Download important Mathematical Iequalities PDF based on previously asked questions in IBPS Clerk and other Banking Exams. Practice Mathematical Inequalities for IBPS Clerk Exam.<\/p>\n
Download Mathematical Inequalities Questions For IBPS Clerk<\/a><\/p>\n Get 25 IBPS Clerk mocks for Rs. 149. Enroll here<\/a><\/p>\n Take Free IBPS Clerk Mock Test<\/a><\/p>\n Download IBPS Clerk Previous papers PDF<\/a><\/p>\n Go to Free Banking Study Material<\/a> (15,000 Solved Questions)<\/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 Question 1:\u00a0<\/b>I.\u00a0 $x^{2}-3x-88=0$ 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 $5x^{2}+29x+20=0$ 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 $2x^{2}-11x+12=0$ 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 IBPS Clerk Online Mock Test<\/a><\/p>\n Question 4:\u00a0<\/b>I.\u00a0 $3x^{2}+10x+8=0$ 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 $2x^{2}+21x+10=0$ 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 Question 6:\u00a0<\/b>I.\u00a0 $x^{2}+13x+42=0$ 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 IBPS Clerk Previous Papers<\/a><\/p>\n Question 7:\u00a0<\/b>I.\u00a0 \u00a0$x^{2}-15x+56=0$ 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$x^{2}+7x+12=0$ 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$x^{2}-22x+120=0$ 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 IBPS Clerk Important Questions PDF<\/a><\/p>\n Free Banking Study Material (15,000 Solved Questions)<\/a><\/p>\n Question 10:\u00a0<\/b>I.$x^{2}+12x+32=0$ 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. (1)if x < y (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. $30 x^{2} + 11x + 1 = 0$ 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. $x^{2}- x – \\sqrt{2}x + \\sqrt{2}=0$ 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 13:\u00a0<\/b>I.$x^{2}-2x-\\sqrt{5}x+2\\sqrt{5}=0$ 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$ 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.$9x^{2}+3x-2=0$ 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 Daily Free Banking Online Tests<\/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 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 Get 25 IBPS Clerk mocks for Rs. 149. Enroll here<\/a><\/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 Highly Rated Free Preparation App for Banking Exams<\/a><\/p>\n
\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
\nII.\u00a0$y^{2}+8y-48=0$<\/p>\n
\nII.\u00a0$25y^{2}+25y+6=0$<\/p>\n
\nII.\u00a0$2y^{2}-19y+44=0$<\/p>\n
\nII.\u00a0$3y^{2}+7y+4=0$<\/p>\n
\nII.\u00a0$3y^{2}+13y+14=0$<\/p>\n
\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
\nII.\u00a0$y^{2} +19y+90=0$<\/p>\n
\nII.\u00a0$y^{2} -23y+132=0$<\/p>\n
\nII.\u00a0$y^{2} +6y+8=0$<\/p>\n
\nII.\u00a0$y^{2} -26y+168=0$<\/p>\n
\nII.\u00a0$y^{2} +17y+72=0$<\/p>\n
\nSolve these questions and answer<\/p>\n
\n(2) if x \u2264 y<\/p>\n
\nII. $42 y^{2} + 13y + 1 = 0$<\/p>\n
\nII.$y^{2}-3y+2=0$<\/p>\n
\nII.$y^{2}-\\sqrt{3}y-\\sqrt{2}y+\\sqrt{6}=0$<\/p>\n
\nII.$y^{2}=16$<\/p>\n
\nII.$8y^{2}+6y+1=0$<\/p>\n