2$<\/p>\n

=> $y = \\pm 4$<\/p>\n

$\\therefore$ $x < y$<\/p>\n

15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Statement I : $9x^{2}+3x-2=0$<\/p>\n

=> $9x^2 + 6x – 3x – 2 = 0$<\/p>\n

=> $3x (3x + 2) – 1 (3x + 2) = 0$<\/p>\n

=> $(3x – 1) (3x + 2) = 0$<\/p>\n

=> $x = \\frac{1}{3} , \\frac{-2}{3}$<\/p>\n

Statement II : $8y^{2}+6y+1=0$<\/p>\n

=> $8y^2 + 4y + 2y + 1 = 0$<\/p>\n

=> $4y (2y + 1) + 1 (2y + 1) = 0$<\/p>\n

=> $(4y + 1) (2y + 1) = 0$<\/p>\n

=> $y = \\frac{-1}{4} , \\frac{-1}{2}$<\/p>\n

$\\therefore$ No relation can be established.<\/p>\n

Download High Rated Free Preparation App<\/a><\/p>\n

Daily Free Banking Online Test<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

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 […]<\/p>\n","protected":false},"author":32,"featured_media":38340,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[277],"tags":[266],"class_list":{"0":"post-38338","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-ibps-so","8":"tag-ibps-so"},"better_featured_image":{"id":38340,"alt_text":"syllogism questions for ibps so","caption":"syllogism questions for ibps so\n","description":"syllogism questions for ibps 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