{"id":42643,"date":"2020-09-13T10:09:46","date_gmt":"2020-09-13T04:39:46","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=42643"},"modified":"2020-09-13T10:09:46","modified_gmt":"2020-09-13T04:39:46","slug":"quadratic-equation-questions-for-ibps-po-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/quadratic-equation-questions-for-ibps-po-pdf\/","title":{"rendered":"Quadratic Equation Questions for IBPS PO PDF"},"content":{"rendered":"

Quadratic Equation Questions for IBPS PO PDF<\/strong><\/span><\/h2>\n

Download Top-15 Banking Exams Quadratic Equation Questions PDF. Banking Exams Quadratic Equation questions based on asked questions in previous exam papers very important for the Banking \u00a0exams.<\/p>\n

Download Quadratic Equation Questions for IBPS PO PDF<\/a><\/p>\n

105 IBPS PO Mocks for just Rs. 149<\/a><\/p>\n

Download IBPS PO Previous Papers PDF<\/a><\/p>\n

Take a IBPS PO free mock test<\/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

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 the following questions two equations numbered I and II are given. You have to solve both equations and
\nGive answer If<\/p>\n

a. x \u02c3 y
\nb. x \u2265 y
\nc. x \u02c2 y
\nd. x \u2264 y
\ne. x = y or the relationship cannot be established<\/p>\n

Question 6:\u00a0<\/b>I. ${x^2}$ – 7x + 10 = 0
\nII. ${y^2}$ + 11y + 10 = 0<\/p>\n

a)\u00a0x \u02c3 y<\/p>\n

b)\u00a0x \u2265 y<\/p>\n

c)\u00a0x \u02c2 y<\/p>\n

d)\u00a0x \u2264 y<\/p>\n

e)\u00a0x = y or the relationship cannot be established<\/p>\n

Take a free mock test for IBPS-PO<\/a><\/p>\n

790 Mocks (cracku Pass) Just Rs. 299<\/a><\/p>\n

Question 7:\u00a0<\/b>I. ${x^2}$ + 28x + 192 = 0
\nII. ${y^2}$ + 16y + 48 = 0<\/p>\n

a)\u00a0x \u02c3 y<\/p>\n

b)\u00a0x \u2265 y<\/p>\n

c)\u00a0x \u02c2 y<\/p>\n

d)\u00a0x \u2264 y<\/p>\n

e)\u00a0x = y or the relationship cannot be established<\/p>\n

Question 8:\u00a0<\/b>I.2x – 3y = – 3.5
\nII. 3x + 2y = – 6.5<\/p>\n

a)\u00a0x \u02c3 y<\/p>\n

b)\u00a0x \u2265 y<\/p>\n

c)\u00a0x \u02c2 y<\/p>\n

d)\u00a0x \u2264 y<\/p>\n

e)\u00a0x = y or the relationship cannot be established<\/p>\n

Question 9:\u00a0<\/b>I. ${x^2}$ + 8x + 15 = 0
\nII. ${y^2}$ + 11y + 30 = 0<\/p>\n

a)\u00a0x \u02c3 y<\/p>\n

b)\u00a0x \u2265 y<\/p>\n

c)\u00a0x \u02c2 y<\/p>\n

d)\u00a0x \u2264 y<\/p>\n

e)\u00a0x = y or the relationship cannot be established<\/p>\n

Question 10:\u00a0<\/b>I. x = $\\sqrt {3136} $
\nII.$ {y^2}$ = 3136<\/p>\n

a)\u00a0x \u02c3 y<\/p>\n

b)\u00a0x \u2265 y<\/p>\n

c)\u00a0x \u02c2 y<\/p>\n

d)\u00a0x \u2264 y<\/p>\n

e)\u00a0x = y or the relationship cannot be established<\/p>\n

Take a free mock test for IBPS-PO<\/a><\/p>\n

IBPS PO\u00a0 previous year papers<\/a><\/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 11:\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 12:\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 13:\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 14:\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

Question 15:\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

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

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(A)<\/strong><\/p>\n

I.$x^{2} – 7x + 10 = 0$<\/p>\n

=> $x^2 – 5x – 2x + 10 = 0$<\/p>\n

=> $x (x – 5) – 2 (x – 5) = 0$<\/p>\n

=> $(x – 5) (x – 2) = 0$<\/p>\n

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

II.$y^{2} + 11y + 10 = 0$<\/p>\n

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

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

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

=> $y = -10 , -1$<\/p>\n

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

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

I.$x^{2} + 28x + 192 = 0$<\/p>\n

=> $x^2 + 16x + 12x + 192 = 0$<\/p>\n

=> $x (x + 16) + 12 (x + 16) = 0$<\/p>\n

=> $(x + 16) (x + 12) = 0$<\/p>\n

=> $x = -16 , -12$<\/p>\n

II.$y^{2} + 16y + 48 = 0$<\/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

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

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

I : $2x – 3y = -3.5$<\/p>\n

II : $3x + 2y = -6.5$<\/p>\n

Multiplying eqn(I) by 2 and eqn(II) by 3, and then adding both equations, we get\u00a0:<\/p>\n

=> $(4x + 9x) + (-6y + 6y) = (-7 -19.5)$<\/p>\n

=> $13x = -26.5$<\/p>\n

=> $x = \\frac{-26.5}{13} \\approx -2$<\/p>\n

=> $y = \\frac{3x + 6.5}{2} = 0.25$<\/p>\n

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

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

I.$x^{2} + 8x + 15 = 0$<\/p>\n

=> $x^2 + 5x + 3x + 15 = 0$<\/p>\n

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

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

=> $x = -5 , -3$<\/p>\n

II.$y^{2} + 11y + 30 = 0$<\/p>\n

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

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

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

=> $y = -6 , -5$<\/p>\n

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

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

I. $x = \\sqrt {3136} $<\/p>\n

=> $x = 56$<\/p>\n

II.$ {y^2} = 3136$<\/p>\n

=> $y = \\sqrt{3136} = \\pm 56$<\/p>\n

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

11)\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

12)\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

13)\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

14)\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

15)\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

Top Rated Banking Study Material<\/a><\/p>\n

DOWNLOAD APP FOR IBPS Clerk MOCKS<\/a><\/p>\n

We hope this Quadratic Equation Questions for IBPS PO\u00a0 will be highly useful for your preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Quadratic Equation Questions for IBPS PO PDF Download Top-15 Banking Exams Quadratic Equation Questions PDF. Banking Exams Quadratic Equation questions based on asked questions in previous exam papers very important for the Banking \u00a0exams. Download IBPS PO Previous Papers PDF Take a IBPS PO free mock test Instructions In the following questions two equations numbered […]<\/p>\n","protected":false},"author":49,"featured_media":42678,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3167,169,125,229,228,1127,2071,2067,2069],"tags":[171,48,4147,214],"class_list":{"0":"post-42643","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads-en","8":"category-downloads","9":"category-featured","10":"category-ibpsclerk","11":"category-ibpspo","12":"category-ibps-rrb","13":"category-ibps-rrb-clerk-en","14":"category-ibps-rrb-clerk","15":"category-ibps-rrb-po","16":"tag-banking","17":"tag-ibps-po","18":"tag-quadratic-equations","19":"tag-quant"},"better_featured_image":{"id":42678,"alt_text":"Quadratic Equation Questions for IBPS PO 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